A refrigerator using Helium Gas operates on a reversed Brayt

A refrigerator using Helium Gas operates on a reversed Brayton Cycle with a pressure ratio of 5. Prior to compression the gas occupies 100cm3 at a pressure of 150kPa and a temperature of -23degC. The volume at the end of the expansion is 80cm3 . What is the refrigerators CoP and its power input if it operates at 60 cycles per second ?

Solution

prior to compression temperature will be:

PV = nRT

n = 150*10^3*10^-6/(0.0821*250) = 0.0072 moles

V1/T1 =V2/T2

T1 = 80*250/100 = 200 K

afterc compression adiabatic process

P2*V2^y = P3*V3^y

V3 = (P2/P3)^(1/y)*V2 = (1/5)^(1/1.6)*100 = 36.57 cm^3

T3 = P3V3/RT = 36.57*10^-4*750*10^3/(0.0821*0.0072)= 463.99 K

for adiabatic process prior to compression:

P1V1^y = P4V4^y

V4 = (1/5)^(1/6)*80 = 29.25 cm^3

T4 = 29.25*10^-4*750*10^3/(0.0821*0.0072) = 371.11 K

Qh = n*Cp*(T3 - T4) = 0.0072*2.5*8.314*(463.99 - 371.11) = 13.899

Qc = 0.0072*2.5*8.314*(200- 150) = 7.48

w = dQ = 13.899 - 7.48 = 6.419 J

COP = Qh/w = 13.899/6.419 = 2.165

power = w*60 = 6.419*60 = 385.14 W