Air at 1 atm, 60 F is compressed to 4 atm, after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle both have efficiency of 90% and kinetic energy in/out of the compressor can be neglected. Find the actual compressor work and its exit temperature and find

Question

Air at 1 atm, 60 F is compressed to 4 atm, after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle both have efficiency of 90% and kinetic energy in/out of the compressor can be neglected. Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity.

Accepted Answer

Steady state separate control volumes around compressor and nozzle. For ideal compressor we have inlet : 1 and exit : 2
Adiabatic : q = 0.
Reversible: [latex]s _{ gen }=0[/latex]

Energy Eq.: [latex]h _{1}+0= w _{ C }+ h _{2}[/latex];

Entropy Eq.: [latex]s _{1}+0 / T +0= s _{2}[/latex]

Ideal compressor: [latex]w _{ c }= h _{1}- h _{2}, \quad s _{2}= s _{1}[/latex]

The constant s from Eq. 6.23 gives

[latex]\begin{aligned}T _{2} &= T _{1}\left( P _{2} / P _{1} ight)^{\frac{ k -1}{ k }}=(459.7+60) R imes(4 / 1)^{0.2857}=772 R \\Rightarrow- w _{ C } &= h _{2}- h _{1}= C _{ P }\left( T _{2}- T _{1} ight)=0.24(772-519.7)=60.55 Btu / lbm\end{aligned}[/latex]

Actual compressor: [latex]w_{ c , AC }= w _{ c , s } / \eta _{ c }=-67.3 B tu / lbm = h _{1}- h _{3}[/latex]

[latex]\Rightarrow \quad T _{3}= T _{1}- w _{ c , AC } / C _{ P }=519.7+67.3 / 0.24= 8 0 0 R[/latex]

Ideal nozzle: [latex]s _{4}= s _{3}[/latex] so use Eq.6.23 again

[latex]\Rightarrow T _{4}= T _{3} imes\left( P _{4} / P _{3} ight)^{\frac{ k -1}{ k }}=800(1 / 4)^{0.2857}=538.4 R[/latex]

[latex]V _{ s }^{2} / 2= h _{3}- h _{4}= C _{ P }\left( T _{3}- T _{4} ight)=0.24(800-538.4)=62.78 Btu / lbm[/latex]

[latex]V _{ AC }^{2} / 2= V _{ s }^{2} imes \eta_{ NOZ } / 2=62.78 imes 0.9=56.5 Btu / lbm[/latex]

[latex]V _{ AC }=\sqrt{2 imes 56.5 imes 25037}= 1 6 8 2 \mathrm { ft } / \mathrm { s }[/latex]

Remember conversion [latex]1 Btu / lbm =25037 ft ^{2} / s ^{2}[/latex] from Table A.1.

.............................................

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

Question 7.235E: Air at 1 atm, 60 F is compressed to 4 atm, after which it is...

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