Air flows into a heat engine at ambient conditions 14.7 lbf/in.^2, 540 R, as shown in Fig. P8.111. Energy is supplied as 540 Btu per lbm air from a 2700 R source and in some part of the process a heat transfer loss of 135 Btu per lbm air happens at 1350 R. The air leaves the engine at 14.7

Question

Air flows into a heat engine at ambient conditions 14.7 [latex]lbf / in .^{2}[/latex], 540 R, as shown in Fig. P8.111. Energy is supplied as 540 Btu per lbm air from a 2700 R source and in some part of the process a heat transfer loss of 135 Btu per lbm air happens at 1350 R. The air leaves the engine at 14.7 [latex]lbf / in .^{2}[/latex], 1440 R. Find the first- and the second-law efficiencies.

Accepted Answer

C.V. Engine out to reservoirs

[latex]h _{ i }+ q _{ H }= q _{ L }+ h _{ e }+ w[/latex]

Table F.5: [latex]h _{ i }=129.18 Btu / lbm , s _{ Ti }^{ o }=1.63979 Btu / lbm R[/latex]

[latex]h _{ e }=353.483 Btu / lbm , s _{ Te }^{ o }=1.88243 Btu / lbm R[/latex]

[latex]w _{ ac }=129.18+540-135-353.483=180.7 Btu / lbm[/latex]

[latex]\eta_{ TH }= w / q _{ H }=180.7 / 540= 0 . 3 3 5[/latex]

For second law efficiency also a q to/from ambient

[latex]s _{ i }+\left( q _{ H } / T _{ H } ight)+\left( q _{0} / T _{0} ight)=\left( q _{ ext {loss }} / T _{ m } ight)+ s _{ e }[/latex]

[latex]\begin{aligned}q _{0} &= T _{0}\left[ s _{ e }- s _{ i }+\left( q _{ ext {loss }} / T _{ m } ight)-\left( q _{ H } / T _{ H } ight) ight] \&=540\left(1.88243-1.63979+\frac{135}{1350}-\frac{540}{2700} ight)=77.02 Btu / lbm\end{aligned}[/latex]

[latex]w _{ rev }= h _{ i }- h _{ e }+ q _{ H }- q _{ loss }+ q _{0}= w _{ ac }+ q _{0}=257.7 Btu / lbm[/latex]

[latex]\eta_{ II }= w _{ ac } / w _{ rev }=180.7 / 257.7= 0 . 7 0[/latex]

Question 8.177E: Air flows into a heat engine at ambient conditions 14.7 lbf/...

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