Question

Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The volumetric flow rate of the refrigerant entering is 4 m3 / min. The work input to the compressor is 60 kJ per kg of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW

Answer 1

heat transfer rate is -15.71 kW

Step-by-step explanation:

given data

Initial pressure = 4 bar

Final pressure = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1 = m2

and mass flow equation is express as

m1 =
`(A1* V1)/(v1)` .......................1

m1 =
`(4* (1)/(60))/(0.05397)`

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) +
`(v1^2-v2^2)/(2)` + g (z1-z2) ] ....................2

so here Qcv will be

Qcv = m × [
`(Wcv)/(m) + (h2-h1)` ] ......................3

put here value and we get

Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv = -15.7130 kW

so here heat transfer rate is -15.71 kW