Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
[tex]\Delta S=mRln\frac{P_1}{P_2}\\\Rightarrow \Delta S=2\times 287.058ln\frac{300}{600}\\\Rightarrow \Delta S=-397.95\ J/K[/tex]
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
[tex]W=mRTln\frac{P_1}{P_2}\\\Rightarrow W=2\times 287.058\times 200ln\frac{300}{600}\\\Rightarrow W=-79589.37\ J[/tex]
∴ Amount of lost work is 79.589.37 kJ
