Respuesta :
Answer:
Explanation:
Given
[tex]T_h=250^{\circ}C\approx 523\ K[/tex]
[tex]T_L=30^{\circ}C\approx 303\ K[/tex]
[tex]Q_1=6 kW[/tex]
From Clausius inequality
[tex]\oint \frac{dQ}{T}=0[/tex] =Reversible cycle
[tex]\oint \frac{dQ}{T}<0[/tex] =Irreversible cycle
[tex]\oint \frac{dQ}{T}>0[/tex] =Impossible
(a)For [tex]P_{out}=3 kW[/tex]
Rejected heat [tex]Q_2=6-3=3\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K[/tex]
thus it is Impossible cycle
(b)[tex]P_{out}=2 kW[/tex]
[tex]Q_2=6-2=4 kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K[/tex]
Possible
(c)Carnot cycle
[tex]\frac{Q_2}{Q_1}=\frac{T_1}{T_2}[/tex]
[tex]Q_2=3.47\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3.47}{303}=0[/tex]
and maximum Work is obtained for reversible cycle when operate between same temperature limits
[tex]P_{out}=Q_1-Q_2=6-3.47=2.53\ kW[/tex]
Thus it is possible
