Table of Contents

In this article, we were discussing the Carnot cycle, Carnot cycle efficiency, and other related terms for the Carnot heat engine.

This article is quite long, but you should have a complete understanding of the terms which were used later in the Carnot cycle.

Let us begin…..

**Terms used in Carnot cycle – A look-up**

1. Laws of thermodynamics that are involved in the Carnot cycle. |

2. What are the Spontaneous and Non-spontaneous processes? |

3. What are the Reversible and irreversible processes? |

4. what is the heat engine? |

5. Heat engine efficiency. |

6. What is a Carnot heat engine and a Carnot cycle? |

7. Complete Carnot cycle processes with diagrams and derivations. |

8. Carnot cycle PV diagram. |

9. Carnot heat engine efficiency with examples. |

10. Carnot theorem. |

11. The refrigerator as a reversed Carnot cycle. |

12. Co-efficient of performance. |

13. Entropy and Carnot cycle. |

14. Carnot cycle TS diagram. |

As you know, **thermodynamics** is the branch of science which deals with the study of the conversion of heat energy into other forms of energy and gives information about this conversion quantitatively.

**Isothermal process – Carnot Cycle **

In thermodynamics, if the temperature change in a thermodynamics process is constant, then the process is called an isothermal process.

As in Carnot heat engine, **ideal gas** as a working substance is used, then for an ideal gas isothermal process,

**PV = Constant**

Here **P** is the pressure of the gas,

**V** is the volume of the gas and,

the temperature **T** is constant.

**Adiabatic process**

In a thermodynamical system, no heat transfer from the system to the surrounding or from surrounding to the system means all heat contained in the system remains constant.

If any work is done in the adiabatic process, then this can only happen when the system’s internal energy (**U**) changed.

For an ideal gas,

**P(V) ^{γ }**= Constant

Here Gamma (**γ**)is the ratio of two specific heats of the gas at constant volume and constant pressure.

**Laws of thermodynamics**

Carnot heat engine obeys the first law and also the second law of thermodynamics.

While the Carnot cycle laid the foundation of the second law of thermodynamics.

let us discuss these laws.

**The first law of thermodynamics**

According to the first law of thermodynamics, if a certain amount of work is done, an equivalent amount of heat is produced and if a certain amount of heat is absorbed, an equivalent amount of work is done.

It means **W = Q **or** Q = W**

**Q** is the heat and **W** is the work done.

The First law of thermodynamics is also the law of conservations of energy.

**Law of conservation of energy**

Energy can neither be created nor be destroyed.

On behalf of this law, you can say that whole energy in the universe is conserved.

These conversions of energies are interconvertible.

**The second law of thermodynamics**

Many statements are given by scientists for the **2nd** law of thermodynamics. Such as the statement by Lord Kelvin, Clausius, Planck-Kelvin, etc.

In thermodynamics, it was seen later that, all these statements are equivalent to each other.

**Kelvin Planck’s statement for the second law of thermodynamics**

It is impossible to construct a heat engine operating in cycles, which extract heat from a single thermal reservoir and converts the whole of it into work.

The second law is the most fundamental law of nature.

The limitations of the first law encouraged scientists to find the second law of thermodynamics.

As you know the first law only tells us the conversion of energies into other forms of energy.

But this law did not give information about the quantity and direction-flow of conversions and many other reverse parameters.

You can say the second law of thermodynamics is the law of nature.

**Spontaneous and Non-spontaneous**

Before you moved to the Carnot cycle, you should know about the spontaneous and non-spontaneous processes?

Carnot heat engine and all other heat engines follow the second law of thermodynamics.

The second law of thermodynamics is correlated with the spontaneous and non-spontaneous processes.

In a simple way, you can say that by the study of **2nd** law you may tell more about the spontaneous and non-spontaneous process.

**Spontaneous process**

A spontaneous process, which has a tendency to do occur by itself in nature means no need for input energy from outside sources.

Also, a process is spontaneous if some initial effort is done and sometimes it is conditional.

**For example**

waterfalls (water itself falls from upper to lower surface) or if we move a ball on an inclined plane from top to bottom,

Iron rust, mixing sugar in water,

Melting of ice at different conditions,

The process of changing diamond into graphite at certain conditions, the process of decay of radioactive isotopes, etc.

