Question

Consider a Carnot-type engine driven by two heat reservoirs at 450 K and 280 K, respectively. Heat can flow from the hot reservoir to the engine at 80,000 BTU/hr (BTU=British Thermal Unit= 1054 J). What is the maximum efficiency of this engine? What is the maximum power, in watts, that can be generated by the engine?

Answer 1

The maximum efficiency of a Carnot engine is given by = (T_H-T_C)/T_H = W/Q_H

Where T_H and T_C are the absolute temperatures of the hot and cold reservoir respectively, W is work done by the system and Q_H is heat energy put into the system.

Thus = (450-280)/450 =0.378

Now heat entering the system = 80,000 BTU per hour = 22.222 BTU per second = 23445.45 J per second (1BTU = 1055.06 J)

W = x Q_H = 0.378 x 23445.45 = 8862.38 J

Thus Maximum power = maximum work done per second= 8862.38 watt (watt =J/s)