The effectiveness for a heat exchanger is 0.6428.
Explanation:
a) Sketch the temperature profiles for a 50% and 80% effectiveness heat exchanger.
The heat transfer type was not mentioned.
Assume the heat exchanger which is counter flow and the heat exchanger performance is high.
In the cold fluid stream, the 80% of effective heat exchanger compares the 50% of heat exchanger.
The diagram was attached below.
b) The effectiveness for a heat exchanger is calculated.
Given data
[tex]m_{h}[/tex] = 2 lb/s
[tex]T_{hi}[/tex] = 160 F
[tex]m_{c}[/tex] = 4 lb/s
[tex]T_{ae}[/tex] = 110 F
[tex]T_{ai}[/tex] = 20 F
ε = [tex]\frac{Actual heat transfer}{Maximum possible heat transfer}[/tex]
From tables,
[tex](C_{p} )water[/tex] = 1.001 Btu / lb ° F
[tex]C_{water} = C_{p} water[/tex]× [tex]m_{h}[/tex]
= 1.001 × 2
= 2.002
[tex](C_{P}) air[/tex] = 0.24 Btu / lb ° F
[tex]C_{air}[/tex] = [tex](C_{P}) air[/tex] × [tex]m_{c}[/tex]
= 0.24 × 4
= 0.96
ε = [tex]\frac{ C_{air} }{C_{air} }[/tex] [tex]\frac{(T_{ae} - T_{ai} ) }{(T_{hi} - T_{ai} )}[/tex]
= [tex]\frac{(110-20)}{(160-20)}[/tex]
ε = 0.6428