22–95E In a one shell pass and eight tube pass heat exchanger, water is to be heated using hot air at 600F. For the given
value of convection heat transfer coefficient on the outer surface of the tubes and the fouling resistances, the heat exchanger
= 0.999 Btu/1bm·F 1 Btu/1bm·F
Analysis The heat gained by water from hot air is,
Btu/h 104F70)–)(150Btu/lbm (1lbm/h)000,50()( 6
,, FTTcmQ incoutcpcc
From energy balance we have, heat lost by air = heat gained by water.
lbm/h 53,333.33
F300)–F)(600Btu/lbm(0.25
Btu/h104
)(
)(
)()(
6
,,
,,
,,,,
outhinhph
incoutcpcc
h
incoutcpccouthinhphh
TTc
TTcm
m
TTcmTTcm
Now the logarithmic temperature difference for a counter flow heat exchanger is calculated as
21
21
,/ln TT
TT
TCFlm
F 450150600 o
,,1 outcinhTTT
and
F23070300 o
,,2 incouthTTT
F 78.327
230/450ln
230450 o
,
CFlm
T
In order to determine correction factor we first need to find the temperature ratios P and R as follows:
566.0
60070
600300
11
12
tT
tt
P
and
266.0
600300
15070
12
21
tt
TT
R
From Figure 22-19 (a), the correction factor is
The overall heat transfer coefficient is to be calculated from the given value of convection heat transfer coefficient on the
266.0
000,50
33.13333
max
min C
C
c