A gas power cycle compresses air at 1 bar and 300 K (state 1) through an adiabat
ID: 1857396 • Letter: A
Question
A gas power cycle compresses air at 1 bar and 300 K (state 1) through an adiabatic, reversible compressor to 6 bar (state 2). Heat is added to air in a constant pressure heat exchanger such that air leaves the heat exchanger at 6 bar and 1400 K (state 3). Air then expands through an adiabatic, reversible turbine to 1 bar. Heat is rejected from air in another constant pressure heat exchanger to complete the cycle. Mass flow rate of air through all devices is 80 kg/s. Determine the net power (kW) produced by the power plant. What is the heat transfer (kW) into the power plant? Calculate thermal efficiency (%) of the power plant. Show the cycle on T-s diagram, indicating temperature values and lines of constant pressure. Do not show entropy values. A gas power cycle compresses air at 1 bar and 300 K (state 1) through an adiabatic, reversible compressor to 6 bar (state 2). Heat is added to air in a constant pressure heat exchanger such that air leaves the heat exchanger at 6 bar and 1400 K (state 3). Air then expands through an adiabatic, reversible turbine to 1 bar. Heat is rejected from air in another constant pressure heat exchanger to complete the cycle. Mass flow rate of air through all devices is 80 kg/s. Determine the net power (kW) produced by the power plant. What is the heat transfer (kW) into the power plant? Calculate thermal efficiency (%) of the power plant. Show the cycle on T-s diagram, indicating temperature values and lines of constant pressure. Do not show entropy values.Explanation / Answer
Adiabatic compresion:
T2 / T1 = (P2 / P1)^((gamma - 1) / gamma)
T2 / 300 = (6/1)^(0.4/1.4)
T2 = 500 K
Work done W1 = R(T2 - T1) / (1 - gamma)
W1 = 0.287*(500 - 300) / (1 - 1.4)
W1 = -143.5 kJ/kg
Heat transfer Q1 = 0 (since adiabatic)
Constant pressure heat addition:
Work done W2 = R (T3 - T2) = 0.287*(1400 - 500) = 258.3 kJ / kg
Heat transfer Q2 = W + Cv (T3 - T2) = 258.3 + 0.718*(1400 - 500) = 904.5 kJ / kg
Adiabatic expansion:
T4 / T3 = (P4 / P3)^((gamma - 1) / gamma)
T4 / 1400 = (1 / 6)^(0.4 / 1.4)
T4 = 839 K
Work done W3 = R (T3 - T4) / (gamma - 1)
W3 = 0.287*(1400 - 839) / (1.4 - 1)
W3 = 402.5 kJ / kg
Heat transfer Q3 = 0 (since adiabatic)
Constant pressure heat rejection:
Work done W4 = R (T1 - T4) = 0.287*(300 - 839) = -154.7 kJ / kg
Heat transfer Q4 = W4 + Cv (T1 - T4) = -154.7 + 0.718*(300 - 839) = -541.7 kJ / kg
a)
Net power = m(W1 + W2 + W3 + W4)
Net power = 80*(-143.5 + 258.3 + 402.5 - 154.7) = 29008 kW
b)
Heat transfer "into" powerplant = m*Q2 = 80*904.5 = 72360 kW
c)
Efficiency = Net Power / heat input
Efficiency = 29008 / 72360 = 0.4 or 40 %
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