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---

CHAPTER TWO
3 wk

7d

24 h 3600 s 1000 ms

= 18144 × 10 9 ms

...
1 ft / s 0
...
98 mi / h ⇒ 26
...
2808
ft
1 h

2
...
2 (a)

554 m 4
1d
1h
d ⋅ kg 24 h 60 min

1 kg 108 cm 4
= 3
...
0006214 mi 3600 s
1 m3
= 57
...
3145 ft 3

(b)

921 kg 2
...
37 × 10 3 kJ 1 min 1000 J
min 60 s
1 kJ

1
...
93 hp ⇒ 120 hp
1
J/s

2
...
For a
classroom with dimensions 40 ft × 40 ft × 15 ft :
40 × 40 × 15 ft 3 (12) 3 in 3 1 ball
n balls =

...

2
...
3 light yr 365 d 24 h
1 yr

3600 s 1
...
2808 ft
0
...
5 Distance from the earth to the moon = 238857 miles
238857 mi

1

m

0
...
001 m

= 4 × 1011 reports

2
...
0006214 mi
1000 L
= 44
...
17 gal
Calculate the total cost to travel x miles
...
25 1 gal
gal 28 mi

x (mi)

= 14,500 + 0
...
25
1 gal
x (mi)
= 21,700 + 0
...
7 mi

Equate the two costs ⇒ x = 4
...
7
5320 imp
...
965 g

220
...
gal

1

cm

1 kg
3

1 tonne

1000 g

1000 kg

tonne kerosene
plane ⋅ yr

= 1
...
02 × 109 tonne crude oil 1 tonne kerosene

plane ⋅ yr

7 tonne crude oil 1
...
8 (a)
(b)
(c)

2
...
10
2
...
1714 ft / s 2

25
...
1714 lb m ⋅ ft / s 2
25 N

1 kg ⋅ m/s 2

1
9
...
5493 kg ⇒ 2
...
66 cm / s 2

1000 g
ton

85
...
20462 lb m

1 dyne
1 g ⋅ cm / s 2

2
...
3145 ft 3
1 m3

500 lb m

= 25
...
5 kg

= 9 × 10 9 dynes

32
...
5 × 10 6 lb f
2
2
1 s
32
...
1 cm)(100 g / cm 3 )

...
53 g / cm 3
H
30 cm
ρ H (30 cm)(0
...

(30 cm - 20
...
12

Vs =

πR 2 H
3

⇒ Vf =

; Vf =

3

−

πr 2 h
3

;

πh Rh

2

2

f

H
H−

h3
H2

3

2

= ρs

H3
= ρs
H 3 − h3

h

3

r

2

2

ρf
h

R r
R
= ⇒r = h
H h
H

FG IJ = πR FG H − h IJ
−
3
3 H HK
3 H
H K
πR F
h I
⇒ρ
GH H − H JK = ρ πR3 H
3

πR H
2

ρ f V f = ρ sVs
⇒ ρ f = ρs

πR 2 H

H

H

2

ρs

s

R

1

1−

FG h IJ
H HK

2-2

3

ρf

2
...
879 g 10 6 cm 2
cm

3

1m

E Substitute for A
L
W b N g = 3
...
14

3

1 kg

9
...
45 × 10 4 A

g g0

2

b g π OPQ
2

+ sin −1 h − 1 +

1 lb f = 1 slug ⋅ ft / s 2 = 32
...
174 lb m
1
1 poundal = 1 lb m ⋅ ft / s 2 =
lb f
32
...
44 slugs
32
...
174 ft
1 poundal
W=
= 5
...
44 slugs
32
...
174 ft
1 poundal
W=
= 938 poundals
2
6
s 1 lb m ⋅ ft / s 2
(b) F = ma ⇒ a = F / m =

355 poundals
25
...
135 m / s 2

2-3

1 slug
32
...
2808 ft

2
...
174 ft / s 2 )

⇒

FG 1IJ = 5
...
3623 bung ⋅ ft / s 2

3 bung 32
...
3623 bung ⋅ ft / s 2
On the earth: W = (3)( 32
...
3623 = 18 fern

(b) On the moon: W =

2
...
7)(8
...
365 + 125
...
5
2
...
0 × 10−4
≈ 1× 10−5
40
(3
...
0 × 10−6
≈

≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4
4
...
5 ⇒ 3810 ⇒ 3
...
18 (a)
A: R = 731 − 72
...
7 o C

...

72
...
6 + 72
...
0
= 72
...
4 − 72
...
8) 2 + (72
...
8) 2 + (72
...
8) 2 + (73
...
8) 2

...
3o C
B: R = 1031 − 97
...


...
3 + 1014 + 98
...
4

...

= 100
...
3 − 100
...
2) 2 + (98
...
2) 2 + (1031 − 100
...
4 − 100
...


...
3o C

(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate
...
19 (a)

12

X=

∑X

12

i

i =1

C min=

= 73
...
5 − 2(1
...



...


12 − 1

C max= = X + 2 s = 735 + 2(12) = 75
...


...

(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor
temperature (failure of reactor control system), problems with the color measurement
system, operator carelessness
2
...
9
Stdev(X) 2
...
5
Min
136
...
5
127
...
5
127
...
5
127
...
5
127
...
5
127
...
5
127
...
5
127
...
9
131
...
9
131
...
9
131
...
9
131
...
9
131
...
9
131
...
9
131
...
4
136
...
4
136
...
4
136
...
4
136
...
4
136
...
4
136
...
4
136
...
An overhaul would have been reasonable after Run 12
...
21 (a) Q ' =

2
...
20462 lb 3
...
2 × 10−6 lb ⋅ ft 2 / s
3 × 103

Q 'exact =1
...
00000156 lb ⋅ ft 2 / s

2-5

2
...
583 J / g ⋅ o C

1936 lb m

0
...
2808 ft

ft ⋅ h

o

3600 s

1000 g

m 2
...
The calculator solution is 163 × 10 3

...

−1
3
2
(3 × 10 )(4 × 10 )(2)
3

3

3

2
...
24 (a)

Duρ

μ

=

0
...
067 in

1 m

1 kg 10 6 cm 3

0
...
2808 ft 0
...
37 in

cm 3

1000 g

1 m3

(5 × 10 −1 )(2)(8 × 10 −1 )(10 6 ) 5 × 101− ( −3)
≈
≈ 2 × 10 4 ⇒ the flow is turbulent
3
(3)(4 × 10)(10 3 )(4 × 10 −4 )

kg d p y
D

1/ 3

⎛ μ ⎞
= 2
...
600 ⎜
⎟
⎝ ρD ⎠

⎛ d p uρ ⎞
⎜
⎟
⎝ μ ⎠

1/ 2

1/ 3

⎡
⎤
1
...
00 + 0
...
00 kg/m )(1
...
00500 m)(0
...
426 ⇒
= 44
...
00 × 10−5 m 2 / s

1/ 2

⎡ (0
...
0 m/s)(1
...
00 × 10−5 N ⋅ s/m 2 )
⎣
⎦
= 0
...

(c)
dp (m)

y

0
...
010
0
...
005
0
...
1
0
...
1
0
...
1

D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s)
1
...
00E-05
1
10
1
...
00E-05
1
10
2
...
00E-05
1
10
1
...
00E-05
1
10
1
...
00E-05
1
20

kg
0
...
620
1
...
796
1
...
25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2

200 crystals 0
...
4 mm
10 crystals
0
...
0 crystals / s
60 s
min

(25
...
4 mm
crystals
= 25
...
4 D ′ − 10 25
...
7 D ′ − 108 D ′

2

2
...
5 lb m / ft 3 ; 8
...
27 × 10−7 in 2 9 × 106 N
14
...
5 lb m / ft 3 )exp ⎢
⎥
lbf
m 2 1
...
57 lb m 35
...
13 g/cm3
3
3
6
3
ft
m 10 cm 2
...
593 g

1 ft 3

0
...
37 2 in 2

1N

d

= 62
...


id

i

d

i

⇒ 62
...
5 exp 8
...
45 × 10 −4 P ' ⇒ ρ ′ = 113 exp 120 × 10 −10 P '

...


P ' = 9
...
20 × 10 −10 )(9
...


...
39V ' ; t bsg = 3600t ′bhr g
1728
⇒ 16
...
06102 expb3600t ′ g

2
...

2
...
00 mol / L, 2
...
00 exp[(-2
...
00 mol / L

t = 1 ⇒ C = 3
...
00)(1)] = 0
...
406 − 3
...
6 min:
(0
...
00 = 14 mol / L

...
00 exp[(-2
...
6)] = 0
...
10 mol/L:

t int =
t exact

1− 0
(010 − 3
...


...
406 − 3
1
C
1 0
...
70 min
2
...
00
2 3
...
5
C exact vs
...
5
2

(t=0
...
4)

1
...
12, C=0
...
5
0
0

1

2

t (min)

2-7

p* =

2
...
2) + 20 = 42 mm Hg
199
...
2

c

MAIN PROGRAM FOR PROBLEM 2
...
1, 10X, F5
...
LE
...
AND
...
LT
...
EQ
...
5
1
...
8
5
...
0
1
...
5
100
...
0
1
...
0

98
...
30 (b) ln y = ln a + bx ⇒ y = ae bx
b = (ln y 2 − ln y1 ) / ( x 2 − x1 ) = (ln 2 − ln 1) / (1 − 2) = −0
...
63(1) ⇒ a = 4
...
00e −0
...
0 − ln 40
...
0 − 1
...
0 − 3ln(2
...
30 (cont’d)
(e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2) b ]1/ 2

b = [ln( y 2 / x ) 2 − ln( y 2 / x ) 1 ] / [ln( x − 2) 2 − ln( x − 2) 1 ]

...
0 − ln 40
...
0 − ln 10) = 4
...
0 − 4
...
0) ⇒ a = 40
...
2( x − 2) 4
...
34 x 1/ 2 ( x − 2) 2
...
31 (b) Plot y 2 vs
...
Slope = m, Intcpt = − n
(c)

1
1 a
1
= +
x ⇒ Plot
vs
...
axes], slope =

a
1
, intercept =
b
b

(d)

1
1
= a ( x − 3) 3 ⇒ Plot
vs
...
axes], slope = a , intercept = 0
2
2
( y + 1)
( y + 1)
OR
2 ln( y + 1) = − ln a − 3 ln( x − 3)
Plot ln( y + 1) vs
...
] or (y + 1) vs
...


x [rect
...


x [semilog ], slope = a, intercept = b

(f) log10 ( xy ) = a ( x 2 + y 2 ) + b
Plot log10 ( xy ) vs
...
] ⇒ slope = a, intercept = b

(g)

x
b
x
1
vs
...
], slope = a , intercept = b
= ax + ⇒ = ax 2 + b ⇒ Plot
y
x
y
y
OR

b
1
1
b
1
1
vs
...
] , slope = b, intercept = a
= ax + ⇒
= a + 2 ⇒ Plot
y
x
xy
xy
x
x

2-9

2
...
R is a line through ( R = 5 , y = 0
...


...
18
0
...
14
0
...
1
0
...
06
0
...
02
0
0

20

40

60

80

100

R

y=aR+b

U
|
V
|
W


...
011
= 2
...
11 × 10 −3 R + 4
...
011 − 2
...
50 × 10

a=

ib g

d

ib g

d

(b) R = 43 ⇒ y = 2
...
50 × 10 −4 = 0
...
092 kg H O kgg = 110 kg H O h
2

2

2
...

ln a = ln T − b ln φ = ln 210 − ( −119) ln(25) ⇒ a = 9677
...
6φ −1
...


b

(b) T = 9677
...
19 ⇒ φ = 9677
...
6 / 85

0
...
6 / 290g
T = 175o C ⇒ φ = 9677
...
8403


...
8403

= 29
...
8403

= 19
...
The value of T=290°C
is probably the least likely to be correct, because it is farthest away from the date range
...
34 (a) Yes, because when ln[(C A − C Ae ) / (C A0 − C Ae )] is plotted vs
...

0

50

100

150

200

0
-0
...
5
-2
t (m in)

Slope = -0
...
3 × 10-3 min −1

(b) ln[(C A − C Ae ) /(C A0 − C Ae )] = − kt ⇒ C A = (C A0 − C Ae )e − kt + C Ae

C A = (0
...
0495)e − (9
...
0495 = 9
...
300 × 10-2 g 30
...
317 L
= 10
...
4805 gal

2
...
t2 in rectangular coordinates, slope=2 and intercept= ln(353 × 10 −2 ) ; or

...
t2 in semilog coordinates, slope=2, intercept= 353 × 10−2

...


...
36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k lnV
8
...
5
7
6
...
5

3

lnP = -1
...
736

3
...


...
736 ⇒ C = e12
...
40 × 105 mm Hg ⋅ cm4
...
4 8 3 5 ln C - 1 0
...
37 (a)

2
1
0
-1
3
...
5

5

ln C

2-11

5
...
37 (cont’d)

m = slope = 2
...
045 ⇒ K L = 4
...
483

G − 180 × 10 −3

...
340 × 10 −5 (475) 2
...

3
...


(b) C = 475 ⇒

2
...
1)

...

ln(2
...


...
68
⇒ c = −1
...
7 volts ⋅ kPa 1
...
678

ln(3
...


...
lnV
...
5
1
0
...
5

0

lnZ = 0
...
0035

0
...
5

lnV

b = slope = 0
...


When V is constant (runs 5 to 7), plot lnZ vs
...
Slope=c, Intercept= ln a + c lnV
2

lnZ

1
...
5
0
1
...
7

lnZ = -0
...
4551

1
...
1

2
...
997 ⇒ 10

...
4551

Z

Plot Z vs V b P c
...
05

Z = 31
...
1

0
...
2

Vb Pc

a = slope = 311 volt ⋅ kPa / (L / s)
...


The results in part (b) are more reliable, because more data were used to obtain them
...
39 (a)

sxy =
sxx =

a=
b=

∑x y

1
n

∑x

1
n

sx =

n

1
n

i i

= [(0
...
3) + (2
...
2)] / 3 = 4
...


...
32 + 19 2 + 3
...
647

...
3 + 1
...
2) / 3 = 18; s y =

...
4 + 2
...


...
677 − (18)(1
...

= 0
...
647 − (18) 2

...
647)(1867) − (4
...


...
182
2
4
...


=


...
936 x + 0182
(b) a =

sxy
sxx

=

4
...
0065 ⇒ y = 1
...
647

4

y

3

y = 0
...
182

2

y = 1
...
40 (a) 1/C vs
...
Slope= b, intercept=a

a = Intercept = 0
...
5
2
1
...
5
0

2
1
...
477 L / g ⋅ h;

1
0
...
4771t + 0
...
082 + 0
...
2 g / L

t = (1 / C − a ) / b = (1 / 0
...
082) / 0
...
5 h
(d) t=0 and C=0
...

(e) The concentration of the hazardous substance could be enough to cause damage to the
biotic resources in the river; the treatment requires an extremely large period of time; some
of the hazardous substances might remain in the tank instead of being converted; the
decomposition products might not be harmless
...
41 (a) and (c)

y

10

1
0
...
1684ln x + 1
...
5
1
0
...
168

0
-1

0

1

2
ln x

3

4

5

Intercept = ln a = 11258 ⇒ a = 3
...


2
...
t in rectangular coordinates
...
0062t

100

400

500

t

Lab 1

600

400

600

-4

-6
ln(1-Cp/Cao) = -0
...
0111 s-1

800

0

0
ln(1-Cp/Cao)

ln(1-Cp/Cao)

200

300

-2

k = 0
...
0064t

-6
ln(1-Cp/Cao) = -0
...
0063 s-1

Lab 4

k = 0
...
k = 0
...


2-14

2
...
44

i

i =1

− axi

g

2

dφ
⇒
=0=
da

∑ 2b y
n

i

i =1

g

− axi xi ⇒

n

∑y x

i i

i =1

−a

n

∑x

2
i

i =1

n

n

∑

∑by
n

yi xi /

∑x

2
i

i =1

DIMENSION X(100), Y(100)
READ (5, 1) N
C
N = NUMBER OF DATA POINTS
1FORMAT (I10)
READ (5, 2) (X(J), Y(J), J = 1, N
2FORMAT (8F 10
...
0
SY = 0
...
0
SXY = 0
...
3, 3X 'INTERCEPTb -- b8b =', F7
...
0
DO 200J = 1, N
YC = A * X(J) + B
RES = Y(J) - YC
WRITE (6, 5) X(J), Y(J), YC, RES
5FORMAT (3X 'Xb =', F5
...
2, 5X 'Y(FITTED)b =', F7
...
3)
200SSQ = SSQ + RES ** 2
WRITE (6, 6) SSQ
6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10
...
0 2
...
5
5
...
0
8
...
5
12
...
0 15
...
536, b = −4
...
45 (a) E(cal/mol), D0 (cm2/s)
(b) ln D vs
...

(c) Intercept = ln D0 = -3
...
05 cm2 / s
...
0E-03

2
...
8E-03

2
...
6E-03

2
...
4E-03

2
...
2E-03

2
...
0E-03

Slope = − E / R = -3666 K ⇒ E = (3666 K)(1
...
0
ln D

-11
...
0
-13
...
0

ln D = -3666(1/T) - 3
...
2
396
...
7
447
...
2

D
1
...
50E-06
4
...
52E-06
1
...
00E-05

1/T
2
...
67E-03
2
...
38E-03
2
...
12E-03
Sx
Sy
Syx
Sxx
-E/R
ln D0

lnD (1/T)*(lnD)
-13
...
03897
-12
...
03447
-12
...
03105
-11
...
02775
-11
...
02495
-10
...
02296
2
...
1
-3
...
16E-06
-3666
-3
...
05

2-16

(1/T)**2
8
...
14E-06
6
...
65E-06
4
...
50E-06

CHAPTER THREE
3
...
68 qt cm

3

≈

4 × 106

b3 × 10gd10 i
3

≈ 1 × 102 g / s

(c) Weight of a boxer ≈ 220 lb m
12 × 220 lb m 1 stone
Wmax ≥
≈ 220 stones
14 lb m

dictionary

(d)

V=
≈

πD 2 L
4

=

2

314 4
...

4

d

2

800 miles 5880 ft 7
...
5 ft 28,317 cm3
≈ 3 × 3 × 104 ≈ 1 × 105 cm3
3
1 ft

(ii) V ≈

1 ft 3

28,317 cm3

62
...


3
...
028317 m3
= 62
...
45359 kg
1 ft

(a) (i)

(ii)

995 kg / m3

62
...
12 lb m / ft 3

(b) ρ = ρ H2 O × SG = 62
...
7 = 360 lb m / ft 3
3
...
70 × 103 kg

1 m3

m3 103 L
1150 kg
min
10 gal

= 35 kg

m3 1000 L 1 min
= 27 L s
0
...
481 gal

0
...
43 lb m
1 ft 3

≅ 29 lb m / min

3-1

3
...
82dg kerosenei
d0
...
82idg blendi = 0
...
70Vg g gasoline
3

g

3

g

=

g

Volumetric ratio =

3
...
S
...
5

0
...
78
3
= 0
...
78 − 0
...
50 cm3
3
3
=
3 = 0
...
0 kg

L 5 Fr
$1
= $68
...
7 × 10 kg 1L 5
...

50
...
20
= $22
...
70 × 10 kg 3
...


V B ( ft 3 / h ), m B ( lb m / h )
V ( ft 3 / h), SG = 0
...
850 × 62
...


d i 0
...
43 lb = 54
...
659 × 62
...


mB =
mH

700 lb m / h

VB ft

3

m
3

H

B

H

VB + VH = 1319 ft / h

...
88VB + 4114VH = 700 lb m

...


VH = 1
...
6 lb m / h hexane
(b) – No buildup of mass in unit
...
)
– Volumes of benzene and hexane are additive
...


3-2

3
...
5 kg H 2SO 4

1 kg solution

L

0
...


...
5 kg H 2 SO 4

L
18255 × 1
...

195
...
65 kg H 2 O
L
+
= 470 L
0
...
000 kg
470 − 445
% error =
× 100% = 5
...
7

b gE

b

Buoyant force up = Weight of block down

g

Mass of oil displaced + Mass of water displaced = Mass of block

b

g

b

g

ρ oil 0
...
542 V = ρ c V
2


...
1: ρ c = 2
...
325 g / cm3
moil = ρ oil × V = 3
...
3 cm3 = 117
...
4 g + 124
...
8

b g

b

Buoyant force up = Weight of block down

g

⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp
...
1: ρ w 15 A g = ρ B 2 A g ⇒ ρ B = ρ w ×

...
00 g/cm3

bg

15

...
75 g / cm3 ⇒ SG

B

= 0
...
2: ρ soln A g = ρ B 2 A g ⇒ ρ soln = 2 ρ B = 15 g / cm3 ⇒ SG

...
9

= 15

...
Note: ρ A > ρ w (object sinks)

d

Volume displaced: Vd 1 = Ab hsi = Ab hp1 − hb1

hs 1

WA + WB

ρ wVd 1 g

Archimedes ⇒

hρ1

hb1

d

Subst
...
3 for h p 1 in

b 2 g, solve for h

b1

hb1 =

WA + WB
V
W + WB
⇒ h p1 = w + A
pw g
Ap
pw gAp

b

W + WB
Vw
+ A
Ap
pw g

g LM 1
NM A

−

p

3-3

1
Ab

OP
QP

(1)

= WA + WB

weight of displaced water

Before object is jettisoned

subst
...
9 (cont’d)
hs 2

WB
WA

WA
ρ Ag

(5)

Volume displaced by boat: Vd 2 = Ab h p 2 − hb 2

(6)

Let V A = volume of jettisoned object =

d

hρ2

h b2

Archimedes ⇒ ρ WVd 2 g = WB

After object is jettisoned

Subst
...
8

⇒

bg

for h p 2 in 7 , solve for hb 2

(a)

p2

(7)

Volume of pond water: Vw = Ap h p 2 − Vd 2 − V A
solve for

i

hb 2 =

WB
W
− A
pw g p A g

(8)

Vw
WB
WA
WB
+
+
−
Ap pw gAp p A gAp pw gAb

(9)

Change in pond level
( 8 ) − ( 3) W ⎡ 1
1 ⎤ WA ( pW − p A ) ρW < ρ A
A
hp 2 − hp1 =
−
⎯⎯⎯⎯ < 0
→
⎢
⎥=
Ap g ⎣ p A pW ⎦
p A pW gAp
⇒ the pond level falls

(b)

Change in boat level

LM 1
A gMp A
N

b 9 g−b 4 g WA

h p 2 − h p1 =

p

A

⇒ the boat rises
3
...
93 kg CaCO 3
L CaCO 3

(b) Wbag = ρ bulkVg =

L
O
OP b=gF V I MM1 + F p F A − 1I I PP > 0
1
+
p A P G A JM G p H A
Q H K M H G JK JK PP
N
Q
>0

−
p

1
pW Ap

5

A

W

0
...
05 kg / L

2
...
807 m / s2

1N

= 100 × 103 N

...

2

(c) The limestone would fall short of filling three bags, because
– the powder would pack tighter than the original particles
...


3-4

3
...
5 kg 9
...
0 N)
1 kg ⋅ m / s2
Wb − WI
=
= 119 L
ρwg
0
...
807 m / s2
1N
m
122
...

Vb
119 L

Vb =

m f + mnf = mb

(b)

xf =

mf
mb

(1)

⇒ m f = mb x f

(2)

d

(1),(2) ⇒ mnf = mb 1 − x f
V f + Vnf = Vb ⇒

b2 g,b 3g

⇒ mb

(c) x f =

Fx
GH ρ

+

f

1 / ρ f − 1 / ρ nf

ρf

b

ρ nf

1 / ρ b − 1 / ρ nf

+

I=m
JK ρ

1− xf

f

mf

i

mnf

ρ nf

=

⇒ xf

b

=

(3)
mb

ρb

F1
GH ρ

−

f

1

ρ nf

I= 1 − 1
JK ρ ρ
b

⇒ xf =

nf

1 / ρ b − 1 / ρ nf
1 / ρ f − 1 / ρ nf

1 / 103 − 1 / 1
...

= 0
...
9 − 1 / 1
...
03 − 11JK − GH
...
5
...

⇒x =
=
= 0
...
9 11K

...
(g Ile/100 g H2O)

3
...
5
4
3
...
5
2
1
...
5
0
0
...
5x - 539
...
9992

0
...
991

0
...
995

0
...

From the plot above, r = 5455ρ − 539
...
9940 g / cm3 ,

mIle =

150 L

0
...
197 g Ile / 100g H 2 O
1000 cm3

3
...
197 g sol 1000 g

= 4
...
9940 g ILE/cm3 solution at 50oC
...

Presuming that the dependence of solution density on T is the same as that of pure water,
the solution density at 47oC would be higher than 0
...
The ILE mass flow
rate calculated in Part (b) is therefore too low
...
13 (a)

Mass Flow Rate (kg/min)

1
...
00

y = 0
...
1523
R 2 = 0
...
80
0
...
40
0
...
00
0
...
0

4
...
0

8
...
0

12
...
13 (cont’d)

b g

From the plot, R = 5
...
0743 5
...
55 kg / min

...
5
300
6
0
...
5
371
8
0
...
5
440
10
0
...
297
0
...
454
0
...
600
0
...
742
0
...
880
0
...
004
0
...
004

0
...
012
0
...
004 + 0
...
004 + 0
...
026 = 0
...

95% confidence limits: (0
...
610 ± 0
...
592 kg / min
and 0
...

3
...
0 kmol C 6 H 6
15
...
114 kg C 6 H 6

...

kmol
lb - mole
= 33
...
6 mol
6 mol C
1 mol C 6 H 6

15,000 mol C 6 H 6

= 90,000 mol C

6 mol H
= 90,000 mol H
1 mol C 6 H 6

90,000 mol C 12
...
08 × 106 g C

90,000 mol H 1
...
07 × 104 g H
mol H
15,000 mol C 6 H 6

6
...
03 × 1027 molecules of C 6 H 6

3-7

3
...
866 kg
3

L

1h
60 min

= 2526 kg / min

2526 kg 1000 mol 1 min
= 457 mol / s
min 92
...
16 (a)

200
...

kmol CH 3OH 1000 mol
= 936 mol CH 3OH
kg mix
32
...
17

M=

mN 2

100
...
25 mol N 2

74
...
850 lb m MA
28
...
75 mol H 2

mol N 2
3000 kg kmol 0
...
52 kg
kmol feed

3
...
02 g H 2

= 8
...
02 kg N 2
= 2470 kg N 2 h
kmol N 2

M CaCO 3 = 215 g − 65 g = 150 g

(a) V = 455 mL min , m = 500 g min
(b) ρ = m / V = 500 g / 455 mL = 110 g mL

...
300 g CaCO 3 g suspension
3
...

mC2 H 5OH =

10
...
07 g C 2 H 5OH

= 461 g C 2 H 5OH
mol C 2 H 5OH
75
...
1 g C 4 H 8 O 2
= 6608 g C 4 H 8 O 2
mC4 H 8O 2 =
mol C 4 H 8O 2
15
...
05 g CH 3COOH
mCH 3COOH =
= 901 g CH 3COOH
mol CH 3COOH
461 g
= 0
...
8291 g C 4 H 8 O 2 / g mix
xC 4 H 8 O 2 =
461 g + 6608 g + 901 g
901 g
= 0113 g CH 3COOH / g mix

...
7 g / mol
100 mol
25 kmol EA 100 kmol mix 79
...
20 (a)
Unit
Crystallizer
Filter
Dryer

Function
Form solid gypsum particles from a solution
Separate particles from solution
Remove water from filter cake
0
...
35 kg C aSO 4 ⋅ 2 H 2 O
L slurry

(b) m gypsum = 1 L slurry

L CaSO4 ⋅ 2H2O
= 0151 L CaSO4 ⋅ 2H2O

...
32 kg CaSO4 ⋅ 2H2O
0
...
15 kg CaSO 4
CaSO 4 in gypsum: m =
= 0
...
18 kg gypsum

Vgypsum =

0
...
: m =


...
35 kg gypsum

(c) m =

% recovery =

1
...
209 kg CaSO 4
L

100
...
209 g CaSO 4
0
...
84 × 10 -5 kg CaSO 4
0
...
209 g sol

0
...
84 × 10 -5 g
× 100% = 99
...
277 g + 0
...
21
CSA:
FB:

= 0
...
8 L
0
...
2 L
0
...
5496
75 kg
kmol
= 0
...
5496
min
⇒
= 1
...
4600
min

She was wrong
...

3
...
07 g EtOH
= 6910 g EtOH
mol EtOH
0
...
400 g EtOH

6910 g EtOH

789 g EtOH

SG =

(b) V ′ =

L

(6910 +10365) g
19
...
903

L
= 18
...
5 L
935
...
123 − 18
...
5%
18
...
123 L ⇒ 19
...
23

0
...
04 g 0
...
0 g Air
+
= 27
...
090 kmol CH 4
= 2
...
83 kg
1
...
264 kmol CH 4
0
...
89 kmol air h
h
0
...
264 kmol CH 4
h

Dilution air required:

0
...
01 kmol air h
0
...
01 - 22
...
20 kmol Air 29 kg Air
Product gas: 700 kg +
= 1286 kg h
h

h

kmol Air

43
...
00 kg O2

h

h

3
...
21 kmol O2
1
...


xi =

B:
1

ρ

xi

∑

mi mi
1
=
M Vi
M

mi Vi
1
1
V
=
∑ Vi = M = ρ Correct
...
60
0
...
15
=
+
+
= 1
...
917 g / cm 3
ρ i 0
...
049 1
...
225

=

∑M

3
...
15 m ol C O / m ol , x C O 2 =
= 0
...
09 mol CH 4 / mol , x N 2 =
= 0
...
30 × 44 + 0
...
46 × 28 = 32 g / mol

...
26 (a)
Samples Species

MW

k

Peak
Mole
Mass
moles
Area Fraction Fraction
3
...
156
0
...
540
2
...
233
0
...
804
2
...
324
0
...
121
1
...
287
0
...
991

mass

1

CH4
C2H6
C3H8
C4H10

16
...
07
44
...
12

0
...
287
0
...
583

2

CH4
C2H6
C3H8
C4H10

16
...
07
44
...
12

0
...
287
0
...
583

7
...
4
5
...
4

0
...
146
0
...
050

0
...
123
0
...
081

1
...
689
2
...
233

18
...
712
115
...
554

3

CH4
C2H6
C3H8
C4H10

16
...
07
44
...
12

0
...
287
0
...
583

3
...
5
2
...
8

0
...
371
0
...
134

0
...
304
0
...
212

0
...
292
1
...
466

8
...
835
53
...
107

4

CH4
C2H6
C3H8
C4H10

16
...
07
44
...
12

0
...
287
0
...
583

4
...
5
1
...
2

0
...
332
0
...
054

0
...
324
0
...
102

0
...
718
0
...
117

11
...
575
26
...
777

5

CH4
C2H6
C3H8
C4H10

16
...
07
44
...
12

0
...
287
0
...
583

6
...
9
4
...
3

0
...
333
0
...
197

0
...
262
0
...
299

0
...
267
2
...
341

15
...
178
98
...
933

(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST
INTEGER N, ND, ID, J
READ (5, *) N
CN-NUMBER OF SPECIES
READ (5, *) (MW(J), K(J), J = 1, N)
READ (5, *) ND
DO 20 ID = 1 , ND
READ (5, *)(A(J), J = 1, N)
MOLT = 0
...
0
DO 10 J = 1, N
MOL(J) =
MASS(J) = MOL(J) * MW(J)
MOLT = MOLT + MOL(J)
MASST = MASST + MASS(J)
10
CONTINUE
DO 15 J = 1, N
MOL(J) = MOL(J)/MOLT
MASS(J) = MASS(J)/MASST
15
CONTINUE
WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N)
20 CONTINUE
1 FORMAT (' SAMPLE: `, I3, /,
∗ ' SPECIES MOLE FR
...
', /,

3-11

8
...
164
49
...
603

3
...
3), /), /)
END
$DATA
∗
4
16
...
150
30
...
287
44
...
467
58
...
583
5
3
...
8 2
...
7
7
...
4 5
...
4
3
...
5 2
...
8
4
...
5 1
...
2
6
...
9 4
...
3
[OUTPUT]
SAMPLE:
1
SPECIES MOLE FR MASS FR
1
0
...
062
2
3
4
SAMPLE: 2
(ETC
...
27 (a)

0
...
324
0
...
7 × 10 6 × 0
...
173
0
...
412

44 kg CO 2
= 1
...
9 × 105 kmol CO 2
12 kg C

(11 × 10 6 × 0
...

12 kg C
( 3
...
10) kg C

m=

= 6
...
38 × 10 4 kmol CO

16 kg CH 4
= 5
...
17 × 10 3 kmol CH 4
12 kg C

(1
...
67 × 10 5 + 5
...
915 × 44 + 0
...
01 × 16 = 42
...
28 (a) Basis: 1 liter of solution
1000 mL

1
...
525 mol / L ⇒ 0
...
08 g H 2 SO 4

3-12

3
...
7854 L
(b) t = V =
gal

V

55 gal

min

60 s

87 L

min

3
...
03 g
gal

1L

= 144 s

0
...
59 g

mL

= 23
...
513 m / s
A min 1000 L
60 s (π × 0
...
513 m / s

(c) u =

3
...
180 mol C6H14/mol
0
...
050 mol C6H14/mol
0
...
50 L C6H14(l)/min
n3 (mol C6H14(l)/min)

n3 =


...
659 kg 1000 mol
min

L

86
...

= 1147 mol / min

...
3 mol / min
W S
|
T

Hexane balance: 0
...


...
820n1 = 0950n2 (mol N2 / min)

...
30

n3
1147

...


...
030 mol

172 g

103 mL

lL

1 mol

= 0155 g Nauseum

...
31 (a) kt is dimensionless ⇒ k (min -1 )
(b) A semilog plot of CA vs
...
4137x + 0
...
9996

0
...
0
t (m in)

10
...
414 min −1

ln CAO = 02512 ⇒ CAO = 1286 lb - moles ft 3

...


FG 1b - molesIJ = C′ mol 28
...
26462 lb - moles

3

1000 mol

= 0
...
06243C A = 1334 exp −0
...

′

drop primes

⇒

b

g

b

C A mol / L = 214 exp −0
...


g

t = 200 s ⇒ C A = 5
...
32 (a)

(b)

2600 mm Hg

14
...
3 psi
760 mm Hg

275 ft H 2 O 101
...
0 kPa
33
...
00 atm 101325 × 105 N m2

...
4 N cm2
2
2
1 atm
100 cm
(d)

280 cm Hg 10 mm 101325 × 106 dynes cm2 1002 cm2

...
733 × 1010
2
2
m2
1 cm
760 mm Hg
1 m

(e) 1 atm −

20 cm Hg 10 mm
1 atm
= 0
...
32 (cont’d)
(f)

(g)

b

25
...
696 psig

b25
...
696gpsi

g = 1293 mm Hg bgaugeg
b g

760 mm Hg
= 2053 mm Hg abs
14
...
(3
...
0 lbf 144 in
=
ρg
in2 1 ft 2 1
...
43 lbm

s2

32
...
174 ft

s ⋅ lbf

3
...
33 (a) Pg = ρgh =

0
...
81 m / s2
m3

h (m)

1N
1 kPa
2
1 kg ⋅ m / s 103 N / m2


...
55 m

...
92 × 1000

IJ FG
K H

IJ
K

kg
16 2 3
× 7
...

4
m3

(b) Pg + Patm = Ptop + ρgh

b

g b g

68 + 101 = 115 + 0
...
81 / 103 h ⇒ h = 5
...
34 (a) Weight of block = Sum of weights of displaced liquids
ρ h + ρ 2 h2
(h1 + h2 ) Aρ b g = h1 Aρ 1 g + h2 Aρ 2 g ⇒ ρ b = 1 1
h1 + h2
(b)

Ptop = P + ρ1gh0 , P
atm
bottom = P + ρ1g(h0 + h ) + ρ2 gh2 , W = ρb (h + h2 ) A
atm
1
b
1

⇒Fdown = ( P + ρ1gh0 ) A + ρb (h1 + h2 ) A , Fup = [ P + ρ1g(h0 + h1) + ρ2 gh2 ]A
atm
atm
Fdown = Fup ⇒ ρb (h1 + h2 ) A = ρ1gh1 A + ρ2 gh2 A ⇒ Wblock = W displaced
liquid

3-15

3
...
36


...
8066 m 150 m 12 m2
1N
s2
1002 cm2 1 kg ⋅ m / s2

FG
H

IJ
K

022481 lb f

...

= 2250 lb f
1N
cm2

14 × 62
...

1 ft 3
2
...
69 × 107 lb m
3
ft
7
...

1 lb f
12 ft 2
lb f 14 × 62
...
174 ft 30 ft
= 14
...
174 lb m ⋅ ft / s2 12 2 in 2
= 32
...

— Tank strength inadequate for that much force
...
37 (a) mhead =

π × 24 2 × 3 in 3

W = mhead g =

4
392 lb m

1 ft 3
8
...
43 lb m
= 392 lb m
3
3
12 in
ft 3
1 lb f
32
...
174 lb m ⋅ ft / s 2

⎡( 30 + 14
...
7 lbf π × 242 in 2
− 392 lb f = 7
...


7
...
174 lb m ⋅ ft/s 2
1 lb f

= 576 ft/s 2

(b) Vent the reactor through a valve to the outside or a hood before removing the head
...
38 (a)

a

Pa = ρgh + Patm , Pb = Patm
If the inside pressure on the door equaled Pa , the force on
the door would be F = Adoor ( Pa − Pb ) = ρghAdoor
Since the pressure at every point on the door is greater than
Pa , Since the pressure at every point on the door is greater
than Pa , F >ρghAdoor

b

2m
1m

(b) Assume an average bathtub 5 ft long, 2
...

V 5 × 25 × 2 ft 3

...


...

10 min
t

For a full room, h = 7 m

(i)

⇒F >

1000 kg 9
...
4 ×105 N

The door will break before the room fills
(ii)

d i
dP i

3
...
3145 ft 3
5 × 15 × 10 m3
V
t fill = room =
V
12
...


b

tap

=

25 m H 2 O

junction

g

1h
= 31 h
60 min

101
...
33 m H 2 O
25 + 5 m H 2 O
101
...
33 m H 2 O

b

g

(b) Air in the line
...
)
(c) The line could be clogged, or there could be a leak between the junction and the tap
...
40

Pabs = 800 mm Hg
Pgauge = 25 mm Hg
Patm = 800 − 25 = 775 mm Hg

3-17

b

g

3
...
792g g 981 cm 30
...
792g g

...

cm
cm
s
N
F 1 dyne I F
I

...
01325101325dynes / cm JK = 123
...
42

3

6

OP
Q

981 cm 24
...

(i) Hg: ρ t = 0
...
6, h = 150 cm ⇒ R = 238 cm

...
866, ρ m = 100, h = 150 cm ⇒ R = 2260 cm

Use mercury, because the water manometer would have to be too tall
...

Advantages of using mercury: smaller manometer; less evaporation
...

3
...
23 g
F P − ρ gIJ b26 cmg
P − P = d ρ − ρ i gb26 cmg = G
H 7
...
23 m ⇒ ρ f =

atm

atm

a

b

f

w

w

F756 mmHg 1 m −1000 kg 9
...
23 m 100 cm m

2

=

3

Ib g
JK

1N
760 mmHg
1m
26 cm
1 kg⋅ m/s2 1
...


3
...

14
...
696 psi
760 mm Hg

3-18

= 6
...
45 (a) h = L sin θ
(b) h = 8
...
3 cm H 2 O = 23 mm H 2 O

b

g b g

3
...
81 m / s2
= 765 − 365 −
m3
= 393 mm Hg

0
...
01325 × 105 N / m2

(b) — Nonreactive with the vapor in the apparatus
...

— Low rate of evaporation (low volatility)
...
47 (a) Let ρ f = manometer fluid density 110 g cm 3 , ρ ac = acetone density

...
791 g cm h
3

d

i

Differential manometer formula: ΔP = ρ f − ρ ac gh


...

1
1
g b110− 0791gg 981 cm h (mm) 10 cm 1 g⋅dynes
cm
s
mm
cm/
= 0
...
114 0
...
341 0
...
568 0
...
5

y = 0
...
2068

5
4
...
5

-2

-1
...
5

0

b g


...

From the plot above, ln V = 04979 ln ΔP + 52068
ml s
⇒ n = 04979 ≈ 05 , ln K = 5
...


...
5

760 mm Hg
1
...
47 (cont’d)

b

b

gb g

g

(c) h = 23 ⇒ ΔP = 0
...
523 mm Hg ⇒ V = 183 0
...
791 g
s

mL

= 104 g s

104 g
s

1 mol
58
...
5

= 132 mL s

= 180 mol s

...


...
48 (a) T = 85° F + 4597 = 544° R / 18 = 303 K − 273 = 30° C

...

85° C 18° F

...
8° R
= 85° K;
= 153° F;
= 153° R

...

10° C
1° C
10° C

150° R 1° F
150° R 1
...
0 K
= 150° F;
= 83
...
3° C
1° R
1
...
8° R

3
...
0940 × 1000 FB + 4
...
0 C ⇒ T = 98
...
8 + 32 = 208 F
(b) ΔT ( C) = 0
...
94 C ⇒ ΔT (K) = 0
...
94 C 1
...


...
0 C

(c) T1 = 15 C ⇒ 100 L ; T2 = 43 C ⇒1000 L

T ( C) = aT ( L) + b
a=

⇒

b43 − 15g C = 0
...

b = 15 − 0
...
0311T ( L) + 11
...
0940T (o FB)+4
...
9 ⎤ = 3
...
0311 ⎣

(d) Tbp = −88
...
6 K ⇒ 332
...
4 F ⇒ −9851 FB ⇒ −3232 L

...
0 L ⇒ 1
...
6 FB ⇒ 156 K ⇒ 2
...
8 R

...
50

bT g = 100° C bT g
(a) V b mVg = aT b° Cg + b
b H 2O

m AgCl

5
...
05524 mV ° C

24
...
2539 mV
V mV = 0
...
2539

b g

b g

⇓
T ° C = 1810V mV + 4
...


b g

b g


...


...

(b) 100 mV →136 mV ⇒1856° C →2508° C ⇒
3
...


...

dt
20 s

T = KR n

g = 1184

...
0 110
...
0 20
...

ln K = ln 1100 − 1184(ln200) = 1154 ⇒ K = 3169 ⇒T = 3169R1184

...


...


...


1/1
...
3

(c) Extrapolation error, thermocouple reading wrong
...
52 (a) PV = 0
...

14696


...

28317 ft 3
, V L = V ′ ft ×
L
3

T ′( F) − 32
453
...

, T( K) =
+ 27315
lb − moles
1
...
696g × V ′ × 28317 = 0
...
59 × L(T ′ − 32) + 27315O

...
P
MN 1
...
696
1
Q

⇒

b

g

b

g

⇒ P ′ + 14
...
08206 × 14
...
59
× n ′ × T ′ + 459
...
317 × 18

...
696 V ′ = 1073n′ T ′ + 459
...


3-21

g

3
...

b500 + 14
...
308 lb - mole
10
...
7g

(b) ntot =
′

mCO =

(c) T ′ =

0308 lb - mole 0
...

= 2
...

b3000 + 14
...
7 = 2733 F
10
...
308

b g

b

g

3
...

0 = 23624a + b

...

b = −25122

...


...


1 min V ′
=
60 s 60

b g d
b g

i


...

V′
n′ 12186 P′
=
⇒ n′ =

...

760 T ′ + 27316 60
(c) T = 10
...


r1 = 26159 ⇒ T1 = 26
...

⇒ r2 = 26157 ⇒ T2 = 26
...

r3 = 44
...

P (mm Hg) = h + Patm = h + (29
...
9
H 29
...
9 mm Hg
1

...
9 mm Hg

3-22

3
...
016034gb987
...
8331 kmol CH
26
...


...
016034gb9119gb195g = 9
...
93 + 27316

...
33 kmol min
(e) V3 =

(f)

b

g b

gb

g

n3 T2 + 27316

...
33 2251 + 27316

...

=
= 387 m3 min
0
...
016034 829
...
8331 kmol CH 4 16
...
21× 9
...
0 kg O2

min
xCH4 =

kmol O2

g

kg CH 4
min
0
...
501 kmol N2 28
...
36

+

min

13
...
0465 kg CH 4 kg
(13
...
15
READ 5, ∗ (TIME (J), CA (J), J = 1 , N)
DO 1 J=1, N
CA J = CA J / MW

3
...
/CAbJg

1

CONTINUE
CALL LS (X, Y, N, SLOPE, INTCPT)

b g

K IT = SLOPE

WRITE (E, 2) TK (IT), (TIME (J), CA (J), J = 1 , N)
WRITE (6, 3) K (IT)
10 CONTINUE
DO 4 J=1, NT
X J = 1
...
54 (cont’d)

4 CONTINUE
CALL LS (X, Y, NT, SLOPE, INTCPT)
KO = EXP INTCPT

b

2

3
5

10

g

E = −8
...
2, /
* ' TIME CA', /,
* ' (MIN) (MOLES)', /
* 100 (IX, F5
...
4, /))
FORMAT (' K (L/MOL – MIN): ', F5
...
4, /, ' E (J/MOL): ', E 12
...
0
94
...
0
20
...
0
40
...
0

4
6
8
...
3
3
...
2
1
...
15
TIME CA
(MIN) (MOLS/L)
10
...
1246
20
...
0662
30
...
0462
40
...
0338

3-24

3
...
0

1
...
00 0
...
00 0
...
707
110
...
0
20
...
0
40
...
0
60
...
5
1
...
2
0
...
73
0
...


6

bat 94°Cg

TEMPERATURE (K): 383
...
758

b
g
E bJ / MOLg: 0
...
2329E + 10
ETC

3-25

CHAPTER FOUR
a
...


4
...
00
c
...
00
=
⇒
= 3
...

= 333 s
1 m3 3
...


Continuous, Steady State

b
...


4
...
3

k = ∞ ⇒ CA = 0

3

A0

A

3

b

a
...
550 kg B / kg
0
...
850 kg B / kg
0
...
106 kg B / kg
0
...
0 kg / h = mv + ml

(2) Benzene Balance: 0
...
0 kg B / h = 0
...

Solve (1) & (2) simultaneously ⇒ mv = 59
...
3 kg h
b
...
The balance equations are also identical (initial input = final output)
...


Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state,
the feed composition is incorrect, the flow rates are not what they are supposed to be, other
species are in the feed stream, measurement errors
...
4

b
...
500 mol N 2 mol
0
...


100
...


3

b

g

g

100 x E g C 2 H 6 1 lb m lb - mole C2 H 6 3600 s
s
h
453593 g 30 lb m C2 H 6

...
21 lb - moleO lb - mole DA U
|
S
| 0
...
21n2 ( lb-mole O 2 / s )

n1 lb - mole H 2 O s
2

xH2O =

2

2

e
...
45x E lb - mole C 2 H 6 / h

8

4

g

0
...
014 n kg N 2
mol N 2 1000 g

xO2 =

n ( mol )

n1 ⎛ lb-mole H 2 O ⎞
n1 + n2 ⎜ lb-mole ⎟
⎝
⎠

0
...
600 − yNO2 ⎤ ( mol N 2 O4 )
⎣
⎦

0
...
600 − yNO2 ( mol N 2 O 4 mol )
4
...


Basis: 1000 lbm C3H8 / h fresh feed
(Could also take 1 h operation as basis flow chart would be as below except
that all / h would be deleted
...
02 lb m C3H8 / lb m
0
...
97 lb m C3H8 / lb m
0
...


b
b

n3 lb m CH 4 / h
n4 lb m H 2 / h

g

g

b

n5 lb m / h

g

Stripper

Absorber

b
b
n blb

n1 lb m C3H 8 / h

g
g

n2 lb m C3H 6 / h
5

4- 2

g
g

n2 lb m C3H 6 / h

6

m

oil / h

g

4
...
Overall objective: To produce C3H6 from C3H8
...

Reactor function: Convert C3H8 to C3H6
...

Stripping tower function: Recover the C3H8 and C3H6 from the solvent
...

4
...


3 independent balances (one for each species)

b
...


y2 = 1 − x2

FG kg A IJ = m + b1200gb0
...
60m FG kg BIJ
B Balance: 0
...
70 − y4
4
...


3 independent balances (one for each species)

b
...
885 g H 2O mR g 0
...

b400gb0115g FGH g CH OOH IJK = 0
...

b0115gb400g − b0
...
096gb461g FGH min IJK ⇒ 44 g min = 44 g min

Overall Balance: mC + 400
c
...
096mE

R

4- 3

E

C

4
...


CH 3COOH

H 2O
some CH3COOH
CH3COOH
H 2O
C 4 H9OH

Extractor

C4 H9OH
CH3COOH

Distillation
Column

C4 H 9OH

4
...

120 eggs/min
0
...
70 unbroken egg/egg

X-large: 25 broken eggs/min
35
45 unbroken eggs/min
Large: n 1 broken eggs/min
n 2 unbroken eggs/min

b
...
30 )(120 ) = 25 + n1
⎪ n2 = 39
⎭

c
...
22

b

g

d
...
9

22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand)
are broken
...
Therefore, Fred is right-handed
...


m1 lb m strawberries

b

015 lb m S / lb m

...
85 lb m W / lb m

c

m2 lb m S sugar

g

b

m3 lb m W evaporated

g

1
...
667 lb m S / lb m

h

0
...


c
...
15 m1 + m 2 = 0
...
49 lb m strawberries, m 2 = 0
...
10

a
...
750 lb m C 2 H 5OH / lb m

m3 lb m

0
...
600 lb m C 2 H 5OH / lb m
0
...
400 lb m C 2 H 5OH / lb m
0
...


m1 =

300 gal

1ft 3
0
...
4 lb m
= 2195 lb m
7
...
750m1 + 0
...
600m3
Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841 lb m
V40 =
4
...


b

1646 lb m

n1 mol / s

ft 3 7
...
952 × 62
...
0403 mol C3H 8 / mol

b

n3 mol / s

0
...
0205 mol C3H 8 / mol

g

0
...
21 mol O 2 / mol
0
...


b

g

Propane feed rate: 0
...
0403n1 = 0
...


>
...


4- 5

4
...


b

m kg / h

g

0
...
500 kg CH 3OH / kg
0
...
040 kg H 2O / kg
673 kg / h

b

g
1 − x b kg H O / kgg
x kg CH 3OH / kg
2

b
...
500 1000 = 0
...
276 kg CH 3OH / kg
Molar flow rates of methanol and water:
673 kg 0
...
80 × 103 mol CH3OH / h
h
kg
kg 32
...
724 kg H 2O 1000 g mol H 2O
= 2
...
80 × 103
= 0176 mol CH 3OH / mol

...
80 × 103 + 2
...


4
...


a
...
000145R

1
...
1

0
...
13 (cont’d)
b
...
000145 388 1
...
494 kg P / kg
p

b g
Product: x = 0
...
861 kg P / kg
Waste: x = 0
...

0
...
8%
0
...
3645

p

1
...


Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg
P balance on purifier:
Input: 0
...
861 kg P / kggb1239 kgg + b0123 kg P / kggb1014 kgg = 1192 kg P

...
Analyzer readings are wrong; impure feed; extrapolation
beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady
state; additional reaction occurs in purifier; normal data scatter
...
14

a
...

00100 lb- mole H2O/ lb- mole

...

0100 lb- mole H2O/ lb- mole

g


...
9 − 40
...
626
=
R2 − R1
50 − 15

b g

Intercept: b = v a − aR1 = 40
...
626 15 = 15
...
626 95 + 15
...
4 lb m lb - mol
= 589 lb - moles H 2 O / h
h
ft 3
18
...
9900n1 = 0
...


Bad calibration data, not at steady state, leaks, 7% value is wrong, v − R relationship is not
linear, extrapolation of analyzer correlation leads to error
...
15

a
...
900 kg E / kg

0
...


0
...
350 kg H 2 O / kg

b

m kg / s

g

3 unknowns ( m, xE , xS )
– 3 balances
0 DF

b
g
x b kg S / kg g
1 − x − x b kg H O / kg g
x E kg E / kg
S

E

b
...


Overall balance: 100 = 2m ⇒ m = 50
...
050 100 = xS 50 ⇒ xS

b g

b g b g
kg E in bottom stream 0
...
25
kg E in feed
0
...
600 100 = 0
...
300 kg E / kg

c
...
400 / 0100g

...

lnb R / R g
lnb38 / 15g
lnbag = lnb x g − b lnb R g = lnb0100g − 1491lnb15g = −6
...


...

x = aRb ⇒ ln x = ln a + b ln R
2

1

2

1

1

−3

1

x = 1764 × 10−3 R1
...


F x I F 0
...

1
b

−3

d
...
491


...
Calibration curve deviates from linearity at high mass
fractions – measure against known standard
...

Mixture is not all liquid – check sample
...
System is not at steady state – take more measurements
...


4- 8

4
...


b
...
00 mol H 2SO 4 0
...
323 kg H 2SO 4 / kg solution
L of solution
mol H 2SO 4 1213 kg solution

...
200 kg H 2SO 4 / kg

3

0
...
323 kg H 2SO 4 / kg

SG = 1139

...
677 kg H 2 O / kg
SG = 1
...
600 kg H 2SO 4 / kg
0
...
498

U ⇒ m = 44
...
800b100g + 0
...
677m W
Overall mass balance: 100 + m2 = m3

2

2

v1 =

100 kg

v2 =

44
...
80 L20%solution
1139 kg

...
64 L 60% solution
1498 kg

...
80
L 20%solution
=
= 2
...
64
L 60% solution
c
...
17

1250 kg P 44
...


b g

m1 kg @$18 / kg
0
...
75 kg H2O / kg

100 kg

...

0
...

0
...


(1)

(2)
b g
Solve (1) and (2) simultaneously ⇒ m = 0
...
615 kg12% paint
Cost of blend: 0
...
00g + 0
...
00g = $13
...
08g = $14
...



...


...
25m1 + 012m2 = 017 100
1

4- 9

2

4
...


gb

m1 kg H 2O 85% of entering water

100 kg
0
...
200 kg H 2O / kg

g

b g
m b kg H Og
m2 kgS
3

2

b

gb g
Sugar balance: m = 0
...
0 kgS

85% drying: m1 = 0
...
200 100 = 17
...
0361 kg H 2O / kg
3 + 80 kg

b

g

m1
17 kg H 2O
=
= 0
...


g

1000 tons wet sugar
3 tons H 2 O
= 30 tons H 2 O / day
day
100 tons wet sugar
1000 tons WS 0
...
15 365 days
= $8
...


b

g

1
x w1 + x w 2 +
...
0504 kg H 2 O / kg
10
1
2
2
SD =
x w1 − x w +
...
00181 kg H 2 O / kg
9
Endpoints = 0
...
00181
xw =

b

g

b

b

g

g

Lower limit = 0
...
0558
d
...
19

The evaporator is probably not working according to design specifications since
x w = 0
...
0450
...


v1 m 3

c h
m b kg H O g
2

1

SG = 1
...
48

5 unknowns ( v1 , v2 , v3 , m1 , m3 )
– 1 mass balance
– 1 volume balance
– 3 specific gravities
0 DF

400 kg galena
SG = 7
...
19 (cont’d)

Assume volume additivity:

b g

m1 kg

b g

m kg
m3
400 kg m 3
m3
+
= 3
(2)
1000 kg
7440 kg
1480 kg

Solve (1) and (2) simultaneously ⇒ m1 = 668 kg H 2O, m3 = 1068 kg suspension
v1 =

668 kg

m3
= 0
...

c
...
20

Specific gravity of coal < 1
...
Stir the suspension
...
00 < Specific gravity of coal < 1
...


b

n1 mol / h

g

b

n2 mol / h

b

g

g

0
...
960 mol DA / mol

1 − x mol DA / mol

b

b

n3 mol H 2O adsorbed / h

g

g

97% of H 2O in feed

Adsorption rate: n3 =


...
40g kg

b

5h

g

mol H 2O
= 1556 mol H 2O / h

...
0180 kg H 2O

97% adsorbed: 156 = 0
...
04n1 ⇒ n1 = 401 mol / h

...


...
54 mol / h

Water balance: 0
...
1) = 1
...
54 ) ⇒ x = 1
...


4
...
Eventually the mole fraction
will reach that of the inlet stream, i
...
4%
...


300 lb m / h
0
...
45 lb m H 2O / lb m

b

mB lb m / h

mC lb m / h

g

g

0
...
90 lb m H 2SO 4 / lb m

0
...

Overall balance: 300 + mB = mC

(1)

H2SO4 balance: 0
...
90mB = 0
...
21 (cont’d)
b
...
78 RA − 44
...
0 RB − 100
60 − 20
ln 100 − ln 20
ln x − ln 20 =
Rx − 4 ⇒ ln x = 0
...
841e 0
...

10 − 4
300 + 44
...
3, mB = 400 ⇒ RB =
= 333,

...
78
15
...
78
ln
0
...
841

b
b

mA − 150 =

g
g

b

g

FG
H

c
...
01xmA + 0
...
75mC = 0
...
75 − 0
...
841e

⇒ 15
...
59 − 0
...
2682 Rx

iR

0
...
78 R

A

b

n A kmol / h

100 kg / h

0
...
2682 7
...

j44
...


...
2682 7
...


nB kmol / h

015

...
2682 Rx − 813

...


e

a
...
4

b

g

0
...
80 kmol N 2 / kmol

0
...
50 kmol N 2 / kmol

b

g

b

g

MW = 0
...
016 + 0
...
012 = 22
...


Check: RA = 44
...
78 ⇒ RB = 2
...
236e

4
...
75 − 0
...
38 kmol / h
h 22
...
38

b g


...
50 nB = 0
...
38


...
29 kmol / h, n B = 110 kmol / h

4- 12

(1)
(2)

4
...


nP =

mP
22
...
813
x m
H2 balance: x A n A + x B nB = P P
22
...


d
...
23

Trial
1
2
3
4
5
6
7
8
9
10
11
12

nA =

b
b

mP x B − x P
22
...
10
0
...
10
0
...
10
0
...
10
0
...
10
0
...
10
0
...
50
0
...
50
0
...
50
0
...
50
0
...
50
0
...
50
0
...
10
0
...
30
0
...
50
0
...
10
0
...
30
0
...
50
0
...
813 x B − x A

mP
100
100
100
100
100
100
250
250
250
250
250
250

g
g

nA
4
...
29
2
...
10
0
...
10
10
...
22
5
...
74
0
...
74

nB
0
...
10
2
...
29
4
...
48
0
...
74
5
...
22
10
...
70

The results of trials 6 and 12 are impossible since the flow rates are negative
...

Results are the same as in part c
...
0 ml / min
1
...
0 ml / min
1
...


v ml / min

Water removal rate: 200
...
0 = 5
...


...
0 − 175 195
...
8 mg urea / min
b
...


v = 1500 + 5
...
8 mg urea/min
c=
= 0
...

b2
...
8 mg removed

4- 13

10 3 ml 5
...
4 h)

4
...


b

n1 kmol / min

g
b

20
...
023 kmol CO 2 / kmol

0
...
0 kg CO 2
kmol
= 0
...
0 kg CO 2
Overall balance: 0
...
455 + 0
...
023n3
Solve (1) and (2) simultaneously ⇒ n2 = 55
...

n1 =

b
...
33 m / s
18 s

1
561 kmol

...

4
min 0123 kmol 60 s 8
...


b

g

Spectrophotometer calibration: C = kA ====> C μg / L = 3
...
25

A = 0
...
600 μg / L

...


...
60 cm 3

1L
5
...
0 μ g
103 cm 3
1L
1 mg

g bg

⇒ 3
...
600 μ g / L ⇒ V = 5
...
26

a
...
26 (cont’d)

8 unknowns ( n1 , n3 , v1 , m2 , m4 , x4 , y1 , y3 )
– 3 material balances
– 2 analyzer readings
– 1 meter reading
– 1 gas density formula
– 1 specific gravity
0 DF

b
...
h is a line through the points h1 = 100, V1 = 142 and h2 = 400, V2 = 290
...
515
lnbh h g lnb400 100g
ln a = ln V − b ln h = lnb142g − 0
...
58 ⇒ a = e

b=

i

d

b

ln V2 V1
2

1

1

1

2
...
2 ⇒ V = 13
...
515

Analyzer calibration:
ln y = bR + ln a ⇒ y = aebR
b=

b

ln y 2 y1
R2 − R1

ln a = ln y1 − bR1

E

a = 5
...


U
|
90 − 20
|
|
= lnb0
...
0600b20g = −7
...
00 × 10
|
|
|
W


...
00166g = 0
...
0600 R

b g = 207
...
2g b150 + 14
...
7 batmg = 0
...
460 kmol / m
=

...
2 210

ρ feed gas

−4

0
...
3 m 3 0
...
34 kmol min
min
m3

b
g
expb0
...
00100 kmol SO

...
4 ⇒ y1 = 500 × 10−4 exp 0
...
4 = 0
...

R3 = 116 ⇒ y3 = 500 × 10−4

...

m2 =

1000 L B 130 kg

...
26 (cont’d)
A balance: 1 − 0
...
34 = 1 − 0
...
7 kmol min

b

SO2

gb g b
g
balance: b0
...
0 kg / kmol) = b0
...
7g( 64) + m x

...
245 kg SO2 absorbed / kg
SO2 removed = m4 x4 = 422 kg SO 2 / min
d
...
27

Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a
higher rate of transfer of SO2 from the gas to the liquid phase
...


V2 m 3 / min

b
g
y b kmolSO / kmol g
1 − y b kmol A / kmol g

d
i
m b kg B / min g

n3 kmol / min

3

d
i
n b kmol / min g
y b kmolSO / kmol g
1 − y b kmol A / kmol g
3

R3

V1 m / min

b
g
x b kgSO kg g
1 − x b kg B / kg g
m4 kg / min

1

1

2

3

2

2

4

1

2

4

P , T1 , R1 , h1
1

b
...
00 × 10−4 e0
...
060 R3

y3 = 5
...
2h10
...
2 P + 14
...
7
1

...


4- 16

V1

(7)

b g

(8)

m2 kg
m3
h
1300 kg

4
...


75 °F

y1

0
...
26 kmol/h

R1

82
...
10
0
...
10
0
...
10
0
...
20
0
...
20
0
...
050
0
...
010
0
...
001
0
...
025
0
...
005
0
...
89
1
...
56
2
...
92
0
...
87
1
...
23
1
...
25
90
...
48
89
...
68
93
...
86
89
...
03
88
...
45
2813
...
78
3982
...
72
641
...
86
1847
...
28
2105
...
11
2532
...
31
3584
...
65
513
...
49
1477
...
03
1684
...
y3
3
...
0 0
2
...
0 0
1
...
0 0
0
...
0 0
0
...
0 2 0

0
...
06 0

y 3 ( k m o l S O 2 /k m o l)
x4 = 0
...
20

For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed
rate ( V2 )
...

d
...
28

Answers are the same as in part c
...
29

a
...
300 mol B / mol
0
...
450 mol X / mol

b

n2 mol / h

b

g

b
g
Column 1 x b mol T / molg
1 − x − x b mol X / molg
n b mol / h g

n4 mol / h

0
...
060 mol T / mol

x B 2 mol B / mol
T2

B2

Column 2

T2

3

0
...
980 mol X / mol

b

n5 mol / h

b
g
x b mol T / molg
1 − x − x b mol X / molg
T5

Column 1
96% X recovery: 0
...
450 100 = 0
...
300 100 = x B 2 n2

(3)

T 2 n2

+ 0
...
97 x B 2 n2 = 0
...
940 n4 + x B5 n5

(7)

T balance: xT 2 n2 = 0
...


(8)

(1) ⇒ n3 = 44
...
9 mol / h

(3) ⇒ x B 2 = 0
...
431 mol T / mol

(5) ⇒ n4 = 30
...
96 mol / h

(7) ⇒ x B5 = 0
...
892 mol T / mol

b

g

0
...
95
× 100% = 97%
0
...
892b24
...
250b100g

Overall benzene recovery:

T5

Column 2:
4 unknowns ( n3 , n4 , n5 , y x )
– 3 balances
– 1 recovery of B in top (97%)
0 DF

Column 1
4 unknowns ( n2 , n3 , x B 2 , xT 2 )
–3 balances
– 1 recovery of X in bot
...
250b100g = x

g

x B5 mol B / mol

B5

b

4- 18

g

4
...


b
b

g

100 kg / h

m3 kg / h

0
...
965 kg H 2O / kg

x3 kg S / kg

b

mw kg H 2O / h

b
...


g

b g
x b kg S / kgg
1 − x b kg H O / kgg

4

g

4

4

g

b

m10 ( kg / h)
0
...
950 kg H2O/kg

mw ( kg H 2 O / h)

b g

Salt balance: 0
...
050m10
Overall balance: 100 = mw + m10
H2O yield: Yw =

b
b

g

mw kg H 2O recovered
96
...
035 kg S/ kg
0
...


b g

Salt balance: 0
...


g

b
g
4 × 0100m b kg H O / hg

...


Overall process
100 kg/h
0
...
965 kg H2O/kg

b

m4 kg / h

g

Yw = 0
...
0398

4- 19

2

g

m10 kg / h

10

g

0
...
950 kg H 2O / kg

b

0100mw kg H 2O / h

...
31

b g

a
...
97 mol B / mol
0
...
50 mol B / mol
0
...
2% of Bin feed )

0
...
97 mol B / mol
0
...
03 mol T / mol

Still

b gb
y b mol B / molg
1 − y b mol T / molg

n4 mol 45% of feed to reboiler

g

B

B

b g
z b mol B / molg
1 − z b mol T / molg
n2 mol
B

b g
x b mol B / molg
1 − x b mol T / molg
n3 mol

Reboiler

B

B

B

Overall process:

Condenser:

3 unknowns ( n1 , n3 , x B )
Still: 5 unknowns ( n1 , n2 , n4 , y B , z B )
– 2 balances
– 2 balances
– 1 relationship (89
...
25 ratio & 45% vapor)
3 DF

Begin with overall process
...


Overall process
89
...
892 0
...
97 n1

b gb g

Overall balance: 100 = n1 + n3

b g

B balance: 0
...
97 n1 + x B n3
Reboiler

e j = 2
...
45n2

(1)

Mole balance: n2 = n3 + n4

(2)

(Solve (1) and (2) simultaneously
...
31 (cont’d)
c
...


Moles of overhead: n1 = 46
...
32

Moles of bottoms: n3 = 54
...

b1 − x gn × 100% = b1 − 010gb54
...
50b100g
0
...


b

m3 kg H 2O

g

Bypass

Mixing point

b g

Basis: 100 kg
100 kg
0
...
88 kg H2O / kg

b g

b g

m5 kg

0
...
42 kg H 2O / kg

012 kg S / kg

...
88 kg H2O / kg

m4 kg

0
...
58 kg H2O / kg

b g

m2 kg

012 kg S / kg

...
88 kg H2O / kg

Overall process:

Evaporator:

2 unknowns ( m3 , m5 )
– 2 balances
0 DF

3 unknowns ( m1 , m3 , m4 )
– 2 balances
1 DF

Bypass:

Mixing point:

2 unknowns ( m1 , m2 )
– 1 independent balance
1 DF

3 unknowns ( m2 , m4 , m5 )
– 2 balances
1 DF

b g


...
42m5
Overall mass balance: 100 = m3 + m5
Mixing point mass balance: m4 + m2 = m5

(1)

Mixing point S balance: 0
...
42m5

...


m1 = 90
...
95 kg, m3 = 714 kg, m4 = 18
...
6 kg product

...


m2
= 0
...


4- 21

4
...


b

m4 kg Cr / h

b

m1 kg / h

g

0
...
9485 kg W / kg

b

m2 kg / h

g

g

b g
x b kg Cr / kgg
1 − x b kg W / kgg
m5 kg / h

0
...
9485 kg W / kg

Treatment
Unit

b

m3 kg / h

5

5

b g
x b kg Cr / kgg
1 − x b kg W / kgg
m6 kg / h
6

6

g

0
...
9485 kg W / kg

b
...
95 0
...
2 kg Cr / h
Mass balance on treatment unit: m5 = 4500 − 220
...
8 kg / h
0
...
2
= 0
...
8
Mixing point mass balance: m6 = 1500 + 4279
...
8 kg / h

b

Mixing point Cr balance: x6 =

c
...
0515 1500 + 0
...
8
= 0
...
8

m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h)
1000
1000
0
48
...
9
1902
3000
3000
0
147
2853
4000
4000
0
196
3804
5000
4500
500
220
4280
6000
4500
1500
220
4280
7000
4500
2500
220
4280
8000
4500
3500
220
4280
9000
4500
4500
220
4280
10000
4500
5500
220
4280

4- 22

m 6 (kg/h)
x5
0
...
00271
1902
0
...
00271
3804
0
...
00271
5780
0
...
00271
7780
0
...
00271
9780

x6
0
...
00271
0
...
00271
0
...
0154
0
...
0247
0
...
0301

4
...
x 6
0
...
03000
0
...
02000
0
...
01000
0
...
00000
0

2000

4000

6000

8000 10000 12000

m 1 (kg/h)

d
...
34

Cost of additional capacity – installation and maintenance, revenue from additional
recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon,
regulatory limits on Cr emissions
...


b

175 kg H 2O / s 45% of water fed to evaporator

b

m1 kg / s

g

b
b

m4 kg K 2SO 4 / s

0196 kg K 2SO 4 / kg

...
804 kg H 2O / kg

g

Crystallizer
Filter

Filter cake

b

10 m2 kg K 2SO 4 / s

g

b
g
R0
...
600 kg H O / kg V
W
T
m2 kgsoln / s
2

Filtrate

b

m3 kg / s

4

2

g

0
...
600 kg H 2 O / kg

Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s
Overall process: 2 unknowns ( m1 , m2 )
- 2 balances
0 DF
Evaporator:

Mixing point:

4 unknowns ( m4 , m5 , m6 , m7 )
– 2 balances
– 1 percent evaporation
1 DF

Crystallizer:

4 unknowns ( m1 , m3 , m4 , m5 )
- 2 balances
2 DF

4 unknowns ( m2 , m3 , m6 , m7 )
– 2 balances
2 DF

U verify that each
| chosen subsystem involves
|
V
Balances around mixing point ⇒ m , m | no more than two
Balances around evaporator ⇒ m , m | unknown variables
W

Strategy: Overall balances ⇒ m1 , m2
% evaporation ⇒ m5

3

6

4- 23

4

7

4
...

0196m1 = 10m2 + 0
...
450m5
W balance around mixing point: 0
...
600m3 = m5
Mass balance around mixing point: m1 + m3 = m4 + m5
K balance around evaporator: m6 = m4
W balance around evaporator: m5 = 175 + m7
Mole fraction of K in stream entering evaporator =

b
...

Production rate of crystals = 10m2 = 416 kg K s s
Recycle ratio:

c
...
3
kg recycle
=
= 160

...
8
kg fresh feed
m1 kg fresh feed s

b

g

Scale to 75% of capacity
...
75(398 kg / s) = 299 kg / s
46
...
7% W

d
...
Principal costs are likely to be the heating cost for the evaporator and the dryer and
the cooling cost for the crystallizer
...
35

a
...

Absorber function: Separates CO2 from CH4
...


b
...
The liquids are heavier than
the gases so the liquids fall through the columns and the gases rise
...


b

n1 mol / h

g

b
g
n b mol CO / h g
n5 mol N 2 / h

0
...
990 mol CH 4 / mol

100 mol / h
0
...
005 mol CO 2 / mol

0
...
995 mol CH 3OH / mol

b
g
n b mol CH OH / h g
n3 mol CO 2 / h
4

Overall:

Stripper:

3

3 unknowns ( n1 , n5 , n6 )
– 2 balances
1 DF

Absorber:

4 unknowns ( n1 , n2 , n3 , n4 )
– 3 balances
1 DF

4 unknowns ( n2 , n3 , n4 , n5 )
– 2 balances
– 1 percent removal (90%)
1 DF

b0
...
990n
Overall mole balance: 100b mol / h g = n + n

Overall CH4 balance:

4

1

1

6

Percent CO2 stripped: 0
...
005n2
Stripper CH3OH balance: n4 = 0
...


n1 = 70
...
55 mol CO 2 / h, n4 = 647
...

n6 = 29
...
0 − 0
...
976 mol CO 2 absorbed / mol fed
Fractional CO2 absorption: f CO 2 =
30
...
35 (cont’d)

Total molar flow rate of liquid feed to stripper and mole fraction of CO2:
n3
n3 + n4 = 680 mol / h, x3 =
= 0
...


Scale up to 1000 kg/h (=106 g/h) of product gas:

b

g

b

g

MW1 = 0
...
99 16 g CH 4 / mol = 16
...


...
28 g / molg = 6142 × 10 mol / h

...
71 mol / h) = 8
...


g
...
36

Ta < Ts The higher temperature in the stripper will help drive off the gas
...

The methanol must have a high solubility for CO2, a low solubility for CH4, and a low
volatility at the stripper temperature
...


Basis: 100 kg beans fed

e

m kg C H
5
6 14

e

m kg C H
1
6 14

j

300 kg C 6 H14

Ex

j

Condenser

b g
b
g
y b kg oil / kgg
1 − x − y b kg C H
m2 kg

2

2

2

6

b g
b
g
1 − y b kg C H
m4 kg

F

x2 kg S / kg

14 / kg

13
...
0 kg S

Ev

y4 kg oil / kg
6

4

g

14

/ kg

g

g

b g

m3 kg

0
...
25 − y3 kg C 6 H14 / kg

Overall:

b

m6 kg oil

4 unknowns ( m1 , m3 , m6 , y3 )
– 3 balances
1 DF

Extractor:

g

3 unknowns ( m2 , x2 , y2 )
– 3 balances
0 DF

2 unknowns ( m1 , m5 )
Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 )
– 1 balance
– 2 balances
1 DF
2 DF
Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 )
– 3 balances
– 1 oil/hexane ratio
3 DF
Mixing Pt:

Start with extractor (0 degrees of freedom)
Extractor mass balance: 300 + 87
...
0 kg = m2

4- 26

4
...
0 kg S = x2 m2

Extractor oil balance: 13
...
0 kg S = 0
...
25 − y3

=

y2
1 − x2 − y2

Filter oil balance: 13
...
Hexane balance: m1 + m5 = 300 kg C6 H14
Evaporator oil balance: y4 m4 = m6
b
...

=
= 0118 kg oil / kg beans fed

...
28 kg C 6 H14 / kg beans fed
100 100 kg beans fed
m
272 kg C 6 H14 recycled
Recycle ratio = 5 =
= 9
...


g

g

Lower heating cost for the evaporator and lower cooling cost for the condenser
...
37

g

m lb m dirt
1
98 lb m dry shirts
3 lb m Whizzo

100 lbm
2 lbm dirt
98 lbm dry shirts

b

m lb m Whizzo
2

g

Tub

b g

Filter

b g

m lb m
3
0
...


0
...
87 lb m Whizzo / lb m

b g

m lb m
5
0
...
08 lb m Whizzo / lb m

b g
b
g
blb Whizzo/ lb g

m6 lbm
1 − x lb m dirt / lbm
x

m

m

Strategy
95% dirt removal ⇒ m1 ( = 5% of the dirt entering)
Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling
the chart) ⇒ m2 , m5 (solves Part (a))

4- 27

4
...
Balances around tub ⇒ m3 , m4
Balances around mixing point ⇒ m6 , x (solves Part (b))
a
...
0 = 010 + b0
...
065 lb dirt

...

Overall Whizzo balance: m = 3 + b0
...
065g blb Whizzog = 317 lb

95% dirt removal: m1 = 0
...
0 = 010 lb m dirt

...


m

m


...

