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sattari [20]
3 years ago
11

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol

umetric 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.
Engineering
1 answer:
sleet_krkn [62]3 years ago
4 0

Answer:

Q=15.7Kw

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=\frac{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

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