Jump to a key chapter

## Thermal Efficiency of an Engine

Throughout this article, we will be considering the thermodynamic model of a heat engine in order to explain the concept of thermal efficiency. So before looking more closely at thermal efficiency, let's recap exactly what we mean by a heat engine.

A **Heat Engine** is a thermodynamic system that converts heat into work.

The general model of a heat engine is of a system, usually, a gas, held between two heat reservoirs, one of which is at a higher temperature than the system and one of which is at a lower temperature. In the context of a heat engine, we call these reservoirs a heat source and sink respectively. The system is able to do work, denoted \(W\) on the environment in some way, for example by pushing up a piston as the gas expands. It's this work \(W\) that makes an engine useful, for example in a car engine this work is used to turn an axle making the wheels turn. Due to the temperature difference, heat is transferred from the heat source to the system. This is denoted \(Q_\mathrm{H}\) in Figure 1. Some of this heat will be used to do useful work on the environment, whilst some of the heat will increase the internal energy of the gas, lifting its temperature. The difference in temperature between the system and the heat sink will then cause heat to flow from the system to the sink, with this energy denoted \(Q_\mathrm{C}\).

## Thermal Efficiency Definition

Heat engines work by converting energy transferred as heat into useful work. However, as we shall see, no heat engine is perfectly efficient and so some heat is always lost to the environment. We can define this lost heat as waste heat, the less heat that is lost, the more effective the engine is. So, the effectiveness of an engine is determined by the amount of useful work done per unit of heat input. Thermal efficiency can then be used to quantify this effectiveness.

**Thermal Efficiency \(\eta\), **or the **Coefficient of Performance,** is the percentage of the heat input \(Q\) entering a system that is transformed into work \(W\) by the system.

\[\eta=\frac{W}{Q}.\]

Thermal efficiency can only take values between \(0\) and \(1\), \(0\%-100\%\), as can be seen by applying the first law of thermodynamics to the definition for \(\eta\). Consider the heat input of an engine \(Q_\text{in}\). Assuming no energy is lost within the engine itself, any heat not converted to work will be lost as heat \(Q_\text{out}\) to the surroundings. Hence we can define the work done to be \[W=Q_\text{in}-Q_\text{out}.\]

Plugging this into the definition of thermal efficiency we find

\[\begin{align}\eta&=\frac{W}{Q_\text{in}}\\&=\frac{Q_\text{in}-Q_\text{out}}{Q_\text{in}}\\&=1-\frac{Q_\text{out}}{Q_\text{in}}.\end{align}\]

Thermal efficiency can only take values between \(0\) and \(1\), \(0\%-100\%\), as can be seen by applying the first law of thermodynamics to the definition for \(\eta\). Consider the heat input of an engine \(Q_\text{in}\). Assuming no energy is lost within the engine itself, any heat not converted to work will be lost as heat \(Q_\text{out}\) to the surroundings. Hence we can define the work done to be \[W=Q_\text{in}-Q_\text{out}.\]

Plugging this into the definition of thermal efficiency we find

\[\begin{align}\eta&=\frac{W}{Q_\text{in}}\\&=\frac{Q_\text{in}-Q_\text{out}}{Q_\text{in}}\\&=1-\frac{Q_\text{out}}{Q_\text{in}}.\end{align}\]

The first law of thermodynamics ensures that the heat lost by the system cannot be greater than the heat supplied to the system and so \(0\leq\eta\leq1\).

**The First Law of Thermodynamics **states that the change in internal energy of a system \(\Delta U\), must always be equal to the difference between the heat supplied to the system \(Q\), and the work done by the system on its environment \(W\).

\[\Delta U=Q-W.\]

Let's look into an example.

If a heat engine, which operates by taking in \(500\,\mathrm{J}\) of heat, loses \(200\,\mathrm{J}\) of heat to the environment during its cycle what is the thermal efficiency of this engine?

Answer:

Using the formula given above:\[\begin{align}\nu&=1-\frac{Q_\text{out}}{Q_\text{in}}\\&=1-\frac{200\,\mathrm{J}}{500\,\mathrm{J}}\\&=0.6 \quad \text{or}\quad 60\%.\end{align}\]

As you will know if you've ever seen a car engine overheat, most heat engines are extremely inefficient. For example, the average diesel engine operates at an efficiency of around \(25\%\), and even the most efficient of heat engines can only reach \(50\%-60\%\). Much of this inefficiency comes from the heat loss and frictional forces within an engine as well as losses during the combustion process. However, as we shall see when investigating the Carnot cycle, perfect efficiency is not possible even for idealized **Reversible **engines due to the 2^{nd} Law of Thermodynamics.

