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Air enters a two-stage compressor operating at steady state at 520ºR, 14 lbffin 2 The overall pressure ratio across the stages is 12 and each stage operates isentropically. Intercooling occurs at constant pressure at the value that minimizes compressor work input as determined in Example 9.10, with air exiting the intercooler at 520°R. Assuming ideal gas behavior, with k = 1.4. Determine the work per unit mass of air flowing for the two-stage compressor, in Btu per lb of air flowing.

User Marsei
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1 Answer

1 vote

Answer:

106.335 Btu/Ib

Step-by-step explanation:

Given data :

T1 = 520°R = 288.89K

P1 = 14 Ibf/in^2 = 96526.6 pa

r = 12

k = 1.4

R = 287 J/kg-k

Calculate work done per unit mass of air flowing ( two-stage compressor )

we will apply the equation below

W = 2k / K-1 * ( RT₁ ) *
[ r^{(k-1)/(2k) } - 1 ]

input values into equation above

W = 247.336 KJ/kg = 106.335 Btu/Ib

User Greg Finzer
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