In these examples some of the processes happen only by themselves, Such as waterfalls, mixing of sugar in water, a ball rolling top to down, etc.

These processes are naturally occurring and also condition-based.

If we move water with a spoon, the mixing rate of sugar is more than by itself.

Similarly in salt.

In the case of a ball, if we give a little push to the ball, after then the ball start continues to rolling itself on the inclined plane from top to down.

**Non-spontaneous process**

If we reverse the spontaneous process, which is not possible by itself, then it called a non-spontaneous process.

These processes need external effort in the form of energy or work.

**Example**

The water does not move by itself from down to top.

You need a water pump to continue this process.

**Reversible and Irreversible process**

A Carnot heat engine can work in a reverse direction, so here, we take a short view of reversible and irreversible processes.

**Reversible process**

A process is reversible if the system and surroundings came back to their original state in the opposite direction, after passing through the same intermediate conditions as during the direct process.

There are infinitesimal changes during the reversible process.

A reversible process can take an infinite amount of time to complete.

In actual practice, a perfect reversible process is impossible.

**Examples**

Some processes can be considered as a reversible process,

such as an **isothermal expansion and compression** in the case of an ideal gas at a constant temperature **in a cyclic process.**

If we **compress a spring** at a slow rate, then the work done by spring during expansion is equal to the applied work during compression.

**Irreversible process**

Any process, which is not reversible is called an irreversible process.

**Examples**

There are lots of examples of irreversible processes.

Some of these are,

**Transfer of heat by radiation.****The heat produced during the current flow through a resistance.****The heat produced during relative motion with friction.****Throttling****The plastic deformation****Diffusion, etc.**

Now you understood a few concepts and terms used in heat engines or Carnot cycle.

Let us talk about the heat engine and the Carnot heat engine process, by keeping these terms in mind.

**What is a heat engine?**

A heat engine is a device or machine, which converts heat energy into useful work.

**Heat engine examples**

The internal combustion engine,

The thermal power plant,

A steam locomotive engine, etc.

**Heat engine construction and working**

In a heat engine, a source (**Heat reservoir**) is used to transfer heat to Engine (**E**)at high-temperature **T _{1.}**

Let **Q**_{1 }be the heat absorbed by the engine from the source.

**Work** is performed by the heat engine = **W**

According to the second law of thermodynamics,

“All work can be converted into heat, but all heat can not be converted into work”

It means some amount of heat is always lost.

Let the heat loss is equal to **Q _{2}**.

Therefore work done is, **W = Q _{1}-Q_{2 }**

To full fill, this requirement a sink (**Heat reservoir**) at low-temperature **T _{2 }**is used in a heat engine.

As you know heat engines obey both the **first** law of thermodynamics and the **second** law of thermodynamics.

From the above figure, it is clear that a heat engine is a device, which converts heat into work.

The term “law of conversation of energy” is not violated by heat engines.

On another hand, we can say that the heat engine obeys the first law of thermodynamics.

According to the **first** law of thermodynamics, conversions of energies are interconvertible.

Hence, **W** =** Q** **⇒** Work is done = Heat produced.

According to the Clausius, the derivation of the **first** law,

As in heat engines, the working is a cyclic process, therefore,

internal energy change (**ΔU**) = **0**

This is because for a cyclic process change in internal energy (**ΔU**) is always Zero (**0**).

Hence proved, **Q** = **W **

**Heat engine efficiency**

The efficiency of a heat engine is the ratio of output mechanical energy to the input heat energy.

The efficiency of heat engines is denoted by a symbol named as, **eta** = **η**

The formula of heat engine efficiency (**η**)** = W/Q _{1}**

Here **W** = output mechanical work during the complete cycle.

**Q _{1} **= Heat absorbing from the source at a high temp during the complete cycle.

As we know **W** = **Q** ** ⇒** ** Q _{1}-Q_{2 }**

after using this, **η = Q _{1}−Q_{2}/Q_{1}**

_{}

**What is a Carnot heat engine?**

The Carnot heat engine is a hypothetical heat engine, which is given by A French mechanical engineer **Nicolas Léonard Sadi Carnot.**

He was a **physicist, military scientist** in the **French army**.

