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3 years ago

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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 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.

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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