Answer to Question #133239 in Molecular Physics | Thermodynamics for Mearls

Question

During the power stroke of an engine, the work produced is 2520 kj/kg and the heat rejected through the cylinder is 570 kj/kg. If the internal energy increased by 120 kj/kg, determine the amount of heat received in kj/s if the working fluid flow at the rate of 3 kg/s.

Answer 1

According to the first law of thermodynamics, the heat received comes to the increase in the internal energy and work:

$$Q=\Delta U+W,\\ Q=120+2520=2640\text{ kJ/kg}.$$

Also, 570 kJ/kg rejected, which means that heat received was actually greater:

$$Q_r=Q+570=3210\text{ kJ/kg}.$$

Convert this to kJ/s:

$$q_r=Q_r\cdot r=3210\text{ kJ/kg}\cdot3\text{ kg/s}=9630\text{ kJ/s}.$$