Answer to Question #219633 in Molecular Physics | Thermodynamics for Unknown346307

Question

A gas – turbine plant operates on the ideal cycle. The air enters the compressor at 1 bar, 15°C, and is delivered at 6 bar; the temperature at turbine inlet is 650°C. Determine per kg of air.

a) the compressor work

b) the turbine work

c) the heat supplied

d) the heat rejected

e) the thermal efficiency of the plant

Assume the working fluid to have the properties of air throughout, and take cp = 1.005 kJ/kg K and γ = 1.4.

Answer 1

Solution;

Pressure ratio for compressor and turbine;

$r_p=\frac61$ =6

Therefore;

$(r_p)^{\frac{\gamma -1}{\gamma}}$ =$(6)^\frac{0.4}{1.4}=1.67$

Temperature of air leaving the compressor,T₂=288×1.67=481K

Temperature of air leaving the turbine T₄=$\frac{923}{1.67}$ =552.695K

a) Compressor work.

W_c=mC_p(T₂-T₁)=1×1.005×(481-288)=193.965kJ

b)Turbine work.

W_T=mC_p×(T₃-T₄)=1×1.005×(923-552.7)=372.1515kJ

c)Heat supplied.

Q_s=m×C_p×(T₃-T₂)=1×1.005×(923-481)=444.21kJ

d)Heat rejected.

Q_R=m×C_p×(T₄-T₁)=1×1.005×(552.7-288)=266.02kJ

e) thermal efficiency.

$\eta$ =1-$\frac{1}{(r_p)^{\frac{1-\gamma}{\gamma}}}$=1-$\frac{1}{1.67}$ =0.401197

$\eta$ =0.40=40%