Tub dirt balance: 2 + 0
...
97m3 = 3 + 0
...
4 lb m , m4 = 19
...
mass balance: 317 + m6 = 20
...
3 lb m

...
Whizzo balance:

m

Whizzo
(1)
(2)

3
...
3) = ( 0
...
4 ) ⇒ x = 0
...
38

a
...
2 + C ⇒
0
...

|
6138 + F = F + C V ⇒

...
01C = F
W
0
...

C1L = 6138 kg L
F2 L = 6
...
7 kg L

1L

3L

3L

2L

F1L = 6
...

F3L = 61 kg L
613
...
1+ C3L ⇒ C3L = 607
...
38 (cont’d)

Solvent
m f 1:

b g
3300 = 495 + F ⇒
U
015b495 + F g = C

...

2720 + C = F + C |
W
015 3300 = C1S ⇒

...
6 kg S

2S

F2 S = 2734
...
4 kg S

3S

F3S = 2722
...
2 + 6
...
4 kg leaf
2805 + 2734
...


b g

5540 kgS
0165 kg E / kg

...
835 kg W / kg

b g

Q0 kg

QR kg


...
15kg F / kg
Unit

b g
Q b kg Fg

Steam
Stripper

b g

b g
Q b kg D g
Q b kg Fg
QE kg E

F

0
...
774 kg W / kg
QB kg

0
...
200 kg E / kg

0
...
987 kg W / kg

F

b

Q3 kg steam
Mass of D in Product:

1 kg D
1000 kg leaf

620 kg leaf

g

= 0
...
835 5540 = 0
...


...


g

b

g

b

g

F balance around stripper
0
...
026Q0 ⇒ Q0 = 3121 kg mass of stripper overhead product

b

g

b

g

E balance around stripper
013 5410 = 0
...
013QB ⇒ QB = 6085 kg mass of stripper bottom product

...
855 5410 + QS = 0
...
987 6085 ⇒ QS = 3796 kg steam fed to stripper
4
...


C2 H 2 + 2 H 2 → C2 H 6
2 mol H 2 react / mol C 2 H 2 react
0
...
39 (cont’d)
b
...

= 15 < 2
...


...
75 mol C 2 H 2 required (theoretical)

...
75 mol required
% excess C 2 H 2 =

...
75 mol required
c
...
00 kg H 2
yr
300 days 24 h 3600 s tonne 30
...
6 kg H 2 / s

d
...


4 NH 3 + 5 O 2 → 4 NO + 6 H 2O
5 lb - mole O 2 react
= 125 lb - mole O 2 react / lb - mole NO formed

...


4
...


dn i

O 2 theoretical

=

100 kmol NH3 5 kmol O 2
= 125 kmol O 2
4 kmol NH 3
h

d i

40% excess O 2 ⇒ nO 2
c
...

= 140 125 kmol O 2 = 175 kmol O 2

b50
...
94 kmol NH

...
0 kg O gb1 kmol O / 32 kg O g = 3125 kmol O
F n I = 3125 = 106 < F n I = 5 = 125
GH n JK 2
...
GH n JK 4
...

= 2
...
94 − 2
...
6% excess NH 3
2
...
41

a
...
625 kmol = 625 mol

...
0 kg NO

...
0 kg NO
5 kmol O 2 1 kmol NO

By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed
...
It also may reduce labor costs
...
41 (cont’d)
b
...
00 × 10 2 kmol 0
...
5 kmol SO 2 / h
h
kmol
2 kmol H 2 S

c
...
2 0

X (mol H 2 S/mol)

1
...
8 0
0
...
4 0
0
...
0 0
0
...
0

4 0
...
0

8 0
...
0

R a (m V )

X = 0
...
0605

b

d
...
0 kmol / h, R = 25
...
0 kmol / h, R c = 10
...
0119 Ra − 0
...
0119 Ra − 0
...
00 × 10 2 kmol / h ⇒ R f =

g

3
n f = 45 mV
20

4- 31

g

4
...


gb g

10
5
45 0
...
5 − 0
...
9 mV
7
7
7
5
⇒ nc = 53
...
4 kmol / h
3
3
Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond
range of calibration data, system had not reached steady state yet
...
42
165 mol / s

b

x mol C 2 H 4 / mol

b

b

n mol / s

g

1 − x mol HBr / mol

g

0
...

0
...
310 2 + n 0
...
173)(1) + n ( 0
...
2) + (Eq
...
77 mol / s, x = 0
...
455 mol HBr / mol
Since the C2H4/HBr feed ratio (0
...
455) is greater than the stoichiometric ration (=1),
HBr is the limiting reactant
...
455 mol HBr / molg = 75
...
08 − ( 0
...
8 )
75
...
749 mol HBr react/molfed

= 75
...
545molC2 H 4 /mol ) = 89
...
93 − 75
...
8%
75
...
8 )( 0
...
2 mol/s

4- 32

4
...


2HCl +

1
O 2 → Cl 2 + H 2 O
2

Basis: 100 mol HCl fed to reactor

b
g
n b mol O g
n b mol N g
n b mol Cl g
n b mol H Og

100 mol HCl

b

n1 mol air

n2 mol HCl
3

0
...
79 mol N 2 / mol
35% excess

2

6

mol O
bO gstoic = 100 mol HCl 0
...
21n b mol O fed g = 135 × 25 ⇒ n = 160
...

2

2

2

1

2

1

85% conversion ⇒ 85 mol HCl react ⇒ n2 = 15 mol HCl
n5 =

85 mol HCl react

b gb g

1 mol Cl 2
= 42
...
5 mol H 2O
N 2 balance:

b160
...
79g = n

4

⇒ n4 = 127 mol N 2

O balance:

b160
...
21g mol O

2

42
...
5 mol O 2
1 mol H 2O
1 mol O 2

Total moles:
5

∑nj

= 239
...
063
, 0
...
530
,
239
...


b
...

mol
mol

As before, n1 = 160
...
5 mol

4- 33

4
...
5 mol O
= 0
...
7g = 127 mol N

O 2 : n3 = 0
...
7 −
N 2 : n4

2

2

Cl 2 : n5 = ξ = 42
...
5 mol H 2 O

c
...
44

These molar quantities are the same as in part (a), so the mole fractions would also be the
same
...
The cheaper process will be the process of
choice
...
90 kg TiO 2

12
...


1 kmol FeTiO 3
= 12
...
89 kmol FeTiO 3 dec
...
06 kmol FeTiO3

= 14
...
90 kg Ti

1 kmol FeTiO 3

b

1 kmol Ti

kmol Ti

g

= 6735 kg Ti fed

...
5 kg Ti / M kg ore = 0
...
06 kmol FeTiO3 + n kmol Fe 2O 3 (Assumption!)

n = 2772 kg ore −
638
...
06 kmol FeTiO3 151
...

kmol FeTiO3

kmol Fe 2O 3
= 4
...
69 kg Fe 2O 3

14
...
00 kmol FeTiO3 3 kmol H2SO4
+
= 4012 kmol H2SO4

...


...

50% excess: 15 4012 kmol H 2SO 4 = 6018 kmol H 2SO 4 fed
Mass of 80% solution:
5902
...
18 kmol H 2SO 4

a

98
...
4 kg H 2SO 4
1 kmol H 2SO 4

bkg solng = 0
...
45

a
...
R (linear scale) on semilog paper, get straight line through

d R = 10, C
1

1

i

= 0
...
67 g m 3

IK

ln C = bR + ln a ⇔ C = ae br
b=

b

g = 0
...
67g − 0
...
78 ⇒ a = e

ln 2
...
30

−1
...
0575 R

...
6 g 35
...


= 16,020C ′

i

16,020C ' = 0169e 0
...
0575 R

...

b
...


gb g = 8
...
0575 37

−5

lb m SO 2 ft 3

8
...
012 < 0
...


S + O 2 → SO 2
1250 lb m coal 0
...
06 lb m SO 2
min

1 lb m coal

32
...
9 lb m SO 2 generated min

2867 ft 3 60 s 886 × 10−5 lbm SO2

...

s
1 min
ft3
air
1250 lb m coal/min
62
...


furnace
ash

b124
...
2g lb

scrubbing fluid
stack gas
124
...
2 lb m SO2 /min
liquid effluent
(124
...
2) lbm SO2 (absorbed)/min

SO 2 scrubbed min
× 100% = 88%
124
...
The new regulation prevents this since the mass of SO2 emitted per mass of coal
burned is independent of the flow rate of air in the stack
...
46

a
...
87 ( nT ’s cancel)
y A yB
n A0 − ξ c nB0 − ξ c

b

c

gh b

g

387ξ 2 − nC0 + nD0 + 487 nA0 + nB0 ξ c − nC0nD0 − 487nA0nB0 = 0

...


...

1
∴ξ c =
−b ± b2 − 4ac where b = − nC0 + nD0 + 487 nA0 + nB0

...


e

b
...
87

g

nA0 = 1 nB0 = 1 nC0 = nD0 = nI 0 = 0

Basis: 1 mol A feed

ξe =

b

j

b = −9
...
87

(

1
9
...
87 )

( 9
...
87 )( 4
...
688

(ξ e 2 = 1
...


nA0 − nA ξ e1
=
= 0
...
87 =

b gb g
b
gb g

70 70
y C y D nC n D
=
⇒
= 4
...
6 mol methanol fed
y A y B n A nB
n A0 − 70 10

4- 36

4
...
6 − 70 = 100
...
401 mol CH3OH mol

B

= 0
...
279 mol CH3COOCH 3 mol

D

= 0
...
6 mol
d
...
47

Cost of reactants, selling price for product, market for product, rate of reaction, need for
heating or cooling, and many other items
...


CO + H 2O ← → CO 2 + H 2
⎯
(A)

(B)

(C)

(D)

b
g
n b mol H O g
n b mol CO g
n b mol H g
n b mol Ig


...
20 mol CO / mol

...
40 mol H 2O / mol

D

0
...


Since two moles are prodcued for every two moles that react,
ntotal out = ntotal in = 100 mol

...
20 − ξ
nB = 0
...

nD = ξ
n I = 0
...

At equilibrium:

b

y D = nD = ξ = 0110 mol H 2

...


b

gb g

FG
H

IJ
K

010 + ξ ξ

...

=
=
= 0
...
20 − ξ 0
...


4- 37

g

4
...


T (K)
1223
1123
1023
923
823
723
623
673
698
688

x (CO)
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5

x (H2O)
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5

1123
1123
1123
1123

0
...
4
0
...
5

x (CO2)

0
...
2
0
...
4

Keq
Keq (Goal Seek) Extent of Reaction
0
...
6610
0
...
8858
0
...
2424
1
...
2569
0
...
9240
1
...
2905
3
...
2661
0
...
4187
6
...
3585
15
...
6692
0
...
7017
9
...
3785
7
...
8331
0
...
5171
8
...
3724

0
0
0
0
0
0
0
0
0
0
0
...
1
0
0

0
...
8858
0
...
8858

0
...
8857
0
...
8867

0
...
1100
0
...
2156

y (H2)
0
...
242
0
...
291
0
...
358
0
...
378
0
...
372
0
...
110
0
...
216

The lower the temperature, the higher the extent of reaction
...

4
...


A + 2B → C

b g
g = lnd10
...
316 × 10 i = 11458

ln K e = ln A0 + E T K

b

ln K e1 / K e 2
E=
1 T1 − 1 T2

−4

1 373 − 1 573

ln A0 = ln K e1 − 11458 T1 = ln 10
...
37 ⇒ A0 = 4
...
79 × 10 −13 exp 11458 T K atm−2 ⇒ Ke (450K ) = 0
...


n A = n A0 − ξ
nB = nB 0 − 2ξ
nC = nC 0 + ξ
nT = nT 0 − 2ξ

U
|
|
V
|
|
W

b
b
b

gb
gb
gb

g
g
g

y A = n A0 − ξ nT 0 − 2ξ
y B = nB 0 − 2ξ nT 0 − 2ξ
⇒
yC = nC 0 + ξ nT 0 − 2ξ
nT 0 = n A0 + nB 0 + nC 0

b

g

At equilibrium,

b
b

gb
gb

n + ξ e nT 0 − 2ξ e
yC 1
= C0
2
2
y A yB P
n A0 − ξ e nB 0 − 2ξ e
c
...
)
e
P2

Basis: 1 mol A (CO)
n A0 = 1 nB 0 = 1 nC 0 = 0 ⇒ nT 0 = 2 , P = 2 atm , T = 423K

b

ξ e 2 − 2ξ e

g

2

b1 − ξ gb1 − 2ξ g
e

e

2

b g

1

...
278 atm -2 ⇒ ξ 2 − ξ e + 01317 = 0
e
4 atm 2

4- 38

4
...
In general, the
equation will be cubic
...
844

...



...
500
y = c1 − 2b0156gh c2 − 2b0156gh ⇒ y = 0
...


...
092

...

n −n
ξ
=
Fractional Conversion of CO b Ag =
n
n
y A = 1 − 0156

...


A0

Use the equations from part b
...


i)
ii)
iii)
iv)

*

1

2

Fractional conversion decreases with increasing fraction of CO
...

Fractional conversion decreases with increasing temperature
...

REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI,
FN, FDN, NT, CON, YA, YB, YC
INTEGER NIT, INMAX
TAU = 0
...
79E–13
E = 11458
...
– 4
...
* YA0)
C2 = 4
...
– P2KE * (YA0 + YB0))
C3 = 4
...
+ P2KE)
EK = 0
...
0 )
NIT = 0
FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2
...
* C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT
...
INMAX)
GOTO 4 IF (ABS((EKPI – EK)/EKPI)
...
TAU) GOTO 2 EK =
EKPI GOTO 1
NT = 1
...
* EKPI
YA = (YA0 – EKPI)/NT
YB = (YB0 – 2
...
48 (cont’d)

CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP
WRITE (6, 5) INMAX, EKPI
FORMAT (' YA YB YC CON', 1, 4(F6
...
3) END
$ DATA 0
...
5
0
...

2
...
500, YB = 0
...
092, CON = 0
...
49

a
...
50 mol CH 4 / mol
0
...


n1 = 50 − ξ1 − ξ 2
n2 = 50 − ξ1 − 2ξ 2

b
...
900 ⇒ n
1

1

50

= 5
...
855 ⇒ n3 = 42
...
0500 mol CH 4 / mol
Equation 3 ⇒ ξ 1 = 42
...
0275 mol O 2 / mol
Equation 1 ⇒ ξ 2 = 2
...
4275 mol HCHO / mol
Equation 2 ⇒ n2 = 2
...
4725 mol H 2 O / mol
Equation 4 ⇒ n4 = 47
...
0225 mol CO 2 / mol
Equation 5 ⇒ n5 = 2
...
75 mol HCHO/s)/(2
...
0 mol HCHO/mol CO 2

4- 40

a
...
Low conversion and excess ethane
make the second reaction unlikely
...


4
...


5 unknowns
–3 atomic balances
2 D
...


100 mol C2H5Cl
n3 (mol C2H6)
n4 (mol HCl)
n5 (mol C2H5Cl2)

Selectivity: 100 mol C 2 H 5 Cl = 14n5 (mol C 2 H 4 Cl 2 ) ⇒ n5 = 7
...
3 mol C H in
g
|
V
2n = 2b100g + 2n + 2b7
...
3 mol C H out
W
6b714
...
3g + n + 4b7
...
1 mol HCl
2n = 100 + 607
...
143g ⇒ n = 114
...
15 n1 = n3
C balance:
H balance:
Cl balance:

1

1

6

3

3

2

2

6

4

4

2

2

2


...
3 mol Cl 2 / 714
...
3 mol Cl 2 1 mol C 2 H 5 Cl
n max =
= 114
...
875
114
...


4
...


C2H4 + H2O → C2H5OH, 2 C2H5OH → (C2H5)2O + H2O
Basis: 100 mol effluent gas
100 mol
0
...
025 mol C 2 H 5OH / mol
0
...
093 mol I / mol

n 3 (mol I)

-2 independent atomic balances
-1 I balance
0 D
...


0
...
433 + 2∗0
...
0014

b

g

g

(2) H balance: 4n1 + 2n2 = 100 4∗0
...
025 + 10∗0
...
4476

b

g

(3) O balance: n2 = 100 0
...
0014 + 0
...
(1)∗2 + Eq
...
(2) ⇒2 independent atomic balances
(4) I balance: n3 = 9
...
51 (cont'd)
b
...
08 mol C 2 H 6

(3) ⇒ n2 = 47
...
3 mol I
% conversion of C2H4:

U
| ⇒ Reactor feed contains 44
...
1% H O, 9
...
08 − 43
...
0%
46
...
08 mol

= 2
...
08 = 0
...
5 mol C 2 H 5 OH
= 17
...
14 mol (C 2 H 5 ) 2 O
c
...
52

Keep conversion low to prevent C2H5OH from being in reactor long enough to form
significant amounts of (C2H5)2O
...


bg

bg

bg

bg

CaF2 s + H 2 SO 4 l → CaSO 4 s + 2HF g

1 metric ton acid

1000 kg acid
0
...
96 kg CaF 2 /kg
0
...
93 H2 SO4 kg/ kg
0
...
04 (100 ) kg SiO 2

28
...
1 kg SiO 2

=

n3 (kg H 2SiF6 )

28
...
59 kg H 2SiF6
144
...
96 (100 ) kg CaF2

38
...
1 kg CaF2

+

9
...
0 kg F
144
...
0 kg F
20
...
2 kg HF

600 kg HF 100 kg ore diss
...
2 kg HF
0
...


4-42

4
...


C 6 H 6 + Cl 2 → C 6 H 5 Cl + HCl
C 6 H 5 Cl + Cl 2 → C 6 H 4 Cl 2 + HCl
C 6 H 4 Cl 2 + Cl 2 → C 6 H 3 Cl 3 + HCl
Convert output wt% to mol%: Basis 100 g output
species
C6 H 6
C 6 H 5 Cl
C 6 H 4 Cl 2
C 6 H 3 Cl 3

g
65
...
0
2
...
5

Mol
...

78
...
56
147
...
46

mol
0
...
284
0
...
003

mol %
73
...
0
1
...
3

total 1
...
0 mo l C6 H6
73
...
0 mo l C6 H 5 Cl
25
...
5 mo l C 6 H 4 Cl2
1
...
5 mol C6 H 3 Cl 3
0
...


3

4 unknowns
-3 atomic balances
-1 wt% Cl 2 in feed
0 D
...


3

C balance: 6n1 = 6 ( 73
...
0 + 1
...
3) ⇒ n1 = 100 mol C 6 H 6
H balance: 6 (100 ) = 6 ( 73
...
0 ) + 4 (1
...
3) + n4 ⇒ n4 = 28
...
9 + 25
...
5 ) + 3 ( 0
...
9 mol Cl2
Theoretical C 6 H 6 = 28
...
9 mol C6 H 6

(100 − 28
...
9 ×100% = 246% excess C6 H 6
Fractional Conversion: (100 − 73
...
268 mol C 6 H 6 react/mol fed
Excess C 6 H 6 :

Yield: (25
...
9 mol C6 H 5 Cl maximum)=0
...
268
g liquid
⎛ 78
...
9 mol Cl2 70
...
98 g Cl2

c
...
Large excess of C 6 H 6 ⇒ Cl 2 much more likely to encounter C 6 H 6
than substituted C 6 H 6 ⇒ higher selectivity
...


Dissolve in water to produce hydrochloric acid
...


Reagent grade costs much more
...


4-43

4
...


2CO 2 ⇔ 2CO + O 2
O 2 + N 2 ⇔ 2NO

2A ⇔ 2B + C
C + D ⇔ 2E

bn
bn
bn
bn
bn

n A = n A 0 − 2ξ e1
nB
nC
nD
nE

yA =
= n B 0 + 2ξ e 2
yB =
= nC 0 + ξ e1 − ξ e 2 ⇒ y C =
= n D0 − ξ e2
yD =
= n E 0 + 2ξ e 2
yE =

bn

ntotal = nT 0 + ξ e1

T0

− 2ξ e1
+ 2ξ e1

A0
B0

C0

D0

E0

g bn
g bn

T0

T0

+ ξ e1
+ ξ e1

gb

g
g

+ ξ e1 − ξ e 2 nT 0 + ξ e1
− 1ξ e 2 nT 0 + ξ e1
+ 2ξ e 2 nT 0 + ξ e1

gb
gb

g
g

g

= n A0 + n B 0 + nC 0 + n D 0 + n E 0

g

Equilibrium at 3000K and 1 atm
y2
A

bn

=

g bn + ξ − ξ g = 01071

...
01493
yC yD ( nA0 + ξ e1 − ξ e 2 )( nD 0 − ξ e 2 )
2

E

b

f 1 = 01071 n A 0 − 2ξ e1

...
01493 nC 0 + ξ e1 − ξ e 2

D0

− ξ e2

2

C0

e1

e2

2

E0

1

1

2

e2

2

1

2

b
...


nA0 = nC0 = nD0 = 0
...
0593, ξe2 = 0
...
2027, yB = 0
...
3510, yD = 0
...
0393
d
...
)

...
57'///)
READ (5, *) NA0, NB0, NC0, ND0, NE0
IF (NA0
...
0
...
54 (cont’d)
2

3

100
4
120

FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6
...
1
X2 = 0
...
1071 * NAS * NT – NBS * NC
F2 = 0
...
4284 * NA * NT * 0
...
0 * NB * NC – NBS
A12 = NBS
A21 = 0
...
01493 * (NC + ND) – 4
...
5, 6X, 'X1C, X2C =', * 2F10
...
LT
...
0E–5
...
ABS(D2/X2C)
...
1
...
4///)
GOTO 30
END
$DATA
0
...
00 0
...
3333 0
...
50
0
...
0
0
...
0
0
...
20 0
...
20
0
...
54
NA0, NB0, NC0, ND0, NE0 = 0
...
00 0
...
10000 0
...
06418 0
...
05969 0
...
33 0
...
06418
X1C, X2C = 0
...
05937

0
...
02986
0
...
54 (cont’d)

ITER = 4 X1A, X2A = 0
...
05931
ITER = 6 X1A, X2A = 0
...
02213
0
...
02083

YA, YB, YC, YD, YE =

2
...
9501E − 01

NA0, NB0, NC0, ND0, NE0 = 0
...
10000
↓
ITER = 7 X1A, X2A = –0
...
55

X1C, X2C = 0
...
05930
X1C, X2C = 0
...
20

0
...
1197 E − 01
3
...
20
0
...
02086
0
...
02083

3
...
20
X1C, X2C = 0
...
08339 X1C, X2C = –0
...
00037
–0
...
5051E − 01 1
...
6693E − 01
2
...
3991E − 02
(B)

a
...
of R

(A)

(P)

m0 (kg/h)

(1 − f ) m0 (kg/h)

m1 (kg/h)

mP (kg/h)

xRA (kg R/kg)

xRA (kg R/kg)

xR1 (kg R/kg)

0
...
01(1 − f )m0 xRA

(2)

Mass balance on reactor:
99% conversion of R:

m1 + fm0 = mP

Mass balance on mixing point:
R balance on mixing point:

m1 xR1 + fm0 xRA = 0
...

b
...
01(1 − f )m0 xRA
m1 + fm0 = mP
m1 xR1 + fm0 xRA = 0
...
0500

4-46

m0 = 2780 kg/h
f = 0
...
55 (cont’d)

mP
4850
4850
4850
4850
4850
4850
4850
4850
4850

xRA
0
...
03
0
...
05
0
...
07
0
...
09
0
...
54
0
...
31
0
...
21
0
...
16
0
...
13

mP
2450
2450
2450
2450
2450
2450
2450
2450
2450

xRA
0
...
03
0
...
05
0
...
07
0
...
09
0
...
54
0
...
31
0
...
22
0
...
16
0
...
13

f v s
...


0
...
50
0
...
30
0
...
10
0
...
00

0
...
04

0
...
08

0
...
12

4
...


900 kg HCHO 1 kmol HCHO
= 30
...
03 kg HCHO

n (kmol CH OH / h)
1
3

30
...
0
= 0
...
0 kmol CH 3 OH / h
n1

b
...
0 kmol HCHO / h

30
...
0 (1) ⇒ n1 = 30
...
0
= 0
...
0 kmol CH 3OH / h
n1 + n3

n1 + n3 = 50
...


4
...
The plot would resemble a concave upward parabola with a minimum
around xsp = 60%
...


Convert effluent composition to molar basis
...
6 g H 2

1 mol H 2
2
...
0 g CO

= 5
...
01 g CO

25
...
28 mol CO

1 mol CH 3 OH
32
...
793 mol CH 3OH

4-48

H : 0
...
274 mol CO / mol
CH OH: 0
...
57 (cont’d)
n4 (mol/min)
0
...
896 - x ) (mol H 2 / mol)

350 mol/ min

Reactor

n3 (mol CH 3OH(l)/min)

0
...


0
...
0953 mol H / mol
2

Condenser

Overall process

3 unknowns (n3 , n4 , x)

2 unknowns (n1 , n2 )

–3 balances

–2 independent atomic balances

0 degrees of freedom

0 degrees of freedom

Balances around condenser
⎫ n3 = 32
...
274 = n4 ∗ x
⎪
⎪
H : 350 ∗ 0
...
996 − x) ⎬ ⇒ n4 = 318
...
301 molCO/mol
CH OH : 350 ∗ 0
...
004 ∗ n4 ⎪
3
⎭

Overall balances
C: n1 =n3 ⎫ n1 = 32
...
16 mol/min H 2 in feed
Single pass conversion of CO:

Overall conversion of CO:
b
...
08 + 318
...
3009) − 350 ∗ 0
...
07%
(32
...
72 ∗ 0
...
08 − 0
× 100% = 100%
32
...
(Recalibrate measurement instruments
...
(Re-analyze feed
...
(Check for it
...
58

a
...


Solvent

Absorb
n3 (kmol CH 4 /h)
n4 (kmol HCl /h)

n4 (kmol HCl /h)

5n5 (kmol CH 3Cl /h)
Still
5n5 (kmol CH 3Cl /h)
n5 (kmol CH 2Cl 2 /h)

n5 (kmol CH 2Cl 2 /h)

Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D
...

Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D
...

Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D
...

Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D
...

Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D
...

Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D
...

Atomic balances around reactor:
⎫
1) C balance : 80 = n 3 + 5n 5 + n 5
⎪
2) H balance : 320 = 4n 3 + n 4 + 15n 5 + 2n 5 ⎬ ⇒ Solve for n 3 , n 4 , n 5
⎪
3) Cl balance : 40 = n 4 + 5n 5 + 2n 5
⎭
CH4 balance around mixing point: n1 = (80 – n3)

Solve for n1

Cl2 balance: n2 = 20
b
...
1 kmol CH4/h

n4 = 20
...
0 kmol Cl2/h

5n5 = 14
...
9 kmol CH4/h
c
...
49 kg) = 19
...
81 kmol CH 3Cl/h
14
...
366

n tot = 50
...
1)(1
...
3 kmol CH 4 /h ⎫
⇒
Fresh feed: 1
⎬
n 2 = ( 20
...
366) = 27
...
0 mol% CH 4 , 54
...
9)(1
...
9 kmol CH4 recycled/h

4-50

4
...


Basis: 100 mol fed to reactor/h ⇒ 25 mol O2/h, 75 mol C2H4/h
n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)

Seperator
separator

reactor
nC2H4 ( mol C 2H 4 /h)
nO2 (mol O 2 /h)

75 mol C 2H 4 //h
25 mol O 2 /h

n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h)
n5 (mol H 2O /h)

n3 (mol C 2H 4O /h)

n4 (mol CO 2 /h)
n5 (mol H 2O /h)

Reactor
5 unknowns (n1 - n5)
-3 atomic balances
-1 - % yield
-1 - % conversion
0 D
...

Strategy: 1
...
Solve balances around mixing point to find nO2, nC2H4
(1) % Conversion ⇒ n1 =
...
200)(75) mol C 2 H 4 ×

90 mol C 2 H 4 O
= n 3 (production rate of C 2 H 4 O)
100 mol C 2 H 4

(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4
(4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5
(5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5
(6) O2 balance (mix pt): nO2 = 25 – n2
(7) C2H4 balance (mix pt): nC2H4 = 75 – n1
Overall conversion of C2H4: 100%
n5 = 3
...
25 mol O2/h

n3 = 13
...
0 mol C2H4/h

n4 = 3
...


n1 = 60
...
75 mol O2 /h

b
...
363
h
44
...
5 mol C 2 H 4 O
mol / h

nC2H4 = (3
...
0) = 50
...
363)(11
...
8 lb-mol O2/h

4-51

4
...


Basis: 100 mol feed/h
...

100 mol/h

n1 (mol /h)

32 mol CO/h
64 mol H 2 / h
4 mol N 2 / h


...


500 mol / h
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)

n3 (mol / h)
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)
Purge

Mixing point balances:
total: (100) + 500 = n1 ⇒ n1 = 600 mol/h
N2: 4 + x1 * 500 =
...
148 mol N2/mol
Overall system balances:
N2: 4 =
...
3 + 2*(1-0
...
3 mol CH3OH/h

x2 = 0
...
284(27)]/32 = 76%
Single pass CO conversion: 24
...
284*500) = 14%
b
...

Purge: to prevent buildup of N2
...
61

a
...
61 (cont’d)
At mixing point:

N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1
I: XI0 + (1-yp) n2 = n2
Total moles fed to reactor: nr = 4n1 + n2
Moles of NH3 produced: np = 2fspn1
Overall N2 conversion:

b
...
01 fsp = 0
...
10
n1 = 0
...
636 mol fed

n2 = 0
...
3536 mol NH3 produced
N2 conversion = 71
...


Recycle: recover and reuse unconsumed reactants
...


d
...
Increasing fsp
results in decreasing nr, increasing np, and increasing fov
...

Optimal values would result in a low value of nr and fsp, and a high value of np, this would
give the highest profit
...
01
0
...
10
0
...
01
0
...
10
0
...
10

fsp
0
...
20
0
...
30
0
...
50
0
...
20
0
...
10
0
...
10
0
...
10
0
...
20
0
...
40

nr
3
...
893
4
...
776
2
...
900
3
...
379
1
...
354
0
...
321
0
...
430
0
...
250
0
...
173

fov
71
...
4%
71
...
1%
87
...
9%
55
...
5%
38
...
62

a
...
25 i-C4 H10
0
...
25 C4 H 8

n 3 (i-C 4 H 10)

Calculate moles of feed

b gb

g b gb

g

M = 0
...
50 M n − C4 H10 + 0
...
75 5812 + 0
...


...
6 kg kmol

b

gb

g

n0 = 40000 kg 1 kmol 57
...
50 694 = 347 kmol n - C 4 H 10 in product
C 8 H 18 balance:

n1 =

b0
...

1 mol C 4 H 8

b

At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 ⇒ n mol i - C 4 H 10

b

g = b5gb0
...
5 kmol
A

i -C 4 H 10 at

moles C 4 H 8 at
A=173
...
5 at (A), (B) and (C) and in feed

b gb g

i - C 4 H 10 balance around first mixing point ⇒ 0
...
5
⇒ n3 = 694 kmol i - C 4 H 10 recycled from still

At C, 200 mol i - C4 H10 mol C4 H 8

b

⇒ n mol i - C4 H10


...
62 (cont’d)

i - C 4 H 10 balance around second mixing point ⇒ 867
...
12 kmol IJK
F
...
5 + 33800gb kmol i - C H g G 5812

...
kmol K
kg I
F
+ 8460 kmol C H G 114
...
00 × 10 kg
...
00 × 10 kg HC 2 kg H SO baq g
=
band leaving reactor g
1 kg HC
= 8
...
00 × 10 b H SO leaving reactor g n + n b n - C H leaving reactor g
⇒ m = 7
...


b
g
d
ib g

...
09gkg H O b1 kmol 18
...
91 kg H 2 SO 4 1 kmol 98
...

5

2

2

Summary: (Change amounts to flow rates)
Product: 173
...
84 × 10 6 kg h 91% H 2 SO 4 ⇒ 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h

4-55

4
...


A balance on ith tank (input = output + consumption)
v L min C A, i −1 mol L = vC Ai + kC Ai C Bi mol liter ⋅ min V L

b

g

b

g

b

gbg

E ÷ v, note V / v = τ

C A , i −1 = C Ai + kτ C Ai C Bi
B balance
...


C Bi − C Ai = C B 0 − C A 0 ⇒ C Bi = C Ai + C B 0 − C A 0
...


b

g

2
0
C A, i −1 = C Ai + kτ C Ai C Ai + C B 0 − C A0
...

Ai
2
C Ai

b

g

kτ + C AL 1 + kτ C B 0 − C A0 − C A , i −1 = 0
⇒α

2
C AL

Solution: C Ai =

b

g

+ β C AL + γ = 0 where α = kτ , β = 1 + kτ C B 0 − C A 0 , γ = − C A, i −1

− β + β 2 − 4αγ

(Only + rather than ±: since αγ is negative and the
2α
negative solution would yield a negative concentration
...

k=
v=
V=
CA0 =
CB0 =
alpha =
beta =

36
...
0567
0
...
48
1
...
670E-02
-2
...
512E-02
-8
...
076E-03
-3
...
837E-03
-1
...
830E-04
-4
...
565E-04
-1
...
667E-05
-5
...
791E-02
1
...
631E-03
5
...
038E-03
1
...
118E-03
6
...
182E-04
2
...
574E-04
9
...
939E-05
3
...
5077
0
...
8478
0
...
9464
0
...
9803
0
...
9926
0
...
9972
0
...
9990
0
...
50, N = 1), (xmin = 0
...
90, N = 4), (xmin = 0
...
99, N = 9), (xmin = 0
...

As xmin → 1, the required number of tanks and hence the process cost becomes infinite
...


(i) k increases ⇒ N decreases (faster reaction ⇒ fewer tanks)
(ii) v increases ⇒ N increases (faster throughput ⇒ less time spent in reactor
⇒ lower conversion per reactor)

(iii) V increases ⇒ N decreases (larger reactor ⇒ more time spent in reactor
⇒ higher conversion per reactor)

4-56

4
...


Basis: 1000 g gas
Species

m (g)

MW

n (mol)

mole % (wet)

mole % (dry)

C3H8

800

44
...
145

77
...
5%

C4H10

150

58
...
581

11
...
5%

H2O

50

18
...
775

11
...
501

100%

100%

Total moles = 23
...
74 mol
Ratio: 2
...
726 = 0
...


C3H8 + 5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Theoretical O2:
C3H8:
C 4 H 10 :

5 kmol O 2
100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8
= 9
...
09 kg C 3 H 8 1 kmol C 3 H 8
6
...
68 kmol O 2 / h
h
100 kg gas 58
...
07 + 1
...
75 kmol O2/h
Air feed rate:

10
...
3 kmol air fed
= 66
...
21 kmol O 2 1 kmol air required

The answer does not change for incomplete combustion
4
...
659 kg C 6 H 14 1000 mol C 6 H 14
= 38
...
684 kg C 7 H 16 1000 mol C 7 H 16
= 27
...
3 mol C 6 H 14

9
...
36 mol C 7 H 16
+
mol C 6 H 14

11 mol O 2
= 665 mol O 2 required
mol C 7 H 16

O2 fed: (4000 mol air )(
...
3% excess air
665

4-57

4
...
500 kmol N2/kmol
x (kmol CO/mol)
(0
...
5 mol O 2 / mol fuel for complete combustion, we can
calculate the air feed rate without determining x CO
...


b

g

A plot of x vs
...
0, x1 = 0
...
7, x 2 = 10
...

ln x = b ln R + ln a

@

x = a Rb
R = 38
...
288

b

g b

g
ln a = lnb10g − 1303 lnb99
...
00

...

a = expb −6
...
49 × 10 −3

b = ln 10 0
...
7 10
...
303

...

⇒ x = 2
...
288 kmol CO 0
...
75
h
kmol
kmol H
h
2
43
...
2 kmol air fed
kmol air
2
Air fed:
= 250
h
0
...
67

a
...
5 kmol O

17% excess air
na (kmol air/h)
0
...
79 N2

→ 4CO 2 + 5H 2 O

0
...
944 CH4
0
...
0060 C3H8
0
...
0060 100 kmol C 3 H 8
h
= 207
...
212 kmol H

175 kmol

b g

2 kmol O 2 0
...
5 kmol O 2
1 kmol C 2 H 6

0
...
5 kmol O 2
1 kmol C 4 H 10

b g

4-58

4
...
0 kmol O 2
h

Air feed rate: n f =

b

1 kmol air
0
...
17 kmol air fed
= 1153 kmol air h
kmol air req
...


na = n f 2 x1 + 35x 2 + 5x 3 + 6
...
21

...


n f = aR f , (n f = 75
...

na = bRa , (na = 550 kmol / h, Ra = 25) ⇒ na = 22
...
7
305
...
2

A2
19
...
57
16
...
35
2
...
78

A4
1
...
70
2
...
5
103
...
0

x1
0
...
945
0
...
0715
0
...
0523

x3
0
...
0079
0
...
0054
0
...
0066

na
934
1194
1592

d
...
68

Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the
flowmeter calibration formulas might not be linear, or the stack gas analysis could be
incorrect
...


C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Basis:

100 mol C4H10

nCO2 (mol CO2)
nH2O (mol H2O)
nC4H10 (mol C4H10)
nO2 (mol O2)
nN2 (mol N2)

Pxs (% excess air)
nair (mol air)
0
...
79 N2
D
...
analysis
6 unknowns (n, n1, n2, n3, n4, n5)
-3 atomic balances (C, H, O)
-1 N2 balance
-1 % excess air
-1 % conversion
0 D
...


4-59

Ra
42
...
3
72
...
68 (cont’d)
b
...
5 mol O2/mol C4H10) = 650 mol O2

n air = (650 mol O 2 )(1 mol air / 0
...
1% N
b gb
g
|
= b100 mol C H react gb4 mol CO mol C H g = 400 mol CO V12
...
9% H O
W

n N2 = 0
...
2(3095) = 3714 mol (780 mol O2, 2934 mol N2)
Exit gas:
400 mol CO2

10
...
6% H2O

130 mol O2

3
...
0% N 2

iii) 90% conversion ⇒ nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed)
20% excess: nair = 1
...
3% C4H10

10 mol C4H10
360 mol CO2

9
...
4% H2O

195 mol O2
2934 mol N2

4
...


4
...
3% N 2

C3H8 + 5 O2 → 3 CO2 + 4 H2O

H2 +1/2 O2 → H2O

C3H8 + 7/2 O2 → 3 CO + 4 H2O
Basis: 100 mol feed gas
100 mol
0
...
25 mol H2

n1 (mol C3H8)
n2 (mol H2)
n3 (mol CO2)
n4 (mol CO)
n5 (mol H2O)
n6 (mol O2)
n7 (mol N2)

n0 (mol air)
0
...
79 mol N2/mol
Theoretical oxygen:

75 mol C 3 H 8

5 mol O 2
25 mol H 2 0
...
5 mol O 2
mol C 3 H 8
mol H 2

4-60

4
...
5 mol O 2 1 kmol air 1
...
5 mol air
h
0
...



...
5 mol C 3 H 8
(67
...

85% hydrogen conversion ⇒ n2 = 0150(25 mol C 3 H 8 ) = 3
...
95(67
...
4 mol CO 2
5% CO selectivity ⇒ n3 =

0
...
5 mol C 3 H 8 react) 3 mol CO generated
= 101 mol CO

...
5 mol C 3 H 8 )(8) + (3
...

mol O
) = (192
...


...


O balance: (0
...
5 mol O 2 )(2

N 2 balance: n7 = 0
...
5) mol N 2 = 1822 mol N 2
Total moles of exit gas = (7
...
75 + 192
...
1 + 291
...
3 + 1822) mol
= 2468 mol
CO concentration in exit gas =

b
...

× 10 6 = 4090 ppm
2468 mol

If more air is fed to the furnace,
(i)

more gas must be compressed (pumped), leading to a higher cost (possibly a larger
pump, and greater utility costs)

(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the
product gas temperature decreases and less steam is produced
...
70

a
...
0027 mol C5H12/mol DPG
0
...
091 mol CO2/mol DPG
0
...
76n2 (mol N2)

3 unknowns (n1, n2, n3)
-3 atomic balances (O, C, H)
-1 N2 balance
-1 D
...
⇒ Problem is overspecified
b
...
76 n2 = 0
...
69 mol O2
C balance: 5 n1 = 5(0
...
091)(100) ⇒ n1 = 2
...
0027)(100) + 2n3 ⇒ n3 = 10
...
053)(2) + (0
...
38 mol O = 39
...


c
...
00304 mol C5H12/mol DPG
0
...
102 mol CO2/mol DPG
0
...
76n2 (mol N2)

N2 balance: 3
...
836 (100) ⇒ n2 = 22
...
00304 + 0
...
34 mol C5H12
H balance: 12 n1 = 12(0
...
2 mol H2O
O balance: 2n2 = 100[(0
...
102)(2)] + n3 ⇒ 44
...
4 mol O √
Fractional conversion of C5H12:

2
...
00304
= 0
...
344

Theoretical O2 required: 2
...
75 mol O2
% excess air:

22
...
75 mol O 2 required
× 100% = 18
...
75 mol O 2 required

4-62

4
...


12 L CH 3 OH 1000 ml 0
...
6 mol CH 3 OH / h
h
L
ml 32
...
0045 mol CH3OH(v)/mol DG
0
...
0181 mol CO/mol DG
x (mol N2/mol DG)
(0
...
6 mol CH3OH(l)/h

n1 (mol O 2 / h)
3
...
F
...


Theoretical O2: 296
...
5) = 444
...
6 = n 2 (0
...
0903 + 0
...
6) = n 2 (4*0
...
6 mol H2O / h

O balance : 296
...
0045 + 2(0
...
0181 + 2(0
...
6
N2 balance: 3
...
3 mol O 2 / h, x = 0
...
6 − 2627(0
...
960 mol CH 3 OH react/mol fed
296
...
3 − 444
...
1%
444
...
6 mol H 2 O
= 0
...
6) mol

c
...
72

Fire, CO toxicity
...


G
...
Say ns mols fuel gas constitute the sample injected into the G
...
If xCH 4 and xC2 H 6 are
the mole fractions of methane and ethane in the fuel, then

b g b
n b molg x b mol CH

gb
g = 20
molgb1 mol C 1 mol CH g
85

ns mol xC2 H 6 mol C 2 H 2 mol 2 mol C 1 mol C 2 H 6

E

s

CH 4

b
bmol CH

4

4

xC2 H 6 mol C 2 H 6 mol fuel
xCH 4

4

mol fuel

g

g = 01176 mole C H

...
72 (cont’d)

b1
...
02 gg = 0126

...
50 mol product gas
mole product gas
Basis: 100 mol product gas
...

2

Condensation measurement:

CH 4 + 2O 2 → CO 2 + 2H 2 O
7
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O
2
100 mol dry gas / h

n1 (mol CH4 )
0
...
126 mol H O / mol
2
0
...
119 mol CO / mol D
...

2
x (mol N / mol)
2
(0
...
G
...
356 mol CH 4 in fuel

...

⇒ 0
...
356) = 0
...
356 + 2 0
...
874 0119 ⇒ n2 = 3
...

Composition of fuel: 5
...
630 mol C 2 H 6 , 3
...
548 CH 4 , 0
...
388 CO 2

b gb

g

N 2 balance: 3
...
874 x

b gb

g

b gb

g b gb

gb g

b

O balance: 2 3
...
126 + 100 0
...
881 − x

...
86 mols O 2 fed , x = 0
...
356 mol CH 4 2 mol O 2 0
...
5 mol O 2
Theoretical O 2 :
+
1 mol CH 4
1 mol CH 4

= 12
...
356 + 0
...
784) mol fuel 7 mol air 0
...
36 mol O2
1 mol fuel
mol air

Desired % excess air:

b
...
36 − 12
...
92

18
...
92
× 100% = 46%
12
...
86 / 0
...
77 mol feed

4-64

4
...


C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Basis 100: mol product gas
n1 (mol C3H8)
n2 (mol C4H10)

100 mol
0
...
526–x) (mol O2/mol)

n3 (mol O2)

Dry product gas contains 69
...
4
=
⇒ x = 0
...
526 − x 30
...
F
...
6 ⇒ n3 = 76
...
1 mol C 3 H 8
C balance : 3 n1 + 4 n 2 = 36
...
1% C 3 H 8 , 34
...
8⎭ n 2 = 3
...


nc=100 mol (0
...
474 mol H2O/mol)(2mol H/mol H2O)=94
...
8%C, 72
...
10 mol C 3 H 8

3
...
10 mol C 3 H 8 11 mol (C + H) 3
...
74

Basis: 100 kg fuel oil
Moles of C in fuel:

100 kg 0
...
08 kmol C
kg
12
...
12 kg H 1 kmol H
= 12
...
017 kg S 1 kmol S
= 0
...
064 kg S

1
...
8% C

4
...
08 kmol C
12
...
053 kmol S
1
...
76 n1 (kmol N2)

C + O2 → CO2
C + 1/2 O2 → CO
2H + 1/2 O2 → H2O
S + O2 → SO2

n2 (kmol N2)
n3 (kmol O2)
n4 (kmol CO2)
(8/92) n4 (kmol CO)
n5 (kmol SO2)
n6 (kmol H2O)

Theoretical O2:
7
...
5 kmol O 2 0
...
133 kmol O 2
1 kmol S
2 kmol H
1 kmol C

20 % excess air: n1 = 1
...
133) = 12
...
16) = 2 (6
...
5664 + 2 (0
...
3102 kmol O2
C balance: 7
...
514 mol CO2
⇒ 8 (6
...
566 mol CO
S balance: n5 = 0
...
00 kmol H2O
N2 balance: n2 = 3
...
16) = 45
...
514 + 0
...
053 + 6
...
310 + 45
...
16 kmol
⇒ 10
...
92% CO, 0
...
8% H 2 O, 3
...
8% N 2

4
...
Basis: 5000 kg coal/h; 50 kmol air min = 3000 kmol air h
5000 kg coal / h
0
...
17 kg H / kg
0
...
06 kg ash / kg

C + 02 --> CO2
2H + 1/2 O2 -->H2O
S + O2 --> SO2
C + 1/2 O2 --> CO

3000 kmol air / h
0
...
79 kmol N / kmol
2

n1 (kmol O2 / h)
n2 (kmol N2 / h)
n3 (kmol CO / h)
2
0
...
75 5000 kg C

b g

1 kmol C

1 kmol O 2

h

12
...
2 kmol O 2 h

4
...
17 5000 kg H 1 kmol H 1 kmol H 2 O
h

101 kg H

...
02 5000 kg S

b g

1 kmol S

1 kmol O 2

h

32
...
4 kmol O 2 h

= 3
...
2+210
...
1) kmol O2/h = 525
...
21 3000 = 630 kmol O 2 h
Excess air:

630 − 525
...
8% excess air
525
...
Balances:
0
...
75 5000 kg C react 1 kmol C
= n3 + 0
...
01 kg C

b gb gb g

⇒ n3 = 266
...
7 kmol CO h

...

b017gb5000g kg H

S:

(from part a)

N2 :

O:

h

1 kmol H 1 kmol H 2 O
101 kg H

...
1 kmol O 2 for SO 2

g

= n5 ⇒ n5 = 420
...

= n4 ⇒ n4 = 31 kmol SO 2 h

b0
...

b0
...
8g + 1b26
...
8g
2

2

2

2

1

⇒ n1 = 136
...
7 3224 = 8
...
6 × 10 −4 mol SO 2 mol

...


1
SO 2 + O 2 → SO 3
2
SO 3 + H 2 O → H 2SO 4
3
...
08 kg H 2SO 4
= 304 kg H 2SO 4 h
kmol H 2SO 4

4
...


Basis: 100 g coal as received (c
...
r
...
d
...
denote air-dried coal; v
...
denote volatile
matter
100 g c
...
r
...
147 g a
...
c
...
207 g c
...
r
...
03 g a
...
c

= 95
...
97 g H 2 O lost by air drying


...
234 − 1204g g H O = 2
...
234 g a
...
c
...
97 g + 2
...
28 g moisture
95
...
d
...

b1347 − 0
...
m
...
31 g H O = 3550 g volatile matter

...
347 g a
...
c
...
03 g a
...
c

0
...
175 g a
...
c
...
98 g ash

b

g

Fixed carbon = 100 − 7
...
98 g = 48
...


7
...
24 g fixed carbon

7
...
2% fixed carbon
35
...
5% volatile matter
8
...
0% ash
100 g coal as received
b
...


C + CO 2 → CO 2 :
2H +

1 mol O 2
1 mol C

1 mol C 10 3 g
1 mol air
= 396
...
01 g C 1 kg 0
...
5 mol O 2
1
O2 → H 2O :
2
2 mol H

Air required:

1 mol H 10 3 g
1 mol air
= 1179 mol air kg H
1
...
21 mol O 2

1000 kg coal 0
...
5 mol air
kg coal
kg C
1000 kg 0
...
m
...
5 mol air
kg
7 kg v
...

kg C
1000 kg 0
...
m
...
72 × 10 5 mol air
kg
7 kg v
...

kg H
+

4-68

4
...


Basis 100 mol dry fuel gas
...
720 mol CO / mol
2
0
...
000592 mol SO / mol
2
0
...
720) + 0
...
000592) + 2 (0
...
20) (74
...
0592 + 0
...
6 mol H, n4 = 144
...
8 mol H2O
Total moles in feed: 258
...
9% C, 71
...
023% S

4
...
5 ppm SO 2)

x n 3 mol SO 2
(N2 , O2 , CO2 , H 2 O)

0
...
87 g C/g
0
...
03 g S/g
n 1 mol O2
3
...
87 100 g C

scrubber

0
...
244 mol O 2
1 mol C 1 mol CO 2
⇒ n4 = 7
...
01 g C

b g

FG
H

IJ
K

2
...
10 100 g H 1 mol H 1 mol H 2 O
⇒ n6 = 4
...


4-69

IJ
K

4
...
0956 mol O 2
1 mol S 1 mol SO 2
⇒ n5 = 0
...
06 g S 1 mol S

0
...
244 + 2
...
0936 ⇒ 12
...


b

g

O 2 balance: n3 = 12
...
244 + 2
...
0936 mol O 2 consumed
= 2
...
76 12
...


b

g

SO 2 in stack SO 2 balance around mixing point :

F
H

I
K

b gb

g

b

x 0
...
0936 = 0
...
0842 x mol SO 2

...
244 + 2
...
00936 + 0
...
0842 x mol dry gas

...


b CO g b O g
2

bN g

2

bSO g

2

b

2

g

612
...
00936 + 0
...
5
=
⇒ x = 0
...
0842 x

...

Basis: 100 mol stack gas

4
...
76 n 4 (mol O2 )

a
...
7566 N 2
0
...
0827 H 2 O
0
...
000825 SO 2

g
gb g

C balance: n1 = 100 01024 = 10
...

mol C
10
...
62
H balance: n2 = 100 0
...
54 mol H
mol H
16
...
25, and that of C2 H 6 is 0
...
62, so the fuel could not be the natural gas
...


b gb

g

S balance: n 3 = 100 0
...
0825 mol S

b10
...
0 g 1 molg = 122
...
88 = 7
...
54 mol Hgb1
...
71 g H| ⇒ 2
...
71
b0
...
07 g 1 molg = 2
...
24 × 100% = 1
...
4 fuel oil

4
...


Basis: 1 mol CpHqOr
1 mol CpHqOr
no (mol S)
Xs (kg s/ kg fuel)

C + 02 --> CO2
2H + 1/2 O2 -->H2O
S + O2 --> SO2

P (% excess air)
n1 (mol O2)
3
...
76 n1 (mol N2)
n5 (mol H2O (v))

Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C)
q (mol H) (1 g / mol) =

q (g H)

⇒ (12 p + q + 16 r) g fuel

r (mol O) (16 g / mol) = 16 r (g O)
S in feed:
n o=

(12 p + q + 16r) g fuel

Theoretical O2:

X s (g S)
X (12 p + q + 16 r)
1 mol S
= s
(mol S) (1)
(1 - X s ) (g fuel) 32
...
07(1 - X s )

p (mol C) 1 mol O 2 q (mol H) 0
...
71, q= 1
...
003, Xs = 0
...
00616 mol S

(5) ⇒ n3 = 0
...
16 mol O2 fed

(6) ⇒ n4 = 0
...
71 mol CO2

(4) ⇒ n5 = 0
...
76*1
...
36 mol N2
Total moles of dry product gas = n2 + n3 + n4 + 3
...
246 mol dry product gas
Dry basis composition
yCO2 = (0
...
246 mol dry gas) * 100% = 13
...
170 / 5
...
2% O2
yN2 = (4
...
246) * 100% = 83
...
00616 / 5
...
1

Assume volume additivity
1

Av
...
5
...
400
0
...
719 kg L
0
...
730 kg L

m

A

= mt + m0 ⇒ m =

A

mass of tank
at time t

mass of
empty tank

A

ρO
a
...
28 kg min bm = mass flow rate of liquidg
b10 − 3g min

14
...
9 L min
min
0
...
m0 = m(t) - mt = 150 − 14
...
2

b

g

void volume of bed: 100 cm3 − 2335 − 184 cm3 = 50
...

porosity: 50
...
274 cm3 void cm3 total
bulk density: 600 g 184 cm3 = 3
...
5 cm3 = 4
...
3
C 6 H 6 (l )

m B (kg / min)
VB = 20
...
5 m) 2 015 m

...
0594 m3 / min
Δt
4
60 min
4 Δt
Assume additive volumes
VT = V - VB = 59
...
0 L / min = 39
...
879 kg 20
...
866 kg 39
...

+
= 517 kg / min
L
min
L
min
m
(0
...
0 L / min)
xB = B =
= 0
...

m = ρ B ⋅ VB + ρ T ⋅ VT =

5-1

a
...


1

ρ sl

=

xc

+

ρc

F IF I
e jge jhbmgGH 11 N JK GH 11 Pa JK = ρ gh
kg

m
s2

m3

kg⋅m
s2

sl

N
m2

b1 − x g ⇒ check units!
c

ρl

1
kg crystals / kg slurry kg liquid / kg slurry
=
+
kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid
L slurry L crystals L liquid
L slurry
=
+
=
kg slurry kg slurry kg slurry kg slurry
ΔP
2775
=
= 1415 kg / m3
c
...
) ρ sl =
9
...
200
gh
ii
...

K

IJ
ρ K
1

l

1

3

F
I = 0
...
3d10001kg / m i − 12d10001kg / m iJJ

...
) Vsl =

=

ρ sl

3

175 kg
1000 L
= 123
...
) mc = x c msl = 0
...
3 kg crystals
v
...
3 kg CuSO 4 ⋅ 5H 2 O 1 kmol
1 kmol CuSO 4
159
...
4 kg CuSO 4
249 kg 1 kmol CuSO 4 ⋅ 5H 2 O 1 kmol

b

g

b

gb

g

vi
...
684 kg liquid / kg slurry 175 kg slurry = 120 kg liquid solution
vii
...
2
1200
2300
2353
...
00

ρl

=

120 kg

1000 L
= 100 L
m3

b1
...

2411
...
05
1229
...
80
0
...
27

2602
...
2
1326
...
84
0
...
02

2772
...
316
1413
...
35
0
...
87

3093
...
5
1577
...
6
Solids Fraction

5
...
5
0
...
3

ΔP = 2775, ρ = 0
...
2
0
...
00

2500
...
00

2900
...
00

5
...


b

g d

i

Basis: 1 kg slurry ⇒ x c kg crystals , Vc m3 crystals =

b1- x gbkg liquid g, V dm
c

l

3

b

g

x c kg crystals

d

ρ c kg / m

3

i

liquid
i b1-ρx dgbkg/ m i g
kg

liquid =

c

3

l

ρ sl =

1 kg

bV + V gdm i
3

c

5
...


29
...
0064 m air

V=

b

ρl

g

1 kg

mol 103 g

= 4
...
00 mol 0
...
2 K
=
= 3
...
06L - 2
...
3%
2
...

Assume Patm = 1013 bar
a
...



...
0 m3
kmol ⋅ K 28
...
2 K 0
...
8

+

0
...
2 K 1 kmol
= 0
...
0 atm 103 mol

1 mol

b
...
7

xc

ρc

1
1 − xc

Assume Patm = 1 atm
PV = RT ⇒ V =

5
...
0 m3

273K
298
...

b10 + 1013gbar
1
...
02 kg N 2
= 249 kg N 2
3
kmol
22
...


R=

Ps Vs 1 atm 22
...
21 × 10 −2
n sTs 1 kmol 273 K
kmol ⋅ K

b
...
05 ft 3
torr ⋅ ft 3
=
= 555
n sTs 1 lb - mole 1 atm
492 R
lb - mole ⋅ R

5-3

5
...

2
10 cm 10
...
0 m3
= 0
...
2 K , V =

b

g

m=

a
...


b

g

g
mol

L
PV
1
...
02
⋅ MW =
L⋅atm
RT
0
...
2 K

400

L
min

28
...
2 K 22
...
10 Assume ideal gas behavior:

2

2

2

=

1

2
nR T2 P1 D1
⋅ ⋅ ⋅ 2
nR T1 P2 D 2

(1
...
013) bar ( 7
...
53 + 1
...
00 cm )
2

T P D 2 60
...
2K
u 2 = u1 2 1 1 =
T1P2 D 2
sec
300
...


...
406 mol
L⋅atm
0
...
0 g 0
...
0 g mol ⇒ Oxygen

5
...
12 Assume ideal gas behavior: Say m t = mass of tank, n g = mol of gas in tank

b
gU ⇒ n = 0
...
440 g = m + n b44
...
0256 g
|
b37
...
0256gg = 3
...
289 g = m t + n g 28
...
009391 mol

5
...


b g

Vstd cm3 STP min =

b g

Δ V liters
Δ t min

763 mm Hg 103 cm3

296
...
031EV + 0
...
4 litersbSTPg 10 cm
= 224 cm

φ

139
268
370

1L

Vstd cm STP min

5
...
0
12
...


273K

Vstd =

3

0
...
031 224 cm3 / min + 0
...
9
5-4

3

3

/ min

= 925
...
14 Assume ideal gas behavior ρ kg L =

d

12

i d

V2 cm s = V1 cm
3

3

1

= V1 P1M1T2 P2 M 2 T1

12

2

LM
N

cm3 758 mm Hg 28
...
2K
s 1800 mm Hg 2
...
2K

OP
Q

12

a
...


M = 0
...
75M C3H 8 = 0
...
05 + 0
...
11 = 37
...
02gb323
...
10gb298
...
15 a
...


n CO 2

d

2
πR 2 Δh π 0
...

12 m 60 s
= 11 × 10 −3 m3 / min

...
4 s min

755 mm Hg
1 atm
1
...
044 mol/min
3
m ⋅atm
300 K
1 kmol
0
...
16
m air = 10
...
0 m3 / h
n CO (kmol / h)
2

150 o C, 1
...
0 kg 1 kmol
n air =
= 0
...
0 kg air
n CO 2 =

PV
15 bar

...
0 m3 / h
=
= 0
...
314 m ⋅kPa 1 bar
423
...
853 kmol CO 2 / h
× 100% = 712%

...
853 kmol CO 2 / h + 0
...
17 Basis: Given flow rates of outlet gas
...
70 kg H 2 O / kg
0
...
12 kmol H 2 O / kmol
0
...


n3 =

1 atm
356
...
08206 m3 ⋅ atm

H 2 O balance : 0
...
64 kmol 0
...
02 kg
kmol
kmol
min

⇒ m1 = 32
...
64 kmol min

b

g

S olids balance: 0
...
2 kg min = m4 ⇒ m4 = 9
...
88 (10
...
36 kmol min air

V2 =

9
...
08206 m3 ⋅ atm
kmol ⋅ K

min

440K

(1033 − 40 ) cm H 2O

1033 cm H 2 O
1 atm

= 352 m3 air min
Vair (m3 / s) 352 m3 1 min
u air (m/min)=
=
A (m 2 )
min 60 s

π

4

⋅ (6 m) 2

= 0
...
If the velocity of the air is too high, the powdered milk would be blown out of the reactor
by the air instead of falling to the conveyor belt
...
18 SG CO 2

5
...


ρ CO2
=
=
ρ air

PM CO2
RT
PM air
RT

=

M CO 2
M air

=

44 kg / kmol
= 152

...
75 x air = 1 − 0
...
25
Since air is 21% O 2 , x O 2 = (0
...
21) = 0
...
25 mole% O 2

b
...
5 × 3 m3 0
...
01 kg CO 2
1 atm
=
=12 kg
m3 ⋅atm
kmol
kmol CO 2
298
...
08206 kmol⋅K

More needs to escape from the cylinder since the room is not sealed
...
19 (cont’d)
c
...
Measures that would reduce hazards are:

1
...

2
...

3
...

4
...


5
...
7 kg

1 kmol
= 0
...
01 kg

b

Assume ideal gas behavior, negligible temperature change T = 19° C = 292
...


g

P1V
n1RT
n1
P
102kPa
=
⇒
= 1 =
P2 V
n1 + 0
...
357 P2 3
...
0115 kmol air in tank
b
...


n1RT 0
...
2 K 8
...
5 mol air ⋅ (29
...
5 g / L
274 L

CO 2 sublimates ⇒ large volume change due to phase change ⇒ rapid pressure rise
...


b

gb

g

5
...
9 in
...
9 ft H 2 O + 28
...
Hg = 37
...
Hg
...
3 in
...
20
0
...


...
967

cm 3
1
...
20 lb 5 × 10 −4 ton 10 6 cm 3
⇒ ρ sl =
= 4
...
17 gal
cm 3 1000 g

a
...
0 ft 3 (STP) 534
...
9 in Hg
= 2440 ft 3 / hr
hr
4
...
1 in Hg

b
...
1
3
⇒ D 3 = 1
...
3

b2
...
0g mm × 100 = 10%
2
...
2 mm

5
...

n inj = moles of gas injected

n B , n air , n He = moles of benzene and air in the container and moles of helium added
n BGC , m BGC = moles, g of benzene in the GC
y B = mole fraction of benzene in room air
a
...
078 mol = n air + n B
kPa
101
...
08206 L ⋅ atm

P2 V2 = n 2 RT2 (2 ≡ condition when charged with He): P2 = 500 kPa, T2 = 306K

n2 =

500 kPa 2 L
mol ⋅ K
= 0
...
3 atm 306 K
...
325 mol = (n air + n B + n He ) − n inj
kPa
101
...
08206 L ⋅ atm

n inj = n 2 − n 3 = 0
...
393 mol m BGC (g B) 1 mol
=
= 0
...
068 mol
78
...
0741 ⋅ m BGC
× 106 =
× 106 = 0
...
078

U
|
|
) = 0
...
is below the PEL
|
) = 0
...
950 × 106 )(0
...
623 ppm
1 pm: y B = (0
...
788 × 10 −6
5 pm: y B = (0
...
910 × 10 −6

b
...
Waiting a day allows the gases to mix
sufficiently and to reach thermal equilibrium
...
(i) It is very difficult to have a completely evacuated sample cylinder; the sample may
be dilute to begin with
...
A reading should be taken on Friday
...
(iv) The benzene may
not be uniformly distributed in the laboratory
...


5- 8

b g

4
3
5
...


492° R 3 atm

4189 m3

1 kmol

b g = 515
...
4 m3 STP

He in balloon:

b

gb

g

m = 515
...
003 kg kmol = 2065 kg He
mg =

b
...
807 m
1N
= 20,250 N
2
s 1 kg ⋅ m / s2

dP
dP

iV = n RT ⇒ n
iV = n RT

gas in balloon
air displaced

Fbuoyant

gas

air

=

air

Pair
1 atm
⋅ n gas =
⋅ 515
...
0 kmol
Pgas
3 atm

Fbuoyant = Wair displaced =

172
...
0 kg 9
...
807 m 1 N
= 27,20
s2 1 kg2⋅m
s

c
...
3 m s2
2065 + 150 kg
N

g

d
...

Need to know how density of air varies with altitude
...
The balloon expands, displacing more air ⇒ buoyant force increases ⇒ balloon rises
until decrease in air density at higher altitudes compensates for added volume
...
24 Assume ideal gas behavior, Patm = 1 atm

a
...
7 atm 400 m / h
PN VN = Pc Vc ⇒ Vc =
=
= 240 m3 h
9
...


Mass flow rate before diversion:
400 m3
h

273 K 5
...
09 kg

22
...
24 (cont’d)

Monthly revenue:

( 4043
c
...
60 kg ) = $1,747,000 month

Mass flow rate at Noxious plant after diversion:
400 m3

1 kmol

44
...
8 atm

22
...
25 a
...
35 ⋅ (2
...
70 atm

PCH 4 = y CH 4 ⋅ P = 0
...
00 atm) = 0
...
45 ⋅ (2
...
90 atm
b
...
00 mole gas
4
...
35 mol He
= 140 g He

...
05 g IJ = 3
...
22 g ⇒ mass fraction CH
0
...
02 g IJ = 12
...
45 mol N G
|
H mol K
W
FG
H

IJ
K

4

2

4

4

=

3
...

= 0186
17
...
2 g / mol
mol
P MW
2
...
2 kg / kmol
m n MW
= =
=
=
= 115 kg / m3

...
08206 kmol⋅K 363
...


MW =

d
...
26 a
...

If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the
C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard
...


fuel-air mixture
n1 ( mol / s)
y C3H 8 = 0
...
0205 mol C 3 H 8 / mol

diluting air
n 2 ( mol / s)
n1 =

150 mol C 3 H 8
mol
= 3722 mol / s
s
0
...
0205 ⋅ n 3 ⇒ n 3 = 7317 mol / s

5- 10

5
...


b g

n 2 = 13 n 2

...
314 m3 ⋅ Pa 398
...
41
V1
m3 fuel gas
3722 mol 8
...
2 K
3
= 83
...

n1 + n 2
3722 mol / s + 4674 mol / s

b

g

d
...
03%
...

5
...
206 O 2
0
...
020 H 2 O

a
...
151 O 2
0
...
750 N 2
0
...
246 mol min

3

min

10 mL 297K 22
...
774 0
...
750n out ⇒ n out = 0
...
246gb0
...
254gb0
...
0 g mol

= 0
...
254gb0
...
01 g mol

= 0
...
254gb0
...
246gb0
...
02 g mol
2

= 0
...
27 (cont’d)

PVin
n RT
= in in
PVout n out RTout

FG IJ FG T IJ = FG 0
...
246 mol min K H 297K K
...

b0
...
394 g O gained ming = 0
...


Vout
n
= out
Vin
n in

out
in

2

2

2

STACK

5
...


M s = 0
...
1 + 0
...
0 + 0
...
0 = 310 g mol , Ts = 655K ,

...


− ρgL

Pa M a
PM
P gL M a M a
gL − a s gL = a
−
RTa
RTs
R
Ta
Ts

a
...
0 g mol , Ta = 294 K , L = 53 m
D=

755 mm Hg

1 atm

53
...
807 m

760 mm Hg

×

s

2

kmol - K
0
...
0 kg kmol − 31
...
013 × 10 N m
2

2

2

5

2

= 3
...


b g

P MW MWCCl2O = 98
...
91
=
= 3
...
0
RT
ρ air
Phosgene, which is 3
...

2
π D in L π
2
Vtube =
= 0
...
0559 cm 15
...
22 cm3
4
4

5
...
ρ =

=======>

b g

b

m CCl O = Vtube ⋅ ρ CCl O =
2

c
...
22 cm3

gb

g

1L
1 atm
98
...
0131 g
3
L⋅atm
296
...
08206 mol⋅K
3

3
...


...
0446 mol CCl 2 O
cm3 98
...
29 (cont’d)

PV 1 atm 2200 ft 3 28
...
2K

...
0446
= 17
...
4 ppm
2563

The level of phosgene in the room exceeded the safe level by a factor of more than 100
...

d
...
He also should have been working under a
hood and should have worn a gas mask
...
30 CH 4 + 2O 2 → CO 2 + 2 H 2 O
7
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O
2
C 3 H 8 + 5O 2 → 3CO 2 + 4 H 2 O
1450 m 3 / h @ 15o C, 150 kPa
n1 (kmol / h)

0
...
08 C 2 H 6 , 0
...
21 O 2 , 0
...
2K

b101
...
2K

101
...
4 m3 STP

b g = 152 kmol h

Theoretical O 2 :

LM F
NM GH

IJ
K

FG
H

IJ
K

108 349
...


1 kmol Air

FG
H

152 kmol
2 kmol O 2
3
...
08
+ 0
...
86
h
kmol CH 4
kmol C 2 H 6
kmol C 3 H 8
Air flow: Vair =

b

g

h

0
...
6 kmol h O
K QP

b g = 4
...
4 m3 STP
kmol

4

b g

2

m3 STP h

5
...
0; R = 14g , bT = 35
...
77R + 14
...
0, R = 6i ⇒ P bkPag = 3
...
0 × 10 , R = 10i ⇒ V dm hi = 200R
dV = 0; R = 0h , dV = 1
...
21 mol O 2 / mol
0
...
2K

dP + 101
...
2gK 101
...
12031V d P + 101
...
4 m3 STP

g

Theoretical O 2 :

dn i

o 2 Th

b

c

gh

= n F 2x A + 3
...
5 x D + x E kmol O 2 req
...


1 kmol air

h

0
...
4 m bSTPg kmoli = 22
...
762 1 +

VA = n a

b1 + P

x

g

100 kmol feed

1 kmol req
...
0 7
...
81 0
...
05 0
...
02
1450 22506
...
63
23
...
0 7
...
2 183
...
74
64
...
8 0
...
31 0
...
05 0
...
8 7
...
5 20
...
3
23 78
...
4 1325
...
6 2
...
00 0
...
65 0
...
10
490 22022
...
51
46
...
0 15
...
1 155
...
1
10
...
2 7
...
4
3
6 0
...
4 0
...
1 0
...
0 248
...
9
13
...
4 7
...
9
4
7 0
...
12 0
...
16 0
...
3 238
...
3
16
...
2 8
...
5
5
9 0
...
3 0
...
04 0
...
4 266
...
8
20
...
5 0
...
1 0
...
06
2000 37196
...
3
27 35
...
8 303
...
6

5- 14

1
5
...
20 mol NO / mol
0
...
21 O 2
0
...


n1 (mol NO)
n 2 (mol O 2 )
n 3 (mol N 2 )
n 4 (mol NO 2 )
Pf (kPa)

U
|
V
|
W

Basis: 1
...
10(0
...
020 mol NO ⇒ NO reacted = 0
...
5 mol O 2

...
0780 mol O 2
mol NO
N 2 balance: n 3 = 0
...
79) = 0
...
80(0
...

= 018 mol NO 2 ⇒ n f = n1 + n 2 + n 3 + n 4 = 0
...

1 mol NO
0
...
022
0
...
086

mol O 2
mol N 2
mol NO 2
y N 2 = 0
...

mol
mol
mol

FG
H

IJ
K

Pf V n f RT
n
0
...


n f = n0

Pf
360 kPa
= (1 mol)
= 0
...
20 − ξ
n 2 ( mol O 2 ) = (0
...
80) − 0
...
79)(0
...
5ξ = 0
...

⇒ n1 = 010 mol NO, n 2 = 0118 mol O 2 , n 3 = 0
...


...
665, y NO 2 = 0105

...


...


NO conversion =
P (atm) =

Kp =

b0
...
20

360 kPa
= 355 atm

...
3 kPa
atm
(y NO 2 P)

( y NO P)( y O 2 P)

0
...
5

( y NO )( y O 2 ) P

0
...
105

(0105) 0
...


g b3
...
5

1

0
...


5
...
2 kg M 1 kmol
= 0
...
6 kg

100 kg liquid ⇒

0
...
6 kg D 1 kmol
= 0
...
0 kg

⇒ 0
...
2 kg B 1 kmol
= 0
...
12 kg

0
...
909 kmol
a
...
920 HCl
0
...
12 kg)

n 0 (kmol Cl 2 )

n 2 (kmol)
0
...
481 C 6 H 5Cl
0
...
298 × 6 + 0
...
221 × 6
1 kmol C 6 H 6

⇒ n 2 = 100 kmol

...
920 (1)
1 kmol C 6 H 6

+ n 2 0
...
481 × 5 + 0
...

V1 =

n1RT 100 kmol 1013 kPa 0
...
2 K

...

= 217 m3
=

...


217 m3

...
278 m3 / kg B

...
278 m
s 1 min 10 cm2
= 5
...
43 ⋅ m B0

4

1
2

c
...


5- 16

5
...
74 SCMM

U
|
V
|
W

n a ( mol / min)
0
...
780 N 2 O

SiH 2 Cl 2(g) + 2 N 2 O (g) → SiO 2(s) + 2 N 2(g) + 2 HCl (g)
a
...
74 m3 (STP)
103 mol
=167 mol / min
min
22
...
220 mol DCS IJK b167 mol / ming = 14
...
60gb0
...
04 mol DCS reacted / min
min

b

60% conversion: n1 = 1- 0
...
780 167
−
N 2 balance: n 3 =

2

22
...

mol DCS
min

22
...
08 mol N 2 / min
min
mol DCS

HCl balance: n 4 =

22
...
08 mol HCl / min
mol DCS
min

n B = n1 + n 2 + n 3 + n 4 = 189 mol / min
⇒ VB =
b
...
36 L ⋅ torr 0
...
29 × 104 m3 / min
P
min
mol ⋅ K
L
0
...
7 mol DCS/min
P=
⋅ 604 mtorr=47
...
2 mol N 2 O/min
= x N 2O ⋅ P= 2 P=
⋅ 604 mtorr=275
...
16 × 10-8 ⋅ p DCS ⋅ p N 2O 0
...
16 × 10-8 ( 47
...
h(A)=r ⋅ t ⋅

MW

ρSiO2

=

)( 275
...
65 = 5
...
7 × 10−5 mol SiO 2 60 s 120 min 60
...
25 × 106 g/m3 1 m
(Table B
...
1 × 10 A
5

The films will be thicker closer to the entrance where the lower conversion yields higher
pDCS and p N 2 O values, which in turn yields a higher deposition rate
...
35

Basis: 100 kmol dry product gas
n1 (kmol C x H y )
m1 (kg C x H y )

R100 kmol dry gasU
|0
...
053 O
V
|0
...
21 O 2
0
...


n 3 (kmol H 2 O)

N 2 balance: 0
...
842(100) ⇒ n 2 = 106
...
21n 2 = 100 2 0105 + 2 0
...


...
5

...
17

H balance: n1y = 2n 3 ====> n1y = 26
...
34 = 2
...
5
fed: 0
...
6 kmol air g = 22
...
3 kmol ⇒ Theoretical O = b22
...
3g kmol = 17
...


V2 =
m1 =

5
...
1 kmol O 2

106
...
4 m3 (STP) 1013 kPa 303 K

...
0 kg n1y kmol H 101 kg n1x=10
...

+
=====> m1 = 152
...
34
kmol

V2
2740 m3 air
m3 air
= 18
...
6 kg fuel
kg fuel
5
...
0 ≤ x ≤ 1
b
...
82 kg 1 kmol
= 128 kmol

...
06 kg

LM 6x kmol H + b1 + 2xg kmol N + b4 − 4xg kmol NH OP
3 kmol N H
N 3 kmol N H 3 kmol N H
Q
128

...

b6x + 1 + 2x + 4 − 4xg = 1707x + 2
...


2

2

2

4

5-18

2

4

3

2

4

5
...
13
2
...
47
2
...
81
2
...
15
3
...
50
3
...
84

Vp (L)
15447
...
93
17923
...
95
20399
...
97
22875
...
99
25352
...
01
27828
...
00

25000
...
00
V (L)

x
0
0
...
2
0
...
4
0
...
6
0
...
8
0
...
00

10000
...
00

0
...
1

0
...
3

0
...
5

0
...
7

0
...
9

1

x

c
...


