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

Answer:

heat transfer rate is -15.71 kW

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 = $\frac{A1\times V1}{v1}$       .......................1

m1 = $\frac{4\times \frac{1}{60}}{0.05397}$  

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + $\frac{v1^2-v2^2}{2}$ + g (z1-z2) ]      ....................2

so here Qcv will be

Qcv =  m × [  $\frac{Wcv}{m} + (h2-h1)$  ]    ......................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