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3 kg of gas (cv = 0.81 kJ/kg K) initially at 2.5 bar and 400 K receives 600 kJ of heat from an infinite source at 1200 K. If the surrounding temperature is 290 K, find the loss in available energy due to above heat transfer.

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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.

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