5
...
(i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in
cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste
containing A poured into sink, A used as cleaning solvent
...
mA

c
...


yA

FG kg A IJ
H hK

FG kg A IJ
H hK
C e j⋅ V
mol A
=
=
mol air M e j ⋅ n
= mA

in

dV h

air min

out

gA
A m3
gA
A mol

yA = 50 ×10−6

= Vair

air

F m I C FG kg A IJ
GH h JK H m K
3

A

3

===================>
m
PV
CA = A ; nair =
k⋅Vair
RT

yA =

mA RT
k ⋅ Vair M A P


...
314 mol⋅Pa
9
...
5 50 × 10 101
...
14 g / mol
3

d

i

Concentration of styrene could be higher in some areas due to incomplete mixing (high
concentrations of A near source); 9
...

e
...
T in the numerator of expression for Vair : At higher T, need
a greater air volume throughput for y to be < PEL
...
38 Basis: 2 mol feed gas

U
|
V
|
W

n p (mol C 3 H 8 )
(1 - n p )(mol C 3 H 6 ) n 2 = n p + 2(1 − n p ) = 2 − n p
(1 - n p )(mol H 2 )

1 mol C 3 H 6
1 mol H 2
25 C, 32 atm

235 C, P2

a
...
0 atm
P2 V n 2 RT2
n T
=
⇒ P2 = 2 2 P1 =
= 27
...
P2 = 35
...
1 atm 298K 2 mol
P2 T1
n1 =
= 1
...
0 atm 508K
P1 T2

1
...
71 mol C 3 H 8 produced

b

g

⇒ 1- 0
...
29 mol C 3 H 6 unreacted ⇒ 71% conversion of propylene
c
...
009
1
...
083
1
...
156
1
...
211
1
...
248
1
...
285
1
...
431
1
...

0
...
9724
0
...
899
0
...
8256
0
...
77055
0
...
73385
0
...
6421
0
...
532

%conv
...
075
97
...
735
89
...
395
82
...
89
77
...
22
73
...
55
64
...
87
53
...
5
28
...
5
30
...
5
32
...
0
33
...
0
34
...
0
37
...
0
40
...


60

40

20

0
25
...
0

29
...
0

33
...
0

37
...
0

41
...
39

Basis: 100 g ⇒

b
b

U
V
W

g
g

97
...
04 g = 5
...
8 mol % C 2 H 6
5 g C 2 H 6 1 mol 30
...


500 m3 / h

n 2 (kmol CO 2 / h)
n 3 (kmol H 2 O / h)

n1 (mol / h)
0
...
028 C 2 H 6
40 C, 1
...

= 211 kmol h
0
...

P1V1 11 bar 500 m3
=
RT1 313K
h

7
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O
2

CH 4 + 2O 2 → CO 2 + 2 H 2 O
Theoretical O 2 =

LM
N

21
...
972 kmol CH 4
kmol
h
+

Air Feed:

0
...


...

h
Q
22
...
5 kmol O 2
1 kmol C 2 H 6

1 kmol Air
0
...
40 Basis: 1 m3 gas fed @ 205° C, 1
...
1 bar

n 3 (kmol), 10 C, 40 bar

condenser

n1 (kmol)
y1 (kmol Ac / kmol)
(1 - y1 )(kmol air / kmol)
p AC = 0
...
379 bar

n 2 (kmol Ac(l))

a
...
00 m3

273K

110 bars

...
0277 kmol

478K 10132 bars 22
...


0
...
379 bar
= 0
...
47 × 10 −3 kmol Ac kmol
1
...
0 bars

b

gb

g

Air balance: 0
...
910 = (1 − 9
...
0254 kmol

Mole balance: 0
...
0254 + n 2 ⇒ n 2 = 0
...
0023 kmol Ac 58
...

1 kmol Ac

5-21

5
...


b g

0
...
4 m3 STP


...
0 bars

= 0
...
0 m3 effluent 0
...
0909 kmol Ac 58
...
0149 m3 effluent
kmol feed
kmol Ac

5
...
00 × 10 6 gal
...
Neglect evaporation of water
...
00 × 10 6 gal / day

Effluent gas: 68 F, 21
...
03n1 (lb-moles NH3 /day)

n2 (lb-moles air/day)
n3 (lb-moles NH3 /day)

300 × 10 6 ft 3 air / day

68 F, 21
...


Effluent liquid

n1 (lb-moles H2O/day)
n4 (lb-moles NH3 /day)

Density of wastewater: Assume ρ = 62
...
02 lb m 0
...
03 lb m ⎤
+
×
⎢
day
1 lb-mole
day
1 lb-mole ⎥ 62
...
00 × 106

7
...
50 × 105 lb-moles H 2 O fed day , 0
...
35 × 104 lb-moles NH 3 fed day

n2 =

300 × 106 ft 3
day

492 R

21
...
7 R 14
...
13 × 106 lb-moles air day

93% stripping: n3 = 0
...
13 × 10

6

)

+ 12555 lb-moles day

1
...
129 × 10

= 0
...


5
...
=

2
...
12 g NaOH 1 mol 0
...
1 mol Cl 2
mol
= 0
...
0 g mol NaOH 2 mol NaOH
min
n 2 (mol air / mol)

2 L / min @ 23 C, 510 mm H 2 O

n1 (mol / min)
y (mol Cl 2 / mol)
(1 - y)(mol air / mol)

0
...
33 + 0
...
84 m H O

Assume Patm = 10
...
84 m H 2 O

2

1 mol

b g = 0
...
33 m H 2 O 22
...
0864y = 0
...


5
...
235 mol C 2 H 6 / mol DG
0
...


Hygrometer Calibration ln y = bR + ln a
b=

b

ln y1 y 2
R 2 − R1

dy = ae i
bR

g = lnd0
...
08942
−4

90 − 5

bg

ln a = ln y1 − bR1 = ln 10 −4 − 0
...
395 × 10 −5 ⇒ y = 6
...
08942R

b
...
315 mol min wet gas
min 298K 101 kPa 22
...
156 mol min wet air
min 348K 101 kPa 22
...
0 → y1 = 0
...
8 → y 2 = 2
...
43 (cont’d)

b

C 2 H 6 balance: n C2 H 6 = 5
...
140g mol DG IJK FGH 0
...
07 mol C 2 H 6 min

b gb gb g
Dry air balance: n = b14
...
00 × 10 i = 14
...
315gb0
...
156gd1
...
746 mol H O min
n
= b1
...
50 + 14
...
72 mol min ,
n
= b18
...
746g = 19
...
47 mol min 22
...
315 0
...
765 = 3
...


p H 2O = y H 2Ol ⋅ P =

273K

FG 1
...
7% C H , 18
...
72 K
2

6

2

4

0
...
03832 atm
19
...
03832 ⇒ R =

FG
H

IJ
K

1
0
...
5
0
...
395 × 10−5

5
...
92 kmol CO 2 h
h
1273K 22
...
92 kmol CO 2 1 kmol CaCO 3 100
...
17 kg clay
= 279 kg clay h
h
0
...
07
× 100% = 18% Fe 2 O 3

...
92 44
...
95 kg CaCO 3

5
...
7 g C b1 mol 12
...
0 mol C
|116
...
01 gg = 115
...
5 g S b1 mol 32
...
4211 mol S
|5
...
3 g I
n 1 (mol CO 2 )
n 2 (mol CO)
n 3 (mol H 2 O)
n 4 (mol SO 2 )
n 5 (mol O 2 )
n 6 (mol N 2 )

72
...
3 mol H
0
...
3 g I

C + O 2 → CO 2
1
C + O 2 → CO
2
S + O 2 → SO 2
1
2H + O 2 → H 2 O
2

n a (mol), 0
...
79 N 2
15% excess air
175 C, 180 mm Hg (gauge)

a
...
0 mol C 1 mol O

2+
1 mol C

+

115
...
25 mol O

2

1 mol H

0
...
2 mol O
2
1 mol S

(

1
...
2 mol O
Air Fed:

2

)

1 mol Air
0
...
4 liter ( STP )
mol

1 m3
3

10 liter

= 554 mol Air = n

a

2

448K

760 mm Hg

273K

940 mm Hg

= 16
...
S balance: n 4 = 0
...
3 = 2n 3 ⇒ n 3 = 57
...
95 72
...
4 mol CO 2 ⇒ 0
...
0) = n 2 = 3
...
79 ( 554 ) =n 6 ⇒ n 6 = 437
...
21( 554 ) 2=57
...
6+2(68
...
4211) +2n 5 ⇒ n 5 = 16
...
6 mol CO
527 mol
3
...
8 × 10 −3

mol CO
mol

,

× 106 = 6150 ppm CO ,

5-25

dry: 527 mols

0
...
2 × 10−4

0
...
46 Basis: 50
...
4 L C5 H12 (l ) / min
n1 (kmol C5 H12 / min)

heater

n1 , n 2

Combustion
chamber

15% excess air, Vair (L / min)
n 2 kmol air
0
...
79 N 2
336 K, 208
...


n 4 (kmol O 2 / min)
n 5 (kmol N 2 / min)
n 6 (kmol CO 2 / min)

Vliq (L/min)
m=3
...
4 L 0
...
440 kmol min
min
L
72
...

= 0
...
15 kg

frac
...
440 - 0
...
440

0
...
15 ( 8 kmol O 2 )
1 mol air
= 19
...
21 mol O 2

Vair =

b g

19
...
4 L STP
min

mol

336K

101 kPa 1000 mol
= 173000 L min
273K 309
...
21)(19
...
90)(0
...
882 kmol O 2 / min
min

n5 =

19
...
79 kmol N 2
= 15
...
90(0
...

= 198 kmol CO 2 / min
min
kmol C5 H 12

Vgas =

0
...
23+1
...
4 L(STP) 275 K 1000 mol
min

mol

5-26

273 K

kmol

= 4
...
46 (cont’d)

n7 =

0
...
440 kmol C5 H 12 ) 6 kmol H 2 O
= 2
...
044 kmol 72
...
04 L min
min
kmol
0
...
38 kmol 18
...
89 L min

Assume volume additivity (liquids are immiscible)

Vliq = 5
...
89 = 47
...

C5 H 12 (l )

bg

C 5 H 12 l

bg

H 2O l

bg

H 2O l

5
...
21 O 2
0
...
20 kmol H 2S / mol
0
...
0 m 3 / min @ 380 C, 205 kPa
n exit (kmol / min)
n 3 (kmol N 2 / min)
n 4 (kmol H 2 O / min)
n 5 (kmol CO 2 / min)
n 6 (kmol S / min)

n exit =

PV
205 kPa 10
...
377 kmol / min
m3 ⋅kPa
RT 8
...
20 n 0 / 3 = 0
...
133n 0

5-27

5
...
0667 n 0 (kmol H 2 S fed) 15 kmol O 2 1 kmol air

...
21 kmol O 2

= 0
...
4764 n 0 (kmol air) 0
...
3764 n 0 ( kmol N 2 / min)
(min)
min

0
...
200n 0 (kmol S / min)
(min)
1 kmol H 2S

Overall CO 2 balance: n 5 = 0
...
200n 0 (kmol H 2 S) 2 kmol H
n kmol H 2 O 2 kmol H
= 4
(min)
1 kmol H 2 S
min
1 kmol H 2 O

⇒ n 4 = 0
...
376 + 0
...
200 + 0
...
377 kmol / min ⇒ n 0 = 0
...
4764(0
...


5
...
0 kg FeS2 (s), 18
...


b

gb

g

n FeS2 fed = 82
...
0 kg = 0
...

18 kg I

Vout m3 (STP)
n 2 (kmol SO 2 )
n 3 (kmol SO 3 )
n 4 (kmol O 2 )
n5 (kmol N 2 )

40% excess air
n 1 (kmol)
0
...
79 N 2
V1 m 3 (STP)

m6 (kg FeS2 )
m7 (kg Fe 2 O 3 )
18 kg I

2 FeS2(s) + 11 O 2(g) → Fe 2 O 3(s) + 4SO 2(g)
2
2 FeS2(s) + 15 O 2(g) → Fe 2 O 3(s) + 4SO 3(g)
2

a
...
6833 kmol FeS2 7
...

= 17
...
21 kmol O 2
kmol air req'd

b

gb

g

V1 = 17
...
4 SCM / kmol = 382 SCM / 100 kg ore
n2 =

(0
...
40)0
...
4646 kmol SO 2
2 kmol FeS2

5-28

5
...
85)( 0
...
6833 kmol FeS2 4 kmol SO 2
= 0
...
21 × 17
...
4646 kmol SO

2

5
...
697 kmol SO 7
...
641 kmol O
2
4 kmol SO
3
n 5 = 0
...
08 kmol N 2 = 13
...
4646+0
...
641+13
...
4 SCM (STP)/kmol]
⎣
⎦
= 365 SCM/100 kg ore fed
ySO2 =

0
...
9%; ySO3 = 4
...
1%; y N 2 = 82
...
285 kmol

b
...
4646 kmol SO 2
0
...

13
...
4646 − ξ
n SO3 = 0
...
641 − 1 ξ
2
n N2 = 13
...
29- 1 ξ
2
K p (T)=

⎫
0
...
697 + ξ
⎪ ySO2 =
, ySO3 =
1
16
...
29- 1 ξ
⎪
2
⎬⇒
1
1
...
49
⎪
y O2 =
, y N2 =
1
⎪
16
...
29- 1 ξ
2
⎭

P ⋅ ySO2 (P ⋅ yO2 )

(0
...
29 − 1 ξ ) 2
2
1

P ⋅ ySO3
1
2

⇒

(0
...
641 − ξ )
1
2

1
2

-1

⋅ P 2 = K p (T)

P=1 atm, T=600o C, K p = 9
...
1707 kmol
-1

⇒ n SO2 = 0
...
4646 − 0
...
4646 kmol SO 2 fed

reacted

= 0
...
4548 kmol
-1

⇒ n SO2 = 0
...
979
The gases are initially heated in order to get the reaction going at a reasonable rate
...


5-29

5
...


SO 3 leaving converter: (0
...
4687) kmol = 1
...
156 kmol SO 3 1 kmol H 2 SO 4 98 kg H 2 SO 4
= 113
...
683 kmol FeS 2 2 kmol S 32
...

kmol FeS2 kmol

113
...
59
43
...
of S:

0
...
9 kg H 2SO 4
kmol FeS2
1 kmol S
kmol

133
...
06
43
...

5
...
00 V

n0 =

b
...
00 atmgb2
...
08206 L ⋅ atm mol - Kg = 0
...


FG n IJ P ⇒ K = n P
n bn + n g
Hn +n K
Ideal gas equation of state ⇒ PV = b n + n gRT ⇒ n + n = PV / RT b1g
p NO 2 = y NO 2 P =

FG n IJ P , p
Hn +n K

2

1

1

N 2O4

=

2
1

2

p

2

1

1

2

2

2

1

1

2

2

Stoichiometric equation ⇒ each mole of N 2 O 4 present at equilibrium represents a loss
of two moles of NO 2 from that initially present ⇒ n1 + 2n 2 = 0
...
103

b2 g

b4 g

n 2 = 0
...
103g dP
n b0
...
272
0
...
097
-0
...
088568 5
...
080821 2
...
069861 0
...
063037 0
...
002857 1
...
002985 0
...
003175 -0
...
003333 -1
...
747
R2 = 1

1
0
-1
-2
0
...
003

0
...
37 Pg + 1

Variation of Kp with Temperature

ln Kp

Kp =

0
...
49 (cont’d)
c
...


1
is a straight line
...
567 × 109 atm
7367
−7367
+ 22
...
567 × 109 exp
⇒
T
T
b = 7367K
10
...
50

5
...
00 kmol S / h

n 4 (kmol A / h)
n 5 (kmol H 2 / h)
Vrcy (SCMH)

A + H2

S

Overall A balance: n1 =

5
...
00 kmol A / h
h
1 kmol S form
5
...
00 kmol H 2 / h
h
1 kmol S form

Overall H 2 balance: n2 =

Extent of reaction equations: n i = n i0 + ν iξ
A + H2 ↔ S
A:

n 4 = 3n 3 − ξ

H 2 : n5 = n3 − ξ
S:

U
|
|⇒ p
V
= 5
...
00|
W

5
...
00
n 5 = n 3 − 5
...
00
P=
10
...
00

3

p H2 = y H2 P =
pS = yS P =
Kp =

b

g

n5
n - 5
...
0
n tot
4 n 3 − 5
...
00
10
...
00

5
...
00
pS

...
94 kmol H 2 / h
p A p H 2 10
...
00 n 3 − 5
...
94) - 5
...
82 kmol A / h
n 5 = 1194 − 5
...
94 kmol H 2 / h

...
82 + 6
...
4 m 3 (STP) / kmol = 846 SCMH

5-31

5
...
Balances on reactor ⇒ 4 equations in n3 , n4 , n5 , and n6
...
0% XS H2: n3 =
C balance:

100 kmol CO fed 2 kmol H 2 reqd 1
...
6
⎛
⎞
⎜ 21
...
492ln ( 500K )
⎟
−7
-2
K p ( T=500K ) = 1
...
11 × 10 kPa
⎜ +4
...
161 × 10-8 ( 500K )2 ⎟
⎝
⎠
n6
(1) − ( 3)
310 − 2 n 6
yMP
yM
Kp =
⇒ KpP2 =
====>
2
2
2
100 − n 6 210 − 2 n 6
y CO P y H 2 P
y CO y H 2
310 − 2 n 6 310 − 2 n 6 2
−4

d

K p P = 9
...
775 =

b

b
b

b

gb
gb

n 6 310 − 2 n 6

g

g

2

b100 − n gb210 − 2n g
6

Solving for n6 ⇒ n6 = 75
...
3 kmol CO/h
n5 = 210 − 2n6 = 58
...
7 kmol CO/h
Overall H balance: n2 (2) = n6 (4) ⇒ n 2 = 151 kmol H 2 /h
Vrec = ( n4 + n5 )

22
...
51 (cont’d)
b
...
5
5 5 28 0 9 E 1
5
...
1 -0
9 0 22 0 2 E 1

...
0

...
7
6 2 16 6 9 E0
3
...
1 + 1
9
...
1
...
6
5 8 18 4 2 E0
5
...
3 + 1
1
...
8

...
3
3 5 13 0 2 E0
5
...
3 + 1
7
...
6
...
7
7 7 14 5 2 E0
6
...
3 + 1

K (T 8
p )E
9 E0

...
1 + 1
9 E0

...
1 + 4
9 E0

...
6 + 0
9 E0

...
1 + 1
9 E0

...
9
2
...
1
11
7 4
...
7
28
01

...
7
28
2
...
7
28

K P - n (k o
p ^2 1 ml
C /h
O)
K c ^2
pP
1E5

...
5
55
2 E0

...
0
4E3

...
7
62
3E8

...
9
83
3E3

...
6
58
-2 E 4

...
5
48
9E3

...
3
35
3E3

...
6
58
-3 E 3

...
7
77

n (k o n (k o n (k o
3 m l 4 ml 5 m l
H /h
2)
H /h
2)
C /h
O)
20
1
7
...
9
20
1
9
...
0
20
1
1
...
5
66
20
1
17

...
1
25
20
1
2
...
6
84
20
1
8
...
8
20
0
2
...
3
30
20
1
2
...
6
84
20
2
2
...
4
45
n (k o
2 ml
H /h
2)
5
...
0
80
13 4
7
...
8
11 6
5
...
1
96
16 0
4
...
3
15 5
5
...
Increase yield by raising pressure, lowering temperature, increasing Hxs
...

d
...

e
...


5
...
0 mol CO 2
1
...
0 mol N 2
T = 3000 K, P = 5
...
3272 atm

1/2

p CO 2
1
1
O 2 + 2 N 2 ⇔ NO
2
p NO
K2 =
= 0
...
52 (cont’d)
nA = 1 − ξ1
nB = ξ1
1
1
nC = 1 + ξ1 − ξ 2
2
2
1
nD = 1− ξ2
2
nE = ξ2
1
6 + ξ1
n tot = 3 + ξ 1 =
2
2

U
| y = n n = 2 b1 − ξ g b 6 + ξ g
|
| y = 2ξ b6 + ξ g
| y = b2 + ξ − ξ g b6 + ξ g
V
| y = b2 − ξ g b6 + ξ g
| y = 2ξ b6 + ξ g
|
|
|
W
A
B

A

1

tot

1

C

1

D

2

2

1

b

g b5g
K =
p
b gb g
⇒ 0
...
236ξ b2 + ξ − ξ g
y y1 2 1+ 1 −1 2ξ 2 + ξ 1 − ξ 2
= B C pb 2 g = 1
yA
2 1 − ξ1 6 + ξ1

p CO p1 2
O2

1

CO 2

12

12

12

12

1

K2 =

d

p NO
p O2 p N2

b

i

12

=

⇒ 01222 2 + ξ 1 − ξ 2

...
3272

12

1

yE
12 12
yC y D

pi = yiP

1

1

2

E

1

1

1

p1−1 2 −1 2 =

g b2 − ξ g
12

12

2

1

b2 + ξ

(1)

2

2ξ 2
1 − ξ2

g b2 − ξ g
12

12

= 01222

...

Solve (1) and (2) simultaneously with E-Z Solve ⇒ ξ 1 = 0
...
2574 mol CO
1

2

mol

y B = 0
...
3030 mol N 2 mol
y E = 0
...
3355 mol O 2 mol
n 4 (kmol / h)

5
...


0
...
96 N 2

PX=C8 H10 , TPA=C8 H 6 O 4 , S=Solvent
V3 (m 3 / h) @105o C, 5
...
0 atm
n 2 (kmol / h)
0
...
79 N 2

condenser

n 3W (kmol H 2 O(v) / h)
V3W (m 3 / h)

reactor

n1 (kmol PX / h)
( n1 + n 3p ) kmol PX / h
ms (kg S / h)
3 kg S / kg PX
n 3p (kmol PX / h)
ms (kg S / h)

5-34

n 3p ( kmolPX / h)
100 kmol TPA / h
m s (kg S / h)

separator

100 mol TPA / s

5
...
Overall C balance:
n1

c
...
04n 4 ⎪
n 2 = 1694 kmol air/h
h
⎬ ⇒
n 4 = 1394 kmol/h
⎪
Overall N 2 balance: 0
...
96n 4
⎭
100 kmol TPA 2 kmol H 2 O
Overall H 2 O balance: n 3W =
= 200 kmol H 2 O / h
h
1 kmol TPA

Overall O 2 balance: 0
...
08206 m3 ⋅ atm 298 K
=
= 6
...
0 atm

( n 3W + n 4 ) RT = ( 200+1394 ) kmol

V3W =

P

h

0
...
5 atm

200 kmol H 2 O (l) 18
...
60 m3 H 2 O(l) / h leave condenser
h
kmol 1000 kg

d

i

n1 =100

d
...
10 n1 + n 3p ====> n 3p = 111 kmol PX / h

...
1) kg PX 106 kg PX 3 kg S 11
...
65 × 104 kg/h
e
...
N2 does not react with anything but enters with O2 in the air
...

f
...


5
...
10n 2 (kmol H 2 / h)

Separator
n 6 (kmol CO / h)
n 7 (kmol H 2 / h)
n 8 (kmol CO 2 / h)
2 kmol N 2 / h

0
...
300 kmol CO / kmol
0
...
020 kmol N 2 / kmol
0
...
54 (cont’d)

CO + 2H 2 ⇔ CH 3OH(M)
CO 2 + 3H 2 ⇔ CH 3OH + H 2 O
a
...
65
63
FG J
n Hn K
FG n IJ FG n IJ
n
(2)

...
27 kmol / h ξ 2 = 0
...
0 − 25
...
73 kmol CO / h

9
...
0 − 2(25
...
0157) = 12
...
2% H 2

n 3 = 5
...
0157 = 4
...
5% CO 2

n 4 = 25
...
0157 = 25
...
0157 = 0
...
4 kmol / h

53
...
03% H 2 O

U
V
W

n 6 = 25
...
3 kmol / h
⇒
n 8 = 0
...
44 mol / s
H balance: 2n 7 = 2(0
...
7 ⇒ n 7 = 618 mol H 2 / s

...
(n 4 ) process = 237 kmol M / h
⇒ Scale Factor =

237 kmol M / h
25
...
54 (cont’d)

237
m (STP)
gFGH 25
...
4 kmol IJK = 18,700 SCMH
kmol
F 237 kmol / h IJ = 444 kmol / h
Reactor effluent flow rate: b 49
...
3 kmol / sK
F kmol IJ FG 22
...
4 + 618 + 0
...
0

...


V=


...
2 K 1013 kPa
= 354 m 3 / h
h
273
...
8 L / mol
n 444 kmol / h m 3 1000 mol
(5
...
In addition, the reaction equilibrium expressions are probably strictly
valid only for ideal gases, so that every calculated quantity is likely to be in error
...
55 a
...
1 for ethane: Tc = 305
...
2 atm
From Table 5
...
098
0
...
422
= −0
...
083 − 1
...
083 −
1
...
2 K
305
...


...
0270
B1 = 0139 − 4
...


...
2
Tr
308
...
4K
RTc
0
...
4 K
B(T) =
−0
...
098 0
...
2 atm
= −01745 L / mol

...
0 atm
mol ⋅ K 2
− V- B=
V − V + 0
...
2K 0
...
395 mol / L 01745 L / mol

...
395 mol / L

g

g = 2
...
188 L / mol

Videal = RT / P = 0
...
2 / 10
...
53, so the second solution is
likely to be a mathematical artifact
...


z=

c
...
0 atm 2
...
926
L⋅atm
RT 0
...
2K
V
V

MW =

1000 L
mol 30
...
8 kg / h
h 2
...
56

b

RTc
PV
B
= 1+ ⇒ B =
Bo + ωB1
RT
Pc
V

g

b
g
T bC H g = 369
...
0 atm
From Table 5
...
559, ω bC H g = 0
...
1 Tc CH 3OH = 513
...
50 atm
c

3

8

c

3

3

8

0
...
422
= 0
...
619
1
...
6
Tr
373
...
2K
0
...
422
= −0
...
083 − 1
...
083 −
1
...
2K
369
...


...


...
516
B1 ( CH 3OH) = 0139 − 4
...
2
Tr
373
...
2K
0172

...


...

= −0
...
2 = 0139 −
4
...
2K
369
...
083 −

e

j

e

j

e

j

e

b

j

g

0
...
2K
−0
...
559 0
...
4868
mol ⋅ K 78
...
08206 L ⋅ atm 369
...
333 − 0152 0
...
2436

...
0 atm

b

c

i

b

g

L
mol

g

∑∑y y B
i

b

c

=

j

ij

d

⇒B ij = 0
...
5 −0
...
2436 L / mol = -0
...
30 0
...
4868 + 2 0
...
70 −0
...
70 0
...
2436
= −0
...
0 atm
mol ⋅ K 2
− V - B mix =
V − V + 0
...
2K 0
...
326 mol / L 0
...
326 mol / L

g

g = 2
...
359 L / mol

RT 0
...
2 K
=
= 3
...
70 L / mol
P
mol ⋅ K 10
...
70 L / mol

15
...
30 kmol CH 3OH / kmol 1 kmol 1000 L

5-38

5
...


van der Waals equation: P =

d

i

RT

−

d V - bi

a2
V2

Multiply both sides by V 2 V - b ⇒ PV 3 − PV 2 b = RTV 2 − aV + ab

b

g

PV 3 + -Pb - RT V 2 + aV - ab = 0
c 3 = P = 50
...
0 atm 0
...
08206

...
Videal =


...

c 0 = − ab = - 133 atm ⋅ L2 / mol 2 0
...
0487

L⋅atm
mol⋅K

g

atm ⋅ L3
mol 3

RT 0
...
366 L / mol
P
mol ⋅ K 50
...

T(K)
223
223
223
223
223

P(atm)
1
...
0
50
...
0
200
...
0
10
...
0
100
...
0

c1

-18
...
6654
-20
...
9594
-25
...
33
1
...
33
1
...
33

-0
...
0487
-0
...
0487
-0
...
2994 18
...
8299
1
...
3660
0
...
1830
0
...
0915
0
...
0000
0
...
0008
-0
...
0002

d
...
in 1 unknown - use Newton-Raphson
...

b1g ⇒ gdVi = 50
...
1294gV + b133gV-
...
259 V + 1
...
(A
...
(A
...
3660 L / mol
...
3660
0
...
33137
0
...
33714
0
...
33114
0
...
2
2
...
5
19
...
6

b

d

5
...
9 K

PC = 42
...
26 × 106 Pa

5
...
09 kg

1 kmol

75 kg

Specific Volume

1 kmol

103 mol

i

ω = 0152

...
93 × 10 −3 m3 mol

Calculate constants
a=
b=

d8
...
9 Kg
2

3

0
...
08664

4
...
949 m6 ⋅ Pa mol 2

d8
...
9 Kg = 6
...
26 × 106 Pa

b

g

b

m = 0
...


...


...
717 1 − 298
...
9

j

2

g

2

−5

m3 mol

= 0
...


SRK Equation:

d8
...
2 Kg −
d2
...
25 × 10 i m mol 2
...
949 m6 ⋅ Pa mol 2

...
93 × 10 −3 + 6
...
40 × 106 Pa ⇒ 7
...
314 m ⋅ Pa mol ⋅ K 298
...
46 × 106 Pa ⇒ 8
...
93 × 10 m mol
V

(8
...
30) atm
× 100% = 14
...
30 atm

Percent Error:

TC = 304
...
9 atm ω = 0
...
2 K

5
...
0 atm ω = −0
...
0 L / 50
...
70 L mol

...
08206 L ⋅ atm mol ⋅ K )
L2 ⋅ atm
L
, m = 0
...
0297
, α = 1 + 0
...
2
2
mol
mol
L2 ⋅ atm
L
a = 137

...
479 , b = 0
...
479 1 − T 151
...
65
Ar:

bg

f T =

e

RT
a
−
1 + m 1 − T TC
V−b V V+b

d

i

j

2

−P=0

Use E-Z Solve
...
70

...
08206
= 435
...
4 K ,

bT g

max Ar

IJ
K

= 431
...
60 O 2 : TC = 154
...
7 atm ; ω = 0
...
2 K 65° C ; P = 8
...
08206 L ⋅ atm mol ⋅ K

...
0221 L mol ; m = 0
...
840
a
d i dVRTbi − VdVα+ bi − P = 0=====> V = 2
...
01 L
= 15,700 L h
h
32
...
61

∑F

y

= PCO 2 ⋅ A - W = 0

e

where W = mg = 5500 kg 9
...


PCO 2 =

W
A piston

=

53900 N
π
4


...
SRK equation of state: P =

1 atm
= 301 atm

...
013 × 105 N / m2
RT

−

αa

dV - bi VdV + bi

For CO 2 : Tc = 304
...
9 atm , ω = 0
...

a = 3
...
02967 m3 / kmol, m = 0
...

e0
...
2 Kg − b1016ge3
...

eV - 0
...
02967i
m3 ⋅atm
kmol⋅K

m6 ⋅atm
kmol 2

m6
kmol 2

m3
kmol

E-Z Solve

=====> V = 0
...
030 m

V before expansion = 0
...

g b15 mg = 0
...

4

2

3

44
...
0565 m3
= 3
...
675 m / kmol kmol
V

mCO 2 (initially) =

PV
1 atm
0
...
01 kg
MW =
= 0
...
08206 kmol⋅K 298
...
68 - 0
...
63 kg

5-41

5
...


W = 53,900 N

V

h

add 3
...
63 (kg)
Final: V = Vo +
, n = no +
= o + 0
...
0825
RT
W
RT
RT
53,900
αa
αa
P=
=
−
⇒ 2
=
−
A piston V - b V V + b
πd / 4 V - b V V + b
V=

i
d i
Substitute expression for V in b1g ⇒ one equation in one unknown
...


5
...
Using ideal gas assumption:

Pg =

35
...
73 ft 3 ⋅ psia 509
...
7 psia = 2400 psig
V
32
...
5 ft 3

b
...
5 ft 3 32
...
27
35
...
9 o R, Pc = 730
...
021
a = 52038

...
518, α 50o F = 0
...
3537
lb - mole
lb - mole 2

d

i


...
73
jd509
...
667ge52038
b2400 + 14
...
3537
Vd V + 0
...
139 ft 3 / lb - mole

5-42

j

ft 6
lb-mole 2

5
...
0 lb m
V
2
...
4 lb m
3
2
...
It calls for charging less O2 than the tank can
safely hold
...
1
...

3
...


Pressure gauge is faulty
The room temperature is higher than 50°F
Crack or weakness in the tank
Tank was not completely evacuated before charging and O2 reacted with something in
the tank
5
...
The tank was mislabeled and did not contain pure oxygen
...
63 a
...

Problem 5
...
2
R=0
...
9
0
...
653924 m^6 atm/kmol^2
0
...
826312

f(V)=B14*E14^3-0
...
08206*A14)*E14-C14*$B$7*$B$8
T(K)
200
250
300
300
300

P(atm)
6
...
3
6
...
5
50
...
3370
1
...
0115
1
...
0115

V(ideal)
2
...
6679
3
...
1450
0
...
1125
1
...
4972
1
...
3392

f(V)
0
...
0001
0
...
0000
0
...
E-Z Solve solves the equation f(V)=0 in one step
...

d
...
08206/
READ (5, *) GAS
WRITE (6, *) GAS
10 READ (5, *) TC, PC, W

5-43

5
...
LT
...
) STOP
R = 0
...
08664 *R*TC/PC

...

M = 0
...
+ M∗ 1 − T / TC ∗∗0
...

VP = R∗ T / P
DO 20 I = 7, 15
V = VP
F = R * T/(V – B) – ALP * A/V/(V + B) – P
FP = ALP * A * (2
...

VP = V – F/FP
IF (ABS(VP – V)/VP
...
0
...
1, 'K', 3X, F5
...
2, 'LITER/MOL')
GOTO 10
END
$ DATA
CARBON
304
...
0
250
...
0
–1

72
...
8
12
...
5
0
...
225

RESULTS
CARBON DIOXIDE
200
...
8 ATM
250
...
3 ATM
300
...
8 ATM
300
...
5 ATM
300
...
0 ATM
5
...


b
...
11 LITER/MOL
1
...
50 LITER/MOL
1
...
34 LITER/MOL

b

g

Tr = 40 + 273
...
2 = 2
...
2 K
⇒
10 atm
40 MPa
PC = 33
...

= 1178
33
...
013 MPa

b

gb
g

g

U
|
V
|
W

Tr = −200 + 273
...
26 + 8 = 5
...
26 K
⇒
PC = 2
...
26 + 8 = 34
...
5
...

⇒ z = 12

U
|
V
|
W

Fig
...
4-4

⇒ z = 1
...
65 a
...


m (kg) (MW)P
=
RT
V (m3 )
30 kg kmol
9
...
8 kg m3
m3 ⋅atm 1
...
08206 kmol⋅K

U
V
P = 9
...
5 = 2
...

Tr = 465 310 = 15

Fig
...
4-3

⇒ z = 0
...
8 kg m3
=
= 831 kg m3

...
84

5
...
01 lb m CO 2

= 2
...
2 K ⎫
(1600 + 14
...
507
⎬ ⇒ Pr = P PC =
72
...
7 psi
PC = 72
...
0 ft 3
72
...
80
3
RTC 2
...
2 K 0
...
8 °R
Fig
...
4-3: Pr = 1507 , Vr = 0
...
85

...
0 ft 3
lb - mole⋅° R
1 atm
PV 1614
...
85
2
...
7302 ft ⋅ atm 14
...
7 atm

Pr1
Tr2
Pr2

V2 = V1
V2 =

Uz = 1
...
5
...
7 = 0
...
4 = 2
...
61 bFig
...
4 - 4g
V
= 1000 49
...
12 W
|

Tr1 = 298 154
...
93

5
...
4 K
2
C

1

2

z 2 T2 P1
z1 T1 P2

127 m3 1
...
246 m3 h
h
1
...
2 154
...
94
Pr1 = 175 49
...
52 ⇒ z1 = 0
...
68 O 2 : TC = 154
...
7 atm

(Fig
...
3-2)

Pr2 = 1
...
7 = 0
...
00
n1 − n 2 =

FG
H

IJ
K

FG
H

IJ
K

10
...

−
=
−
= 74
...
2 K 0
...
95
RT z1 z 2
1
...
69 a
...
0 mL 44
...

n
5
...
06 mL ⋅ atm
1000 K
P=
=
= 186 atm
mol ⋅ K 440
...
For CO 2 : Tc = 304
...
9 atm
T 1000 K
Tr =
=
= 3
...
2 K

Vr ideal =

VPc 4401 mL 72
...

mol ⋅ K
=
= 1
...
2 K 82
...
4-3: Vr ideal = 1
...
29 ⇒ z=1
...


zRT 1
...
06 mL ⋅ atm
mol 1000 K
=
= 190 atm
ˆ
mol ⋅ K 440
...
654 × 10 6 mL2 ⋅ atm / mol 2 , b = 29
...
8263, α (1000 K ) = 0
...
06 hb1000 Kg − b0
...
654 × 10
P=
b440
...
67g
440
...
1 + 29
...
70 a
...

b
...
Since the O2 is so dilute at this

condition, the properties of the gas will be that of N2
...
2 K, Pc = 335 atm, Tr = 2
...


n initial = n1 =

1 atm
5000 L
PV
=
= 204
...
08206 mol⋅K 298
...
3 mol air
n O2
n2

FG 0
...
9 mol O
H mol air K
2

2

= 10 × 10 −6 ⇒ n 2 = 4
...

= 116 × 10 -3 L / mol
4
...


...

=
= 38 × 10 −3
Vr ideal =
RTc
mol 0
...
2 K
⇒ not found on compressibility charts
V=

Ideal gas: P =

RT
V

=

0
...
2 K
= 2
...


The pressure required will be higher than 2
...
5
...


ib

d

g

n added = 4
...
3 ≅ 4
...
028 kg N 2 / mol = 120 × 105 kg N 2

...
70 (cont’d)
c
...


143 kmol N 2

...
204 kmol
y O = 0
...

y1

143 kmol N 2

...
4-2

N 2 at 700 kPa gauge = 7
...
⇒ Pr = 0
...
36 =======> z = 0
...
91 atm
5000 L

...
99 0
...
2 K

y1 =

0
...
204
y init n init
=
= 0
...

1
...

1
...
0033

FG y IJ
FG n IJ ⇒ n = H y K = 4
...

F n IJ
lnG
H 1
...

= 5b143 kmol N gb28
...
Multiple cycles use less N2 and require lower operating pressures
...


5
...
m = MW
b
...
9 K = 665
...
85
Pc = 42
...

m = 60
...
09 lb m / lb - mol SPV
⇒ Cost ($ / h) = mS = MW
=
= 60
...
7302 lb-mol⋅o R
Fig
...
4-2

⇒ z = 0
...

zT
z

⇒ Delivering 10% more than they are charging for (undercharging their customer)

5- 47

5
...


For N 2 : Tc = 126
...
16o R, Pc = 335 atm

...
7 o R
= 2
...
16o R
⇒ z = 1
...
2
Pr =

...
7 psia

After heater: Tr =

n=

150 SCFM
= 0
...
418 lb - mole 10
...
7 o R

...
65 ft 3 / min
P
min
lb - mole ⋅o R 600 psia

b
...
418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks
min
h
day week
0
...
4 lb m / ft 3

b g

= 4668 ft 3 = 34,900 gal
5
...


For CO: Tc = 133
...
5 atm
300 K
= 2
...
0 K
2514
...
0
34
...
7 psia

Initially: Tr1 =

U
|
|
V
|
|
W

Fig
...
4-3

⇒ z = 1
...
7 psia 150 L 1 atm
mol ⋅ K
= 1022 mol
1
...
7 psia 0
...
26
133
...
7 psia 1 atm
Pr1 =
= 4
...
5 atm 14
...


Fig
...
4-3

⇒ z = 1
...
7 psia 150 L 1 atm
= 918 mol
1
...
7 psia 0
...

60 h

n 2 = y 2 n air = y 2
t min =

1 atm
30
...
25 mol
L⋅atm
RT
mol air
0
...
25 mol
=
= 014 h

...
73 mol / h
⇒ t min would be greater because the room is not perfectly sealed

c
...


5- 48

5
...
7 K , Pc = 458 atm

...
4 K , Pc = 48
...
5 atm

...
20gb45
...
30gb48
...
50gb50
...
9 atm
U
b90 + 273
...

|
| ⇒ z = 0
...

V
200 bars
1 atm
Reduced pressure:
P =
= 4
...
9 atm 1
...
20gM
+ b0
...
50gM
= b0
...
04g + b0
...
07g + b0
...
05g


...

Pseudocritical temperature: Tc′ = 0
...
7 + 0
...
4 + 0
...
4-3

r

CH 4

C2 H 6

C2 H 4

= 26
...
71 10 kg 1 kmol 0
...
25 kg
kmol ⋅ K

b90 + 273g K = 0
...
10b309
...
90b126
...
5 K
N O: T = 309
...
7 atmV P ′ = 0
...
7g + 0
...
5g = 37
...
10b44
...
90b28
...
62
n = 5
...
62 kgg = 0
...
T = b24 + 273
...
5 = 2
...
97bFig
...
4 - 3g
30 L
37
...
56|
169 mol 144
...
08206 L ⋅ atm
W

5
...
2 K, PC = 33
...


0
...
2 K 0
...
14 bFig
...
4 - 3g
|
V
V = 0
...
g |
W
Pr = 273 37
...
32
r

T=

273 atm 30 L
mol ⋅ K
= 518 K ⇒ 245° C
1
...
08206 L ⋅ atm

5- 49

b g
b g

U
V
W

b

g
g

5
...
0 K, Pc = 34
...
60 133
...
40 33 + 8 = 96
...
8 atm
Pc′ = 0
...
5 + 0
...
8 + 8 = 29
...
01
1 atm
V ⎯
= 4
...
7 psi
W
g

Tr = 150 + 273
...
2 = 4
...
0 atm

Fig
...
4-1

Turbine exit: Tr = 373
...
2 = 3
...
0

Pr = 1 29
...
03

Pin Vin
z nRTin
P z T
ft 3 14
...
01 423
...
00 373
...
01 would instead be 1
...
97b133
...
03b304
...
1 K
V
CO : T = 304
...
9 atmW P ′ = 0
...
5g + 0
...
9g = 35
...
8 psi
Initial: T = 303
...
1 = 2
...
97
P = 2014
...
8 = 3
...
77 CO: Tc = 133
...
5 atm
2

c

c

c
c

r

Fig
...
4-3

1

r

Final: Pr = 1889
...
8 = 3
...
97
Total moles leaked:

FG P − P IJ V = b2000 − 1875gpsi
0
...
0 L

1

n1 − n 2 =

2

303 K 14
...
08206 L ⋅ atm

= 10
...
97 10
...
3 mol CO
Total moles in room:

Mole% CO in room =

24
...
4 mol

303 K 22
...
3 mol CO
× 100% = 10% CO

...
4 mol

5- 50

CO + 2H 2 → CH 3OH

5
...
5 kmol CH 3OH h

n1 (kmol CO / h)
2n1 (kmol H 2 / h)
644 K
34
...
5 kmol CH 3OH (l ) / h

a
...
5 kmol CH 3OH 1 kmol CO react
1 kmol CO fed
= 218 kmol h CO
h
1 kmol CH 3OH 0
...
0 K

Pc = 34
...
8 atm
⇓ Newton’s corrections

b

g b

g

b g b

g

Tc′ =

1
2
133
...

3
3

Pc′ =

1
2
34
...
8 + 8 = 25
...
7 = 8
...
5 MPa
10 atm
⎯Fig
...
4-4 → z 1 = 1
...
45
24
...
013 MPa
Vfeed =

1
...
08206 m3 ⋅ atm 1
...
5 MPa
kmol ⋅ K
10 atm

Vcat =

120 m3 h

1 m3 cat
25,000 m3 / h

= 0
...
8 L)

b
...
5 kmol CH 3OH (l ) / h

Overall C balance ⇒ n 4 = 54
...
5 kmol CO h
109
...
5 kmol feed gas h

1
...
5 kmol
644 K
0
...
013 MPa
= 29
...
5 MPa
kmol ⋅ K
10 atm

5- 51

5
...
3 + 8) K = 41
...
6 K

Pc = (12
...
8 atm

Pc = 39
...
15(413 K) + 0
...
6 K) = 362
...


U
V
P ' = 0
...
89

Pc ' = 0
...
8 atm) + 0
...
7 atm) = 36
...
86

r

znRT 0
...
08206 m3 ⋅ atm 323 K 1 h
=
= 1
...
80

Fig
...
4-2

2

d

i FG 100 cmIJ = 10
...

4V
⇒d=
=
πu
π 150 m / min

b

Tc = 190
...
8 atm

C2 H 4 : Tc = 283
...
5 atm
C2 H 6 : Tc = 305
...
2 atm

U
|
V
P ' = 015( 458 atm) + 0
...
5 atm) + 0
...
2 atm) = 49
...


...
W
T=90o C

Tc ' = 015(190
...
60(283
...
25(305
...
8 K ====>

...


P=175 bar

c

r

⎯Fig
...
4-3→ z = 0
...
02 mg
GH s JK H s K
H s K H min K 4
3

V

2

n=
5
...


m3
min

PV 175 bar 1 atm
kmol ⋅ K 0188 m3 / min

...

3
363 K
zRT
0
...
013 bar
...
2 K = 227
...


acetonitrile: Tc = 548 K = 986
...
7 atm
Fig
...
4-3

Tank 1 (acetonitrile): T1 = 550 o F, P1 = 4500 psia ⇒ Tr1 = 1
...
4 ⇒ z 1 = 0
...
200 ft 3
=
z 1 RT1
0
...
7 o R

lb - mole ⋅ o R
= 0
...
7302 ft 3 ⋅ atm
Fig
...
4-3

Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm ⇒ Tr2 = 4
...
4 ⇒ z 2 = 1
...
0 atm 2
...
00 1009
...
027 lb - mole

...
81 (cont’d)


...
4 R + FG 0
...
16 R = 830 R
H 0131K
H 0131 K

...


...
7 atm + FG 0
...
5 atm = 44
...


...


c

dV i
r

P=

=

ideal

Fig
...
4-2
VPc '
2
...
8 atm lb - mole ⋅ o R
=
= 1
...
85
RTc ' 0
...
7302 ft 3 ⋅ atm

znRT 0
...
7302 ft 3 ⋅ atm 1009
...

=
= 37
...
2 ft 3

5
...
48 g C a H b O c , 26
...
9 kPa
n c (mol C), n H (mol H), n O (mol O)

1 L @483
...
387 mol CO 2 / mol
0
...
355 mol H 2 O / mol

n O 2 (mol O 2 )
26
...
9 kPa

a
...
42 g 1 cm3 159 g = 2
...

O 2 in Charge:
n O2

d

1
...
15 cm 3 10 −3 L km 3
=
L ⋅ atm
0
...
9 kPa

1 atm

300 K

101
...
200 mol O 2

Product
1
...
310 mol product
756
...
3 kPa
0
...
200 + n O = 0
...
387 + 2 0
...
355 ⇒ n O = 0
...
387 0
...
120 mol C in sample

b

gb

g

H: n H = 2 0
...
310 = 0
...
120 0
...
1 b = 0
...
110 = 2
Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20),
etc
...


b g

b g

MW = 12
...
01b + 16
...
01 1
...
01 2c + 16
...
23c
300 < MW < 350 ⇒ c = 10 ⇒ C 11 H 20 O 10

5- 53

bg

C5 H 10 +

5
...
21 O 2
0
...
2 L, Po (bar)

a
...
0 mL C5 H 10 l

n 3 (mol CO 2 )
n 4 mol H 2 O(v)
n 5 (mol N 2 )
75
...
745 g

1 mol

mL

n1 =

70
...
1062 mol C5 H 10

0
...
5 mol O 2
1 mol C 2 H 10

1 mol air
= 3
...
21 mol O 2

nRT 3
...
08314 L ⋅ bar 300K
= 8
...
2 L
mol ⋅ K
V

(We neglect the C5 H 10 that may be present in the gas phase due to evaporation)
Initial gauge pressure = 8
...
44 bar
b
...
531 mol CO 2
1 mol C 5 H 10
0
...
052 mol product gas
n4 =
= 0
...
79 3
...
99 mol N 2
n3 =

0
...
531 / 4
...
131 mol CO 2 / mol, Tc = 304
...
9 atm
H 2 O: y 4 = 0
...
052 = 0
...
99 / 4
...
738 mol N 2 / mol,

Tc = 647
...
3 atm
Tc = 126
...
5 atm


...
2 K) + 0
...
4 K) + 0
...
2 K) = 217
...


...
9 atm) + 0
...
3 atm) + 0
...
5 atm) = 62
...
2 L 62
...
7 ⇒ z ≈ 1
...
5
...
052 mol 217
...
08206 L ⋅ atm

b

g

75
...

PV
mol ⋅ K
=
= 2439 K - 273 = 2166o C
znR
1
...
052 mol 0
...
1

a
...
T = 100 o C
...


b
...
00 g/mL
and VB = 10 mL

Point C: H2O (v, 100°C)
n=

10 mL

1
...
555 mol
mL 18
...
2

nRTC 0
...
08206 L ⋅ atm 373 K
=
= 17 L
1 atm
mol ⋅ K
PC

a
...
Since liquid is still present, the pressure and temperature must lie on the

vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the
species at the system temperature
...
Assuming ideal gas behavior for the vapor,

m(vapor) =
m(liquid) =

(3
...
010) L
mol ⋅ K 243 mm Hg
1 atm
119
...
59 g
(30 + 273
...
08206 L ⋅ atm
760 mm Hg
mol
10 mL

1
...
89 g

m total = m(vapor) + m(liquid) = 19
...
3

a
...
59
= 0
...
48

log 10 p ∗ = 7
...
71
= 2
...
370 = 234
...
ln p = −
+ B⇒−
=
=
1
1
R T
R
− T1
T2
∗

*
B = ln( p1 ) +

b

g

ln 760 / 118
...
0 + 273
...
3 +
= 18
...
5 + 273
...
5+ 273
...
3 (cont’d)

ln p ∗ (45o C) = −

b

4151
+ 18
...

45 + 273
...

2310 − 234
...

234
...


p∗ =

FG 118
...
5g + 118
...
7 mm Hg
H 29
...
7 − 234
...
7% error
234
...
scale) on semilog paper
T + 273
...


Plot p ∗ log scale vs

b

g

ln p ∗ mm Hg =

b

7076 K
Δ Hv
= 7076K ⇒ Δ H v =
R

g

OP
Q

8
...
8 kJ mol
mol ⋅ K 10 3 J

ln p* = A/T(K) + B
p*(mm Hg)
5
20
40
100
400
760

1/T(K)
0
...
002639
0
...
002410
0
...
002125

ln(p*) p*(fitted)
1
...
03
2
...
01
3
...
26
4
...
05
5
...
81
6
...
13

7
6
5
4
3
2
1
0

1/T

6-2

0
...
0026

0
...
0022

y = -7075
...
666

0
...
7
105
...
0
141
...
5
197
...
5

LM
N

−7076
−7076

...

+ 2167 ⇒ p ∗ mm Hg = exp
+ 2167
o
T ( C) + 273
...
2
o

0
...
4

T(oC) p*(fitted)
50
0
...
12
110
24
...
00
230 2000
...
6

a
...
9 + hright -hleft
-3
42
...
17×10
34
...
9 3
...
9
-3
68
...
93×10
122
...
9 2
...
9
88
...
76×10-3
282
...
3 2
...
9
105
...
64×10-3
524
...

b
...
855
RT
T

T(°C)

1/T(K)

At the normal boiling point, p∗ = 760 mmHg ⇒ Tb = 116° C
Δ Hv =

8
...
8 K

1 kJ

mol ⋅ K

10 3 J

= 42
...
Yes — linearity of the ln p∗ vs 1 / T plot over the full range implies validity
...
7

a
...
2 + b ⇒ y = ax + b

b

y = ln p∗ ; x = 1 T + 273
...
5° C , p1 ∗ = 400 mm Hg ⇒ x1 = 31980 × 10 −3 , y1 = 5
...

T2 = 56
...
0331 × 10 −3 , y 2 = 6
...
0941 × 10 −3
x − x1
y 2 − y1 = 6
...
39588 = 599 mm Hg
y = y1 +
x 2 − x1

FG
H

IJ b
K

g

b

Cox chart
b
...


6
...
02447 −

b

g

12 psi 760 mm Hg
= 625 mm Hg
14
...

= 2
...
7872 = 613 mm Hg
50 + 224

g

Estimate p∗ 35° C : Assume ln p∗ =

a
+ b , interpolate given data
...
1 + 25
...
630
|
35 + 273
...
1
a
b = ln p ∗− = lnb50g +
= 25
...
5 mm Hg
p∗ b35° Cg = e
|
25 + 273
...
2

= −6577
...
2

4
...
0 × 10

b
−4

273 K
102
...
2 K 760 mm Hg 10 mL 22
...
9

a
...


Two intensive variable values (e
...
, T & P) must be

specified to determine the state of the system
...
6
b
...
97421 −
= 2
...
5107 = 324 mm Hg
55 + 216
...
27 > 0115 The vessel does not constitute an explosion

...


6
...
The solvent with the lower flash point is easier to ignite and more dangerous
...
The
other one should be kept from any heating sources when contacted with air
...
At the LFL, y M = 0
...
06 × 760 mm Hg = 45
...
11
Antoine ⇒ log 10 45
...
87863 ⇒ T = 6
...
The flame may heat up the methanol-air mixture, raising its temperature above the flash point
...
11 a
...
1 = 50 mm Hg ⇒ T = 38
...
3
...
VH2O =

30
...
100 mol H2 O 18
...

(50 + 273) K 760 mm Hg 22
...
(iv) (the gauge pressure)

6-4

6
...


b
g
= 110° C , p ∗ = 755 mm Hg − b577 − 222gmm Hg = 400 mm Hg

T1 = 58
...
2

−

1
58
...
2

= −46614

...

= ln 60 +
= 18156

...
3 + 273
...
2 K
−46614

...

T
ln p∗ 130° C = 6
...
595 = 7314 mm Hg

...


g

b

g

Basis: 100 mol feed gas CB denotes chlorobenzene
...
3°C, 1atm
y1 (mol CB(v)/mol) (sat’d)
(1-y1) (mol air/mol)

100 mol @ 130°C, 1atm
y0 (mol CB(v)/mol) (sat’d)
(1-y0) (mol air/mol)

n2 mol CB (l)

b

g

Saturation condition at inlet: y o P = pCB ∗ 130° C ⇒ y o =

731 mm Hg
= 0
...
0789 mol CB mol
g
760 mm Hg
Air balance: 100b1 − y g = n b1 − y g ⇒ n = b100gb1 − 0
...
0789g = 4
...
126 = 9587 mol CBbl g

...
3° C ⇒ y1 =
o

1

1

1

% condensation:

2

1

2

95
...
7%
0
...
Assumptions: (1) Raoult’s law holds at initial and final conditions;
(2) CB is the only condensable species (no water condenses);
(3) Clausius-Clapeyron estimate is accurate at 130°C
...
13 T = 78° F = 25
...
9 in Hg = 759
...
87 p∗ 25
...
3

yH 2 O =

0
...
559 mm Hg )

( )

759
...
0281( 759
...
34 mm Hg

6-5

Table B
...
0281 mol H 2 O mol air

Tdp = 23
...
13 (cont’d)

hm =

0
...
0289 mol H 2 O mol dry air
1 − 0
...
0289 mol H 2 O 18
...
0180 g H 2 O g dry air
mol dry air
mol H 2 O
29
...
0289
× 100% =
× 100% = 86
...
559 [ 759
...
559]
p ∗ ( 25
...
56°C ) ⎤
⎣
⎦

6
...
1° C), 1 atm, hr = 50%

b

hr = 50% ⇒ y H 2 O P = 0
...
1° C
Table B
...
50 × 18
...
012
760
...
012 mol H 2 O 18
...
988 mol dry air 29
...
87 g

b g b273
...
1gK = 24
...
4 L STP
1 mol

273
...
87 g
= 1196 g L

...
13 L

Basis II: 1 mol humid air @ 70° F (21
...
80 p H 2O ∗ 21
...
3

Mass of air:

y H 2O =

mol H 2 O
0
...
765 mm Hg
= 0
...
0 mm Hg
mol

0
...
02 g
1 mol

Volume of air:
Density of air =

1 mol

g

+

0
...
0 g
1 mol

b g b273
...
1gK = 24
...
4 L STP
1 mol

273
...
78 g
= 1193 g L

...
13 L

Basis III: 1 mol humid air @ 90° F (32
...
80 p H 2 O ∗ 32
...
3

y H 2O =

g

mol H 2 O
0
...
068 mm Hg
= 0
...
0 mm Hg
mol

6-6

= 28
...
14 (cont’d)

Mass of air:

0
...
02 g 0
...
0 g
+
= 28
...
2 + 32
...
04 L

1 mol

22
...
2K

28
...

25
...
Therefore, the statement is wrong
...
15 a
...
50 p H 2 O ∗ 90° C
Table B
...
50 × 525
...
346 mol H 2 O / mol
760
...
346 760 = 262
...
3

Tdp = 72
...
7 = 17
...
Basis:

1 m 3 feed gas 10 3 L 273K
m

Saturation Condition: y1 =

3

mol

b g = 33
...
4 L STP

b

*
p H 2 O 25° C

g = 23
...
0313 mol H O mol

760
P
Dry air balance: 0
...
6 ) = n1 (1 − 0
...
7 mol

2

Total mol balance: 33
...
7+n2 ⇒ n2 = 10
...
y H 2 O P = p∗ 90° C ⇒ P =

p * (90° C) 525
...
00 atm
y H 2O
0
...
16 T = 90° F = 32
...
7 in Hg = 754
...
43 lb m

7
...
02 lb m

= 4
...
4oC), 754 mm Hg

y1 (lb-mol H2O (v)/lb-mol)
(1-y1) (lb-mol DA/lb-mol)
hr=95%, 90oF (32
...
7 in Hg (754 mm Hg)

4
...
95 p∗ 32
...
3

1 lb-mol

3

b

g

0
...
068 mm Hg
754
...
4°C )

g = 0
...
3

y2 =

6
...
00817 lb-mol H 2 O lb-mol
754
...
631

⎫ ⎧n1 = 124
...
045n1 = 0
...
631⎭ ⎩n2 = 120
...
7 lb-moles 359 ft 3 (STP) (460+90)o R 760 mm Hg
min

lb-moles

492o R

754 mm Hg

= 5
...
17 a
...


p final =

760 mm Hg

(15 + 273) K
(200 + 273) K

= 462
...
20 × 462
...
6 mm Hg

p * (15° C) = 12
...
Impossible ⇒ condensation occurs
...
80 × 760) mm Hg ×
= 370
...
2 + 12
...
Basis:

mol
1 L 273 K
= 0
...
4 L (STP)

6-8

6
...
1 mm Hg
y1 (mol H2O (v)/mol) (sat’d)
(1-y1) (mol dry air/mol)

0
...
200 H2O mol /mol
0
...
79 mm Hg = 0
...


g b

g

Dry air balance: 0
...
0258 = n1 1 − 0
...
02135 mol

c
...
0258 = 0
...
00445 mol
Mass of water condensed =

0
...
02 g
= 0
...
18 Basis: 1 mol feed
n2 (mol), 15
...
10 mol H 2O (v)/mol
0
...
6°C, 3 atm

Saturation: y 2 =

b

*
p H 2O 15
...
3

P

y2 =

bg b
g
H O mol balance: 0
...
00583b0
...
29 mm Hg
atm
= 0
...
90 1 = n2 1 − 0
...
9053 mol
2

Fraction H 2 O condensed:

3

⇒ n3 = 0
...
0947 mol condensed
= 0
...

0100 mol fed

b

g

hr =

y 2 P × 100% 0
...

p∗ 100° C
1 atm

V2 =

0
...
4 L STP
mol

b

g

b g

b g

1 mol 22
...
24 × 10 −3 m 3 outlet air @ 100° C
273K 3 atm 10 3 L

363K 1 m 3
= 2
...
24 × 10 −3 m 3 outlet air
=
= 0
...
98 × 10 −2 m 3 feed air

6-9

6
...


1 kmol
18
...


23
...
0208 mol H 2 O mol air
15 × 760 mm Hg

...
0208
= 0
...
0208

b g

15 L STP
1 mol
= 0
...
4 L STP
0
...
0212 mol H 2 O
= 0
...
387 kmol 10 3 mol
1h
min
= 1628 h
kmol
0
...
20 a
...
Δp =

b g

π
4

⋅ 30 2 ⋅ (18 − 8) =

7
...
8 daysg

7
...
288 × 10 4 gal / day
3
ft

5
...
703 × 62
...
481 gal
ft 3
= 0
...
10 × 10 5 lb m C 8 H 18 / day

0
...
43 lb m
ft

32
...
174

(18-8) ft

lb m ⋅ft
s

2

1 ft 2
144 in

29
...
696 lb f / in 2

= 6
...
696 psi
= 0
...


*
c
...
4: pC8 H18 (90 o F) =

Volume:

5
...
74 mm Hg

1 ft 3
= 7069 ft 3
7
...
0 + 14
...
77 lb - moles
RT
10
...
40 psi
Mole fraction of C 8 H 18 : y = 8 18 =
= 0
...
0 + 14
...
0130(36
...
479 lb - mole ( = 55 lb m = 25 kg)
d
...


6-10

*
*
6
...
Antoine equation ⇒ ptol (85o F) = ptol (29
...
63 mmHg = ptol

Mole fraction of toluene in gas: y =

ptol 35
...
0469 lb - mole toluene / lb - mole
760 mmHg
P

yPV
RT
0
...
7302

1 ft 3

900 gal

1 atm
ft 3 ⋅ atm
lb - mole ⋅ o R

(85 + 460) o R 7
...
13 lb m tol
lb - mole

= 1
...

Basis: 1mol
0
...
9531 mol G/mol

nV (mol)
y (mol C7H8(v)/mol)
(1-y) (mol G/mol)
T(oF), 5 atm

Assume G is
noncondensable

nL [mol C7H8 (l)]
90% of C7H8 in feed

90% condensation ⇒ n L = 0
...
0469)(1) mol C 7 H 8 = 0
...
0422 ⇒ nV = 0
...
0469(1) = y (0
...
0422 ⇒ y = 0
...
004907)(5 × 760) = 18
...
773 − 219
...
95805 − log10 18
...
11o C=62
...
95805 − log10 18
...
22 a
...
53 kmol / h
3
82
...
Antoine Equation:

1175
...
26601
100+224
...
88555-

0
...
00) atm 760 mm Hg
*
= 228 mm Hg < p Hex ⇒ not saturated
atm
1175
...
88555= 2
...
8°C
T+224
...
22 (cont’d)
c
...
53 kmol/h
0
...
85 N2

nV (kmol/h)
y (kmol C6H14 (v)/kmol), sat’d
(1-y) (kmol N2/kmol)
T (oC), 2 atm

nL (kmol C6H14 (l)/h)
80% of C6H14 in feed

80% condensation:
Mole balance:
Hexane balance:
Raoult’s law:
Antoine equation:

n L = 0
...
53 kmol / h) = 0
...

6
...
7836 ⇒ nV = 5
...
53) = y (5
...
7836 ⇒ y = 0
...

*
p Hex = yP = (0
...
82 mmHg = p Hex (T )
1175
...
82 = 6
...
52o C
T + 224
...
23 Let H=n-hexane
n1 ( kmol / min)
0
...
95 kmol N2/kmol
T (oC), 1 atm

a
...
sat’n

Condenser

1
...
500 p H (80 o C)

Table B
...
500)(1068 mmHg)
= 0
...
05 P = p * (T1 ) ⇒ p H (T1 ) = 0
...
88555 −

1175
...
26o C
T1 + 224
...

n0 = 2
...
703) n0 = 0
...
682 kmol / min
N2 volume: V N 2 =

(0
...
682 kmol 22
...
5 SCMM

6
...


Assume no condensation occurs during the compression

2
...
703 H(v)
0
...
682 kmol/min
0
...
95 N2
T1 (oC), 10 atm

V0 ( m 3 / min)
2
...
703 H(v)
0
...
S
...
5 kmol H(l)/min

50% relative saturation at condenser inlet:
*
*

...
500 p H (T0 ) = 0
...
050(7600 mmHg) = 380 mmHg = p H (T1 )

Volume ratio:

Antoine

Antoine

T0 = 187 o C
T1 = 48
...
2) 0
...
22 3
V0 n0 RT0 / P n0 (T0 + 273
...
18 kmol/min 460 K
m in

c
...
26o C (installed cost of condenser + utilities and other operating
costs) vs
...

6
...


Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes
...
718 × 1
...
25 kg

= 0
...
818 kmol
3
298 K 22
...
084 kmol C 9 H 20
=
= 010 kmol C 9 H 20 / kmol (10 mole%)

...
818 kmol
n gas

As the nonane evaporates, the mole fraction will pass through the explosive range (0
...
9%)
...

The nonane will not spread uniformly—it will be high near the sump as long as liquid is present
(and low far from the sump)
...

b
...
00 mmHg

...
0 o C = 339 K, p2 = 40
...
24 (cont’d)

ln(40
...
00)
5269
5269
⇒ A = 5269, B = ln(5
...
23 ⇒ p * = exp(19
...
008 kmol C 9 H 20 / kmol ⇒ p * ( T ) = yP = (0
...
08 mm Hg

Formula for p*

T = 302 K = 29 o C

c
...
Using steam rather
than air is to make sure an explosive mixture of nonane and oxygen is never present in the
tank
...


6
...
79 mol N 2/mol
y 1 (mol H 2 O/mol)
+ O2 , CO2

Lungs

O2

Air inhaled: n1 =

37°C, 1 atm
n2 (mol), saturated
0
...
4 liter STP

1 hr

1 day

= 356 mol inhaled day

Inhaled air - -10% r
...
: y1 =
Inhaled air - -50% r
...
: y1 =

b

g = 010b2107 mm Hgg = 2
...


...

P

mol H 2 O
mol

−2

mol H 2 O
mol

760 mm Hg

b

g = 0
...


...
50 p∗ H 2 O 23° C
P

−3

760 mm Hg

H 2 O balance: n0 = n2 y 2 − n1 y1 ⇒ (n0 ) 10% rh − ( n0 ) 50% rh = (n1 y1 ) 50% − (n1 y1 ) 10%

FG
H

= 356

mol
day

IJ L(0
...
00277) mol H O OFG 18
...
0619, and the rate of weight loss by breathing at 23oC and 50% relative humidity is
n0 (18) = (n2y2 - n1y1)18 = 329 g/day
...
26 a
...

b
...
A=acetone
n 1 mol @ To C, 1 at m
y1 mol A(v)/ mol (sat’d)
(1-y1 ) mol N2 /mo l

1 mo l @ 90o C, 1 atm
0
...
80 mol N2 /mo l

n 2 mol A(l)

For cooling water at 20oC

(

)

log10 p* 20 o C = 7
...
595
= 2
...
6 mmHg
A
20 + 229
...
6
= 0
...
2 , so no saturation occurs
...
11714 −
A

(

)

1210
...
89824 ⇒ p* −35 o C = 7
...
664

(Note: -35oC is outside the range of validity of the Antoine equation coefficients in Table
B
...
An alternative is to look up the vapor pressure of acetone at that temperature in a
handbook
...
)
7
...
0100
d i
760
N mole balance: 1b0
...
01g ⇒ n = 0
...

Total mole balance: 1 = 0
...

d
...
192
× 100% = 96%
2

Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant
The condenser temperature could never be as low as the initial cooling fluid temperature
because heat is transferred between the condenser and the surrounding environment
...


6-15

6
...
5 mol / h
h
22
...
5 (mo l/h) @ 20o C, 103 KPa
y1 [mol H2O(v)/mol]
y1 mol (mol DA/mol)
1– y1 H2 O(v)/ mol (sat’d)
(1-y1 ) mol DA/mo l

n2[mol H2O(l)/h]
n 2 mol H2O(l)/h

Inlet: yo =

(

*
hr ⋅ pH 2O 35 o C

P

Outlet: y1 =

*
p H 2O

) = 0
...
175 mmHg

101325 Pa
= 0
...
535 mmHg 101325 Pa = 0
...
04913 no = 1 − 0
...
5 ⇒ no = 543
...
2 mol 22
...
2 = 528
...
7 mol / h

Inlet air:

14
...
02 g H 2 O 1 kg
= 0
...
28

Basis:

10000 ft 3 1 lb - mol 492 o R 29
...
82 lb - mol / min
min 359 ft 3 (STP) 550 o R 29
...
8 in
...
82 lb-mole/min
90oF, 29
...
Hg
y0 [lb-mole H2O(v)/mol
1- y0 (lb-mole DA/mol)
hr = 88%

n1 lb-mole/min
65oF, 29
...
Hg
y1 [lb-mole H2O(v)/lb-mole]
1- y1 (lb-mole DA/lb-mole)

n2 [lb-mole H2O(l)/min]

Inlet: y o =

d

i = 0
...
07 mmHgg

*
hr ⋅ p H 2 O 90 o F

Outlet: y1 =

P

29
...
274 mmHg

*
p H 2 O 40 o F

P

29
...
0419 lb - mol H 2 O / lb - mol
25
...
00829 lb - mol H 2 O / lb - mol
25
...
82 1 − 0
...
00829 ⇒ n1 = 23
...
82 = 23
...
84 lb - mole / min

6-16

6
...
84 lb - mol 18
...
48 gal
= 181 gal / min

...
4 lb m 1 ft 3

Condensation rate:

Air delivered @ 65oF:

23
...
92 in Hg
= 9223 ft 3 / min
o
min
1 lb − mol 492 R 29
...
29 Basis: 100 mol product gas
100 mol, T1, 1 atm
y1 mol H2O(v)/mol, (sat’d)
(1-y1) mol DA/mol

no mol, 32oC, 1 atm
yo mol H2O(v)/mol
(1-yo) mol DA/mol
hr=70%

100 mol, 25oC,1 atm
y1 mol H2O(v)/mol,
(1-y1) mol DA/mol
hr=55%

n (mol HO(l))
n22lb-molH22O(l)/min

Outlet: y1 =

d

*
hr ⋅ p H 2 O 25o C

i = 0
...
756g = 0
...
70b35
...
0328 mol H O / mol

*
Saturation at T1 : 0
...
07 = p H 2 O T1 ⇒ T1 = 15
...
0328 = 100 1 − 0
...

Total balance: 1016 + n2 = 100
...
6 mol (i
...
removed)

...
02 g 1 kg

...
0288 kg H 2 O
1 mol 1000 g

g

100 1 − 0
...
0 g 1 kg
= 2
...
0288
= 0
...
85

6-17

6
...


Room air − T = 22° C , P = 1 atm , hr = 40% :

y1 P = 0
...
40 )19
...
01044 mol H O
2

760 mm Hg

mol

Second sample − T = 50° C , P = 839 mm Hg , saturated:
y2 P = p ∗H2 O ( 50°C ) ⇒ y2 =

92
...
1103 mol H 2 O mol
839 mm Hg


...
01044, H1 = 5 , y 2 = 01103, H 2 = 48

b=

b

ln y 2 y1


...
01044g = 0
...
054827 H g

b

g

ln a = ln y1 − bH1 = ln 0
...
054827 5 = −4
...
8362 = 7
...
937 × 10 −3
b
...
4m 3 STP

g

b g

10 3 mol
= 4131 mol air delivered

...
31 mol, C,1 at m
41
...
31 mol, T, 1 at m
n o mol, 35o C, 1 at m

mol 2O(v)/mol,
0
...
09896 mo l DA/ mol
0
...
0104 mo l H2 O(v)/mo l
0
...
09896 mo l DA/ mol
0
...
01044 )( 760 mm Hg ) = 7
...
8°C (from Table B
...
0411 mol H 2 O mol
Overall dry air balance: n0 (1 − y0 ) = 41
...
9896 ) ⇒ n0 =

( 41
...
9896 )
= 42
...
0411)

Overall water balance: n0 y0 = n1 + ( 41
...
0104 ) ⇒ n1 = ( 42
...
0411) − ( 41
...
0104 )
= 1
...
32 mol H 2 O 18
...
024 kg H 2 O condensed m 3 air delivered

6-18

6
...


Basis: n0 mol feed gas
...
3 for n1

⇒

n1 = n0

b gb g

0

0

do

Po

S balance: n0 y 0 = n1 y1 + n2
(1) - (4)

b g = LMn
P
NM

n0 p∗ Tdo

0

−

o

⇒

b g OPFG p∗ dT iIJ + n fp∗ bT g
P
QPGH P JK p

n0 fp∗ Tdo

f

o

f

do

f

do

o

do

o

b1 − f g p∗ bT g = LM1 − fp∗ bT g OP p∗ dT i
P
NM P QP P
o

0

LM fp∗ eTdo j OP
p∗ d T i 1 −
MM Po PP
Q
N
=
p∗ T
b1 − f g eP do j
f

⇒ Pf

f

o

b
...
9565
B= 1423
...
09
=
R
un T0
P0 Td0
f
1
2
3
4

50
50
50
50

765
765
765
765

40
40
40
40

0
...
95
0
...
95

Tf

p* (Td0) p*(Tf)

45
40
35
20

21
...
472
21
...
472

6-19

27
...
47
16
...
07

Pf
19139
14892
11471
4902

C
refr C p C
com
tot
2675
4700
8075
26300

107027 109702
83329 88029
64239 72314
27582 53882

(3)

(4)

6
...
Decreasing temperature and increasing pressure both to
c
...
When you decrease Tf, less compression is required to
achieve a specified fractional condensation
...


6
...


A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr)
...
Similarly, running at a higher Tf lowers the refrigeration cost but raises
the compression cost
...

Basis : 120 m 3 min feed @ 1000 o C(1273K), 35 atm
...


b g P batmg bT g

Cmpd
...


c corr

H2

33
...
8

41
...
0
304
...
5
72
...
7


...
8

CO
CO 2

bApply Newton' s corrections for H g

−

∑y T
P′ = ∑ y P

Tc′ =

i ci

c

i

ci

b g b g b g b g
= 0
...
8g + 0
...
5g + 0
...
9g + 0
...
3 atm

...

= 0
...
35 133
...
20 304
...
05 190
...
4 K

Feed gas to cooler
Tr = 1273 K 133
...
54

⎫ Generalized compressibility charts (Fig
...
4-3)
⎬
Pr = 35
...
3 atm = 0
...
02

V =

1
...
314 N ⋅ m
mol ⋅ K

120 m 3

m ol
3
...
04 × 10 − 3 m 3 mol
101325 N m 3

= 39
...
4 K = 2
...
5
...
0 atm 37
...
94 ⎭
⇒ z = 0
...
98

8
...
5 kmol 103 mol
min

1 kmol

283 K

1 atm
101325 N m

6
...
7

= 6
...
32 (cont’d)

1
...
5) kmol/min

n1 (kmol/min), 261 K, 35 atm

39
...
40 mol H2/mol
0
...
20 mol CO2/mol
0
...
97 × 10
y McOH =

n MeOH

n MeOH
+ n H 2 + nCH 4

A

= input

b
...
02 of input

g = 10

b

7
...
11 −12 + 2300

b

g

mm Hg

g

35 atm 760 mm Hg atm

−4

+ nCO

mol MeOH mol
=

A
= input

n MeOH

n MeOH + 39
...
40 + 0
...
05) + 0
...
0148 kmol min MeOH in gas

The gas may be used as a fuel
...


