Question 8.178E: Air enters a compressor at ambient conditions, 15 psia, 540 ...

Ideal (isentropic, Eq.6.23)

Actual exit temperature

Eq.6.16:        s _{2}- s _{1}=0.24 \ln (1055 / 540)-(53.34 / 778) \ln 8=0.01817   Btu / lbm – R

Exergy, Eq.8.23

2nd law efficiency, Eq.8.32 or 8.34 (but for a compressor):

……………………………………..

Eq.6.23 : \frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}

Eq.6.16 : s_{2}-s_{1}=C_{p 0} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}

Eq.8.23 :
\psi_{i}-\psi_{e}=\left[\left(h_{\text {tot } i}-T_{0} s_{i}\right)-\left(h_{0}-T_{0} s_{0}+g Z_{0}\right)\right]-\left[\left(h_{\text {tot } e}-T_{0} s_{e}\right)-\left(h_{0}-T_{0} s_{0}+g Z_{0}\right)\right]=\left(h_{\text {tot } i}-T_{0} s_{i}\right)-\left(h_{\text {tot } e}-T_{0} s_{e}\right)

Eq.8.32 :  \eta_{2nd  law }=\frac{\dot{m}_{1}\left(\psi_{2}-\psi_{1}\right)}{\dot{m}_{3}\left(\psi_{3}-\psi_{4}\right)}

Eq.8.34 :  \eta_{2 \text { nd law }}=\frac{\dot{\Phi}_{\text {wanted }}}{\dot{\Phi}_{\text {source }}}=\frac{\dot{\Phi}_{\text {source }}-\dot{I}_{ c.v }}{\dot{\Phi}_{\text {source }}}