CARNOT CYCLE: STAGES, APPLICATIONS, EXAMPLES, EXERCISES - PHYSICAL - 2024

The Carnot cycle is the sequence of thermodynamic processes that take place in a Carnot engine, an ideal device that consists only of reversible-type processes; that is, those that having taken place, can return to the initial state.

This type of motor is considered ideal, since it lacks the dissipation, friction or viscosity that arise in real machines, converting thermal energy into usable work, although the conversion is not carried out 100%.

Figure 1. A steam locomotive. Source: Pixabay

An engine is built starting from a substance capable of doing work, such as gas, gasoline or steam. This substance is subjected to various changes in temperature and in turn experiences variations in its pressure and volume. In this way it is possible to move a piston within a cylinder.

What is the carnot cycle?

The Carnot cycle takes place within a system called the Carnot engine or C, which is an ideal gas enclosed in a cylinder and provided with a piston, which is in contact with two sources at different temperatures T ₁ and T ₂ as the shown in the following figure to the left.

Figure 2. On the left a diagram of the Carnot machine, on the right the PV diagram. Left figure source: From Keta - Own work, CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=681753, right figure Wikimedia Commons.

There, the following processes happen roughly:

1. A certain amount of heat Q _input = Q ₁ is supplied to the device from the high temperature thermal reservoir T ₁.
2. Carnot's engine C performs work W thanks to this supplied heat.
3. A part of the heat used: the waste Q _output, is transferred to the thermal tank that is at a lower temperature T ₂.

Stages of the Carnot cycle

The analysis is carried out using a PV (Pressure –Volume) diagram, as shown in figure 2 (right figure). The purpose of the motor may be to keep the thermal reservoir 2 cool, extracting heat from it. In this case it is a cooling machine. If, on the other hand, you want to transfer heat to thermal tank 1 then it is a heat pump.

The pressure-temperature changes of the motor under two conditions are shown in the PV diagram:

- Keeping the temperature constant (isothermal process).

- No heat transfer (thermal insulation).

The two isothermal processes need to be connected, which is achieved by thermal insulation.

Point

You can start at any point in the cycle, in which the gas has certain conditions of pressure, volume and temperature. The gas undergoes a series of processes and can return to the starting conditions to start another cycle, and the final internal energy is always the same as the initial one. Since energy is conserved:

The area within this loop or loop, in turquoise in the figure, is precisely equivalent to the work done by the Carnot engine.

In figure 2 points A, B, C and D are marked. We will start at point A following the blue arrow.

First stage: isothermal expansion

The temperature between points A and B is T ₁. The system absorbs heat from the thermal tank 1 and undergoes an isothermal expansion. Then the volume increases and the pressure decreases.

However, the temperature remains at T ₁, since when the gas expands it cools. Therefore, its internal energy remains constant.

Second stage: adiabatic expansion

At point B the system begins a new expansion in which the system does not gain or lose heat. This is achieved by putting it in heat insulation as indicated above. Therefore it is an adiabatic expansion that continues to point C following the red arrow. The volume increases and the pressure decreases to its lowest value.

Third stage: isothermal compression

It begins at point C and ends at D. The insulation is removed and the system comes into contact with thermal tank 2, whose temperature T ₂ is lower. The system transfers waste heat to the thermal reservoir, the pressure begins to increase and the volume to decrease.

Fourth stage: adiabatic compression

At point D, the system goes back to thermal insulation, the pressure increases and the volume decreases until it reaches the original conditions of point A. Then the cycle repeats again.

Carnot's theorem

Carnot's theorem was first postulated in the early 19th century by the French physicist Sadi Carnot. In the year 1824 Carnot, who was part of the French army, published a book in which he proposed the answer to the following question: under what conditions does a heat engine have the maximum efficiency? Carnot then established the following:

The efficiency η of a heat engine is given by the quotient between the work done W and the heat absorbed Q:

In this way, the efficiency of any heat engine I is: η = W / Q. While the efficiency of a Carnot R motor is η´ = W / Q´, assuming both motors are capable of doing the same work.

Carnot's theorem states that η is never greater than η´. Otherwise it falls into contradiction with the second law of Thermodynamics, according to which a process in which the result is that heat comes out of a lower temperature body to go to a higher temperature without receiving external help is impossible. Thus:

η _< η ^'

Proof of Carnot's theorem

To show that this is so, consider the Carnot engine acting as a cooling machine driven by an I engine. This is possible since the Carnot engine works by reversible processes, as specified at the beginning.

Figure 3. Proof of Carnot's theorem. Source: Netheril96

We have both: I and R working with the same thermal reservoirs and it will be assumed that η _> η ^'. If along the way a contradiction is reached with the second law of thermodynamics, Carnot's theorem is proved by reduction to the absurd.

Figure 3 helps you follow the process. The engine I takes an amount of heat Q, which it divides in this way: doing work on R equivalent to W = ηQ and the rest is the transferred heat (1-η) Q to the thermal reservoir T ₂.