6-21

6
...
sat
...
0oC

1500 kg/min wet pulp

m1 (kg/min wet pulp)

0
...
75) kg H2O/kg
1/1
...
0015 kg H2O/kg
0
...
0015) ⇒ m1 = 858 kg / min
1 + 0
...
sat’n at inlet: y1 P = 0
...
50(28
...
0187 mol H 2 O/mol
40 C dew point at outlet: y 2 P =
o

*
p H 2 O (40 o C)

⇒ y 2 = (55
...
0718 mol H 2 O / mol

Mass balance on dry air:
n0 (1 − 0
...
0718)

(1)

Mass balance on water:
n 0 ( 0
...
0 kg / kmol ) + 1500 ( 0
...
75) = n1 ( 0
...
0015) ( 2 )
Solve (1) and (2) ⇒ n0 = 622
...
4 kmol / min
Mass of water removed from pulp: [1500(0
...
75)–858(
...
8 kmol 22
...
538 × 10 4 m 3 / min

6
...
61 lb m H2 O(l) / lb m

500 lb m / h
0
...
94 lb m L / lb

0
...
39m0 = 0
...
50 p∗ H 2 O 130° F ⇒ y1 =

b gb

g

a

H 2 O balance: 0
...
06 ) 500 lb

b

E

m

mol H 2 O
0
...
0756
760 mmHg
mol

fb

hr +

g

0
...
02 lb m
hr
1 lb - mole

n1 = 517
...
09 × 10

Dry air balance: n0 = 1 − 0
...
5) lb - moles hr = 478
...
35 a
...
4 lb - moles
hr

3

5

492° R

1 lb - mole

ft 3 hr

Basis: 1 kg dry solids
n1 (kmol)N 2, 85°C

n2 (kmol) 80°C, 1 atm
y 2 (mol Hex/mol)
(1 – y 2 (mols N 2 /mol)
)
70% rel
...


dryer
1
...
78 kg Hex

condenser

n3 (kmol) 28°C, 5
...
05 kg Hex
1
...
78 − 0
...
47 × 10 −3 kmol Hex
86
...


↓

70% rel
...
: y2 =

0
...
70 )106
...
817 (80+ 224
...
984 mol Hex mol

6
...
47 × 10 −3 kmol Hex

1 kmol
= 0
...
984 kmol Hex

b

g

N 2 balance on dryer: n1 = 1 − 0
...
0086 = 1376 × 10 −4 kmol

...


↓

Saturation at outlet: y3 =

p ∗hex ( 28°C )
P

=

10

6
...
817 ( 28 + 224
...
0452 mol Hex mol

g

Overall N 2 balance: 1
...
0452 ⇒ n3 = 144 × 10 −4 kmol

...
0086 = 144 × 10 −4 + n4 ⇒ n4 = 0
...

Fractional hexane recovery:

0
...
86
...
939 kg cond
...
78 kg feed
kmol

b
...
9n

3

heater

0
...
0 atm

y3
(1 – y3)

n 1 (kmol)N 2
85°C
dryer
1
...
78 kg Hex

y 3 (mol Hex/mol) sat'd
(1 – y 3 (mol N 2/mol)
)

n 2 (kmol) 80°C, 1 atm
y 2 (mol Hex/mol)
(1 – y 2 (mols N2 /mol)
)
70% rel
...


condenser

n3 (kmol)
y3
(1 – y 3
)

0
...
05 kg Hex
1
...
47x10-3 + 0
...
0452) = n2(0
...
9n3 (1 − 0
...
984)

( 2)

Overall N2 balance: n1 = 0
...
0452)

(3)

⎧n1 = 1
...
00861 kmol
⎪
−4
⎩n3 = 1
...
376 × 10-4 − 1
...
376 × 10−4
−5

Introducing the recycle leads to added costs for pumping (compression) and heating
...
36 b
...
2
= 0167 lb m T(l) / lb m

...
2
0
...
02 / (102) = 0
...

0
...
s
...
2 atm
D=dry solids

Heater

n (lb-mole/h)
3

y3 (lb-mole T(v)/lb-mole)
(1-y3) (lb-mole N2/lb-mole)

Condenser

Eq
...
488 lb - mole / h
W T

Toluene Balance: 300 × 0167 = m1 × 0
...
13

...
833 = m1 × 0
...
56O C)=10

(6
...
70 pC7 H8 (150 O F) ⇒ y1 =

1346
...
56 + 219
...
70 pC7 H8

P

=

= 172
...
70)(172
...

12 × 760

...
22O C)=10

y3 =

*
pC7 H8

P

=

(6
...
773
)
65
...
693

= 40
...
90
= 0
...
488 + n3 × 0
...


...
0538)

6-25

U ⇒ Rn
V Sn
W T

1

= 5
...
387 lb - mole / h

6
...
875 × (1 - 0
...
097 lb - mole

lb - mole

h

28
...


Vinlet =

6
...
387 lb - moles 359 ft 3 STP
hr

(200 + 460)° R
492° R

1 lb - mole

C 6 H 14 +

Basis: 100 mol C 6 H 14

= 2590 ft 3 h

19
O 2 → 6CO 2 + 7H 2 O
2

100 mol C 6H 14

n1 (mol) dry gas, 1 atm
0
...
G
...
069 mol CO2/mol D
...

0
...
G
...
021 mol CO/mol D
...

0
...
G
...
00265 mol C6H14/mol
0
...
G
...
00265 mol(mol N /mol D
...

(0
...
21 mol O 2/mol
0
...
069 + 0
...
00265gP ⇒ n = 5666 mol dry gas
NM b g b g b g QP
100 − 0
...
0%
100 mol fed
H balance: 14b100g = 2n + 5666b14gb0
...
2 mm Hg ⇒
595 + 5666 760 mm Hg
2

1

CO 2

CO

2

1

C 6 H 14

2

2

2

Table B
...
79n0 = 5666(0
...
21(n0 )(2) = 5666[(0
...
021 + 2 x) + 595

Solve simultaneously to obtain n0 = 5888 mol air, x = 0
...
21 mol O 2

5888 − 4524
× 100% = 30
...

Tdp = 451° C

6
...
76n1 (mol N 2 / min)

CH 4 + 2O 2 → CO 2 + 2H 2 O

pCO 2 = 80 mmHg ⇒ y1 =

C2 H 6 +

7
O 2 → 2CO 2 + 3H 2 O
2

80 mmHg 101325 Pa
= 01016 mol CO 2 / mol

...

C balance: no yo + 2no 1 − yo = 01016

(2)


...
0770 mol
| y = 0
...

|n = 01912 mol O
|y = 01793 mol H O / mol
T
...

p dT i =
= 141
...
39 Basis: 100 mol dry stack gas
n P (mol C 3 H 8)
n B (mol C 4H10 )
n out (mol)
0
...
79 N2

P = 780 mm Hg
Stack gas: Tdp = 46
...
000527 mol C 3 H 8/mol
0
...
0148 mol CO/mol
0
...
8 o C Table B
...
39 (cont’d)

b

g

Dew point = 46
...
5° C ⇒ y w =
But yw =

mol H 2 O
77
...
0995
780 mm Hg
mol

nw
= 0
...
05 mol H 2 O (Rounding off strongly affects the result)
100 + nw

b gb

gb g b

gb g

C balance: 3n p + 4n B = 100 0
...
000527 4 + 0
...
0712
⇒

b1g

3n p + 4n B = 8
...
000527 )( 8 ) + ( 0
...
05 )( 2 )
⎣
⎦

( 2)

⇒ 8n p + 10nB = 23
...
25 mol C3 H8 ⎫
⎪
⎪
⎪
Solve (1) & ( 2 ) simultaneously: ⇒ ⎨
⎬ ⇒⎨
⎪nB = 1
...
40 a
...
07 (lb - mole C 3 H 8 / lb - mole)
0
...
985gb1gb0
...
93 = G1 1 − y1 ⇒ G1 1 − y1 = 0
...
5% propane absorption ⇒ G1 y1

b1g & b2g ⇒ G

1

1 1

= 1
...
93105 lb - mol h , y1 = 1128 × 10 −3 mol C 3 H 8 mol

...
1-4), p *C3 H8 (80 o F ) = 160 lb / in 2 = 10
...
07 p ⇒ x 2 =


...
07gb10 atmg = 0
...
89 atm

Propane balance: 0
...
726 lb - mole h

b

gd h b
dL / G h

b

gd

mol

0
...
93105 1128 × 10 −3

...
006428

gb

g

Decane balance: L1 = 1 − x 2 L2 = 1 − 0
...
726 = 10
...
7 mol liquid feed / mol gas feed

6-28

6
...
The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed
rate and fractional absorption], or
nC 3 H 8 =

10
...
006428 lb - mole C 3 H 3
h

lb - mole

= 0
...
2 x 10
...
8 lb-moles C10H22/h
⇒ x2 =

b

0
...
00536 lb - mole C 3 H 8 / lb - mole
0
...
8 lb - moles h

g

c
...

All three costs would have to be determined as a function of the feed ratio
...
41 a
...
5 mol B/s
87
...
125 mol/s
0
...
5 mol HC/s

p * (30 o C) ≅ 41 lb / in 2 = 2120 mm Hg (from Figure 6
...

B
=
= 0
...
3487 = 12
...
95 ⇒ n4 = 34
...
06 + 88
...
18 mol/s
⇒

mol gas fed
22
...
222 mol gas fed/mol liquid fed
mol liquid fed
100 mol/s

b
...
8 × 0
...
2790 , following the same steps as in Part (a),

b

g b gb g

95% n-butane is stripped: n4 ⋅ 0
...
5 0
...
56 mol / s
Total mole balance: 100 + n3 = 42
...
68 mol / s

...
68 mol/s
=
= 0
...
When the N2 feed rate is at the minimum value calculated in (a), the required column length
is infinite and hence so is the column cost
...
To determine the optimum gas/liquid feed ratio, you would
need to know how the column size and cost and the N2 purchase and compression costs
depend on the N2 feed rate and find the rate at which the cost is a minimum
...
42 Basis: 100 mol NH 3
Preheated
air
100 mol NH 3
780 kPa sat'd

N2
O2

converter

n3
n4
n5
n6

n 1 (mol) O2
3
...
5

a
...
09 atm
Antoine:
log 10 6150 = 7
...
711 Tsat + 247
...
4° C = 291
...
1 ⇒
VNH 3 =

g

U
V
W


...

Pc = 1113 atm ⇒ Pr = 8
...
073
⇒ z = 0
...


...

Tc = 4055K ⇒ Tr = 2916 / 4055 = 0
...
92 100 mol

g

(Fig
...
3-1)

8
...

= 0
...
02094 =

g = 0
...
02094

...
76( 200 )

n2 +

⇒ n2 = 20
...
76 mol air mol O 2 )

Vair =

b g

b g

4
...
36 mol 22
...
2 m 3 air

ii) Reactions: 4 NH 3 + 5O 2 → 4 NO + 6H 2 O , 4 NH 3 + 3O 2 → 2 N 2 + 6H 2 O

Balances on converter
NO: n3 =

97 mol NH 3

4 mol NO
= 97 mol NO
4 mol NH 3

6-30

6
...
76 2
...
5 mol N 2

4 mol NH 3
−

3 mol NH 3

3 mol O 2
4 mol NH 3

H 2 O: n6 = 20
...
5 mol O 2

6 mol H 2 O

100 mol NH 3

4 mol NH 3

= 170
...
5 + 170
...

= 1097 mol converter effluent
8
...
7% N 2 , 7
...
5% H 2 O
iii) Reaction: 4 NO + 3O 2 + 2 H 2 O → 4 HNO 3
HNO 3 bal
...
02 g HNO 3
mol

= 277
...
02 g H 2 O

gb gb

H balance on absorber: 170
...
6 2 mol H

g

⇒ n7 = 155
...


bg

155
...
02 g H 2 O 1 cm 3
1 m3
= 2
...
02 g HNO3 277
...
02 g H 2 O
+
mol
mol

= 11115 g = 11
...
997 × 10
11
...
997 × 10 id24
...
18 × 10 m air
= d8
...
81 × 10 m H Oi = 253 m H Obl g

VNH 3 = 8
...
272 m 3 NH 3 = 2
...
43 a
...
10 mol NH3 /mol
0
...
Basis: 100 g solution
Perry, Table 2
...
2-99: T = 10oC (50oF), ρ = 0
...
120 g NH3/g solution

⇒

12
...
0 g H 2 O
= 4
...
706 mol NH 3 ,
(18
...
0 g / 1 mol)

⇒ 12
...
4 mole% H 2 O(l)
Composition of gas effluent

gU
|

...
7 psia
W
b


...

⎯
T = 50 F, x A = 0126 ⎯ ⎯ →
Perry

o

pH 2O
p total


...
7 = 0
...

⇒ yW = 0155 / 14
...
0105 mol H 2 O mol
y G = 1 − y A − yW = 0
...
907g = 99
...
2gb0
...


...
90 = n2 y G ⇒ n2 = 100 0
...
84 mol absorbed
× 100% = 18
...

100 010 mol fed

b gb g

b
...

6
...


15% oleum: Basis - 100kg
15 kg SO 3 +

85 kg H 2 SO 4

1 kmol H 2 SO 4
98
...
4% SO 3

6-32

1 kmol SO 3
1 kmol H 2 SO 4

80
...
4 kg
1 kmol SO 3

6
...


Basis 1 kg liquid feed
n o (mol), 40o C, 1
...
2 at m

0
...
10 mol G/ mol

y1 mol SO3 /mo l
(1-y1 ) mol G/ mol
Equilib riu m @ 40o C

1 kg 98% H2 SO4

m1 (kg) 15% oleu m

0
...
02 kg H2 O

0
...
85 kg H2 SO4 /kg

b

pSO 3 40° C, 84
...


...
02 kg H
2
...
98 kg H 2 SO 4
0
...
08 kg H 2 SO 4
18
...
85 m1 H 2 SO 4
2
...

98
...
28 kg SO 3 10 3 g 1 mol
SO 3 absorbed = 128 − 1
...

= 350 mol

...
07 g
⇒ 3
...

G balance: 0
...
89 mol
n1 = 0
...
89 mol
22
...
2 atm 10 3 L

= 8
...
45 a
...

b
...

c
...

∗
6
...
89272 − 1203
...
888 ) ) = 1350
...
95805 − 1346
...
693) ) = 556
...
40 (1350
...
60 ( 556
...
0711 mol Benzene mol

= 0
...
0711 − 0
...
885 mol N 2 mol

6-33

6
...
2 - 127, Table 2 - 138

b

g

⇒ H N 2 80° C = 12
...

3551 mm Hg
1 atm
H O - Raoult' s law: p b80° Cg =
= 0
...
997gb0
...
466 atm
⇒ p
⇒ p N 2 = x N 2 H N 2 = 0
...
6 × 10 4 = 378 atm
∗
H 2O

2

H 2O

H 2O

∗
H 2O

Total pressure: P = p N 2 + p H 2 O = 378 + 0
...
5 atm
Mole fractions: yH 2O = pH2O P = 0
...
5 = 1
...
999 mol N 2 mol gas

b

g

∗
6
...
7 mm Hg

b

1 atm
= 0
...
3075

Methane − Henry' s law: p m = x m ⋅ H m
Total pressure: P = p m + p H 2 O = x m ⋅ 6
...
3075) = 10
⇒ x m = 1
...
49 a
...
02 g

= 55
...
334) × 14
...
4 L (STP) 1000 cm 3

= 4
...
334) ⋅ 14
...
4 L (STP) 1000 cm

3

= 2
...
192 × 10−4
xN 2 =
=
= 7
...
49 + 4
...
102 × 10
mol
xO2 =

nO2
nH 2O + nN2 + nO2

=

2
...
788 × 10−6 mol O 2 / mol
55
...
192 × 10−4 + 2
...
49 (cont’d)
Henry' s law

Nitrogen: H N 2 =
Oxygen: H O 2 =
b
...
79 ⋅ 1
= 1
...
554 × 10 −6

0
...

3
...
102 × 10 -4 mol 32
...


6
...


mol
0
...
726 × 10 −3 g

min

1 L blood
6
...
9°C
...
0
⎯(SG) H2O =⎯→
⎯⎯⎯

18
...
0555 mol H 2 O

b g

0
...
0901

⎯ ⎯⎯⎯⎯ →
⎯

1 mol
3

b g

22,400 cm STP

= 4
...
022 × 10 i mol CO = 7
...
0555 + 4
...
246 × 10
For simplicity, assume n
≈ n b molg
x
=p
H = b35 atmg b13800 atm mole fractiong = 2
...

−6

2

pCO 2 = 1 atm ⇒ x CO 2 =

CO 2

b
...
8 oz

10 3 g H 2 O 1 mol H 2 O
1L

18
...
536 × 10 −4 mol CO 2

44
...
220 g CO 2
c
...
220 g CO 2

b g b273 + 37gK = 0127 L = 127 cm

...
4 L STP

44
...
51 a
...

– From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in
the liquid, which increases with time
...


b

g

b

b
...
07 gg = 0
...
01561r F mol SO I
⇒x =
J
G
5
...
01561r H mol K

g

Basis: 100 g H 2 O 1 mol 18
...
55 mol H 2 O
2

2

2

SO 2

From this relation and the given data, pSO 2 = 0 mmHg ⇔ xSO 2 = 0 mol SO 2 mol
1
...
8 x 10–3
4
...
6 x 10–3

42
85
129
176

A plot of pSO 2 vs
...
Fitting the line using the method of least squares
(Appendix A
...


mm Hg
mole fraction

c
...
00 × 10−4 mol SO 2
2
mol
106 mols gas

(

)

⇒ pSO2 = ySO2 P = 1
...
0760 mm Hg
Henry's law ⇒ xSO 2 =

Since xSO 2

H SO 2

=

0
...
136 × 104 mm Hg mole fraction

= 2
...
78 × 103 moles

7
...
40 × 10 −6 mol SO 2
3

1 mol solution

= 0
...
0187 mol SO 2 dissolved

...
824 mm Hg)
= 0
...
958

mol dry air
mol

d
...
Add a base to the
solution to react with the absorbed SO2
...
52

Raoult’s law + Antoine equation (S = styrene, T = toluene):
∗
yS P = xS pS ⇒ xS =

0
...
06623 − 1507
...
985 )

0
...
95334 − 1343
...
377 )

0
...
06623 − 1507
...
985 )

0
...
95334 − 1343
...
377 )

10
10
⇒ T = 86
...
65(150)
10

7
...
434 ( 86
...
985 )

= 0
...
853 = 0
...
89272 − 1205
...
888 )

= 881
...
95805 − 1346
...
693)

= 345
...
53 pB ( 85°C ) = 10
∗
pT ( 85°C ) = 10

∗
Raoult's Law: yB P = xB pB ⇒

yB =
yT =

0
...
6 )
10 ( 760 )

0
...
1)
10 ( 760 )

= 0
...
0295 mol Toluene mol

yN 2 = 1 − 0
...
0295 = 0
...
54 a
...
50(140) + 0
...
20(51) = 91 psia ⇒ 76 psig
P < 200 psig, so the container is technically safe
...
From the Cox chart, at 140° F, pP = 300 psig, pnB = 90 psig, piB = 120 psig

Total pressure P = 0
...
30( 90) + 0
...

∗
6
...
Antoine: pnp (120°C ) = 10
∗
pip (120°C ) = 10

6
...
793 (120 + 231
...
73457 − 992
...
541)

= 6717 mm Hg

= 7883 mm Hg

(Note: We are using the Antoine equation at 120oC, well above the validity ranges in Table B
...
To get more accuracy, we would need to find a vapor pressure correlation valid at
higher temperatures
...
55 (cont’d)

xnp = 0
...
500 mol i -C5 H12 (l)/mol

*
*
Total pressure: P =xnp ⋅ pnp +xip ⋅ pip = 0
...
50(7883) = 7300 mm Hg
*
xnp ⋅ pnp

ynp =

P

=

0
...
46 mol n-C5 H12 (v)/mol
7300

yip = 1 − ynp = 1 − 0
...
54 mol i -C5 H12 (v)/mol

When the last drop of liquid evaporates,
ynp = 0
...
500 mol i -C5 H12 (v)/mol
=

0
...
500 P
+
= 1 ⇒ P = 7291 mm Hg
6725
7960

0
...
54 mol n-C5 H12 (l)/mol
6717 mm Hg

xip = 1 − xnp = 1 − 0
...
46 mol i -C5 H12 (l)/mol
b
...
500 mol n - C5 H 12 (v) / mol

yip = 0
...
500 P 0
...
84471−1060
...
541) + 6
...
019 /(T + 231
...
1o C
∗
pnp = 10
∗
pip = 10

xnp =

6
...
793 ( 63
...
541)

6
...
019 ( 63
...
541)

= 1758 mm Hg

= 2215 mm Hg

0
...
56 mol n-C5 H12 /mol
*
pnp (63
...
56 = 0
...
500 mol n - C5 H 12 (l) / mol

xip = 0
...
5)10

xnp ⋅

6
...
793 (T + 231
...
6o C)

+ (0
...
73457 −992
...
541)

0
...
44 mol n-C5 H12 (v)/mol
P
1960
yip = 1 − ynp = 1 − 0
...
56 mol i -C5 H12 (v)/mol

ynp =

=

6-38

⇒ T = 62
...
56

nv ( mol / min) at 80o C, 3 atm
10 L(STP)/min

yN2 (mol N2/mol)
yB [mol B(v)/mol]

nN 2 ( mol / min)
nN 2 =

xB [mol B(l)/mol]
xT [mol T(l)/mol]

yT [mol T(v)/mol]

10
...
4464 mol N 2 / min
22
...
89272 −1203
...
888 )

= 757
...
95805 −1346
...
693)

= 291
...
Initially, xB = 0
...
500
...
4464 mol N 2 / min = nv (1 − 0166 − 0
...
5797 mol / min

...

JK GH 0166 mol BIJK = 0
...
5797
J G 0
...
0370 mol T(v)
H
min K H
mol
min

⇒ nB0 = 0
...
Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (=
1–xB) increases
...
Since xB decreases, yB (= xBpB*/P) also decreases
...

6
...
P =

∗
xhex phex

dT i +
bp

∗
xhep phep

dT i
bp

, yi =

d i

xi pi∗ Tbp
P

, Antoine equation for pi∗

6
...
817 /(Tbp + 224
...
500 ⎡106
...
828 /(Tbp + 216
...
500 ⎡10
⎣
⎦
⎣
⎦
E-Z Solve or Goal Seek ⇒ Tbp = 80
...
713, yhep = 0
...


xi =

yi P
pi∗

dT i
dp

⇒

∑x
i

i

=P

∑

i

yi

pi∗

dT i

=1

dp

0
...
30
⎡
⎤
760 mmHg ⎢ 6
...
817 /(T + 224
...
90253−1267
...
823) ⎥ = 1
dp
dp
10
⎣10
⎦


...
279, xhep = 0
...
58

a
...

Calculation of Bubble Points
A
B
Benzene
6
...
531
Ethylbenzene 6
...
543
Toluene
6
...
773

C
219
...
091
219
...
226
0
...
226

xEB
0
...
226
0
...
09
96
...
48

xT
0
...
331
0
...


C6 H 6

d i

When x EB = 1 pure ethylbenzene , Tbp = Tbp

b

g

d i

When xT = 1 pure toluene , Tbp = Tbp

C7 H8

pB
pEB
pT
f(T)
378
...
2 233
...
086
543
...
6 165
...
11
344
...
3 348
...
07

C8 H 10

= 136
...
6o C

Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp)
than Mixture 2, and so (Tbp)1 > (Tbp)2
...
Mixture 3 contains
more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3 >
(Tbp)2

6-40

a
...
0 L/s vapor mixture

6
...
600 mol B(v)/mol
0
...
500 mol B(v)/mol
0
...

Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law
for butane and hexane
b
...
0 L 273 K
mol
= 4
...
4 L (STP)

Raoult's law for butane: 0
...
82485−943
...
711)
6
...
817 /(T + 224
...
400(1100)=(1-x 2 ) ⋅ 10
Mole balance on butane: 4
...
5)=n1 ⋅ 0
...
652(0
...
4 + n 2 ⋅ (1 − x 2 )
c
...
6)
1100(0
...
453
1175
...
82485 −
) 10 **(6
...
711
T + 224
...
0 ° C
x2 =

1100(0
...
82485 − 943
...
0 + 239
...
149 mol butane /mol

Solving (3) and (4) simultaneously ⇒ n1 = 3
...
03 mol C6 H14 /s
d
...
60

Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure;
(2) Raoult’s law is accurate;
(3) Ideal gas law is valid
...
0 kmol/h, T1a (oC), 1 atm

85
...
98 mol P(l)/mol
0
...
45 kmol P(l)/kmol
0
...
60 (cont’d)
a
...
45)(0
...
98) ⇒ n 0 = 195 kmol / h

Total mole balance : 195 = 85
...
45) = 85
...
98) + 110 ⋅ x 2 ⇒ x 2 = 0
...
Dew point of column overhead vapor effluent:

Eq
...
4-7, Antoine equation
0
...
02(760)
⇒
+ 6
...
817 /(T + 224
...
3o C
6
...
793/(T1a + 231
...
Assuming ideal gas behavior,
Vvapor =

170 kmol 0
...
2 + 37
...

Table B
...
621 g / mL, ρ H = 0
...
98(85) kmol P 72
...
621 kg P
+

L
0
...
17 kg H
= 9
...
659 kg H

c
...

0
...
84471−1060
...
541) + 0
...
88555−1175
...
867) = 760 ⇒ T2 =66
...

y2 =

*
x2 pP (66
...
04 ⋅ 106
...
793/(66
...
541)
=
= 0
...
898 mol H(v) / mol
d
...
39 m (39 cm)
π 10 m / s 3600 s

Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation,
constant temperature and pressure in the pipe connecting the column and the condenser,
column operates at steady state
...
61 a
...
96 mol butane/mol
R (mol)
x 1 (mol butane/mol)

T
P

Partial condenser: 40° C is the dew point of a 96% C 4 H 10 − 4% C 5 H 12 vapor mixture at
P = Pmin
Total condenser: 40° C is the bubble point of a 96% C 4 H 10 - 4% C 5 H 12 liquid mixture at
P = Pmin

Dew Point: 1 =
(Raoult's Law)

∑x =∑ p
i

yi P

∗
i

Antoine Eq
...
82485− 943
...
711K

Antoine Eq
...
793 ⎞
⎛
⎜ 6
...
541 ⎠
⎝

g

= 2830
...
22 mmHg

b

1
= 2595
...
96 2830
...
04 867
...
96b2830
...
04b867
...
16 mm Hg b total condenser g

Bubble Point: P =

∑y P=∑x p
i

i

∗
i

b
...
5 kmol / h

...


Feed and product stream compositions are identical: y = 0
...
5 = 187
...


Total balance as in b
...
5 kmol / h

U P = 2596 mm Hg
gb gV x = 0
...
5b0
...
96b75g ⇒ x = 0
...
96 P = x1 2830
...
04 P = 1 − x1 867
...
5x 0

6
...


b
...
434 ⎞
⎛
⎜ 7
...
985 ⎠
1423
...
95650 −
⎟
85+ 213
...
5 kmol / h

1203
...
89272 −
⎟
85 + 219
...
95 mm Hg
= 151
...
59 mm Hg

6-43

6
...
95
p* 881
...
725 , α B,EB = *B =
= 5
...
69
p*
pEB 151
...


c
...
α B , EB = 5810 ⇒ y B =

...

, P = x B p * + (1 − x B ) p *
B
EB
1 + 4
...
63 a
...
0
0
...
4
0
...
8
10 mol B l mol

...
0 0
...
795 0
...
959 10 mol B v mol

...
For ideal stages, the temperature of each stage corresponds to the bubble point
temperature of the liquid
...


b
...
55 mol B mol ⇒ 0
...
65 mol B mol ⇒ 0
...
400 × 760) mmHg = ( 0
...
89272−1203
...
888) + ( 0
...
06623−1507
...
985)
1
E-Z Solve
⎯⎯⎯⎯ T1 = 67
...
55 508

= 0
...
080 mol S mol
P
0
...
910 mol B mol ⇒ 0
...
400 × 760) mmHg = 0
...
090 p S ( T2 ) ⎯E-Z Solve → T2 = 55
...
991 mol B mol ⇒ 0
...
910 3310

...
400
B balance: y1nv + x 3 nl = y 2 nv + x 2 nl ⇒ x 3 ≈ 1 mol B mol ⇒ ≈ 0 mol S mol
c
...
In general, the
calculation of part b would be repeated until (1–yn) is less than the specified fraction
...
64 Basis: 100 mol/s gas feed
...

nGN (mol/s)

200 mol oil/s

nL (mol/s)
xi–1 (mol H/mol)

yN (mol H/mol)
(1–yN) (mol N2/mol)
100 mol/s
0
...
95 mol N2/mol

a
...
5% of H in feed)
(1–x1) (mol oil/mol)

nL (mol/s)
xi (mol H/mol)

nG (mol/s)
yi+1 (mol H/mol)

⎫ nGN = 95
...
5% absorption: 0
...
005 ) = y N nGN ⎪ y N = 2
...
95 (100 ) = (1 − y N ) nGN

Mole Balance: 100 + 200 = 95
...
05 (100 ) = 2
...
025 ) + x1 ( 204
...
0243 mol H(l) mol
nL =

1
1
( 200 + 205) ⇒ nL = 202
...
025 ) ⇒ nG = 97
...


b

g

b

g

∗
y1 = x1 p H 50° C / P = 0
...
73 / 760 = 0
...
00643 mol H(l) mol
c
...

d
...
05
95
...
88555

T
30

p*(T)
187
...
43E-02
3
...
86E-04

PR=
x1=
nL1=
B=

1
0
...
98
1175
...
63E-04
nG= 97
...
867

T
50
y(i)
5
...
98E-03
7
...
44E-04

p*(T)
405
...
43E-02
6
...
88E-03
7
...
99E-04

6-45

202
...
00E-02
1
...
45E-03
1
...
74E-04
2
...
5546

i
0
1
2
3
4
5

...
00E-02
2
...
53E-02
1
...
29E-02
6
...
69E-03
3
...
58E-03
1
...
02E-03

...

4
...
56E-04

6
...
If the column is long enough, the liquid flowing down eventually approaches equilibrium
with the entering gas
...
56x10–4 mol H/mol, which is
insufficient to bring the total hexane absorption to the desired level
...
63x10–4 or less
...

6
...
Intersection of vapor curve with y B = 0
...
T = 100°C ⇒ xB = 0
...
46 mol B mol ( vapor )
n V (mol vapor)
0
...
24 mol B(l)/mol

Basis: 1 mol
0
...
30 = 0
...
24n L

L

= 0
...
273 mol

⇒

nV
mol vapor
= 0
...
Intersection of liquid curve with x B = 0
...
66 a
...
50 mol B(v) mol

b
...
15 mol B(l) mol

c
...
43 mol B(v) mol, xB = 0
...
43 mol B/mol
nL (mol)
0
...
: 10 = nV + n L
B bal
...
43nV + 0
...


L

= 6
...
46
mol liquid
nl

Answers may vary due to difficulty of reading chart
...
i)

P = 1000 mm Hg ⇒ all liquid
...

V=

3 mol B 78
...
879 g B

+

7 mol T 92
...
Assume liquid volume negligible

6-46

mol T

10 −3 L
0
...


6
...
16 mol vapor 0
...
6 L = 97
...
6 L)

iii) 600 mm Hg

v=

10 mol vapor 0
...
67 a
...
4, x = 0
...
62 ⇒ f =

U⇒ x
V
W

F nV

+ x F n L = ynV + xn L ⇒ f =

0
...
23
= 0
...
62 − 0
...
Tmin = 75o C, f = 0 , Tmax = 87 o C, f = 1
6
...


Txy diagram
(P=1 atm)
80
75
T(oC)

70

Vapor

65
liquid

60
55
50
0

0
...
4

0
...


x A = 0
...
66

6-47

0
...
68 (cont’d)
c
...
34; y A = 0
...
: 1 = nV + nL
A bal
...
762 mol vapor, nL = 0
...
50 = 0
...
34nL ⎪
⎭
⇒ 76
...
791 g/cm3 , ρ E(l) = 0
...
790 g/cm3
(To be more precise, we could convert the given mole fractions to mass fractions and
calculate the weighted average density of the mixture, but since the pure component
densities are almost identical there is little point in doing all that
...
08 g/mol, M E = 46
...
34 )( 58
...
34 )( 46
...
15 g/mol

Basis: 1 mol liquid ⇒ (0
...
238 mol liquid) = 3
...

= 63
...
790 g / cm )

Vapor volume:
Vv =

3
...
9 volume% vapor
88747 + 63
...
For a basis of 1 mol fed, guess T, calculate nV as above; if nV ≠ 0
...

T
65 °C
64
...


xA
0
...
36

yA
0
...
56

fV
0
...
200

*
Raoult' s law: y i P = x i pi* ⇒ P = x A p * + x E p E
A

760 = 0
...
11714 −1210
...
664)

8
...
864/(Tbp + 226
...
5 × 10

⇒ Tbp = 66
...
5 × 107
...
595/(66
...
664)
y= A =
= 0
...
25 − 61
...
8o C ⇒
=
× 100% = 7
...
8
Tbp (real)
yA = 0
...
696 − 0
...
3% error in yA
yA (real)
0
...
There is consequently no reason to expect Raoult’s
law to be valid for acetone mole fractions that are not very close to 1
...
69 a
...
At 1 atm, (Tbp)B = 80
...
0oC

The Txy diagram should look like Fig
...
4-1, with the curves converging at 80
...
0oC when xC = 1
...
)
b
...
90328 1163
...
4
Chloroform
6
...
531
219
...
05
0
...
15
0
...
25
0
...
35
0
...
45
0
...
55
0
...
65
0
...
75
0
...
85
0
...
95
1

T
80
...
92
77
...
66
75
...
53
73
...
52
71
...
62
69
...
82
67
...
11
66
...
48
64
...
93
63
...
45
61
...
084
0
...
236
0
...
370
0
...
488
0
...
593
0
...
686
0
...
770
0
...
844
0
...
911
0
...
972
1

p1
0
63
...
65
179
...
10
281
...
61
371
...
18
450
...
27
521
...
15
585
...
02
641
...
76
692
...
27
738
...
13
636
...
34
527
...
59
432
...
79
347
...
20
272
...
38
205
...
10
145
...
36
92
...
35
43
...
33
0

p1+p2
760
760
...
93
759
...
96
759
...
91
759
...
03
759
...
07
760
...
98
760
...
96
760
...
93
760
...
03
760
...
1

0
...
3

0
...
5

0
...
7

M ole fraction of chloroform

6-49

0
...
9

1

6
...

Txy diagram
(P=1 atm)
85

T(oC)

80

yc

xc

75
70
x

y

65
60
0

0
...
2 0
...
4 0
...
6 0
...
8 0
...
58 ⇒

71 − 75
...
7% error in Tbp
75
...
58 − 0
...
33% error in y
0
...
There is consequently no reason to expect
Raoult’s law to be valid for chloroform mole fractions that are not very close to 1
...
70 P ≈ 1 atm = 760 mm Hg = x m pm Tbp + 1 − x m p * Tbp
P

760 = 0
...
87863−1473
...
60 × 10

7
...
686 /(Tbp +198
...
9o C
→

We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and
surface tension effects on the boiling point are negligible
...
9 °C, where boiling will commence
...
The increasing fraction of the less volatile component in the residual liquid will
cause the boiling temperature to rise
...
71 Basis: 1000 kg/h product
nH4 (mol H 2 /h)
E = C2 H5 OH ( M = 46
...
05)

scrubber
n3 (mol/h)
y A3 (mol A/mol), sat'd
y E3 (mol E/mol), sat'd
y H3 (mol H 2 /h)
vapor, –40°C

P = 760 mm Hg
Fresh feed
n0 (mol E/h)
nA1 (mol A/h)
nE1 (mol E/h)
280°C

reactor

nA2 (mol A/h)
nE2 (mol E/h)
nH2 (mol H 2/h)

condenser

nC (mol/h)
0
...
450 E
liquid, –40°C

Scrubbed
Hydrocarbons
nA4 (mol A/h)
nE4 (mol E/h)

still

Product
1000 kg/h
np (mol/h)
0
...
03 E

nr (mol/h)
0
...
95 E

Strategy

d i

•
•

Calculate y A3 and y E3 from Raoult’s law: y H3 = 1 − y A3 − y E3
...

Overall atomic balances on C, H, and O now involve only 2 unknowns ( n0 , n3 )

•

Overall C balance
⇒ n0 , n3
Overall H balance

•
•
•
•
a
...


U
V
W

U
V
W

Molar flow rate of product

b gb

g b gb

g

M = 0
...
03 M E = 0
...
05 + 0
...
05 = 44
...
67 kmol h
h
44
...
4 (Antoine) ⇒

*
pA ( −40°C ) = 44
...
360 mm Hg

Note: The calculations that follow can at best be considered rough estimates, since we are
using the Antoine correlations of Table B
...

Raoult’s law ⇒ yA3 =

*
0
...
550(44
...
03242 kmol A/kmol
760

6-51

6
...
450 pE ( −40 °C )

yH3

0
...
360)
= 2
...
9674 kmol H 2 kmol
=

Mole balance about still: nc = n p + nr ⇒ nc = 22
...
550nc = 0
...
67) + 0
...
5 kmol / h recycle

c

= 52
...
03242n3

(1)

E balance about scrubber: nE4 = n3 yE3 = 2
...
9764n3

(3)

Overall C balance:

b gb g b gb g d

ib g d

ib g

n0 (mol E) 2 mol C
= nA4 2 + n E4 2 + 0
...
03n p 2
h
1 mol E

⇒ n0 = nA 4 + n E 4 + 22
...
97 4 + 0
...
4 kmol E/h (fresh feed), nH 4 = 22
...
3 kmol/h, nA 4 = 0
...
00496 kmol E/h
A balance about feed mixing point: nA1 = 0
...
475 kmol A h
E balance about feed mixing point: nE1 = n0 + 0
...
5 kmol E h
E balance about condenser: nE2 = n3 yE3 + 0
...
5 kmol E h
Ideal gas equation of state :
Vreactor feed =
b
...
47 + 51
...
4 m3 ( STP ) ( 273+280 ) K

h

1 kmol

n0 − 0
...
4 − ( 0
...
67 )

= 2
...
4
n −n
51
...
5
× 100% = 54%
Single-pass conversion = E1 E2 × 100% =
nE1
51
...
76 kmol A/h

Feed rate of E to scrubber:

nE4 = 0
...
72 a
...
0 × 10 6 SCF / d
4 × 80 = 320 lb m W / d
n1 (lb - mole G / d)
n 2 [lb - mole W(v) / d]

n5

Distillation
Column

6

FG lb - mole TEGIJ
H d K
F lb - mole WIJ
n G
H d K
n5
8

Overall system D
...
analysis:

Water feed rate: n2 =

5 unknowns (n1 , n2 , n3 , n4 , n7 )
− 2 feed specifications (total flow rate, flow rate of water)
− 1 water content of dried gas
−2 balances (W, G)
0 D
...


320 lb m W 1 lb - mole
= 17
...
0 lb m

Dry gas feed rate:
4
...
78
= 1112 × 104 lb - moles G / d
n1 =

...

Overall G balance: n1 = n3 ⇒ n3 = 1112 × 10 4 lb - moles G / d
Flow rate of water in dried gas:
n4 =

(n3 + n4 ) lb - moles
d

359 SCF gas 10 lb m W 1 lb - mole W
lb - mole 10 6 SCF
18
...
112 ×10
⎯n⎯⎯⎯ → n4 = 2
...
78 − 2
...
0 lb m

d

1 lb - mole

6-53

= 280

lb m W
×
d

F 1 ft I = 4
...
4 lb JK
d
3

3

m

6
...
Mole fraction of water in dried gas =

yw =

n4
2
...
99 × 10 −4
4
lb - mole
n3 + n4 (2
...
112 × 10 ) lb - moles / d

Henry’s law: ywP = Hwxw ⇒

( x w ) max =

(199 × 10 −4 )(500 psia)(1 atm / 14
...

lb - mole dissolved W
= 0
...
398 atm / mole fraction
lb - mole solution

c
...
0 lb m W
n5
lb - mole TEG
=
= 4
...
2 lb m TEG 1 lb m W
n2 − n4
lb - mole W absorbed
⇒ n5 = 4
...
78 − 2
...
0 lb - moles TEG / d
(xw)in = 0
...
0170) = 0
...
0
=
⎯⎯⎯⎯ n8 = 0
...
951 lb - moles W 18
...
0 lb - moles TEG
+
d
lb - mole
d

150
...
04 × 104 lb m / d
W balance on absorber
n6 = (17
...
95 − 2
...
51 lb-moles W/d
16
...
19 lb-mole W/lb-mole
(16
...
9) lb-moles/d

d
...

6
...
999 H 2
0
...
04 H2 S, sat'd
0
...
8 atm
absorber
stripper
L2
0°C
L1
40°C
n3 (mol/h)
n4 (mol/h)
0
...
998 solvent
0°C
heater

6-54

G4
200 mol air/h
n2 mol H 2S/mol
0
...
73 (cont’d)

b gb

g

Equilibrium condition: At G1, p H 2S = 0
...
072 atm

...
072 atm
= 2
...
998n4 = solvent flow rate

b100gb0
...
999n ⇒ n
Overall H S balance: b100gb0
...
001n + n
Overall H 2 balance:

1

2

1

1

= 961 mol h

...
1

⇒ n2 = 3
...
90gmol
h

b g b273 + 40gK = 5240 L hr

22
...
04g + 0
...
001n
0
...
67 × 10 i
4

1

+ n3 x 3 ⇒ n4 = 1
...
998n4 = 5820 mol solvent h
6
...
4 g NaHCO 3

Sat'd solution @ 30°C
100 g H 2 O
11
...
4 = 111 + ms ⇒ ms = 5
...

% crystallization =

5
...
3%
16
...
75 Basis: 875 kg/h feed solution
m1 (kg H 2 O(v )/h)

875 kg/h
x 0 (kg KOH/kg)
(1 – x0) (kg H 2O/kg)

Sat'd solution 10°C
m2 (kg H2 O(1)/h)
1
...
75 (cont’d)

Analysis of feed: 2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O
x0 =

bg

22
...
85 mol H 2 SO 4

2 mol KOH

56
...
427 g KOH g feed
60% recovery: 875 ( 0
...
60 ) = 224
...
2 kg KOH 92
...
2 kg KOH ⋅ 2H 2 O h (143
...
11 kg KOH

KOH balance: 0
...
2 + 1
...
1 kg h
Total mass balance: 875 = 368
...
03 (145
...
76 a
...
200 0
...
R ⇒ CA = R / 150

CA =

500 mol 1
...
2 °C
...
2 °C =
= 0
...
2° C

...
5 150 g A 1 mL soln
At 0°C, R = 17
...
106 g A g soln
mL soln
110 g soln

...
106 g A & 0
...

0106 g A

...


...
894 g S
390 g S 11
...
Mass of solution:

bg

g A remaining in soln
g A initial

c
...
5 × 390 g S

bg

11
...
0 g A s
100 g S

6
...
Table 6
...
0oF), the salt that crystallizes is MgSO 4 ⋅ 7 H 2 O , which

contains 48
...

b
...


m0 (g/h) sat’d solution @ 130oF

m1 (g/h) sat’d solution @ 50oF
0
...
77 g H2O/g

0
...
65 g H2O/g

1000 kg MgSO4·7H2O(s)/h

6-56

6
...
35m0 = 0
...
488(1000) kg MgSO 4 / h ⇒ m = 1150 kg soln / h
1
The crystals would yield 0
...
78

kg anhydrous MgSO 4
h

Basis: 1 lbm feed solution
...
5-1 ⇒ a saturated KNO3 solution at 25oC contains 40 g KNO3/100 g H2O
⇒ x KNO 3 =

40 g KNO 3
= 0
...
286 lb m KNO 3 / lb m x
(40 + 100) g solution
1 lbm solution @ 80oC
0
...
50 lbm H2O/lbm

m1(lbm) sat’d solution @ 25oC
0
...
714 lbm H2O/lbm soln
m2 [lbm KNO3(s)]

Mass balance: 1 lb m = m1 + m2
KNO3 balance: 0
...
286m1 + m2

⇒

m1 = 0
...
300 lb m crystals / lb m feed

0
...
429 lb m crystals / lb m solution
0
...
79 a
...


Figure 6
...
281 g NaCl / g = 0
...
281 kg NaCl/kg soln
0
...
100 kg NaCl/kg
0
...
100 kg NaCl = 0281m1 + m2

...
700 lb m solution / lb m feed
m2 = 0
...
300 lb m crystals / lb m feed
= 0
...
700 lb m solution / lb m feed

The minimum feed rate would be that for which all of the water in the feed evaporates to
produce solid NaCl at the specified rate
...
79 (cont’d)

0100(m0 ) min = 1000 kg NaCl / h ⇒ (m0 ) min = 10,000 kg / min

...


m2 [kg H 2 O(v) / h]

m1 (kg/h) sat’d solution @ 80oC
0
...
719 kg H2O/kg soln
1000 kg NaCl(s)/h

m0 (kg/h) solution
0
...
900 kg H2O/kg

40% solids content in slurry ⇒ 1000

kg NaCl
kg
= 0
...
100m0 = 0281(2500) ⇒ m0 = 7025 kg / h

...
80 Basis: 1000 kg K 2 Cr2 O 7 ( s) h
...


Composition of saturated solution:

0
...
20 kg K
⇒
= 01667 kg K kg soln

...
20 kg soln

b

g

n2 (mol / h)
y2 (mol W(v) / mol)

me [kg W(v) / h)

(1− y2 )(mol A/ mol)

...
210 kg K/ kg

0
...
10 kg soln / kg
0
...
790 kg W(l) / kg

0
...
1667 kg K / kg
0
...
2° C ⇒ y 2 =

53
...
0698 mol W mol
760 mm Hg

Overall K balance: 0
...
80 (cont’d)

b

gb

g


...

K balance on dryer: 0
...
10 × 1090g kg

h not recycled

95 kg recycled
5 kg not recycled

= 2070 kg h recycled
Water balance on dryer


...
8333gb010gb1090g kg W h = 0
...
01 × 10

−3

kg mol

2

⇒ n2 = 7
...
0698g7
...
4 L STP

...
81
...
Assume Patm=1 atm
n 2w (kmol H 2 O )(sat' d)

100 kg Feed
0
...
93 kg H 2 O / kg

n 2c (kmol CO 2 )
n 2a (kmol Air)

Reactor
Reactor

e

70 o C, 3 atm(absolute)

n1 (kmol)
0
...
30 kmol Air / kmol

Rm (kg solution) U
|0
...
976 kg H O / kg V
|
T
W
4

3

2

Filtrate
m5 (kg)

Filter

0
...
976 kg H 2 O / kg
Filter cake
m 6 (kg)
0
...
14 kg solution U
|0
...
976 kg H O / kg V
|
T
W
3

2

Degree of freedom analysis:
Reactor
6 unknowns (n1, n2, y2w, y2c, m3, m4)
–4 atomic species balances (Na, C, O, H)
–1 air balance
–1 (Raoult's law for water)
0 DF

Filter
2 unknowns
–2 balances
0 DF

Na balance on reactor

100 kg 0
...
038 = 0
...
024m4 )

23 kg Na
84 kg NaHCO 3

(1)

Air balance: 0
...
024m4 ) kg NaHCO 3

( 2)

C balance on reactor :

n1 (kmol) 0
...
07 kg Na 2 CO 3
12 kg C
+
kmol
1 kmol CO2
kg
106 kg Na 2 CO 3
= (n2c )(12) + (m3 + 0
...
40n1 + 0
...
024m4 )

...
024m4 )( ) + 0
...
33 = 2n2 w + 0
...
024m4 ) + 01084m4

...
93)(

6-60

(3)

6
...
700)(932) + 100 (0
...
93)( )
106
18
48
16
= (n2 w )(16) + n2 c (32) + (m3 + 0
...
976m4 ( )
84
18
⇒ 22
...
5714(m3 + 0
...
8676m4

...


n2 w
233
...

n1 = 0
...
2426 kmol air,
n2c = 0
...
0848 kmol H 2 O(v),
m3 = 8
...
50 kg solution

NaHCO3 balance on filter:
m3 + 0
...
024m5 + m6 [0
...
024)]

...
874

11
...
024m 5 + 0
...
50


...
874 + 92
...
09 kg filtrate
m6 = 10
...
86)(10
...
867 kg NaHCO 3 (s)

500 kg / h
= 56
...
867 kg

(a) Gas stream leaving reactor

n2w = (0
...
39) = 4
...
500)(56
...
2 kmol O 2 / h
n2a = (0
...
39) = 13
...
7 kmol / h)(0
...
7kmol / hH O(v) / kmol
| ⇒ |0
...
604 kmol CO / kmol
| S
W |0
...
39 × 0
...
4 m 3 (STP)
1h
(b) Gas feed rate: V1 =
= 17
...
81(cont'd)
(c) Liquid feed: (100)(56
...

(d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual
solution would contain less than 2
...

(e)
Henry's law

Benefit: Higher pressure ⇒ greater pCO2

higher concentration of CO 2 in solution

⇒ higher rate of reaction ⇒ smaller reactor needed to get the same conversion ⇒ lower cost
Penalty: Higher pressure ⇒ greater cost of compressing the gas (purchase cost of compressor,
power consumption)
6
...
90 MgSO4 ⋅ 7H 2 O Dissolution
Tank
Tank
Tank
010 I

...
32 kg MgSO / kgV
S
|0
...
23 lb m MgSO 4 / lb m
0
...
23 lb
|
S0
...
32 MgSO |
S
|
|0
...
32 MgSO 4
0
...
05m
|0
...
77 lb
T

4

(lb m soln)
m MgSO 4 / lb m
m H 2 O / lb m

U
|
V
|
W

a
...

(b) Strategy: Do D
...


6-62

6
...
tank overall mass balance
Diss
...
31 + 32
...
00) = 120
...
37 + 7 * 18
...
44
Overall MgSO4 balance:

60,000 lb m
h

0
...
37 lb m MgSO 4
lb m
246
...
32 lb m MgSO 4 / lb m ) + m4 (120
...
44) + 0
...
23)
⇒ m4 = 5
...
257 x104 lb m / h

Overall mass balance: 60,000 + m1 = 6300 + 105m4

...


Diss
...
tank MgSO 4 balance:
⇒

60,000 + m1 + m6 = m2 + 6000
54,000(120
...
44) + 0
...
32m2

U
V
W


...
575x10 4 lb m / h recycle

Recycle/fresh feed ratio =

9
...
83 a
...
96 kg CaSO 4 (s) / kg
0
...
83 (cont’d)
b
...

c
...
10 ⇒ mw = 8000 kg H 2 O / h
(2000 + mw ) kg / h

Overall S balance:
1000 kg H 2 SO 4

32 kg S

h

98 kg H 2 SO 4
+

=

m5 (kg / h) (0
...
04 X a ) (kg CaSO 4 )

32 kg S

kg

136 kg CaSO 4

32 kg S
m8 (kg / h) X a (kg CaSO 4 )
kg
136 kg CaSO 4

⇒ 3265 = 0
...
96 + 0
...
2353m8 X a

...
04m5 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 )
kg

m8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 )

28 kg N
164 kg Ca(NO 3 ) 2

28 kg N

kg

164 kg Ca(NO 3 ) 2

⇒ 222
...
00683m5 (1 − 501X a ) + 0171m8 (1 − 501X a )

...
96 + 0
...
04m 5 (kg / h)
kg

40 kg Ca
136 kg CaSO 4

40 kg Ca
164 kg Ca(NO 3 ) 2

m 8 (kg / h) X a (kg CaSO 4 )
kg

40 kg Ca
136 kg CaSO 4

m 8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 )
kg

40 kg Ca
164 kg Ca(NO 3 ) 2

⇒ 0
...
294m5 (0
...
04 X a ) + 0
...
294m8 X a + 0
...
01m0 = n1

( 4)

6-64

12 kg C
1 kmol C

6
...
04m5 (kg / h) 500 X a (kg H 2 O)
kg

2 kg H
18 kg H 2 O

⇒ 92517 = 2
...


...
(1)-(5) simultaneously, using E-Z Solve
...
5 kg CaCO 3 (s) / h,

m5 = 1428
...
9 kg soln / h,

n1 = 18
...
00173 kg CaSO 4 / kg

Recycle stream = 2 * m0 = 3625 kg soln / h

R 0
...

|
| 500 * 0
...
5% H O |
V S
S
|
|(1 − 501* 0
...
3% Ca(NO ) V
W |
T
|
W
T
4

4
2

2

3 2

d
...
1, for CO2:
Tc = 304
...
9 atm
T (40 + 273
...

Tc
304
...
411
72
...
5
...
86 ⇒ V =

zRT 0
...
08206 L ⋅ atm 313
...
737
mol ⋅ K
30 atm
P
mol CO 2

Volumetric flow rate of CO2:
V = n1 * V =

e
...
1 kmol CO 2

0
...
33x10 4 L / h

Solution saturated with Ca(NO3)2:
⇒

1 − 501X a (kg Ca(NO 3 ) 2 / kg)
= 1
...
00079 kg CaSO 4 / kg
500Xa (kg H 2 O / kg)

Let m1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without
crystallization
...
83 (cont’d)

Overall S balance:
1000 kg H2SO4

32 kg S

h

98 kg H2SO4
+

=

m5 (kg / h) (0
...
04)(0
...
00079 (kg CaSO4 )
32 kg S
kg
136 kg CaSO4

⇒ 326
...
226m5 + 0
...
04m5 (kg / h) (1 − (501)(0
...
00079)) (kg Ca(NO3 ) 2 )
28 kg N
+
kg
164 kg Ca(NO3 ) 2

...


...
04m5 (kg / h) 500(0
...
00079) (kg H 2 O)

kg
kg

2 kg H
18 kg H 2 O

2 kg H
18 kg H 2 O

⇒ 909
...
0159m1 = 0
...
0439m8

(3' )

Solve eqns (1')-(3') simultaneously using E-Z solve:
m1 = 1155x10 4 kg / h;

...

m 5 = 1424 x10 3 kg / h;

Maximum ratio of nitric acid to sulfuric acid in the feed
=

1155x10 4 kg / h

...

1000 kg / h

6-66

m 8 = 2
...
84

550
...
879 g

Moles of benzene (B):
⇒ x DP =

bg

U
|
|
V
1 mol
= 6
...
11 g
W

56
...
363 mol
154
...
363
= 0
...
19 + 0
...
945 120
...
0 mm Hg

ΔTm =
ΔTbp =

2
RTm0

ΔH m
2
RTb0

ΔH v

x DP

b

g b0
...
6 K = 3
...
314b273
...

=

...
0554g = 185 K = 1
...
314 273
...


2

o

m


...

= 55 − 3
...
0 ° C

...


6
...
0o C, ΔTm = 4
...
6K
Eq
...
5-5
⎯⎯⎯ xu =
→
Table B
...
(6
...
6K)(600
...
314 J/mol ⋅ K)(273
...
0445 mol urea/mol

RTb 0 2
(8
...
2) 2
xu =
0
...
3K = 1
...
06 g/mol

(1000 − mu1 )(g)
18
...
0445 =

(mol urea)
60
...
06 + 18
...
0o C = 3
...
0K)(40,656 J/mol)
(8
...
2K) 2

mu 2

= 0
...
06
xu 2 = 0
...
06 + 18
...
86 x a =


...
5150 gg b1101 g molg
b0
...
1 g molg+ b100
...
10 g molg = 0
...
00523g mol solvent
0
...
87 a
...
49° C
= 0
...
41° C
mol solution

95
...
50 g solute mol

g b0
...
38 kJ / mol
2

ΔH vI
ΔH vII
*
+ B , ln p s Tbs = −
+B
RTb 0
RTbs

b g

Assume ΔH vI ≅ ΔH vII ; T0 Ts ≅ T02

IJ
K
ΔH Δ T
p bT g = b1 − x g p bT g ⇒ lnb1 − x g ≈ − x = −
RT

b g

b g

⇒ ln Ps* Tb 0 − ln P0* Tbs = −

b
...
00438

94
...
4460 g solute

2
8
...
2 − 5
...
49
ΔTm

*
ln p s Tb 0 = −

II
ΔTm

I
xs

*
s

*
0

b0

FG
H

ΔH v 1
ΔH v Tbs − Tb 0
1
−
≅
R Tb 0 Tbs
R
Tb20
v

bs

2
b0

6
...
19
100 + m2
90 + m1

IJ
K

solve simultaneously

m1 = 25
...
4 g styrene in ethylene glycol phase

6-68

RTb2
0
⇒ ΔTb =
x
ΔH v

6
...


A=oleic acid; C=condensed oil; P=propane

100 kg / h
0
...
0 kg C / h
m 2 kg A / h

0
...
90% extraction: m3 = (0
...
05)(100 kg / h) = 4
...
05)(100) = m2 + 4
...
5 kg A / h
Equilibrium condition:

015 =

...
5 / (n1 + 0
...
2 kg P / h
4
...
5 + 95)

b
...
1-4 ⇒ p propane = 500 psi = 34 atm
c
...

6
...
Benzene is the solvent of choice
...


Basis: 100 kg feed
...
30 kg A / kg

m1 (kg)
0
...
70 kg W / kg

0
...
70 = m1 * 0
...
8 kg

Balance on A:

100 * 0
...
8 * 010 ⇒ m2 = 22
...


Equilibrium for H:
KH =

m2 / (m2 + mH ) 22
...
2 + mH )
=
= 0
...

xA
010

...
2 / (22
...
098 ⇒ m B = 2
...


(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and
environmental considerations
...
91

a
...
A = acetone, H = n - C 6 H 14 , W = water
40 g A
60 g W

e 1 (g A)
60 g W

25°C

100 g H

75 g H

100 g H
r 1 (g A)

b

25°C

xA in H phase / xA in W phase = 0
...
8 g acetone
b
g = 0
...
2 g acetone
V
|
e b60 + e g
W
27
...
2 g acetone
|⇒
r b75 + r g
= 0
...
6 g acetone
|
e b60 + e g
W

40 = e1 + r1
r1 100 + r1

1

1

1

Balance on A − stage 2:

e 2 (g A)
60 g W

1

2

2

2

Equilibrium condition − stage 2:

2

2

% acetone not extracted =

2

2

2

20
...

40 g A fed

b
...


40 g A
60 g W

U r = 17
...
343| ⇒ e = 22
...
0 = e1 + r1
r1 175 + r1
e1

1

1

22
...

40 g A fed
19
...
6 g A
m (g H)

m (g H)

Equilibrium condition:

1

20
...
6)
= 0
...
4 / (60 + 19
...
Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane
over process lifetime) – (cost of an equilibrium stage x number of stages)
...


6-70

6
...
P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution

Broth
Mixing tank

100 kg
0
...
985 Ac

m1 (kg BA)
Extraction Unit I

Acid

D
...
analysis:
Extraction Unit I
3 unknown (m1, m2p, m3p)
–1 balance (P)
–1 distribution coefficient
–1 (90% transfer)
0 DF

m3P (kg P)
98
...
1
m4 (kg Alk)

Extraction
II
m6P (kg P)
m1 (kg BA)

m5P (kg P)
m4 (kg Alk)
pH=5
...
In Unit I, 90% transfer ⇒ m3 P = 0
...


...


...


...

135 / (135 + m1 )
⇒ m1 = 34
...
1 ⇒ K = 25
...


...
5)


...
90(m3 P ) = 1215 kg P
P balance:
m3 P = 1215 + m6 P ⇒ m6 P = 0135 kg P

...

m / (m6 P + 34
...
65 kg Alk

...
8 ⇒ K = 010 = 6 P
1215 / (1
...

m1 34
...
3416 kg butyl acetate / kg acidified broth
100 100 kg broth
m4 29
...
2965 kg alkaline solution / kg acidified broth
100 100 kg broth
Mass fraction of P in the product solution:
m5 P
1
...
394 kg P / kg
m4 + m5 P (29
...
215) kg
c
...
The first transfer (low pH) separates most of the P from the other broth constituents,
which are not soluble in butyl acetate
...

(ii)
...

(iii)
...


6-71

6
...
20
x A = 0
...
47

U
| ⇒
V
|
W

Figure 6
...
07, x A = 0
...
58
Phase 2: x W = 0
...
25, x M = 0
...
2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2
...
95 kg in MIBK - rich phase
12 * 0
...
07m1 + 0
...

⇒
m2 = 0
...
33 = 0
...
25m2

...
94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK

Overall system composition:

U
b
g
|
V
3500 g b20 wt% A, 80 wt% M g ⇒ 700 g A, 2800 g M |
W
2200 g A U
|
⇒ 3500 g WV ⇒ 25
...
2% W, 32
...
6
...

H 2 O Balance:
Acetone balance:
6
...
06m1 + 0
...
31m1 + 0
...
0 lb m / h
x A,1 , x W,1 , 0
...
6-1⇒ Phase 1: x M = 0
...
05; x A,1 = 0
...
81; x A,2 = 0
...
03
Overall mass balance:
MIBK balance:

U
V
W


...
0 lb m / h + m1 = 410 lb m h + m2

...

m1 = 410 * 0
...
03
m2 = 19
...
96 a
...
375 mol A, x m,org = 0
...
075 mol W
x a,aq = 0
...
050 mol M, x w,aq = 0
...

Mass balance:
maq ,1 + morg ,1 = 100
⇒
Acetone balance: maq ,1 * 0
...
375 = 33
...
3 kg
System 2: x a,org = 0
...
870 mol M, x w,org = 0
...
055 mol A, x m,aq = 0
...
055 + morg ,2
b
...
375
= 136;

...
275

K a ,2 =

x a ,org ,2
x a ,aq ,2

=

= 0
...

W

aq,2

= 22
...
8 kg

0
...

0
...

c
...
375 / 0
...
040

...
3; β
=
= 418

...
055 / 0
...
275 / 0
...


aw

→∞

Organic phase= extract phase; aqueous phase= raffinate phase

β a ,w =

( x a / x w ) org

=

( x a / x w ) aq

( x a ) org / ( x a ) aq
( x w ) org / ( x w ) aq

=

Ka
Kw

When it is critically important for the raffinate to be as pure (acetone-free) as possible
...
97 Basis: Given feed rates: A = acetone, W = water, M=MIBK

e1

e2

(kg / h)

(kg / h)

x1A (kg A / kg)

x 2A (kg A / kg)

x1W (kg W / kg)

x 2W (kg W / kg)

x1M (kg M / kg)

x 2M (kg M / kg)

200 kg / h
0
...
70 kg M / kg

r1 (kg / h)
y1A (kg A / kg)
y1W (kg W / kg)
y1M (kg M / kg)

300 kg W / h

6-73

Stage II
IIS
300 kg W / h

r2 (kg / h)
y 2A (kg A / kg)
y 2W (kg W / kg)
y 2M (kg M / kg)

6
...
30g = 60 kg A h U 500 kg h
|
200 − 60 = 140 kg M h V ⇒
12% A, 28% M, 60% W
300 kg W h |
W
Figure 6
...
095, x1W = 0
...
025
Raffinate: y1A = 015, y1W = 0
...
815

...
095e1 + 015r1

...

b227gb015g = 34 kg A h
b227gb0
...
5% A, 35
...
4% W
V
b227gb0
...
6-1 ⇒

Extract: x 2 A = 0
...
94, x 2 M = 0
...
085, y 2 W = 0
...
89

Mass balance:

527 = e2 + r2

Acetone balance:

34 = 0
...
085r2

⇒

Re
|
Sr
|
T

2

= 240 kg / h

2

= 287 kg / h

Acetone removed:
[60 − ( 0
...
59 kg acetone removed / kg fed
60 kg A / h in feed
Combined extract:
Overall flow rate = e1 + e2 = 273 + 240 = 513 kg / h
Acetone:

( x1 A e1 + x 2 A e2 ) kg A

=

0
...
04 * 240
= 0
...
88 * 273 + 0
...
908 kg W / kg
e1 + e2
513

MIBK:

( x1 M e1 + x 2 M e2 ) kg M 0
...
02 * 240
=
= 0
...
98
...


1
...
50 L / min)
PV
=
= 0
...
08206 L ⋅ atm / mol ⋅ K)(298 K)

r
...
=25%⇒

pH 2 O
*
pH2O (25o C)

= 0
...
5

*
0
...
5 * 0
...


g H 2 O ads
100 g silica gel

0
...
756 mm Hg)
mol H 2 O
= 0
...
06134 mol 0
...
01 g H 2 O
= 0
...
00863 g H 2 O / min)(120min) = 1
...
035 g H 2 O
3
...
1 g silica gel
M (g silica gel) 100 g silica gel

Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and
T are constant
...
Humid air is dehumidified by being passed through a column of silica gel, which absorbs a
significant fraction of the water in the entering air and relatively little oxygen and nitrogen
...
If air were still fed to the column past this point, no further
dehumidification would take place
...

6
...


Let c = CCl4
Relative saturation = 0
...
30 * (169 mm Hg) = 50
...
Initial moles of gas in tank:

n0 =

1 atm
50
...

=
= 1985 mol
RT0 0
...
7 mm Hg
× 1985 mol = 0
...

n0 =
760 mm Hg
P0

6-75

6
...
500nc 0 = 0662 mol CCl 4 (= nads)


...
0662) mol = 1
...
Assume T = T0 and V = V0
...
08206)(307)

...
0

yC =

nc
0
...
0345
n tot
1
...
0345(760 mm Hg) = 26
...
Moles of air in tank: na = n0 − nc 0 = (1
...
853 mol air

...

= 0
...

nc + 1853
mol


...

50
...
001

0

tot

0

0
...
710 mm Hg
X*

FG g CCl IJ = 0
...
096 p
4

⇒ X* =

c

c

Mass of CCl4 adsorbed

mads = (nc 0 − nc )( MW ) c =

g CCl 4 adsorbed
0
...
710)
= 0
...
096(0
...
001854) mol CCl 4

...
85 g
1 mol CCl 4

= 20
...
3 g CCl 4 ads
= 400 g carbon
Mass of carbon required: mc =
g CCl 4 ads
0
...


β
β
X * = K F p NO2 ⇒ ln X * = ln K F + β ln p NO2

ln(PNO2)

6
...
406x - 1
...
5
1
0
...
5
-1
-1
...
100 (cont’d)
ln X * = 1406 ln p NO2 − 1
...
965 p 1
...
406

...

NO2
NO2
K F = 0
...
406 ; β = 1406

...
Mass of silica gel : mg =

π * (0
...
75 kg gel
1m 3

L

= 5
...
010(760 mm Hg) = 7
...
140(7
...
406 kg NO 2
100 kg gel

5
...
143 kg NO 2

Average molecular weight of feed :
MW = 0
...
99( MW ) air = (0
...
01) + (0
...
0) = 29
...
00 kg 1 kmol 0
...
17 kg
kmol
tb =

Breakthrough time:

46
...
126
kmol NO 2
h

0
...
13 h = 68 min
0
...
The first column would start at time 0 and finish at 1
...
13+1
...
63 h
...
13 h and finish
at 2
...
Since the first column would still be in the regeneration stage, a third column
would be needed to start at 2
...
It would run until 3
...
The first few cycles are shown below on a
Gantt chart
...
13

2
...
39

4
...
02

Column 2
1
...
26

3
...
52

5
...
26

6-77

3
...
89

5
...
78

Let S=sucrose, I=trace impurities, A=activated carbon

Add m A (kg A)

mS (kg S)

mS (kg S)
m I (kg I)

m I0 (kg I)
R0 (color units / kg S)

R (color units / kg S)

Come to equilibrium

V (L)

V (L)
m A (kg A)

m IA (kg I adsorbed)

Assume

no sucrose is adsorbed
• solution volume (V) is not affected by addition of the carbon
m
a
...
500
9
...
ln R should be linear: slope = β ;

ln v

6
...
4504x + 8
...
500
8
...
000 1
...
000 3
...
101 (cont’d)

ln υ = 0
...
0718 ⇒ υ = e 8
...
4504 = 3203R 0
...
4504

b
...
025(20
...
50 color units / kg sucrose

'
υ = K F R β = 3203(0
...
4504 = 2344

⇒ 2344 =

97
...
0 kg carbon
m A / mS
m A / 480

6-79

CHAPTER SEVEN

7
...
80 L 35 × 10 4 kJ 0
...

1h
1 kW
= 2
...
3 kW
h
L
1 kJ heat
3600 s 1 k J s

2
...
341 × 10 −3 hp
1 kW
7
...
1 hp

...
174 lbm ⋅ ft / s2

9
...
7376 ft ⋅ lb f

(b)

3 × 108 brakings 715 Btu 1 day
1h
1W
1 MW
= 2617 MW
−4
day
braking 24 h 3600 s 9
...
3

(a) Emissions:
1000 sacks
Paper ⇒
Plastic ⇒

2000 sacks

1000 sacks

sack 16 oz
( 0
...
0146) oz

(724 + 905) Btu
sack

Plastic ⇒

1 lb m

= 6
...
0510 + 0
...
39 lb m

= 163 × 10 6 Btu

...
30 × 10 6 Btu
sack

(b) For paper (double for plastic)

Materials
for 400 sacks

Raw Materials
Acquisition and
Production

Sack
Production and
Use

7- 1

1000 sacks

400 sacks

Disposal

7
...
0510 oz

1 lb m

sack 16 oz

+

1000 sacks

0
...
5 lb m

⇒ 30% reduction
Plastic ⇒

800 sacks

0
...
0146 oz

1 lb m

sack 16 oz

= 2
...


724 Btu 1000 sacks
+
sack

800 sacks

905 Btu
= 119 × 10 6 Btu; 27% reduction

...
08 × 10 6 Btu; 17% reduction
sack

1h

24 h

649 Btu

1J

1 MW

1 sack 9
...

Savings for recycling: 017(2,375 MW) = 404 MW
(d) Cost, toxicity, biodegradability, depletion of nonrenewable resources
...
4

1 ft 3

(a) Mass flow rate: m =

3
...
4805 gal

Stream velocity: u =

3
...
4805 gal Π 0
...
330 lb m
=
2
s

d

(0
...
43) lb m


...


7- 2


...
70 × 10−3
2
s
2 32
...


...
7376 ⋅
/

= 7
...
330 lb m s

−5

hp

7
...
0 m π 0
...
9 g 1 kg
=
2
2
s 1000 g

130 kPa

1 mol

b g

573 K 101
...
4 L STP
42
...
9 g 1 mol 673 K 101
...
4 L ( STP )
s

Ek =

29 g

273 K

130 kPa

mu 2 127
...
32 2 m2
s

1N

4

L π (0
...
9 g s

= 113 J s

N⋅m

1 m3
103

29 g

= 49
...

= 1558 J / s

N⋅m

ΔE k = E k (400 C) - E k (300 C) = (155
...
8 J / s ⇒ 43 J / s
(c) Some of the heat added goes to raise T (and hence U) of the air
7
...

1 lbf
1 ft 3
62
...

= −834 ft ⋅ lb f
3
7
...
174 lbm ⋅ ft / s2

b g

b g

mu 2
= mg − Δz ⇒ u = 2 g − Δz
2

12

LM FG
NH

= 2 32
...
4

ft
s

(c) False
7
...

ΔE p ⇒ negative The gas exits at a level below the entrance level
...
5 cm 2
s
2

1 m3
10 4 cm 2

273 K
10 bars
1 kmol
303 K 1
...
4 m 3 STP

b g

16
...
0225 kg s
PoutVout nRT
V
u (m/s) ⋅ A(m2)
P
P
=
⇒ out = in ⇒ out
= in
2)
PinVin
nRT
Vin Pout
uin (m/s) ⋅ A(m
Pout
⇒ uout = uin

Pin
10 bar
= 5( m s)
= 5
...
5(0
...
5552 − 5
...
0659 W
ΔE p = mg ( zout − zin ) =

0
...
8066 m -200 m
s
s

= −44
...
8

ΔE p = mgΔz =

105 m3 103 L 1 kg H2O 981 m −75 m
1N
1 J 2
...

h 1 m3
1L
s2
1 kg ⋅ m/ s2 1 N ⋅ m
1J


...
04 × 10 4 kW ⋅ h 24 h 7 days
= 3
...
9

(b) Q − W = ΔU + ΔE k + ΔE p

b
g
= 0 b no height changeg

ΔE k = 0 system is stationary
ΔE p

Q − W = ΔU , Q < 0, W > 0
(c) Q − W = ΔU + ΔE k + ΔE p

b

g

b

g

Q = 0 adiabatic , W = 0 no moving parts or generated currents
ΔE k = 0 system is stationary
ΔE p = 0 no height change

b
b

g

g

ΔU = 0
(d)
...
If the
temperature went up in the adiabatic reactor, heat must be transferred
from the system to keep T constant, hence Q < 0
...
10 4
...
00 bar ⇒ V (L), T (°C), 8
...
Closed system:

ΔU + ΔE k + ΔE p = Q − W

RΔE
SΔE
|
T

k
p

b
b

= 0 initial / final states stationary
= 0 by assumption

g

g

ΔU = Q − W
(b)

Constant T ⇒ ΔU = 0 ⇒ Q = W =

−7
...
314 J
= −765 J
0
...
65 L ⋅ bar > 0, Tfinal > 30° C

7- 4

transferred from
gas to
surroundings

7
...
83 × 10 −3 m 2
10 4 cm 2
(a) Downward force on piston:
A=

cm 2

Fd = Patm A + mpiston+weight g
=

1 atm 1
...
83 × 10 −3 m2
atm

+

24
...
81 m

d

s

i d

Upward force on piston: Fu = APgas = 2
...
83 × 10 −3 m2 ⋅ P0 = 527 ⇒ P0 = 186 × 10 5 N m 2 = 186 × 10 5 Pa

...


V0 =

303 K
1
...
08206 L ⋅ atm
nRT 1
...
677 L
=
P0
28
...
86 × 105 Pa
1 atm
mol ⋅ K

(b) For any step, ΔU + ΔE k + ΔE p = Q − W ⇒ ΔU = Q − W
ΔE k = 0
ΔE p = 0

Step 1: Q ≈ 0 ⇒ ΔU = −W
Step 2: ΔU = Q − W As the gas temperature changes, the pressure remains constant, so
that V = nRT Pg must vary
...

Overall: Tinitial = Tfinal ⇒ ΔU = 0 ⇒ Q − W = 0
In step 1, the gas expands ⇒ W > 0 ⇒ ΔU < 0 ⇒ T decreases

b gd

i b gb gb g

id


...

(c) Downward force Fd = 100 101325 × 10 5 2
...
50 9
...

−3
2
A 2
...

186 × 10 5 Pa
= 1
...
677 L
Pf

...
08 − 0
...
83 × 10 −3 m2


...

