Answer:
[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]
Explanation:
Hello,
In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:
[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]
Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:
[tex]s_2=s_1[/tex]
Thus, the liquid and liquid-vapor entropies are included to compute the quality:
[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]
Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:
[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]
Then, by using the first law of thermodynamics, the maximum specific work is computed via:
[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]
Best regards.
