Answer in Molecular Physics | Thermodynamics for SUSHIL SHENDE #144953
February 21st, 2023
As per the given question,
Temperature of the space “(T)=298K”
Power of heat engine “(P)=1600 J/s”
Flow of vapor=3.9kg/min
Inlet temperature “(T_i)=333K”
Outlet temperature “(T_o)=40^circ C=(273+40)=313K”
Entering temperature of evaporate = 285K
Quality =15%
Specific heat of saturated vapor is represented by “h_1” and specific heat of compressed vapor is represented by “h_2”
a)“dfrac{dW}{dt}=frac{dm}{dt}(h_2-h_1)times 0.15”
Now, substituting the values in the above equation,
“dfrac{dm}{dt}=frac{1600}{0.15(423.9-392.7)}=341.88 kg/s”
b)“frac{dQ}{dt}= frac{3.9kg}{60s}(392.7-241.7)=9.81J/s”
c)“E_f=1-frac{T_c}{T_h}”
“E_f=1-frac{313}{333}=frac{20}{333}”
“Rightarrow COP= frac{1}{E_f}”
“Rightarrow COP = frac{333}{20}=16.65”
