Our values are given by,
[tex]T_1 = 320\°cT_2 = 160\°c\\P_1 = 5 bar = 5000kPa\\P_2 = 1bar = 1000kPa\\\dot{V} = 0.65m^3/s\\[/tex]
From steam tables A5 we can calculate the values for State 1, that is,
[tex]\nu_1 = 0.5416m^3/kg\\s_1 = 7.5308kJ/KgK\\h_1 = 3105.6kJ/kg[/tex]
Here is easily to find the flow mass rate, through
[tex]\dot{m} = \frac{\dot{V}}{\nu_1}[/tex]
[tex]\dot{m} = \frac{0.65}{0.5416} = 1.2kg/s[/tex]
We apply a similar process to State 2, then
[tex]s_2 = 7.6597kJ/kgK\\h_2 = 2796.2kJ[/tex]
With this values we can now calculate the Power, which is given by,
[tex]P=\dot{m} (h_1-h_2) = 1.2(3105.6-2796.2)) = 371.28kW[/tex]
The total rate in entropy, that is
[tex]\Delta S = \dot{m}(s_2-s_1) = 1.2 (7.6597-7.5308) = 0.1547kW/K[/tex]
To find the efficiency we know that in a isentropic state, the entropy [tex]s_{2s} = s_1[/tex] then
[tex]\frac{h_{2s}-h_{120c}}{h_{160c}-h_{120c}} = \frac{s_{2s}-s_{120c}}{s_{160c}-s_{120c}}[/tex]
Replacing to find [tex]h_{2s}[/tex]
[tex]\frac{h_{2s}-2716.6}{2796.2-2716.6}=\frac{7.5308-7.4668}{7.6597-7.4668}h_{2s}=2743.01kJ/kg[/tex]
Then the efficiency is given by,
[tex]\eta = \frac{h_1-h_2}{h_1-h_{2s}} = \frac{3105.6-2796.2}{3105.6-2743.01}[/tex]
[tex]\eta = 0.85 = 85\%[/tex]
