This Question is incomplete.The complete question is
Combustion gases enter a gas turbine at 627∘C and 1.2 MPa at a rate of 2.5 kg/s and leave at 527∘C and 500 kPa. It is estimated that heat is lost from the turbine at a rate of 20 kW. Using air properties for the combustion gases and assuming the surroundings to be at 25∘C and 100 kPa, determine (a) the actual and reversible power outputs of the turbine. (b) the exergy destroyed within the turbine, and (c) the second-law efficiency of the turbine.
Answer:
(a) W=257.5kW Wrev=367.3kW
(b) X=109.8kW
(c) n=0.7
Explanation:
For Part(a)
Actual power output is determined from the energy balance. The final and initial enthalpy are taken from A-17 for the given temperature
mh₁=W+Q+mh₂
W=m(h₁-h₂)-Q
W=2.5(939.93-821.95)-20
W=257.5kW
The reversible power output is determined from the rate of energy destruction
Wrev=W+X
Wrev=W+T₀(m(s₂-s₁-Rln(p₂/p₁)+Q/T₀)
Wrev=257.5+298( 2.5 (2.71787-2.84856-0.287*ln(500/1200)+20/298)
Wrev=367.3kW
For part(b)
X=Wrev-W
X=367.3-257.5
X=109.8kW
For part (c)
Second-law efficiency is determined from the ratio of the actual and reversible power output:
n=W/Wrev
n=257.5/367.3
n=0.7