Page 347 - Air Pollution Control Engineering

06_chap_wang.qxd 05/05/2004 4:10 pm Page 325
Condensation 325

1. The logarithmic mean temperature difference (∆T )
LM
2. The condenser (heat exchanger) surface area (A )
con
3. The coolant flow rate (Q )
coolant
4. The refrigeration capacity (Ref)
5. The quantity of recovered product (Q )
rec
The following data are given:
T = 90ºF
e
T = 20ºF
con
H = 239,000 Btu/h
load
Condenser = countercurrent flow
U = overall heat transfer coefficient
2
U = 20 Btu/h-ft ºF (assumed)
HAP = 0.06633 lb-mol/min
e,m
HAP = 0.006633 lb-mol/min
o,m
Solution
1. Determine ∆T using Equation (9a) for counter-current flow.
LM
T = 90 ° F
e
T con = 20 ° F
−
5
T = 20 15 = °F (9c)
cool, i
=
T = T + 25 30 F °
cool, o cool, i (9d)
e (
T − T o) − T ( − T )
∆T = cool, con cool, i (9a)
LM e ( [ T ( T )]
ln T − T cool, o) con − cool, i
∆T = (90 − 30 ) − (20 − ) 5
LM − )]
ln ( [ 90 − 30 ) (20 5
∆T = 32 F °
LM
2. Determine A using Eq. (8).
con
∆T = 32 F °
LM
,
H = 239 000 Btu h
load
2
°
U = 20 Btu h-ft F (assumed)
A con = H load U ∆T LM (8)
× )
A con = 239 000 (20 32
,
A con = 370 ft 2
3. Determine Q using Eq. (10).
coolant
,
H load = 239 000 Btu h
T = 20 − 15 = F ° 5 (9c)
cool, i
T = T + 25 = 30 F ° (9d)
cool, o cool, i
°
C = . 065 Btu lb F (ref. 3)
p coolant
Q = H load [ C T ( − T )]
coolant p coolant cool, o cool, i (10)
− )]
Q = 239 ,000 [ .0 65 (30 5
coolant
Q =14 ,700 lb h
coolant