5. 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

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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.

(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

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that would be because of both gravity and friction. so     c