⇒ W = Fd = 331 N 0142 m = 47 N ⋅ m = 47 J
Since work is done by the gas on its surroundings, W = +47 J ⇒ Q = +47 J
Q −W = 0

(heat transferred to gas)
32
...
684 cm3 103 L
= 01499 L mol

...
64 atm 0
...
314
J / (mol ⋅ K)
H = U + PV = 1706 J mol +
= 2338 J mol
mol
0
...
12 V =

7- 5

7
...
310 bar

(a)

(b) ΔU = U final − U initial = 0
...
24 = −28
...
24 kJ mol +

b

0
...
0516 − 79
...
314 J

1 kJ

mol 0
...
7 kJ mol

g

Δ H = nΔ H = 5
...
7 kJ / mol = −15358 kJ ⇒ −154 kJ

...
205 bar = U 300 K, 0
...
24 kJ mol

d

i b

g

U 340 K, Pf = U 340 K, 1
...
62 kJ mol

ΔU = U final − U initial

E

ΔU = 29
...
24 = 1380 kJ mol

...
At constant temperature ⇒ PV = P' V' ⇒ V' = PV / P'
(0
...
94 L / mol)
V' (T = 300K, P = 0
...
88 L / mol
0
...
00 L
1 mol
n=
= 0
...
88 L

b

gb

g

ΔU = nΔU = 0
...
0571 kJ

...
0571 kJ
0

0

0

(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is
being neglected; internal energy is not completely independent of pressure
...
14 (a) By definition H = U + PV ; ideal gas PV = RT ⇒ H = U + RT

b g bg

b g bg

bg

U T , P = U T ⇒ H T , P = U T + RT = H T independent of P
(b) Δ H = ΔU + RΔT = 3500

b

gb


...
5 mol 3599 cal / mol = 8998 cal ⇒ 9
...
15 ΔU + ΔE k + ΔE p = Q − Ws

b
b

g

Δ E k = 0 no change in m and u
Δ E p = 0 no elevation change
Ws = PΔV since energy is transferred from the system to the surroundings

g

b

ΔU = Q − W ⇒ ΔU = Q − PΔV ⇒ Q = ΔU + PΔV = Δ (U + PV ) = ΔH

7- 6

g

b
b

g

7
...
(a) Δ E k = 0 u1 = u2 = 0
Δ E p = 0 no elevation change

g

ΔP = 0 (the pressure is constant since restraining force is constant, and area is constrant)
Ws = PΔV the only work done is expansion work

b


...
8 J

...
0295 mol
RT 8
...
0295 mol 34980 + 35
...
5(400K) (J / mol)
83
...
0295 35
...
5(400) ⇒ T2 = 480 K
nRT 0
...
314 m3 ⋅ Pa 106 cm3 480 K
= 941 cm3
=
125 × 105 Pa
1 m3
mol ⋅ K
P
125 × 105 N (941 - 785)cm3 1 m3
= 19
...
5 J = 64
...

i) V =

(b) ΔEp = 0
7
...
" (If heat losses could
occur, the temperature would drop during these periods
...
90 × 1
...
26 J s

U (J ) = 126 t

...
10 L
25 + 273 K

g

1
mol ⋅ K
= 0
...
08206 L ⋅ atm

U
126t (J)

...
67t
n 0
...
51

...
249
T =100 , E = 5
...
67t
0 440 880 1320
T = 181E + 4
...

(c) To keep the temperature uniform throughout the chamber
...

(e) In a closed container, the pressure will increase with increasing temperature
...


bg

Ideality could be tested by repeating experiment at several initial pressures ⇒ same
results
...
18 (b) ΔH + ΔE k + ΔE p = Q − Ws (The system is the liquid stream
...

(e) ΔH + ΔE k + ΔE p = Q − Ws (The system is the reaction mixture)

c

Δ E k = Δ E p = 0 given

c

h

ΔWs = 0 no moving parts or generated current

h

Δ H = Q, Q pos
...
depends on reaction

7
...
25 m3 273 K
min

122 kPa

1 mol

b g

423 K 101
...
4 L STP

103 L
1 m3

= 43
...
37 mol 1 min
min 60s

3640 J
kW
= 2
...
The change in kinetic energy would depend on the
cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet
and outlet pipes would be needed to answer this question
...
20 (a) H = 1
...
0 − 25 = 9
...

H in = 1
...
0 − 25 = 5
...
36 − 5
...
16 kJ kg

Δ H + Δ E k + Δ E p = Q − Ws

c

h
Ws = 0 c no moving partsh

Δ E k = Δ E p = 0 assumed

Q
=1

...
25 kW
=
= 10
...
16 kJ
kW
1 kg 28
...
7 mol 22
...


mol

273 K

110 kPa

= 2455 L / s ⇒ 246 L s

...


U
|
V
|
W


...
21 (a) H = aT + b a = H 2 − H1 = 129
...
2
⇒ H kJ kg = 5
...
2
T2 − T1
50 − 30
b = H1 − aT1 = 258 − 5
...
2

...
2
= 25° C
5
...
1 ⇒ S
...


b

b

g

bg

C 6 H 14 l

= 0
...


U kJ kg = H − PV = 5
...
2 kJ / kg
−

b

1 atm 1
...
52 × 10 −3 m3
1

atm

1

kg

1J

1 kJ

1 N ⋅ m 103 J

g

⇒ U kJ kg = 5
...
4

(b) Energy balance: Q = ΔU =
ΔE k , Δ E p , W = 0

20 kg [(5
...
4) - (5
...
4)] kJ

Average rate of heat removal =

1
6240 kJ 1 min
= 20
...
22

m (kg/s)
260°C, 7 bars
H = 2974 kJ/kg
u0 = 0

m (kg/s)
200°C, 4 bars
H = 2860 kJ/kg
u (m/s)

ΔH + ΔE k + ΔE p = Q − Ws
ΔE p = Q = Ws = 0

ΔE k =− ΔH ⇒

d

d

mu 2
=− m Hout − Hin
2

i

u 2 = 2 Hin − Hout =

b

i

g

(2) 2974 − 2860 kJ 103 N ⋅ m 1 kg ⋅ m / s2
kg

7
...
28 × 105

m2
⇒ u = 477 m / s
s2

5 L/min
100 mm Hg (gauge)

0 mm Hg (gauge)

Qout

Qin

Since there is only one inlet stream and one outlet stream, and min = mout ≡ m ,
Eq
...
4-12) may be written
m
mΔU + mΔ PV + Δ u 2 + mgΔz = Q − Ws
2
ΔU = 0 (given )

d i

d i

a

f

mΔPV = mV Pout − Pin = VΔP
Δu = 0 (assume for incompressible fluid )
2

Δz = 0
Ws = 0 (all energy other than flow work included in heat terms)
Q = Qin − Qout

VΔP = Qin − Qout
(b) Flow work: VΔP =

5L

b100 − 0gmm Hg

1 atm

8
...
08206 liter ⋅ atm
5 ml O 2 20
...
7 J min
=
× 100% = 66%
101 J min
Qin

7- 10

= 66
...
24 (a) ΔH + ΔE k + ΔE p = Q − Ws ; ΔE k , ΔE p , Ws = 0 ⇒ ΔH = Q

b
g
H b100° C, sat' d ⇒ 1 atmg = 2676 kJ kg (Table B
...
7)

100 kg H 2 O(v)/s
100o C, saturated

100 kg H 2 O(v)/s
400o C, 1 atm
Q (kW)

Q=

100 kg

b3278 − 2676
...
02 × 10 7 J s

(b) ΔU + ΔE k + ΔE p = Q − W ; ΔE k , ΔE p , W = 0 ⇒ ΔU = Q
kJ
m3
ˆ
ˆ
ˆ
Table B
...
673
= V ( 400 ° C, Pfinal )
kg
kg

ˆ
Interpolate in Table B
...
673 at 400oC, and then interpolate again
ˆ
to find U at 400oC and that pressure:
⎛ 3
...
673 ⎞
o
ˆ
ˆ
V = 1
...
0 + 4
...
3 bar , U (400 C, 3
...
11 − 0
...
59 × 10 7 J

The difference is the net energy needed to move the fluid through the system (flow work)
...
)

bg

c

h

7
...
9 kJ kg (Table B
...
2 k J kg

(Table B
...
65(813 kW) = 528 kW

(a) ΔH + ΔE k + ΔE p = Q − Ws ; ΔE k , ΔE p , Ws = 0 ⇒ ΔH = Q
ΔH = mΔH

m=

Q
ΔH

b

=

528 kW

gd

kg

b2797
...
9 gkJ

1 kJ / s
1 kW

3600 s
1

h

= 701 kg h

i

(b) V = 701 kg h 0
...
7 m 3 h sat' d steam @ 20 bar

A

Table B
...
4 K 0
...
5 m3 / h
18
...

(d) Most energy released goes to raise the temperature of the combustion products, some is
transferred to the boiler tubes and walls, and some is lost to the surroundings
...
26 H H 2 O l , 24° C, 10 bar = 100
...
5 for saturated liquid at 24oC; assume H

independent of P)
...
2 kJ kg (Table B
...
2 − 100
...
6 kJ kg

m [kg H2O(l)/h]

m [kg H2O(v)/h]
15,000 m3/h @10 bar (sat'd)

24oC, 10 bar
Q(kW)
m=

15000 m 3

kg

4
01943 m 3 = 7
...


h

A

b Table 8
...
72 × 10 4 kg

2

=

2

ΔEk = Ekfinal

d15,000 m h i
3

0
...
96x10 J/s

3

A =π D 2 4
= 5
...
72 × 10 4 kg 2675
...
96 × 10 5 J 1 kJ
+
h
kg 3600 s
s 10 3 J

= 57973 kJ s = 5
...
27 (a)

228 g/min
T(oC)
Q( kW)

=0

b g

Energy balance: Q = ΔH ⇒ Q W =
ΔE x , ΔE p , Ws =0

228 g 1 min ( Hout − Hin ) J
min
60 s
g

b g

b g

⇒ H out J g = 0
...
263Q W

b g

b

g

(b) H = b T − 25

b g

25 26
...
47

27
...
0 32
...
28 13
...
8

Fit to data by least squares (App
...
1)

b g

b=

∑H
i

i

bT − 25g ∑ bT − 25g
i

i

i

2

= 3
...
34 T ° C − 25
(c) Q = Δ H =

b

g

kW ⋅ s

g

10 3 J

350 kg 10 3 g 1 min 3
...


7- 12

7
...
5 bar

me [ kg C 2 H 6 / min]
93 o C, 2
...
50 bar
2 6
e
min
m 3 289 K

1

K - mol 30
...
08314 L - bar

1 kg

mol 1000 g

= 2
...
487 × 103

LMb
N

g OPQ

2
...
47 × 103 kW
min
min 60 s
kg

b
g

...
5)

(b) H s1 3
...
7 kJ kg (Table B
...
47 × 10 3 kW

d

Energy balance on H 2 O: Q = ΔH = m H s2 − H s1
⇒m=

Q
H s2 − H s1

b

=

−5
...

b1131 − 2724
...
09 kg s steam

⇒ Vs = 2
...
606 m 3 kg = 1
...
6

Too low
...

(c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it

would require heat flow from the ethane to the steam over some portion of the exchanger
...
29

250 kg H2 O(v )/min
40 bar, 500°C
H1 (kJ/kg)

Heat
exchanger

250 kg/min
5 bar, T 2 (°C), H2 (kJ/kg)

Turbine

W s =1500 kW

b
g
H Obv , 5 bar, 500° Cg: H

250 kg/min
5 bar, 500°C
H3 (kJ/kg)

Q(kW)

H 2 O v , 40 bar, 500° C : H1 = 3445 kJ kg (Table B
...
7)

(a) Energy balance on turbine: ΔE p = 0, Q = 0, ΔE k ≅ 0

d

i

ΔH = −Ws ⇒ m H 2 − H1 = −Ws ⇒ H 2 = H1 − Ws m
=

3445 kJ 1500 kJ
−
s
kg

min 60 s
= 3085 kJ kg
250 kg 1 min

H = 3085 kJ kg and P = 5 bars ⇒ T = 310° C (Table B
...
0864 m 3 kg (Table B
...
5318 m 3 kg (Table B
...
0864 m 3
min

60 s

kg

250 kg min 0
...
5 π 4 m 2
2

1
0
...

= 183 m s

= 113 m s

...

b11
...
26 kW << 1500 kW

7- 14

2

s

2

2

m2

1 kW ⋅ s

1N
1 kg ⋅ m / s

2

10 3 N ⋅ m

b

g

b

g


...
30 (a) ΔE p , ΔE k , Ws = 0 ⇒ Q = ΔH ⇒ − hA Ts − To = −300 kJ h ⇒ 18h Ts − To = 300
(b) Clothed:

kJ
h

h = 8 ⇒ To = 13
...
2

Nude, immersed: h = 64 ⇒ To = 316° C (Assuming Ts remains 34
...

Ts =34
...
(Stagnant air acts as a thermal
insulator —i
...
, in the absence of wind, h is low
...

7
...
= 125
...
g = 376
...
5 - neglecting effect of P on H )
Energy Balance: Q - W = ΔU + ΔE p + ΔE k

b

g

Q =W = ΔE p = ΔE k = 0

b

ΔU = 0

g

⇒ 3U f − (1 kg) 125
...
9 kJ / kg = 0
⇒ U f = 293
...
05° C (Table B
...
=
7
...
05 − 70
...
07% (Any answer of this magnitude is acceptable)
...
05


...

0
...
15 kg H2 O( l)/kg
5 bar, saturated, T(oC)
P = 5 bars

(a) Table B
...
5 m3 H2O(v)/h

...


Q (kW)

T = 1518° C , H L = 6401 kJ kg , H V = 2747
...


...
375 m 3 / kg ⇒ m =

b gb

g

52
...
375 m3

(b) H 2 O evaporated = 015 140 kg h = 21 kg h

...

b2747
...
33 (a) P = 5 bar

Tsaturation = 1518 o C
...


Table B
...
7

Table B
...
0665 m 3 / kg
H out = 314
...
03 × 10 -3 m 3 / kg

3
Vin 200 kg 1 min 0
...

π (0
...
00103 m 3 / kg
= 175 m / s

...
05) 2 / 4 m 2

m 2
2
(u2 − u1 )
2
(314-3095) kJ
200 kg 1 min (1
...
182 ) m 2
+
kg
2 min 60 s
s2

Energy balance: Q − Ws ≈ ΔH + ΔE k = m( H 2 − H1 ) +
Q − Ws =

200 kg

1 min

min

60 s

= −13, 460 kW ( ⇒ 13,460 kW transferred from the turbine)
7
...
00 × 10 kJ 1 min = 167 kJ s
min 60 s
100 kg oil/min
135°C
m (kg H2O(v)/s)
25 bars, sat'd

d

Energy balance on H 2 O: Q = ΔH = m H out − H in

i

100 kg oil/min
185°C
m (kg H2O(l)/s)
25 bars, sat'd

ΔE p , ΔE k , Ws = 0
H (l , 25 bar, sat' d ) = 962
...
9 kJ kg (Table B
...
0 − 2800
...
091 kg s
1200 g
1s
1 kg
= 13 s discharge
discharge 0
...
9 − 83
...
9486 Btu
= $2
...
091 kg stream 0
...
6 × 10 −3$ 3600 s 24 h 360 day
trap ⋅ s
kg stream
kg lost
h
day
year
= $7
...
35 Basis: Given feed rate

200 kg H2O(v)/h
10 bar, sat’d, H = 2776
...
6 (saturated steam) or Table B
...
2 − n2 3052 , Q in kJ h
ΔE K , ΔE p , W = 0

(a) n3 = 300 kg h
(b) Q = 0

(1), (2)

(1)

( 2)

n2 = 100 kg h

(2)

Q = 2
...
36 (a) Tsaturation @ 1
...
6 °C ⇒ T f = 99
...
0 bar, sat' d) ⇒ H l = 417
...
4 kJ / kg
H 2 O (60 bar, 250 C) = 1085
...
8 kJ / kg) = 0
(1,2)

ml = 70
...
6 kg ⇒ y v =

(2)

29
...
296
100 kg
kg

(b) T is unchanged
...
The system undergoes expansion, so assuming the same pipe diameter, ΔEk > 0
...

(c) Pf = 39
...
8 bar the temperature cannot increase,
because a higher temperature would increase the enthalpy
...
8 bar , the product
is only liquid ⇒ no evaporation occurs
...
36 (cont’d)
0
...
3
0
...
1
0
0

20

40

60

300
250
200
150
100
50
0

80

1

5

10

Pf (bar)

15

20

25

30

36 39
...
37 10 m3, n moles of steam(v), 275°C, 15 bar ⇒ 10 m3, n moles of water (v+l), 1
...
0 m3 H2O (v)

10
...
5 bar
, 15 bar

mv [kg H2O (v)]
ml [kg H2O (l)]
Q

(a) P=1
...
2 bar, saturated

Table B
...
8 C

1 kg
= 55 kg
0
...
0
Volume additivity: Vv + Vl = 10
...
001048 m 3 / kg)

...
0 kg, ml = 48
...
7 ⇒ U in = 2739
...
1818 m 3 / kg

R
|
S
|
T

3
Table B
...
2 kJ / kg; Vl = 0
...
1 kJ / kg;
Vv = 1
...
0)(2512
...
0)( 439
...
2)] kJ
= −1
...
38 (a) Assume both liquid and vapor are present in the valve effluent
...
0 bar, saturated

(b) Table B
...
3o C ⇒ Tin = 348
...
7 ⇒ H in = H (348
...
6 ⇒ H l (1
...
5 kJ / kg; H v (1
...
4 kJ / kg

7-18

7
...
5) + (1 − ml )( 2675
...
(For any value
in this range, the right-hand side would be between 417
...
4)
...

min = mout = 1
3149 kJ / kg = H (1 bar, Tout )
(c) Energy balance ⇒ mout H out = min H in
Table B
...

7
...
3 psig, 86°F
H = 27
...
8 Btu / lb m , H l = 9
...
8 Btu 40 1 − X v lb m R l
+
lb m
min

E

∑n H − ∑n H
i

i

out

9
...
8 Btu
=0
lb m

g

X v = 0
...
7% evaporates

(b) Evaporator coil
40 lb /min
m
0
...
733 R( l )
11
...
8 Btu/lbm , Hl = 9
...
8 psig, 5°F
H = 77
...
8 Btu 40 0
...
733 lb m R l
77
...
6 Btu
lb m

7
...
Let us first note that the net rate of heat input to the system is

Q = Qevaporator − Qcondenser = 2000 − 2500 = −500 Btu min
and the compressor work Wc represents the total work done on the system
...
Consider a time interval Δt min
...
The
total heat input is QΔt , the work input is Wc Δt , and (Eq
...
3-4) yields
QΔt − Wc Δt = 0 ⇒ Wc = Q =

−500 Btu 1 min
1
...

= 118 hp
min
60 s 9
...
40 Basis: Given feed rates
n1 (mol / h)
nC 3H 8 (mol C 3 H 8 / h)
nC 4 H10 (mol C 4 H 10 / h)
227 o C

0
...
8 C 4 H 10
0o C, 1
...
40 C 3 H 8
0
...
1 atm

Q (kJ / h)

Molar flow rates of feed streams:
300 L 1
...
7 mol h
n1 =
hr 1 atm 22
...
1 atm
1 mol
= 9
...
4 L STP

b g

14
...
20 mol C 3 H 8 9
...
40 mol C 3 H 8
+
h
mol
h
mol
= 6
...
7 + 9
...
54) mol C 4 H 20 h = 17
...
40 × 9
...
54 mol C 3 H 8
h

b

20
...
16 mol C 4 H 10
+
mol
h

g

1
...
60 × 9
...
442 kJ
mol

2
...
4 L STP

b g

7
...


n0 (kmol/min)
38°C, h r = 97%
y 0 (mol H 2 O/mol)
x
(1 –y0) (mol dry air/mol)

1 kmol
= 214 kmol min

...
4 kmol/min
18°C, sat'd
y 1 (mol H 2 O/mol)
(1 – y 1) (mol dry air)


...
97 ( 49
...
0634 mol H 2 O mol

15
...
0204 mol H 2 O mol
760 mm Hg

g

Dry air balance: 1 − 0
...
0204 214 ⇒ no = 22
...


b

g

b

g

Water balance: 0
...
4 = n2 + 0
...
4 ⇒ n2 = 0
...
98 kmol 18
...
Enthaphies: H air 38° C = 0
...
3783 kJ mol

b g
b g
1
bv, 38° Cg = 2570
...
02 g = 46
...
5 kg 10 kg 18
...
67 kJ mol |Table B
...
5 kJ
1 kg 18
...
36 kJ mol |
|
W
H air 18° C = 0
...
204 kJ mol

H H 2O
H H 2O
H H 2O

3

3

3

Energy balance:
ΔE , W = 0, ΔE ≅ 0
p

s

gd
ib g
+b0
...
67g + d0
...
0634gd22
...
3783g

...

−b0
...
4 × 10 ib46
...
67 × 10 kJ min
k

Q = ΔH =

∑n H − ∑n H
i

i

i

out

i

b

⇒ Q = 1 − 0
...
204

...
67 × 10 kJ 60 min 0
...
42 Basis: 100 mol feed
n2 (mol), 63
...
98 A( )
v
0
...
5n2 (mol)
0
...
02 B(l )

100 mol, 67
...
65 A(l )
0
...
8°C

n5 (mol), 98
...
544 A( )
v
0
...
5n2 (mol)
0
...
02 B(l )

n5 (mol), 98
...
155 A(l )
0
...
5n2 + n5
n2 = 120 mol
A: 0
...
98 0
...


b g

b

g

U
V
W

b g
b g
0
...
2 mol A
0
...


Product flow rates: Overhead 0
...
98 = 58
...
5 120 0
...
2 mol B

Bottoms

Overall energy balance: Q = ΔH =
ΔE , W = 0, ΔE ≅ 0
p

2

∑n H − ∑n H
i

i

out

x

i

i

in

interpolate in table

↓

interpolate in table

↓

⇒ Q = 58
...
2 ( 0 ) + 6
...
8 (1312 ) − 65 ( 354 ) − 35 ( 335 ) = 1
...
4 mols B

...
8 = 117
...
6 0 − 7322 + 2
...
77 × 10 5 cal heat removed from condenser
Assume negligible heat transfer between system & surroundings other than Qc & Qr

(

)

Qr = Q − Qc = 1
...
77 × 105 = 8
...
43

1
...
0 bar, T1

2
...
0 bar, T3=250oC

1
...
0 bar, T2
Q= 0

7-22

7
...
0 bar, sat' d steam) = 165
...
0 bar, T = 250 o C) = 2954 kJ kg (Table B
...
0 bar, sat' d) = 2760 kJ kg (Table B
...
96 H 3 − 196 H1 − 10 H 2 ⇒ 196 H1 = 2
...
0 kg(2760 kJ / kg)

...


...
0 bar, T1 ) = 3053 kJ / kg ⇒ T1 ≅ 300 C
(b) The estimate is too low
...

7
...
0 bar, sat' d
...
5 C

Vapor

Vl ( P = 3
...
) = 0
...
0 bar, sat' d
...
606 m 3 / kg

(a)

Liquid

3

0
...
2 L
kg
m3
Vspace = 200
...
2 L = 22
...
8 L

m=165
...
0 L
Pmax=20 bar

3

1 m
1 kg
= 0
...
606 m 3

(b) P = Pmax = 20
...
0 + 0
...
04 kg

T1 = T ( P = 20
...
) = 212
...
0 bar, sat' d
...
001177 m 3 / kg; Vv ( P = 20
...
) = 0
...
0 L

1 m 3 = m kg(0
...
04 - m ) kg(0
...
98 kg; mv = 0
...
001177 m 3
kg

mevaporated =

1000 L 164
...
2 L;
m3

Vspace = 200
...
2 L = 5
...
06 - 0
...
0 bar, sat' d) − U ( P = 3
...
0 bar, sat' d
...
2 kJ / kg; U v ( P = 20
...
) = 2598
...
0 bar, sat' d
...
1 kJ / kg; U v ( P = 3
...
) = 2543 kJ / kg
Q = 0
...
2 kJ / kg) + 164
...
2 kJ / kg) - 0
...
0 kg (561
...
70 × 10 4 kJ
Heat lost to the surroundings, energy needed to heat the walls of the tank

7-23

7
...

(e) – Using an automatic control system that interrupts the heating at a set value of pressure
– A safety valve for pressure overload
...

7
...
97 kg H2 O(v)
0
...
7g
|
V
|
W
= H = 0
...
2g + 0
...
6g

b
g
H bl , 20 bars, sat' d g = 908
...
2 kJ kg
Energy balance on condenser: ΔH = 0 ⇒ H 2
ΔE , ΔE , Q , W =0
p

K

1

3

⇒ H 2 = 2740 kJ / kg

Table B
...
When it reaches
its saturation temperature at 1 atm, it begins to condense, so that T = 100° C
...


bg

bg

1 quart
1 m3
1000 kg
= 0
...
46 Basis:

8 oz H 2 O l

0
...
2365) (kg H2O (l))
4°C

Assume P = 1 atm

Enthalpies (from Table B
...
5 kJ/kg; H ( H 2 O(l ), 4°C ) = 16
...
2365) kg(16
...
2365 kg(75
...
038 kg = 38 g ice

7-24

7
...
13dT − 175ical 4
...
0 g

f

g

Table B
...
9j J = 1000eU dT i − 83
...

dT i − 160 × 10 = f dT i = 0

1
...
1 × 10

d i

f

H 2O

H 2O

f

5

f

30

f Tf

f

cal

f

40
4

+2
...
6° C

7
...
1) mm Hg
Tf
⇒ 1
...
8°C (Table 8
...

1044

...
0
2506
...

g II b108 bar, 101
...

1576
426
...
6

Initial vapor volume: VvI = 20
...
0 L −

b

50 kg

g

1L
8
...
4 L H 2 O v

bg

I
Initial vapor mass: mv = 14
...
61 × 10 −3 kg H 2 O v

b
g
= 0
...
8g = 36
...
0 L 1
...
79 kg H 2 O l
II
Final energy of bar: U b

Assume negligible change in volume & liquid ⇒ VvII = 14
...
4 L 1576 L kg = 9
...
14 × 10 −3 2508
...
79 426
...
0 36
...
61 × 10 −3 2506
...
79 419
...
0 kg
= 441 kJ kg

...
1 kJ / kg
(a) Oven Temperature: To =
= 122
...
36 kJ / kg ⋅ o C
I
Ub =

II
I
H 2 O evaporated = mv − mv = 9
...
61 × 10 −3 kg = 5
...
53 g

I
(b) U b = 44
...
3 5
...

To = 458 0
...
2° C

...
49 (a) Pressure in cylinder =
P=

30
...
807 N

400
...

105 N m2

+

1 atm 1
...

= 108 bar


...

=

7
...
6 − 83
...
0 kg

gi

[0
...

= 2630 kJ
kg

kg
2630 kJ < 3310 kJ ⇒ Sufficient heat for vaporization
(b) T f = Tsat = 1018° C
...
5 ⇒

...
0 kg H 2 O(l )
H = 426
...
046 L / kg

Vl = 1046 L kg , U l = 426
...

Vv = 1576 L kg , U v = 2508
...
6 kJ/kg

T ≡ 101
...
08 bars

1
...
6 kJ/kg
ml (kg H 2 O(l ))

Q (kJ)

W (kJ)

(Since the Al block stays at the same temperature in this stage of the process, we can
ignore it -i
...
, U in = U out )
Water balance: 7
...
08 bar g 1576m + 1
...
046gb7
...
314 J / mol ⋅ K
1 kJ
×
= b170
...
113m − 0
...
08314 liter - bar / mol ⋅ K 10 J

LM w + P
NA

atm

v

3

l

v

l

Energy balance: ΔU = Q − W
ΔU

Q

bg

W

⇒ 2508
...
6m L − 426
...
2mv + 0
...
7908)
⇒ 2679mv + 426
...
302 kg , ml = 6
...
302 kggb1576 L kgg = 476 L vapor

...
01 + 476 − b7
...
698 kg 1046 L kg = 7
...


10 3 cm3

1

= 1190 cm
1 L 400 cm2
⇒ All 3310 kJ go into the block before a measurable amount is transferred to the
A

(c) Tupper

b

g b

g

water
...
0 kg 0
...
In fact, the bar would melt at 660 C
...
50

100 L H 2 O( v ), 25o C

...
at
V is vaporized
...
m = kg H O vaporized
...
00 L H 2 O(l ), 25 C
m L1 (kg)

Q=2915 kJ

Initial conditions: Table B
...
8 kJ kg , V L1 = 1
...
0317 bar

b

gb

T = 25° C, sat' d ⇒ U v1 = 2409
...
00 l 43400 l kg = 2
...
00 l
Energy balance:

d

i d i b

g b1
...
988 kg

g d i d

ib

ΔU = Q ⇒ 2
...
988 − me U L T f − 2
...
9

b

g

− 3
...
8) = 2915 kJ

d

g d i
i dEi b
3333 − d2
...
988U
=

⇒ 2
...
988 − me U v T f = 3333
−5

⇒ me

F
GG
H

v

L

Uv − U L

I
J d i b
AJ A
K
5
...
304 × 10 iV − 3
...
304 × 10 −5 + me V L T f + 3
...
00 L
kg

liters kg

−5

⇒ me

bg b g

d i

1 − 2 ⇒ f Tf =

v

Vv − V L

d

i d i

d

Uv − U L

i

5
...
304 × 10 −5 Vv − 3
...
8 856
...

198
...
4 842
...
0 2590
...
5
196
...
6

...
7
131
...
7
136
...
5

Procedure: Assume T f

bg

d i

3333 − 2
...
988U L T f

−

Eq 1

b2g

L

f

d i

such that f T f = 0

VL
f
1159 −512 × 10 −2

...

1154 −1
...

1149 134 × 10 −2

...

1151 −4
...
4° C, Pf = 14
...


me = 2
...
6 g evaporated

or Eq 2

7-28

g

7
...


Basis: 1 mol feed
B = benzene
T = toluene

nV (mol vapor)
y B(mol B(v)/mol)
(1 – y B ) (mol T(v)/mol)

1 mol @ 130°C
z B (mol B(l)/mol)
(1 – z B )(mol T(l)/mol)

in equilibrium
at T(°C), P(mm Hg)

nL (mol liquid)
x B(mol B(l)/mol)
(1 – x B ) (mol T(l)/mol)

(a) 7 variables: (nV , y B , n L , x B , Q, T , P)
–2 equilibrium equations
–2 material balances
–1 energy balance
2 degrees of freedom
...

(b) Mass balance: nV + n L = 1 ⇒ nV = 1 − n2
Benzene balance: z B = nV y B + n L x B

(1)
(2)

bg d
i d
i
C H bv g: dT = 80, H = 4161i , dT = 120, H = 45
...
25

...

C H bl g: dT = 0, H = 0i , dT = 111, H = 18
...

C H bv g: dT = 89, H = 49
...
05i ⇒ H = 01304T + 37
...

C 6 H 6 l : T = 0, H = 0 , T = 80, H = 10
...

6

6

7

8

7

8

BV

TL

TV

Energy balance: ΔE p , Ws = 0, neglect ΔE k

b

g
g H bT g

b

g

bg

b g

Q = ΔH = nV y B H BV + nV 1 − y B HTV + n L x B H BL + n L 1 − x B HTL − 1 z B H BL TF

b gb

− 1 1 − zB

Raoult' s Law:

TL

(3)
(4)
(5)
(6)

(7)

F

*
y B P = x B pB

(8)

*
(1 - y B ) P = (1 − x B ) pT
Antoine Equation
...
89272−1203
...
888)] = 1021 mmHg
*
pT (90o C) = 10[6
...
773/(90 + 219
...
7 mmHg

Adding equations (8) and (9) ⇒
*
*
P = x B pB + (1 − x B ) pT ⇒ x B =

*
P − pT
*
pB

−

*
pT

=

*
P − pT
*
pB

x B p*
B

−

*
pT

=

652 − 406
...
399 mol B(l) / mol
1021- 406
...
399(1021 mmHg)
=
= 0
...
5 − 0
...
446 mol vapor
y B − x B 0
...
399
n L = 1 − nV = 1 − 0
...
554 mol liquid
yB =

7-29

(9)

7
...
446(0
...
25] + 0
...
625)[01304(90) + 37
...


...
554(0
...
554(1 − 0
...
5[ 01356(130)]

...


...


...
5[01674(130)] ⇒ Q = 814 kJ / mol
(c)
...
If P>Pmax, all the output is liquid
...
The higher the pressure,
there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium
vapor: enthalpy out < enthalpy in
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
7
406
...
7
406
...
7
406
...
7
406
...
7
406
...
7
406
...
7
406
...
7
406
...
399
0
...
285
0
...
315
0
...
347
0
...
380
0
...
412
0
...
445
0
...
477
0
...
Pmax = 714 mmHg, Pmin = 582 mmHg
nV vs
...
8
0
...
4
0
...
5 @ P ≅ 640 mmHg

7-30

yB
0
...
715
0
...
516
0
...
554
0
...
589
0
...
622
0
...
653
0
...
682
0
...
710

nV
0
...
001
0
...
925
0
...
758
0
...
605
0
...
460
0
...
318
0
...
176
0
...
029

nL
0
...
001
0
...
075
0
...
242
0
...
395
0
...
540
0
...
682
0
...
824
0
...
971

Q
8
...
09
26
...
8
21
...
3
15
...
3
10
...
60
6
...
04
1
...
50
-2
...
14

ΔP

Δu 2
+
+ gΔz = 0
ρ
2

7
...
Bernoulli equation:

ΔP

ρ

=

d0
...


m3
1
...
7

m2
s2

gΔz = (9
...
8 m 2 / s 2
Δu 2
2
2
= 46
...
8 m 2 / s 2 ⇒ u2 = u1 + 2 −12
...
00 m 2 / s 2 − (2)(12
...
800 m 2 / s 2 ⇒ u2 = 0
...
Since the fluid is incompressible, V m 3 s = π d 12 u1 4 = π d 2 u2 4

b g

u2
= 6 cm
u1

0
...
54 cm
5
...
53 (a)
...
Bernoulli equation (Δz = 0)

d

2
2
ρ u2 − u1
Δu 2
+
= 0 ⇒ Δ P = P2 − P1 = −
ρ
2
2

ΔP

i

Multiply both sides by − 1
Substitute u 2 = 16u1
2

2

2

Multiply top and bottom of right - hand side by A1
note V

P1 − P2 =

d

i

(c) P1 − P2 = ρ Hg − ρ H 2 O gh =
V =
2

b g

2 π 7
...
8066 m

108 cm4

s2

⇒ V = 0
...
6 − 1g = 1955 × 10

...
54 (a)
...
P1 = 31 bar , z1 = +7 m , u1 = 0 m s

...
P2 = 1 atm , z 2 = 0 m , u2 = ?
Δρ

ρ

=

gΔz =

b =1
...

b1
...

m 2 ⋅ bar 0
...
8066 m
s2

Bernoulli equation ⇒

= −68
...
5 + 68
...
1 m 2 s 2
2
ρ

b

Δu = u 2 − 0
2

2

g

2

2
u2 = 2(332
...
2 m 2 s 2 ⇒ u2 = 258 m / s

...


2580 cm 1 L
60 s
= 122 L / min
1
s 10 3 cm 3 1 min

4

(b) The friction loss term of Eq
...
7-2), which was dropped to derive the Bernoulli equation,
becomes increasingly significant as the valve is closed
...
55 Point 1 - surface of lake
...
P2 = 1 atm , z 2 = z ft

u2 =

1 ft 3
7
...
041b2 z g ft ⋅ lb lb

Pressure drop: Δ P ρ = 0
Friction loss:

1
2

...
5 × 1049 in 2

1

g

144 in 2
1 ft 2

1 min
= 35
...
0822 z (ft ⋅ lb f lb m )

W -8 hp 0
...
341 × 10 −3 hp

95 gal

1 ft

3

1 ft 3

60 s

62
...
4805 gal

1 min

= −333 ft ⋅ lb f lb m
Kinetic energy: Δ u 2 2 =

b35
...
7
...
174 ft
s2

− 0 2 ft 2
s

2

bg

z ft

1

lb f

32
...
4 ft ⋅ lb f lb m

b

1 lb f
= z ft ⋅ lb f lb m
32
...
4 + z + 0
...
56 Point 1 - surface of reservoir
...
P2 = 1 atm (assume), u2 = ? , z 2 = 0

ΔP ρ = 0

d h

2
V A
Δu 2 u2
=
=
2
2
2

2

=

V 2 (m 6 / s 2 )

= 3
...
8066 m
s2

gΔz =

b g
bN ⋅ m kgg
π 35

(2)

10 8 cm 4

2 2

cm 4

1 m3
= 800 V N ⋅ m kg
1000 kg

b

d i
Mechanical energy balance: neglect F b Eq
...
7 - 2g
ρ

+

1 kg ⋅ m / s 2

1N
= −637 N ⋅ m kg
1 kg ⋅ m / s 2

6
s
Ws 0
...

+ gΔz =
⇒ 3
...
2 m 3 min
s
1 min
2
m
V

Include friction (add F > 0 to left side of equation) ⇒ V increases
...
57 (a)
...
P2 = 1 atm , z 2 = 0

u2 =
ΔP

ρ

600 L
min π 4
...
807 m
s

10 3 cm 3 1 min
1m
= 7
...
96 m sg
=b
2

2

c0 − Hbmgh

2

Bernoulli Equation:

2

ΔP

ρ

+

1 J
= 31
...
807 H (J / kg)

Δu 2
+ gΔz = 0 ⇒ H = 3
...
Point 1: Fluid in washing machine
...
P2 = 1 atm , u2 = 7
...
23 m
ΔP

ρ

= 0;

J
J
J
Δu 2
; gΔz = 9
...
23 − 0 = 317
= 317

...

; F = 72
kg
2
kg
kg

b

Mechanical energy balance: Ws = − m
⇒ Ws = −

600 L 0
...
7 + 31
...
30 kW
kg 10 3 J s
(work applied to the system)

Rated Power = 130 kW 0
...
7 kW

...
58 Basis: 1000 liters of 95% solution
...

xi 0
...
05
1
l
Density of 95% solution:
=
=
+
= 0
...

ρ
ρ i 1
...

kg
b Eq
...
1-1g

∑

Density of 35% solution:

Mass of 95% solution:

1

ρ

=

0
...
65
l
+
= 0
...

126 100

...

kg

1000 liters 1
...
95 G
0
...
60 G
0
...
35 G
0
...
D
...
95 1240 + 0
...
60 m2
Volume of 35% solution added =

1740 kg

b

g

1

2

1L
= 1610 L
1
...
Surface of fluid in 35% solution storage tank
...
Exit from discharge pipe
...
5 cm 2

b g

b g

2
2
Δu 2 Δu2
1
...
8066 m
s2

Mass flow rate: m =

23 m

1740 kg 1 min
= 2
...
552 N ⋅ m kg
1 kg ⋅ m / s 2

1N
= 225
...
7
...
051 m s
1 m2

OP
Q

+ gΔz + F = −

g

2
...
552 + 225
...
62 kW ⇒ 0
...


7-34

1J

1 kW

1 N ⋅ m 10 3 J s

CHAPTER EIGHT
8
...


U (T ) = 25
...
02134T 2 J / mol
U (0 o C) = 0 J / mol

U (100 o C) = 2809 J / mol

Tref = 0 o C (since U(0 o C) = 0)

b
...
U (100 o C) is just the change from U (0 o C) to
c
...

Q − W = ΔU + ΔE k + ΔE p
ΔE k = 0, ΔE p = 0, W = 0

Q = ΔU = (3
...


Cv =

F ∂U I
GH ∂T JK

z

=
V

dU
= [25
...
04268T ] J / (mol⋅ o C)
dT

T2

ΔU =

z

F
GG
H

100

Cv (T )dT =

(25
...
04268T )dT = 25
...
04268

0

T1

T2
2

OP
QP

100

0

I
JJ J / mol
K

ΔU = (3
...
0 mol) ⋅ [25
...
02134(100 2 − 0)] (J / mol) = 8428 J ⇒ 8400 J

8
...


b

g

b

gb

Cv = C p − R ⇒ Cv = 35
...
0291T [J / (mol⋅° C)] − 8
...
0 + 0
...


ˆ
ΔH =

∫ C p dT = 35
...
0291
100

25

z

100

c
...

8
...
314 100 − 25 = 2160 J mol

H is a state property

a
...
0252 + 1547 × 10 −5 T − 3
...

n=

PV
(2
...
00 L)
=
= 0
...
08206[atm ⋅ L / (mol ⋅ K)](298 K)

z

1000

Q1 = nΔU 1 = (0
...
0252 dT ( kJ / mol) = 6
...
245) ⋅

[0
...
91 kJ

...
245) ⋅

[0
...
012 × 10 −9 T 2 ] dT = 7
...


25

6
...
67
× 100% = −215%

...
67
7
...
67
% error in Q2 =
× 100% = 313%

...
67
% error in Q1 =

8-1

8
...


C p = Cv + R

C p [ kJ / (mol⋅ o C)] = (0
...
012 × 10 −9 T 2 ) + 0
...

= 0
...
012 × 10 −9 T 2

...
245 mol) ⋅

[0
...
012 × 10 −9 T 2 ] dT [kJ / (mol⋅ o C)] = 9
...


25

Piston moves upward (gas expands)
...

a
...


8
...


dC i

( 40 C ) = 0
...
4 × 10 ( 40 ) = 0
...
07406 + 32
...
20 × 10 b40g
−8

2

b g

+ 77
...
08684 [kJ / (mol⋅ o C)]

c
...


e
...
5


...
01118 + 1095 × 10 b313g − 4
...
009615 [ kJ / (mol ⋅ K)]

32
...
57 × 10 −12 4

...
07406T +
T −
T +
T
3
2
4

1095 × 10 −5 2

...
01118T +
T + 4
...

40

573

= 3
...


H = 2926 kJ kg − 2676 kJ kg = 250 kJ kg

z

350

b
...
03346 + 0
...
7604 × 10 −8 T 2 − 3593 × 10 −12 T 3 dT

...
845 kJ mol ⇒ 491
...
The numerical difference
is ΔH for H 2 O v, 350° C, 1 atm → H 2 O v, 350° C, 100 bar

b

8
...


g

b

z

g

80

dC i

p n − C H (l)
6 14

= 0
...
2163] dT = 1190 kJ / mol

...
90 kJ/mol
c
...
85 × 10 −5 T − 23
...
66 × 10 −12 T 3

...
85 × 10 −5 T − 23
...
66 × 10 −12 T 3 ] dT = −110
...


500

The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110
...
The
specific enthalpy of hexane vapor at 0oC relative to hexane vapor at 500oC is –110
...


8-2

8
...
5556T ′b° Fg − 17
...

C bcal mol⋅° Cg = 6
...
001436 0
...
78 = 6
...
0007978T ′b° Fg
cal
453
...

=
mol⋅° C 1 lb - mole 252 cal 1
...
864 + 0
...
8


...


...

bT g = 01031 + b0158810001031g T = 01031 + 0
...
0 L 789 g 1 mol F

...
000557 T OPQ
s
1 L 46
...
5

2

20

kJ mol

= 941
...
636 kJ / s = 7193 kW

8
...


b

g

Q = ΔH = 5,000 mol s ⋅

z

kJ mol

200

0
...
473 × 10 −12 T 3 dT

...


100

= 17,650 kW

b
...
0 kmol ⋅ 8
...


8
...

b
...

Qadditional = heat needed to raise temperature of vessel wall + heat that escapes from wall to
surroundings
...
e
...

Q
mΔT
(16
...
14) kJ

Q = mC p ΔT ⇒ C p =
Cp =

Q
1 L 86
...
223 kJ / (mol ⋅ K)
mΔT (2
...
10 K) 659 g 1 mol
1 kJ

Table B
...
216 kJ / (mol⋅ o C) = 0
...
11

a∂ ∂T f

P

H = U + PV =====> H = U + RT =====>

p

But since U

p

8-3

+ R ⇒ Cp =
p

FG ∂U IJ
H ∂T K

+R
p

≡ Cv ⇒ C p = Cv + R
V

g

8
...


dC i

= 75
...
4 kJ/(kmol
...
3 kmol
M
1 L 18 kg

zd

T2

n⋅
Q
Q= =
t

b
...
3 kmol 75
...

8h
3600 s
kmol⋅ o C

Qtotal = Qto the surroundings + Qto water , Qto the surroundings = 1967 kW

...
3 kmol 75
...
245 kW
3h
3600 s / h

Qtotal = 7
...
212 kW × 3 h = 21
...


Cost heating up from 29 o C to 40 o C = 21
...
10 / (kW ⋅ h) = $2
...
967 kW × 13 h × $0
...
56
Costtotal = $2
...
56 = $4
...
If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher
cost
...
13 a
...


ΔH H

2 (800

c
...


ΔH O

8
...


o

o

= HH

F) → H 2 (77 o F)

2 (300

2 (970

= HN

o

C) → CO 2 (1250o C)

m = 300 kg / min n =

Q = n ⋅ ΔH = n ⋅

z

T2

2 (77

o

o

2 (0

o

F)

C)

F)

= H CO

= HO

F) → O 2 (0o F)

2 (700

− HN

− HH

2 (1250

− HO

o

b
g
= b0 − 5021g = −5021 Btu / lb - mol
−H
= b63
...


...
59 − 0 = 20
...
5 mol / s
min 60 s 1 kg 28
...
5 mol / s) ⋅

b

[0
...
411 × 10 −5 T + 0
...
22 × 10 −12 T 3 ] dT [kJ / mol]

450

g

= (178
...
076 [kJ / mol] = −2,156 kW

b
...
15 a
...
5 mol / s)(0
...
815[kJ / mol]) = −2,157 kW

n = 250 mol / h

Q = nΔH =

i)

Q = nΔH = n ⋅

z

T2

T1

ii)
=

250 mol (2676 − 3697) kJ 1 kg
1 h 18
...
278 kW
h
1 kg
1000 g 3600 s 1 mol

C p dT

250 mol 1 h
h
3600 s

z

100

[003346 + 06880 × 10−5 T + 07604 × 10−8 T 2 − 3593 × 10−12 T 3 ] = −1
...


...


...
15 (cont’d)

b

g

250 mol
⋅ 2
...
91 [kJ / mol] = −1276 kW

...
Method (i) is most accurate since it takes into account the dependence of enthalpy on pressure and
(ii) and (iii) do not
...
The enthalpy change for steam going from 10 bar to 1 atm at 600oC
...
16 Assume ideal gas behavior, so that pressure changes do not affect ΔH
...

= 0
...
6125

8
...


b

g

lb - mole
) ⋅ (2993 − 0) [Btu / lb - mole] = 1833 Btu / h
h

50 kg 1
...


(C )

p Na CO
2
3

≈ 2 (C p )

Na

+ ( C p ) + 3 ( C p ) = 2 ( 0
...
0075 + 3 ( 0
...
1105 kJ mol ⋅ °C
C

50,000 g 0
...
18

dC i
dC i

p C H O(l)
6 14

O

1 mol
105
...
6% error
2280

b

g b

g b

g

= 6 0
...
018 + 1 0
...
349 kJ / (mol⋅ o C) (Kopp’s Rule)

p CH COCH (l)
3
3

= 01230 + 18
...


Assume ΔH mix ≅ 0
↓ C 6 H 14 O

↓ CH 3 COCH 3

0
...
1230+18
...
70 ( 0
...
08 g
mol ⋅ °C
102
...
003026 + 9
...
003026 + 9
...
07643 kJ g
45

8
...
08 kJ / mol, ΔH = dC i dT = 14
...
49 kJ / molg + b10
...
68 g K
Mw =
O2

b

g
1
2
16
...
00 = 26
...
20 n =

1000 m 3 1 min 273 K
min

1 kmol

b g = 0
...
4 mol / s

303 K 22
...
8 for ΔH

Q = ΔH = nΔH

Solar energy required =
Area required =
8
...
4 mol 0
...
4 kW heating 1 kW solar energy
0
...
4 kW

= 1631 kW

1631 kW 1000 W 1 m 2
= 1813 m 2
1 kW 900 W

C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O

n fuel =


...

= 103 × 10 4

...
211b - mol O 2
h
T2

Q =ΔH =n ⋅ ∫ C p dT
T1

lb − mol ⎞
⎛
= ⎜1
...
02894 + 0
...
3191 × 10−8 T 2 − 1
...
03 × 10 lb-mol 8
...
593 mol 9
...
97 × 107 Btu/h
h
mol
lb-mol
kJ
4

=
8
...


n O 2 = 125 ⋅

...

= 259
...
4-95(2)-5(3
...
9 mol O 2

N 2 : 3
...
4)=975 mol N 2

Energy balance (enthalpies from Table B
...
845 − 42
...
09 kJ / mol
2

ΔH H 2 O = H (H
ΔH O 2 = H (O

2 O,

2,

ΔH N 2 = H (N

2,

2

450o C)

− H (H

450o C)

− H (O

450o C)

− H (N

2,

2 O,

900o C)


...
32 = −18
...
375 − 28
...
51 kJ / mol

900o C)

= 12
...
19 = −14
...
09) + 205(-18
...
9(-15
...
49)
Q = 21,200 kJ / 100 mol feed

b
...
5: H liq (40 o C) = 167
...
2 kJ / kg;
Q = n ⋅ ΔH = n(2794
...
5) = 21200 ⇒ n = 8
...
22 (cont’d)
c
...
07 kg steam is produced per 100 mol feed
1250 kg steam 01 kmol feed 1 h

...
30 × 10 −3 kmol / s
h
8
...
30 mol feed 1336
...
314 Pa ⋅ m 3
= 3
...
01325 × 10 Pa

d
...
Without the waste heat boiler, the steam required will have to be
produced with additional cost to the plant
...
23

d i

Kopp’s rule: C p
ΔH C10 H12 O2 =
ΔH C6 H6 =

C10 H12 O2

e

j

e

= 10(12) + 12(18) + 2(25) = 386 J mol⋅ o C = 2
...
0 L 1021 g 1 kJ 2
...
0 L 879 g 1 mol
⋅
L 78
...
06255 + 23
...
24 a
...
0 bar

mw kg H2 O(l, sat‘d) @ 5
...
References: H2O (l, 0
...
2)

...
7); H out = 640
...
6)
c
...
36 kJ/mol, ΔH w = (640
...
798 kg

b

g

From Table B
...
0 bar, 300° C = 0
...
008314 m ⋅ kPa (mol ⋅ K) 313 K
= 0
...
798 kg steam 0
...


e
...
798 kg steam
100 mol C3 H8

3

1 mol C3 H8
0
...
400 m 3 steam m 3 C3 H8

1 mol C3 H8

kg steam 0
...


8-7

8
...

5500 L(ST P)/ min CH3 OH (v) 65o C

n 2 mol/min CH3 O H (v) 260o C

n 2 (mol/ min)
mw kg/ min H2 O(l, s at‘d) @ 90o C

mw kg/ min H2 O(v, s at‘d) @ 300o C

Vw 2 (m3 /min )

n2 =

Vw 1 (m3 /min )

5500 L(STP)
1 mol
= 245
...
4 L(STP)

An energy balance on the unit is then written, using Tables B
...
6 for the specific enthalpies
of the outlet and inlet water, respectively, and Table B
...

The only unknown is the flow rate of water, which is calculated to be 1
...

b
...
26 a
...
13
⎟⎜
⎟ ⎜ 2373
...
7 kW
min ⎠ ⎝
kg ⎠ ⎝ 60 sec ⎠ ⎝ 1 kJ/s ⎠
⎝
100 mol/s (30o C)
0
...
100 mol CO/ mol
0
...
020 mol H2 O(v)/ mo l
y 2 mol CO/s
(mol CO/mol)
(0
...
002 /1
...
000 /1
...


U
|
V
|
W

(1) CO balance: (100)(0
...


...
800) = n 2 (1 − y 2 )
1000

...

(100)(0
...
020)(18)
0
...

1000

...
8 )
substance
n out ( mol / s)
n in ( mol / s)
H in (kJ / mol)
(3) Dry air balance: m3

H2O(v)
CO
CO2

10
10
80

0
...
146
0
...
84(0
...
169
0
...
193

H2O(v)
dry air

m3(0
...
002)(1000/18)
m3(1
...
002) (1000/29)

0
...
727

m4y4(1000/18)
m4(1-y4) (1000/29)

0
...
672

8-8

8
...

FG 0
...
847) + m FG 1000 IJ FG 1000 IJ (0
...


...
672)FG 1000 IJ

...
84(0
...
779)G
H 18 K
H 29 K


...
(3)–(5) simultaneously ⇒ m3 = 2
...
70 kg/s, y4 = 0
...
55 kg humid air / s
kg humid air
= 0
...


=
...
0564) kg dry air kmol DA 18 kg H 2 O
⇒

Relative humidity:

29 kg DA 1 kmol H 2 O

0
...


p H 2O
*
p H 2O

e48 Cj
o

= 00878

...
0878)(760 mm Hg)
× 100% = 79
...
71 mm Hg

=

c
...

8
...


y H 2O =

b

p * 57° C
P

b g

28
...
82 mm Hg = 0171 mol H O mol

...
2 mol H 2 O h
0
...
91 kg H O hg

R 89
...
5 mol CO h
mol dry gas
|
1270 − 217
...
3 mol O h
|847
...
8); H 2 O l , 0
...
5
110
...
3
847
...
91
m

bg
H Obl g

H in
18
...
60
19
...
03

3749
83
...
5
110
...
3
847
...
03
17
...
54
11
...
91 + m 3330
---

2

8-9

U n in mol h
|
V H in kJ mol
|
W
U n in kg h
V H in kJ kg
W

2

8
...
62 kg h

in

b
...


b gb

g


...
28 2°C, 15% rel
...
294 mm Hg = 0
...

b0
...
2438 mol air inhaled min

3

min

275 K 10 ml 22
...
067
= 0
...
2438 mol/min 2oC

n2 kmol/min 37oC

1
...
999 dry gas

0
...
9381 dry gas
n1 mol H2O(l)/min 22o C

Mass of dry gas inhaled (and exhaled) =

b0
...
999gmol dry gas

29
...
999gb0
...
9381 n

Dry gas balance:

= 7
...
2596 mols exhaled min

2


...
2438ge1045 × 10 j + n = b0
...
0619g ⇒ n = 0
...
063
0
...
285

bg
H Obl g
2

Q = ΔH =

mout

H in
0
2505
92
...
063
0
...
75
2569
—

m in g min

mH2 O = 18
...
4
ˆ
H dry gas = 1
...
8 J 60 min 24 hr
min

1 hr

8-10

1 day

= 1
...
29 a
...
07 g

e

j
1 mol
= 3054 mol H Obl g
18
...
557 × 10 −3 T kJ / (mol⋅ o C) (fitting the two values in Table B
...


bg

55 L H 2 O l

1000 g
liter

2

b

(C p ) H 2 O = 0
...
0oC)
1284 mol C2H5 OH (l) (To C)
3054 mol H2O(l) (20
...
1031 + 0
...
0754 ) dT

Q = ΔU ≅ ΔH ( liquids ) ⎫
⎪
⎬ ⇒ ⇓ Integrate, solve quadratic equation
Q = 0 ( adiabatic )
⎪
⎭ T=44
...
1
...

3
...

5
...

Heat loss to the outside (not adiabatic)
Heat absorbed by the flask wall & thermometer
Evaporation of the liquids will affect the final temperature
...
Mistakes in measured volumes & initial temperatures of feed liquids
7
...
30 a
...

n1 + n 2 =

mol ⋅ K
1515 L 835 mm Hg
= 26
...
36 L ⋅ mm Hg

n2
p * (30 o C) 31824 mmHg

...
0381 mol H 2 O / mol air
835 mmHg
n1 + n 2
Ptotal
⇒ n1 = 25
...


8-11

8
...
2
14
...
0

dC i
dC i

100

p H O ( l ) dT
2

25
500

100

H2O(l)

n out (mol / s)

25
...
1

+ H vap

p H O ( v ) dT
2

6
...
2gFGH dC i dT IJK + b7
...
2gb14
...

H
K
T

25

p air

z

100

25

p H O(l )
2

vap

100

25

p H O(l )
2

vap

z

T

100

p H O(v )
2

500

100

p H O(v )
2

Integrate, solve : T = 139 o C

b
...
2 ) ∫

139

500

(C )

p air

dT − (1
...

c
...


8-12

8
...
48 mol NH 3 s
h
1 kg 17
...
48 mol NH 3 /s
25°C
n 1 (mol air/s)
T °C

n 2 (mol/s)
0
...
900 air
600°C

Q = –7 kW

NH 3 balance: 8
...
8 mol s

...
900 84
...
3 mol air s

bg

References for enthalphy calculations: NH 3 g , air at 25° C
NH 3

Hin = 0
...
2
3

= 25
...
02894 + 0
...
3191 × 10−8 T 2 − 1
...
4913 × 10−12 T 4 + 0
...
20735 × 10−5 T 2 + 0
...
7248 ) ( kJ mol )

600
ˆ
H out = ∫ C p dT = 17
...
48 mols NH 3 s )( 25
...
3 mols air s )(17
...
48 )( 0
...
3) ( −0
...
1064 × 10−8 T 3 + 0
...
02894T − 0
...
32 a
...
Let M represent methane, and E for ethane
Stack gas (900oC)

100 mol/s
0
...
05 mol E/mol

Furnace

n3
n4
n5
n6

mol
mol
mol
mol

CO2/s
H2O/s
O2/s
N2/s

Stack gas (ToC)

Heat
Exchanger

n3
n4
n5
n6

mol
mol
mol
mol

CO2/s
H2O/s
O2/s
N2/s

air (245oC)

20 % excess air (20oC)

n1 mol O2/s
n2 mol N2/s

n1 mol O2/s
n2 mol N2/s

CH 4 + 2O 2 → CO 2 + 2H 2 O

b g

C 2 H 6 + 7 / 2 O 2 → 2CO 2 + 3H 2 O

8-13

8
...
76 mol air 5 mol E 3
...
76 mol air ⎤
nair = 1
...
21 × 1185 = 249 mol O 2 /s, n2 = 0
...
5 mol O 2
+
= 41
...
649
⎟ = 7879 ( = 7879 kW)
s ⎠⎝
mol air ⎠
s
⎝

Energy balance on stack gas:
6

(

Q = −ΔH = −∑ ni ∫
i =3

−7879 = n3 ∫

T

900

T

900

(C )

p CO
2

( C ) dT )
p i

dT + n4 ∫

T

900

(C )

p H O (v )
2

dT + n5 ∫

T

900

(C )

p O
2

dT + n6 ∫

T

900

(C )

p N
2

dT

Substitute for the heat capacities (Table B
...


350 m 3 (STP)
mol
1000 L 1 h
= 4
...
4 L(STP) m 3 3600 s
4
...
0434
100 mol / s
Q ′ = 0
...
33 a
...


...
4 + 42
...
7 + 40
...
9 = 23100 J mol
3
150 mol 23100 J 1 kW
Q = ΔH = nΔH =
= 3465 kW
s
mol 1000 J / s
ΔH =

600

0

C p dT =

b
...
0334 + 1732 × 10−5 T ° C kJ (mol ⋅° C) ⇒ Q = 150

...
0334 + 1732 × 10−5 T dT = 3474 kW

...

8
...


e

j

ln C p = bT 1 2 + ln a ⇒ C p = a exp bT 1 2 ,
b=

ln C p 2 C p1
T2 − T1

T1 = 7
...
329 ,

= 0
...


−1
...
235|
|
W

p

T2 = 17
...
533

e

= 0
...
0473T 1 2

j

8
...


z

150

e

j

0
...
0473T 1 2 dT =

1800

1

20

30

40
2
200

b0
...
473T jLT
S
MN
0
...
0473

OPU
QV
W

150

= −1730 cal g

1800

DIMENSIONS CP(101), NPTS(2)
WRITE (6, 1)
FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8
...
0 – 1800
...
0
DO 20 J = 1, N
CP (J) = 0
...
0473*SQRT(T))
T = T + DT
SUMI = 0
...
0
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + CP (J)
DH = DT*(CP(1) + 4
...
0 = SUM2 + CP(N))/3
...
4,'bCAL/G')
CONTINUE
STOP
END

Solution: N = 11 ⇒ ΔH = −1731 cal g
N = 101 ⇒ ΔH = −1731 cal g

Simpson's rule with N = 11 thus provides an excellent approximation
8
...


U
|
175 kg 1000 g 1 mol 56
...
W
...
07 g / mol V ⇒ Q = ΔH =
= 2670 kW
min
kg 62
...
9 kJ / mol W

m = 175 kg / min

v

b
...

c
...
The stream goes from state A to state B as shown
in the following phase diagram
...
1 ⇒ Tb = 68
...
85 kJ / mol

8
...


Assume: n - hexane vapor is an ideal gas, i
...
ΔH is not a function of pressure

bC H g
B ΔH
bC H g
6

6

1

14 l, 68
...
74

20
200
68
...
74 o C

0
...
54 kJ / mol

...
85 × 10 −5 T − 23
...
66 × 10 −9 T 3 dT

ΔH2 = 24
...
54 + 24
...
85 = 64
...


ΔH = −64
...


U 200 o C , 2 atm = H − PV

e

j

Assume ideal gas behavior ⇒ PV = RT = 3
...
05 − 3
...


b g

Tb = 100
...
656 kJ mol

e
j
B ΔH
H O el, 100 Cj

8
...
77 kJ mol

25

ΔH2 =

25

C pH 2 Ob v g dT = −169 kJ mol

...
1

B

g

ΔHv 50° C = 3
...
656− 169 = 42
...


Steam table:

( 2547
...
8) kJ

18
...
0 kJ mol

The first value uses physical properties of water at 1 atm (Tables B
...
2, and B
...
5 is for a pressure of 0
...
12 atm)
...
1234 bar to 50oC and 1 atm plus ΔH for water vapor going from
50oC and 1 atm to 50oC and 0
...

8
...
75 m3
2
...
1 mol/s
78
...
765 kJ mol

...
38 (cont’d)
⎯
e
j ⎯→
B ΔH
−Δ H
C H e v, 80
...
1 Cj

C 6 H 6 l, 25o C

C6H 6

o

80
...
23 kJ mol

580

ΔH2 =

298

C pC 6 H 6 b l g dT = −7
...
1

d i

...
1 mol / sgb −115
...
7 kJ / mol

...
39

U
V
W

b

B

g

35° C
P ∗ 25° C
176
...

= 015

...
0347 mol CCl 4 mol
15% relative saturation
1 atm
760 mm Hg
( ΔH v ) CCl 4

Table B
...
0

10 mol 0
...
0 kJ
= 10
...
40 g CCl 4
g carbon

8
...


b

g

1 mol CCl 4
153
...
0347 mol CCl 4

g

CO 2 g, 20° C → CO 2 s, − 78
...
4

20

1 min
= 45
...
4° Cg

dC i

p CO g
2

In the absence of better heat capacity data; we use the formula given in Table B
...

−78
...
03611 + 4
...
887 × 10 −8 T 2 + 7
...
184 × 10 −3 kJ

mol

1 cal

Q = ΔH = nΔH =

= −28
...
66 kJ removed

h

1 kg

44
...
23 × 10 7 cal hr or 72
...
According to Figure 6
...
4

−56

p CO (l) dT
2

−78
...


C p = a + bT

8
...


U
| ⇒ C bJ mol ⋅ Kg = 4512 + 0
...

V
|
a = 53
...
01765gb500g = 4512 |

...
01765T gdT OP J + 30
...

Q mol mol
b=

53
...
41
= 0
...
44 × 10

Q = ΔU = n

b
...
44 g

1 kg
t=

c
...
42

2
...
85 × 3000 kJ 10 3 J

= 2
...
98 kJ mol , Tb = 136
...
4 K , Pc = 37
...
7 K (from Table B
...
088Tb = 0
...
4 K = 36
...

Chen's rule:

LM
NM

g

FG T IJ − 0
...
0297 log P OP
HT K
QP = 35
...
7% error)
ΔH ≈
FT I
107 − G J

...
7 − 373
...
2 kJ mol
Watson’s correlation : ΔH b100° Cg ≈ 35
...
7 − 409
...
0331

v

b

c

10

c

b
c

0
...
43

b g b g
Trouton's Rule ⇒ ΔH b200° Cg = 0
...
2g = 41
...
012 + 12 0
...
033 = 0
...
333(200 − 25) mol
p

v

25

8-18

+ 41
...
44 a
...

− 220
...

6
...
6 − 299
...
765G

...
6 − 3531 K

...
38

= 33
...
Antoine equation: Tb 50 mm Hg = 118° C ; Tb 150 mm Hg = 35
...


Clausius-Clapeyron: ln p = −
ΔH v = −0
...


R
|
S
|
T

b

Δ Hv (80
...
1°C)

zd
zd

i

Δ H2

C6 H 6 (v , 80
...
50 kJ mol

26
...
1°C)

Δ H1

80
...
3 kJ mol
mol ⋅ K 1 308
...
0 K

C 6 H 6 ( l , 26
...
1

l

i

C p dT = −4
...
1

b

v

g

ΔHv 261° C = 7
...
765 − 4
...
4 kJ mol

...
45 a
...
3oC
...
1 is 49
...

b
...


Ideal gas equation of state
nf =

1550 L 273 K
s

1 mol

423 K 22
...
66 mol C 5 H 10 (v) / s

55% condensation: n l = 0
...
66 mol / s) = 24
...
66 − 24
...
10 mol C 5 H 10 (v) / s
Reference: C5H10(l) at 49
...
56

0

C5H10 (v)

44
...
10

Hv

H i = ΔH v +

z

Ti

49
...
45 (cont’d)

Substituting for ΔH v from Table B
...
2
⇒ H f = 38
...
30 kJ / mol
Energy balance: Q =

8
...


∑n

out H out

−

∑n

in H in

= −116 × 10 3 kJ / s = −116 × 10 3 kW

...


Basis: 100 mol humid air fed
n 2 (mol), 20o C, 1 atm

Q(kJ)

y 2 (mol H2 O/ mol), sat’d
1-y 2 (mol d ry air/ mo l)

100 mol
y 1 (mol H2 O/ mol)
1-y 1 (mol d ry air/ mo l)
50o C, 1 at m, 2o superheat

n 3 (mol H2 O(l))

There are five unknowns (n2, n3, y1, y2, Q) and five equations (two independent material
balances, 2oC superheat, saturation at outlet, energy balance)
...

b
...


3

g

b

References: Air 25° C , H 2 O l, 20° C
Substance

n in

Air

100 ⋅ 1 − y1

H 2O v

H2 =
=

zd i z
z d i
z
z
zd i
z d i
50

25

Cp

100

20

100

20
50

H2 =

20

25

H 2 O(l)

25

n2 ⋅ 1 − y 2

H2
−

g

b

g

H3

n in mol

n2 ⋅ y 2

H4

H in kJ mol

n3

0

0
...
4147 × 10 −5 T + 0
...


e

j

dT + ΔHv 100o C +

z

50

100

dC i

p H O(v) dT
2

0
...
656 +

Cp

100

20

Cp

50

H1

b

H out

0
...
688 × 10 −5 T + 0
...


100

H3 =

air

dT =

n out

−

2

H in

100 ⋅ y1

bg
H Obl g

H1 =

g

air

Cp

dT

H 2 O(l)

e

j

dT + ΔHv 100o C +

z

20

100

dC i

8-20

p H O(v) dT
2

8
...


Q = ΔH = ∑ ni Hi − ∑ ni Hi
out

Vair =

in

100 mol 8
...


∑

∑

ni H i −
ni H i
Q
out
in
⇒
=
Vair 100 mol 8
...


d
...
71 mm Hg = 0110 mol H O mol

...
535 mm Hg = 0
...
023g + n ⇒ n

...


dry air balance: 100 1 − 0110 = n2 1 − 0
...


...
90 mol H 2 O 0
...


Q = ΔH = ∑ ni Hi − ∑ ni Hi = −480
...
314 Pa ⋅ m 3
323 K
= 2
...


⇒
⇒

0160 kg H 2 O condensed

...
65 m air fed
−480
...
0604 kg H 2 O condensed / m 3 air fed

= −181 kJ / m 3 air fed

2
...

f
...

Q=

−181 kJ 250 m 3 air fed 1 h 1 kW
= −12
...
47 Basis:

226 m 3

10 3 mol

273 K

b g

309 K 22
...
DA = Dry air

Q( kJ / min)

8908 mol / min
y 0 [ mol H 2 O(v) / mol]
(1- y 0 )(mol DA / mol)

n1 ( mol / min)
y1 [ mol H 2 O(v) / mol]
(1- y1 )(mol DA / mol)

36 o C, 1 atm, 98% rel
...


10 o C, 1 atm, saturated

n 2 [ mol H 2 O(l) / min], 10 o C

a
...


B

Table B
...
98(760
...
0575 mol H O(v) mol
mm Hg
Outlet air: y = p (10 C) / P = b9
...
0121 mol H O(v) mol
Air balance: b1 − 0
...
0121gn ⇒ n = 8499 mol / min
F mol IJ = 0
...
0575G 8908
H min K
min
References: H Obl, triple point g, air b77° Fg

b
...
98 p w

∗

0

2

o

2

1

1

2

1

2

2

2

2

ˆ
ˆ
Substance nin
H in
nout
H out
Air
8396 0
...
4352 n in mol min

...
2
103
45
...
741

Air: H from Table B
...
5 × (0
...
50 ×104 kJ 60 min 9
...
001 kJ
−12000 Btu h

8-22

8
...
7 m 3 outlet gas / h 3 atm

Basis:

1 kmol

b g

1 atm 22
...
0 kmol / h

100 kmol/h @ 0o C, 3 atm
yout (kmol C6 H14 (v)/kmol), sat'd
(1 − yout )(kmol N 2 /kmol)

o

n1 (kmol/h) @ 75 C, 3 atm
yin (kmol C6 H14 (v)/kmol), 90% sat'd
(1 − yin )(kmol N 2 /kmol)

n2 (kmol C6 H14 (l)/h), 0o C

Antoine:
∗
log pv = 6
...
817
224
...
24 mm Hg, pv ( 75°C ) = 920
...
24 = 0
...
90 p b75° Cg b0
...
44g
kmol C H
y =
=
= 0
...
363g = 100b1 − 0
...
9 kmol h
C H balance: b153
...
363g = b100gb0
...


...
363 × 153
...
5%
y out =

∗
p v 0° C

6

14

∗
v

6

14

in

2

6

1

1

14

2

2

6

14

References: N2(25oC), n-C6H14(l, 0oC)
Substance

n in

H in

N2

98000


...
726

55800 44
...
33

53800

n in mol h

0
...
7

g

H out

N 2 : H = C p T − 25 , n − C 6 H 14 (v): H =

H in kJ mol

b g

C p dT + ΔH v 68
...
7

Energy balance: Q = Δ H = ( −2
...
49 Let A denote acetone
...
2 kW

Q( kW)

n1 (mol / s) @ − 18 o C, 5 atm
y1 [mol A(v) / mol], sat' d
(1 − y1 )( mol air / mol)

142 L / s @ 150 o C, 1
...
Degree of freedom analysis:

b
...
973 − 4
...
05 g]
[(3
...
2

(7)

nout

Tab le B
...
2

H air (T ) from Table B
...
2 kJ / s)

8-24

8
...


(2) ⇒ y1 = 6
...
32 mol feed gas/s

(3) ⇒ y0 = 0
...
57 mol outlet gas/s

(5) ⇒ n2 = 0
...
1 kJ/mol, H A1 = 34
...
666 kJ/mol, H a1 = −1
...
1 kW

8
...
Feed:

3 m π (35)2 cm2

1 m2
4

s

273 K
2

10 cm

850 torr

1 kmol

103 mol

3

(273+40)K 760 torr 22
...
3

mol
s

Assume outlet gas is at 850 mm Hg
...
3 mol/s @ 40 C, 850 mm Hg
yo (mol C6 H14 (v)/mol)
(1 − yo )(mol air/mol)

n1 (mol C6 H14 (l)/s), T (o C)

Tdp = 25o C

60% of hexane in feed

Degree-of-freedom analysis
6 unknowns ( y0 , n1 , n2 , n3 , T , Q )
– 2 independent material balances
– 2 Raoult’s law (for feed and outlet gases)
– 1 60% recovery equation
– 1 energy balance
0 degrees of freedom ⇒ All unknowns can be calculated
...
Let H = C6H14

(T )

dp feed

Antoine equation, Table B
...
600

P

=

151 mm Hg
= 0
...
3)( 0
...
37 mol H ( l ) s

Hexane balance: (0
...
3) = 5
...
58 mol H ( v ) s

8-25

8
...
3)(1 − 0
...
3 mol air s
Mole fraction of hexane in outlet gas:

pH ( T )
n2
3
...
8 mm Hg
n2 + n3 ( 3
...
3) 850 mm Hg
Table B
...
8 mm Hg ⎯⎯⎯⎯ T = 7
...
8° C , air (25°C)
Substance

n in

C6 H14 ( v )
C6 H14 ( l )
Air

C6 H14 ( v ) : H =

n out

8
...
5

3
...
7

—

—

5
...
3

0
...
3

–0
...
74

b

g

T

C pl dT + Δ H v 68
...
8

z

n in mol/s
H in kJ/mol

C pv dT ,

68
...
2
Δ H v from Table B
...
8
Energy balance: Q = ΔH =

∑n H −∑n H
i

i

i

out

c
...
0 m / s
2

8-26

−1 kJ s

= 257 kW

8
...
0 atm
0
...
500 mol H(l) / mol

Q( kJ / s)
n l ( mol / min) @ 65o C, P0 (atm)
0
...
59 mol H(l) / mol

a
...
4) ⇒ p * (65 o C) = 1851 mm Hg, p * (65 o C) = 675 mm Hg
P
H
Raoult' s law for pentane and hexane
0
...
656 mol P(v) / mol

⇒

0
...
52 atm)

Total mole balance: 100 mol = n v + n l
Pentane balance: 50 mole P = 0
...
410n l
Ideal gas equation of state: Vv =

n v RT 36
...
6 mol vapor / s
n l = 63
...
08206 L ⋅ atm
mol ⋅ K

P(v)

−

P(l)

50

H(v)

−

H(l)

50

H in

n out

−

H out

24
...
33

2
...
0
−

0

n in mol s
H in kJ / mol

12
...
05

3
...
4

Vapor: H (T ) =

z
z

Tb
o

65 C

0

C pl dT + ΔH v (Tb ) +

z

T

Tb

C pv dT

T

Liquid: H (T) =

65o C

C pl dT

Tb and ΔH v from Table B
...
2

Energy balance:


...
656 × 36
...
480
50
...
52 a
...
735 mol B/ mo l
0
...
500 mol B/ mo l
0
...
425 mol B/ mo l
0
...
500) = n 2 (0
...
425)
n 3 = 1001 mol / s

References: B(l, 25oC), T(l, 25oC)
Substance

n in (mol / s) H in ( kJ / mol) n out (mol / s) H out ( kJ / mol)

B(l)
B(v)
T(l)
T(v)

660
-660
--

0
-0
--

425
234
576
85

ˆ
ˆ
Q = ∑ ni H i − ∑ ni H i = 2
...


9
...
91
11
...
06

in

e

j

e

j

*
Antoine equation (Table B
...
9 torr
B

Raoult' s law
Benzene:
Toluene:

b0
...
735g P ⇒ P = 680 torr U ⇒ P ≠ P'
|
V
0
...
9g = b0
...


Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the
samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is
invalid at the system conditions (not likely)
...
(8
...
1 kJ mol

...
(8
...
050gb52 + 273g = 16
...
53 Kopp’s rule (Table B
...
5 + 12 9
...
4 m STP

10 3 mol

1h

1 kmol

3600 s

= 2
...


b

g

b

g
Obs, 25° Cg

b

C 5 H 12 O v , 113° C → C 5 H 12 O l , 113° C → C 5 H 12 O v , 52° C

b

g

→ C 5 H 12 O s, 52° C → C 5 H 12

8-28

g

8
...
2
− b301gb61g + b170gb27g
×
= −813 kJ mol

...
1

kJ
mol

3

Required heat transfer: Q = ΔH = nΔH =

2
...

s

1 kW

mol

1 kJ s

= −167 kW

8
...


