Consider that the air in the previous problem after the compressor flows in a pipe to an air tool and at that point the temperature has dropped to ambient 540 R and an air pressure of 110 psia. What is the second law efficiency for the total system?

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C.V. Compressor

Energy: [latex]0=h_{1}-h_{2}+w_{C}[/latex]; Entropy: [latex]0=s_{1}-s_{2}+0[/latex]

[latex]T _{2}= T _{1}\left(\frac{ P _{2}}{ P _{1}} ight)^{\frac{ k -1}{ k }}=540\left(\frac{120}{15} ight)^{0.2857}=978.1 R[/latex]

Exergy increase through the compressor matches with the ideal compressor work

[latex]w _{ C s }= h _{2}- h _{1}= C _{ P }\left( T _{2}- T _{1} ight)=0.24(978.1-540) Btu / lbm =105.14 Btu/lbm[/latex]

The actual compressor work is

[latex]w _{ C a c }= w _{ C s } / \eta_{ C s }=105.14 / 0.85=123.7 Btu / lbm[/latex]

From inlet, state 1 to final point of use state 3.

[latex]\begin{aligned}\psi_{3}-\psi_{1} &=\left( h _{3}- h _{1} ight)- T _{0}\left( s _{3}- s _{1} ight)=0- T _{0}\left[0- R \ln \left( P _{3} / P _{1} ight) ight] \&= T _{0} R \ln \left( P _{3} / P _{1} ight)=537 R imes(53.34 / 778) Btu / lbmR imes \ln (110 / 15) \&=73.36 Btu / lbm\end{aligned}[/latex]

So then the second law efficiency is the gain [latex]\left(\psi_{3}-\psi_{1} ight)[/latex] over the source [latex]w _{ C ac }[/latex] as

[latex]\eta_{ II }=\frac{\psi_{3}-\psi_{1}}{ w _{ C a c }}=\frac{73.36}{123.7}= 0 . 5 9[/latex]

Question 8.179E: Consider that the air in the previous problem after the comp...

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