Respuesta :
Answer:
Heat supplied = 208.82 BTU/s
Heat rejected = 66.82 BTU/s
Carnot thermal efficiency = 0.68
Explanation:
Data
Hot temperature,[tex] T_H [/tex] = 1200 F + 459.67 = 1659.67 R
Cold temperature,[tex] T_C [/tex] = 70 F + 459.67 = 529.67 R
Engine power, [tex] \dot{W} = 200 hp \times 0.71\frac{BTU/s}{hp} = 142 \frac{BTU}{s} [/tex]
Carnot thermal efficiency is computed by
[tex] \eta = 1 - \frac{T_C}{T_H} [/tex]
[tex] \eta = 1 - \frac{529.67 R}{1659.67 R} [/tex]
[tex] \eta = 0.68 [/tex]
Efficiency is by definition
[tex] \eta = \frac{\dot{W}}{\dot{Q_{in}}} [/tex]
[tex] \dot{Q_{in}} = \frac{\dot{W}}{\eta} [/tex]
[tex] \dot{Q_{in}} = \frac{142 \frac{BTU}{s}}{0.68} [/tex]
[tex] \dot{Q_{in}} = 208.82 \frac{BTU}{s} [/tex]
where [tex] \dot{Q_{in}} [/tex] is the heat supplied
Energy balance in the engine
[tex] \dot{Q_{in}} = \dot{W} + \dot{Q_{out}} [/tex]
[tex] \dot{Q_{out}} = \dot{Q_{in}} - \dot{W} [/tex]
[tex] \dot{Q_{out}} = 208.82 \frac{BTU}{s} - 142 \frac{BTU}{s} [/tex]
[tex] \dot{Q_{out}} = 66.82 \frac{BTU}{s} [/tex]
where [tex] \dot{Q_{out}} [/tex] is the heat rejected
