Question

What is the maximum possible thermal efficiency of a gas power cycle when using thermal energy reservoirs at \(1100^{\circ} \mathrm{F}\) and \(80^{\circ} \mathrm{F} ?\)

Short answer

Answer: The maximum possible thermal efficiency of a gas power cycle using thermal energy reservoirs at 1100°F and 80°F is approximately 65.36%.

Step-by-step solution

Convert Temperatures to Kelvin

To convert the given temperatures in Fahrenheit to Kelvin, use the following conversion formula: \(K = \frac{(°F - 32)}{1.8} + 273.15\) For the high temperature reservoir at \(1100^{\circ}\mathrm{F}\): \(K_{high} = \frac{(1100 - 32)}{1.8} + 273.15\) For the low temperature reservoir at \(80^{\circ}\mathrm{F}\): \(K_{low} = \frac{(80 - 32)}{1.8} + 273.15\)

Calculate the Kelvin Temperatures

Now calculate the Kelvin temperatures using the conversion formula: \(K_{high} = \frac{(1100 - 32)}{1.8} + 273.15 = 866.85\,K\) \(K_{low} = \frac{(80 - 32)}{1.8} + 273.15 = 300.372\,K\)

Calculate the Maximum Efficiency using Carnot Efficiency Formula

The formula for Carnot efficiency is given by: \(Efficiency_{Carnot} = 1 - \frac{T_{low}}{T_{high}}\) Using the given Kelvin temperatures, calculate the maximum possible thermal efficiency: \(Efficiency_{Carnot} = 1 - \frac{300.372}{866.85}\)

Calculate the Maximum Thermal Efficiency

Now calculate the maximum possible thermal efficiency: \(Efficiency_{Carnot} = 1 - \frac{300.372}{866.85} = 1 - 0.3464 \approx 0.6536\) To express this as a percentage, multiply by 100: \(Efficiency_{Carnot} = 0.6536 \times 100\% \approx 65.36\%\) The maximum possible thermal efficiency of a gas power cycle using thermal energy reservoirs at \(1100^{\circ}\mathrm{F}\) and \(80^{\circ}\mathrm{F}\) is approximately \(65.36\%\).