Question

1 mol of an Ideal Gas expands through a turbine adiabatically to produce work. Assuming steady flow conditions, how much work is lost due to the entropy generated, if the surroundings are at 300 K? Give the value of lost work to the nearest J. Inlet conditions: 494.8 K and 6.3 bar Outlet conditions: 383.5 K and 1.7 bar Cp = (9/2)R where R = 8.314 J/mol-K.

Answer 1

W = 1.8 KJ

Step-by-step explanation:

turbine adiabatically: Q = 0

∴ Lost work:

• W = To [ ( S2 - S1 ) + ΔSo ]

∴ To = 300 K

ideal gas:

• S2 - S1 = Cp Ln (T2/T1) - R Ln (P2/P1)

⇒ S2 - S1 = (9/2)*(8.314) Ln (383.5/494.8) - (8.314) Ln (1.7/6.3)

⇒ S2 - S1 = 1.36 J/mol.K

entropy generated in the sourroundings:

• ΔSo = Q/To

∴ Q = ΔU = Cv*ΔT

∴ Cv = 3/2*R = 12.5 J/mol.K.....ideal gas

∴ ΔT = 494.8 - 383.5 = 111.3 K

⇒ Q = 12.5 * 111.3 = 1391.25 J/mol

⇒ ΔSo = 1391.25/300 = 4.638 J/mol.K

⇒ W = ( 300 K ) * [ 1.36 J/mol.K + 4.638 J/mol.K ]

⇒ W = 1799.25 J/mol * 1 mol * ( KJ/1000J )

⇒ W = 1.799 KJ ≅ 1.8 KJ