95 kg dry film

0
...
5 kg acetone exit in gas phase

95 kg DF
0
...
5 kg C3 H 6 O( v) (40% sat'd)
Ta2 = 49°C, 1
...
01 atm

*
Antoine equation (Table B
...


1 kmol

⇒y=

10 3 mol

58
...
5 kg C 3 H 6 O

kmol

b

bg

= 77
...
6 mol 22
...
40 59118 mm Hg

...
5
=
⇒ n1 =
77
...


= 405

...
1

0

8
...
5

32
...
6

0
...
6

Air

25

z

86

H A(v) =

p l

35

p air dT

zd

49
v

Cp

i dT ,
v

H out

d
0
...
33 T f 2 − 35

b

f2

i

i

i

p air dT

=

out

g

z

25

c
...


...
4 T f 2 − 35 + 111(T f 2 − 35) + 2623
...
5 T f 2 − 35 + 2623
...


d

i

= 2
...
8° C

8-29

Ta1

25

dC i

p air dT

=0

8
...
T f 2 = 34° C ⇒ Ta1 = 506° C , T f 2 = 36° C ⇒ Ta1 = 552° C
e
...
55

In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed
phase drops
...
If the feed air temperature is above about 530 °C, enough heat is
transferred to keep the film above its inlet temperature of 35 °C; otherwise, the film temperature
drops
...
6
...
Basis:

3
...
357 lb ⋅ mole h C 3 H 8
h
359 SCF

8
...
357 lb-mole C3H8(l)/h
200 psia, 100oF

Q

m(lb - mole H 2 O(l) / h
70oF

m(lb - mole H 2 O(l) / h
85oF

The outlet water temperature is 85oF
...

b
...
1

B
Btu
Q = ΔH = − nΔH v = 8
...
77 kJ 0
...
593 mol = −6
...

Q = ΔH = mC p ΔT ⇒ m =

6
...
0 Btu 15 °F
h

8
...
200 kg solids/kg
0
...
350 kg solids/kg
0
...
6 bar, sat' d

a
...
35m3

b


...
65 5714

m1 [ kg H 2 O(l) / h], 1
...


bg

⇒ m2 = 428
...
56 (cont’d)

References: Solids (0
...
01oC)
Substance
min
mout
H

H out

in

bg
H Obv g

200
800
—

62
...
7
—

200
571
...
6

209
...
1
2676

H 2 O , 1
...
2

m1

m ( kg h ) H H 2 O from steam tables

475
...
B
...
315 × 106 − 2221m1 = 0 ⇒ m1 = 592 kg steam h

in

b
...
0 − 428
...


The cost of compressing and reheating the steam vs
...


8
...


kg h additional steam

A = acetone, B = acetic acid, C = acetic anhydride
Q c (kJ/h)
2 n 1 (kg A(v )/h)
329 K

15000 kg/h
0
...
27 B
0
...


b gb gb

g

n 2 = 0
...
46 15,000 kg h = 69 kg A h

b gb g
Acetic anhydride balance: n = b0
...
46gb15,000g = n + 69 ⇒ n = 6831 kg h `
Acetic acid balance: n 3 = 0
...
8% acetone
Bottoms product: 69 + 4050 + 4050 kg h =
49
...
6% acetic anhydride

b

g

b
...
57 (cont’d)

b

g

b

g

b

C 3 H 6 O v , 329 K → C 3 H 6 O l , 329 K → C 3 H 6 O l , 303 K

z

g

303

b

g C dT = −520
...
3gb−26g = −580
...
4 kJ = −7
...


6

h

kg

Overall process energy balance
Reference states: A(l), B(l), C(l) at 348 K (All H m = 0 )
Substance
n in H
n out
H
in

out

b
g — 0 6831 –103
...
0
69
0
—
A bl, 398 Kg
4050 109
...
1 g 1 kg 10 3 J

= 2
...
93 × 10
i

i

6

out


...
)
d
...
6)
Qr = n H 2 O ΔH v ⇒ n H 2 O =

813 × 10 6 kJ h

...
58 Basis: 5000 kg seawater/h
a
...
2 bars
2610 kJ/kg

n 2 (kg H 2 O(v )/h @ 0
...
6 bars)
0
...
945 H 2 O(l )
360 kJ/kg

5000 kg/h @ 300 K
0
...
965 H 2O( l)
113
...
2 bars)
x (kg S/kg)
(1 – x) (kg H2 O(l )/hr)
252 kJ/kg
n 2 (kg H 2 O(l )/h @ 0
...
S balance on 1st effect: 0
...
055n1 ⇒ n1 = 3182 kg h

Mass balance on 1st effect: 5000 = 3182 + n 2 ⇒ n 2 = 1818 kg h

8-32

j

+ 2
...
58 (cont’d)

Energy balance on 1st effect:

b gb

g b gb g b gb
= 2534 kg H Obv g h

g b gb g

ΔH = 0 ⇒ n 2 2654 + n1 360 + n5 605 − 2738 − 5000 1131 = 0

...


2

Mass balance on 2nd effect: 3182 = n 3 + n 4 (1)

bΔH = 0g
bn gb2610g + bn gb252g + bn gb360 − 2654g − bn gb360g = 0
E n = 3182, n = 1818

Energy balance on 2nd effect:
4

3

2

1

1

2

5
...
035 5000 = 1267 x ⇒ x = 0138 kg salt kg

...
The entering steam must be at a higher temperature (and hence a higher saturation pressure) than
that of the liquid to be vaporized for the required heat transfer to take place
...


n 5 (kg H 2 O(v )/h)
2738 kJ/kg

3733 kg/h H 2 O(v ) @ 0
...
2 bar
252 kJ/kg

5000 kg/h
0
...
965 H 2 O(l )
113
...

b3733gb2610g + b1267gb252g + n b605 − 2738g − b5000gb1131g = 0
⇒ n = 4452 kg H Obv g h

Energy balance:

5

5

2

Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or
the construction and maintenance of the second effect?

8-33

8
...


b0
...
30
Fresh water produced: n L 7 − n L1 = 5000 − 583 = 4417 kg fresh water h

b
...

c
...


Fresh steam
Effect 1
Effect 2
Effect 3
Effect 4
Effect 5
Effect 6
Effect (7)

P
(bar)
2
...
9
0
...
5
0
...
2
0
...
0

T
(K)
393
...
9
363
...
5
342
...
3
319
...
0

nL
(kg/h)
--584
1518
2407
3216
3950
4562
5000

8-34

xL
--0
...
1153
0
...
0544
0
...
0384
0
...
7
405
...
8
340
...
3
251
...
8
113
...
3
2670
...
1
2646
...
4
2609
...
8
---

8
...


dC i = dC i
p v

p l

b g ≈ dC i

= 20 cal (mol⋅° C) ; Cv

p v

v

b

− R ≈ 10 − 2

cal
g mol⋅° C = 8 cal (mol⋅° C)

b
...
00 L@ 93 C, 1 atm

n2 [mol A(v)]
85oC, P(atm)

n1 (mol A(l)

n3 [mol A(l)]
o

85oC, P(atm)

0
...
00 L

n1 =

70
...

= 0100 mol N 2
273 + 93 K 22
...
90 g 1 mol

...


i

=0

in

b g b gb

g

References: N 2 g , A l 85° C, 1 atm

Substance n in U in n out U out
010 39
...


...

U in cal mol
Av
n 2 20050
−
−

bg
bg

b
g
b
g
b g
Abv , 85° Cg: U
= 20b90 − 85g + 20,000 + 10b85 − 90g = 20050 cal mol
ΔU = 0 ⇒ n b20050g − b010gb39
...
012 mol A evaporate

...

A l , 93° C and N 2 g , 93° C : U = Cv 93 − 85
A( v )

v1

v1

0
...
51 g evaporate

d
...
112 mol
3
...
08206 L ⋅ atm
= 1097 atm

...
012 mol 1097 atm

...
117 atm
0
...
3 mmHgg

8
...
4553 − 3
...
600 kg ⇒ bSGg
2
...
600

g

ii) Expt 2 ⇒ Mass of gas = 3
...
2551 kg = 0
...
0 g

Moles of gas =

b763 − 500gmm Hg

2
...
0232 mol
22
...
0232 molg = 86 g mol

Molecular weight = 2
...
1 ⇒ n =

b liquid g

2
...
600 g 1 mol
= 14 mol
cm 3
86 g

Energy balance: The data show that Cv is independent of temperature
Q = ΔU = nCv ΔT

b g

⇒ Cv

liquid

b g

⇒ Cv

liquid

=

Q
800 J
=
= 24 J mol ⋅ K@284
...
4 K

b
gb g
800 J
=
b14 molsgb2
...
2 K

≡ 24 J mol ⋅ K

bg

Expt
...
0232 mol from ii

b vapor g

Cv = a + bT ⇒ Q = 0
...
0232 a (T2 − T1 ) + (T22 − T12 )
2

b
130 J = 0
...
9 - 363
...
9 2 − 363
...

2
b
1
...
0232 a(492
...
0) + (492
...
02 )
2

b g

⇒ Cv

vapor

bg

OP U
Q | ⇒ a = −4
...
05052
|
Q|
W

(J / mol ⋅ K) = −4
...
05052T K

iv) Liquid: C p ≈ Cv ≡ 24 J mol ⋅ K
Vapor: Assuming ideal gas behavior, C p = Cv + R = Cv + 8
...
245 + 0
...
3 ⇒ T = 315K , p ∗ = 763 − 564 mm Hg = 199 mm Hg
∗

T = 334 K , p = 401 mm Hg
T = 354 K , p ∗ = 761 mm Hg
T = 379 K , p ∗ = 1521 mm Hg

8-36

OP
Q

8
...
1 T (linear scale); straight line fit yields
−3770
ln p ∗ =
+ 17
...

T K

b

bg

g

b g

1 17
...
824 × 10 −3 K −1 ⇒ Tb = 354 K
A v Tb
3770
Part

vi) p ∗ = 760 mm Hg ⇒

vii)

ΔH v
= 3770 K ⇒ ΔH v = 3770 K 8
...
5 L feed 273 K
1 mol
= 0
...
4l STP
Let A denote the drug

b g

(b) Basis:


...
0836 mol/s @ 510 K
0
...
80 N 2

n 1 [mol A(v)/s]

...


Q(kW)

n 3 (mols A(l )/s), 90% of A in feed
T(K)

b

gb
g
90% condensation: n = b0
...
200 × 0
...
01505 mol Abl g s
n = b0100gb0
...
0836g = 167 × 10 mol Abv g s

...

N 2 balance: n2 = 0
...
0836 mol s = 0
...

167 × 10 −3 mol
(760 mm Hg) = 18
...
0686 mol
n1 + n2

bg

g

E Part (a) - (v)
1 17
...
5g
=
= 3
...
0669

7286

0
...
0167 37575 1
...
01505

8-37

0

8
...
8

B

= [6
...
05)] kJ / mol = 7
...
245 + 0
...
62 a
...
M
...
72 H 2O( s)
0
...
M
...
M
...
96 0
...
56 kg H 2 O v min

b
g
bg
m = 0
...
72 × 50 kg ming = 144 kg H O bl g min

...
28gb50g = 14
...
M
...
0
0
14
...
0 −390
−
−

...
56 2599
−
−

g

kJ
b g 1
...


8
...
01 kJ

−26

ΔH = − ΔH m 0° C + C p dT =

1 mol 10 3 g 2
...
02 g 1 kg + kg⋅° C

mol

A

0

−26° C

= −390 kJ kg

Table B
...
0754 kJ
50
ΔH = C p dT = mol ° C

1 mol 1000 g
18
...
2

b

g

b

g

b

g

b

g

b

H 2 O v , 60° C : H 2 O l , 0° C → H 2 O l , 100° C → H 2 O v , 100° C → H 2 O v , 60° C

0
...
656

Ad

Table B
...
1 ΔH v

46
...
02 g

1 kg

zd
60

100

i

Cp

i

H 2 O(v)

A

g

dT

Table B
...
06 × 10 5 kJ 1 min
min

in

60 s

1 kW
1 kJ s

= 1760 kW

8
...
S = solids in juice
...


...
12 solids(S)
0
...
(kg/h)

...
45 kg S/kg

...
55 kg W/kg
(1 – x 2) (kg W/kg)
0
...

0
...
45 S, 0
...
(kg/h), 0°Ckg W( l )/kg
m3
separator
0
...
45 kg W( l )/kg

(a) 10% ice in slurry ⇒

20000 10
=
⇒ m4 = 180000 kg h concentrate leaving freezer
90
m4

U
V
W

m1 = 27273 kg h feed
Overall S balance: 012m1 = 0
...

⇒
m5 = 7273 kg h concentrate product
Overall mass balance: m1 = m5 + 20000
Mass balance on filter: 20000 + m4 + m5 + 20000 + m6

⇒

m4 =180000
m5 = 7273

m6 = 172730 kg h recycle

Mass balance on mixing point:
27273 + 172730 = m2 ⇒ m2 = 2
...
63 (Cont’d)

S balance on mixing point:
012 27273 + 0
...
000 × 105 X 2 ⇒ X 2 ⋅ 100% = 40
...


b gb

g b gb

g

(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:
fresh feed and recycle streams; output; slurry leaving freezer)

bg

Refs: S, H 2 O l at −7° C
substance

min

Hin

mout

Hout

12% soln

27273

108

−

−

45% soln 172730

28

180000

0

−

20000

−337

bg

−

H 2O s

b g
H b kJ kgg
m kg h

bg
b g kJ kg
H = − ΔH b − T ° Cg ≈ − ΔH b0° Cg

Solutions: H T = 4
...
0095 kJ mol ⇒ −337 kJ kg

D Table B
...
B
...

h

in

1h

1 kW

3600 s 1 kJ s

d

= −4030 kW

8
...
B=n-butane, I=iso-butane, hf=heating fluid
...
62 kJ / kg⋅ o C
p hf
24
...
35 kmol B(l)/h

i

24
...
35 kmol B(l)/h

Q( kW)
m (kg HF / h), T( o C)

m (kg HF / h), 215 o C

From the Cox chart (Figure 6
...
5 psi

b
...
01325 bar IJ = 1
...
696 psi K

i

d

Hv
H1
B l, 10 o C ⎯Δ⎯→ B v, 10 o C ⎯Δ⎯→ B v, 180 o C
⎯
⎯

d

i

d

i

d

Hv
H2
I l, 10 o C ⎯Δ⎯→ I v, 10 o C ⎯Δ⎯→ I v, 180 o C
⎯
⎯

i

i

Assume temperature remains constant during vaporization
...
e
...


8-40

8
...
21
-41
...
21 kJ / mol

...
21g − 15825b41
...

c
...
015 × 10 6 kJ / h = mhf 2
...



...
62 kJ / dkg⋅ Ci b215 − 45g C = 1131 × 10
o

o

6

kJ / h

Heat transfer rate = 1131 × 10 6 − 1015 × 10 6 = 116 × 10 5 kJ / h

...


...
The heat loss leads to a pumping cost for the additional heating fluid and a greater
heating cost to raise the additional fluid back to 215oC
...
Adding the insulation reduces the costs given in part (e)
...
The final decision would depend on how long it would take for
the savings to make up for the cost of buying and installing the insulation
...
65 (a) Basis: 100 g of mixture, SGBenzene=0
...
866

50 g
50 g
+
= (0
...
542) mol = 1183 mol

...
11 g / mol 92
...
6 cm3
3
0
...
866 g / cm3

ntotal =

Vtotal

dx i
f

C6 H 6

=

0
...
541 mol C 6 H 6 mol
1
...
5 m3 106 cm3
h

1 m

3

1183 mol mixture

...
6 cm mixture 3600 s

= 9319 mol / s

...
1-1)

b

gb g b

gb g

∗
∗
Raoult' s law: ptot = x C6 H 6 pC6 H 6 + x C7 H 8 pC7 H 8 = 0
...
459 407

=

739
...
973 atm ⇒ P0 > 0
...
65 (cont’d)
∗
∗
(b) T = 75° C ⇒ pC6 H 6 = 648 mm Hg , pC7 H 8 = 244 mm Hg (from Table 6
...
439 648 + 0
...
554 atm
yC 6 H 6 =

bg

284 mm Hg
= 0
...
675 C 6H 6 (v )
0
...
325 C 7H 8 (v )
n L (mol/s), 75°C
0
...
541 C7H8 (l )

93
...
541 C 6H 6( l )
0
...
439n V n
W

v

Mole balance: 9319 = nv + n L

...
541 9319 = 0
...


bg

L

= 40
...
92 mol liquid s

bg

(c) Reference states: C 6 H 6 l , C 6 H 6 l at 75° C

Substance

nin

H in

nout

H out


...
18 310 n in mol s
0
C H bl g 50
...
16 23
...
09 35
...
78 2
...
69
C H bl , 90° Cg: H = b0
...
16 kJ mol
C H bl , 90° Cg: H = b0
...
64 kJ mol
C H bv , 75° Cg: H = b0144gb801 − 75g + 30
...
074 + 0
...


...
1°C

80
...


b

g

b

gb

g

C 7 H 8 v , 75° C : H = 0176 110
...
47 +

...
6

0
...
380 × 10 −3 T dT

= 35
...

(e) Heat is required to vaporize a liquid and heat is lost from any vessel for which T>Tambient
...
To run the experiment
isothermally, a greater heating rate is required
...
66 a
...

Tref(deg
...
84471
6
...
793
1175
...
541
224
...
C)
P(mm Hg)
HAF(kJ/mol)
HBF(kJ/mol)

0
...
6
18
...
5
110
1000
16
...
4

0
...
4
27
...
C)
pA*(mm Hg)
pB*(mm Hg)
x
y
nL(mol/s)
nV(mol/s)
HAL(kJ/mol)
HBL(kJ/mol)
HAV(kJ/mol)
HBV(kJ/mol)
DH(kJ/s)

51
...
395
0
...
598
0
...
2
5
...
4
42
...
00

60
...
412
0
...
648
0
...
8
7
...
5
43
...
00

al
0
...
216

62
...
349
0
...
394
0
...
3
8
...
8
44
...
00

8-43

av
0
...
137

bv
3
...
09E-04

Tbp
36
...
74

DHv
25
...
85

8
...

C*

C*
1
C*

C*

C*

C*

2

20
25
3
30

PROGRAM FOR PROBLEM 8
...
4)
READ (5, 1) TRA, TRB
ARBITRARY REFERENCE TEMPERATURES (DEG
...
) FOR A AND B
READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2
READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2
CP(LIQ, KS/MBL-DEG
...
), NORMAL BOILING POINT (DEG
...
, KJ/MOL-DEG
...
C)
READ (5, 1) XF, TF, P
MOLE FRACTION OF A IN FEED, FEED TEMP
...
C), EVAPORATOR
PRESSURE (MMHG)
WRITE (6, 2) TF, XF, P
FORMAT (1H0, 'FEEDbATb', F6
...
CbCONTAINSb', F6
...
4, 'bMMbHG'/)
ITER = 0
DT = 0
...
0 – XF)*HBF
F2 = CAL*(TBPA – TRA) + DHVA – CAV1*TBPA – 0
...
5*CBV2*TBPB**2
T = TF
INTER = ITER + 1
IF(ITER – 200) 30, 30, 25
WRITE (6, 3)
FORMAT (1H0, 'NO CONVERGENCE')
STOP
PAV = 10
...
0** (A2 – B2/(T + C2))
XL = (P – PBV)/(PAV – PBV)
XV = XL*PAV/P
NL = (XV – XF)/(XV – XL)
NV = 1
...
LE
...
OR
...
GE
...
0
...
NL
...
0
...
OR
...
GE
...
0) GO TO 45
HAL = CAL*(T – TRA)
HBL = CBL*(T – TRB)
HAV = F2 + CAV1*T + 0
...
5*CBV2*T**2

8-44

8
...
0 – XL)*HBL) + NV*(XV*HAV + (1
...
1, 3X' NLb=', F7
...
4, 3X'DELHb
=',* E11
...
1, 3X' XL, HAL, HBLb=', F7
...
4,3X' XV, HAV, HBVb=', F7
...
4/)
IF (DELH) 50, 50, 40
40
DHOLD = DELH
TOLD = T
45
T = T – DT
GO TO 20
50
T = (T*DHOLD – TOLD*DELH)/(DHOLD – DELH)
PAV = 10
...
0**(A2 – B2/(T + C2))
XL = (P – PBV)/(PAV – PBV)
XV = XL * PAV/P
NL = (XV – XF)/(XV – XL)
NV = 1
...

*1//3X' LIQUIDbPRODUCTb--', F6
...
3,
'bMOLEbA/
*MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6
...
3,
*'bMOLEbA/MOLEb TOTAL')
STOP
END
$DATA (Fields of 10 Columns)
Solution:
Tevaportor = 52
...
552 mol, x C5H12
nv = 0
...
383 mol C 5 H 12 mol liquid

vapor

= 0
...
67 Basis:

2500 kmol product 1 kmol condensate
h


...


m1 , (kg/h) at T1
1090 kmol/h C 3 H 8 (v )
7520 kmol/h i -C 4H10 (v )
1390 kmol/h n -C 4H10 (v )
saturated vapor at Tf, P

1090 kmol/h C 3 H 8 ( l )
7520 kmol/h i -C 4H10 ( l )
P (mm Hg) 1390 kmol/h -C H ( )
n 4 10 l
T out


...


Antoine constants
C3H 8
i − C 4 H 10
n − C 4 H 10

A
7
...
78866
6
...
65
899
...
453

C
283
...
942
239
...

P = ∑ xi pi* ( 0°C ) = 0
...
752 (1176 mm Hg ) + 0
...
T f = Tdp ⇒ f (T f ) = 1 − P ∑
i

bg

yi

p (T f
*
i

)

= 0 trial & error to find T f ⇒ T f = 5
...
1
substance
C3H 8
i − C 4 H 10

n − C 4 H 10
Refrigerant

nin

Hin

1090 19110
7520 21740
1390 22760

m1

0

nout
1090
7520
1390

0
0
0

m1

151

↓

Hout
n (kmol/h)
H (kJ/kmol)

U H bvapor g = ΔH b0° Cg +
|
V C dT bTable B
...
95

p

0

m (kg/h)
H (kJ/kmol)

U H = ΔH
V
W

v

E
...
:
ΔH = ∑ ni Hi − ∑ ni Hi = 0 ⇒ 151m1 − 2
...

out

in

8-46

8
...
109 (11,877 ) + 0
...
139 ( 2831) = 4667 mm Hg
i

f (T f ) = 1 − P ∑

bg

i

T +E
yi
= 0 ⇒ T f = 45
...

ΔH = 0 ⇒ 37
...
17 × 10 = 0 ⇒ m1 = 5
...

8
...
1 bars, sat'd

n 3 (mol O 2 )
3
...
199 mol HCHO/mol
n 4 (mol H 2 O( v))
0
...
303 mol N 2/mol
n 1 (mol CH3 OH(l )) n 2 (mol CH3 OH(l ))
0
...
050 mol H 2/mol
m w1 (kg H 2 O(l )) m w2 (kg H 2 O(l ))
n 8 (mol CH3 OH(l ))
0
...
1 bars, sat'd
30°C
Q (kJ)
CH 3 OH( l ), 1 atm, sat'd

2
...
37 g HCHO/g (x 1 mol/min)
Absorber off-gas
m w3 (kg H 2 O(l ))
n 5a (mol N 2 )
30°C
0
...
82 g H 3 O/g (x 3 mol/min)
n 5b (mol O 2 )
n 5 c (mol H 2 )
n 5d (mol H 2 O(v )), sat'd
n 5 e (mol HCHO(v )), 200 ppm
27°C, 1 atm

a
...
68 (cont’d)

HCHO balance on absorber ⇒ n6a , CH 3 OH balance on absorber ⇒ n6b
Wt
...
B
...
9 mol HCHO + 8
...
24 mol CH 3 OH
N 2 balance on conversion reactor:
3
...
3 ⇒ n3 = 8
...
76 × 8
...
3 mol N 2 feed
H balance on conversion reactor:

bg

bg

bg

bg bg

bg

n4 2 + 28
...
9 2 + 8
...
6 2 ⇒ n4 = 20
...
1 mol O in, 65
...
Accept (precision error)
N 2 balance on absorber: 30
...
3 mol N 2
O 2 balance on absorber: 0
...
83 mol O 2
H 2 balance on absorber: 5
...
00 mol H 2
H 2 O saturation of off - gas:

yw =

b

*
p w 27° C

P

g = LM 26
...
3 + 0
...
00 + n
5d

g U
|
|
200 ppm HCHO in off gas:
V
n
200 |
⇒
=
2|
|
3613 + n + n

...
03518 3613 + n5d + n5e 1

...

n5e = 7
...
46 mol off - gas
HCHO balance on absorber: 19
...
49 × 10 −3 ⇒ n6a − 19
...
34 = n6b ⇒ n6b = 8
...

x1 = 0
...
0 g HCHO ⇒ 1232 mol HCHO
1
...
031 mol CH 3 OH ⇒ x 2 = 0
...
732 mol H 2 O mol
62
...
441 mol H 2 O

8-48

8
...
89 = 0
...
9 mol product
CH 3 OH balance on distillation column:

b g

8
...
006 75
...
88 mol CH 3 OH
CH 3 OH balance on recycle mixing point:

n1 + n8 = n2 ⇒ n1 = 28
...
83 = 20
...
4 mol CH 3 OH l fresh feed
n2 = 75
...
88 mol CH 3 OH l recycle
n4 = 37
...
9
8
...
3
0
...
0
35
...
55
32
...
39
18
...
81
20
...
9
8
...
3
0
...
0
35
...
51
3
...
47
4
...
2

mw1 2726
...
B
...
84 kg 3
...
68 (cont’d)

b

g

Gas cooler: Same refs
...
9
8
...
3
0
...
0
35
...
51
3
...
47
4
...
9
8
...
3
0
...
0
35
...
78
–2
...
19
2
...
16
2
...
76

N2
O2
H2
H 2O
H 2O
(coolant)
E
...
ΔH =

∑n H − ∑n H
i

i

i

out

i

n (mol)
H (kJ/mol)

m (kg)
H (kJ/kg)

H = 4
...
6mw 2 = 0 ⇒ mw 2 = 2
...


in

b gb g
Q = − nΔH b1 atmg = −b27
...
27 kJ molg

Condenser: CH 3 OH condensed = n8 + 2
...
5 7
...
58 mol CH 3 OH condensed
E
...
:

b
...
6 × 10 4 tonne / y

10 6 g

1 yr

1d

1 metric ton 350 d 24 h

...
37gd4
...
01gd4
...
286 × 10
b0
...
286 × 10 i = 2
...
286 × 10 6 g h product soln

6

6

g HCHO h ⇒ 5
...


⇒ 2
...
016 × 105 mol h
= 2657 h −1
75
...
69 (a) For 24°C and 50% relative humidity, from Figure 8
...
0093 kg water / kg DA, Humid volume ≈ 0
...
2) kJ / kg DA = 47
...
Tskin ≈ 13o C (Twb )
...
H
...


8-50

Vroom = 141 ft 3
...


8
...

= 101 lb m DA
lb - mol 550 o R
0
...
205 lb m H 2 O
= 0
...


From the psychrometric chart, Tdb = 90 o F, ha = 0
...
5o F

Tdew point = 77
...
71

Tdb = 35° C
Tab = 27° C

8
...


H = 44
...
9 Btu / lb m

...
8
...
Mass of dry air: mda =

Mass of water:
c
...
3 ft 3 / lb m DA

⇒

hr = 33%, ha = 0
...


1 m 3 1 kg dry air
= 2
...
92 m 3
↑ from Fig
...
4-1

2
...
2 × 10 −3 kg dry air 0
...
033 g H 2 O
1 kg dry air
1 kg

b
g b
g
H b20° C, saturated g ≈ 57
...
8
...
2 × 10 kg dry air b57
...
4g kJ 10 J
ΔH
=
= −44 J

H 40° C, 33% relative humidity ≈ 78
...
65 kJ kg dry air = 77
...
Energy balance: closed system

n=

2
...
033 g H 2 O 1 mol
+
= 0
...
078 mol 8
...
44 kg water

= 10
...
56 kg air
10 kg H 2 O min
(b) ha =
= 0
...
73 (a)

min

Fig
...
4-1

b

g

H = 116 − 11 = 115 kJ kg dry air , Twb = 33° C, hr = 32%, Tdew point = 28
...


(c) Tdb = 10° C , saturated ⇒ ha = 0
...
5 kJ kg dry air

(d)

b0
...
0077g kg H O = 6
...
5

—

—

6
...
4

J

1 mol

mol⋅° C

b10 − 0g° C

1 kJ

103 g

18 g

= 42 kJ kg
103 J 1 kg
−34027
...
To calculate (Tair)in, you would need to know the
flow rate, heat capacity and temperature change of the solids
...
74 a
...
0226 lb m H 2 O lb m D
...
,

H = 455 − 0
...
A
...


...
0075 lb m H 2 O lb m D
...
,
H = 26
...
02 = 26
...
A
...
0075 lb m H 2 O lb m D
...

⇒ H = 214 − 0
...
A
...
07 ft 3 lb m D
...


...

Dry air delivered:

1,000 ft 3 1 lb m D
...

= 76
...
A
...
07 ft 2

H 2 O condensed:
76
...
A
...
0226 − 0
...
A
...


8
...
Since ha remains constant in the
second step, the condition of the air following the cooling step must lie at the intersection
of the ha = 0
...
8
...
5 45
...
5 21
...
A
...
2 17
...
A
...


...
5g 214 − 455 + 1
...
0) (Btu)
min

60 min 1 ton cooling
−12,000 Btu h
1h

= 9
...

6
7

(76
...
0075 lb m H 2 O/lb m DA
75o F, 26
...
5 lb m DA/min)

hr = 80%, ha = 0
...
5 Btu/lb m DA

76
...
0075 lb m H 2 O/lb m DA

Lab

55o F, 21
...
5
⎜ 76
...
0226
min ⎠ ⎝
lb m DA ⎠ 7
⎝

)( 0
...
5)(0
...
165 kg H 2 O condensed/min

8-53

2

8
...
8
...
93 45
...
57 26
...
4

—
o

Condensed water (49 F)

—

76
...
165 17
...
3 Btu 60 min 1 ton cooling
= 2
...
1 tons − 2
...
1 tons

Once the system reaches steady state, most of the air passing through the conditioner is
cooler than the outside air, and (more importantly) much less water must be condensed
(only the water in the fresh feed)
...
Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup
of carbon dioxide in the laboratory
...
75 Basis: 1 kg wet chips
...
6 m3(STP), Tdb=100oC
m1a (kg DA)

1 kg wet chips, 19oC
0
...
60 kg DC/kg

m3c (kg dry chips)
m3w [kg H2O(l)]
T (oC)

(a) Dry air: m1a =

b g

11
...
4 m STP

29
...
02 kg DA = m 2a

Outlet air:
Fig
...
4-1
→ˆ
(Tdb = 38°C, Twb = 29°C) ⎯⎯⎯⎯ H 2 = (95
...
48) = 94
...
0223 15
...
335 kg H 2 O
(b) H 2 O balance: 0
...
335 kg + m3w ⇒ m3w = 0
...
0223

kg H2 O
kg DA

8
...
065 kg water
× 100% = 9
...
600 kg dry chips + 0
...


substance

min

Air
H 2O l
dry chips

Hin

15
...
2
0
...
5
0
...
9

bg

mout

Hout

15
...
8 mair in kg DA, Hair in kJ/kg DA
0
...
184T m in kg DC, Hin in kJ/kg DC
0
...
10T

Energy Balance:
ˆ
ˆ
ΔH = ∑ mout H out − ∑ min H in =0 ⇒ −136
...
532T = 0 ⇒ T = 89
...
Twb = 210° C

...
76 a
...
77

Tas = Twb = 210° C

...
8° C
15 kg air
1 kg D
...

min
1
...
3 m3 1 kg D
...

0
...
8
...
0142 − 0
...

2

2

1 kg D
...


V = 0
...
A
...
0050 kg H 2 O kg D
...


= 12
...
A
...
0142 kg H 2 O kg DA

Fig
...
4-1

Inlet air: Tdb = 50° C
Tdew pt
...
0059 kg H 2 O kg DA

Fig
...
4-1

12
...
A
...
0165 kg H 2 O kg D
...


b0
...
0050g kg H O = 014 kg H O min

...
A
...
78 a
...
0050 kg H 2 O kg D
...


Twb = 20
...
908 m 3 kg D
...


Fig
...
4-1

b g

Twb = Tas = 20
...
0151 kg H 2 O kg D
...


b
...
A
...
0050 kg H2O/kg DA

m1 (kg D
...
)
0
...
05 kg S/kg
0
...
20 kg S/kg
0
...
0050g + b1gb0
...
0151g + b0
...
80g
Sugar balance: 0
...
20 m2 ⇒ m2 = 0
...
A
...
A
...
A
...
908 m 3
= 67 m 3
1 kg D
...


74 kg dry air

B

A

8
...
8
...
Inlet of spray chamber (B):

1 lb m D
...

ha3(lb m H 2O)
T d = 70°F
h r = 35%

ha1 ≈ 0
...
A
...
2 ft 3 lb m D
...


Fig
...
4-2

U
V
W

Coil
bank

ha 3 = 0
...
A
...
0017 lb m H 2 O lb m D
...

⇒ Twb = 49
...
5° F
...
0054 lb m H 2 O lb m D
...

⇒ hr = 52%
Twb = 49
...
5° F

b
...
0054 − 0
...
0 × 10
12
...
79 (cont’d)

( 20 - 6
...
QBA = ΔH = H B − H A ≅

QDC = ΔH = H D − H C ≅

12
...
2 ft / lb m dry air

= 1
...
25 Btu / ft 3

d
...
5

20

70

75

8
...
A
...


1 kg D
...

ha1(kg H 2 O/kg D
...
)
Tdb = 40°C, Tab = 18°C

1 kg D
...

ha2(kg H 2 O/kg D
...
)
20°C,

m w kg H2 O

Tdb = 40° C

⇒ ha1 = 0
...
A
...
0122 kg H 2 O kg D
...

Outlet air:
Twb = 18° C adiabatic humidification

Inlet air:

b

b gb g b gb g

g

b

g

Overall H 2 O balance: mw + 1 ha1 = 1 ha 2 ⇒ mn = 0
...
0039 kg H 2 O kg D
...

= 0
...
A
...


ma (lb m H2 O/h)
T=15o C, sat’d

1250 kg/h
T=37o C, h r=50%

mc (lb m H2 O/h)
liquid, 12°C
Qc (Btu/h)

8-57

8
...
0198 kg H O kg DA
SH = b88
...
5g kJ kg DA = 88
...
8
...
0198 kg

Outlet air: Tdb = 15° C, sat' d

Rh = 0
...
1 kJ kg DA
T
1226 kg DA b0
...
0106g kg H O
=
Fig
...
4-1

2

a

⇒

2

Overall water balance ⇒ mc

2

h

kg DA

= 113 kg H 2 O h withdrawn

...
(Cp)H2O(l) = 4
...
3 kJ / kg
0

Overall system energy balance:
Qc = ΔH =

∑m H − ∑m H
i

i

out

=

i

i

in

LM113 kg H O 50
...
1 − 88g kJ OPFG 1 h IJ FG 1 kW IJ

...

ΔH =

8
...
82 a
...
2 kJ
mol NH 3

b

g

b

= −31,280 kJ

g

b

g

HCl g , 25° C , H 2 O l , 25° C → HCl 25° C, r = 5
...
05 + 7514g kJ mol HCl = 11
...


ΔH = ΔH s 25° C, r = 5 ⎯Table B
...
05 kJ mol HCl
⎯⎯
⎯
b
...
83 Basis: 100 mol solution ⇒ 20 mol NaOH, 80 mol H2O

80 mol H 2 O
= 4
...
00g
2

b

H in

nout

H out

20
...
0

−

−

n in mol

80
...
0

−

−

H in kJ mol

−

20
...
43 ← n in mol NaOH

−

g

H NaOH, r = 4
...
43 kJ mol NaOH (Table B
...
43) =
out

Q=

−688
...
486 × 10 −4 Btu
10 −3 kJ

in

b

−653
...
0 40
...
0 18
...
20462 lb m

= −653
...
3 Btu lb m product solution

8
...
09808 kg
= 0
...
230 kg
= 1
...

b1230 − 0
...
5 mol H 2 O

18
...
49 mol H 2 O
= 11
...
6, 25o C + H 2 O l , 25o C
ΔH1 = ΔH s ( r = 11
...

2

=

4

R
S
T

Table B
...
6 + 96
...
6

z

60

25

OP
Q

kJ
mol H 2 SO 4

C p dT kJ

n H 2SO ( mol H 2 SO 4 )

4 mol H 2 SO 4
1
4 mol H 2 SO 4

4

28
...
230 kg
+
mol H 2 SO 4

= 60
...
00 kJ
kg⋅° C

b60 − 25g° CU
V
W

d

i

8
...
30 2
...
67 mol H 2 O ⇒ r =
a
...
33 =

b
...
08 g H 2SO 4
mol

b

g
b280
...
6 kJ +

Basis:

−44
...
33 + m

8
...


mol H 2 O
4
...
33
mol H 2SO 4
2

z

= −88
...
67 mol H 2 O 18
...
2 g

C p dT = 0

3
...
12 10 3 g
= 1120 g solution
L

1 L 8 mol HCl 36
...
0 mol H2 O(l, 25°C)

8
...
0 mol H 2 O
18
...
0 mol H 2 O
= 5
...
0 mol HCl

Assume all HCl is absorbed
Volume of gas:

b g

b g

8 mol 293 K 760 mm Hg 22
...
Ref: 25° C
nin

substance

bg
bg

H 2O l

HCl g

b

g

HCl n = 5
...
0 0
...
0 −015 −

...
0

Hout
−
−
−59
...
86 (cont’d)

b

g

b

g

H HCl, n = 5
...
75 +
= −64
...
66 cal
g⋅° C

8 mols

b40 − 25g° C

4
...

0
...
9715 × 10 −8 T 2 − 4
...
15 kJ / mol
Q = ΔH = −471 kJ L product

c
...


...
87 +

b

g

o
1120 g 0
...
184 J 1 kJ
8 mol g⋅o C
cal 1000 J

T = 192 o C

8
...



...
1 bars

n a (mol air/min)

...
001 NaOH
0
...


n 2 (mol/min) @ 50°C
0
...
95 H 2O

b

gb g
H O balance: b0
...
95b3
...
51 mm Hg ⇒
n +n
NaOH balance: 0
...
05n2 ⇒ n2 = 3
...
4

∗
H 2O

1

H 2O

Vinlet air =

1

n1 =147
P = 760

a

b g

1061 mol 22
...

= 37,900 L min
273 K 1
...
47 kJ mol NaOH

References for enthalpy calculations: H 2 O l , NaOH s , air @ 25° C

999 mol H 2 O Table B
...
1% solution @ 25°C: r =

5% solution @ 50°C: r =
Solution mass: m =

b

g

b

b

g

95 mol H 2 O 19 mol H 2 O
kJ
=
⇒ ΔH s 25° C = −42
...
0 g 19 mol H 2 O 18
...
81

s

z

50

25

C p dT

382 g
4
...
85 kJ
103 J

8
...
8 ⇒ H = 515 kJ mol

...
8 ⇒ H = 0
...
5 ⇒ H =

b2592 − 104
...
0 g

3

kg

10 g 1 mol

substance

b g
H Obv g

nin

Hin

nout

Hout

NaOH aq

0
...
47

0
...
85 n in mol min

−

−

147

44
...

515

1061

= 44
...
73

2

H in kJ mol

Energy balance: Q = ΔH = ∑ ni Hi − ∑ ni Hi = 1900 kJ min transferred to unit

b neglect ΔE g

out

n

in

8
...
Basis: 1 L 4
...
G
...
231)

4
...
3 = 838
...
3 g H 2 SO 4
L
= 46
...
11→ ΔH s = −67
...


b

g

b

Ref: H 2 O l , 25° C , H 2 SO 4 25° C
substance
H 2O l
H 2 SO 4 l
H 2 SO 4 25° C, n = 1164

...
57 0
...
00
0
−
−

b

g

g

b

g

gb

nout H out
n in mol
−
−
H in kJ mol
−
−
4
...
6

g

Q = ΔH = 0 = 4
...
6 − 46
...
0754 T − 25 ⇒ T = −52° C
(The water would not be liquid at this temperature ⇒ impossible alternative!)

b

g

b

b
...

H 2 SO 4 25° C, n = 1164

bg
bg

b

b

g

g

nin
H in
0
...
01 + 0
...
00
0

b

g

b

g

g

nout
H out
n in mols
−
−
H in kJ mol
−
−
−
−
4
...
61

ΔH m H 2 O, 0° C = 6
...
1

U⇒n
V
ΔH = 0 = 4
...
61g − n b −1885g − b46
...
895gW n

...
3 g H Ob sg@0° C

...
57

l

2

l

2

8-62

l
s

= 1618 mol liquid H 2 O

...
39 mol ice

8
...


wt% P2 O 5 =

b

g × 100% ,

n 14196

...
00g × 100%

2n

wt% H 3 PO 4 =

B

g H 3 PO 4 mol

mc

A

g total

where n = mol P2 O5 and mt = total mass
...
00
14196

...


b
...
67 wt% H 3 PO 4
m1 (lbm H2 O(v )), T , 3
...
3867 lb m 3
H PO 4
0
...
5800 lb m 3
H PO 4
/lb
0
...
3867 = 0
...
667 lb m solution

bg

Total balance: 1 = m1 + m2 ⇒ m1 = 0
...
3333 lb m H 2 O v lb m feed solution
c
...
255 bar

...
6

⇒ Tsat = 654 o C =149 o F, Vliq =

...
00102 m3 353145 ft 3 / m3

...
0163
lb m H 2 O(l)
kg
2
...
3 lb m / min
1 ton
day
3 lb m
(24 × 60) min

V=

46
...
0163 ft 3

7
...
65 gal condensate / min

Heat of condensation process:

46
...
3lbm H2O(l)/min

(149+37)°F, 3
...
7 psia


...
89 (cont’d)

R
|H
|
|
Table B
...
4 o C) = (274 kJ / kg) 0
...
3

I
JJ = 1141 Btu / lb
kJ
kg K

Btu

o
o
H2 O ( v ) (186 F = 85
...
4303

OP = −47,360 Btu / min
Q

⇒ 4
...
Refs: H 3 PO 4 l , H 2 O l @77° F

substance
H 3 PO 4 28%
H 3 PO 4 42%
H 2O v

min
Hin
mout
Hout
100 13
...

m in lb m
−
−
0
...
13 H in Btu lb m
−
−
0
...
705 Btu
lb m ⋅° F

1 lb - mole H 3 PO 3
98
...
95 Btu lb

0
...
00 lb m H 3 PO 4

b186
...
13 Btu lb

b g b

0
...
00 lb m soln

m

0
...
00 lb m sol
...
7 psia, 186° F − H l , 77° F = 2652 − 104
...
6 psia (=1
...
6 ⇒ ΔHv = 2206 kJ / kg = 949 Btu / lb m
ΔH = ∑ ni Hi − ∑ ni Hi = 375 Btu = msteam ΔHv ⇒ msteam =
out

⇒
⇒

in

375 Btu
= 0
...
395 lb m steam 100 × 2000 lb m H 3 PO 4

1 day

lb m 28% H 3 PO 4

24 h

day

= 3292 lb m steam / h

3292 lb m steam
lb m steam

...
3 × 60) lb m H 2 O evaporated / h
lb m H 2 O evaporated

8-64

8
...
A = NaC 2 H 3 O 2


...
9% of H 2O
in feed
200 kg/h @ 60°C

...
20 A
0
...

n 2 (kmol A-3H 2 O(v )/h)

...
154 A
0
...
Average molecular weight of feed solution: M = 0
...
800 M H 2 O

b

gb g b

gb g

= 0
...
0 + 0
...
0 = 30
...
49 kmol h
h
30
...
20gb6
...
16
...
80 6
...
877 kmol H 2 O v h

...


...
80 6
...
877 +

1 mole A ⋅ 3 H 2 O

+ 0154n3

...
846n3 ⇒ 3n2 + 0
...
315

bg b g

g

3 moles H 2 O
1 mole A ⋅ 3 H 2 O

bg

b2g

Solve 1 and 2 simultaneously ⇒ n2 = 113 kmol A ⋅ 3H 2 O s h

...

Mass flow rate of crystals

bg

1
...
877 18
...


bg

bg

References for enthalpy calculations: NaC 2 H 3 O 2 s , H 2 O l @25° C

b

g

Feed solution: nH = n A ΔH s 25° C + m

nH =

b0
...
49 kmol A
h

z

60

25

C p dT (form solution at 25° C , heat to 60° C )

−171 × 104 kJ 200 kg 3
...

+
hr
kg⋅° C
kmol A

8-65

b60 − 25g° C = 2300 kJ h

8
...
154g1
...
13 kmol A ⋅ 3H 2 O s
h
= −36700 kJ h

=

b

z

LM
N

g

H 2 O v , 50° C : nΔH = n ΔH v +
=

OP
Q

50

25

b neglect ΔE g

C p dT (vaporize at 25° C , heat to 50° C )

b gb

i

g

4
...
4 50 − 25 kJ

∑n H − ∑n H
i

out

R

b50 − 25g° C

−3
...
2 kJ
+
h
kg⋅° C
kmol

0
...
5 kJ

...

84
...

= 84
...
678 mol H Oblg|
|
mL
W

8
...


= 917 g H 2SO4 ⇒ 0935 mol H 2SO4

...


2

2

2

2

2

Ref: H 2 O , H 2SO 4 @ 25 °C
H ( H 2 O(l ), 15o C) = [0
...
754 kJ / mol

b

g

H H 2 SO 4 , r = 5
...
03

(91
...
2) g
kJ
+
mol 0
...
43 J

bT − 25g° C

1 kJ

g⋅° C

10 3 J

= ( −69
...
457T )( kJ / mol H 2 SO 4 )
substance

nin

bg

nout

Hin

Hout

H 2O l

4
...
754 —
—
n in mol
0
...
0
—
—
H 2SO 4
H in kJ/mol
—
—
0
...
46 + 0
...
00

b

g b

b

g

b

g

b

g

g

Energy Balance: ΔH = 0 = 0
...
46 + 0
...
678 −0
...


8-66

4

8
...

mA (g A) @ TA0 (oC)
nA (mol A)

nS (mol solution) @ Tmax (oC)

mB (g B) @ TB0 (oC)
nB (mol B)
Refs: A(l), B(l) @ 25 °C
substance nin H in nout

H out

A

nA H A

—

—

n in mol

B

nB H B

—

—

H in J / mol

S

—

nA

H S (J mol A)

—

mA (g A)
m
, nB = B
M A (g A / mol A)
MB

Moles of feed materials: n A (mol A) =
Enthalpies of feeds and product

H A = m A C pA ( T A 0 − 25 o C), H B = m B C pB ( TB 0 − 25 o C)
r (mol B mol A) = n B n A =

HS

FG J IJ =
H mol A K n
⇒ HS =

A

mB / M B
mA / M A

LMn
1
M
( mol A) M
MM+
N

A ( mol

FG J IJ
H mol A K
F J I × (T
)( g soln) × C G
H g soln ⋅ C JK

A) × Δ H m ( r )

(m A + m B

ps

max

o

− 25)( o

OP
PP
C) P
QP

1
n A Δ H m ( r ) + ( m A + m B ) C ps ( Tmax − 25)
nA

Energy balance
ΔH = n A H S − n A H A − n B H B = 0

bg

b

g

b

g

mA
ΔHm r + (m A + mB )C ps (Tmax − 25) − m A C pA TA0 − 25 − mB C pB TB 0 − 25 = 0
MA
m
m A C pA TA0 − 25 + mB C pB TB 0 − 25 − A ΔHm r
MA
⇒ Tmax = 25 +
(m A + mB )C ps
⇒

b

g

b

g

bg

Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container,
negligible dependence of heat capacities on temperature between 25oC and TA0 for A, 25oC
and TB0 for B, and 25oC and Tmax for the solution
...


m A = 100
...
0 g

b

gU
V
|
W

M A = 40
...
00
M B = 18
...
18 J (g⋅° C)
mol NaOH

C ps = 3
...
00 = −37,740 J mol A ⇒ Tmax = 125° C

8-67

8
...


1
r

M A C pA T0 − 25
M w C pw T0 − 25
—

b g

b
b

—

g
g

nout

—
—
1

Hout

—
n in mol
H in J/mol
—
ΔH m r + M A + rM w C ps Ts − 25

bg b

g b

g

(J/mol H2SO4)

bg b
g b g
b g
= ΔH br g + b98 + 18r gC bT − 25g − (98C + 18rC )bT
1
( 98C + 18rC )bT − 25g − ΔH br g
⇒ T = 25 +
(98 + 18r )C

b

g

ΔH = 0 = ΔH m r + M A + rM w C ps Ts − 25 − M A C pa T0 − 25 − rM w C pw T0 − 25
m

ps

s

s

pa

pa

pw

pw

0

0

g

− 25

m

ps

c
...
5
1
1
...
4
185
...
2
1
...
58
1
...
89
1
...
1
2
...
43
3
...
56
3
...
9
174
...
2
205
...
8
184
...
5
121
...
0
59
...
0

250

Ts

200
150
100
50
0
0
...
Some heat would be lost to the surroundings, leading to a lower final temperature
...
94 a
...


b

nB 0 RT
c0 + c1T
Vg

n Al =

g

LM1 + n RT bc + c T gOP
NM V
QP
n
=
LM1 + n RT bc + c T gOP
QP
NM V

(5)

B0

0

1

g

n Av

Ao

(6)

B0

0

1

g

Ideal gas equation of state
P=

n Av RT ( 6 )
n A 0 RT
=
Vg
Vg + nB 0 RT c0 + c1T

b

g

(7)

b g bg

Refs: A g , B l @ 298 K

nin

nB0

bg
Bb l g

Ag

Solution
U 1 = ΔU s +

U in

n Ao

substance

U eq

M A CvA T0 − 298

n Av

M A CvA T − 298

M B CvB

—

—

n Al

U 1 (kJ/mol A)

—

b
bT

0

g
− 298g

—

b

g b

b

g

n in mol
U in kJ/mol

g

1
n Al M A + n B 0 M B Cvs T − 298
n Al

E
...
: ΔU = 0 =

∑n U − ∑n U
i

out

c

neq

i

i

i

in

b

g hb
g
b
d−ΔU i + bn C + n C gbT − 298g
n C + bn M + n M gC

gb

g

0 = n Av CvA + n Al M A + nB M B Cvs T − 298 + n Al ΔU s − n Ao CvA + n B CvB T0 − 298
⇒ T = 298 +

n Al

s

Av

Ao

vA

Al

vA

B

A

8-69

vB

B

0

B

vs

8
...

Vt
20
...
0

CvA
0
...
0

CvB
3
...
76

Vl
3
...
0
3
...
0
3
...
0
3
...
0

T0
300
300
300
300
330
330
330
330

P0
1
...
0
10
...
0
1
...
0
10
...
0

Vg
17
...
0
17
...
0
17
...
0
17
...
0

nB0
203
...
1
203
...
1
203
...
1
203
...
1

nA0
0
...
453
6
...
811
0
...
139
6
...
555

c0
c1
0
...
60E-06
T
301
...
0
313
...
6
331
...
4
342
...
3

nA(v)
0
...
624
5
...
414
0
...
359
4
...
381

Dus
-174000

Cvs
3
...
164
0
...
671
3
...
155
0
...
569
3
...
8
3
...
9
16
...
8
3
...
8
16
...


C*

REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS
REAL NA0, T, DEN, P, NAL, NAV, NUM, TN
INTEGER K
R = 0
...
LT
...
1 * T0
K=1
10
DEN = VG/R/T/NB + C + D * T
P = NA0/NB/DEN
NAL = (C + D * T) * NA0/DEN
NAV = VG/R/T/NB * NA0/DEN
NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298)
DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS
TN = 298 + NUM/DEN
WRITE (6, 901) T, P, NAV, NAL, TN
IF (ABS(T – TN)
...
0
...
LT
...
)'/
*
'
(K)
(atm) (mols) (mols)
(K)')
901
FORMAT (F9
...
3, 2X, F7
...
3, 2X, F7
...
0
15
...
54E–3
–2
...
0
18
...
0291
0
...
2E–03

8-70

Tcalc
301
...
0
313
...
6
331
...
4
342
...
3

8
...
0
–1

50
...
0

Program Output
T (assumed)
(K)
321
...
54
296
...
0
0
...
54E–3
0
...
6E–6
4
...
019
7
...
416

Nav
(mols)
4
...
571
4
...
703
1
...
711

T(calc
...
542
296
...
568

P
(atm)
40
...
676
39
...
895
22
...
885

Nal
(mols)
8
...
523
8
...
)
(K)
316
...
942
316
...
10
316
...
94

8
...
78

30% H2SO4
ms(g), T(oF)

H2O, Vw(mL),
mw(g), 60 oF

a
...

178 g
1 mL feed


...
85(70 / 30) − 015 g H 2 O added 1 mL water
g feed
1 g water

= 1140 mL H 2 O
b
...
8
...
78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g
Energy Balance:

c
...
9 Btu/lb m
1765

ˆ
T ( H = −18
...
When acid is added slowly to water, the rate of temperature change is slow: few isotherms
are crossed on Fig
...
5-1 when xacid increases by, say, 0
...
On the other hand, a change
from xacid=1 to xacid=0
...


8-71

8
...
2
...
0 wt% H 2 SO 4

@ 77 F ⇒ H1 = −10 Btu / lb m
o

U
|
|
| ⎯⎯⎯⎯⎯→ m ( lb
⎯
V
|
|
|
W
adiabatic mixing

3

m2 (lb m ) 80
...
0 wt% H 2 SO 4 @ T o F, H 3

U ⇒ |m
| R
V S
mass balance: 2
...
150g + m b0
...
600) W |m
| T

Total mass balance:
H 2 SO 4

m)

2

2

3

= 517 lb m (80%)

...
30 + m2 = m3

= 7
...
Adiabatic mixing ⇒ Q = ΔH = 0


...
47gH − b2
...


E Figure 8
...


d

i

H 60 wt%, 77 o F = −130 Btu / lb m

d

i

b

gb

g

Q = m3 H 60 wt%, 77 o F − H 3 = 7
...

d
...
The rate of temperature rise is

much lower (isotherms are crossed at a lower rate) when moving from left to right on
Figure 8
...

8
...


b
...
30

Fig
...
5-2

y NH 3 = 0
...
90 lb m liquid

⇒ 010 lb m vapor

...
27 + 0
...
30

0
...
63 lb m H 2 O

x NH 3 = 0
...
096 lb m NH 3
0
...
37 lb m NH 3 lb m

1 − 0
...
63 lb m H 2 O lb m
Enthalpy: H =

0
...
10 lb m vapor
670 Btu
−25 Btu
+
= 44 Btu lb m
1 lb m
1 lb m
1 lb m liquid
1 lb m vapor

8-72

8
...
8
...
14


...
80 x NH3

Mass Balance: mv + mL = 250

...
80m g + 014mL = ( 0
...
5 g NH , 64
...


Vapor: mNH 3 = 0
...
99 Basis: 200 lb m feed h

mv (lb m h)
xv(lbm NH3(g)/lbm)
H v ( Btu lb m )

200 lbm/h
0
...
30 lbm H2O(l)/lbm

ml (lb m h)

H f = −50 Btu lb m

in equilibrium
at 80oF

xl[lbm NH3(aq)/lbm]
H l ( Btu lb m )

Q( Btu h)

Figure 8
...
96 lb m NH 3 lb m
Mass fraction of NH 3 in liquid: xl = 0
...
70gb200g = 0
...
30m V m = 80 lb h liquid
W
200 = mv + ml

Mass balance:
Ammonia balance:

m

v

v

l

l

m

Energy balance: Neglect ΔE k
...
1

ΔH ro = −904
...


When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25°C and 1 atm react to form 4 g-moles of
NO(g) and 6 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -904
...


b
...
The reactor must be cooled to keep the temperature constant
...
The energy required to break the molecular bonds of the
reactants is less than the energy released when the product bonds are formed
...


5
O 2 (g) → 2NO(g) + 3H 2 O(g)
2
Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half
...
7
ΔH ro = −
= −452
...
Also reducing the stoichiometric
coefficients to one-fourth reduces the heat of reaction to one-fourth
...
7)
ΔH ro = −
= +226
...


e
...
03 g

Q = ΔH=

ˆ
n NH3 ΔH o
r

ν NH

=

= 20
...
0 mol NH 3
s

3

−904
...
52 × 103 kJ/s

The reactor pressure is low enough to have a negligible effect on enthalpy
...


Yes
...


9
...


When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25°C and 1 atm react to form 9 g-moles of
CO2(g) and 10 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -6124 kJ
...


Exothermic at 25°C
...
The temperature
would increase under adiabatic conditions
...


c
...
0 mol C 9 H 20
s

−6124 kJ

1 kW

1 mol C 9 H 20 1 kJ / s

9-1

= −153 × 105 kW

...
2 (cont'd)
Heat Output = 1
...

The reactor pressure is low enough to have a negligible effect on enthalpy
...


C 9 H 20 (g) + 14O 2 (g) → 9CO 2 (g) +10H 2 O(l)
ΔH ro

(1)

= −6171 kJ / mol

C 9 H 20 (l) + 14O 2 (g) → 9CO 2 (g) +10H 2 O(l)
ΔH ro

(2)

= −6124 kJ / mol

(2) − (1) ⇒ C 9 H 20 (l) → C 9 H 20 (g)
o
ΔH v (C 9 H 20 ,25 C) = − 6124 kJ / mol − ( −6171 kJ / mol) = 47 kJ / mol

e
...
3

a
...
Pure n-nonane can only exist as vapor at 1 atm above 150
...


Exothermic
...
The temperature
would increase under adiabatic conditions
...


b
...

2
C H bgg → C H blg b3g ΔH = −e ΔH j
= −13,550 Btu lb - mole
H Oblg → H Obgg b4g ΔH = e ΔH j
= 18,934 Btu lb - mole

...
4

M O2 =32
...


o
r

2

3

4

n = 3
...


3
...
672 × 106 Btu
= −6
...
5
1 lb-mole O 2

bg

bg

bg

bg

CaC 2 s + 5H 2 O l → CaO s + 2CO 2 g + 5H 2 g , ΔH ro = 69
...

Endothermic
...
The temperature
would decrease under adiabatic conditions
...

b
...
314 J
− RT M ∑ ν − ∑ ν P = 69
...
0 kJ mol

9-2

1 kJ 298 K
103 J

b7 − 0g

9
...


c
...
0 kJ
= 121
...
10 g 1 mol CaC 2

Heat must be transferred to the reactor
...
5
a
...


Given reaction = (1) – (2)

Hess's law

⇒

b

g

ΔH ro = ΔH ro1 − ΔH ro2 = −121,740 + 104,040 Btu lb - mole
= −17,700 Btu lb - mole

⎛
⎝

Hess's law

9
...


ˆ
Reaction (3) = 0
...
5 ⎜ −326
...
8
⎟ = 122
...


9
...


a
...


n − C5 H 12 g +

bg

bg

bg

e j

=2

ΔH fo

NO(g)

bg
bg
bg
+ 6e ΔH j
ΔH = 5e ΔH j
b g − e ΔH j
bg

...

= b5gb −110
...
4g kJ mol = −21212 kJ mol
19
C H blg + O bgg → 6CO bgg + 7H Obgg
2
ˆ = 6 ( ΔH ) + 7 ( ΔH )
ˆ
ˆ
ˆ
ΔH
− ( ΔH )
( )
()
o
r

c
...
37 kJ molJ = 180
...
1

ΔH ro

6

o
f

11
O 2 g → 5CO g + 6H 2 O l
2

CO(g)

14

2

o
r

o
f

o
f

o
f

H 2O l

2

2

o
f

CO 2

n − C5 H 12 g

o
f

H2 O g

C6 H14 l

= ⎡( 6 )( −393
...
83) − ( −198
...


Na 2SO 4 (l) + 4CO(g) → Na 2S(l) + 4CO 2 (g)

e j

− 4e ΔH j
e j − e ΔH j

...
2 + 6
...
5 + 24
...
52

ΔH ro = ΔH fo

Na 2S( l )

+ 4 ΔH fo

CO 2 ( g )

o
f

9-3

Na 2SO 4 ( l )

o
f

CO(g )

kJ mol = −138
...
8

a
...
76 + 52
...
48 kJ mol
e j
ΔH
= −276
...
31 + 333
...
03 kJ mol
b g + e ΔH j b g − e ΔH j
Given reaction = b1g + b2g ⇒ −385
...
03 = −420
...


c
...
79 kJ

h

mol

C 2 H 2 Cl 4 ( l )

C 2 H 2 Cl 4 (l )

b

= −126 × 105 kJ h = −35 kW

...

9
...


5
O 2 (g) → 2CO 2 (g) + H 2 O(l)
2

C 2 H 2 ( g) +

o
ΔHc = −1299
...
5 g-moles of O2(g) at 25°C and 1 atm react
to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25°C and 1 atm is -1299
...

b
...
1

b

=

c
...
6 mol

2 −3935 + −28584 − 226
...


...
1

=

d

i

B

Table B
...

b−84
...
75g mol = −3114 mol

(ii) ΔH ro = ΔH co

d
...

b−1299
...
84) − b−1559
...
6 kJ mol

o
(2) ΔHc 2 = −28584 kJ mol

...
9 kJ mol

The acetylene dehydrogenation reaction is (1) + 2 × (2) − (3)
Hess's law

⇒

o
o
o
ΔH ro = ΔHc1 + 2 × ΔHc 2 − ΔHc3

b

g


...

= −1299
...
9) kJ mol = −3114 kJ / mol

9-4

9
...


bg

C8 H 18 l +

bg

bg

25
O 2 (g) → 8CO 2 g + 9H 2 O g
2

ΔH ro = −4850 kJ / mol

When 1 g-mole of C8H18(l) and 12
...

b
...
2 = 75
...
C
− Q = mH 2 O (Cp ) H 2 O(l) ΔT =

Q = ΔU ⇒ −89
...
00 kg

75
...
34° C

1 mol
18
...
C

2
...
2 g

= 89
...
314 J 1 kJ 298 K
mol ⋅ K 103 J

b8 + 9 − 12
...


(−5068) − ( −4850)
× 100 = − 4
...
5 ) + 9 ( −241
...
5 kJ/mol
⎦
() ⎣

o
ΔHc = 8 ΔH fo

CO 2 g

o
f

o
ff

H 2O g

o
f

C8 H 18 l

C8 H18 l

There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only
product
...
11

a
...
930 mol n-C4H10

(nn- C4H10)out

0
...
020 mol HCl

0
...
930(1 − 0
...
560 mol
(n i-CH 4 H10 ) out = 0
...
930 × 0
...
420 mol

ξ =

(n n-C4H10 ) out − (n n-C4H10 ) in

0
...
930
= 0
...


ΔH ro = ΔH fo

c
...
5 − b−124
...
1

o
r

n − C 4 H 10
4

kJ mol = −9
...
600

H1

i − C 4 H 10

−

−

0
...
2

149

25

Cp

O kJ
dT P
PQ mol = 14
...
2

149

25

C p dT

OP kJ
PQ mol = 14
...
370 ⎡−9
...
142 ) − (1)(14
...
68 kJ
For 325 mol/h fed, Q =
d
...
8 kJ
325 mol feed
1h
1 kW
= −0
...
68 kJ
ˆ
ΔH r (149°C ) =
= −9
...
370 mol

9-6

9
...

1 m3 at 298K, 3
...
00 torr

n0 (mol)
0
...
8889 mol O2/mol

n1 (mol O2)
n2 (mol SiO2)
n3 (mol H2)

SiH 4 ( g) + O 2 (g) → SiO 2 (s) + 2H 2 (g)

Ideal Gas Equation of state : no =

1 m3

273 K 3
...
1614 mol
298 K 760 torr 22
...
1111(0
...
0179 mol
O 2 : n1 = 0
...
1614 mol) − ξ = 0
...
0179 mol SiO 2
H 2 : n3 =2ξ =0
...


ΔH ro = ( ΔHfo ) SiO 2 (s) − ( ΔHfo ) SiH 4 ( g)
= [ −851 − ( −61
...
1 kJ / mol

References : SiH 4 (g), O 2 (g),SiO 2 (g), H 2 (g) at 298 K

Substance

ˆ
H in

nin

nout

ˆ
H out

(mol h) ( kJ mol) (mol h) ( kJ mol)
SiH 4

0
...
1435

0

0
...
0179
0
...
8

O 2 (g,1375K): H1 = H O 2 (1102 o C) = 3614 kJ / mol

...
18 kJ / mol

298

B

Table B
...
35 kJ / mol
o

c
...
01 kJ/m3 feed
out

Q=

−7
...
5 m
m3
h

3

in

1h
1 kW
= −0
...
13

a
...
111 × 105 Btu lb - mole

Basis:

2000 lb m Fe 1 lb - mole

= 3581 lb - moles Fe produced

...
85 lb m
53
...
9 lb - moles Fe 2 O 3 fed
53
...
9 lb-moles Fe2O3 (s)
77° F

35
...
72 lb-moles C
77° F

53
...


bg bg bg

Substance

b

nin

g

Fe 2 O 3 s,77° F

b g
Febl,2800° Fg
CObg,570° Fg

Fe(l,2800 F): H1 =

nFe ΔHro

ν Fe

Hout

−

−

53
...


H1

−

z

0
−

53
...
91
0
−
−

C s,77° F

c
...
9 IK
from Table

∑n H − ∑n H
out

g

dT + ΔHm 2794° F +

i

i

i

in


...
111 × 10 j + b3581gb28400g + b53
...
98 × 10
=

...


6

Btu / ton Fe produced

Effect of any pressure changes on enthalpy are neglected
...

Specific heat of Fe(l) is assumed to remain constant with temperature
...

No vaporization occurs
...
14

a
...


substance nin

nout

Hin

Hout

b g bkJ molg b g bkJ molg
mol

C 7 H 16
C7 H 8
H2

mol
−
1
4

H1
−
−

1
−
−

−
H2
H3

⎡ 400 ↓
⎤
ˆ
ˆ
C7 H16 ( g,400°C ) : H1 = (ΔH f )C7 H16 (g) + ⎢ ∫ C p dT ⎥
⎢ 25
⎥
⎣
⎦
= ( − 187
...
0) kJ/mol= −96
...
2427

b

g

C 6 H 5 CH 3 g,400° C : H 2 = ( ΔH f ) C6 H 5CH 3 ( g)

L
+M
MN

z

B

Table B
...
2) kJ / mol = 110
...
8

g

H 2 g,400° C : H3 = H H 2 (400 C) = 10
...


ˆ
ˆ
Q = ΔH = ∑ ni H i − ∑ ni H i
out

in

= ⎡(1)(110
...
89 ) − (1)( −96
...


ˆ
ΔH r (400 C)=

251 kJ
= 251 kJ/mol
1 mol C7 H16 react

9-9

9
...


bCH g Obgg → CH bgg + H bgg + CObgg
3 2

4

2

Moles charged: (Assume ideal gas)

b g

2
...
01286 mol CH 3 2 O
873 K 760 mm Hg 22
...
01286(1 – x ) mol (CH3 )2O
0
...
01286 x mol H2
0
...
01286 mol
(CH 3)2 O
600°C, 350 mm Hg

b g

600°C
875 mm Hg

b

g

Total moles in tank at t = 2h = 0
...
01286 1 + 2 x mol
Pf V
P0V

b
...
01286 1 + 2 x
0
...
75 ⇒ 75% decomposed
350

References: C ( s ) , H 2 ( g ) , O 2 ( g ) at 25 C
substance

bCH g Obgg
CH bgg
H bg g
CObgg
3 2
4

2

nin
nout
Hin
Hout
(mol) ( kJ / mol)
( mol)
(kJ / mol)
0
...
25 × 0
...
75 × 0
...
75 × 0
...
75 × 0
...

PQ mol × 10 J = (−18016 + 62
...
85 + 29
...
39 kJ mol
⎢ 25
⎥
⎣
⎦
Table B
...
8

H 2 (g,600 C): H3 = H H 2 (600 C) = 16
...
1

Table B
...


Table B
...
52 + 17
...
95 kJ mol

For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are :

ˆ
ˆ
ΔH = ∑ nout H out − ∑ nin H in = [ −1
...
5175) ] kJ = −0
...
15 (cont’d)

ξ=

(n CH4 )out − (n CH4 )in

ν CH

=

4

0
...
01286
mol = 0
...
0340 kJ
ˆ
ˆ
ΔH = ξ ΔH r ( 600°C ) ⇒ ΔH r ( 600°C ) =
= −3
...
009645

b

g

b

g

ΔU r 600° C = ΔH r 600° C − RT [

∑ν

i

d
...
53 kJ mol −

8
...
0 kJ mol

ˆ
Q = ξ ΔU r ( 600°C ) = (0
...
0 kJ/mol) = −0
...
16

a
...
07 kg SO 3

= 1249 mol SO 3 min

mol SO /min
0
n0 (molnSO 2 /min),2450o C
450°C
100% excess excess o C
100% air, 450
n1 (mol ni1 2 mol O2 /min
O /min)
3
...
76n1 (mol N 2 /min)
450°C

1249 mol SO /min
1249 mol SO 33/min
s
nn2 (molSO22/min
mol SO /min)
0
nn mol OO/min
3 (mol 2 /min)
3
2
3
...
76n1 (mol N 2 / min)
550°C / i
550o C


...

mw (kg H 2O(l) /h)
40°C

Assume low enough pressure for H to be independent of P
...
65 mol SO 2 react 1 mol SO 3 produced

bGeneration = output g

min

1 mol SO 2 fed

1 mol SO 2 react

= 1249

mol SO 3
min

⇒ n0 = 1922 mol SO 2 / min fed

100% excess air: n1 =

1922 mol SO 2

0
...
76 1922 = 7227 mol / min in & out

g

b1 + 1g mol O

2

fed

1 mol O 2 reqd

g
b2gb1922g + b2gb1922g = b3gb1249g + b2gb673g + 2n

= 1922 mol O 2 min fed

b

65% conversion : n2 = 1922 1 − 0
...
16 (cont’d)

...



...


Extent of reaction : ξ =

ν SO 2

=

673 − 1922
= 1249 mol / min
1

B

Table B
...
9) = −99
...


bg bg bg

bg

References : SO 2 g , O 2 g , N 2 g , SO 3 g at 25 C
nin

Substance

Hin

nout

Hout

( mol / min) ( kJ / mol) ( mol / min) ( kJ / mol)
1922
H1
H4
673

SO 2
O2

1922

H2

1298

H5

N2
SO 3

7227
−

H3
−

7227
1249

H6
H7

SO 2 (g,450 C) : H1 =

z

B

Table B
...
62 kJ / mol

25

B

Table B
...
36 kJ / mol

B

Table B
...
69 kJ / mol

Out :
Table B
...
79 kJ/mol

B

Table B
...
71 kJ / mol

B

Table B
...

SO 3 (g,550 C) : H7 =

z

B

Table B
...
34 kJ / mol

ˆ
ˆ
ˆ
Q = Δ H = ξΔH ro + ∑ ni H i − ∑ ni H i
out

in

= (1249 )( −98
...
796 ) + (179
...
711) + ( 7227 )(15
...
336 ) − (192
− 1922 (13
...
691)
=

c
...
111 × 104 kJ 1 min 1 kW
= −1350 kW
min 60 s 1 kJ/s

Assume system is adiabatic, so that Qlost from reactor = Qgained by cooling water

9-12

9
...
5 − 104
...

=m G
H min K
min
kg
w

w

w

Table B
...


Table B
...


bg

CO(g) + H 2 O v → H 2 (g) + CO 2 (g) ,

9
...
1

ΔH ro

a
...

= − 4115

bg

H 2O v

b g

kJ
mol

b g


...
5 m 3 STP product gas h 1000 mol 22
...
6 mol/h
condenser
0
...
40 mol CO 2/mol
0
...
40g + b2gb0
...



...
40 1116 mol h = 44
...
64 mol CO 1 mol H 2 O
h

1 mol CO

bg

= 66
...
64 mol H 2 O

b66
...
64g mol h × 100% = 50% excess steam
44
...
40gb1116 mol hg = 44
...


CO 2 balance on condenser : n3 = 0
...
64 mol CO 2 h

...
788 mm Hg
yH 2O = w
⇒
=
⇒ n5 = 153 mol H 2 O v h

...
64 + 44
...
6gb0
...

2

2

6

⇒ n6 = 20
...
374 kg / h

9-13

bg

9
...
Energy balance on condenser

References : H 2 (g), CO 2 (g) at 25 C, H 2 O at reference point of steam tables

nin
nout
H in
H out
mol / h kJ / mol mol / h kJ / mol
H1
H4
44
...
64
H2
H5
44
...
64

...
32
153
−
−
H7
20
...
8

CO 2 (g,500 C) : H1 = H CO 2 (500 C) = 2134 kJ / mol

...


FG
H

IJ
K

18 kg
kJ
×
= 62
...
552 kJ/mol
H 2 (g,15 C) : H 5 = H H 2 (15 C) = −0
...
9

Q = ΔH =

out

i

IJ
K

FG
H

IJ
K

18
...

kg 10 3 mol

∑n H − ∑n H
i

FG
H

18
...

kg
10 3 mol

i

i

=


...
22 − 29718g kJ
h

in

bheat transferred from condenserg
c
...
812 kW

Energy balance on reactor :
References : H 2 (g), C(s), O 2 (g) at 25° C
Substance
CO(g)
H 2 O( v )
H2 g
CO 2 g

bg
bg

nin
nout
Hin
Hout
( mol / h) ( kJ / mol) ( mol / h) ( kJ / mol)
44
...
96
H2
22
...
64
H4
−
−
44
...
1

=

−110
...
1, B
...
56 kJ mol

9
...


=

−224
...
8

=

1383 kJ / mol

...
1, B
...
1, B
...
16 kJ / mol

1h
1 kW
−2101383 − ( −20839
...

= −0
...


9-15

9
...

References : FeO(s), CO(g), Fe(s), CO 2 (g) at 25o C
Substance
FeO
CO

nin
nout
Hin
Hout
( mol) ( kJ / mol) ( mol) ( kJ / mol)
1
...

⇒ n1 = 1 − X

...
02761 ( T0 − 298) + 2
...
451 + 0
...
51 × 10 −6 T02 ) kJ / mol
H1 = 0
...


...
0814 + 0
...


...
02761 (T − 298) + 2
...
451 + 0
...
51 × 10 −6 T 2 ) kJ / mol
H 3 = 0
...
335 × 10 −5 (T 2 − 298 2 )

...
335 + 0
...
04326(T − 298) + 0
...

⇒ H 4 = ( −16145 + 0
...
573 × 10 −5 T 2 + 8
...


