Question

A Carnot heat engine receives heat at \(900 \mathrm{K}\) and rejects the waste heat to the environment at \(300 \mathrm{K}\). The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at \(-15^{\circ} \mathrm{C}\) at a rate of \(250 \mathrm{kJ} / \mathrm{min}\) and rejects it to the same environment at 300 K. Determine ( \(a\) ) the rate of heat supplied to the heat engine and \((b)\) the total rate of heat rejection to the environment.

Short answer

Answer: (a) The rate of heat supplied to the heat engine is 6250 W. (b) The total rate of heat rejection to the environment is 7199.45 W.

Step-by-step solution

Calculate the efficiency of the Carnot heat engine

First, we need to calculate the efficiency of the Carnot heat engine using the following expression: Efficiency = \(1 - \frac{T_{L}}{T_{H}}\) where \(T_{L}\) is the temperature of the low-temperature reservoir (300 K) and \(T_{H}\) is the temperature of the high-temperature reservoir (900 K). Efficiency = \(1 - \frac{300}{900}\) = \(1 - \frac{1}{3}\) = \(\frac{2}{3}\) = 0.6667 The efficiency of the Carnot heat engine is 0.6667, or 66.67%.

Calculate the rate of work output from the heat engine

We are given the rate at which the refrigerator removes heat from the cooled space, \(Q_{refrigerator} = 250\;\text{kJ/min}\). The work output of the heat engine is used to drive this refrigerator. Let's convert the given heat transfer rate to watts (W). \(Q_{refrigerator}\;\text{rate} = \frac{250\;\text{kJ}}{1\;\text{min}} \times \frac{1000\;\text{J}}{1\;\text{kJ}} \times \frac{1\;\text{min}}{60\;\text{s}} = 4166.67\;\text{W}\) The heat engine's work output rate = \(Q_{refrigerator}\;\text{rate} = 4166.67\;\text{W}\)

Calculate the rate of heat supplied to the heat engine

We can now use the efficiency of the heat engine to calculate the rate of heat supplied: \(\text{Efficiency} = \frac{\text{Work}\;\text{output}\;\text{rate}}{\text{Heat}\;\text{input}\;\text{rate}}\) Rearrange to solve for Heat input rate: \(\text{Heat}\;\text{input}\;\text{rate} = \frac{\text{Work}\;\text{output}\;\text{rate}}{\text{Efficiency}}\) Heat input rate = \(\frac{4166.67\;\text{W}}{0.6667}\) = 6250 W Therefore, the rate of heat supplied to the heat engine is 6250 W.

Calculate the coefficient of performance for the refrigerator cycle

We can now determine the coefficient of performance (COP) for the refrigerator cycle using the temperatures supplied: \(COP_{refrigerator} = \frac{T_{L}}{T_{H} - T_{L}} = \frac{255\;\text{K}}{300\;\text{K} - 255\;\text{K}} = 5.1\) The coefficient of performance for the refrigerator cycle is 5.1.

Calculate the total rate of heat rejection to the environment

We can now determine the total rate of heat rejection to the environment by considering the heat rejected by both the engine and the refrigerator. \(Q_{rejected} = Q_{input,engine} - W_{engine} + Q_{refrigerator}(1 + 1/COP)\) Substituting known values: \(Q_{rejected} = 6250\;\text{W} - 4166.67\;\text{W} + 4166.67\;\text{W}(1 + 1/5.1)\) \(Q_{rejected} = 2083.33\;\text{W} + 4166.67\;\text{W}(1 + 1/5.1)\) = \(2083.33\;\text{W} + 5116.12\;\text{W}\) = 7199.45 W The total rate of heat rejection to the environment is 7199.45 W. In conclusion, (a) the rate of heat supplied to the heat engine is 6250 W, and (b) the total rate of heat rejection to the environment is 7199.45 W.