The exit nozzle in a jet engine receives air at 2100 R, 20 psia with negligible kinetic energy. The exit pressure is 10 psia and the actual exit temperature is 1780 R. What is the actual exit velocity and the second law efficiency?

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Accepted Answer

C.V. Nozzle with air has no work, no heat transfer.

Energy eq.: [latex]h _{ i }= h _{ e }+\frac{1}{2} V _{ ex }^{2}[/latex]

Entropy Eq.: [latex]s _{ i }+ s _{ gen }= s _{ e }[/latex]

[latex]\frac{1}{2} V _{ ex }^{2}= h _{ i }- h _{ e }=532.57-444.36=88.21 Btu / lbm[/latex]

[latex]V _{ ex }=\sqrt{2 imes 88.21 imes 25037}= 2 1 0 2 ~ f t s ^{-1}[/latex]

Recall [latex]1 Btu / lbm =25037 ft ^{2} / s ^{2}[/latex] . This was the actual nozzle. Now we can do the reversible nozzle, which then must have a q.

Energy eq.: [latex]h _{ i }+ q = h _{ e }+\frac{1}{2} V _{ ext {ex rev }}^{2}[/latex]

Entropy Eq.: [latex]s _{ i }+ q / T _{ o }= s _{ e } \quad \Rightarrow q = T _{ o }\left( s _{ e }- s _{ i } ight)[/latex]

[latex]\begin{aligned}q &= T _{ o }\left[ C \ln \frac{ T _{ e }}{ T _{ i }}- R \ln \frac{ P _{ e }}{ P _{ i }} ight]=536.7\left[0.24 \ln \frac{1780}{2100}-\frac{53.34}{778.17} \ln \frac{10}{20} ight] \&=4.198 Btu / lbm\end{aligned}[/latex]

[latex]\frac{1}{2} V _{ ext {ex rev }}^{2}= h _{ i }+ q - h _{ e }=88.21+4.198=92.408 Btu / lbm[/latex]

[latex]\eta_{ II }=\frac{1}{2} V _{ ex }^{2} / \frac{1}{2} V _{ ext {ex rev }}^{2}=88.21 / 92.408= 0 . 9 5[/latex]

Notice the reversible nozzle is not isentropic (there is a heat transfer).

Question 8.186E: The exit nozzle in a jet engine receives air at 2100 R, 20 p...

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