How to calculate the thermal efficiency of a Carnot Cycle

Method 1

• 1

  Determine the heat in and heat out if not already known. For example, if the heat in is calories, we can use heat in of 150 calories and heat out of 50 calories.

• 2

  Subtract the heat in by the heat out. This is the heat flow. The heat flow in our example is 150 calories minus 50 calories, for a heat flow of 100 calories.

• 3

  Divide the heat flow found in step 2 by the heat in. The result is the thermal efficiency of the Carnot cycle. For our example, it will be 100 calories divided by 150 calories for a thermal efficiency of 0.666 or 66.7%.

Method 2

• 4

  Determine the work in, which is the work performed by the pump in the Carnot cycle.

• 5

  Determine the work out, which is the work performed by the turbine. This will often be measured in Kilowatt-hours, Joules, calories, BTUs or foot-pounds.

• 6

  Subtract the work performed by the pump from the work performed by the turbine. This is the net-work out of the Carnot cycle.

• 7

  Divide the net work out by the heat in. The result is the thermal efficiency of the Carnot cycle.

Method 3

• 8

  Determine the highest temperature and lowest temperature of the Carnot cycle. For example, in a steam turbine, the highest temperature is 2000 Kelvin and the lowest temperature is 300 Kelvin.

• 9

  Subtract the lowest temperature from the highest temperature. This is the heat flow. In the example, this heat flow will be 2000 - 300 = 1700 Kelvin.

• 10

  Divide the heat flow found in step 2 of this method by the highest temperature of the Carnot cycle. The result is the thermal efficiency of the Carnot cycle. For the example, it will be 1700 Kelvin divided by 2000 Kelvin for a thermal efficiency of 0.85 or 85%.

Method 4

• 11

  Determine the heat put out and the heat put into the Carnot cycle. Heat can be measured in Calories, Joules or British Thermal Units. For example, heat in of 1500 BTU and heat out of 500 BTU.

• 12

  Divide the heat out by the heat in. This should result in a ratio less than 1. For example, 500 BTU out divided by 1500 BTU in for a result of 0.333.

• 13

  Subtract the ratio found in step 2 from the number 1. The percentage remaining is the thermal efficiency of the Carnot cycle. In this example, 1 - 0.333 = 0.666 or a thermal efficiency of 66.6%.

Method 5

• 14

  Determine the highest and lowest temperatures found in the Carnot cycle. For example, use the highest temperature of 1000 Celsius and lowest temperature of 200 Celsius.

• 15

  Divide the lowest temperature by the highest temperature. For example, 200 degrees divided by 1000 degrees for a result of 0.200.

• 16

  Subtract the ratio found in step 2 from the number 1. The percentage remaining is the thermal efficiency of the Carnot cycle. In this example, the equation becomes 1 - 0.20 = 0.80 or 80% thermal efficiency.