Question

A Carnot engineoperates between a heat source at

1200 F and a heat sink at 70 F.The engine delivers 200 hp. Compute
the heat supplied (Btu/s), theheat rejected (Btu/s), and the
thermal efficiency of the heatengine.

Answer 1

Heat supplied = 208.82 BTU/s

Heat rejected = 66.82 BTU/s

Carnot thermal efficiency = 0.68

Step-by-step explanation:

Data

Hot temperature,
`T_H` = 1200 F + 459.67 = 1659.67 R

Cold temperature,
`T_C` = 70 F + 459.67 = 529.67 R

Engine power,
`\dot{W} = 200 hp * 0.71(BTU/s)/(hp) = 142 (BTU)/(s)`

Carnot thermal efficiency is computed by

`\eta = 1 - (T_C)/(T_H)`

`\eta = 1 - (529.67 R)/(1659.67 R)`

`\eta = 0.68`

Efficiency is by definition

`\eta = \frac{\dot{W}}{\dot{Q_(in)}}`

`\dot{Q_(in)} = \frac{\dot{W}}{\eta}`

`\dot{Q_(in)} = (142 (BTU)/(s))/(0.68)`

`\dot{Q_(in)} = 208.82 (BTU)/(s)`

where
`\dot{Q_(in)}` is the heat supplied

Energy balance in the engine

`\dot{Q_(in)} = \dot{W} + \dot{Q_(out)}`

`\dot{Q_(out)} = \dot{Q_(in)} - \dot{W}`

`\dot{Q_(out)} = 208.82 (BTU)/(s) - 142 (BTU)/(s)`

`\dot{Q_(out)} = 66.82 (BTU)/(s)`

where
`\dot{Q_(out)}` is the heat rejected