9- 16

9
...

n0 = 2
...
700 mol FeO reacted/mol FeO fed

⇒ n1 = 1 − 0
...
3, n2 = 2 − 0
...
3, n3 = 0
...
7, ξ = 0
...
520 kJ/mol, H1 = 13
...
494 kJ/mol,
ˆ
ˆ
H 3 = 7
...
87 kJ/mol
ˆ
ΔH o = −16
...
7)(−16
...
3)(13
...
3)(7
...
7)(7
...
7)(10
...
520)
⇒ Q = 11
...

no

To
400
400
400
400
400
400
400
400

X T
1
298
1
400
1
500
1
600
1
700
1
800
1
900
1 1000

Xi

1
1
1
1
1
1
1
1
no
1
1
1
1
1
1
1
1

To
298
400
500
600
700
800
900
1000

X
1
1
1
1
1
1
1
1

T
700
700
700
700
700
700
700
700

Xi

no

To
400
400
400
400
400
400
400
400
400
400
400

X
T
0 500
0
...
2 500
0
...
4 500
0
...
6 500
0
...
8 500
0
...
5
0
...
5
0
...
5 400
0
...
8 400
1
...
995
2
...
995
2
...
995
2
...
995
2
...
335
10
...
254
21
...
555
33
...
159

0
2
...
982
9
...
11
15
...
43
21
...
713
5
...
839
12
...
033
20
...
295

0
4
...
553
13
...
113
23
...
339
33
...
48
-12
...
279
2
...
941
19
...
895
38
...
995
5
...
019
12
...
24
18
...
67

H1
21
...
864
21
...
864
21
...
864
21
...
864

H2
12
...
11
12
...
11
12
...
11
12
...
11

H3
12
...
303
12
...
303
12
...
303
12
...
303

H4
18
...
113
18
...
113
18
...
113
18
...
113

Q
13
...
941
7
...
917
1
...
308
-4
...
733

1
0
...
8
0
...
6
0
...
4
0
...
2
0
...
9 0
...
1
0
...
2 0
...
7 0
...
3
0
...
4 0
...
5 0
...
5
0
...
6 0
...
3 0
...
7
0
...
8 0
...
1 0
...
9
0
1
1

H0
2
...
995
2
...
995
2
...
995
2
...
995
2
...
995
2
...
737
10
...
737
10
...
737
10
...
737
10
...
737
10
...
737

H2
5
...
55
5
...
55
5
...
55
5
...
55
5
...
55
5
...
643
5
...
643
5
...
643
5
...
643
5
...
643
5
...
643

H4
8
...
533
8
...
533
8
...
533
8
...
533
8
...
533
8
...
72
11
...
92
8
...
12
4
...
32
0
...
48
–3
...
28

0
0
...
2
0
...
4
0
...
6
0
...
8
0
...
5
0
...
5
0
...
5
0
...
5
0
...
0 0
...
5
0
...
5 0
...
3 0
...
5
0
...
5 0
...
995
2
...
995
2
...
335
5
...
335
5
...
995
2
...
995
2
...
713
2
...
713
2
...
121
4
...
121
4
...
653
-3
...
653
-3
...
18 (cont'd)
400
400
400
400
400

0
...
5
0
...
5
0
...
5
0
...
5
0
...
5

0
...
5
0
...
5
0
...
7
0
...
1
1
...
5

0
...
5
0
...
5
0
...
5
0
...
5
0
...
5

2
...
995
2
...
995
2
...
335
5
...
335
5
...
335

2
...
995
2
...
995
2
...
713
2
...
713
2
...
713

10
5
0
-5
-10
500

1000

0

1500

500

0
-1

-3

Q

Q

-2

-4
-5
-6
0

0
...
4

0
...
8

0
-0
...
5
-2
-2
...
5
-4

1

0

0
...


-3
...
653
-3
...
653
-3
...
19

4
...
121
4
...
121
4
...
2
1
...
6
1
...
0

1
...
5

no

Fermentor capacity : 550,000 gal
Solution volume : (0
...
071 lb C H OH / lb solution
|
Final reaction mixture : S0
...
86 lb H O / lb solution
T
m

2

5

m

m

2

Mass of tank contents :

Mass of ethanol produced :

solution

m

1 ft 3

65
...
4805 gal

495,000 gal

m

1 ft 3

= 4335593 lb m

4
...
071 lb m C 2 H 5OH

⇒

3
...
1 lb m C 2 H 5OH

= 6677 lb - mole C 2 H 5OH

1 ft 3 C 2 H 5OH

7
...
67 lb m C 2 H 5OH

1 ft 3

9- 18

= 3
...
19 (cont’d)
Makeup water required : 495,000 gal −

b
...
6 gal C 2 H 5OH

= 4
...

Acres reqd
...
6 gal C2 H5OH 101 bu 8 h 1 day 1 year
year

ΔHco = −56491 kJ / mol

...

C12 H 22 O11 (s) + H 2 O(l) → 4C 2 H 5OH(l) + 4CO 2 (g)

c
...
5kJ / mol

...
6 mol 0
...
811 × 104 Btu / lb - mole
1 mol
1 lb - mole
1 kJ
Moles of maltose :
4
...
071 lb C 2 H 5OH 1 lb - mole C 2 H 5OH 1 lb - mole C12 H 22 O11
1 lb m solution

46
...
336 × 106 lb m )(0
...
9 × 107 Btu ( heat transferred from reactor)

= (1669 lb - moles)( −7
...


Brazil has a shortage of natural reserves of petroleum, unlike Venezuela
...
20

a
...
1

( ΔH fo ) NH 3

B

=

−46
...
20 (cont’d)

b

g

b

Table B
...
67 kJ mol

NO g, 700° C : H3 = 90
...
1,
Table B
...
97 kJ mol

−216
...
59 kJ mol

Table B
...


B

=

C p dT

Table B
...
1,Table B
...
86 kJ mol

∑n H − ∑n H
i

i

out

i

i

= −4890 kJ min × (1 min / 60s) = −815 kW

...


9
...
5-1b used to determine ΔH
...


a
...
537 C2H4 (v)
0
...
096 N2(g)

Products at 310°C
n1 (mol C2H4 (v))
n2 (mol H2O(v))
0
...
537 0
...
02685 mol C 2 H 4 consumed

b gb

g

⇒ n1 = 0
...
537 = 0
...
02685 mol C 2 H 4 consumed 0
...
02417 mol C 2 H 5OH

g

b

g

C balance : 2 0
...
510 + 2 0
...


...
367 = n2 + 0
...
3414 mol H 2 O

9
...
537
H

H2O
N2

0
...
096

ˆ
H2
0

0
...
096

C2 H 5 OH

−

−

0
...


g

bC H g Obg, 310° Cg: H = eΔH j
2

5 2

o
f

4

(C 2 H 5 )O(l)

z

310

25

Energy balance: Q = ΔH =

C p dT

b

⇒
Table B
...
8

⇒
Table B
...
2

g

+ ΔH v 25° C +

= −204
...
1 for ΔH fo
Table B
...
415 × 10

310

25

b

0
...
28 + 16
...
69

kJ mol


...

b−24183 + 9
...

b−235
...
16g = −21115 kJ mol

310

25

b

C p dT = −272
...
05 + 42
...
3 kJ ⇒ 1
...
Separation of unconsumed reactants from products and
recycle of ethylene
...
22

C 6 H 5CH 3 + O 2 → C 6 H 5CHO + H 2 O
C 6 H 5CH 3 + 9O 2 → 7CO 2 + 4H 2 O

Basis: 100 lb-mole of C 6 H 5CH 3 fed to reactor
...
76n 0 (lb-moles N 2 )
350°F, 1 atm
V0 (ft3 )

reactor
Q(Btu)
jacket

mw(lbm H2 O( l )), 80°F

Vp (ft3 ) at 379°F, 1 atm
n1 (lb-moles C6 H5 CH3 )
n2 (lb-moles O 2 )
3
...
The calculated quantities will then be scaled to the known flow rate of water in

b

g

the product gas 29
...


9- 21

9
...
5% CO 2 formation ⇒ n4

E
...
on reactor ⇒ Q

C balance ⇒ n1

E
...
on jacket ⇒ mw

H balance ⇒ n5
O balance ⇒ n2

Scale V0 , V p , Q, mw by n5 actual / n5 basis

100% excess air:
n0 =

100 lb - moles C 6 H 5CH 3

b

g

b g

b1 + 1gmole O

1 mol O 2 reqd
1 mole C 6 H 5CH 3

2

fed

b g

b g

= 200 lb - moles O 2

1 mol O 2 reqd

N 2 feed & output = 3
...
13 mole C6H5CH3 react 1 mole C6H5CHO formed
1 mole C6H5CH3 fed
1 mole C6H5CH3 react

= 13 lb-moles C6H5CHO
0
...
005glb - moles C H CH
6

5

3

react

7 moles CO2
= 35 lb - moles CO2

...
5fa1f ⇒ n = 86
...
5gb8g + b13gb6g + 2n ⇒ n = 15
...

b200gb2glb - moles O = 2n + b13gb1g + b35gb2g + b15gb1g ⇒ n = 182
...
218 × 10

359 ft 3 STP

Ideal gas law – outlet:
C7 H 8

Vp

O2

C 7 H 8O

CO 2

H 2O

N2

IJ
K

13 + 35 + 15 + 752 lb - moles

...
5+ 182
...
443 × 10
492 R

5

ft 3

9
...
5
182
...

35
15

H7
H8
H9

substance

blb - molesg bBtu lb - moleg blb - molesg bBtu lb - moleg

Enthalpies:

LB
O
430
...
1

6

5

o
f

3

C 6 H 5CH 3 (g,350 F): H1 = 2
...
088 × 104 Btu lb - mole

bg

b

g

C 6 H 5CHO(g,T): H T = −17200 + 31 T − 77 F Btu lb - mole
⇒ H7 = −7
...
9

e

j

O 2 g,350 F : H2 = HO 2 (350 F) = 1972 × 103 Btu / lb − mole

...
9

e

j


...
9

e

j

O 2 g,379 F : H5 = HO 2 (379 F) = 2
...
9

e

j

N 2 g,379 F : H6 = H N 2 (379 F) = 2
...
1 and B
...


Table B
...
9

9

= ( ΔH f ) H 2 O( g) + H H 2 O (379 F)

=

− 1016 × 105 Btu / lb − mole

...
376 × 106 Btu

in

Energy balance on cooling jacket:

Q = ΔH = mw

⇓

z

105

80

dC i

b g dT

p H O l
2

Q = + 2
...
0 Btu (lb m ⋅ F)

b g

2
...
0

b

g

bg

Btu
× 105 − 80 F ⇒ mw = 9
...
22(cont’d)

bn g
bn g

5 actual

Scale factor:

=

29
...


d
id
i
= d6
...
02711 h i = 175 × 10

...


= 0
...
376 × 10 6 Btu 0
...
44 × 10 4 Btu / h

id

mw = 9504 × 10 Btu 002711 h

...


9
...
016 lb m H 2 O 15
...
218 × 10 5 ft 3 0
...


Vp
b
...
4805 gal

62
...

= 515 gal H 2 O min
60 min

CaCO 3 (s) → CaO(s) +CO 2 (g)

CaO(s)
900°C
CaCO3(s)
25°C

CO2(g)
900°C
Q (kJ)

10
...
0 kmol CaCO 3 ⇒ 10
...
100 kg
10
...
1

CaCO 3 (s, 25 C) : H1 =
o

B

( ΔH fo ) CaCO 3 ( s)

z

=

− 1206
...
1,
Table B
...
6 + 48
...
06 kJ / mol

298
Table B
...
8

CO 2 (g, 900o C) : H3 = ( ΔH fo ) CO 2 ( g) + HCO 2 (900o C)

Energy balance: Q = ΔH =

B

=

F n H − n H I = 2
...
94) kJ / mol = −350
...

6

kJ

9
...
Basis : 1000 kg CaCO3 fed ⇒ 10
...
75 N2
0
...
090 CO
0
...
0 kmol CaO
n2 = (014)(200) +

...
0 kmol CaCO 3 react 1 kmol CO 2 4 kmol O 2 react 2 kmol CO 2
+
= 46 kmol CO 2
1 kmol O 2
1 kmol O 2

n3 = (0
...
0)(1) + (200)(0
...
14)(1) = 46(1) + n4 (1) ⇒ n4 = 10
...
0
H1
−
−
CaO
10
−
−
−587
...
56
−350
...
0
H
−
−
2

3

N2

ˆ
H4

150

ˆ
ˆ
CaCO3 (s, 25 C) : H1 = (ΔH fo ) CaCO3 (s)
o

ˆ
H4

150

Table B
...
9 kJ/mol

ˆ
ˆ
ˆ
CO(g, 900o C) : H1 = (ΔH fo )CO(g) + H CO (900o C)

Table B
...
8
↓

=

(−110
...
49) kJ/mol = −83
...
8
↓

ˆ
ˆ
O 2 (g, 900 C) : H 2 = H O2 (900o C) = 28
...
8
↓

ˆ
ˆ
N 2 (g, 900 C) : H 3 = H N2 (900 C) = 27
...
44 × 10
GH ∑ ∑ JK
i

out

i

i

i

kJ

in

% reduction in heat requirement =

c
...
7 × 106 − 0
...
7 × 106

× 100 = 838%

...
Additional thermal energy is provided by the combustion of CO
...
24

a
...
5 × mol C consumed = (1 2) × (mol A consumed − mol C out)
⇒ nD = (1 2)( x AO f A − nC )

Balance on B: mol B out = mol B in − mol B consumed in (1) + mol B generated in (2)
= mol B in − mol A consumed in (1) + mol D generated in (2)
⇒ n B = x BO − x AO f A + n D
Balance on I: mol I out = mol I in ⇒ n I = x IO
b
...
537
0
...
096

Q(kJ) =

c
...
28
-241
...
31
-246
...
537

a
0
...
03346
0
...
08945
0
...
15E-04
6
...
57E-04
4
...
20E-05

xB0
0
...
096

c
-6
...
60E-09
-8
...
24E-07
5
...
31

H(in)
(kJ/mol)
68
...
9
-211
...
2
9
...
510
0
...
024
0
...
096

fA
0
...
77E-11
-3
...
98E-11
0
-2
...
90

H(out)
(kJ/mol)
68
...
9
-211
...
2
9
...
90 kJ
...


9- 26

9
...


CH 4 ( g) + O 2 ( g) → HCHO(g) + H 2 O(g)

n3 (mol HCHO)
n4 (mol H2O)

10 L, 200 kPa
n0 (mol feed gas) at 25°C
0
...
15 mol O2 /mol

n5 (mol CH4)
T (°C), P(kPa), 10L
Q (kJ)

Basis : n0 =

200 kPa 1000 Pa 10 L 10 −3 m 3
1 kPa

1L

1 mol K
8
...
8072 mol feed gas mixture
0
...
85)(0
...
6861 mol CH 4 ,
⇒ (0
...
8072) = 0
...
1211 mol O 2 fed
= 01211 mol CH 4

...
6861 − 01211) mol CH 4 = 0
...

HCHO produced : n3 =
H 2 O produced : n4 =

1 mol HCHO

01211 mol CH 4 consumed

...


1 mol CH 4 consumed

Extent of reaction : ξ =

( nO 2 ) out − ( nO 2 ) in

ν O2

=


...



...


References : CH 4 (g), O 2 (g), HCHO(g), H 2 O(g), at 25o C
Substance
CH 4

nin

U in

nout

U out
mol kJ mol mol kJ mol
0
...
5650
U1

O2

z
T

Ui =

25

01211

...

01211

...
2 and R = 8
...

U 1 = (0
...
7345 × 10 −5 T 2 + 01220 × 10 −8 T 3 − 2
...
6670) kJ / mol
U 2 = (0
...
1340 × 10 −5 T 2 − 2
...
6623) kJ / mol
U 3 = (0
...
3440 × 10 −5 T 2 + 0
...
8983 × 10 −12 T 4 − 0
...
25 (cont’d)

Q=

100 J 85 s
s

1 kJ
1000 J

= 8
...
1

ΔHro = (ΔHfo ) HCHO + (ΔHfo ) H2O − (ΔHfo ) CH4

B

=


...

b(−11590) + (−24183) − (−74
...
88 kJ / mol
ΔU ro = ΔH ro − RT (

∑

νi −

gaseous
products

= −282
...
314 J

1 kJ
10 3 J

mol K

= −282
...
1211)(−28288 kJ / mol) +0
...


...

Substitute for U 1 through U 3 and Q

0 = 0
...
09963 × 10 −8 T 3 − 1926 × 10 −12 T 4 − 43
...


...
8072 mol 8
...


Add heat to raise the reactants to a temperature at which the reaction rate is significant
...


Side reaction : CH 4 + 2O2 → CO2 + 2H 2 O
...


9- 28

9
...


bg

1
O 2 (g) → C 2 H 4 O g
2
C 2 H 4 + 3O 2 → 2CO 2 + 2H 2 O

C2 H 4 g +

Basis: 2 mol C 2 H 4 fed to reactor

n 6 (mol CO2 )
n 7 (mol H 2 O(l ))
25°C

Qr (kJ)
heat
n1 (mol C 2H 4)
n2 (mol O 2 )
25°C

reactor

2 mol C2 H4
1 mol O2
450°C

n 3 (mol C 2H 4)
n 4 (mol O 2 )

separation
n 3 (mol C 2H 4)
process
n 4 (mol O 2 )
n 5 (mol C 2H 4O)
n 6 (mol CO2 )
n 7 (mol H 2 O)
450°C

n 5 (mol C 2H 4O(g))
25°C


...
500 mol C 2 H 4 consumed ⇒ n 3 = 150 mol C 2 H 4
70% yield ⇒ n5 =

0
...
700 mol C 2 H 4 O

b gb g b gb g b gb

g

1 mol C 2 H 4

= 0
...

C balance on reactor: 2 2 = 2 150 + 2 0
...
300 mol CO 2

Water formed: n 7 =

0
...
300 mol H 2 O

b gb g
= n + 2n = 0
...
350g ⇒ n = 0
...
300g + b0
...
350g ⇒ n = 0
...
350 + 2 0
...
300 ⇒ n 4 = 0
...
4% C 2 H 4 , 55
...
7% C 2 H 4 , 33
...
0% C 2 H 4 , 20
...
1% C 2 H 4 , 13
...
4% C 2 H 4 O, 10
...
6% H 2 O

Mass of ethylene oxide =
b
...
350 mol C 2 H 4 O 44
...
0154 kg

References for enthalpy calculations : C s , H 2 g , O 2 g at 25° C

bg

H i T = ΔH o +
fi

z
z

= ΔH 0 +
f

T

25

C p dT for C 2 H 4

T + 273

298

C p dT for C 2 H 4 O

= ΔH o + H i ( table B
...
26 (cont’d)
Overall Process

nin

Substance

Reactor

nout

Hin

Hout

(mol) (kJ / mol) (mol) (kJ / mol)
−
−
0
...
28

C2 H4

nin

C2 H 4

Hin

−

−

O2

1

−

− 0350

...


C2 H 4 O

CO2

−

− 0300

...


H2 O l

−

− 0300

...


O2


...
375


...
350

−19
...
300

−374
...
300

−226
...

1337

= −248 kJ

i

in

∑n H −∑n H
i

out

c
...

2
79
...
26

substance

i

i

i

= −236 kJ

in

Scale to 1500 kg C 2 H 4 O day :
C 2 H 4 O production for initial basis = (0
...
05 kg
10 3 mol

) = 0
...
73 × 10 4 day −1
0
...
500gb28
...
625gb32
...
025 × 10 kg
W
kgje9
...
4% C H , 55
...
500 mol C 2 H 4
0
...
025 × 10 −3
Qprocess =

b−248 kJ ge9
...
73 × 10

4

4

−1

2

−1

24 hr 3600 s 1 kJ s

4

day −1

j

1 day

1 hr

1 kW

24 hr 3600 s 1 kJ s

9- 30

= −265 kW

4

2

9
...


1200 lb m C 9 H 12 1 lb - mole
= 10
...
75 C 3H 6
0
...
0 lb-moles C H12 /h
9

bg

bg

bg

b

C 3 H 6 l + C 6 H 6 l → C 9 H 12 l ,
Benzene balance: n2 =

10
...
0 lb - moles C 6 H 6

78
...
75n1 = n3 +

b input = output + consumption g

= 781 lb m C 6 H 6 h

10
...
67 lb - moles h
U
gV ⇒ n = 2
...
75gb16
...
08 lb C H
feed =

⇒ 0
...
20 0
...
25gb16
...
12 lb m C 4 H 10
= 768 lb m h
1 lb - mole

6

1 mole fresh feed

16
...
75 C3H6
0
...
0 lb-moles C9H12/h
2
...
17 lb-moles C4H10/h
30
...
0 lb-moles C6H6/h

6
...
5% Cmoles h
H
V
hW
62
...
50 lb - moles C 3 H 6 h
4
...
0 lb-moles C9H12 /h
2
...
17 lb-moles C 4H10 /h
30
...
0 lb-moles C6H6 /h
77°F

T (°F)

9- 31

6

4

b
...
0 lb - moles fresh feed

Overhead from T1 ⇒

6

h

Reactor :

Benzene feed rate =

3

10

46
...
4% C9H12
5
...
9% C4H10
64
...
27 (cont'd)

Energy balance: ΔH = 0 ⇒

∑ n eH
i

i , out

j ∑ n C bT

− Hi , in =

i

pi

out

− Tin

g

=0

i

(Assume adiabatic)

LM10 lb - moles C H
h
N
9

C 4 H10

B

b gb

C 3H 6

OPe
Q

B
120 lb m
0
...
50 42
...
57 200 F − 400
1 lb - mole 1b m ⋅ F

12

gb ge

j b A gb

j b gb

gb ge

gb ge

j

+ 4
...
55 200 F − 400 F + 30
...
45 200 F − 400 F

...


b A gb

C6H6 in
effluent

gb ge

j

+ 40
...
45 T − 77 F = 0 ⇒ T = 323° F

...
0
0
2
...
17
0
4
...
0
8650
30
...
0
15530

Substance
C3H 6
C 4 H10
C6H6
C 9 H12

H in

Energy balance on reactor :
nC H ΔH ro
Q = ΔH = 9 12
+
ni Hi −
vC 9 H12
out

∑

∑n H
i

i

in

b10
...
50gb7750g + b4
...
0gb11350g + b10
...
0gb8650g = −183000 Btu h b heat removalg
=

9
...


100 kg C8 H 8 10 3 g
h

1 kg

1 mol
104
...
28 (cont'd)
Reactor :

...
35n3 mol C8 H 10 react
h
⇒ n3 = 2740 mol C 8 H10 h fed to reactor

35% 1-pass conversion ⇒

a

g

1 mol C8 H 8
= 960 mol C8 H 8 h
1 mol C8 H 10

f

⇒ Recycle rate = 2740 − 960 = 1780 mol C 8 H10 h recycled
Reactor feed mixing point
2740 mol C8H10(v)/h
500oC
2740 mol C8H10(v)/h
n4 [mol H2O(v)/h]
600oC

n4 [mol H2O(v)/h]
700oC

b g

Energy balance: ΔH = 2740 ΔHC H + n4 ΔHH O = 0 kJ h
8 10
2

b Neglect Q, ΔE g
k

ΔHC8 H10
ΔH H 2 O

LM
OP J 1 kJ
=M
b118 + 0
...
3 kJ mol
MN
PQ

z

600

3

500

Table B
...
9 kJ mol

a2740fa28
...
9f = 0 ⇒ n
Ethylbenzene preheater bA g :
4

b
...
99 × 10 4 mol H 2 O / h

bg

960 mol fresh feed 1780 mol recycled 2740 mol EB l
at 25° C
+
=
h
h
h
2740 mol EB v
at 500° C
⇒
h

bg

ΔH =

z

a

136

25

f

C pi dT + ΔH v 136° C +

QA = ΔH =

z

a

500

136

f

C pv dT = 20
...
0 + 77
...
9 kJ mol

2740 mol C 8 H10

133
...
67 × 10 5 kJ h preheater

bg
19400 mol h H Obl, 25° Cg → 19400 mol h H Ob v, 700° C, 1 atmg
Table B
...
8 kJ kg ;
Table B
...
28 (cont'd)
QF = ΔH =

19400 mol H 2 O 18
...
34 × 10

b

6

kJ h steam generator

a3928 − 104
...
8)

Substance
C 8 H10
H 2O
C8H8
H2

nin

nout

Hin

Hout

(mol h ) (kJ mol) (mol h ) (kJ mol)
2740
0
1780
−11
...
56
960
−
−
−10
...

960
−
−
−119

Energy balance :
Qc = ΔH =

960 mol C 8 H 8 produced
124
...
61 × 10 4 kJ h reactor

c
...

n f (mol/h) at 145°C, 1 atm
0
...
58 mol air/mol
0
...
79 mol N2 /mol air
n s mol H2 O(v )/h
saturated at 145°C

b
...
The reactor effluents are cooled to 25 C , and
then all but the hydrogen are reheated after separation
...

CH 3OH → HCHO + H 2 ,

9
...
37 kg HCHO/h
n 5 (mol H 2 /h)
0
...
1 bars
30oC

In the absence of data to the contrary, we assume that the separation of methanol from
formaldehyde is complete
...


9- 34

9
...
42 760 mmHg = 319
...
2 mmHg ⇒ Tm = 44
...


Moles HCHO formed :
=

36 × 106 kg solution 0
...
03 kg HCHO

24 h

= 52
...
80 kmol HCHO h
70% conversion :
52
...
70 kmol CH 3OH react 0
...
59 kmol h

Methanol unreacted:
n1 =

b0
...
59gkmol CH OH fed b1 − 0
...
63 kmol CH OH
3

3

h

b

1 kmol CH 3OH fed

3

h

gb gb g

N 2 balance: n3 = 179
...
58 0
...
29 kmol N 2 h

Four reactor stream variables remain unknown — ns , n2 , n5 , and n6 — and four relations are
available — H and O balances, the given H 2 content of the product gas (5%), and the energy
balance
...


b

gb gb g

b

gb g b gb g

H balance: 179
...
42 4 + 2ns = 22
...
8 2 + 2n5 + 2n6
⇒ n s = n5 + n 6 − 52
...
6 0
...
6 (0
...
21 2 + n s = (22
...
80)(1) + n 6
⇒ n s = 2n 2 + n 6 − 43
...
05 ⇒ 19n5 − n 2 − n 6 = 157
...
63 + n 2 + 82
...
89 + n5 + n 6

H 2 content:

bg b g b g b g

References : C s , H 2 g , O 2 g , N 2 g at 25° C
H=

ΔH fo

+

z

Table B
...
8 for O 2 , N 2 and H 2

9- 35

(3)

9
...
43

−195220

22
...

2188
82
...
29

18410
17390

H 2O

ns

−237740

n6

−220920

HCHO

−

−

52
...
406 × 106

(4)

in

We now have four equations in four unknowns
...

ns =

bg

58
...
02 kg

h

1 kmol

= 1060 kg steam fed h

n2 = 2
...
00 kmol H 2 O h

...
63 kmol CH3OH/h, 2
...
29 kmol N2/h, 52
...
58 kmol H2/h, and 98
...


272 kmol h product gas
8% CH 3 OH, 0
...
Since we have already calculated specific enthalpies of all
components of the product gas at the boiler inlet (at 600°C), and for all but two of them at the
boiler outlet (at 145°C), we will use the same reference states for the boiler calculation

bg b g b g b g
H Oblg at triple point for boiler water

Reference States: C s , H 2 g , O 2 g , N 2 g at 25° C for reactor gas
2

Substance
CH 3 OH
O2
N2
H 2O
HCHO
H2
H 2O

nin

ˆ
H in

nout

ˆ
H out

kmol/h

kJ/kmol
− 163200
18410
17390
− 220920
− 88800
16810
125
...
1
(kJ/kg)

22
...
26
82
...
02
52
...
58
mb
(kg/h)

9- 36

22
...
26
82
...
02
52
...
58
mb
(kg/h)

9
...
7 − 4
...

⇒ mb = 1892 kg steam h

9
...


C 2 H 4 + HCl → C 2 H 5Cl

Basis:

bg

1600 kg C 2 H 5Cl l
h

n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4 ( g)/h)
condenser
n 5 (mol C 2H 6 ( g)/h)
n 6 (mol C 2H 5 Cl( g)/h)
50°C

A
n 1 (mol HCl(g)/h)
0°C

C
n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4 ( g)/h)
n 5 (mol C 2H 6 ( g)/h)
0°C

6 (mol 2H Cl(l)/h
n 6n(mol CCH 55Cl( g)/h)
2

reactor

B

103 g 1 mol
= 24800 mol h C 2 H 5Cl
1 kg 64
...
93 C 2H 4
0
...
015n1

b

b1g
b2 g
b3g

g

n4 = 0
...
93n2 = 0
...
07n2

Overall Cl balance :

b

n1 mol HCl h

g

b gb g b

gb g

b4g

1 mol Cl
= n3 1 + 24800 1
1 mol HCl

Solve (4) simultaneously with (1) ⇒ n1 = 25180 mol h = 2518 kmol HCl fed / h

...
93 2 + n2 0
...
(2) and (3) ⇒ 2n2 0
...
07 − 0
...
07 = 2 24800
n2 = 27070 mol fed h = 27
...


U 2
...
01395b27070g = 378 mol C H hV
14
...
3% C H , 71
...
07b27070g = 1895 mol C H h |
W

n3 = 378 mol HCl h
n4
n5

2

2

4

6

9- 37

2

4

2

6

9
...


bg

bg

bg

bg

References : C 2 H 4 g , C 2 H 6 g , C 2 H 5 Cl g , HCl g at 0 C

e

z

j

50

C 2 H 4 g, 50 C : H = C p dT

Table B
...
181 kJ mol

0

50
ˆ
C2 H 6 ( g, 50 C ) : H = ∫ C p dT ⇒ 2
...
2

e

z

j

0

50

HCl g, 50 C : H = C p dT

Table B
...
456 kJ mol

0

e j
C H Cleg, 50 Cj: H =

e j

C 2 H 5Cl l, 0 C : H = − Δ H v 0 C = −24
...
709 kJ mol

0

nin
nout
Hin
Hout
mol
mol kJ / mol
kJ / mol

...
181
1895
0
1895 2
...
7
n6
2
...
5 kJ
+ ( 378)(1
...
181) + (1895)( 2
...
709n6 − ( n6 − 24800)( −24
...
7
h

C2 H 5 Cl recycled =

d
...

ΔH v is independent of temperature
...

Heat of mixing and influence of pressure on enthalpy is neglected
...

No C2H4 or C2H6 is absorbed in the ethyl chloride product
...
31

a
...
7 kJ / mol

9
...


Well-insulated reactor, so no heat loss
No absorption of heat by container wall
Neglect kinetic and potential energy changes;
No shaft work
No side reactions
...


References : NH 3 ( g), O 2 ( g), NO(g), H 2 O(g) at 25o C, 1atm
Substance
NH 3 (g)
O 2 ( g)

nin
nout
Hin
Hout
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
−
−
4
...
00
100
H2
H3

...
00
6
...
2

200

H1 =

−
−

(C p ) NH 3 dT

B

=

Table B
...
74 kJ / mol,

H2 = HO 2 (200 C)
o

B

=

5
...
2 :
ˆ
H3 = (0
...
5790 ×10−5 Tout 2 − 0
...
3278 ×10−12 Tout 4 − 0
...
0295 Tout + 0
...
0975 ×10−8 Tout 3 + 0
...
7400) kJ/mol
ˆ
H5 = (0
...
3440 ×10−5 Tout 2 + 0
...
8983 ×10−12 Tout 4 − 0
...
00) H 3 + (4
...
00) H 5 − (4
...
00) H 2

⇓

o
ˆ
ˆ
ˆ
Substitute for ξ , ΔH r , and H1 through H 6

ΔH = (0
...
28 × 10−5 Tout 2 + 0
...
697 × 10−12 Tout 4 )
− 972
...


If only the first term from Table B
...
03515(200 − 25) = 615 kJ / mol, H 2 = 5
...
0291(Tout − 25),

...
0295(Tout − 25), H5 = 0
...
B
...
00) H 3 + (4
...
00) H 5 − (4
...
00) H 2 = 0

⇓

o
ˆ
ˆ
ˆ
Substitute for ξ (=1 mol/s), ΔH r ( = −904
...
3479 Tout − 969
...


9
...


Basis : 100 lb m coke fed
⇒ 84 lb m C ⇒ 7
...
00 lb - moles CO 2 fed
7
...
00 lb-moles(84 lbm)C/hr
16 lb mash/hr
77°F
585,900 Btu

a
...

−39350 − b2gb −282
...
9486 Btu 453
...
00 x lb - moles C reacted

b g
= 7
...
0 x lb - moles CO
1 lb - mole C reacted

n2 = 7
...
32 (cont'd)
⇒ 3130 Btu lb - mole
b
g
CO bg,1830° Fg: H = H (1830 F) ⇒ 20,880 Btu lb - mole
CObg,1830° Fg: H = H (1830 F) ⇒ 13,280 Btu lb - mole
0
...
9

Table B
...
9

CO

m

m

Mass of solids (emerging)
=

b g

7
...
0 lb m
1 lb - mole

substance

b

g

+ 16 lb m = 100 − 84 x lb m

nin
nout
Hin
Hout
(lb − moles) (Btu lb - mole) (lb − moles) (Btu lb - mole)
7
...
00 1 − x
20,890
14
...
0 x = 2ξ ⇒ ξ (lb - moles) = 7
...
0 x fa13,280f + a100 − 84 x fa 420f − a7
...
0 x (lb - moles)

74,210 Btu
lb - mole

+ 7
...
801 ⇒ 80
...


Advantages of CO
...

Disadvantages of CO
...
Also, it costs something to produce it from coke
...
33

Basis :

17
...
00 atm

1 mol

a f = 3497 mol h feed

3

h
1m
298 K 1
...
4 L STP
CO g + 2 H 2 g → CH 3OH g ,

bg

ΔH ro

bg
=e
j
ΔH fo

bg
b g − e ΔH j
o
f

CH 3OH g

= −90
...
333 mol CO/mol
n 2 (mol CO/h)
0
...
05 kW

Let f = fractional conversion of CO (which also equals the fractional conversion of H 2 , since
CO and H 2 are fed in stoichiometric proportion)
...
333) mol CO feed

f ( mol react )

= 1166 f ( mol CO react)
mol feed
1166 f mol CO react 1 mol CH 3OH
= 1166 f mol CH 3OH h
CH 3OH produced : n1 =
1 mol CO
CO remaining : n2 = 1166 1 − f mol CO h
1166 f mol CO react 2 mol H 2 react
H 2 remaining : n3 = 3497 0
...
8

e

j

CO g,127 C : H1 = HCO (127 C) = 2
...
8

e

j

H 2 g,127 C : H2 = H H 2 (127 C) = 2
...
2

C p dT = 5
...
05 kJ 3600 s
s

b

+ 2332 1 − f

1h

= (1166 f )( −90
...
993g + 1166 f b5
...
99g
g


...
173 × 10 4 ⇒ f = 0
...
33 (cont’d)

b g
= 1166b1 − 0
...
9 mol h
= 2332b1 − 0
...
9 mol h
E

n1 = 1166 0
...
1 mol h
n2
n3

ntot = 1980

9
...


b g

mol
1980 mol 22
...
00 atm

273 K 5
...
0 m 3 h

bg

CH 4 g + 4S g → CS 2 g + 2 H 2S g , ΔHr ( 700° C ) = −274 kJ mol

Basis : 1 mol of feed
1 mol at 700°C
0
...
80 mol S/mol

Product gas at 800°C
n1 (mol CS2)
n2 (mol H 2S)
n3 (mol CH 4)
n4 (mol S (v))

Reactor

Q = –41 kJ

Let f = fractional conversion of CH 4 (which also equals fractional conversion of S, since the
species are fed in stoichiometric proportion)
Moles CH 4 reacted = 0
...
20 f

b

g

n 3 = 0
...
80 mol S fed −

b

0
...
20 f mol CH 4 react

n2 =

0
...
80 1 − f mol S

= 0
...
40 f mol H 2 S

bg

References: CH 4 (g), S g , CS 2 (g), H 2S(g) at 700°C (temperature at which ΔHr is known)
substance nin
CH 4
S

nout

Hin

Hout

bmolg bkJ molg bmolg bkJ molg
0
...
20b1 − f g
H
0
...
80b1 − f g
1

2

CS 2

−

−

0
...
40 f

H4

b

g

CH 4 g, 800° C : H1 = 7
...
64 kJ / mol

b
g
H Sbg, 800° Cg: H

CS 2 g, 800° C : H3 = 3
...
48 kJ / mol

9
...
20b1 − f gb7
...
80b1 − f gb3
...
20 f b3180g + 0
...
480g

...
20 f −274
...
800

b
...
04 mol CH4
0
...
16 mol CS2
0
...
20 mol CH4
0
...
20 mol CH4
0
...
20 mol CH4
0
...
04 mol CH4
0
...
16 mol CS2
0
...
Assume the heat exchanger is adiabatic, so
that the only heat transferred to the system from its surroundings is Q for the preheater
...
20

H1

0
...
04

H2

0
...
80

H3


...

016

H4

0
...


H5

016

...
32

H6

0
...
6 − 200J + ΔH bT g + dC i aT − 444
...
7 kJ mol

b

9- 44

p Sg

9
...
84 kJ / mol
Sbl, 150° Cg: H = − 1
...
83 kJ / mol

b
g
H Sbg, 800° Cg: H = 26
...
7 kJ / mol
Sbg, 700° Cg: H = 100
...
57 kJ / mol
4

CS2 g, 800° C : H5 = 19
...


i

7

8

b g ∑n H − ∑n H

Energy balance: Q kJ =

6

i

i

⇒ Q = 59
...
2 kJ mol feed

in

The energy economy might be improved by insulating the reactor better
...


9- 45

9
...


C2 H 6 ⇔ C2 H 4 + H 2 , K p =

x C2 H 4 x H 2

P = 7
...


P⇒ Kp =

x C2 H 4

p

= f

2

g

1 − f mol C 2 H 6
1+ f
mol
f mol C 2 H 4
C2 H 4 =
1+ f
mol
f mol H 2
H2 =
1 + f mol

ξ (mol) = f
C2 H 6

(1)

2

b1+ f g P
b1− f g
b1+ f g
2

F K I
P⇒ f =G
H P + K JK

=

f2P
f2
P
=
1− f 1+ f
1− f 2

b

gb

12

g

b2 g

p

p

bg

bg

bg

References : C 2 H 6 g , C 2 H 4 g , H 2 g at 1273 K

Energy balance:

b

g ∑n H −∑n H

ΔH = 0 ⇒ ξ ΔH r 1273 K +

eH j
eH j
i

i

i

i

i

out

b

i

in

= 0 inlet temperature = reference temperature

in

out

=

z

g

T

1273

C pi dT

⇓ energy balance

b

g b

f ΔH r 1273 K kJ + 1 − f

z

g dC i
T

1273

p

dT + f

z

T

1273

C2 H 6

dC i

p C H dT
2 4

rearrange, reverse limits and change signs of integrals

1− f
=
f

b

g

ΔH r 1273K −

z

1273

z

dC i
dC i

p C H dT
2 4

T

−

p C H
2 6

dC i

dT

bg

φ T

1− f
1
= φ T ⇒ 1 − f = fφ T ⇒ f =
f
1+ φ T

bg

1273

T

1273

T

z

bg

b g b4 g

9-46

p H dT
2

b3g

+f

z

T

1273

dC i

p H dT
2

=0

9
...


1273

−3

T

1273

T

z

e26
...
167 × 10 T jdT

...

b1135 + 01392T gdT

9
...


3052 + 36
...
05943T 2
127240 − 113T − 0
...


F K I = 1 ⇒ F K I − 1 = ψ bT g = 0
GH 1 + K JK 1 + φbT g GH 1 + K JK 1 + φbT g
φ bT g given by expression of Part b
...
(1)
12

12

p

p

p

p

p

d
...
01
0
...
1
0
...
4
872
...
8
960
...
518
0
...
446
0
...
36
0
...
261

Kp
(atm)
0
...
0141
0
...
0886
0
...
4646
0
...
93152 -0
...
12964 -0
...
24028 0
...
57826
3
...
77566
4
...
42913 -2
...
83692 -7
...
ln P

1100

0
...
5

1000

f

T(K)

0
...
3
0
...
1
0

700
-3

-2

-1

0

1

2

-3

-2

ln P(atm)

e
...
0
TLAST = 0
...
0

9-47

1

2

9
...
AND LOOK FOR A SIGN IN PSI
DO 10I =1, 20
CALL PSICAL (T, PHI, PSI)
IF ((PSIL*PSI)
...
0
...

10
CONTINUE
40
IF (T
...
0
...
NE
...
LT
...
01) GO TO 99
CALL PSICAL (T2, PHIT, PSIT)
IF (PSIT
...
0) GO TO 99
IF ((PBIT*PBIL)
...
0
...
GT
...
0) TLAST = T2
IF ((PSIT*PSI)
...
0
...
GT
...
0) T = T2
50
CONTINUE
IF (I
...
20) WRITE (3, 3)
3
FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS')
93
STOP
END
SUBROUTINE PSICAL (T, PHI, PSI)
REAL KF
PHI = (3052 + 36
...
2*T + 0
...
– 11
...
0636*T**2)
KP = 7
...
/T)
FBI = SQRT((KP/(1
...
/12
...
2, 4X, 'PSI =', E11,4)
RETURN
END
OUTPUT: SOLUTION TO PROBLEM 9-35
T = 1200
...
8226E + 00
T = 1150
...
7048E + 00
T = 1100
...
5551E + 00
T = 1050
...
3696E + 00

...
00 PSI = 01619 E + 00
T = 950
...
3950E − 01

...
80 PSI = −01824 E − 02
T = 960
...
7671E − 04
T = 960
...
3278E − 05
Solution: T = 960
...
360 mol C 2 H 6 reacted mol fed

9-48

2CH 4 → C 2 H 2 + 3H 2

9
...
0 mol CH 4 (g)/s
o

n2 (mol C 2 H 2 / s)

1500 C

n3 (mol H 2 (s)/s)
n4 (mol C(s)/s)
1500o C

975 kW
a
...
600 = 4
...

10
41
...
45
n3
−
−
45
...
45

Energy Balance: Q = ΔH ⇒ 975 kJ / s =

∑n H −∑n H
i

out

i

i

i

(3)

in

n 2 = 2
...
50 mol H 2 / s

...


2
...
417 mol C 2 H 2 mol CH 4 consumed
6
...
0(1 − 0
...
00 mol CH 4 / s
n 3 = 0 ⇒ n 2 = 3
...
00 mol H 2 / s

Yield of acetylene =

3
...
500 mol C 2 H 2 mol CH 4 consumed
6
...
417
= 0
...
500

9-49

9
...
94 m3 at 1400°C, 1 atm
n g (mol)

n g (mol), 900°C

ng =

4
...
4 L

= 35
...

n1 = 0

n4 = ξ 2

n1 = 1 − ξ 1 ⇒ ξ 1 = 1 mol
ξ 1 =1

ξ 1 =1

n 2 = 6 − 3ξ 1 − ξ 2 ⇒ n 2 = 3 − ξ 2

n5 = 7ξ 1 + ξ 2 ⇒ n5 = 7 + ξ 2

ξ 1 =1

n 3 = 3ξ 1 − ξ 2 ⇒ n 3 = 3 − ξ 2

bg bg

References : C(s), H 2 g , O 2 g at 25°C, heating gas at 900°C

z
T

o
H i = ΔH fi + C pi dT

for C 3 H 8

25

= Table B
...
39

mol
0

kJ / mol
−

H 2O

6

−238
...
78

CO

−

−

3−ξ2

−86
...


H2

−

−

7+ξ2

22
...
00

35
...
99

Energy Balance :

∑n H −∑n H
i

out

i

i

i

= 0 ⇒ ξ 2 = 2
...
7 mol % H 2 O, 7
...
4% CO 2 , 69
...
38

a
...
Without the energy released by reaction
(2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic
reactor and hence the reaction rate would drop
...


Basis : 1
...
500 kg H20)
0
...
0 kg coal
0
...
226 kg ash/kg coal
0
...
812 kg C / kg combustible
0
...
054 kg H / kg combustible

R
|
|
S
|
|
T

nf1 (mol C)
nf2 (mol O)
nf3 (mol H)
nf4 (mol H2O)
0
...
00)(0
...
812) kg C][1 mol C / 12
...
23 mol C
n f 2 = (1
...
669)(0
...
0 × 10 −3 = 5
...
00)(0
...
054) / 1
...
77 mol H
n f 4 = [ (0
...
105) kg][1 mol H 2 O / 18
...
58 mol H 2 O

n0 (mol O2) 25°C

Product gas at 2500°C
n1 (mol CO2)
n2 (mol CO)
n3 (mol H2)
n4 (mol H2O)

1 kg coal + H2O, 25°C
45
...
60 mol O
35
...
58 mol H2O
0
...
226 kg slag
2500°C

Reactive oxygen (O) available = (2n0 + 5
...
77 mol H 1 mol O
= 17
...
60) − 17
...
28) mol O
CO2 formed ( C+2O → CO2 ) : n1 =

C balance : 45
...
28) mol O 1 mol CO2
2 mol O

= (n0 − 6
...
14

⇒

n2 = (51
...
60 + 33
...
77+ 2(33
...
14

⇒

n2 = 51
...
06

⇒

n4 = (n0 + 0
...
37 − n0 ) mol H 2

9
...


1 kg coal contains 45
...
77 mol H
⇒ 1 kg coal + nO 2 → 45
...
77 / 2) mol H 2 O (l)
ΔH r = −21,400 kJ = 45
...
77 / 2)( ΔH fo ) H 2 O(l) − ( ΔH fo ) coal
⇒ ( ΔH fo ) coal = −1510 kJ / kg

Re ferences : C(s), O 2 (g), H 2 (g), ash(s) at 25o C
nin

ˆ
H in

nout

ˆ
H out

(mol)

(kJ/mol)

(mol)

CO 2

−

−

n0 − 6
...
37 − n0

H2

−

51
...
06

ˆ
H4

Substance

1

H2O

33
...
266 kg

ˆ
H 5 (kJ/kg)

0

3

ˆ
ˆ
H i = ΔH o + C pi (2500 − 25), i = 1,3
fi
ˆ
H1 = −393
...
0508(2475) = −267
...
52 + 0
...
35 kJ/mol CO
ˆ
H 3 = 0
...
25 kJ/mol H 2
ˆ
H 4 = −241
...
0395(2475)= − 144
...
4(2475) = 710 + 1
...
4 mol O 2

9-52

9
...
064 g

L

1 mL

FG 98
...
941× 10
H 1 mol K

5

g H 2 SO 4

= 3192 × 10 6 g solution

...
941 × 10 5 )g H 2 O(

...

18
...
6 mol H O mol H SO

...
39g

...
7 kJ mol
eΔH j b
g
b
g
bg
mol
5

2

n

2

2

2

f

4

2

o
f

H 2SO 4 aq
...
6

2

4

4

s

H 2SO 4 l

H 2SO 4 aq
...
6

A

A

Table B
...
11

H = (3000 mol H 2 SO 4 )(-884
...
65 × 10 6 kJ

9
...
1, B11

=

∞

− 92
...
45 kJ mol

...
11

ΔH fo

NaOH (aq):

=

e

e

ΔH fo

j

ΔH fo

bg

NaOH s

+

e

ΔH so

j

=

∞

− 426
...
89 = −469
...
1

NaCl (aq):

ΔH fo

b g

=

b g

j

bg

NaCl s

+

e

ΔH so

b g

j

= −4110+ 4
...


...
45 − −469
...
0 kJ mol

...

2

ΔH ro

=

∑

v i ΔH fo −
products

∑

v i ΔH fo
reactants

b

g b

g

= −4110 − 28584 − −92
...
6

...


kJ mol = −177
...
41

a
...

b g

bg

H 2 SO 4 aq + 2NaOH aq → Na 2 SO 4 aq + 2H 2 O l

Basis: 1 mol H 2 SO 4 soln ⇒
⇒

⇒

b

g

0
...
08 g mol = 9
...
90 mol H 2 O × 18
...
22 g H 2 O

U
|
V
|
W

26
...
49 cm 3
127 g

...
10 mol H 2 SO 4

2 mol NaOH
1 mol H 2 SO 4

b g

1 liter caustic soln 10 3 cm 3
= 66
...
41 (cont'd)
Volume ratio =

b
...
67 cm 3 NaOH(aq)

= 3
...
49 cm H 2 SO 4 (aq)

b g

H 2 SO 4 aq : r = 9 mol H 2 O / 1 mol H 2 SO 4

eΔH j
o
f

soln

e

= ΔH fo

j


...
, r = 9g = b−81132 − 65
...

NaOH(aq) : The solution fed contains 66
...
34 g , and

b
g
⇒ b75
...
00g g H O ⇒ b67
...
02 gg = 3
...
2 mol NaOH) 40
...
00 g NaOH
2

2

2

⇒ r = 3
...
20 mol NaOH = 18
...
, r =18
...
6 − 42
...
4

4

o
f

o
f

= ΔH fo

o
f

soln

kJ

o

NaOH s

bg

Na 2SO 4 s

e

+ ΔH fo

j

b g

Na 2SO 4 aq

b

= −1384
...


kJ mol NaOH

kJ
g mol = −1385
...


...
10 mol) −1385
...
55) −(2)( −469
...


9
...
2 kJ
mol

Table B
...


e

NaCl(aq): ΔH fo = ΔH fo

j

bg

NaCl s

e

+ ΔH so

j

B

∞

=


...

b−4110 + 4
...
1

B
e j b g + eΔH j = b−426
...
89gkJ / mol = −469
...
5 − b−4061g − b−28584g kJ mol = 222
...


...


8500 ktonne Cl 2
yr

10 3 tonne 10 3 kg 10 3 g 1 mol Cl 2
1 ktonne 1 tonne 1 kg 70
...
778 × 10 −7 kW ⋅ h 1 MW ⋅ h
1 kJ
1J
10 3 kW ⋅ h

9-54

222
...
5 mol Cl 2

= 148 × 10 7 MW ⋅ h / yr

...
43

a
...
44

ΔH ro2 = +32
...
26 kJ mol
From (1), ΔH = e ΔH j
b
g − eΔH j b g
⇒ e ΔH j
b
g = b−64
...
96g kJ mol = −859
...


ΔH ro1 = −64
...


References : Elements at 25°C

b

z

FG
H

75

g

NH 3 g, 75° C : H =

b

ΔH fo

g

25

e

H 2 SO 4 aq , 25° C : H = ΔH fo

j

o
f

4

b g = −907
...


...
36 kJ mol (Table B
...
2)

NH 4

kJ mol H 2 SO 4 (Ta
...
1)


...
1)

Energy balance:
Q = ΔH =


...
36g − b1gb−907
...


i

i

in

kJ withdrawn
mol NH 4 2 SO 4 produced

b

g

1 mole % (NH 4 ) 2 SO 4 solution ⇒

1 mol (NH 4 ) 2 SO 4
99 mol H 2 O 18 g

132 g
= 132 g (NH 4 ) 2 SO 4
mol

1782 g H 2 O
mol
1914 g solution
The heat transferred from the reactor in part (a) now goes to heat the product solution from
25 C to Tfinal ⇒ 177 kJ =

c
...

1914g

1 kg
3

10 g

=

4
...
1 C

In a real reactor, the final solution temperature will be less than the value calculated in part b, due
to heat loss to the surroundings
...
1oC
...
45

a
...
3 = −884
...

bg
NaOHeaq , r = 19, 25 Cj:
L
O
nH = b2 mol NaOH gMe ΔH j
+ ΔH b r = 19gPb kJ molg + 38e ΔH j
bg
N
Q
= b2g −426
...
8 = −938
...
0 kg
1 kmol

b

nH = 1 mol Na 2 SO 4

gLMNeΔH j
o
f

b

89 kmol H 2 O 18
...


Na 2 SO 4

1 kmol

e

+ ΔH so

j

Na 2SO 4

g

OP + 89eΔH j
Q
o
f

= 1604 kg ⇒ 1746 kg

...


b g + mC p b40 − 25g

H 2O l

o
ΔH f =−1384
...
1
o
ΔHs =−1
...
814 kJ (kg⋅ C)
m=1
...
08 g H 2 SO 4
1 mol

⇒

b
...
84 kJ mol

e

ni H i = 547
...
02 g H 2 O
1 mol

b g = −24
...
981 kg

Q
−24
...
8 kJ / kg acid transferred from reactor contents
M acid 0
...
3 × 10 3 J =

1746 kg 4
...

kg ⋅ C

9-56

dT

f

i

− 40 C

⇒ T f = 43 C

9
...


b g

b g

b g

bg

H 2SO 4 aq + 2NaOH aq → Na 2SO 4 aq + 2H 2 O l

H 2SO4 solution: :
75 ml of 4M H 2SO 4 solution ⇒

4 mol H 2SO 4
1 L acid soln

1L
75 mL
= 0
...

b75 mLgb123 g mLg = 92
...
3 mol H SO )b98
...
42 g H SO
⇒ b92
...
42g g H O ⇒ b62
...
02 gg = 3
...
49 mol H 2 O 0
...


eΔH j
o
f

soln

e j

= ΔH fo

e j

+ ΔH fo

bg

H 2SO 4 l

Table B
...
11

b

B
=

g

H 2SO 4 aq
...
63

kJ

...
42g mol

= −878
...
30 m ol H 2 SO 4 2 m ol N aOH
1 m ol H 2 SO 4

b g

10 3 m L
= 50
...

b50
...
5 g

12 mol NaOH

1L

50 mL

1 L NaOH(aq) 103 mL

b

g

40 g/ mol NaOH

= 0
...
00 g NaOH

⇒ 68
...
00 g H 2 O ⇒ 44
...
02 g = 2
...
47 mol H 2 O 0
...
12 mol H 2 O
mol NaOH


...
, r =4
...
6 − 3510g mol
kJ

o

NaOH s

= −46170 kJ mol NaOH

...

b g + eΔH f j Na SO baq g = b−1384
...
7
kJ

o

Na 2SO 4 s

2

kJ mol Na 2SO 4

4

mtotal = total mass of reactants or products = (92
...
5g NaOH) = 160
...
161 kg
Extent of reaction: (nH 2SO ) final = (nH 2SO 4 ) fed + ν H 2SO 4 ξ ⇒ 0 = 0
...
30 mol
4

Standard heat of reaction

e j

ΔH ro = ΔH fo

b g + 2e ΔH f j H Oblg − eΔH f j H SO baq g − 2eΔH f j NaOHbaq g
o

Na 2SO 4 aq

o

2

o

2

4

Energy Balance : Q = ΔH = ξΔHro + mtotal C p (T − 25) C

F
GH

= (0
...
184

...

b
...

Heat transferred to and through the container wall is negligible
...
47

Basis : 50,000 mol flue gas/h

50,000 mol/h
0
...
997 N 2
50°C

n4 (mol SO2 /h)
n5 (mol N 2 /h)
35°C

n1 (mol solution/h)
0
...
900 H 2O( l )
25°C

n2 (mol NH4 HSO3 /h)
1
...
997fb50,000 mol hg = 49,850 mol N h
N balance:
NH balance: b2gb0100gbn g = n + b15gb2gn ⇒ n = 20n U n = 5400 mol h

...

|⇒
V
S balance:
0100n + b0
...


...
W
270 mol NH HSO produced
1 mol H O consumed
H O balance: n = b0900gb5400g −

...
100 0
...
0 mol SO2 h
2

5

2

+
4

1

2

2

1

1

2

4

2

1

2

4

2

2

3

3

h

2

3

4

3

2

Heat of reaction:
− e ΔH j
e j
b g − e ΔH j b g b g − e ΔH j

...
90g − b −28584g kJ mol = −47
...

SO eg, 50 Cj: H = z dC i dT = 101 kJ mol ( C from Table B
...
40 kJ mol

N 2 g, 50 C : H = 0
...
8)
2

H = 0
...
5 × 270 mol NH 4 2 SO 3 116 g 4725 mol H 2 O 18 g
g
+
= 159,000
h
1 mol
h
h
mol

nH = mC p ΔT =

159,000 g 4 J
h
g⋅° C

a35 − 25f° C

1 kJ
= 6360 kJ / h
10 3 J

Extent of reaction:
(nNH 4 HSO3 ) out = (nNH 4 HSO 3 ) in + ν NH 4 HSO 3 ξ ⇒ 270 mol / h = 0 + 2ξ ⇒ ξ = 135 mol / h

9-58

9
...
48

a
...
3 kJ

135 mol
h

effluent solution

6360

bg

∑n H − ∑n H
i

i

out

i

i

in
N 2 out

a fa f a fa f
mol
−22,000 kJ
− a50,000fa0
...
01f − a 49,850fa0
...
40 + 49,850 0
...
11 kW
3600 s 1 kJ s

bg

CH 4 g + 2O 2 g → CO 2 (g) + 2H 2 O v

at 25° C

Table B
...
36 kJ / mol, LHV =

− 2 ΔH v

H 2O

b

g

= 890
...
01 kJ mol

= 802
...
9 kJ / mol, LHV = 1559
...
01 kJ mol = 1427
...
0 kJ / mol, LHV = 2220
...
01 kJ mol = 2043
...
875 890
...
070 1559
...
020 2200
...
875 802
...
070 1427
...
020 2043
...


g
g
gFGH mol IJK + b0
...
07 mol IJK
g I
g I
1 kg
F
F
+ b0
...
09
H mol JK + b0
...
02 mol JK ] × 10 g = 0
...
875 mol CH 4 16
...


9
...
01800 kg

3

= 46800 kJ kg

The enthalpy change when 1 kg of the natural gas at 25oC is burned completely with oxygen at
25oC and the products CO2(g) and H2O(v) are brought back to 25oC
...
1

bg

e j

o
C s + O 2 (g) → CO 2 (g), ΔHc = ΔH fo

CO 2 ( g)

B

=


...
01 g 1 kg

Table B
...
064
2

SO 2

= −296
...
1

bg

bg

e j

1
o
H 2 g + O 2 (g) → H 2 O l , ΔHc = ΔH fo
2

9-59

M H =1
...


B

⇒ − 141,790 kJ kg H

9
...


x0 (kg O) 2 kg H
H available for combustion = total H – H in H 2 O ; latter is kg coal 16 kg O

FG
H

Eq
...
6-3) ⇒ HHV = 32,764C + 141,790 H −

A

in water

IJ
K

O
+ 9261S
8

This formula does not take into account the heats of formation of the chemical constituents of
coal
...


b

C = 0
...
051, O = 0
...
016 ⇒ HHV

1 kg coal ⇒

g

Dulong

= 31,646 kJ kg coal

0
...
07 kg SO 2 formed

= 0
...
06 kg S burned
0
...

φ=
= 101 × 10 −6 kg SO 2 kJ
31,646 kJ kg coal

c
...
The
dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it
...
36 kJ mol Table B
...
50

bg

7
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O l , HHV = 1559
...
99 kJ mol
2
2
...
2K
2323 mm Hg
1 mol
Initial moles charged:
= 0
...
2 K 760 mm Hg 22
...
wt
...
929 g) (0
...
72 g / mol

c b

g
b
gh
MW = 19
...
04 g mol CH g + x b30
...
01g = 19
...
7 kJ mol ⇒ x b890
...
9g + b1 − x − x gb282
...
7 b2g

Let x1 = mol CH 4 mol gas, x2 = mol C 2 H 6 mol gas ⇒ 1 − x1 − x2 mol CO mol gas
1

4

1

2

2

1

1

2

2

Solving (1) & (2) simultaneously yields

x1 = 0
...
087 mol CO mol

...
51

a
...
36 kJ / mol
7
o
C 2 H 6 (g) + O 2 (g) → 2CO 2 (g) + 3H 2 O(v), ΔHc = −1559
...
51 (cont’d)
1 mol / s fuel gas ⇒ 0
...
15 mol C 2 H 6 / s

Theoretical oxygen =

2 mol O 2

0
...
5 mol O 2

0
...
225 mol O 2 / s

Assume 10% excess O 2 ⇒ O 2 fed = 1
...
225 = 2
...
85gb4g + b015gb6g ⇒ n = 2
...

10% excess O ⇒ n = b01gb2
...
223 mol O / s

...
85 1 + 015 2 ⇒ n 2 = 115 mol CO 2 / s

...

3

3

2

4

2

2

2

Extents of reaction: ξ 1 = n CH 4 = 0
...


bg

bg bg bg

bg

Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C

(We will use the values of

o
ΔH c

bg

given in Table B
...
85
015

...
225

0

0
...


0

H 2O v

−

−

2
...
01 kJ / mol

Energy Balance :

e j

o
Q = nCH 4 ΔHc

b

CH 4

e j

o
+ nC 2 H 6 ΔHc

C2 H 6

+

gb

∑n H − ∑n H
i

out

i

i

i

in

g b
+ b2
...
01 kJ / molg = −896 kW

gb

= 0
...
36 kJ mol + 015 mol / s C 2 H 6 −1559
...


g

2

⇒ − Q = 896 kW (transferred from reactor)

b
...
The flowchart and stoichiometry and material balance calculations are
the same as in part (a), except that amounts replace flow rates (mol instead of mol/s, etc
...
85 mol CH 4 , 0
...
225 mol O 2
Assume 10% excess O 2 ⇒ O 2 fed = 1
...
225 = 2
...
85gb4g + b015gb6g ⇒ n

...
225g mol O

...
85 1 + 015 2 ⇒ n2 = 115 mol CO 2

...

3

2

4

9-61

3

= 2
...
223 mol O 2

9
...

nin
nout
U in
U out
Substance
mol kJ mol mol kJ mol
0
...

−
−
2
...
223
0
O2
115
0
CO 2

...
15
H 2O v
U1
−
−

bg

e

j

e

8
...
01 kJ / mol −

Eq
...
1-5) ⇒

o
ΔU c

=

o
ΔH c

∑ν

− RT (

i

e

o
⇒ ΔU c

j

CH 4

eΔU j
o
c

C2 H 6

b

g

= −890
...


kJ
mol

)

gaseous
reactants

8
...
36

kJ
mol

mol K
10 J
8
...

kJ
= −1559
...

3
mol
mol K
10 J

b

g

Energy balance:

e

j + n eΔU j + ∑ n U − ∑ n U

...

= b0
...
36 kJ molg + b015 mol / s C H gb −156114 kJ molg

...
15 mol / s H Ogb4153 kJ / molg = −902 kJ

o
Q = ΔU = n CH 4 ΔU c

CH 4

o
c

C2 H 6

C2 H 6

i

i

i

out

4

i

in

2

6

2

⇒ − Q = 902 kJ (transferred from reactor)

c
...
52

Since the O2 (and N2 if air were used) are at 25°C at both the inlet and outlet of this process, their
specific enthalpies or internal energies are zero and their amounts therefore have no effect on the
calculated values of ΔH and ΔU
...


o
n fuel ( − ΔHc ) = Ws − Ql (Rate of heat release due to combustion = shaft work + rate of heat loss)

V (gal)

28
...
4805 gal
100 hp

=

0
...
341 × 10 −3 hp

1 kg
1 kJ
10 3 J

g
3600 s
h

+

15 × 10 6 kJ
⇒ V = 2
...


The work delivered would be less since more of the energy released by combustion would go into
heating the exhaust gas
...


Heat loss increases as Ta decreases
...


9-62

9
...


Energy balance: ΔU = 0 ⇒

a

f

b

n lb m fuel burned

b

g

o
ΔU c (Btu)

gb

ga

lb m

b

g

+ mCv Tout − 77° F = 0

f

o
⇒ 0
...
62 lb m 0
...
06° F − 77
...


o
The reaction for which we determined ΔU c is

1 lb m oil + aO 2 (g) → bCO 2 (g) + cH 2 O(v)

(1)

The higher heating value is ΔH r for the reaction
1 lb m oil + aO 2 ( g) → bCO 2 (g) + cH 2 O(l)

(2)

o
o
Eq
...
1-5) on p
...
(9
...
462 ⇒ − ΔHc2 = − ΔHc1 + cΔH v (H 2 O, 77 F)
( HHV )

( LHV )

To calculate the higher heating value, we therefore need
a = lb - moles of O 2 that react with 1 lb m fuel oil
b = lb - moles of CO 2 formed when 1 lb m fuel oil is burned
c = lb - moles of H 2 O formed when 1 lb m fuel oil is burned

9
...


bg

bg

3

e j

o
ΔH ro = ΔHc

CH 3 OH v + O 2 (g) → CO 2 (g) + 2H 2 O l
2

bg

CH 3OH v

= −764
...
1 atm

vaporizer

1 mol CH 3OH( v)
100°C
1 atm

Q 1 (kJ)

reactor

n 0 (mol O2 )
3
...
048 mol CO 2/mol D
...

0
...
G
...
809 mol N 2/mol D
...

n w (mol H 2O)

gb g

= n p 0
...
83 mol dry gas

N 2 balance: 3
...
83 0
...
482 mol O 2

Theoretical O 2 :

% excess air =

b


...


(4
...

× 100% = 200% excess air
15 mol O 2

...
96 mol O = 9
...
)
∗
pw =

d i

nw
2 mol H 2 O
∗
×P=
× 760 mm Hg = 66
...
83 mol
nw + n p

a

f

9-63

Table B
...
1° C

9
...


Energy balance on vaporizer:

LM
Q = ΔH = nΔH = 1 mol M
MN

OP kJ
C dT + ΔH +
C dT
PP mol = 40
...
2

CH 3 OH

2

16
...
7

Table B
...
1

3

n in
(mol)
1
...
7

2

2

2

n out
(mol)
−

H in
(kJ / mol)
3
...
187

16
...


4
...
235

2
...
470

CO 2

−

−

100

...


H 2O

−

−

2
...


N2
O2

af

H T = Hi for N 2 , O 2 , CO 2 (Table B
...
9
...
462, Table B
...
2)

bg

o
(Note: H 2 O l was chosen as the reference state since the given value of ΔHc presumes liquid

water as the product
...
0 + 16
...
482 2
...


bTable B
...
55

a
...
76n0 (mol N2 /s)
n3 (mol CO/s)
10 n3 (mol CO/s)
n4 (mol H2 O/s)

n0 (mol O2 /s)
3
...
10 (1000 ) = 100 mol CH 4 s
Theoretical O2 required = 2000 mol/s

9-64

9
...
1(2000 mol/s)=2200 mol/s

C balance:

(1000 mol CH 4 s )(1 mol C

mol CH 4 ) = (100 )(1) + n3 (1) + 10n3 (1) ⇒ n3 = 81
...
8)(1) + (818)( 2 ) + (1800 )(1) ⇒ n2 = 441 mol O2

s

af b g b g b g

References :C s , H 2 g , O 2 g , N 2 g at 25 C

ˆ
H in

nin

Substance

( mol s ) ( kJ
CH 4
O2
N2
CO
CO 2
H2O

Bo

H = ΔH f +

( mol s ) ( kJ

−74
...
24
2
...
1

mol )

z

ˆ
H out

nout

100
441
8272
81
...
62
11
...
15
−99
...
2
−228
...
2

T

25

C p dT for CH 4

B

Table B
...


i

i

i

= −5
...
)
S
A ⇒ −Q A (reaction to form CO2 has a greater heat of combustion and so releases

(iv)

more thermal energy)
Tstack ⇒ − Q (more energy required to heat combustion products)

(i)
(ii)

Tair

o

2

2

CO 2 CO

A

B

9-65

CH 4 + 2O 2 → CO 2 + 2H 2 O, C 2 H 6 +

9
...
Assume ideal gas behavior
...
0532 mol CO 2 /mol
0
...
0732 mol O2 /mol
0
...
7352 mol N2 /mol

n3 (mol O2 )
3
...


b gb g
C balance: n b1g + n b2g = b100gb0
...
0160gb1gU n = 3
...


...

b3
...
4 LbSTPg 298
...

V =
N 2 balance: 3
...
7352 mol N 2 ⇒ n3 = 19
...
72 mol CH 4

3
...


2 mol O 2 1
...
5 mol O 2
+
= 13
...
9 mole% CH
V
...
60 mol C H W 301 mole% C H
b19
...
04gmol O in excess × 100% = 50% excess air

Fuel composition:

% Excess air:

273
...
04 mol O 2 required

bg b g b g b g

References : C s , H 2 g , O 2 g , N 2 g at 25° C

CH 4

n in
mol
3
...
85

n out
mol
−

H out
kJ / mol
−

C2 H 6


...
67

−

−

O2

19
...
31

7
...
35

N2


...

513


...

2386

CO

−

−


...
39

CO 2

−

−

5
...

−3561

H2O

−

−

12
...
78

Substance

9-66

9
...
1

H = ΔH fo +

z

Table B
...
8

bg

Energy balance:
Q = ΔH =

∑n H − ∑n H
i

out

i

i

in

i

=

−2764 kJ
0
...
13 × 104 kJ m3 fuel

9
...
730gb50000glb

mC

1b - mole C

h

12
...

b0
...
037gb50000g 32
...
7 lb - moles S h
b0
...
02 = 189 lb - moles H O h

...
7 lb-moles S/h
189 lb-moles H 2O/h
5900 lb m ash/h
77°F, 1 atm (assume)
n 1 (lb-moles air/h)
0
...
790 N 2
77°F, 1 atm (assume)

a
...
287 lb mC/lb m
0
...
697 lb mash/lb m

Feed rate of air :

b

g

O 2 required to oxidize carbon C + O 2 → CO 2 =

3039 lb - moles C 1 lb - mole O 2
h
1 lb - mole C

= 3039 lb - moles O 2 h

Air fed: n1 =

1
...
210 mole O 2

= 21710 lb - moles air h

b

g

30% ash in coal emerges in slag ⇒ 0
...
30 5900 lb m h ⇒ m8 = 2540 lb m slag / h

b g

⇒ m7 = 0
...
287 2540 12
...
01

⇒ n 2 = 2978 lb - moles CO 2 h

b

131 × 10 5 lb m CO 2 h

...
5 lb - moles H 2 O h

b

M H 2 O =18
...
44 × 10 4 lb m H 2 O h

g

N 2 balance: n6 = 0
...
02

4
...
7 lb - moles S h = 1 n 4 + 0
...
06
⇒ n 4 = 56
...
2

3620 lb m SO 2 h

gb g b

gb g b

gb g b gb g

O balance: 189 1 + 0
...
5 1 + 56
...
57 (cont'd)

Summary of component mass flow rates
Stack gas at 600° F, 1 atm
2978 lb - moles CO 2 h ⇒

131000 lb m CO 2 h

1352
...
4 lb - moles SO 2 h ⇒

3620 lb m SO 2 h

943 lb - moles O 2 h ⇒

30200 lb m O 2 h

17150 lb - moles N 2 h ⇒

48100 lb m N 2 h

674,350 lbm stack gas/h

4130 lb m fly ash h

b

gb g

Check: 50000 + 21710 29
⇒

b679600g

in

b

⇔ 674350 + 2540

in

g

⇔ 676900

out

out

(0
...


a f

1060° R
= 1
...
063 Btu

lb - mole⋅° F 28
...
90 × 10

1 lb - mole

0
...
92 × 10

b

5

i

i

i

out

=

5 × 104 lb m
h

Btu h

Btu h

g ∑n H −∑n H

o
Energy balance: Q = ΔH = n coal burned ΔH c 77° F +

7

i

in


...
90 × 107 + 2
...

= −811 × 108 Btu h

Power generated =


...
35ge811 × 10 jBtu
8

e

Q = −811 × 108 Btu h

...
486 × 10

h

c
...

= 831 MW

g

coal h = −162 × 104 Btu lb m coal

...
62 × 104 Btu lb m
−Q
= 0
...
80 × 104 Btu lb m

Some of the heat of combustion goes to vaporize water and heat the stack gas
...


− Q HHV would be closer to 1
...


9- 69

9
...


Basis : 1 mol fuel gas/s
n0 (mol O 2 s)
Stack gas, Ts ( o C)

3
...
76n0 (mol N 2 / s)

1 mol / s @ 25o C
x m (mol CH 4 / mol)

nCO (mol CO / s)
rnCO (mol CO 2 / s)

x a (mol Ar / mol)

n H 2 O (mol H 2 O / s)

(1 − x m − x a ) (mol C 2 H 6 / mol)

nAr (mol Ar / s)

CH 4 + 2O 2 → CO 2 + 2 H 2 O
C2 H 6 +

7
2

O 2 → 2CO 2 + 3H 2 O

Pxs
) 2 x m + 35(1 − x m − x a )

...
76no
−
−
−

H2
−
−
−

3
...
2

c
...
85, x a = 0
...
0, Ta = 150 o C, Ts = 700 o C
⇒ n o = 2
...
0955, n H 2 O = 2
...

H1 (kJ / mol) = 8
...
588, H 3 = 0
...
279,
H 5 = 166
...
567, H 7 = −345
...
82
Energy balance: Q =

∑n

out H out

−

∑n

in H in

9- 70

= −655 kW

Xa

Pxs

r

Ta

Ts

Q
0

150
150
150
150
150
150

700
700
700
700
700
700

10
10
10
10
10

150
150
150
150
150

-

700
700
700
700
700

-200 0

996
905
813
722
631
905
869
799
-

0
...
2
0
...
4
0
...
1
0
...
1
0
...
1
0
...
1
0
...
1

5
5
5
5
5
5
5
5
5

1
2
3
4
5
10
20
50
100

150
150
150
150
150
150
150
150
150

700
700
700
700
700
700
700
700
700

-722
-796
-834
-856
-871
-905
-924
-936
-941

0
...
1
0
...
1
0
...
1
0
...
1
0
...
1
0
...
2

0
...
6

0
...
2

80

100

120

0
...
5

0
...
1

0
...
3

-870
Q -880
-890
-900
-910
x

0
0

20

40

60

-200
-400
Q

10
10
10
10
10
10

-600
-800
-1000
r

-800
-850
Q

0
...
0
5
1
0
...
0
5
1
0
...
0
5
1
0
...
1 10
0
...
1 50
0
...
58 (cont'd)
d
...
59

a
...
4 liters 273
...
1 atm
1 mol
= 10
...
2 K 1
...
4 liters STP

b g

H = C 6 H 14 ; M = CH 4
m (kg H 2O( l)/s)
w
25°C

Qc (kW)
n0 (mol/s)
y 0 (mol C 6 14/mol)
H
(1 – y 0 (mol CH 4/mol)
)
60°C, 1
...
0 mol/s at 5°C, 1
...
21 mol O /mol
2
0
...


Tdp = 55° C ⇒ y 0 P = p H 55° C
⇒ y0 =

Stack gas at 400°C, 1 atm
n3 (mol O 2/s)
n4 (mol N 2/s)
n5 (mol CO 2/s)
n6 (mol H 2O( v)/s)

reactor

nb (mol C 6 14( l)/s)
H

α

mw (kg H 2O( v)/s)
10 bars, sat'd

=

483
...
3 mm Hg
= 0
...
470 mol CH 4 mol
1
...
89 mm Hg
H
y2 =
=
= 0
...
93% mol CH 4 mol
P
1
...
0 1 − y2
Hexane balance on condenser: n0 y0 = nb + 10
...
78 mol C 6 H 14 l

⇒

y2 = 0
...
78 mol s

nb = 9
...
78
y0 = 0
...
070

cm 3

86
...
530

1L
3

0
...
1

3600 s
3

1h

Table B
...


e

j

e

References : CH 4 g, 5 C , C 6 H 14 l, 5 C
Substance
CH 4

bg
bl g

C 6 H 14 v

nin

bg

nout

Hin

Hout

(mol / s) (kJ / mol) (mol / s) (kJ / mol)
1985
9
...
30

...
48

41212

...
70

32
...
78

0

C 6 H 14

CH 4 g : H =

j

z

B

Table B
...
1

Tb

5

B

Table B
...
59 (cont'd)
CH 4 + 2O 2 → CO 2 + 2H 2 O , C 6 H 14 +

2

O 2 → 6CO 2 + 7H 2 O

9
...
79b240
...


fed

4

4

+

0
...
5 mol O 2

s

1 mol C 6 H 14

= 25
...
21na = 2 × 25
...
95 mol air s

= 190
...
30 mol CH 4

1 mol C

s

1 mol CH 4

+

0
...


