What is the maximum work that can be obtained from steam at 2 MPa and 700°C in a non-flow process?

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Accepted Answer

Using the data in the previous illustration with the following additional information from the steam tables (note that at the ambient conditions the steam has condensed to liquid water). At 2 MPa and 700°C:

[latex]\begin{aligned}&\hat{U}=2808.6 kJ / kg , \hat{V}=1.3162 m ^{3} / kg ext { so that } \&\hat{ A }=2808.6 \frac{ kJ }{ kg }+1.0 ext { bar } imes 1.3162 \frac{ m ^{3}}{ kg } imes 102.67 \frac{ kJ }{ bar \cdot m ^{3}}-298.15 imes 7.8926 \frac{ kJ }{ kg }=590.55 kJ / kg\end{aligned}[/latex]

and at the ambient conditions 0.10135 MPa and 25°C

[latex]\begin{aligned}&\hat{U}=104.88 kJ / kg , \hat{V}=0.001 m ^{3} / kg , ext { so that } \&\hat{ A }=104.88+1 imes 0.001 imes 102.67-298.15 imes 0.3674=-4.56 kJ / kg\end{aligned}[/latex]

Therefore,

[latex]\frac{\dot{W}_{s, \max }}{\dot{M}_{1}}=\hat{ A }\left(T_{ amb }, P_{ amb } ight)-\hat{ A }\left(T_{1}, P_{1} ight)=-4.56-590.55=-595.11 \frac{ kJ }{ kg }[/latex]

We see that the maximum useful work that can be obtained from this stagnant stream is 595.1 kJ/kg. (As usual, the negative sign indicates that shaft work is being done by the system.)

Question 4.6-3: What is the maximum work that can be obtained from steam at ...

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