Thermodynamics HELP Air at 15psi and 80F enters a compressor

Thermodynamics... HELP?!

Air at 15psi and 80F enters a compressor with a speed of 200 ft/s. The air exits the compressor at 300psi and 1,000F with a speed of 15 ft/s. The mass flow rate of the air is 550 lbm/s. The compressor loses heat to its surroundings at a rate of 3,000 Btu/s. Determine Wcv expressed in kW. Note: 1 Btu/s=1.055kW

Solution

This is a straightforward application of the energy eqn.

dm/dt Cp(DT) + DP + rho( V1^2 -V2^2)/2 = DQ

Here dM/dt = 550 lbm/s

Cp = .24 BTU/lb/deg F

DP (300-15)psi ( convert to ft divide by 144)

rho density of air .0748 lbm/cft

V1 and V2 are given

Calculate in BTU convert to KW