22-18
22.16
22-19
22.16 (cont’d)
Solution to Problem 22.16
22-20
22.17
22-21
22.17 (cont’d)
Solution to Problem 22.17
22.17 (cont’d)
22-23
19.18 (a) Acrylic Acid reboiled at 89°C = 362 K
(b) Ptop Pbottom P
Ptop 15.92 kPa 9.5 kPa
6.42 kPa
(c) T-Q Diagrams
Condenser Reboiler
45°C
44.68°C 120°C
49.68°C
22-24
19.19 (a)
T-Q Diagram
m
m
Q
Q
Q
steam
steam
1
2
1,
2,
1
2
All resistance on tube side 8.0
8.0
1
2
1
2M
m
m
U
U
For M = 1.15
140°C
120°C
(b) 2
vbaP
to get system curve
a
P
sd
P
head
P
P220.57 0.01633«
v
2
35m3h1.15
40.25m3h
at 40 m3/h
P
247.03 kPa No problems
(c) Desuperheater – Q = UA∆T
If Q increases ∆T increases must raise steam pressure or lower column
pressure
22-26
19.20
1 2
With valve closed U = UBC, –∆Pf,BC = P1 – P2, VOL Flow = QBC
With valve open U = U2, –∆Pf,2, Q2
(a) For fully developed turbulent flow, –∆Pf
Q2
(b) 1.69 1.3 0.39 or 39% of new flow through bypass
22-27
19.21 (a) P
gh Pf
(b) Q = UA∆T
1
U1
5000W m2K1
3000W m2KU1875W m2K
(c) Flooding – Will column perform this separation without flooding (may need large
reflux ratio if N is much lower than needed?
22-28
19.22 (a) Tsteam = 160ºC
Tprocess = 70ºC
T
160C70C90C
In order to find Twall need hprocess Trial and Error
Twall
(°C)
∆T
(°C)
hprocess
(W/m2°C)
hsteam
(W/m2°C)
U
(W/m2°C)
Twall
(°C)
110 40 2500 3000 1364 119.09
Twall is approximately 120°C and U = 1301 W/m2°C
19.23 (a) P
12 P2P
1101 kPa
200 kPa
99 kPa
(b)
∆P (kPa) P (kPa) Location/Equipment
101 TK-2901
100(1.5)2=225 Pipe
Pin at E-2904 = 607.3 kPa
Pin at E-2905 = 438.5 kPa
Note to instructor: There is no part c for this problem. What is labeled part c in the book
is the question for part b.
22-30
19.24 (a) 2
mPf
(b) 2
mPf
(c) Exchanger 1
1
U1
1
200W m2K1
hi
1
1000W m2Khi250W m2K
U21
250M0.8


 

1
1000


 




 


1
1
250 2.28 2

0.8







1
1000


 










1
U2 = 217.3 W/m2K
(d) Exchanger 2
22-31
(e) Exchanger 1
2,22
22 ,2
2
111,1 1,11
2.28 1.14
2
p
lm
lm p
mC T
UA T
Q
QUAT mCT


(f) Exchanger 2
2,22
22 ,2
2
111,1 1,11
2.163 1.442
1.5
p
lm
lm p
mC T
UA T
Q
QUAT mCT


(g) P2P
1
40 3.1
2


 


2
25 m2
1.5


 


2
22-32
22.25
The maximum possible scale up is 148/120 = 1.23 or 235
1 = old
2 = new
(a)
3
condensing since 3
11
11
8.0
1
8.0
12
1211
22
11
1
2
Mh
Mhh
hhhh
hh
hh
U
U
po
pii
oooi
oi
oi
22-33
(b)
Now T2 = 153C – solve for Mp
27
155
ln25.71
25153
3
1
4
130
25153
8.0
p
M
M