Answer:
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The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
<u>From pressure table of R 134a refrigerant</u>
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49
Answer:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Explanation:
To calculate the mass of the octane(m):
Number of mole of octane (n) =0.3kmol(given)
Molarmass of octane (M) =114.23kg/kmol
m=n*M
m=(0.3kmol)*(114.23kg/kmol)
m=34.269kg
To calculate for the weight of octane(W):
W=g*m
W=(9.81m/s^2)*(34.269kg)
W=336.18N
b) For specific volumes of Vn and Vm:
Given volume of octane (V) =5m^3
Vm=V/m
Vm=5m^3/34.269kg
Vm=0.1459m^3/kg
And Vn will be :
Vn=V/m=5m^3/0.3kmol
Vn=16.67m/Kmol
Therefore, the answers are:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
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