A gas-turbine power plant operates on the simple Brayton cycle with ai

Step 1 
The enthalpy and the relative pressure at the initial state are taken from A-17 for the given temperature: 
${\displaystyle h_1=310.24\frac{kj}{kg}}$ 
${\displaystyle P_{r1}=1.5546}$ 
The pressure ratio and the relative pressure at state 1 are used to calculate the relative pressure at state 2:
${\displaystyle P_{r2}=\frac{P_2}{P_1}{P}_{r1}}$ 
${\displaystyle =8\cdot 1.5546}$ 
${\displaystyle =12.44}$ 
From this the enthalpy at state 2 for an isentropic process can be obtained using interpolation with data from A-17: 
${\displaystyle h_{2s}=562.58\frac{kj}{kg}}$ 
The relative pressure and the enthalpy at state 3 are determined from A-17 for the given temperature: 
${\displaystyle h_3=932.93\frac{kj}{kg}}$ 
${\displaystyle P_{r3}=75.29}$ 
Inferring from the relative pressure in state 3, one may calculate the relative pressure at state 4.
${\displaystyle P_{r4}=\frac{P_4}{P_3}{P}_{r3}}$ 
${\displaystyle =\frac{1}{8}\cdot 75.29}$ 
${\displaystyle =9.41}$ 
From this value the isentropic enthalpy at state 4 can be obtained using interpolation with data from A-17: 
${\displaystyle h_{4s}=519.3\frac{kj}{kg}}$ 
Step 2 
The power input of the compressor and the power output of the turbine can be used to calculate the mass flow rate:
${\displaystyle W=Wtask,outPSK-W_c}$, in 
${\displaystyle Q=m{n}_T(h_3-h_{4s})-\frac{m}{n_C}(h_2-h_1)}$ 
${\displaystyle m=\frac{W}{n_T(h_3-h_{4s})-\frac{1}{n_C}(h_{2s}-h_1)}}$ 
${\displaystyle =\frac{32000}{0.86(932.93-519.3)-\frac{1}{0.8}(562.58-310.24)}}$ 
${\displaystyle =794\frac{kg}{s}}$

A 32-MW has-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined.
Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with variable specific heats.
Analysis. Using variable specific heats,
${\displaystyle T_1=310K\Rightarrow h_1=310.24\frac{kJ}{kg}}$
${\displaystyle \Rightarrow P_{r1}=1.5546}$
${\displaystyle P_{r2}=\frac{P_2}{P_1}P{r}_1=\left(8\right)\left(1.5546\right)=12.44\Rightarrow h_{2s}=562.26\frac{kJ}{kg}}$
${\displaystyle T_3=900K\Rightarrow h_3=932.93\frac{kJ}{kg}}$
${\displaystyle \Rightarrow P_{r3}=75.29}$
${\displaystyle P_{r4}=\frac{P_4}{P_3}{P}_{r3}=\left(\frac{1}{8}\right)\left(75.29\right)=9.411\Rightarrow h_{4s}=519.32\frac{kJ}{kg}}$
${\displaystyle w_{\ne t,out}=w_{T,out}-w_{c,\in }=n_T(h_3-h_{4s})-\frac{h_{2s}-h_1}{n_C}}$
${\displaystyle =\left(0.86\right)(932.93-519.32)-\frac{562.26-310.24}{0.80}=40.68\frac{kJ}{kg}}$
and ${\displaystyle m=\frac{W_{\ne t,out}}{w_{\ne t,out}}=\frac{32.000\frac{kJ}{s}}{40.68\frac{kJ}{kg}}=786.6k\frac{g}{s}}$

Use the gas table for initial state,
$\begin{array}{}{h}_1=310.24kJ/kg,P_{r1}=1.5546\\ \text{Calculate the relative pressure at state 2}\\ P_{r2}=\frac{P_2}{P_1}\Rightarrow 8\times 1.5546\Rightarrow P_{r2}=12.44\\ \text{use the table at state 2 and 3,}\\ h_{2s}=562.58kJ/kg,h_3=932.93kJ/kg\\ P_{r3}=75.29\\ \text{Calculate the relative pressure at state 4,}\\ P_{r4}=\frac{P_4}{P_3}\times P_{r3}\Rightarrow \frac{1}{8}\times 75.29\Rightarrow 9.41\\ \text{use table at reative pressure of state 4,}\\ h_{4s}=519.3kJ/kg\\ \text{Calculate the mass flow rate.}\\ W=(W_t{)}_{out}-(W_c{)}_{in}\\ m{n}_t(h_3-h_{4s})-\frac{m}{n_c}(h_2-h_1)\\ m=\frac{w}{n_T(h_3-h_4)-\frac{1}{n_c}(h_{2s}-h_1)}\\ =\frac{32000}{0.86(932.93-519.3)-\frac{1}{0.8}(562.56-310.24)}\\ m=794kg/s\end{array}$