A general expression for the efficiency of a heat engine can be written as:

$$\text{Efficiency =}\frac{\text{Work}}{{\text{HeatEnergy}}_{Hot}}$$

We know that all the energy that is put into the engine has to come out either as work or waste heat. So work is equal to Heat at High temperature minus Heat rejected at Low temperature. Therefore, this expression becomes:

$$\text{Efficiency=}\frac{{\text{Q}}_{\text{Hot}}{\text{-Q}}_{\text{Cold}}}{{\text{Q}}_{\text{Hot}}}$$

Where, Q_{Hot} = Heat input at high temperature and Q_{Cold}= Heat rejected at low temperature. The symbol is often (Greek letter eta) used for efficiency this expression can be rewritten as:

$${\eta}^{\prime}\left(\%\right)=1-\frac{{Q}_{Cold}}{{Q}_{Hot}}\times 100$$

The above equation is multiplied by 100 to express the efficiency as percent.

French Engineer Sadi Carnot showed that the ratio of Q_{HighT} to Q_{LowT} must be the same as the ratio of temperatures of high temperature heat and the rejected low temperature heat. So this equation, also called **Carnot Efficiency**, can be simplified as:

Note: Unlike the earlier equations, the positions of T_{cold} and T_{hot} are reversed.

The Carnot Efficiency is the theoretical maximum efficiency one can get when the heat engine is operating between two temperatures:

- The temperature at which the high temperature reservoir operates ( T
_{Hot}). - The temperature at which the low temperature reservoir operates ( T
_{Cold}).

In the case of an automobile, the two temperatures are:

- The temperature of the combustion gases inside the engine ( T
_{Hot}). - The temperature at which the gases are exhausted from the engine ( T
_{Cold}).

Please watch the following 4:40 presentation about how automobile engines work:

Then, why should we operate the automobiles at low efficiencies?

It is not that we cannot achieve high temperatures, but we do not have the **engine materials** that can withstand the high temperature. As a matter of fact, we do not let the engine gases go the maximum that they can go even now and instead try to keep the engine cool by circulating the coolant.

So we are taking the heat out of the gases (thus lowering the T_{hot}) and making the engine operate at cooler temperatures so that the engine is protected - but lowering the efficiency of an automobile.

It's like Taxes. The more money you earn (heat), the more money is taxed (cold), leaving you with less money to take home (efficiency). However, if you could earn more money (heat) and find a way to have less taxes taken out (better engine material), you would have more money to take home (efficiency).

Below are two temperature scales. The first scale, labeled "HOT," shows the range of temperatures for the combustion of gases in a car engine. The second scale, labeled "COLD," shows the range of temperatures at which gases are exhausted from the car engine.

**Instructions:** Select numbers from the range on the "HOT" scale and enter them (one at a time) into the text box labeled "Hot" below. At the same time, select numbers from the range on the "COLD" scale and enter them (one at a time) into the text box labeled "COLD" below. Try various combinations of hot and cold numbers and observe the graph to see the temperatures' effect on efficiency.

### Example

For a coal-fired utility boiler, the temperature of high pressure steam (T_{hot})would be about 540°C and T_{cold}, the cooling tower water temperature, would be about 20°C. Calculate the Carnot efficiency of the power plant:

Solution:

Carnot efficiency depends on high temperature and low temperatures between which the heat engine operates. We are given both temperatures. However, the temperatures need to be converted to Kelvin:

$$\begin{array}{l}{T}_{hot}={540}^{o}C+273=813K\\ {T}_{cold}={20}^{o}C+273=293\text{}K\\ \eta =\left[1-\frac{{T}_{cold}}{{T}_{hot}}\right]\times 100\%\\ \eta =\left[1-\frac{293\text{}K}{813\text{}K}\right]\times 100\%\\ =64\%\end{array}$$

### Practice

For a coal fired utility boiler, the temperature of high pressure steam would be about 540 degrees C and T_{cold}, the cooling tower water temperature, would be about 20 degrees C. Calculate the Carnot efficiency of the power plant.

**Step 1**

Convert the high and low temperatures from Celsius to Kelvin:

$$\begin{array}{l}{T}_{hot}={540}^{o}C+273\\ =813K\end{array}$$

$$\begin{array}{l}{T}_{cold}={20}^{o}C+273\\ =293\text{}K\end{array}$$

**Step 2**

$$\begin{array}{l}\eta =\left[1-\frac{{T}_{cold}}{{T}_{hot}}\right]\times 100\%\\ \\ \eta =\left[1-\frac{293\text{}K}{813\text{}K}\right]\times 100\%\\ \\ =64\%\end{array}$$

From the Carnot Efficiency formula, it can be inferred that a maximum of 64% of the fuel energy can go to generation. To make the Carnot efficiency as high as possible, either T_{hot} should be increased or T_{cold} (temperature of heat rejection) should be decreased.

#### Practice Problem

Use the following link to generate a random practice problem.