Respuesta :
Answer:
a) Possible
b) Possible
c) Possible
Explanation:
Given:
temperature of hot reservoir, [tex]T_H=1000\ K[/tex]
temperature of the cold reservoir, [tex]T_C=500\ K[/tex]
heat rejected to the colder reservoir, [tex]Q_C=5.1\times 10^5\ btu.hr^{-1}[/tex]
We know that:
[tex]1\ hp=2544.43\ Btu.h^{-1}[/tex]
so,
[tex]Q_C=200.438\ hp[/tex]
a)
power rate developed by the power, [tex]W=90\ hp[/tex]
Now the Carnot efficiency of the cycle:
[tex]\eta_c=1-\frac{T_C}{T_H}[/tex]
[tex]\eta_c=1-\frac{500}{1000}[/tex]
[tex]\eta_c=0.5[/tex] .........................(1)
Now the actual efficiency of the cycle:
[tex]\rm \eta=\frac{desired\ effect}{energy\ consumed}[/tex]
[tex]\eta=\frac{W}{Q_C+W}[/tex]
[tex]\eta=\frac{90}{90+200.438}[/tex]
[tex]\eta=0.31[/tex]
Since [tex]\eta_c>\eta[/tex] therefore the cycle is possible
b)
power rate developed by the power, [tex]W=100\ hp[/tex]
Now the actual efficiency of the cycle:
[tex]\rm \eta=\frac{desired\ effect}{energy\ consumed}[/tex]
[tex]\eta=\frac{W}{Q_C+W}[/tex]
[tex]\eta=\frac{100}{100+200.438}[/tex]
[tex]\eta=0.33[/tex]
Since [tex]\eta_c>\eta[/tex] therefore the cycle is possible
c)
power rate developed by the power, [tex]W=110\ hp[/tex]
Now the actual efficiency of the cycle:
[tex]\rm \eta=\frac{desired\ effect}{energy\ consumed}[/tex]
[tex]\eta=\frac{W}{Q_C+W}[/tex]
[tex]\eta=\frac{110}{110+200.438}[/tex]
[tex]\eta=0.35[/tex]
Since [tex]\eta_c>\eta[/tex] therefore the cycle is possible
