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:

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 }

At steady state, the power input of a refrigeration cycle is 500 kW. The cycle operates between hot and cold reservoirs which are at 550 K and 300 K, respectively. a) If cycle's coefficient of performance is 1.6, determine the rate of energy removed from the cold reservoir, in kW. b) Determine the minimum theoretical power required, in kW, for any such cycle operating between 550 K and 300 K