Question

It is proposed to run a thermoelectric generator in conjunction with a solar pond that can supply heat at a rate of \(7 \times 10^{6} \mathrm{kJ} / \mathrm{h}\) at \(90^{\circ} \mathrm{C}\). The waste heat is to be rejected to the environment at \(22^{\circ} \mathrm{C}\). What is the maximum power this thermoelectric generator can produce?

Short answer

Heat Supply Rate: \(7 \times 10^6 \, \mathrm{kJ/h}\) Source Temperature: \(T_H = 90^{\circ} \mathrm{C}\) Sink Temperature: \(T_C = 22^{\circ} \mathrm{C}\) Maximum Power Output: \(\_\_\_\_\, \mathrm{kW}\)

Step-by-step solution

Convert temperatures to Kelvin

To work with absolute temperatures, convert the Celsius temperatures to Kelvin: \(T_H = 90^{\circ}\mathrm{C} + 273.15 \mathrm{K} = 363.15 \mathrm{K}\) \(T_C = 22^{\circ}\mathrm{C} + 273.15 \mathrm{K} = 295.15 \mathrm{K}\)

Determine Carnot efficiency

For a Carnot engine (maximum efficiency), the efficiency is given by: \(\eta_\mathrm{Carnot} = 1 - \frac{T_C}{T_H}\) Plug in the values for \(T_H\) and \(T_C\): \(\eta_\mathrm{Carnot} = 1 - \frac{295.15 \mathrm{K}}{363.15 \mathrm{K}} = 1 - 0.81269 \approx 0.18731\)

Convert heat supply rate to work per second

To find work rate, first convert the given heat supply rate to \(\mathrm{kJ/s}\): \(\text{Heat supply rate} = \frac{7 \times 10^6 \, \mathrm{kJ}}{1\, \mathrm{h}} \times \frac{1\, \mathrm{h}}{3600\, \mathrm{s}} = 1944.44 \, \mathrm{kJ/s}\)

Calculate maximum work output

Maximum work output equals the product of the Carnot efficiency and the heat supply rate: \(\text{Max work output} = \eta_\mathrm{Carnot} \times \text{Heat supply rate}\) \(\text{Max work output} = 0.18731 \times 1944.44 \, \mathrm{kJ/s} \approx 364.14 \, \mathrm{kJ/s}\)

Convert work output to power output

The maximum power output is the maximum work output in kilowatts: \(\text{Max power output} = 364.14 \, \mathrm{kW}\) The maximum power output of this thermoelectric generator is approximately \(364.14 \, \mathrm{kW}\).