Thermal Efficiency

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 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 \(0\leq\eta\leq1\).

Let's look into an example.

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).

Fig. 2 - A fridge can be thought of as a heat engine operating in reverse, extracting heat from a cold reservoir and pumping it into a hot reservoir by inputting work.

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.

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.

Fig. 3 - Sadi Carnot (1796-1832) is dubbed the 'Father of Thermodynamics' thanks to his work on heat engines and thermal efficiency.

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.

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.

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

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

1. 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.

   \[\Delta S_\mathrm{H}=\frac{Q_\mathrm{H}}{T_\mathrm{H}}.\]

2. 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.

3. 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 \(\Delta S_\mathrm{C}=\frac{Q_\mathrm{C}}{T_\mathrm{C}}\).

4. 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.

Fig. 3 - The Carnot Cycle can be represented as a closed curve on a Pressure-Volume graph as shown above. Step 1. follows the curve AB, 2. follows BC, 3 follows CD, and 4 follows DA.

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.

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.

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.

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.