## Thermal Efficiency Example

Many appliances and technologies we need in modern society are based on this simple heat engine model, and improving the efficiency of these appliances can help to reduce energy usage. For example, when buying a fridge it's important to compare the Coefficient of Performance (COP) of different models before making a purchase. A fridge is a sort of heat engine in reverse, whereby work is done by the environment on the system, usually in the form of a compressor, to extract heat from a cold reservoir (the inside of the fridge) and pump it out into a hot reservoir (the outside room).

This means that for fridges, the COP is defined inversely to the thermal efficiency of a heat engine, as in this case, we are interested in how much heat can be extracted per unit of work. Here the 'waste' heat is the heat that is transferred into the system by the environment \(Q_\text{in}\).

\[\begin{align}COP &= \frac{Q_\text{out}}{W_\text{in}} \\&=\frac{Q_\text{out}}{Q_\text{in}-Q_\text{out} }\\&=\frac{Q_\text{out}}{Q_\text{in}}-1\end{align}\]

This means that, unlike thermal efficiency, the COP can take values greater than one, the higher the COP, the more heat is removed per unit of work.

An average modern fridge operates at a COP of \(1.37.\) If the power capacity of the fridge is \(300\,\mathrm{W}\), how much heat is extracted from the interior of the fridge in one minute?

Answer:

First, we need to calculate how much work is done by the fridge in one minute, this can be found from the definition of power.\[\begin{align}\text{Power}&=\frac{\text {Work done}}{\text {Time}}\\implies \text{Work done}=\text{Power}\cdot\text{Time}\\\implies W&=300\,\mathrm{W}\cdot60\,\mathrm{s}\\&=18000\,\mathrm{J}\end{align}\]

Combining this with the definition of COP gives us the heat removed\[\begin{align}Q_\text{out}&=COP\cdot W\\&=1.37\cdot18\,000\,\mathrm{J}\\&=24\,660\,\mathrm{J}\end{align}\]

## Thermal Efficiency of Carnot Cycle

We've touched on the idea that, even for an idealized reversible heat engine, \(100\%\) efficiency is impossible. This was first realized by the French physicist and engineer Sadi Carnot, who established the upper limit of a heat engine's thermal efficiency by considering an ideal thermodynamic process, now known as the Carnot Cycle.

The Carnot Cycle is the most efficient heat engine possible because it is a reversible process. In a reversible process, no energy is lost to the environment or through dissipative forces like friction. The defining feature of a reversible process is that there is no net entropy change in the system at the end of the process.

A **reversible ****thermodynamic process** is a process that takes a system from some initial state, through a cycle of different thermodynamic states, before returning it back to its exact initial state.

Reversible processes are never seen in nature, as it is essentially impossible to prevent frictional forces from arising either between the fluid molecules themselves or within the components of the system such as the piston in a heat engine. As such the Carnot Cycle is not a workable heat engine, however, it offers a simple illustration of the relationship between quantities such as heat, work, and temperature within heat engines. Let's look now at the specifics of the Carnot Cycle.

### Carnot Cycle

The Carnot cycle considers a heat engine as described at the start of this article whereby an ideal gas is held between two thermal reservoirs, one at \(T_\mathrm{H}\) and the other at \(T_\mathrm{C}\), with \(T_\mathrm{H}>T_\mathrm{C}\). Work can be done by the gas on its environment (or vice versa) via a moveable piston. The cycle is comprised of four different thermodynamic processes, Isothermal Expansion, Isentropic Expansion, Isothermal Compression, and Isentropic Compression. That list is quite a mouthful so let's go over some definitions. Firstly, compression and expansion refer to the effect of the process on the gas' volume.

**Compression - **The process of reducing the volume of gas by allowing the environment to do work on it. In the case of a heat engine, this work is done by moving a piston downwards.

**Expansion - **The process of increasing the volume of a gas by allowing it to do work on its environment. In a heat engine, the gas expands by pushing up the piston.

Whilst, isothermal and isentropic refer to the conditions under which the process occurs, and which quantity remains constant throughout.

**Isothermal - **A thermodynamic process during which the system maintains a constant temperature.

**Isentropic **- This is a reversible adiabatic process, meaning that there is no net change in entropy at the end of the process (reversible) and that no heat is exchanged between the system and its environment throughout the process (adiabatic).

With these definitions in mind, let's go through the four stages of the Carnot Cycle.

