Illustration 3.4-1 Continued

Compute the entropy generated by the flow of 1 kg/s of steam at 400 bar and 500◦C undergoing a Joule-Thomson expansion to 1 bar.

Question

Compute the entropy generated by the flow of 1 kg/s of steam at 400 bar and 500◦C undergoing a Joule-Thomson expansion to 1 bar.

Accepted Answer

In Illustration 3.4-1, from the energy balance, we found that

[latex]\hat{H}_{1}=\hat{H}\left(T_{1}=500^{\circ} C , P_{1}=400 bar ight)=\hat{H}_{2}=\hat{H}\left(T_{2}=?, P_{2}=1 bar ight)[/latex]

and then by interpolation of the information in the steam tables that [latex]T_{2}=214.1^{\circ} C[/latex]. From the entropy balance on this steady-state system, we have

so that

[latex]\dot{S}_{ ext {gen }}=\dot{M}_{1}\left(\hat{S}_{2}-\hat{S}_{1} ight)[/latex]

From the steam tables (using interpolation to obtain the entropy of steam at 214.1°C and 1 bar), we have

[latex]\hat{S}_{1}=\hat{S}\left(T=500^{\circ} C , P=400 bar ight)=5.4700 \frac{ kJ }{ kg K }[/latex]

and

[latex]\hat{S}_{2}=\hat{S}\left(T=214.1^{\circ} C , P=1 ext { bar } ight)=7.8904 \frac{ kJ }{ kg K }[/latex]

Therefore

[latex]\dot{S}_{ ext {gen }}=\dot{M}_{1}\left(\hat{S}_{2}-\hat{S}_{1} ight)=1 \frac{ kg }{ s } \cdot(7.8904-5.4700) \frac{ kJ }{ kg K }=2.4202 \frac{ kJ }{ Ks }[/latex]

Since [latex]\dot{S}_{ ext {gen }}>0[/latex], the Joule-Thomson expansion is also an irreversible process.

Question 4.4-2: Illustration 3.4-1 Continued Compute the entropy generated b...

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