Water (cp = 4180 J/kg·K) is to be heated by solar-heated hot air (cp = 1010 J/kg·K) in a double-pipe counter-flow heat exchanger. Air enters the heat exchanger at 102°C at a rate of 0.3 kg/s, while water enters at 25°C at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given as 80 W/m2·K. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm. Determine the outlet temperature of water. (Round the answer to a single decimal place.) The outlet temperature of water is ______ °C.
Heat Exchangers
Heat exchangers are the types of equipment that are primarily employed to transfer the thermal energy from one fluid to another, provided that one of the fluids should be at a higher thermal energy content than the other fluid.
Heat Exchanger
The heat exchanger is a combination of two words ''Heat'' and ''Exchanger''. It is a mechanical device that is used to exchange heat energy between two fluids.
Water (cp = 4180 J/kg·K) is to be heated by solar-heated hot air (cp = 1010 J/kg·K) in a double-pipe counter-flow heat exchanger. Air enters the heat exchanger at 102°C at a rate of 0.3 kg/s, while water enters at 25°C at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given as 80 W/m2·K. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm.
Determine the outlet temperature of water. (Round the answer to a single decimal place.)
The outlet temperature of water is ______ °C.
![Effectiveness can be determined from the following table.
Heat exchanger type
Effectiveness relation
1 Double pipe:
Parallel-flow
Counter-flow
2 Shell-and-tube:
One-shell pass
2, 4,...tube
passes
n-shell passes
2n, 4n, ... tube
passes
3 Cross-flow
(single-pass)
Both fluids
unmixed
Cmax mixed,
Cmin unmixed
Cmin mixed,
Cmax unmixed
4 All heat exchangers
with c-0
8 =
8-
1 - exp [-NTU(1 + c)]
1 + c
1-
exp [-NTU(1
- c)]
c exp[-NTU(1 c)])
1
NTU
1 + NTU
(for c= 1)
e-1-exp
4-2{1+0+v0
-1
&n=
· [( ²₁ - ²)² - ¹ ] [( ¹₁ - )" -] ¹
81
NTU⁰.22
C
(for c < 1)
8-1-exp
& = 1- exp(-NTU)
1+ exp -NTU, V1 + c²
c².
1- exp-NTU, V1+ c²
- [exp(-CNTU078) - 11}
e = (1 - exp(-cf1 - exp (-NTU)]})
p{-11 - exp (-C NTU)]}
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