Kx 
xi 

xCO2 xH 2
xCO xH 2O
nCO2 N H 2
nCO N H 2O
ni
so that nT cancels
nT
Problem 1.15
k 1
k 1


k




P3 k 
kPV
1 1  P2
W
  2  
k  1  P1 
 P2  


1
1




k  1 
k 




p
p3  
dW kPV
1
1
k
p
k


3
1 1 
2

   
     0
dP2 k  1  k  p1   p1 
k  p2   p22  


 p2   1   p3   p3 
      2   0
 p1   p1   p2   p2 

p2
1/ k
p 

1
k
p1 k  p3
1
1
2 k 1
2
1
1
1
k
1
p2 k
  p1 p3  k
1
1
k
2
0
1
p2  p1 p3
2
p2  p1 p3  (1)(4) = 2 atm
Problem 1.16
(a)
C  50  0.1P 
9000
($ / bbl)
P
dC
9000
 0.1  2  0
dP
P
P*  300 bbl/day
12
(b)
f  300  50  0.1P 
9000
($ / bbl)
P
9000  $  Pbbl 

f   300  50  0.1P 



P  bbl  day 

(c)
df
 300  50  0.2 P  0  0
dp
250
P* 
 1250 bbl / day
0.2
(d)
They are different because you can sell more
Problem 1.17
Basis: 1 hr
Heat balance for the gas:
T
q= m
Cp
q  (3000) (0.3) (195  90)  9.45 x104 Btu/hr
Heat balance for cooling water
q  m(1) (To  80)  m(To  80) Btu/hr
so m 
q
70  80
For the heat exchanger
q  UATLM  8 ATLM Btu/hr
so A 
where
TLM 
(195  To )  (90  80)
in 0 F
 195  To 
ln 

 90  80 
q
8TLM
Basis: 1 yr.
Annual cooling water cost ($)
13
 9.45 x104   1  0.2 



 (24 x 365)
 To  80   62.4  1000 
2.6533 x103

To  80
Annual fixed charges for the exchanger ($)
 9.45x104 

 (0.5) 
 8TLM 
5.9063x103
TLM
dc
 1.5C1 D 0.5 L  0.081C2 m3  2 LD 6  0
dD
D
opt
C 
 0.638  2 
 C1 
0.1538
m0.4615  0.3077
For   1cP(2.42 lb/ft hr) , p  60 lb/ft 3 ,
D opt  0.366ft.
Problem 1.18
(a)
C  C1D1.5 L  C2mP / 
where
P  2 V 2 L / Df
0.046 0.2
D0.2V 0.2  0.2
V  4m /  D2
f 
Substituting the expression for f and V into that for P, we get
P  0.1421 1m1.8  0.2 D4.8 L
The cost function in terms of D is now
C  C1D1.5 L  0.1421C2  2m2.8  0.2 D4.8 L
dc
 1.5C1D 0.5 L  0.682C2  2 m2.8  0.2 D 5.8 L = 0
dD
14
Solving this equation for D, we get
C 
D  0.882  2 
 C1 
From this,
0.1587
opt
V
C 
 1.6367  1 
 C2 
opt
C opt  0.828C1
0.3174
0.8413
 0.2596C1
C2
0.7618
(b)
 0.317 m0.444  0.0317
 0.366 m0.112  0.0634
0.1587
C2
 0.4755m0.666  0.0476
0.2382
 04784 m0.6688  0.0478 L
C1  1.42363 x103 $ / hr ft 2.5 

for C in $/hr
2
2 
13
C2  2.7097 x10 $hr /ft lb 

For   1cP(2.42 lb/ft hr),   60 lb/ft 3
D opt  0.384ft, V opt  1151.5ft/hr
For   0.2cP(0.484 lb/ft hr),   50 lb/ft 3
D opt  0.387ft, V opt  1363.2ft/hr
For   10cP(24.2 lb / ft hr),   80 lb/ft 3
D opt  0.377ft , V opt  895.6ft/hr
Problem 1.19
C 
D  0.882  2 
 0.317 m0.444  0.0317
C
 1
 dln Dopt 
Dopt
 0.317
S  

 dln p   ,m,C
0.1587
opt
2
 d ln Dopt 
 0.317
SDopt  

 d ln    ,m,C2
15
 dln D opt 
SmDopt  
 0.444

 dln m   ,  ,C2
 dln D opt 
SCDopt

 0.1587


2
dln
C

2   ,  ,m
C opt  0.828C1
0.8413
C2
 0.2596C1
0.7618
0.1587
C2
 0.4755m0.666  0.0476 L
0.2382
 0.4784 m0.6688  0.0478 L
  ln C opt 
1

 opt (0.4755T1  0.4784T2 )

 p   ,m,C2 C
S
Copt
S
Copt
  ln C opt 
1

 opt (0.476T1  0.0478T2 )

  ln    ,m,C2 C
Copt
 ln C opt 
1

 opt (0.666T1  0.6688T2 )

 ln m   ,  ,C2 C
Sm
SC2
Copt
 ln C opt 
1

 opt (0.1587T1  0.2382T2 )

 ln C2   ,  ,m C
where T1  0.828C1
0.8413
C2
0.1587
 0.4755  0.0476m0.666
 0.4784  0.0478m0.6688 L
For   60lb/ft 3 and   1cp(2.42 lb/ft hr)
and T2  0.2596C1
0.7618
S C
opt
S C
opt
SmC
opt
C2
0.2382
  0.476
 0.476
 0.666
SCC2  0.163
opt
Problem 1.20
The variables selected could be times, but the selection below is easier to use.
Let Xij be the number of batches of product i (i = 1, 2, 3) produced per week on unit j
(j = A,B,C). We want to maximize the weekly profit.
16