H balance:

...
30 mol CH sgb4 mol H mol CH g + b0
...
3 mol O s
2
References : Cbsg, H bgg, O bgg, N bgg at 25° C for reactor side, H Oblg at triple point for
4

4

6

2 remaining

2

2

6

2 excess

2

2 fed

2

3

2

2

steam side (reference state for steam tables)
n in

kJ / mol mol / s
−75553 −

...
70

−170
...
6

5
...
3

1172

...
35


...
35


...


−377
...


−228
...
8

Table B
...
2

B
=

bg

H T

H in

n out

mol / s
9
...
2

T

ΔH fo + C p dT

for CH 4 , C 6 H 14

25
Table B
...
8

B
=

bg

bg

ΔH fo + Hi T for O 2 , N 2 , CO 2 , H 2 O v

Energy balance on reactor (assume adiabatic):
ΔH =

∑n H − ∑n H
i

out

i

i

i

b

g

= 0 ⇒ −8468 + mw 2776
...
8 = 0 ⇒ mw = 3
...
60

a
...
21 kmol O 2 / kmol
0
...
2 kmol O 2 fed
1 kmol air
1 kmol CH 4
1 kmol O 2 req' d 0
...
79 5143 × 10 6 mol h = 4060 kmol N 2 h

...
0805
y
450 kmol CO h U
| y = 0161

...
726
4060 kmol N h |
| y = 0
...
21 5143

2

4

react

2 mol O 2
= 180 kmol O 2 h
1 mol CH 4

CO 2

2

2

H 2O

2

N2

2

O2

5590 kmol / h

Mean heat capacity of stack gas
Cp =

∑y C
i

pi

b

gb

g b

gb

g b

gb

g b

gb

= 0
...
0423 + 0161 0
...
726 0
...
0322 0
...


= 0
...
60 (cont’d)

H p = n2 ( ΔH v ) H

o
2 O(25 C)

+ nstack gas (C p ) stack gas (Tstack gas − 25o C)

3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44
...
63 × 10 kJ / h
o
Q = ΔH = ξ ( ΔH c ) CH 4 +

∑n H −∑n H
i

i

=

i

i

FG 450 kmol IJ FG1000 mol IJ FG −890
...
63 × 10
H h K H kmol K H
mol K
out

0
...
44 × 10 8
h
h

Energy balance on steam boiler

LM FG IJ OP LMb2914 − 105g kJ OP
kg Q
N H KQ N
Table B
...
6

kg
kJ
= mw
Q = mw ΔH w ⇒ + 3
...


b
...
21 O 2
0
...
B
...


b g b g b g b g bg
b
g
H Oalf at triple point (steam table reference) (steam tube side)

References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25 C furnace side
2

Substance

n in
(kmol / h)

H in
(kJ / kmol)

n out H out
(kJ / h)

CH 4
Air
Stack gas

450
5143
−

0
0
−

−
−
Hp

H 2O

H p = n2 ( ΔH v ) H

o
2 O(25 C)

m w ( kg / h) 105 kJ / kg

m w ( kg / h) 2914 kJ / kg

+ nstack gas (C p ) stack gas (Tstack gas − 25o C)

3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44
...
99 × 10 kJ / h

9- 75

0
...
60 (cont’d)
o
ΔH = ξ ( ΔH c ) CH 4 +

∑n H −∑n H
i

i

i

i

=0

FG 450 kmol IJ FG1000 mol IJ FG −890
...
99 × 10 kJ
H h K H kmol K H
mol K
h
L F kg I OL
kJ O
+ Mm G J P Mb2914 − 105g P = 0 ⇒ m = 1
...
0315 C b150 − 300g C = −2
...
64 × 10 kJ / h 1 kmol = 5
...
8


...


7

p

Ta = 199 o C

The energy balance on the furnace includes the term −

∑n

in H in
...
The stack gas is a
logical heating medium since it is available at a high temperature and costs nothing
...
61

Basis: 40000 kg coal h ⇒

b0
...
531 × 106 mol C h
1 kg 12
...


...
05 × 4000gkg H h e10 101j = 198 × 10
b0
...
0j = 2
...

b011 × 40000g = 4400 kg ash h
3

6

3

5

mol H h
mol O h

4400 kg ash/h, 450°C
Q to steam

40,000 kg coal/h
2
...
98 ×10 6 mol H/h
2
...
078 mol CO 2 /mol D
...

0
...
G
...
114 mol O 2/mol D
...

0
...
G
...


air at 30°C, 1 atm, hr = 30%
n1 (mol O2 /h)
3
...
078 CO 2
0
...
114 O 2
0
...
531 × 10 6 = 0
...
012n 3 ⇒ n 3 = 2
...
76n1 = 0
...
812 × 10 7 ⇒ n1 = 5
...
76 5
...
61 (cont'd)

30% relative humidity (inlet air):

B

Table B
...
30 p H 2 O

b30° Cg ⇒ 5
...

b760 mm Hgg = 0
...
24 × 10 7 + n 2

⇒ n 2 = 3
...
95 × 10

6

b g

j

+ 224 × 10 7 + 3
...
4 liters STP
h

Air/fuel ratio:

1 m3
3

1 mol

= 6
...
43 × 10 5 m 3 air h
= 161 SCM air kg coal

...


...
61 × 10 5 mol H h = 2n 4 ⇒ n 4 = 1351 × 10 6 mol H 2 O h
H in coal

H 2 O content of stack gas =

b
...



...
812 × 10 7 mol h

6

× 100% = 4
...
193 × 10 6
0
...
706 × 10 6
22
...
357 × 10 6

H i (T ) from Table B
...
193 × 10 6
0
...
206 × 10 6
72
...
351 × 10 6

H in
kJ mol
4
...
738
6
...
193
6
...
2

B
Cp

dT for outlet

g

= −101 × 10 8 kJ h Heat transferred from stack gas

...
01 ×10 8 kJ/hr

2
...
61 ×10 5 mol H 2O/h
30°C

2
...
61 ×10 5 mol H 2O/h
T a (°C)

(We assume preheater is adiabatic, so that Qstack gas = − Qair )
Energy balance on air:
Q = ΔH ⇒ 101 × 10

...
2

n dry air

z

T

Table B
...
61 (cont'd)
⇒ 101 × 10 8 = 8
...
92(Ta2 − 30 2 ) + 0
...


...


4400 kg ash/h at 450°C

40,000 kg coal/h
25°C

Flue gas at 260°C
2193× 106 mol CO 2 /h
0
...
206 × 106 mo l O2 /h
22
...
351 × 106 mo l H2 O( v
)/h

5
...
24 ×10 7 mo l N /h
2
3
...
8°C)

m (kg H 2 O(l )/h)
50°C

m (kg H 2 O(v )/h)
30 bars, sat'd

bg

References for energy balance on furnace: CO 2 , CO, O 2 , N 2 , H 2 O l , coal at 25° C

bg

(Must choose H 2 O l since we are given the higher heating value of the coal
...
95 × 10 6
O2
N2
2
...
61 × 10 5
H2O

H in
0
−
3
...
655
−
−
48
...
206 × 10 6
2
...
193 × 10 6
0
...


H out
−
n kg h
412
...
193
6
...
738 H kJ mol
6
...
14

b
b
b
b

g

g
g
g

(Furnace only — exclude boiler water)
Heat transferred from furnace
Q = n coal ΔH io +

FG
H

= 4 × 10 4

∑n H −∑n H
i

kg
h

= −8
...
5 × 10 kJ IJ + FG 2
...


A

H of preheated air

I kJ
JJ kg
K

kJ h
8

8

Heat transferred to boiler water: 0
...
76x10 kJ/h) = 5
...
3 − 209
...
02 × 10 kg steam h

Energy balance on boiler: Q kJ h = m
⇒ 5
...
6

9- 78

Table B
...
62

1

o
CO + O 2 → CO 2 , ΔH c = −282
...


2

1 mol CO
n 0 mol O2
3
...


1 mol CO2
(n 0 – 0
...
76n 0 mol N2
1400°C

b

g

1 mol CO react 0
...
5
1 mol CO
References: CO, CO 2 , O 2 , N 2 at 25 C

Oxygen in product gas: n1 = n0 mol O 2 fed −

n in
mol

H in
kJ mol

1
n0
3
...
5
3
...
51 kJ mol
B
CO bg,1400° Cg: H = H (1400 C) = 7189 kJ mol

...
8

O 2 g,1400° C : H1 = HO 2 (1400 C) = 47
...
8

2

2

N2

Table B
...
B
...
99 + 47
...
5 + 44
...
76n0 + 71
...


b

gb

g

Theoretical O 2 = 1 mol CO 0
...
500 mol O 2
1094 mol fed − 0
...


...
500 mol
Increase %XS air ⇒ Tad would decrease, since the heat liberated by combustion would go into
heating a larger quantity of gas (i
...
, the additional N 2 and unconsumed O 2 )
...


9
...


Basis : 100 mol natural gas ⇒ 82 mol CH 4 , 18 mol C 2 H 6
o
CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(v), ΔHc = −890
...
9 kJ / mol
2

82 mol CH4
18 mol C2H6
298 K

Stack gas at T(°C)
n2 (mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
n5 (mol N2)

n0 (mol air) at 423 K
0
...
79 N2

9-79

9
...
2 × 227 mol O 2

1 mol air
0
...
00gb4g + b18
...
5 mol O 2

18 mol C 2 H 6

1 mol C 2 H 6

= 227 mol O 2

= 1297
...
00 1 + 18
...
00 mol CO 2
H balance : 2n3

3

= 218
...
2 227 mol O 2 = 45
...
79 1297
...
63 mol N 2

Extents of reaction: ξ 1 = nCH 4 = 82 mol, ξ 1 = nC 2 H 6 = 18 mol

bg

bg bg bg

bg

Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l at 298 K

bg

o
(We will use the values of ΔHc given in Table B
...
)

bg

b

g

Hi T = C pi T − 298 K for all species but water

b

g

b

g

= ΔH v,H 2 O 298 K + C p , H 2 O T − 298 K for water

Substance

n in
mol

H in
kJ mol

CH 4
C2 H 6
O2
N2
CO 2
H2O v

82
...
00
272
...
63
−
−

0
4
...
91
−
−

bg

n out
mol

H out
kJ mol

−
−
−
−
45
...
0331 T − 298
1024
...
0313 T − 298
118
...
0500 T − 298
218
...
013 + 0
...
40gb0
...
63gb0
...
00gb0
...
00gb0
...
00gb44
...
40)(4
...
63)(3
...
00 mol CH 4 −890
...
00 mol C 2 H 6 −1559
...


Solving for T using E - Z Solve ⇒ T = 2317 K

Increase % excess air ⇒ Tout decreases
...
(lower heat of combustion, but heat released goes
into heating fewer moles of gas
...
4 kJ mol

9
...
4 m STP

60 s

= 1049 mol s feed gas

Stochiometric proportion:

b

g

1 mol C 3 H 6 O ⇒ 4 mol O 2 ⇒ 4 × 3
...
04 mol N 2 ⇒ 1 + 4 + 15
...
04 mol

yC 3 H 6O =

1 mol C 3 H 6 O
mol C 3 H 6 O
4
, yO 2 =
= 0
...
0499
20
...
04 mol
mol
Preheat

Feed gas
1049 mol/s
0
...
1996 2
0
...
7505 N 2
T f (°C), 150 mm Hg
Rel
...
2%

a
...

Relative saturation = 12
...


Reaction

b

gb

0
...
122

g = 61352 mm Hg

...
4

T f = 50
...
1996g = 209
...
7505g = 787
...
34gb3g = 157
...
25 mole% CO
|
⇒ Product contains n = b52
...
0 mol H O sV ⇒ 14
...
5% N
n = 787
...
0499 C 3 H 6 O mol = 52
...
34
209
...
66
17
...
3
−
−

17
...
3
157
...
0

0
...
052 (Ta − 25 )
44
...
040 (Ta − 25 )

Energy balance on reactor:
o
ΔH = nC3 H 6O ΔHc +

∑n H − ∑n H
i

out

i

i

i

b g

= 0 kJ s

in

kJ ⎞
⎛
4
⇒ ( 5234 mol s) ⎜ −1821
...
638(Ta − 25) + 157
...
013) − 2
...
64 (cont'd)
c
...
34
315
52
...
66

...
4
0
...
4
17
...
3
0
...
3
16
...
B
...


in

af

Cooling step
...
B
...
0
157
...
3

142
...

88
...
0
157
...
3

16
...
35
10
...
64 × 10 4 kW

in

Exchange heat between the reactor feed and product gases
...
65

a
...
5 kJ / mol

1 mol C 5 H 12

8 mol O 2
= 8 mol O 2
1 mol C5 H 12

30% excess ⇒ n0 = 13 × 8 = 10
...


b gb g
= b1gb12g ⇒ n

C balance: n2 = 1 5 ⇒ n2 = 5 mol CO 2
H balance: 2n3

3

= 6 mol H 2 O

b gb g

30% excess O2, complete combustion ⇒ n4 = 0
...
4 mol O 2

bg b g

bg

Reference states: C5 H 12 l , O 2 g , H 2 O l , CO 2 (g) at 25o C

bg

0
(We will use the values of ΔHc given in Table B
...
65 (cont'd)
substance
C5 H 12
O2
CO 2
H2O

nin
Hin
mol kJ mol
100
0

...
40
H1
−
−
−
−

z

nout
Hout
mol kJ mol
−
−
2
...
00
H3
6
...
8

B

H1 = H O 2 (75 o C)

=

148 kJ / mol

...
2 :
kJ
mol
kJ
− 0
...
0291 Tad + 0
...
2025 × 10 −8 Tad 3 + 0
...
7311)
H 3 = (0
...
1165 × 10 −5 Tad 2 − 0
...


H4 = 44
...
03346 Tad + 0
...
2535 × 10 −8 Tad 3 − 08983 × 10 −12 Tad 4 − 0838)

...


kJ
mol


...

⇒ H4 = 4317 + (0
...
3440 × 10 −5 Tad 2 + 0
...
5 kJ / mol) + (2
...
00) H3 + (6
...
40)( H1 ) = 0
Substitute for H1 through H 4

ΔH = (0
...
036 × 10 −5 Tad 2 − 3777 × 10 −8 Tad 3 + 4
...
20 kJ / mol = 0

...
20 + 0
...
036 × 10 −5 Tad 2 − 3777 × 10 −8 Tad 3 + 4
...

Check :

−3272
...
727 × 10 −12

= −6
...


Terms
1
2
3

Tad
% Error
7252
64
...
1%
3938 –10
...
65 (cont’d)
c
...


T
7252
5634
4680
4426
4414

f(T)
6
...
73E+03
3
...
41E+01
3
...
74 5634
1
...
22 4426
1
...
11 4414

The polynomial formulas are only applicable for T ≤ 1500°C

9
...
5 L/s at 25°C, 1
...
76 n 2 (mol N2/s) n 4 (mol CO2/s)
150°C, 1
...
76 n 2 (mol N2/s)
n 5 (mol H2O/s)
T(°C), 1
...
1 atm

...
0 atm 22
...
247 mol CH 4 / s

Theoretical O 2 = 2 × 0
...
494 mol O 2 / s

...
494) = 0
...
76 × 0
...
32 mol N 2 / s
Complete combustion ⇒ ξ = n1 = 0
...
247 mol CO 2 / s, n5 = 0
...
6175 mol O 2 fed / s − 0
...

= 0124 mol O 2 / s
Re ferences: CH 4 , O 2 , N 2 , CO 2 , H 2 O at 25 o C
n in
n out
H in
H out
Substance
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
CH 4
O2
N2
CO 2
H2O

ˆ
ˆ
H1 = H (O 2,150o C)
ˆ
ˆ
H 2 = H (N 2,150o C)

0
...
6175
2
...
8

3
...
8

3
...
36 kJ/mol
T

ˆ
H i = ∫ C pi dT , i = 3 − 5
25

9-84

−
0124

...
32
0
...
497

−
H3
H4
H5
H6

9
...


Energy Balance
ˆ
ˆ
ˆ
ΔH = ξ (ΔHco )CH4 + ∑ nout Hout −∑ nin Hin = 0
ˆ
Table B
...
01 kJ/mol
2

0
...
36) + 0
...
01) + 0
...
02 ×10−5 (T 2 − 252 ) + 0
...
61×10−12 (T 4 − 254 ) − 0
...
78) − 2
...
66) = 0

⇒ −211
...
0963Tad + 1
...
305 × 10−8 Tad 3 − 1
...


In product gas,
T = 1832o C, P = 1
...
494 mol/s
= 0
...
124 + 2
...
247 + 0
...
155)(798) = 124 mmHg

Table B
...
superheat = 1832 C − 56 C = 1776 C
9
...


CH 4 (l) + 2O 2 (g) → CO 2 (g) + 2H 2 O(v)
Basis : 1 mol CH 4
2 mol O 2
= 2
...
00) = 2
...
76 × 2
...
78 mol N 2

...
60 mol O2
9
...
00 atm

n2 (mol CO2)
n3 (mol H2O)
n4 (mol O2 )

Complete combustion ⇒ n 2 = 100 mol CO 2 , n 3 = 2
...

2
...
60 − 2
...
60 mol O 2

Internal energy of reaction: Eq
...
1-5) ⇒

o
ΔU c

=

o
ΔH c

F
I
GG ∑ ν − ∑ ν JJ
− RT
GH
JK
i

gaseous
products

e

o
⇒ ΔU c

z

j

CH 4

T

U=

25

FG
H

= −890
...
314 J 298 K (1 + 2 − 1 − 2) 1 kJ
kJ
kJ
−
= −890
...
67 (cont’d)
b
...
1, which are based on H 2 O l as a

combustion product, and so must choose the liquid as a reference state for water
...


kJ mol
0

mol
−

kJ mol
−

O2

2
...
60

U1

N2

9
...
78

U2

CO 2

−

−

100

...
00

U4

Substance

nin

bg

Part a

B

U i = (C p − Rg )(T − 25) for all species except H 2 O(v)

e

j

e

j

= ΔU v 25 o C + (C p − Rg )(T − 25) = ΔH v 25 o C − Rg Tref + (C p − Rg )(T − 25) for H 2 O(v)
Substituting given values of (C p ) i and Rg = 8
...
033 − 8
...
02469T − 0
...
032 − 8
...
02369T − 0
...
052 − 8
...
04369T − 10922) kJ / mol

...
314 × 10 −3
(298 K) + (0
...
314 × 10 −3 )(T − 25)
mol
mol ⋅ K
mol
kJ
kJ
kJ
⇒ U 4 = 4153
+ (0
...
314 × 10 −3 )(T − 25)
= (0
...
74)

...
01

Energy Balance

e

j +∑n U −∑n U = 0
⇒ Q = (100)b −890
...
60)U + (9
...


o
Q = n CH 4 ΔU c

CH 4

i

i

out

i

i

in

1

2

+ (100)U 3 + (2
...


Substituting U 1 through U 4

0
...

Ideal Gas Equation of State ⇒

c
...
00 atm = 34
...
68

b
...
21 mol O2/mol DA
0
...
5 mol O2
1 mol C2 H 6

= ( 2 yCH 4 + 35 yC 2 H 6 + 5 yC 3H 8 )

...
21n a (1 − y wo ) = 1 +

FG
H

⇒ na = 1 +

IJ
K

yC2H6 (mol C2 H 6 )

+

5 mol O2
1 mol C3 H 8

yC3H8 (mol C3 H8 )

IJ
K

Pxs
( 2 y CH 4 + 35 y C 2 H 6 + 5 y C3 H 8 ) mol O 2

...

mol air
100
0
...
21n a (1 − y wo ), (n N 2 ) in = 0
...

100

9-87

9
...

References : C(s), H 2 ( g) at 25o C

z
T

H CH 4 (T) = ( Δ H o ) CH 4 + (C p ) CH 4 dT
f
25

Using ( Δ H fo )CH 4 from Table B
...
2
HCH (T) = −7485 kJ / mol +

...
03431+5
...


...

⇒ HCH (T ) = [ −75
...
431 × 10 −2 T + 2
...
75 × 10 −12 T 4 ] kJ / mol
4

n in

CH 4
C2 H 6
C3H8
O2
N2
CO 2
H2O
7

ΔH =

∑

n out

H out
mol kJ / mol mol kJ / mol
−
−
n1
H1
n2
H2
−
−
n3
H3
−
−
n4
H4
n7
H7
n5
H5
n8
H8
n6
n9
H9
−
n10
H10
−
−

Substance

H in

6

(ni ) out ( Hi ) out −

i =4

∑ (n )

i in ( Hi ) in

i =1
3

Hi = ai + bi T + ci T 2 + di T + ei T 4
6

∑

3

(ni ) in ( Hi ) in =

i =1

∑

6

(ni ) in Hi (Tf ) +

i =1

∑ (n )

i in Hi (Ta )

i =4

7

⇒ ΔH =

∑

3

(ni ) out (ai + bi T + ci T 2 + d i T 3 + ei T 4 ) out −

i =4
7

⇒ ΔH =

7

∑ (n )

i out ai

+

i =1

i out bi T +

i =4

i out ci T

2

i =1

3

−

∑ (n )

+

∑ (n )

i out d i T

i =1

6

∑ (n )

i in Hi (Tf

)−

i =1

∑ (n )

i in Hi (Ta )

i =4

= α 0 + α 1T + α 2 T 2 + α 3 T 3 + α 4 T 4
where α 0 =

α1 =
α3 =

7

6

3

∑

(ni ) out ai −

∑

(ni ) out bi

i =1
7

∑

(ni ) in Hi (Tf ) −

i =1

i =1
7

∑(n )

i out d i

i =1

α2 =
α4 =

∑ (n )

i in Hi (Ta )

i =4

7

∑ (n )

i out ci

i =1
7

∑ (n )

i out ei

i =1


...
68 (cont’d)
d
...
75
yC2H6
0
...
04
Tf
40
Ta
150
Pxs
25
ywo 0
...
04
nN2
11
...
46
HCH4
-74
...
9
HC3H8
-102
...
6
HN2i
3
...
6
nCO2
1
...
75
nO2
0
...
44
Tad 1743
...
4892
alph2 0
...
00E-08
alph4 3
...
00E-07

Species
CH4
C2H6
C3H8
O2
N2
H20
CO2

a
-75
...
95
-105
...
731
-0
...
7
-394
...
86
0
...
04
40
150
25
0
...
84
10
...
43
-74
...
9
-102
...
6
3
...
6
1
...
61
0
...
67
1737
...
9
0
...
00E-04
-3
...
00E-12
9
...
75
0
...
04
150
150
25
0
...
04
11
...
46
-70
-77
-93
3
...
8
-237
...
29
2
...
61
11
...
7
-1057
0
...
0001
-3
...
00E-12
-4
...
75
0
...
04
40
250
25
0
...
04
11
...
46
-74
...
9
-102
...
6
6
...
1
1
...
75
0
...
44
1812
...
4892
0
...
00E-08
3
...
00E-04

Run 5
0
...
21
0
...
0306
4
...
31
0
...
3
-83
...
7
3
...
8
-237
...
29
3
...
44
18
...
5
-1093
0
...
0001
-4
...
00E-12
-1
...
431
4
...
803
2
...
91
3
...
611

c
x 10^5
2
...
96
11
...
11
0
...
344
2
...
122
-1
...
37
0
...
203
0
...
962

e
x 10^12
-2
...
82
7
...
718
0
...
898
1
...
75
0
...
04
40
150
25
0
...
04
11
...
61
-74
...
9
-102
...
6
3
...
6
1
...
9
0
...
44
1633
...
5278
0
...
00E-08
2
...
00E-04

9
...
988 n 9 (mol CO/h)
0
...
006 n 7 (mol C 2H 2/h)

n 13 (mol C(s )/h)
Converter
converter product quench
Tad (°C)

Feed gas, 650°C
n 14 (mol CH4 /h)
0
...
04n 15 (mol N2 /h)

Converter
product
38°C

0
...
96 mol O /mol
2
0
...
991 mol C2 H2 ( g)/mol
0
...
00841 mol CO2 /mol

n 6 (mol CH4 /h)
n 7 (mol C 2H 2/h)
n 8 (mol H 2/h)
n 9 (mol CO/h)
n 10 (mol CO2 /h)
n 11 (mol H 2O/h)
n 12 (mol N 2/h)

Rich solvent

n 1 (mol/h)
0
...
0063 mol CO2 /mol
0
...
00055 mol CH4 /mol
0
...
917 mol DMF/mol

Average M
...
of product gas:

b

g

b

g

b

stripper

Stripper off-gas
n 2 (mol CO/h)
n 3 (mol CH4 /h)
n 4 (mol H 2O( )/h)
v
n 5 (mol CO2 /h)

g

M = 0
...
04 + 0
...
016 + 0
...
01 = 26
...
19 g

24 h

= 7955 mol h

Material balances -- plan of attack (refer to flow chart):
Stripper balances: C 2 H 2 ⇒ n1 , CO ⇒ n2 , CH 4 ⇒ n3 , H 2 O ⇒ n4 , CO 2 ⇒ n5
Absorber balances: CH 4 ⇒ n6 , C 2 H 2 ⇒ n7 , CO ⇒ n9 , CO 2 ⇒ n10 , H 2 O ⇒ n11

R5
...
0155n1 = 0
...
086 × 105 mol h

b

ge
j
CH : b0
...
086 × 10 j = n ⇒ n = 79
...
0596ge5
...
00059gb7955g ⇒ n = 30308 mol H O h
CO : b0
...
086 × 10 j = n + b0
...
00055 5
...
7 mol CO h
5

3

4

3

4

5

4

2

4

2

5

5

2

Absorber balances

b

5

ge

j

CH 4 : n6 = 0
...
00055 5
...
69 (cont'd)

b

ge
j
CO: n = 0
...
00055ge5
...
0068ge5
...
0596ge5
...
0567gn (mol CH )
1 mol C
Soot formation:
⇒ n = 0
...
0155 5
...
006n7 ⇒ n7 = 7931 mol C 2 H 2 h
5

9

9

9

5

2

10

2

5

2

11

2

13

14

4

13

14

4

Converter C balance:

b

gb

g b gb g b

gb g b gb g

n14 = 5595 mol CH 4 h 1 mol C mol CH 4 + 7931 2 + 23311 1 + 3458 1 + n13
⇒ n14 = n13 + 48226

b2 g

bg
= b5595gb4g + b7931gb2g + 2n + b30313gb2g

Solve (1) & (2) simultaneously ⇒ n13 = 2899 mol C s h , n14 = 51120 mol CH 4 h
Converter H balance:

51120 mol CH 4 4 mol H
h
1 mol Ch 4

CH 4

C2 H 2

H 2O

H2

8

⇒ n8 = 52816 mol H 2 h

b

gb g

Converter O balance: 0
...
04 31531 n12 ⇒ n12 = 1261 mol N 2 h

a
...


51120 mol CH 4
h

b

31531 mol O 2 + N 2
h

Gas feed to absorber
5595 mol
7931 mol
23311 mol
3458 mol
30313 mol
52816 mol
1261 mol

b g

0
...
0244 m 3 STP
= 706 SCMH O 2 + N 2
1 mol

g

U
|
|
|
|
V
|
|
|
|
W

CH 4 h
C2H 2 h
CO h
CO 2 h
4
...
4 % C 2 H 2 , 18
...
8% CO 2 , 24
...
4% H 2 , 1
...
2469 × 10 5 mol h

Absorber off-gas
52816 mol H 2 h
1261 mol N 2 h
23031 mol CO h
64
...
5% N 2 , 27
...
5 kmol h, 6
...
06% C H
4
2 2
41
...
2471 × 10 4

U
|
|
|
V
|
|
mol h |
W

9-91

9
...
7 mol CO h
279
...
3 kmol h, 0
...
82% CH 4 , 88
...
9% CO 2
3392 mol CO h
3
...


d
...
991gb7955g mol C H in product gas = 0154 mol C H
Overall product yield =

...
917 5
...
This yield is (1 mol C2H2/2 mol CH4) = 0
...

The ratio of the actual yield to the theoretical yield is 0
...
500 = 0
...

e
...
2
650

B

d i
Cp

25

dT =
CH 4

51120 mol 32824 J
1h
1 kJ
= 466 kW
h
mol
3601 s 10 3 J

Oxygen preheater
Table B
...
8

B

B

QO 2 = ΔH = 0
...
04n15 H ( N 2 ,650 C)

FG
H

IJ LMb0
...
04 × 18
...

KN
mol ⋅ C QH 3600 s K
Cbsg, H bgg, O bgg, N bgg at 25° C
n
H b650° Cg n
H bT g

= 31531

f
...
026

...
988

5595
−

−74
...
52 + C p dT

CO 2

−

−

3458

H2O

−

−

30313 −24183 + C p dT

...
75

Ta

25

9-92

C p dT

C p dT

−3935 + C p dT

...
69 (cont’d)

z

T
0 + C dT
ΔH i
pi
kJ mol 25
kJ mol⋅°C

Hi =

∑n H
i

i

= −1575 × 10 6 kJ h

...
888 × 10 6 kJ h +

in

∑n H
i

out

z

LM5595dC i + 1261dC i + 7931dC i
N
OP 1 kJ dT
+52816dC i + 23311dC i +3458dC i
+ 3013dC i
b g Q 10 J
1 kJ
+
dC i b g × 10 J dT
Tout

p CH
4

25

z

p H
2

p CO

p N
2

p CO
2

p C H
3 2

p H O v
2

3

Tad + 273

298

p C s

3

We will apply the heat capacity formulas of Table B
...


25

Tad + 273

298

∑

z


...
9885 × 10
F 32
...
031744T − 14179 × 10 IdT

...
888 × 10 6 kJ h +

6

2

ni H i = −1000 × 10 7 + 3943Ta + 0
...
5405 × 10 −8 Ta4 +

...


out

Energy balance: ΔH =

∑n H −∑n H
i

out

i

i

i

=0

in

b g

⇒ f Tc = −8
...
6251Tc2 − 1996 × 10 −4 Tc3 − 2
...

E-Z Solve

1418 × 10 6

...


9-93

1418 × 106

...
70 a
...
35 solids, 0
...
sludge/d), 100 o C

INCINERATOR

0
...
25 W (l)

Waste gas

m3 [kg W(v)/d]
4B, sat'd
C

m3 [kg W(l)/d]

Q (kJ / d)
3

m3 [kg W(l)/d] BOILER

4B, sat'd

20 o C

Q1

D

m 6 (kg gas/d)
0
...
10
kmol

Q4 (kJ / d)

m 4 (kg oil/d)

Stack gas

0
...
10 H

125 C

SO 2

0
...
0216 ash

ash

m 7 (kg air/d)
25 o C

Q0 (kJ/d)
E

m5 (kg air/d)
25 o C

9-94

9
...
35 × 24,000 kg / d = 0
...
2 tonnes / d (conc
...
5(100 − 22) = 195
...
1 − 92
...
9 kJ/kg
2

ˆ
H 3 = (2676 − 92
...
5)
Q2 =

∑m H − ∑m H ⇒ Q
i

i

out

Qsteam =

i

i

2

= 356 × 107 kJ day

...

356 × 10
= 6
...
91 × 107 kJ / d
0
...
47 × 107

FG IJ
H K

ΔH v for
H 2 O(sat'd, )

B
kJ
kg
= n3
× 2133
d
d

FG kJ IJ F 1 tonne I ⇒ n
H kg K GH 10 kg JK

3

3

= 30
...
6 − 83
...
04 × 107 kJ / d
d
kg

62% efficiency ⇒ Fuel heating value needed =
⇒ n4 =

130 × 108 kJ / d

...
75 × 104 kJ / kg

8
...

0
...
5 tonnes / day (fuel oil)

Air feed to boiler furnace: C + O 2 → CO 2 , 4H + O 2 → 2H 2 O,

(nO2 ) theo = 3458

S + O 2 → SO 2

kg ⎡
kgC 1 kmol C 1 kmol O 2
1 1
1 1⎤
⎢ (0
...
10)( 1)( 4 ) + (0
...
70 (cont’d)

Air fed (25% excess) = 1
...
76
⇒

kmol air
kmol O2
kmol air
)(338
) = 2011
kmol O2
d
d

2011 kmol 29 kg 1 tonne
⇒ E = 58
...
93 kJ
kJ
ˆ
⇒ Q0 =
= 5
...
8 ⇒ H air (125o C) = 2
...
2 tonne sludge 195 SCM
d

tonne

1 kmol
22
...
5 kmol d

MWgas = 0
...
90)(16) + (010)(30) = 17
...


...
5)(17
...
7 tonne / d (natural gas)
CH 4 + 2O 2 → CO 2 + 2H 2 O, C 2 H 6 +

7
O 2 → 2CO 2 + 3H 2 O
2

Air feed to incinerator:

(air)th, sludge :

11200 kg sludge 0
...
5 m3 (STP) air

(air)th , gas : 97
...
4 m3 (STP)

= 1781

kmol air
d

⎤ ⎛ 4
...
10)(3
...
90
d ⎣
kmol
kmol CH 4
d
⎦ ⎝ 1 kmol O2 ⎠

kmol air
= 5558 kmol air/d
d
5558 kmol air 29
...
8 ⇒ H air (110o C) = 2
...
486 kJ
kJ
= 1
...


Cost of fuel oil, natural gas, fuel oil and air preheating, pumping and compression, piping, utilities,
operating personnel, instrumentation and control, environmental monitoring
...


c
...

Make use of steam from dryer
...


Sulfur dioxide, possibly NO2, fly ash in boiler stack gas, volatile toxic and odorous compounds in gas
effluents from dryer and incinerator
...
1 b
...

An infeasible set:

ln , n , n , x , x , T q
1

2

3

1

2

1

Specifying n1 and n2 determines n3 (from a total mole balance)
c
...
2 10 variables n1 , n2 , n3 , n4 , x1 , x 2 , x 3 , x 4 , T , P

−2 material balances

g

bgb

g b

g bg

*
*
−2 equilibrium relations: [ x 3 P = x 4 PB T , 1 − x 3 P = 1 − x 4 PC T ]

6 degrees of freedom

ln , n , n , x , x , Tq

a
...
An iterative set:

ln , n , n , x , x , x q
1

2

3

1

2

3

Calculate n4 from total mole balance, x 4 from B balance
...

c
...


bg

bg

bg

bg

10
...


100 kg ore, T 0 (K)
xb (kg BaSO4 /kg)

n 1 (kg C)
n 2 (kg BaS)
n 3 (kg CO2 )
n 4 (kg other solids)
T f (K)

n 0 (kg coal), T 0 (K)
xc (kg C/kg)
Pex (% excess coal)

Q (kJ)

d

i

11 variables n0 , n1 , n2 , n3 , n4 , x b , x c , T0 , T f , Q, Pex
−5 material balances C, BaS, CO 2 , BaSO 4 , other solids
−1 energy balance

b

g

+1 reaction
−1 relation defining Pex in terms of n0 , x b , and x c
5 degrees of freedom

n

b
...
3 (cont’d)

l

q

c
...


l

q

d
...
4 2C 2 H 5 OH + O 2 → 2CH 3 CHO + 2H 2 O
2CH 3 COH + O 2 → 2CH 3 CHOOH
n f (mol solution), T 0
x ef (mol EtOH/mol)
1 – x ef (mol H 2 O/mol)

n e (mol EtOH), T
n ah (mol CH 3 CHO)
n ea (mol CH 3 COOH)
n w (mol H 2O)
n ax (mol O 2)
n n (mol N 2)

n w (mol air), Pxs , T 0
0
...
21 n air (mol O 2 )
(Pxs = % excess air)

a
...
Design set: n f , x ef , Pxs , ne , nah , T0 , T

s

Calculate nair from n f , x ef and Pxs ; nn from N 2 balance;

naa and nw from n f , x ef , ne , nah and material balances;
nex from O atomic balance; Q from energy balance

n

c
...
Design set: nair , nn , …
...
5
a
...
However, one relation also was counted
twice: the catalyst material balances on the reactor and flash tank are each n9 = n10
...
The catalyst circulation rate is not included in any equations other than the catalyst balance
(n9 = n10)
...


10- 4

b

10
...
Mixer:

5 variables n1 , n2 , n3 , nr , x r

g

−2 material balances

3 degrees of freedom

b

Reactor:

4 variables n2 , n3 , n3 , n5

g

−2 material balances
+1 reaction
3 degrees of freedom

b

Still:

6 variables n4 , n5 , n6 , x 6 , nr , x r

g

−2 material balances
4 degrees of freedom
Process:

10 Local degrees of freedom
− 6 ties
4 overall degrees of freedom

b
...
115 mol n − C 4 H 10 mol , x r = 0
...

× 100% = 88
...
115 4 + 0
...
85nT = n2

b1g

b1g

35% S
...
conversion: n4 = 0
...
5525nr
Still n – B balance:

b2 g

b

b2g

gb g

n4 = n6 x 6 + nr x r ⇒ 65 + 0
...
85nr ⇒ nr = 179
...


b

Recycle ratio = 179
...

g b100 mol fresh feedg = 179 mol fresh feed

10- 5

10
...


k =1 k = 2 k = 3
100
...
3 1515

...
0 212
...
8

nr
n2 = 100 + 0
...
85

b

g

n4 = 0
...
115n6 + 0
...
w =

q=

U
V
W

15
...
25
79
...
69
⇒
n r = 132
...
85
138
...
21
80
...


22
...
7
102
...
55
163
...
83 − 163
...
3% error
179
...
3

...
595
132
...
0
0
...

0
...
3 + 1 − −1470 1515 = 179
...


...

Error:

179
...
8
× 100 = < 01% error

...
8

e
...
8319 nr = 179
...
8319 nr = 179
...
7
SF

Split

S2

a
...
C)

B

C

0
...
5
51
...
5
31
...
0
7
...
0
0
...
2
21
...
8
0
...
7 (cont’d)
b
...

N=3
SF(1) = 0
...
57*FLOW
SF(3) = 0
...

X1 = 0
...
2,' mols/h n-octane', /,
*10X, F8
...
2,' mols/h inerts', /,
*
10X, F8
...

Program Output: Stream 1 3150 mols h n-octane
51
...
20 mols h inerts
315
...
00 mols h n-octane
34
...
80 mols h inerts
315
...
8
a
...
89272−1211
...
790) (=1350
...
95805−1346
...
693) (=556
...
307) , yBz = xBz pBz / P ( = 0
...
13), nl = 100 − nv (=55
...
571)
Fractional toluene vaporization : fT = nv (1 − yBz ) / 60 (=0
...

Q = ∑ nout H out − ∑ nin H in (= 1097
...
Once the spreadsheet has been prepared, the goalseek tool can be used to determine the
bubble-point temperature (find the temperature for which nv=0) and the dew-point
temperature (find the temperature for which nl =0)
...
9 o C, Tdp = 103
...


C **CHAPTER 10 PROBLEM B
DIMENSION SF(3), SL(3), SV(3)
DATA A1, B1, C1/6
...
033, 220
...
95334, 1343
...
377/
DATA CP1, CP2, HV1, HV2/ 0
...
190, 30
...
47/
COMMON A1, B1, C1, A2, B2, C2, CP1, CP2, NV1, NV2
FLOW = 1
...
30*FLOW
SF(2) = 0
...
0
P = 512
...
4,' mol/s Benzene',/,
* 15X, F7
...
2, 'K')
WRITE (6, 901) Q

10- 8

10
...
2,' kW')
END

C

C

C

C

C

C

SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q)
REAL NF, NL, NV
DIMESION SF(3), SL(3), SV(3)
COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2
Vapor Pressure
PV1 = 10
...
15 + C1))
PV2 = 10
...
15 + C2))
Product fractions
XL1 = (P – PV2)/(PV1 – PVS)
XV1 = XL1*PM/P
Feed Variables
NF = SF(1) + SF(2)
XF1 = SF(1)/NF
Product flows
NL = NF*(XF1 – XV1)/(XL1 – XV1)
NV = NF – NL
SL(1) = XL1*NL
SL(2) = NL – SL(1)
SY(1) = XY1*NY
SY(2) = NV – SY(1)
SL(3) = T
SV(3) = T
Energy Balance
Q = CP1*SF(1)*SF(1) + CP2*SF(2)
Q = Q*(T – SF(3)) + (NV1*XV1 + HV2*(1 – XV1))*NV
RETURN
END

b1g
XF b I g∗ NF = XLb I g∗ NL + XV b I g∗ NV I = 1,2… n − 1 b2g
Energy Balance: Q = bT − TF g∗ ∑ CPb I g∗ c XLb I g∗ NL + XV b I g∗ NV h

10
...
Mass Balance: NF = NL + NV

N

I =1

bg

N

b g b3g

+ NV ∗ ∑ HV I ∗ XV 1
I =1

b g

where: XL N = 1 −

∑ XLb I g

N −1

b g

XV N = 1 −

I =1

b g b4 g
XV b I g∗ P = XLb I g∗ PV b I g
N

∑ XV b I g

N −1
I =1

bg

Raoult’s law: P = ∑ XL I ∗ PV I
I =1

10- 9

I = 1,2, … N − 1

b5g

10
...


C **CHAPTER 10 - - PROBLEM 9
DIMENSION SF(8), SL(8), SV(8)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, NV
DATA A/6
...
87776, 6
...
, 0
...
, 0
...
63, 1171
...
115, 0
...
, 0
...
/
DATA C/232
...
366, 216
...
, 0
...
, 0
...
188, 0
...
213, 0
...
, 0
...
/
DATA NV/25
...
85, 31
...
, 0
...
, 0
...
0
N*3
SF(1) = 0
...
300*FLOW
SF(3) = 0
...
9 (cont’d)

900

901
C

C

100
200

C
300
C

500

400
900
C

FORMAT (A15, F7
...
4,' mols/s n-hexane', /,
*15X, F7
...
2,' K')
WRITE (6, 901) Q
FORMAT ('Heat Required', F7
...

DO 100 I = 1, N
NF = NF + SF(I)
DO 200 I = 1, N
XF(I) = SF(I)/NF
TF = SF (N + 1)
T = SL (N + 1)
TC = T – 273
...
**(A(I) – B(I)/(TC + C(I)))
Find NV -- Initial Guess = NF/2
NVP = NF/2
DO 400 ITER = 1, 10
NV = NVP
F = –1
...

DO 500 I = 1, N
PPM1 = PV(I)/P – 1
...

NVP = NV – F/FP
IF (ABS((NVP – NV)/NVP)
...
TOL) GOTO 600
CONTINUE
WRITE (6, 900)
FORMAT ('FLASHN did not converge on NV')
STOP
Other Variables

10- 11

10
...

Q2 = 0
...
0563 mols
0
...
2011 mols
338
...
2944 mols
0
...
1509 mols
338
...
01 kW

s n-pentane
s n-hexane
s n-heptane
s n-pentane
s n-hexane
s n-heptane

10
...


Q(kW)

nv (mol / s)
x v ( mol A(v) / mol)
1 − x v ( mol B(g) / mol)
T (K), P(mm Hg)

n F (mol / s)
xF

(mol A(v) / mol)

1 − x F (mol B(g) / mol)
TF (K), P(mm Hg)
nl (mol A(l) / s)

10- 12

10
...


References: A(l), B(g) at 25oC
Substance
nin
H in

nout

H out

A(l)

—

—

nl

H3

A(v)

nF x F

H1

nv x v

H4

B(g)

n F (1 − x F )

H2

nv (1 − x v )

H5

Given n F and x F (or n AF and n BF ), TF , P , y c (fractional condensation),
Fractional condensation ⇒ nl = y c n F x F
Mole balance ⇒ nv = n F − nl
A balance ⇒ x v = (n F x F − nl ) / nv
Raoult' s law ⇒ p * = x v P
A
Antoine' s equation ⇒ T =

B
−C
A − log 10 p *
A

Enthalpies: H1 = ΔH v + C pv (TF − 25), H 2 = C pg (TF − 25), H 3 = C pl (T − 25),
H 4 = ΔH v + C pv (T − 25), H 5 = C pg (T − 25)
Energy balance: Q = ∑ n out H out − ∑ n in H in
c
...
704
nV
0
...
050

nBF
0
...
1921
Cpg
0
...
00
A
7
...
02

xF
0
...
11
H2
1
...
2183

P
760
pA*
146
...
41

yc
0
...
8
H5
0
...
6336
Cpl
0
...
7

Greater fractional methanol condensation (yc) ⇒ lower temperature (T)
...
10 ⇒
T = 328oC
...
10 (cont’d)
e
...
87863, 1473
...
0/
DATA CPL, HV, CPV, CPG,/ 0
...
27, 0
...
029/
FLOW = 1
...
704*FLOW
SF(2) = FLOW – SF(1)
YC = 0
...

SF(3) = 333
...
2,' K')
901
FORMAT (A15, F7
...
3, 'mols/s air')
902
FORMAT ('Heat Removal Rate', F7
...

T = B/(A - LOG(N)/LOG (10
...
15
Q = ((CPV * XV + CPG * (1 - XY)) * NV + CPL * NL) * (T - TF) - NL * HV
C
Output Variables
SL(1) = NL
S2(2) = T
SV(1) = XV*NV
SV(2) = NV - SV(1)
SV(3) = T
RETURN
END

10- 14

10
...
Extent of reaction equations:

b g
b g
SPb I g = SF b I g + NU b I g∗ ξ I = 1,2, … N

ξ = −[ SF IX ∗ X ] NU IX

Energy Balance: Reference states are molecular species at 298K
...
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O

Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O
mol ⋅ K

270 m 3 1 atm
h

273K 0
...
348 mol C 3 H 8 s [=SF(1)]

g

3
...
2 5 mol O2
= 20
...

sec
mol C3 H8
X C3 H8 = 0
...
348) = 0
...


b g

b g

ξ = −[ SF IX ∗ X ] NU IX
Nu
nin (SF)
X
Xi
nout (SP)
Cp
Tin
Hin
Tout
Hout
HF
DHr
Q

= –(3
...
90)/(–1) = 3
...
348

2-O2
-5
20
...
54

4-CO2 5-H2O(v)
3
4

0
...
1431

5
...
033

75
...
0308

9
...
0495

12
...
0375

17
...
1

3
...
2

4
...
6
-103
...
8
0

23
...
2
-393
...
2
-241
...
90
3
...
As Tstack increases, more heat goes into the
stack gas so less is transferred out of the reactor: that is, Q becomes less negative
...
11 (cont’d)

C **CHAPTER 10 PROBLEM 11
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
DATA NU/–1
...
, 3
...
, 0
...
/
DATA CP/0
...
0330, 0
...
0495, 0
...
, 0
...
8, 0
...
, –393
...
83, 0
...
/
COMMON CP, HF
SF(1) = 3
...
09
SF(3) = 75
...

SF(5) = 0
...

SP(6) = 1050
...
90
N=5
CALL REACTS (SF, SP, NU, N, X, IX, Q)
WRITE (6, 900) (SP(I), I = 1, N + 1), Q
900
FORMAT ('Product Stream', F7
...
3,' mols/s oxygen', /,
*15X, F7
...
3,' mols/s carbon dioxide', /,
*15X, F7
...
2,'K', /,
Heat required', F8
...

DO 300 I = 1, N
300

HP = HP + SP(I)*CP(I)
HP = HP + (SP(N + 1) – SF (N + 1))
Q = EXT * HR + HP
RETURN
END
10- 16

10
...
Extent of reaction equations:
ξ = − SF IX ∗ X NU IX

b g
b g
SPb I g = SF b I g + NU b I g∗ ξ

I = 1, N

Energy Balance: Reference states are molecular species at feed stream temperature
...
2CO + O 2 → 2CO 2
Temporary basis: 2 mol CO fed

b

g

2 mol CO 1
...
70 mol N 2

...
00 + 1
...
70) mol = 7
...
0

kmol
1h
10 3 mol

...
8036 ⇒ SF (2) = 1
...
95 mol
SF (3) = 3
...
12 (cont’d)
Solution to Problem 10
...
607

2-O2
-1
1
...
777

4-CO2
2
0

0
...
36
0
...
02895
4
...
55E-09
-2
...
642425
3
...
72315
0
...
029 0
...
16E-05 2
...
23E-05
-6
...
72E-09 -2
...
31E-12 -2
...
46E-12

0
...
00E-05
-2
...
57E-12
650
-110
...
5

-566
1560
-4
...

As X increases, T increases
...
)
d
...
, –1
...
, 2
...
, 0
...
/
DATA ACP/ 28
...
10E-3, 29
...
11E-3, 0
...
, 0
...
4110E-5, 1
...
2199E-5, 4
...
, 0
...
/
DATA CCP/ 0
...
6076E-8, 0
...
887E-8, 0
...
, 0
...
220 E-12, 1
...
871E-12, 7
...
, 0
...
/
DATA HF / –110
...
, 0
...
5, 0
...
, 0
...
607
SF(2) = 1
...
777
SF(4) = 0
...

IX = 1
X = 0
...
12 (cont’d)

900

C

C
C
100
C

200
C

300
C

400
900

FORMAT ('Product Stream', F7
...
3, 'mols/s oxygen', /
...
3, 'mols/s nitrogen', /
...
3, 'mols/s carbon dioxide', /,
15X, F7
...
E-6
Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
Solve Material Balances
DO 100 I = 1, N
SP(I) = SF(I) + EXT*NU(I)
Heat of Reaction
HR = 0
DO 200 I = 1, N
HR = HR + HF(I) * NU(I)
HR = HR * EXT
Product Heat Capacity
AP = 0
...

CP = 0
...

DO 300 I = 1, N
AP = AP + SP(I)*ACP(I)
BP = BP + BP(I)*BCP(I)
CP = CP + SP(I)*CCP(I)
DP = DP + SP(I)*DCP(I)
Find T
TIN = SF (N + 1)
TP = TIN
D0 400 ITER = 1, 10
T = TP
F = HR
FP = 0
...
+ T*(CP/3
...
)))
*–TIN*(AP + TIN*(BP/2
...
+ TIN*DP/4
...
LT
...
12 (cont’d)

500

SP(N + 1) = T
RETURN
END

Program Output:
0
...
642 mol/s oxygen
3
...
723 mol/s carbon dioxide
T = 1560
...
13

37
...


Separator
50 mol C2H4
50 mol O2

208
...
6667
18
...
5
8
...
333333

mol C2H4
mol O2
mol C2H4O
mol CO2
mol H2O

8
...
75
8
...
333333

mol C2H4
mol O2
mol CO2
mol H2O

Xsp = 0
...
9
158
...
3333 mol C2H4 (Rc)

Rc-Ra =

0

Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator
...

Use goalseek to find the value of Ra that drives (Rc-Ra) to zero
...


Xsp
0
...
2
0
...
3

Ysp
0
...
75333
1

Yo
0
...
833
0
...
896

no
158
...
33
99
...
25

The second reaction consumes six times more oxygen per mole of ethylene consumed
...
At a certain yield for a specified ethylene conversion, all the oxygen in
the feed is consumed
...


10-21
21

10
...
001
KMAX = 20
IPR = 1
XA(1) = 2
...
0
CALL CONVG (XA, XC, N, KMAX, EPS, IPR)
END
C
SUBROUTINE FUNCGEN(N, XA, XC)
DIMENSION XA(3), XC(3)
XC(1) = 0
...
– XA(2) + (XA(1) + XA(2))**0
...
– 5
...
EQ
...
EQ
...
GE
...
EQ
...
)
IF (Q
...
0
...
5
IF (Q
...
–5) Q = –5
...
– Q)*XCM(I)
GOTO 110
500
WRITE (6, 900)
900
FORMAT (' CONVG did not converge')
STOP
END

10- 22

10
...
EQ
...
NE
...
6)
END
Program Output: K Var
Assumed
Calculated
1
1 0
...
150000E + 01
1
2
0
...
275000E + 01
2
2

1
2

0
...
115578E + 01
0
...
282353E + 01

3
3

1
2

0
...
482384E + 00
0
...
245041E + 01

8

1

0
...
113289E + 01

8

2

0
...
269315E + 01

4
9

1
2

0
...
113180E + 01
0
...
269241E + 01

10- 23

CHAPTER ELEVEN
11
...


The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:

xp =

Mp
M

Therefore, the leakage rate of hydrogen peroxide is m1 M p / M
b
...
2 a
...

Density of H3PO4: ρ = 1834 g / ml
...
00 g / mol
...
0 L 1000 ml 1
...
3743 kmol / min
Input =
min
L
ml
98
...
3743

t = 0, n p0 = 150 × 0
...
5 kmol

z

z

np

b
...
3743 dt ⇒ n p = 7
...
3743t (kmol H 3PO 4 in tank )

7
...


0

np

015 =

...
5 + 0
...
3743t

kmol H 3PO 4
kmol

7
...
3743t
⇒ t = 471 min

...
3743t

11-1

11
...


g b

b

g

b

g

bg

mw = a + bt t = 0, mw = 750 & t = 5, mw = 1000 ⇒ mw kg h = 750 + 50t h

Balance on methanol: Accumulation = Input – Output
M = kg CH 3OH in tank
dM
= m f − mw = 1200 kg h − 750 + 50t kg h
dt

b

g

E

b

dM
= 450 − 50t kg h
dt
t = 0, M = 750 kg

z zb

M

b
...

dt
c
...


−450 ±

b450g + 4b25gb750g ⇒ t = –1
...
54 h
2b −25g
2

3
...
792 kg
= 2693 kg (capacity of tank)
1 m3
1 liter
M = 2693 = 750 + 450t − 25t 2

t=

−450 ±

b450g + 4b25gb750 − 2693g ⇒ t = 719 h,10
...

2b −25g
2

Expressions for M(t) are:

R750 + 450t - 25t b0 ≤ t ≤ 719 and 10
...
54g (tank is filling or draining)

...
81)

...
54 ≤ t ≤ 20
...
3 (cont’d)
3000
2500
M(kg)

2000
1500
1000
500
0
0

5

10

15

20

t(h)

11
...


Air initially in tank: N 0 =

10
...
0258 lb - mole

532° R 359 ft 3 STP

Air in tank after 15 s:
Pf V
P0V

=

N f RT
N 0 RT

⇒ N f = N0

Rate of addition: n =

Pf
P0

=

0
...
7 psia
= 0
...
7 psia

b0
...
0258g lb - mole air = 0
...
Balance on air in tank: Accumulation = input

b

g

dN
= 0
...
0258 lb - mole
dt

z z

N

c
...
0258 + 0
...
0258

0

Check the solution in two ways:
(1) t = 0, N = 0
...


dN
= 0
...
0258 + 0
...


b g

O 2 in tank = 0
...
30 lb - mole O 2

...
5 a
...

Accumulation = Input – Output
1h
dV 540 m 3
=
− ν w m 3 min
dt
h
60 min

e

j

b

t = 0, V = 3
...
00×103

b
...
00 − ν w dt ⇒ V m 3 = 3
...
00t − ν w dt t in minutes

Let ν w i = tabulated value of ν w at t = 10 i − 1
240

g

LM
MN

i = 1, 2, … , 25

OP
PQ

b

g b

24
24
10
10

...
8 + 4 124
...
4
3
3
i = 3, 5, …
i = 2, 4, …

= 2488 m 3

b g

V = 3
...
00 240 − 2488 = 2672 m 3

c
...

The feed rate decreased, or the withdrawal rate increased between data points,
or the storage tank has a leak, or Simpson’s rule introduced an error
...


REAL VW(25), T, V, V0, H
INTEGER I
DATA V0, H/3
...
/
READ (5, *) (VW(I), I = 1, 25)
V= V0
T=0
...
00 * H – 0
...
2, 7X, F6
...
4 11
...
1
11
...
5
11
...
00
10
...
00

VOLUME (CUBIC METERS)
3000
...

2944
...
00
240
...

2674
...
07%
2672

Simpson’s rule is more accurate
...
6 a
...
200V ν out = 20
...
Balance on water: Accumulation = input – output (L/min)
...
0 − 0
...


dV
= 0 = 200 − 0
...
t begins at (t=0, V=300)
...
0 − 0
...
0
...
⇒ dV / dt = 20
...
200V
becomes less negative, approaches zero as t → ∞
...


t

z

z

V

d
...
0 − 0
...
0 − 0
...
0
0
...
5 + 0
...
200t ⇒ V = 100
...
0 exp −0
...

V = 101 100 = 101 L 1% from steady state ⇒

b

g

101 = 100 + 200 exp −0
...
200

g = 26
...
7 a
...
t (rectangular scale) yields a straight line through the points ( t = 1 week,
D = 2385 kg week ) and ( t = 6 weeks, D = 755 kg week )
...
230
6−1
t 2 − t1

b g b

gb g

ln a = ln D1 − bt1 = ln 2385 + 0
...
007 ⇒ a = e 8
...
230t

b
...
230t kg week
dt
t = 0, I = 18,000 kg

g

z z
I

18,000

c
...
8 a
...
230t dt ⇒ I − 18,000 =
0

3000 −0
...
230

t
0

⇒ I = 4957 + 13,043e −0
...
4 m 3 STP
700 m 3
min

273 K

103 mol

b g = 28,920 mol min

295 K 22
...


b g

d
dx
= −0
...

= 330 × 10 −5 mol SO 2 mol

...
The plot of x vs
...
30×10-5)
...
6364 × 330 × 10 −5 = −210 × 10 −5
...
⇒

...

dx dt = −0
...
The curve
is therefore concave up
...
8 (cont’d)

0
t

c
...
30×10 −5

dx
x
= −0
...
6364t ⇒ x = 330 × 10 −5 e −0
...

x 0
330 × 10 −5

...
30 × 10-5 mol SO 2 / mol ⇒ satisfies the initial condition;
dx
(2)
= −0
...
6364t = −0
...


...


CSO 2 =

45,440 moles x mol SO 2
1100 m 3

mol

i)

ii)

e
...


x = 10 −6 ⇒ t =

e

1 m3
103 L

mol SO 2
liter

ln 10 −6 3
...
6364


...

= 4131 × 10 −2 x = 13632 × 10 −6 e −0
...


The room air composition may not be uniform, so the actual concentration of the SO2
in parts of the room may still be higher than the safe level
...


11-7

11
...


Balance on CO: Accumulation=-output
N ( mol ) x ( mol CO / mol) = total moles of CO in the laboratory
Pν p
kmol
)=
h
RT
Pν p
kmol CO
kmol
)x
Rate at which CO leaves: n (
=
x
h
kmol
RT
CO balance: Accumulation = -output
Molar flow rate of entering and leaving gas: n (

FG
H

FG
H

IJ
K

IJ
K

Pν p
d ( Nx )
dx
P
x⇒
=−
=−
ν px
dt
RT
dt
NRT

E PV = NRT

νp
dx
=−
x
dt
V
kmol CO
kmol

t = 0, x = 0
...
01
t

x

b
...


b g

V = 350 m 3
350
tr = −
ln 100 × 35 × 10 −6 = 2
...
The room air composition may not be uniform, so the actual concentration of CO
in parts of the room may still be higher than the safe level
...


Precautionary steps:
Purge the laboratory longer than the calculated purge time
...


11
...


Total mass balance: Accumulation = input – output

b

g

dM
= m − m kg min = 0 ⇒∴ M is a constant = 200 kg
dt

b
...
45

11-8

11
...


0
...

dt

m increases ⇒

z

z

x

d
...
45 exp −
0
...
45
0

IJ
K

Check the solution:
(1) t = 0, x = 0
...
45 ×
exp( −
)=−
x ⇒ satisfies the mass balance
...
45
0
...
35

m = 100 kg / m in

0
...
25
0
...
15
0
...
05
0
0

5

10

15

20

t(m in)

e
...
45
90% ⇒ x f = 0
...
6 min
99% ⇒ x f = 0
...
2 min

99
...
00045 ⇒ t = 138 min

...
11 a
...
If perfectly mixed, Cout = C tank = C

b g = −ν C bkg ming

d VC

dC
ν
=− C
dt
V
m
t = 0, C = 0
V

V is constant

dt

b
...


dC
=−
C

z

t

ν

0

V

dt ⇒ ln

FG C IJ = − νt ⇒ C = m
V
Hm VK V

0

FG νt IJ
H VK

exp −

0

Plot C (log scale) vs t (rect
...
223 × 10 −3 & t = 2, C = 0
...

−

ν
V

=

b

2 −1

E
V = e30 m

11
...


g = −1495 min

...
050 0
...


...
7 psia, V=40
...
7 psia
40
...

ft ⋅ psia
RT
528 o R
10
...
Molar throughout rate:
60 ft 3 492° R 16
...

min 528° R 14
...
35n − xn ⇒ 01038 dx = 01695b0
...
35 − xg

...


...


z

dt

z

t = 0, x = 0
...
35 − x
dx
= 163t

...

0
...
35 − x
0
0
...
21
x

b

g

0
...

⇒
= e −1
...
35 − 014e −1
...

1
0
...
27
x = 0
...
343 min (or 20
...

0
...
21

LM FG
N H

IJ OP
KQ

11-10

11
...


b gb

Mass of isotope at any time = V liters C mg isotope liter

g

Balance on isotope: Accumulation = –consumption

FG IJ b g
H K

b g

dC
= − kC
dt
t = 0, C = C0

Cancel V

mg
d
VC = − kC
V L
L⋅s
dt

Separate variables and integrate

z z
C

C0

dC
=
C

FG C IJ = − kt ⇒ t = − lnbC C g
k
HC K
− lnb0
...
5C0 ⇒ t 1 2

b
...
6 hr ⇒ k =
C = 0
...
267 hr −1
2
...
2 hr

− ln 0
...
267

11
...


Mole balance on A: Accumulation = –consumption

b g = − kC V

d C AV

A

dt

bV constant; cancelsg

t = 0, C A = C A0
⇒

z

CA

CA0

dC A
=
CA

z

t

− kdt ⇒ ln

0

FG C IJ = − kt ⇒ C
HC K
A

A

b g

= C A0 exp − kt

A0

b
...
t (rect
...
The data fall on a straight line (verifies

b

g b

g

assumption of first-order) through t = 213, C A = 0
...
0, C A = 0
...


...
0185 0
...
0 − 213

...


−3

min −1 ⇒ k = 35 × 10 −3 min −1

...
15 2 A → 2 B + C
a
...
15 (cont’d)
b
...
5C A0 ⇒ −
n A = 0
...
5n

n
P
RT
1
1
1
+
= − kt 1 2 ⇒ t 1 2 =
; but C A0 = A0 = 0 ⇒ t 1 2 =
0
...
gb1 mol C 2 mol A react
...
25n

n B = 0
...
2 mol B 2 mol A react
...
5n A0

nC

A0

total moles = 125n A0 ⇒ P1 2 = 125

...

c
...

V

Plot t 1 2 vs
...
Data fall on straight line (verifying 2nd order

d

i d

i


...
683
RT
1060 − 209
=
= 143
...
683

...


t1 2 =

b1015 Kgb0
...
582 L mol ⋅ s
143
...
1 T (rect
...


bg

bg

t 1 2 s , P0 = 1 atm, R = 0
...
0, 1 T = 1 900 &

dt

i

RT = 0
...
6383 74
...


b

R=8
...
49 × 10 5 J mol

g

1
29,940
= ln 0
...
96 ⇒ k 0 = 3
...
204 L (mol ⋅ s)
RT

g

0
...


b0
...


90% conversion

LM
N

OP
Q

LM
N

1 1
1
1
1
1
−
=
−
−3
k C A C A0
0
...


...
4 min

C A = 010C A0 ⇒ t =

...
16 A → B
a
...
Plot t C A − C A0 vs
...
28, −0
...
01, −0
...
01 − 116
...
62 ⇒ k 1 = 2
...
2496 − −0
...
01 + 356
...
2496 = 4100 ⇒ k 2 = 0115 L mol

...

k1

11
...


3
...

b g = 012035 mol gas

...
60b012035 molg 3
...
02407 mol L CO U
|initial concentrations

...
40b012035 molg 3
...
01605 mol L Cl V
|
W
C bt g = 0
...
01605 − C bt g V
|
W
303
...
4 L STP

CO i
Cl 2

2

i

CO

p

Cl 2

p

2

b
...


d1 + 58
...
00 L

8
...
3C p

i

dt

2

=

d

id

2
...
02407 − C p 0
...

d1941 − 24
...
75 0
...
01204 mol L

1
t=
2
...

d1941 − 24
...
02407 − C id0
...
01204

0

p

p

p

11-13

p

2

i

11
...


11
...


REAL F(51), SUM1, SUM2, SIMP
INTEGER I, J, NPD(3), N, NM1, NM2
DATA NPD/5, 21, 51/
FN(C) = (1
...
3 * C) ** 2/(0
...
01605 – C)
DO 10 I = 1, 3
N = NPD(I)
NM1 = N – 1
NM2 = N – 2
DO 20 J = 1, N
C = 0
...

DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
30
CONTINUE
SUM2 = 0
...
01204/FLOAT(NM1)/3
...
0 * SUM1 + 2
...
92
WRITE (6, 1) N, T
10 CONTINUE
1 FORMAT (I4, 'POINTS —', 2X, F7
...
0 MINUTES
21 POINTS — 90
...
4 MINUTES
t = 90
...
Since p A = y A P is constant, C * = p A H is also a constant
...
18 (cont’d)
b
...
020 cm s , S = 78
...
62 × 10
3

2

−3

mol / cm

a fa f d9230 atm ⋅ cm moli = 0
...
62 × 10 I = 9800 s ⇒ 2
...
02 cm sge78
...
65 × 10 JK
C * = y A P H = 0
...
If the liquid is well agitated, S may in fact be much greater than
this value, leading to a significantly lower t than that to be calculated)
11
...


Total Mass Balance: Accumulation = input
dM d ( ρV )
=
= ρv
dt
dt

E

dV
=v
dt
t = 0, V = 0

A Balance: Accumulation = input – consumption
dN A
= C A0 v − ( kC A )V C =N /V
A
A
dt

dN A
= C Ao v − kN A
dt
t = 0, N A = 0

b
...


t

0

dN A
C v
= 0 ⇒ N A = A0
dt
k

0

dV =

NA

0

⇒−

vdt ⇒ V = vt

dN A
=
C A0 v − kN A

z

t

dt

0

FG
H

IJ
K
C v
=
1 − expb− kt g
k

C v − kN A
C v − kN A
1
ln A0
= t ⇒ A0
= e − kt
k
C A0 v
C A0 v

⇒ NA

CA =

A0

t →∞⇒ NA =

N A C A0 [1 − exp( − kt )]
=
V
kt

11-15

C A0 v
k

11
...

NA would never reach the steady value in a real reactor
...

But in a real reactor, the volume is limited by the reactor volume;
(2) The steady value can only be reached at t → ∞
...

d
...
20 a
...
00 L)(100 kg / L) = 3
...

Cv = C p = (0
...
018 kg) = 4
...
0797Q (kJ / s)
dt
t = 0, T = 18 o C

z z
100o C

c
...
21 a
...


0

dT =
o

0
...
287
= 4
...
0797
s

Stove output is much greater
...

Some energy heats the kettle
...

Energy balance: MCv

dT
= Q −W
dt
M = 20
...
0754 kJ / mol ⋅ C)(1 mol / 0
...
184 kJ / (kg ⋅ C)
o

o

a f

Q = 0
...
50) = 2
...
0290 ° C s , t = 0, T = 25° C
dt

The other 3% of the energy is used to heat the vessel or is lost to the surroundings
...


t

dT =

o

25 C

c
...
0290dt ⇒ T = 25° C + 0
...
0290 = 2585 s ⇒ 43
...


11-16

11
...
Energy balance on the bar
MCv

b

dTb
= Q − W = −UA Tb − Tw
dt

g

B

Table B
...
7 g cm j = 462 g
3

Cv = 0
...
050 J (min ⋅ cm ⋅° C)

a fa f a fa f a fa f

2

A = 2 2 3 + 2 10 + 3 10 cm = 112 cm

b

gb

dTb
= −0
...
02635 Tbf − 25 ⇒ Tbf = 25° C
dt

b
...


t
dTb
= −0
...
02635t
H 95 − 25K
⇒ T bt g = 25 + 70 expb−0
...
02635e −0
...
02635(Tb − 25) ⇒ reproduces the mass balance;
dt
(3) t → ∞, Tb = 25o C ⇒ confirms the steady state condition
...
23

12
...
0 kg/min
T (oC)
Q (kJ/min) = UA (Tsteam-T)

a
...
0 kg min

dT / dt = 150 − 0
...

Cv ≈ C p = 2
...


a

f

Tsteam sat' d; 7
...
8° C

b
...
0224Ts ⇒ Ts = 67° C

...


IJ
K

FG
H

z

Tf

dT
1
150 − 0
...

150 − 0
...
0224t )

...
0224T 0

...
0224
0
...
0224
25
t

t = 40 min
...
8° C

d
...
Let x = (UA) new
...
3947 + 0
...
01579 + 5
...
3947 + 0
...
01579 + 5
...
01579 + 5
...
3947 + 0
...
01579 + 5
...
3947 + 0
...
01579 + 5
...
27 kJ / (min⋅o C)

ΔU
Δ(UA)
14
...

=
=
× 100% = 241%

...


11-18

11
...


Energy balance: MCv

dT
= Q −W
dt

...
2W = 40
...
0649tbsg
V T = 40° C ⇒ t = 308 s ⇒ 51 min

...
0649 ° C s
dt
t = 0, T = 20° C

b
...

input will serve to vaporize benzene isothermally
...

Time to reach Tb neglect evaporation : t =
= 926 s
0
...
765 kJ mol 1 mol 78
...


Evaporation rate = 40
...


c
...
25 a
...
Used a dirty flask
...
Use a clean flask
...
Put an open flask on the burner
...

Use a covered container or work under a hood
...
Left the burner unattended
...
Looked down into the flask with the boiling chemicals
...
Wear goggles
...
Rubbed his eyes with his hand
...

6
...
Use lab gloves
...
Put hot flask on partner’s homework
...

60 m 3

273 K 1 kg - mole
= 2
...
4 m 3 STP
dT
Energy balance on room air: nCv
= Q −W
dt
Q = ms ΔHv H 2 O, 3bars, sat' d − 30
...
0 T − T0
dt
N = 2
...
8 kJ (kg - mole⋅° C)
ΔH v = 2163 kJ kg from Table B
...
3ms − 0
...
We simplify the analysis by assuming n is constant
...
25 (cont’d)
0
...
3

b
...
3ms − 0
...
333 kg hr

c
...
3ms − 0
...
333

E

T f = 23°C

t=−

11
...


b

Integral energy balance t = 0 to t = 20 min
Q = ΔU = MCv ΔT =

=t

LM
NM

b g OP = 4
...
4 − 0
...
559 13
...
559 10

g

b60 − 20g° C = 4
...
00 kJ
kg⋅° C

Required power input: Q =

dT

23

10 13
...
559T

dT
=Q
dt

bg

dT
= 0
...
00 kJ kg⋅°C

z z

T

kJ

4
...

20 min
60 s 1 kJ s

b
...
001 Q dT ⇒ T = 20 o C + Qdt

20o C

0

0

Evaluate the integral by Simpson's Rule (Appendix A
...
001 oC / kJ 34830 kJ = 54
...


Past 600 s, Q = 100 +

b

g

10 kW
t − 600 s = t 6
60 s

LM
M
T = 20 + 0
...
001M Qdt +
MM
N
I
0
...
8 +
GH 6 − 600 JK ⇒ tbsg =
6
2

z

z z

600

t

0

0
34830

2

2

t

OP
t P
dt
6 P
PP
Q

600

b

12000 T − 24
...
27 a
...
00ρ − 4
...
00 L / s
dt
t = 0, V0 = 400 L

KCl Balance:
dM KCl
d( CV )

...
00 − 4
...
(i)The plot of V vs
...
The slope (=dV/dt) is 4 (a positive constant)
...
Then the tank begins to overflow
and V stays constant at 2000
...
t begins at (t=0, C=0)
...
02
...
The curve is therefore concave down
...


dV
=4⇒
dt

z

V

z

t

dV = 4 dt ⇒ V = 400 + 4t

400

0

11-21

11
...
5t
C dC
t
dt
C
t
=
⇒ − ln(1 − C ) 0 = 2 ln(50 + 0
...
5t
50 + 0
...
01t )2
50
1
1
⇒
= (1 + 0
...
01t )2

z

z

When the tank overflows, V = 400 + 4t = 2000 ⇒ t = 400 s
1
= 0
...
01 × 400

b

g

11
...
Salt Balance on the 1st tank:
Accumulation=-Output

E

v
d(CS1V1 )
dC
= − CS1v ⇒ S1 = − CS1 = −0
...
08( CS1 − CS 2 )
dt
V2
dt
CS 2 ( 0) = 0 g / L

Salt Balance on the 3rd tank:
Accumulation=Input-Output

E

v
d(CS3V3 )
dC
= CS 2 v − CS 3v ⇒ S3 = ( CS 2 − CS 3 ) = 0
...


CS1, CS2, CS3

3

CS1

CS2
CS3

0
t

11-22

11
...
t begins at (t=0, CS1=3)
...
08 × 3 = −0
...

As t increases, CS1 decreases ⇒ dCS1/dt=-0
...
The curve is therefore concave up
...
t begins at (t=0, CS2=0)
...
08(3 − 0) = 0
...

As t increases, CS2 increases, CS1 decreases (CS2 < CS1)⇒ dCS2/dt =0
...
Then CS2 decreases with increasing t as well
as CS1
...
Therefore, CS2 increases until it reaches a
maximum value, then it decreases
...
t begins at (t=0, CS3=0)
...
04(0 − 0) = 0
...
04(CS2-CS3) becomes positive ⇒ CS2
increases with increasing t until dCS3/dt changes to negative (CS3 > CS1)
...
Therefore, CS3 increases until it reaches a maximum value then it
decreases
...

3

CS1, CS2, CS3 (g/L)

2
...
5
CS2

1

CS3

0
...
29 a
...
2C A
2
(ii) Rate of consumption of B in the 2nd reaction: − rB 2 = r2 = 0
...
Mole Balance on A:
Accumulation=-Consumption

E

d ( C AV )
dC A

...

= −01C AV ⇒
= −01C A
dt
dt

...
2C AV − 0
...
2C A − 0
...
29 (cont’d)
c
...
t begins at (t=0, CA=1)
...


...

As t increases, CA decreases ⇒ dCA/dt=-0
...
CA→0 as t→∞
...

The plot of CB vs
...
When t=0, the slope (=dCB/dt) is 0
...
2
...
2(CA- C B ) becomes less positive

2
until dCB/dt changes to negative ( CB > CA)
...

Finally dCB/dt approaches zero as t→∞
...
CB→0 as t→∞
...
t begins at (t=0, CC=0)
...
2(0) = 0
...
2 C B becomes positive also increases with increasing t
2
⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0
...
Finally CC→2 as t→∞
...


CA, CB, CC (mol/L)

d
...
2
2
1
...
6
1
...
2
1
0
...
6
0
...
2
0

CC

CB
CA

0

10

20

30
t (s)

11-24

40

50

11
...
When x = 1, y = 1
...
Raoult’s Law:

p *C5 H12 (46o C) = 10

Antoine Equation:

(6
...
793
)
46 + 231
...
70, y=0
...
970 =

0
...
70 + b

=

xp *C5 H12 ( 46o C )
P

= 1053 mm Hg

0
...
970
760

(1)

⇒

( 2)

Ra = 1078
|
...
078
|
T

c
...
70

IJ
K

d
...
0 kJ/ mol

t = 0, N L
dN L
= − nV
dt
nV
Q 27
...
0

= 100 mol

nV =

Q

b27
...
0

11
...
0
ax
=− V
−x =−
−x
dt
NL x + b
Qt x + b
100 27
...
70

IJ
K

e
...
9
0
...
5 kJ/s)

0
...
6
x (Q=1
...
5
0
...
3

y (Q=3 kJ/s)

0
...
1
0
0

200

400

600

800

1000 1200 1400 1600 1800

t(s)

f
...
The initial and final mole fraction of pentane in the vapor are 0
...
The
higher the heating rate, the faster x and y decrease

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