**Isothermal Expansion:**Initially, the ideal gas is in thermal contact with the hot reservoir, whilst thermally insulated from the cold reservoir. The gas is at a temperature infinitesimally smaller than the hot reservoir to allow heat transfer to occur without any change in the gas' temperature. This transfer of heat \(Q_\mathrm{H}\) causes the gas to expand with all the heat energy used up as work pushing the piston up, hence there is no temperature change. Due to the Ideal Gas Law, there is a corresponding drop in pressure as the gas expands at a constant temperature. The heat transfer corresponds to an entropy increase in the gas.\[\Delta S_\mathrm{H}=\frac{Q_\mathrm{H}}{T_\mathrm{H}}.\]

**Isentropic Expansion:**The gas is then thermally insulated from both reservoirs so no heat transfer can occur. However, expansion continues due to the increase in pressure causing the gas to do work on the piston. This work done by the gas causes a reduction in its internal energy, hence the gas cools to a temperature that is infinitesimally higher than \(T_\mathrm{C}\). As there is no heat transfer, there is no change in entropy.**Isothermal Compression:**Now the gas is thermally insulated from the hot reservoir but is in thermal contact with the cold reservoir. The piston does work on the gas compressing it, with all this work converted to waste heat \(Q_\mathrm{C}\) lost to the cold reservoir, so there is no change in temperature. The compression at constant temperature causes an increase in pressure. There is a reduction in the entropy of the gas given by \(\Delta S_\mathrm{C}=\frac{Q_\mathrm{C}}{T_\mathrm{C}}\).**Isentropic Compression:**The gas is thermally insulated from both reservoirs again with work continuing to be done on it by the environment. This work done on the gas increases the internal energy of the gas rising the temperature back up to being infinitesimally less than \(T_\mathrm{H}\) and returning the system to its initial state. There is no change in entropy at this point, as there is no heat transfer.

The Carnot Cycle is often represented as a closed path around a pressure-volume graph, as shown in figure 3. The curve AB follows a path at a fixed temperature, known as an *isotherm*, representing the initial isothermal compression. BC represents the adiabatic compression seen by the fact that no heat transfer is annotated on this path. The cycle is then completed by the curve CD following a lower temperature isotherm before DA returns the system to its initial state. Note the work done by the system is given by the area enclosed by the curve.

So, what can this cycle tell us about the thermal efficiency of a heat engine? Well, let's first look at how the entropy of the system changes throughout the process. As a reversible process, there can be no net entropy change in the Carnot cycle. Hence \[\Delta S_\mathrm{H}+\Delta S_\mathrm{C}=\Delta_{\text{net}}=0\]

Applying the definition of entropy, \(\Delta S=\frac{Q}{T}\), we find a relation between the heat transferred between the system and the thermal reservoirs, and the temperature of those thermal reservoirs.\[\begin{align}\frac{Q_\mathrm{H}}{T_\mathrm{H}}+\frac{Q_\mathrm{C}}{T_\mathrm{C}}&=0\\\implies \frac{Q_\mathrm{H}}{T_\mathrm{H}}&=-\frac{Q_\mathrm{C}}{T_\mathrm{C}}\\\implies \frac{Q_\mathrm{C}}{Q_\mathrm{H}}&=-\frac{T_\mathrm{C}}{T_\mathrm{H}}\end{align}\]

Applying this to the definition of efficiency gives\[\eta=1+\frac{Q_\mathrm{C}}{Q_\mathrm{H}}=1-\frac{T_\mathrm{C}}{T_\mathrm{H}}.\]

This is the central property of reversible heat engines, their efficiency is determined *only *by the temperatures of the reservoirs they work between. The greater the ratio between \(T_\mathrm{H}\) and \(T_\mathrm{C}\), with \(T_\mathrm{C}<T_\mathrm{H}\) by definition, the more efficient the heat engine will be. Again this tells us that the thermal efficiency can never be greater than one or less than zero. However, it also tells us that in order for the thermal efficiency to equal \( 1,\) the cold reservoir must be at absolute zero \(T_\mathrm{C}=0\,\mathrm{K}\). This fact has incredibly important consequences not only for real heat engines but for all of thermodynamics.

If a reversible heat engine operates between two thermal reservoirs, one at \(T_1=300\,\mathrm{K}\) and one at \(T_2=400\,\mathrm{K}\), what is its thermal efficiency?

Answer:

Using the equation given above we find the efficiency to be \[\begin{align}\eta&=1-\frac{T_1}{T_2}\\&=1-\frac{300\,\mathrm{K}}{400\,\mathrm{K}} \\&=25\%.\end{align}\]

## 2^{nd} Law Thermal Efficiency

Sadie Carnot realized that, due to the lack of any energy loss in a reversible engine, the efficiency of a reversible heat engine is the maximum possible efficiency of any heat engine. This was summarized in his highly influential theorem.

