Question

Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work that can be delivered by thisturbine.

Answer 1

`w_(out)=319.1(BTU)/(lbm)`

Step-by-step explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

`h_1=1456.0(BTU)/(lbm) \ \ \ s_1=1.6413(BTU)/(lbm*R)`

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:

`s_2=s_1`

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

`x_2=(s_2-s_f)/(s_(fg))=(1.6313-0.39213)/(1.28448)=0.965`

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

`h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9(BTU)/(lbm)`

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

`h_1=w_(out)+h_2\\\\w_(out)=h_1-h_2=1456.0(BTU)/(lbm)-1136.9(BTU)/(lbm)\\\\w_(out)=319.1(BTU)/(lbm)`

Best regards.