A steady flow air compressor takes 5.00 kg/min of atmospheric air at 101.3 kPa a

Question

A steady flow air compressor takes 5.00 kg/min of atmospheric air at 101.3 kPa and 20.0°C and delivers it at an exit pressure of 1.00 MPa. The air can be considered an ideal gas with constant specific heats. Potential and kinetic energy effects are negligible. If the process is not reversible but is adiabatic and polytropic with a polytropic exponent of n = 1.47, calculate:

a. The power required to drive the compressor:
b. The entropy production rate of the compressor:
c. The entrance and exit specific flow availabilities if the ground state local environmental temperature and pressure are 20.0°C and 101.3 kPa

Thank you for any help. If someone could show me how this type of problem is solved, hopefully I'll be better at doing them on my own. Thank you

Explanation / Answer

For polytropic process, T2/T1 = (P2/P1)(1-1/n)

By putting values, T2 = (273+20)*(1000/101.3)(1-1/1.47) = 609.3 K

(a) Work done in polytropic compression W = mRT1[(P2/P1)(1-1/n)-1]/(1-n)

= (5/60)*287*(273+20)[(1000/101.3)(1-1/1.47)-1]/(1-1.47) = -16093.3 W (negative sign indicates that power is required to drive the compressor.)

(b) Entropy production rate = m[Cp ln(T2/T1) - R ln(P2/P1)]

Putting values we get,

Entropy production rate = (5/60)*[1005 ln(609.3/(273+20)) - 287 ln(1000/101.3)] = 6.555 J/K-s

(c) Specific flow availability = (h-h0) -T0(s-s0) = Cp(T-T0) - T0(Cp ln(T/T0) - R ln(P/P0))

Specific flow availability for entrance = 1005*((273+20)-(273+20)) - (273+20)(1005* ln(293/293) - 287 ln(101.3/101.3)) = 0

Specific flow availability for exit = 1005*(609.3-(273+20)) - (273+20)(1005* ln(609.3/293) - 287 ln(1000/101.3)) = 294833 J/kg

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