The Carnot Sadi was born on 1 June 1796 and died at the age of 36 years on 24 August 1832 in France.

He published a book at the age of 27 years in 1824 named “**Reflections on the Motive Power of Fire”.**

The Carnot heat engine is a reversible heat engine working between cycles of operations.

So, in the Carnot heat engine, the **ideal gas**, which is used as a working substance comes back to its initial state at the end of each cycle.

**What is the Carnot cycle?**

The Carnot heat engine works on the four reversible processes.

Isothermal gas expansion

Adiabatic gas expansion

Isothermal gas compression

Adiabatic gas compression

In the Carnot heat engine, the working series of these four processes is called the Carnot cycle.

**The Carnot cycle construction**

As you know that the Carnot heat engine works on the four operations, called cycle.

In the making of this type of cycle, the Sadi Carnot uses two different heat reservoir (Source and sink) at different temperature range, **T _{1}** and

**T**

_{2. }

A cylinder, a piston, and also an insulating stand.

Let us talk about these requirements in detail.

**♦ Cylinder **

A cylinder that has perfectly non-conducting walls and the base of the cylinder is perfectly conducting.

A frictionless, weightless, and Perfectly Non-conducting piston is used to fit inside the cylinder.

An ideal gas is used in the cylinder as a working substance.

**♦ Source**

It is an infinite thermal capacity heat reservoir at a fixed high-temperature **T _{1}**.

Infinite thermal capacity means during adding and subtracting of heat from the reservoir does not change its temperature **T _{1}**.

**♦ Sink**

It is a second heat reservoir, which has also infinite heat capacity at a fixed low-temperature **T _{2}**.

**♦ Perfectly Non-Conducting stand**

A perfectly Non-Conducting stand is used during operations in the Carnot cycle process.

**Carnot cycle PV diagram**

The processes during the Carnot cycle and total work done can be shown by the PV diagram.

In this pressure and volume graph, change in pressure shown vertically, and change in volume shown horizontally directions.

This Carnot cycle PV diagram shows all four reversible processes and their changing values.

The total area ABCDA inclosed by the PV diagram is the work done during a complete Carnot cycle.

**Carnot cycle working**

For changing the pressure during four processes, some weights were added on the piston and some weight removed from the piston.

**1 Reversible Isothermal expansion**

In this process, the cylinder is placed on the source.

As the base of the cylinder is perfectly conducting, then let **Q _{1}** be the heat absorbed from the source at a temperature of

**T**by the gas.

_{1}Now slowly remove the weight.

Due to this gas expand at a constant temperature **T _{1}**, as a result, the volume increases and the pressure decreases.

As you know, if the gas is expanding slowly at a constant temperature and the pressure-volume changes then the process is an isothermal expansion.

In this process, work **W** is done by the gas on the surroundings is equal to **Q**_{ }heat absorbed from the source.

Let in the initial state the gas has **P _{1}**,

**V**,

_{1}**T**are the respectively the pressure, volume, and temperature.

_{1}**See in the PV diagram**

After expansion, the pressure decreases from **P _{1}** to

**P**, the volume increases from

_{2}**V**to

_{1}**V**, and the temperature remains constant at

_{2}**T**.

_{1}The **work done **

**W = Q − ΔU**

But in this process, internal energy (**U**) is not changing, as the temperature is constant, the work is done only due to heat absorbed.

The internal energy is temperature-dependent.

Therefore, the change in internal energy (**ΔU**) is equal to zero (**0**).

**W = Q − 0 ⇒ W = Q**

As we are in the first process of the cycle then,

**ΔU = 0** can be denoted by** ΔU _{1} = 0**

**W _{1} = Q_{1} ⇒** Work done for curve

**AB**on the

**PV**diagram.

Now applying the work done formula for the Isothermal process in the ideal gas case.

Which is **W = nRT log _{e} V_{2}/V_{1 }**

(Where n = no. of moles of gas, **R** is gas constant, and **V _{2}, V_{1}** are volume changes and

**T**is the temperature).

Hence, **W _{1} = Q_{1} =nRT_{1 }log_{e }V_{2}/V_{1} ** (For first process temperature

**T**is

**T**)

_{1}This is the work done in the **first** process.