Since energy is conserved, all of the following are true:

E _input = Q = Work W + heat transferred to T ₂ = ηQ + (1-η) Q = E _output

Now the Carnot refrigerating machine R takes from the thermal reservoir 2 an amount of heat given by:

(η / η´) (1-η´) Q =

Energy must also be conserved in this case:

E _input = ηQ + (η / η´) (1-η´) Q = (η / η´) Q = Q´ = E _output

The result is the transfer to the thermal reservoir T ₂ of a quantity of heat given by (η / η´) Q = Q´.

If η is greater than η´, it means that more heat has reached the thermal deposit of higher temperature than I originally took. Since no external agent, such as another heat source, has participated, the only way that could happen is for the cooler thermal reservoir to give up heat.

This is in disagreement with the second law of thermodynamics. It is then concluded that it is not possible that η ^' is less than η, therefore the engine I cannot have more efficiency than the Carnot R engine.

Corollary of the theorem and limitations

The corollary of Carnot's theorem states that two Carnot machines have the same efficiency if they both operate with the same thermal reservoirs.

That means no matter the substance, the performance is independent and cannot be raised by changing it.

The conclusion from the above analysis is that the Carnot cycle is the ideally achievable top of the thermodynamic process. In practice there are many factors that decrease efficiency, for example the fact that the insulation is never perfect and in the adiabatic stages there is actually heat exchange with the outside.

In the case of a car, the engine block gets hot. On the other hand, the mixture of gasoline and air does not behave exactly like an ideal gas, which is the starting point of the Carnot cycle. This is to mention just a few factors that will cause a drastic reduction in performance.

Examples

A piston inside a cylinder

If the system is a piston enclosed in a cylinder as in figure 4, the piston rises during isothermal expansion, as seen in the first diagram on the extreme left, and also rises during adiabatic expansion.

Figure 4. Movement of a piston inside a cylinder. Source: self made.

It is then compressed isothermally, giving up heat, and continues to compress adiabatically. The result is a movement in which the piston rises and falls inside the cylinder and that can be transmitted to other parts of a particular device, such as a car engine for example, that produces a torque, or a steam engine.

Various reversible processes

In addition to the expansion and compression of an ideal gas inside a cylinder, there are other ideal reversible processes with which a Carnot cycle can be configured, for example:

- Back and forth movements in the absence of friction.

- An ideal spring that compresses and decompresses and never deforms.

- Electric circuits in which there are no resistances to dissipate energy.

- Magnetization and demagnetization cycles in which there are no losses.

- Charging and discharging a battery.

A nuclear power plant

Although it is a very complex system, a first approximation of what is required to produce energy in a nuclear reactor is as follows:

- A thermal source, consisting of a radioactively decaying material such as uranium.

- The cold heat sink or reservoir that would be the atmosphere.

- The “Carnot engine” that uses a fluid, almost always running water, to which heat is supplied from the thermal source to convert it into steam.

When the cycle is carried out, electrical energy is obtained as net work. When being transformed into steam at high temperature, the water is made to reach a turbine, where the energy is transformed into motion or kinetic energy.

The turbine in turn drives an electrical generator that transforms the energy of its movement into electrical energy. In addition to fissile material such as uranium, fossil fuels can of course be used as a heat source.

Solved exercises

-Example 1: efficiency of a heat engine

The efficiency of a heat engine is defined as the quotient between the output work and the input work, and is therefore a dimensionless quantity:

Denoting the maximum efficiency as e _max, it is possible to show its dependence on temperature, which is the easiest variable to measure, as:

Where T ₂ is the temperature of the sink and T ₁ is the temperature of the heat source. Since the latter is higher, the efficiency always turns out to be less than 1.

Suppose you have a heat engine capable of operating in the following ways: a) Between 200 K and 400 K, b) Between 600 K and 400 K. What is the efficiency in each case?

Solution

a) In the first case the efficiency is:

b) For the second mode the efficiency will be:

Although the temperature difference is the same between both modes, the efficiency is not. And even more remarkable is that the most efficient mode operates at a lower temperature.

-Example 2: heat absorbed and heat transferred

A 22% efficient heat engine produces 1,530 J of work. Find: a) The amount of heat absorbed from thermal tank 1, b) The amount of heat discharged to thermal tank 2.

a) In this case, the definition of efficiency is used, since the work carried out is available, not the temperatures of the thermal tanks. A 22% efficiency means that e _max = 0.22, therefore:

The amount of heat absorbed is precisely Q _input, so solving for we have:

b) The amount of heat transferred to the coldest tank is found from Δ W = Q _input - Q _output

Another way is from e _max = 1 - (T ₂ / T ₁). Since temperatures are not known, but they are related to heat, efficiency can also be expressed as:

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 654-657
2. Nuclear energy. Operation of a nuclear power plant. Recovered from: energia-nuclear.net
3. Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Ed. Cengage Learning. 618-622.
4. Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. MacGraw Hill. 414-416.
5. Walker, J. 2008. Physics. 4th Ed. Addison Wesley. 610-630