4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is
1 answer:
Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the
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Answer:
a) 764.45K
b) 210.48 kJ/kg
c) 30.14%
Explanation:
pressure ratio = 10
minimum temperature = 295 k
maximum temperature = 1240 k
isentropic efficiency for compressor = 83%
Isentropic efficiency for turbine = 87%
<u>a) Air temperature at turbine exit </u>
we can achieve this by interpolating for enthalpy
h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air
T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) = 764.45K
<u>b) The net work output </u>
first we determine the actual work input to compressor
Wc = h2 - h1 ( calculated values )
= 626.57 - 295.17 = 331.4 kJ/kg
next determine the actual work done by Turbine
Wt = h3 - h4 ( calculated values )
= 1324.93 - 783.05 = 541.88 kJ/kg
finally determine the network output of the cycle
Wnet = Wt - Wc
= 541.88 - 331.4 = 210.48 kJ/kg
<u>c) determine thermal efficiency </u>
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
<em>equation 1 becomes </em>
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