**2** **Reversible Adiabatic Expansion**

In this process, the cylinder removed from the source and placed on the perfectly Non-conducting stand.

As the stand is perfectly non-conducting, no heat transfer takes place during this process.

Now remove the weight fast and gas start expanding.

Now the work is done due to a decrease in internal energy (**U**) of gas and because of it, the temperature starts falling.

As you know, if the gas expanding and no heat transfer takes place from the system to the surrounding or from the surrounding to the system, then the process is an adiabatic expansion.

Removing of weight is until we reach to the sink’s temperature, which is **T _{2}**.

Let, due to a decrease in internal energy (**U**) the temperature falls from **T _{1}** to

**T**equal to sink temperature.

_{2}The volume and pressure also change, volume increases from **V _{2}** to

**V**and pressure decreases from

_{3}**P**to

_{2}**P**.

_{3}The **work done**

**W = Q − ΔU,**

**W = 0 − ΔU ⇒ W = −ΔU, ** (**Q = 0**, as no heat transfer)

As we are in the second process of the cycle then,

**W _{2} = −ΔU_{2 }⇒ **Work done for curve

**BC**on the

**PV**diagram.

The Work done formula in terms of internal energy change for an adiabatic process in the ideal gas case.

**W = −nR (T _{2}−T_{1})/1−γ**

Here **ΔU = nC _{v}ΔT ⇒ nR (T**

_{low}

**−T**

_{high})

For ideal gas case **⇒ nR (T _{2}−T_{1})/1−γ**

In term of the second process, the work done is,

**W _{2} = −ΔU_{2 }⇒ −nR (T_{2}−T_{1})/1−γ**

This is the work done in the **second** process.

**3 Reversible Isothermal Compression**

In this process, the cylinder removed from the stand and placed on the sink.

Some weights were added on the piston.

In this process, the temperature (**T _{2}**) of the gas inside the cylinder is the same as the temperature (

**T**) of the sink.

_{2}It means no temperature change during the process, hence temperature is constant.

The work is done in the form of heat rejected to the sink.

Let **Q _{2}** be the heat rejected to the sink at constant temperature

**T**

_{2 }as the base of the cylinder is conducting.

During this process, the volume starts to decrease from **V _{3}** to

**V**and the pressure starts to increase from

_{4}**P**to

_{3}**P**.

_{4}We know that if the gas compress at a constant temperature and the value of its pressure and volume changes then the process is Isothermal compression.

As the temperature is constant then there is no change in internal energy (**U**).

The **work done**

W** = Q − ΔU ⇒ W = Q − 0 ⇒ W = Q ** (As the **ΔU** is equal to Zero)

As we are in the third process of the cycle then,

**ΔU = 0** can be denoted by **ΔU _{3} = 0**

**W _{3} = Q_{2 }⇒ nRT_{2 }log_{e }V_{4}/V_{3 }**

**W _{3} = nRT_{2 }log_{e }V_{4}/V_{3 }⇒ **Work done for curve

**CD**on the

**PV**diagram.

This is the work done for the **third** process.

**4 Reversible Adiabatic Compression**

In this process, the cylinder removed from the sink and placed again on the stand.

Weights are added on the piston.

The work is done during this process is due to an increase in the internal energy (**U**) of gas and because of it, the temperature starts increasing.

Let the temperature change from **T _{2}** to

**T**, which is the same as the initial state of the gas.

_{1}As the stand is perfectly Non-conducting, no heat transfer takes place during this process.

As you know, if the gas compress and no heat transfer takes place from the system to the surrounding or from surrounding to the system, then the process is an adiabatic compression.

In this process, the volume decreases from **V _{4}** to

**V**and respectively the pressure increases from

_{1,}**P**to

_{4}**P**and the gas comes back to its initial state.

