A gas turbine power plant operates on the simple Brayton cycle with air. It has

Question

1. A gas turbine power plant operates on the simple Brayton cycle with air. It has a pressure ratio of 12. The compressor inlet temperature is 300K and the turbine has an inlet temperature of 1000K. Determine the required mass flow rate of air for a net power output of 90 MW assuming the turbine and compressor have isentropic device efficiencies of (a) 100% [363.7 kg/s] and (b) the turbine has an isentropic efficiency of 85% [550.2 kg/s]. Assume air is an ideal gas with k = 1.4. Assume the Cp for air in the turbine is 1.1 [kJ / (kg K)] and Cp for air in the compressor 1.005 [kJ / (kg K)]. Assume the combustion chamber is isobaric and P3 / P4 = P2 / P1. Recall the polytropic relationship for isentropic processes.

Explanation / Answer

a)

P3 / P4 = P2 / P1 = 12

T2 / T1 = (P2 / P1)^((k-1)/k)

T2 / 300 = 12^((1.4-1)/1.4)

T2 = 610.2 K

T3 / T4 = (P3 / P4)^((k-1)/k)

1000 / T4 = 12^((1.4-1)/1.4)

T4 = 491.7 K

Turbine power output = m*Cp*(T3 - T4)

= m*1.1*(1000 - 491.7)

= 559.17*m kW

Compressor power input = m*Cp*(T2 - T1)

= m*1.005*(610.2 - 300)

= 311.75*m kW

Net power = 559.17*m - 311.75*m

= 247.42*m kW

90*10^3 = 247.42*m

m = 363.7 kg/s

b)

Now, T4_is = 491.7 K

Isentropic efficiency = (T3 - T4) / (T3 - T4_is)

0.85 = (1000 - T4) / (1000 - 491.7)

T4 = 567.9 K

Turbine output = m*1.1*(1000 - 567.9)

= 475.26*m kW

Net output = 475.26*m - 311.75*m

= 163.51*m kW

90*10^3 = 163.51*m

m = 550.4 kg/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.