**Carnot's Theorem** states that a heat engine operating between two thermal reservoirs \(T_\mathrm{H},T_\mathrm{C}\) can't have an efficiency greater than that of a reversible heat engine (Carnot heat engine) operating between the same two temperatures.

As we have seen the efficiency of a reversible heat engine is determined only by the temperatures of the thermal reservoirs it operates between.

\[\eta=1-\frac{T_\mathrm{C}}{T_\mathrm{H}}.\]

This equation tells us that the only way for \(\eta\) to equal one is if \(T_\mathrm{C}=0\,\mathrm{K}\), known as absolute zero. However, the third law of thermodynamics forbids any system from ever reaching absolute zero, hence we see that there is no way for our reversible heat engine to have perfect efficiency.

**The Third Law of Thermodynamics**states that it is impossible for a thermodynamic system to reach absolute zero in a finite number of steps

So if it is impossible for a Carnot engine to have an efficiency of one, by the third law, and no heat engine can have a greater efficiency than a Carnot engine, by Carnot's Theorem, then all engines must have a thermal efficiency of less than one.\[\begin{align}\eta&<1\\\frac{W}{Q_\text{in}}&<1\\W&<Q_\text{in}.\end{align}\]

This demonstrates that heat energy can never be fully converted into work during a cyclic process such as the heat engine. This fact is known as Kelvin's Statement of the Second Law of Thermodynamics.

**Kelvin's Statement of the 2 ^{nd} Law of Thermodynamics: **It is impossible for heat to be converted completely during a cyclic process.

## Thermal Efficiency - Key takeaways

- A Heat Engine is a thermodynamic system that converts energy transferred to it as heat into useful work. A common model for a heat engine is a gas in thermal contact with two reservoirs, one at a higher temperature than the gas \(T_\mathrm{H}\) and one at a lower temperature \(T_\mathrm{C}\).
- The thermal efficiency of a heat engine is the ratio of work done by the system to the heat input received by the system\[\eta=\frac{W}{Q}.\]
- The thermal efficiency of a heat engine can be written in terms of heat input from the hot reservoir \(Q_\mathrm{H}\) and the waste heat lost to the cold reservoir \(Q_\mathrm{C}\)\[\eta=\frac{Q_\mathrm{H}-Q_\mathrm{C}}{Q_\mathrm{H}}=1-\frac{Q_\mathrm{C}}{Q_\mathrm{H}}.\]
- Combustion engines, fridges, and steam engines are all types of heat engines, with a fridge operating in reverse to a conventional heat engine and doing work to remove heat from a cold reservoir and dump it into a hot reservoir.
- The Carnot Cycle demonstrates that even for reversible engines, where no energy is lost as friction, it is impossible for a heat engine to be perfectly efficient.
- For a reversible engine, the efficiency is determined by the temperature of the reservoirs.\[\eta=1-\frac{T_\mathrm{C}}{T_\mathrm{H}}.\]

## References

- Fig. 1 - Heat Engine, StudySmarter Originals.
- Fig. 2 - Fridge Cycle, StudySmarter Originals.
- Fig. 3 - Sadi Carnot Boilly 1813 (https://commons.wikimedia.org/wiki/File:Sadi_Carnot_Boilly_1813.jpg) by Louis-Leopold Boilly is under Public Domain.
- Fig. 4 - Carnot cycle on pV diagram (https://commons.wikimedia.org/wiki/File:Carnot_cycle_pV_diagram.svg) by Cristian Quinzacara (https://commons.wikimedia.org/wiki/User:Cristian_Quinzacara) is licenced under CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/deed.en)

###### Learn with 10 Thermal Efficiency flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Thermal Efficiency

What is the difference between thermal efficiency and mechanical efficiency?

Mechanical efficiency is the ratio of power delivered by a mechanical system to the power received by the system, where this power can come from any source. Thermal efficiency is the ratio of work done by a heat engine to the heat supplied to the system.

How do you calculate thermal efficiency?

Thermal efficiency is the ratio of work done W to heat received Q. Efficiency=W/Q.

What is thermal efficiency?

Thermal efficiency is the ratio of work done by a heat engine to the heat supplied to the system.

Why is thermal efficiency important?

Thermal efficiency determines how effectively a heat engine performs work. The better a heat engines efficiency the less heat is wasted by the engine. This is very important in the automotive industry.

How to improve thermal efficiency?

The Carnot cycle shows that the efficiency of a reversible heat engine is dependent only on the temperature of the thermal reservoirs it works between. The efficiency of the engine is improved by increasing the difference in temperature of the two reservoirs.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more