The coefficient of performance of a reversible refrigeration cycle is always (a) greater than, (b) less than, (c) equal to the c
1 answer:
Answer:
Option B is correct
Explanation:
Let
Higher temperature =
Lower temperature =
We know that COP is given by
We see that COP is depends only on the temperature difference & Temperature difference is maximum for the Carnot cycle.
Therefore the COP of reversible refrigeration cycle is always less then the COP of an irreversible refrigeration cycle when each operates between the same two thermal reservoirs.
Therefore option B is correct
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Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW