Mass of gas, mg = 3 kg
Initial pressure of gas = 2.5 bar
Initial temperature, T1′ = 400 K
Quantity of heat received by gas, Q = 600 kJ
Specific heat of gas, cv = 0.81 kJ/kg K
Surrounding temperature = 290 K
Temperature of infinite source, T1 = 1200 K
Heat received by the gas is given by,
Q = mgcv (T2′ – T1′)
600 = 3 × 0.81 (T2′ – 400)
T2′ = \(\cfrac{600}{3\times0.81}\) + 400 = 646.9 K say 647 K
Available energy with the source
= area 1-2-3-4-1
= (1200 – 290) × \(\cfrac{600}{1200}\) = 455 kJ
Change in entropy of the gas
Unavailability of the gas = area 3′- 4′- 5′- 6′- 3′
= 290 × 1.168 = 338.72 kJ
Available energy with the gas
= 600 – 338.72 = 261.28 kJ
∴ Loss in available energy due to heat transfer
= 455 – 261.28 = 193.72 kJ.