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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.

Sagot :

okpalawalter8

Answer:

[tex]Q=15.7Kw[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]P_1=4bar[/tex]

Initial Temperature [tex]T_1=20 C[/tex]

Final Pressure  [tex]P_2=12 bar[/tex]

Final Temperature [tex]T_2=80C[/tex]

Work Output [tex]W= 60 kJ/kg[/tex]

Generally Specific Energy from table is

At initial state

 [tex]P_1=4bar \& T_1=20 C[/tex]

 [tex]E_1=262.96KJ/Kg[/tex]

With

Specific Volume [tex]V'=0.05397m^3/kg[/tex]

At Final state

 [tex]P_2=12 bar \& P_2=80C[/tex]

 [tex]E_1=310.24KJ/Kg[/tex]

Generally the equation for The Process is mathematically given by

 [tex]m_1E_1+w=m_2E_2+Q[/tex]

Assuming Mass to be Equal

 [tex]m_1=m_1[/tex]

Where

 [tex]m=\frac{V}{V'}[/tex]

 [tex]m=frac{0.06666}{V'=0.05397m^3/kg}[/tex]

 [tex]m=1.24[/tex]

Therefore

 [tex]1.24*262.96+60)=1.24*310.24+Q[/tex]

 [tex]Q=15.7Kw[/tex]

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