Question

Air is compressed adiabatically in a piston–cylinder assembly from 1 bar, 300 K to 8 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 8 bar? Note that work is positive into the compressor.

Answer 1

See we need to calculate the enropy change for irreversible adiabatic process

Delta S total =Delta S system + Delta S surrounding

since process is adiabtic so no heat transfer with surrounding or delta S surrounding = 0

now for delta S system (See, delta S is atate function and not dependent upon the path so we will calculate delts S for a reversible path for the same first and final state)

so lets consider the process goes isochoric (volume = constant) from 1 bar 300 K to 8 bar 600 K, W = 0 for this assumed process because deltaV = 0

so delta S system = deltaQ/T = delta U/T = nC_VdT/T

By integrating we get Delta S system = nCvlnT2/T1

Cv for air = 0..718, n = 1/29,

delta S syst = 1/29*0.718ln2 = 0.017 kJ/K = delta S total

for adibatic process, PV^y = constant or T2/T1 = (P2/P1)^(y-1)/y

600/300 = (8/1)^(y-1)y or y= 3/2

work done minimum for comprsssion = work done in reversible process = nR(T2-T1)/(1-y)

for 1 kg of air, n = 1000/29, R = 8.314,

so Wmin = 1000/29*8.314*(600-300)/(1-1.5) =-172.014 kJ/kg