Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the amount of entropy produced, in kJ/K per kg of air, for the compression. What is the minimum theoretical work input, in kJ per kg of air, for an adiabatic compression from the given initial state to a final pressure of 10 bar?

Question

Physics

Devyn Sanford

9 May, 02:36

Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the amount of entropy produced, in kJ/K per kg of air, for the compression. What is the minimum theoretical work input, in kJ per kg of air, for an adiabatic compression from the given initial state to a final pressure of 10 bar?

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Alexandra · Answer 1

1) the entropy generated is Δs = 0.0363 kJ/kg K

2) the minimum theoretical work is w piston = 201.219 kJ/kg

Explanation:

1) From the second law of thermodynamics applied to an ideal gas

ΔS = Cp * ln (T₂/T₁) - R ln (P₂/P₁)

and also

k = Cp/Cv, Cp-Cv=R → Cp * (1-1/k) = R → Cp = R / (1-1/k) = k*R / (k-1)

ΔS = k*R / (k-1) * ln (T₂/T₁) - R ln (P₂/P₁)

where R = ideal gas constant, k = adiabatic coefficient of air = 1.4

replacing values (k=1.4)

ΔS = k*R / (k-1) * ln (T₂/T₁) - R ln (P₂/P₁)

ΔS = 1.4 / (1.4-1) * 8.314 J/mol K * ln (600K/300K) - 8.314 J/mol K * ln (10 bar / 1bar)

ΔS = 1.026 J / mol K

per mass

Δs = ΔS / M

where M = molecular weight of air

Δs = 1.026 J / mol K / 28.84 gr/mol = 0.0363 J/gr K = 0.0363 kJ/kg K

2) The minimum theoretical work input is carried out under a reversible process. from the second law of thermodynamics

ΔS = ∫dQ/T = 0 since Q=0→dQ=0

then

0 = k*R / (k-1) * ln (T₂/T₁) - R ln (P₂/P₁)

T₂/T₁ = (P₂/P₁) ^[ (k-1) / k]

T₂ = T₁ * (P₂/P₁) ^[ (k-1) / k]

replacing values

T₂ = 300K * (10 bar/1 bar) ^[0.4/1.4] = 579.2 K

then from the first law of thermodynamics

ΔU = Q - Wgas = Q + Wpiston,

where ΔU = variation of internal energy, Wgas = work done by the gas to the piston, Wpiston = work done by the piston to the gas

since Q=0

Wpiston = ΔU

for an ideal gas

ΔU = n*Cv * (T final - T initial)

and also

k = Cp/Cv, Cp-Cv=R → Cv * (k-1) = R → Cv = R / (k-1)

then

ΔU = n*R / (k-1) * (T₂ - T₁)

W piston = ΔU = n*R / (k-1) * (T₂ - T₁)

the work per kilogram of air will be

w piston = W piston / m = n/m*R / (k-1) * (T₂ - T₁) = (1/M*) R / (k-1) * (T₂ - T₁),

replacing values

w piston = (1/M*) R / (k-1) * (T₂ - T₁) = 1 / (28.84 gr/mol) * 8.314 J/mol K / 0.4 * (579.2 K - 300 K) = 201.219 J/gr = 201.219 kJ/kg