Answer:
790.904 kg/s
Explanation:
From the ideal gas properties table
At T₁ =310K
h₁ =310.240 kJ/kg
[tex]p_{r1}[/tex] = 1.5546
s₁⁰ =1.73498 kJ/kg K
Process 1-2s is isentropic compression (shown in the attached diagram)
[tex]p_{r2s}=\frac{p_2}{p_1}\times p_{r1}[/tex]
[tex]p_{r2s}=\frac{8}{1}\times 1.5546[/tex]
= 8 × 1.5546
= 12.43
now enthalpy
[tex]h_{r2s}[/tex] = 562.46 kJ/kg
Actual enthalpy at state 2 is given by
[tex]\eta_{c}=\frac{h_{2s}-h_{1}}{h_2-h_1}[/tex]
[tex]0.8=\frac{562.45-310.24}{h_2-310.24}[/tex]
[tex]h_2[/tex] = 625.50 kJ/kg
Process 3 - 4s is isentropic expansion
[tex]T_3[/tex] = 900K
[tex]h_3[/tex] = 932.93kJ/kg
[tex]P_{r3}[/tex] = 75.29
hence
[tex]P_{r4s}=\frac{P_4}{P_3}\times P_{r3}[/tex]
[tex]P_{r4s}=\frac{1}{8}\times 75.29[/tex]
[tex]P_{r4s}=9.41[/tex]
Enthalpy at state 4
[tex]\eta_T =\frac{h_3-h_4}{h_3-h_{4s}}[/tex]
[tex]0.86 =\frac{932.93-h_4}{932.93-519.303}[/tex]
[tex]h_4 = 577.21kJ/kg[/tex]
the net work output [tex]W_{NET}[/tex] is given by
[tex]W_{NET}=Q_H-Q_L[/tex]
[tex]= h_3-h_2-(h_4-h_1)[/tex]
[tex]= 932.93-625.50-(577.21-310.24)[/tex]
= 40.46kJ/kg
Power output = mass flow rate × net work output
32 × 10³ = m × 40.46
m = 790.904 kg/s
