Question

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The volumetric flow rate of the refrigerant entering is 4 m3/min. The work input to the compressor is 60 kJ/kg of refrigerant flowing. Neglect kinetic and potential energy effects, determine the heat transfer rate, in kW.

Answer 1

`Q=15.7Kw`

Step-by-step explanation:

From the question we are told that:

Initial Pressure
`P_1=4bar`

Initial Temperature
`T_1=20 C`

Final Pressure
`P_2=12 bar`

Final Temperature
`T_2=80C`

Work Output
`W= 60 kJ/kg`

Generally Specific Energy from table is

At initial state

`P_1=4bar \& T_1=20 C`

`E_1=262.96KJ/Kg`

With

Specific Volume
`V'=0.05397m^3/kg`

At Final state

`P_2=12 bar \& P_2=80C`

`E_1=310.24KJ/Kg`

Generally the equation for The Process is mathematically given by

`m_1E_1+w=m_2E_2+Q`

Assuming Mass to be Equal

`m_1=m_1`

Where

`m=(V)/(V')`

`m=frac{0.06666}{V'=0.05397m^3/kg}`

`m=1.24`

Therefore

`1.24*262.96+60)=1.24*310.24+Q`

`Q=15.7Kw`