_{1,}The **work done**

**W = Q − ΔU,**

**W = 0 − ΔU ⇒ W = −ΔU ** (**Q = 0**, as no heat transfer)

As we in the fourth process of the cycle then,

**W _{4} = −ΔU_{4 }⇒ **Work done for curve

**DA**on the

**PV**diagram.

Here temperature changes from **T _{2}** to

**T**then,

_{1}**W _{4} = −nR (T_{1}−T_{2})/1−γ**

Here **ΔU = nC _{v}ΔT ⇒ nR(T**

_{high}

**−T**

_{low}),

**W _{4} = −ΔU_{4 }⇒ −nR(T_{1}−T_{2})/1−γ**

This is the work done by the fourth process of the cycle.

**Total work done by the Carnot cycle. **

When you study the **PV** diagram of the Carnot heat engine, you see, that the total

area enclosed by the **PV** diagram is equal to the total work done by the Carnot cycle.

Now the total area enclosed by the **PV** diagram = **Area ABCDA.**

The total work done by the Carnot cycle = **W**

**W = W _{1} + W_{2} + W_{3} + W_{4}**

Now adding the terms,

**W _{1} = nRT_{1 }log_{e }V_{2}/V_{1}**

**W _{2} = −nR (T_{2}−T_{1})/1−γ**

**W _{3} = nRT_{2 }log_{e }V_{4}/V_{3}**

**W _{4} = −nR (T_{1}−T_{2})/1−γ**

Here **W _{2 }+ W_{4 }= 0 ** (After solving)

Hence **W = w _{1} + W_{3}**

**W = nRT _{1}log_{e}V_{2}/V_{1} − nRT_{2}log_{e}V_{3}/V_{4 }** (After solving)

After solving the curves **AB, BC, CD, DA** in terms of isothermal and adiabatic processes, we get,

**V _{2}/V_{1 }= V_{3}/V_{4}**

Substitute this, **W = nRT _{1}log_{e}V_{2}/V_{1} − nRT_{2}log_{e}V_{2}/V_{1} **

Hence total work done by the carnot cycle **W = nRlog _{e}V_{2}/V_{1}[T_{1}−T_{2}]**

In this complete cycle net amount of heat absorbed **Q = Q _{1}-Q_{2}**

Hence from the first law of thermodynamics, **W = Q**

Then, **W = nR log _{e}V_{2}/V_{1 }[T_{1}−T_{2}] = Q = Q_{1}-Q_{2}**

**Carnot cycle efficiency**

The efficiency of any heat engine **(η) = W/Q _{1}**

Here, **W = Q = Q _{1}-Q_{2 }**

**= nR log _{e }V_{2}/V_{1 }[T_{1}−T_{2}] **

Putting these valves in the heat engine efficiency formula,

**η = nR log _{e }V_{2}/V_{1 }[T_{1}−T_{2}]/nRT_{1 }log_{e }V_{2}/V_{1}**

**η = [T _{1}−T_{2}]/T_{1} =**

**Carnot cycle efficiency.**

From the above relations,

**η = [T _{1}−T_{2}]/T_{1} = [Q_{1}-Q_{2}]/Q_{1 }**=

**Carnot cycle efficiency.**

**Q _{2}/Q_{1} = T_{2}/T_{1}**

From this equation, we can say that the efficiency of the Carnot heat engine or the Carnot cycle efficiency depends upon the temperature of the source and sink,

it is independent of the nature of the working substance.

**Carnot Cycle efficiency calculator**

**Why no heat engine can have an efficiency of 100% or 1?**

The efficiency of a heat engine **η = [T _{1}−T_{2}]/T_{1} = [Q_{1}-Q_{2}]/Q_{1}**

If you want to gain the efficiency equal to **1** or **100%**,

then you must have **T _{2} = 0** Kelvin and also,

**Q**

_{2}= 0**The T _{2} = 0 Kelvin, only when you have the value of heat Q_{2}= 0**

This means no loss of heat and all heat can be completely converted into work.

Which not possible, as it violates the second law of thermodynamics.

Hence, no **Q _{2} = 0** and thus no

**T**, and no heat engine can have an efficiency of

_{2}= 0**1**or

**100%**.

**Carnot theorem**

The above relation for Carnot cycle efficiency is, **Q _{2}/Q_{1} = T_{2}/T_{1 }**

It can be seen from the relation, that heat engines operating in cycles between the same

two temperatures **T _{1 }**and

**T**can not have efficiency more than the Carnot reversible heat engine.

_{2}This is the Carnot theorem, that is, **the efficiency of a Carnot reversible heat engine is maximum.**

Another form of Carnot theorem in the same sense,

**According to the Carnot theorem,** the efficiency of all reversible heat engines operating between the same two temperatures is the same (equally efficient) and no irreversible heat engines operating between the same two temperatures can have efficiency more than the Carnot heat engine.

