A piston cylinder device contains 5 kg of Refrigerant 134a at 600 kPa and 80 C. The refrigerant is now cooled at constant pressure until it reaches a liquid-vapor mixture state with a quality of 0.3. How much heat was extracted in the process?

The answer is 920 kJ

Explanation:

Solution

Given that:

Mass = 5kg

Pressure = 600 kPa

Temperature = 80° C

Liquid vapor mixture state (quality) = 0.3

Now we find out the amount of heat extracted in the process

Thus

Properties of  RI34a at:

P₁ = 600 kPa

T₁ = 80° C

h₁ = 320 kJ/kg

So,

P₁ = P₂ = 600 kPa

X₂ =0.3

h₂ = 136 kJ/kg

Now

The heat removed Q = m(h₁ -h₂)

Q = 5 (320 - 136)

Q= 5 (184)

Q = 920 kJ

Therefore the amount of heat extracted in the process is 920 kJ

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