A compressor is employed to compress gas flowing steadily through it. The gas en

Question

A compressor is employed to compress gas flowing steadily through it. The gas enters the compressor at a temperature of 16°C, a pressure of 100 kPa, and with an enthalpy of 391.2 kJ/kg. The gas leaves the compressor at a temperature of 245°C, a pressure of 0.6 MPa, and with an enthalpy of 534.5 kJ/kg. There is no heat transfer to or from the gas as it flows through the compressor. Note that power has to be provided to the compressor to compress the gas.

(a) Evaluate the work done per unit mass of gas assuming the gas velocities at entry and
exit are negligible.
(b) Evaluate the work done per unit mass of gas when the gas velocity at entry is 80 m/s
and that at exit is 160 m/s.

Explanation / Answer

Given

T1 = 16 C = 289 K

P1 =100 kPa

h1 = 391.2 kJ/kg

T2 = 245 C = 518 K

P2 = 0.6Mpa

h2 = 534.5 kJ/kg

Also

Qin = 0 (heat transfer is neglizible)

In general the first law for CV is

Qin + m*(h1 + KE1 + PE1) = Wout + m*(h2 + KE2 + PE2)

Qin + m*(h1 + V1^2/2 + g*Z1) = Wout + m*(h2 + V2^2/2+ g*Z2)

In terms of specific terms everthing is divided by m

qin + (h1 + V1^2/2 + g*Z1) = wout + (h2 + V2^2/2+ g*Z2)

a)

When KE and PE are neglibile

qin + (h1) = wout + (h2)

0 + 391.2 = wout + 534.5

wout = -143.3 kJ/kg

Therfore work input to the comprssor is 143.3 kJ/kg

b)

Now V1 = 80 m/s

V2 = 160 m/s

qin + (h1 + V1^2/2) = wout + (h2 + V2^2/2)

0 + 391.2 + 80^2/2000 = wout + 534.5 + 160^2/2000

wout = -143.3 -9.6

wout = -152.9 kJ/kg

Therfore the work input to the compressor is 155.9 kJ/kg

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