A watercooled air compressor takes air in at 70 F, 14 lbf/in.^2 and compresses it to 80 lbf/in.^2. The isothermal efficiency is 88% and the actual compressor has the same heat transfer as the ideal one. Find the specific compressor work and the exit temperature.

Question

A watercooled air compressor takes air in at 70 F, 14 [latex]lbf / in .^{2}[/latex] and compresses it to 80 [latex]lbf / in .^{2}[/latex]. The isothermal efficiency is 88% and the actual compressor has the same heat transfer as the ideal one. Find the specific compressor work and the exit temperature.

Accepted Answer

Ideal isothermal compressor exit 80 psia, 70 F

Reversible process: [latex]dq = T ds \Rightarrow q = T \left( s _{ e }- s _{ i }\right)[/latex]

[latex]q = T \left( s _{ e }- s _{ i }\right)= T \left[ s _{ Te }^{ o }- s _{ T 1}^{ o }- R \ln \left( P _{ e } / P _{ i }\right)\right][/latex]

[latex]=- RT \ln \left( P _{ e } / P _{ i }\right)=-(460+70) \frac{53.34}{778} \ln \frac{80}{14}=-63.3 Btu / lbm[/latex]

As same temperature for the ideal compressor [latex]h _{ e }= h _{ i } \Rightarrow[/latex]

[latex]w = q =-63.3 Btu / lbm \Rightarrow w _{ ac }= w / \eta=-71.93 Btu / lbm , \quad q _{ ac }= q[/latex]

Now for the actual compressor energy equation becomes

[latex]q _{ ac }+ h _{ i }= h _{ e ac }+ w _{ ac } \Rightarrow[/latex]

[latex]h _{ e ac }- h _{ i }= q _{ ac }- w _{ ac }=-63.3-(-71.93)=8.63 Btu / lbm \approx C _{ p }\left( T _{ e ac }- T _{ i }\right)[/latex]

[latex]T _{ e ac }= T _{ i }+8.63 / 0.24= 1 0 5 . 9 F[/latex]

Question 7.233E: A watercooled air compressor takes air in at 70 F, 14 lbf/in...

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