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Hunter-Best [27]
4 years ago
10

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 kg of refrigerant flo

wing through its circuit each cycle. The refrigerant enters the evaporator coils in phase equilibrium, with 54.0 % of its mass as liquid and the rest as vapor. It flows through the evaporator at a constant pressure and when it reaches the compressor 95% of its mass is vapor. In each cycle, how much heat Qc is absorbed by the refrigerant while it is in the evaporator? The heat of vaporization of the refrigerant is 1.50×105 J/kg
Physics
1 answer:
kozerog [31]4 years ago
4 0

Answer:

5.15J

Explanation:

First. 54% of the 7kg refrigerant is liquid

So we find mass of vapour at inlet generator

M1 = ( 1-0.54)*7= 3.2kg

At compressor mass of vapour will be

M2= 0.95*7= 6.7kg

So the Mass of vapour at exit generator is

M2-M1= 3.5kg

So to find heat absorbed by refrigerant in evaporation

Its using

Q= mh

°= 3.5x 1.50×10^5 J/kg

=5.15J

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Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

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Rate of heat = -1.1 kW

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\terxt{time for compression}=\text{time for half revolution}

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\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60

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U=Q-W

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\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)

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\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}

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Hence, The temperature change per compression stroke is 32.48°.

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Answer:

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