**The refrigerator as a reversed Carnot cycle**

As you know, the processes in the Carnot cycle are reversible.

Therefore, the Carnot cycle can be used as a reverse Carnot cycle.

After this, the reversed Carnot cycle worked as a refrigerator.

For a Carnot cycle, the **Q _{1}** heat is absorbed from the source at temperature

**T**and

_{1}**Q**heat is rejected to the sink at temperature

_{2}**T**.

_{2}The second law of thermodynamics not violated here.

But in the case of a refrigerator or reverse Carnot cycle,

the **Q _{2}** heat absorbed from the sink at temperature

**T**and

_{2}**Q**heat is rejected to the source at temperature

_{1}**T**.

_{1}This is a clear violation of the second law of thermodynamics.

As the heat can not transfer by itself from a lower temperature to a higher temperature, without any external agency.

Therefore in the case of a refrigerator, some external work is done to transfer heat from a low temp body to a high temp body.

For refrigerators, this work is performed by the compressor of the refrigerator by the use of electricity.

**An interesting fact in the refrigerator**

In the refrigerator, the amount of rejected heat **Q _{1}** is more than the amount of heat absorbed

**Q**.

_{2}This is because the amount of rejected heat **Q _{1} **is equal to the sum of work done

**W**and heat absorbed

**Q**.

_{2}that is, **Q _{1} = W + Q_{2}**

**Coefficient of Performance COP**

The efficiency of the refrigerator is measured in terms of the Coefficient of performance.

The coefficient of performance is the ratio of the amount of heat absorbed from the sink to the work done by an external agency.

That is, **COP (k) = Q _{2}/W**

From the first law of thermodynamics, **W = Q**

Also,** Q = Q _{1}−Q_{2}**

**k = Q _{2}/Q_{1}−Q_{2}**

**Interesting fact!**

The efficiency of all heat engines cannot be more than **100%** but, the coefficient of performance can be **100%** or more than **100%**.

**Example of Coefficient performance, **in terms of Numerical question.

**Question:**

A Carnot’s refrigerator takes heat from water at **0°C** and leaves it in the room. The temperature of the room is **27°C. ****100** grams of water at **0°C** is to be changed into ice at **0°C**.

Find, how many calories of heat are discarded to the room? Also, find the Coefficient of performance? Given, one gram of heat = **80** calories.

**Answer:**

Her given that the temperature of water **= 0°C = T _{2} =** Sink’s temp.

**T _{2} = 0°C + 273 K = 273** Kelvin

Here, Source temp = Room’s temp **= T1 = 27°C**, Given

**T _{1} = 27°C + 273 K = 300** Kelvin,

Let **Q _{2}** and

**Q**be the heat absorbed from the sink and heat rejected to the source.

_{1}Given, that **Q _{2} = 100** gram of water

**= 100 × 80**calories =

**8000**calories,

Applying carnot cycle efficiency equation,

**Q _{2}/Q_{1} = T_{2}/T_{1 }**

**8000/Q _{1} = 273 K/300 K**

**Q _{1} =** approximately

**8791**calories.

The work is done by the compressor of the refrigerator

**W = Q _{1}−Q_{2 }= 8791−8000 = 791**

Then the coefficient of performance COP,

**K = Q _{2}/W = 8000/791 = 10.1**

Also, **Q _{1} = 8791** calories and

**Q**calories

_{2}= 8000From this example, we can say that the quantity of rejected heat by the refrigerator is much higher than the total quantity of heat absorbed from the sink.

**Entropy and Carnot cycle**

As you know **dS = dQ/T ⇒ dQ = TdS**

The above relation is also the second law of thermodynamics, in fact, it is the mathematical formulations of the second law of thermodynamics.

In the Carnot, cycle entropy is used to draw a **TS** diagram to find Carnot cycle efficiency in terms of temp and entropy.

