Question

A gas-turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 32 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K, and the pressure of air at the compressor exit is 8 times the value at the compressor inlet. Assuming an isentropic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle. Account for the variation of specific heats with temperature.

Answer 1

790.904 kg/s

Step-by-step explanation:

From the ideal gas properties table

At T₁ =310K

h₁ =310.240 kJ/kg

`p_(r1)` = 1.5546

s₁⁰ =1.73498 kJ/kg K

Process 1-2s is isentropic compression (shown in the attached diagram)

`p_(r2s)=(p_2)/(p_1)* p_(r1)`

`p_(r2s)=(8)/(1)* 1.5546`

= 8 × 1.5546

= 12.43

now enthalpy

`h_(r2s)` = 562.46 kJ/kg

Actual enthalpy at state 2 is given by

`\eta_(c)=(h_(2s)-h_(1))/(h_2-h_1)`

`0.8=(562.45-310.24)/(h_2-310.24)`

`h_2` = 625.50 kJ/kg

Process 3 - 4s is isentropic expansion

`T_3` = 900K

`h_3` = 932.93kJ/kg

`P_(r3)` = 75.29

hence

`P_(r4s)=(P_4)/(P_3)* P_(r3)`

`P_(r4s)=(1)/(8)* 75.29`

`P_(r4s)=9.41`

Enthalpy at state 4

`\eta_T =(h_3-h_4)/(h_3-h_(4s))`

`0.86 =(932.93-h_4)/(932.93-519.303)`

`h_4 = 577.21kJ/kg`

the net work output
`W_(NET)` is given by

`W_(NET)=Q_H-Q_L`

`= h_3-h_2-(h_4-h_1)`

`= 932.93-625.50-(577.21-310.24)`

= 40.46kJ/kg

Power output = mass flow rate × net work output

32 × 10³ = m × 40.46

m = 790.904 kg/s