Question

A heat engine operates between a source at \(477^{\circ} \mathrm{C}\) and a \(\operatorname{sink}\) at \(25^{\circ} \mathrm{C}\). If heat is supplied to the heat engine at a steady rate of \(65,000 \mathrm{kJ} / \mathrm{min},\) determine the maximum power output of this heat engine.

Short answer

The maximum power output of this heat engine is approximately 652 kW, given the conditions mentioned.

Step-by-step solution

Convert temperatures to Kelvin

To calculate the Carnot efficiency, we need to have the temperatures in Kelvin. So, let's convert the given temperatures from Celsius to Kelvin. To convert a temperature from Celsius to Kelvin, you can use the formula: K = °C + 273.15 For the heat source, T1 = 477°C = 477 + 273.15 = 750.15 K For the heat sink, T2 = 25°C = 25 + 273.15 = 298.15 K

Calculate Carnot efficiency

Now that we have the temperatures in Kelvin, we can calculate the Carnot efficiency: Carnot efficiency = 1 - (T2 / T1) Here, T1 is the temperature of the heat source, and T2 is the temperature of the heat sink. Carnot efficiency = 1 - (298.15 / 750.15) ≈ 0.6020

Calculate the heat input rate in Watts

We've been given the heat input rate in kJ/min, and we need to convert it to Watts (Joules per second) before we calculate the power output. 1 kJ = 1000 J 1 minute = 60 seconds Heat input rate = 65,000 kJ/min * (1000 J/1 kJ) * (1 min/60 s) = 1,083,333.33 W

Calculate the maximum power output

Now that we have the efficiency and the heat input rate, we can calculate the maximum power output. Maximum power output = Carnot efficiency * Heat input rate = 0.6020 * 1,083,333.33 W ≈ 652,000 W or 652 kW So, the maximum power output of this heat engine is about 652 kW.