Question

A heat engine receives heat from a source at \(1100 \mathrm{K}\) at a rate of \(400 \mathrm{kJ} / \mathrm{s},\) and it rejects the waste heat to a medium at \(320 \mathrm{K}\). The measured power output of the heat engine is \(120 \mathrm{kW}\), and the environment temperature is \(25^{\circ} \mathrm{C}\). Determine \((a)\) the reversible power, (b) the rate of irreversibility, and \((c)\) the second-law efficiency of this heat engine.

Short answer

Question: Determine the reversible power, the rate of irreversibility, and the second-law efficiency of a heat engine with a heat source at 1100 K, a medium temperature of 320 K, a power output of 120 kW, and a heat input of 400 kJ/s. Answer: (a) The reversible power of the heat engine is 283,640 W. (b) The rate of irreversibility is 163,640 W (or 163.64 kW). (c) The second-law efficiency of the heat engine is approximately 42.33%.

Step-by-step solution

Calculate the Carnot efficiency

To find the Carnot efficiency, we use the following equation: \(\eta_\text{Carnot} = 1 - \frac{T_\text{low}}{T_\text{high}}\) Where \(T_\text{low}\) is the temperature of the medium to which waste heat is rejected, and \(T_\text{high}\) is the temperature of the heat source. We are given that \(T_\text{high} = 1100 \, \mathrm{K}\) and \(T_\text{low} = 320 \, \mathrm{K}\). Plugging these values into the equation, we get: \[\eta_\text{Carnot} = 1 - \frac{320}{1100} = 1 - 0.2909 = 0.7091\]

Calculate the actual efficiency

To find the actual efficiency, we use the following equation: \(\eta_\text{actual} = \frac{\text{Power output (work)}}{\text{Heat input}}\) The power output is given as \(120 \, \mathrm{kW} = 120,000 \, \mathrm{W}\) and the heat input is given as \(400 \, \mathrm{kJ/s} = 400,000 \, \mathrm{W}\). Therefore, we get: \[\eta_\text{actual} = \frac{120000}{400000} = 0.3\]

Calculate the reversible power

To find the reversible power, we use the maximum possible efficiency which is the Carnot efficiency, and the given heat input. We can calculate this as follows: \[\text{Reversible power} = \text{Heat input} \times \eta_\text{Carnot}\] Plugging in the values we have: \[\text{Reversible power} = 400000 \, \mathrm{W} \times 0.7091 = 283640 \, \mathrm{W}\] So, the reversible power is \(283,640 \, \mathrm{W}\).

Calculate the rate of irreversibility

To calculate the irreversibility rate, we use the difference between the reversible power and the actual power output, as follows: \[\text{Rate of irreversibility} = \text{Reversible power} - \text{Actual power output}\] Plugging in the values we have: \[\text{Rate of irreversibility} = 283640 \, \mathrm{W} - 120000 \, \mathrm{W} = 163640 \, \mathrm{W}\] So, the rate of irreversibility is \(163,640 \, \mathrm{W}\) (or \(163.64 \, \mathrm{kW}\)).

Calculate the second-law efficiency

To calculate the second-law efficiency, we use the ratio of the actual efficiency to the Carnot efficiency, as follows: \[\eta_{\text{II}}=\frac{\eta_{\text{actual}}}{\eta_{\text{Carnot}}}\] Plugging in the values we have: \[\eta_{\text{II}}=\frac{0.3}{0.7091} = 0.4233 \approx 42.33\%\] So, the second-law efficiency of the heat engine is approximately \(42.33\%\). To summarize, we have determined that: (a) The reversible power is \(283,640 \, \mathrm{W}\). (b) The rate of irreversibility is \(163,640 \, \mathrm{W}\) (or \(163.64 \, \mathrm{kW}\)). (c) The second-law efficiency of the heat engine is approximately \(42.33\%\).