Question

A nuclear power plant operates at 67 % of its maximum theoretical (Carnot) efficiency between temperatures of 685 ∘C and 390 ∘C.

Part A)

If the plant produces electric energy at the rate of 1.4 GWGW , how much exhaust heat is discharged per hour?

Express your answer using two significant figures.

Part B)

A heat engine exhausts its heat at 400 ∘C∘C and has a Carnot efficiency of 35%.

What exhaust temperature would enable it to achieve a Carnot efficiency of 44%?

Express your answer with the appropriate units.

TL2=

Answer 1

The exhaust heat discharged per hour is 3.12 GW and the exhaust temperature to achieve a Carnot efficiency of 44% is 104.31 °C.

Step-by-step explanation:

Part A) To calculate the exhaust heat discharged per hour, we need to calculate the input heat to the power plant. The input heat is equal to the electric energy produced divided by the efficiency of the power plant. The efficiency of the power plant is given by the Carnot efficiency, which is equal to 1 - (Tc/Th), where Tc is the cold temperature in Kelvin and Th is the hot temperature in Kelvin. In this case, Th = 685 °C + 273.15 = 958.15 K and Tc = 390 °C + 273.15 = 663.15 K. Therefore, the Carnot efficiency is 1 - (663.15/958.15) = 0.3097. The input heat is then 1.4 GW / 0.3097 = 4.52 GW. To calculate the exhaust heat, we subtract the input heat from the electric energy produced: 4.52 GW - 1.4 GW = 3.12 GW.

Part B) To find the exhaust temperature that would enable a Carnot efficiency of 44%, we need to solve the Carnot efficiency equation for the cold temperature. The Carnot efficiency is given by 1 - (Tc/Th), where Th is the hot temperature and Tc is the cold temperature in Kelvin. We can rearrange the equation to solve for Tc: Tc = Th - (1 - efficiency) * Th. In this case, Th = 400 °C + 273.15 = 673.15 K, and the efficiency is 0.44. Plugging in the values, we get Tc = 673.15 K - (1-0.44) * 673.15 K = 377.46 K. Converting back to Celsius, the exhaust temperature is 377.46 K - 273.15 = 104.31 °C.