Also, the area of the **TS** diagram of the Carnot cycle is equal to the area of the **PV** diagram for the same Carnot cycle.

**dQ = dW + dU** **⇒ **Frist law of thermodynamics,

**dQ = TdS ** **⇒ **Second law of thermodynamics,

**dW + dU = TdS **

** dU** is zero as internal energy is a state function and change in internal energy for a reversible cycle is **0**.

Hence mathematically,

**∫PdV = ∫TdS**

**Carnot cycle TS digram**

The above diagram represents a graph between temperature **T** and entropy **S** for a reversible Carnot cycle.

Like the **PV** diagram, the processes for a **TS** diagram of a complete Carnot cycle is similar, such as,

Line **AB** = Isothermal expansion, where the temperature **T _{1}** = Constant.

Line **BC** = Adiabatic expansion, where the Entropy **S _{2}** = Constant.

**CD** line = Isothermal compression, where the temperature **T _{2}** = Constant.

and line **DA** = Adiabatic compression, where the Entropy **S _{1} **= Constant.

Both the Lines **AB** (Isothermal expansion) and **CD** (Isothermal compression) are parallel to the entropy axis.

Here change in temperature **ΔT = 0**

The lines **BC** (Adiabatic expansion) and **DA** (Adiabatic compression) are parallel to the temperature axis.

Here change in entropy **ΔS = 0**

Now the quantity of heat absorbed by the source during the complete Carnot cycle = **Q _{1}**

**Q _{1} = +ΔS**

For curve **AB** on the **TS** diagram.

The first process, which is reversible Isothermal expansion **Q _{1} = TΔS**

**Q _{1} = T_{1}(S_{2}−S_{1})**

For the curve **BC** on the **TS** diagram. **⇒ **(Second process, Adiabatic expansion)

**ΔS = 0** (constant)

For the curve **CD** on the **TS** diagram. **⇒ **(Third process, Isothermal compression)

**Q _{2} = −ΔS = −T_{2}(S_{1}−S_{2}) = T_{2}(S_{2}−S_{1})**

For the curve **DA** on the **TS** diagram **⇒ **(Fourth process, Adiabatic compression)

**ΔS = 0** (constant)

Net work done during complete process **W = W _{1} + W_{3} = Q_{1} − Q_{2}**

**W = (T _{1}−T_{2}) (S_{2}−S_{1}) ⇒ **Enclosed Area of complete ABCDA Carnot cycle on the

**TS**diagram.

**Carnot cycle TS diagram efficiency.**

The Carnot cycle efficiency in terms of **TS** diagram

The efficiency of a heat engine **(η) = W/Q _{1}**

**As, W = Q _{1}−Q_{2}**

**η = Q _{1}−Q_{2}/Q_{1}**

**(T _{1}−T_{2}) (S_{2}−S_{1})/T_{1}(S_{2}−S_{1}) = T_{1}T_{2}/T_{1} =** Carnot cycle efficiency.

**Carnot cycle Conclusion**

In the Carnot cycle, all processes are reversible but in nature, no process is perfectly reversible. So, the concept of the Carnot cycle is hard to digest.

The idea of the weightless or frictionless piston is also imaginary (hypothetical).

Hence, you can say that the setup of the Carnot heat engine Purely imaginary concept, and in real life, no heat engine exists.

That’s why the Carnot cycle is not used in power plants, automobile industries, etc.

Although the Carnot heat engine or Carnot cycle is completely imaginary,

but it established a maximum efficiency concept for all heat engines operating between the same **T _{1}**(High) and

**T**(Low) temperatures.

_{2}That’s the End Of The Carnot heat engine!

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**FAQ related to Carnot Cycle**

There are few questions that are somehow co-relate to the Carnot Heat engine or if we say can be asked to know more about the Carnot Cycle.

**1 What is the Carnot cycle in conventional and non-conventional?**

According to us, after study complete the Carnot cycle we can say that (personal opinion) Carnot Cycle is non-conventional. As it broke our known-thermodynamic physics standards.

**References**

- https://www.grc.nasa.gov
- https://en.wikipedia.org
- https://chem.libretexts.org
- http://hyperphysics.phy
- https://www.sciencedirect.com

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