A reciprocating air compressor takes in 2m3 min at 011 MPa 2

A reciprocating air compressor takes in 2m^3 /min at 0.11 MPa, 20C which it delivers at 1.5 MPa, 111C to an aftercooler where the air is cooled at constant pressure to 25C. The power absorbed by the compressor is 4.15 kW. Determine the entropy generation, exergy destruction and second law efficiency of the compressor and aftercooler.

Solution

P1 = 1.1 bar; T1 = 293K

P2 = 15bar; T2 = 384K

V=2m3/min=0.033m3/sV=2m3/min=0.033m3/s ; W = 4.15kW

T3= 298K

Solution:

Part of the power absorbed by the compressor is used to compress the air, some heat is lost in aftercooler and the rest through compressor Power = Work on air + Heat through aftercooler + Heat through compressor

P1V1 = mRT1

1.1 x 10^5 x 0.033 = m x 280 x 293 ( R= 280 J/kg for air)

m = 0.044 kg/s

P = W + Qa

P-W = Qa

Qa = 4150 – mCp(T2-T1)

= 4150 – 0.044 x 1000 ( 384-293 ) = 0.15 kW

Similarly, heat lost in cooler

Q = mCp(T2-T3) = 0.044 x 1000 x (384 – 298) = 3.78 kW