Question

At steady state, a well-insulated compressor takes in air at 60F, 14.2 lbf/in.2 , with a volumetric flow rate of 1200 ft3 /min, and compresses it to 500F, 120 lbf/in.2 Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft3 /min.

Answer 1

Answer:

the compressor power is -223.12 hP   { -ve indicate work done }

Volumetric flow rate at exit = 262.74 ft³/s

Explanation:

Given the data in the question;

p₁ = 14.2 psi = 0.978 bar

p₂ = 120 psi = 8.268 bar

T₁ = 60°F = 288.706 K

T₂ = 500°F = 533.15 K

Q₁ = 1200 ft³/min = 20ft³/s = 0.566 m³/s

δ₁ = p₁/RT₁ = 0.978 × 10⁵ / ( 287 ×  288.706) = 1.18 kg/m³

δ₂ = P₂/RT₂ = 8.268 × 10⁵ / ( 287 × 533.15 ) = 5.4 kg/m³

so

ω₁ = δ₁Q₁ = 1.18 × 0.566 = 0.668 kg/s

we know that; ω₁ = ω₂    { in a steady flow }

ω₂ = δ₂Q₂

Q₂ = ω₂/δ₂

Q₂ = 0.668 / 5.4  = 0.1237 m³/s  

Hence Volumetric flow rate at exit = 262.74 ft³/s

from the steady state energy equation;

ω( h₁ - h₂ ) = dW/dt

{ where h = CpT)

( Cp = 1.005  )  

dW/dt =  ωCp( T₁ - T₂ )

we substitute

dW/dt = 0.668 × 1.005( 288.706 - 533.15 )

dW/dt = 0.67134 × -244.444

dW/dt = -164.105 kW = -223.12 hP

Hence, the compressor power is -223.12 hP   { -ve indicate work done }