Air at 100 kPa and 300 K passes through a battered, squeaky old compressor that

Question

Air at 100 kPa and 300 K passes through a battered, squeaky old compressor

that probably needs oiling. It is well insulated. The air leaves at 500 kPa and

600 K. This is too hot so we next pass the air through a heat exchanger where

the air cools to 300 K and the pressure goes down to 450 kPa. The entire

process is sketched below.

a) What is the entropy change, kJ/(kg K), in going from point 1 to point 2, from

point 2 to point 3, and the overall change from point 1 to point 3? Approximate

answers: delta s

1-2 = 0.2 kJ/(kg K) and delta s

2-3 = -0.7 kJ/(kg K)

b) What is the work needed (kJ/kg) for the compression step (point 1 to point 2)?

Approximate answer: 300 kJ/kg. c) What would be the work needed (kJ/kg) (point 1 to point 2) for a brand new,

adiabatic, reversible compressor compressing the air to 500 kPa? Approximate

answer: 200 kJ/kg.

Explanation / Answer

Part A

from 1 to 2

deltaS=cp*ln(t2/t1)-R*ln(p2/p1)

=1.005ln(600/300)-.287ln(500/100)

= .234704 kj/kg*k

from 2 to 3

deltaS=cp*ln(t2/t1)-R*ln(p2/p1)

deltaS=1.005*ln(300/600)-.287*ln(450/500)

=-.667 kj/kg*k

part B

W comp, in = K*R*(T2-T1)/K-1

= 1.4*.287*(600-300)/(1.4-1)

=301.35 kj/kg

Part C

w comp, in = k*R*T1/(k-1)[(p2/p1)^((k-1)/k)-1]

w=1.4*.287*300/(1.4-1)[(500/100)^((1.4-1)/1.4)-1]

=175.934 kj/kg

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