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

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 =       .......................1

m1 =  

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + + g (z1-z2) ]      ....................2

so here Qcv will be

Qcv =  m × [    ]    ......................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

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explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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