Carnot Engine

The Carnot engine, devised by French physicist Sadi Carnot in 1824, embodies an idealized heat engine operating based on thermodynamic principles. It functions through a reversible cycle between two heat reservoirs, typically employing an ideal gas. Central to its concept is the theoretical maximum efficiency, determined solely by the temperatures of these reservoirs. The engine operates through four stages: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. Despite its theoretical significance, practical Carnot engines face challenges such as frictional losses, heat dissipation, and other inefficiencies, limiting their real-world application.

1.0Operation of Carnot Engine

• It is an ideal reversible heat engine that operates between two temperatures T₁(Source) and T₂(Sink).
• It operates through two isothermal and two adiabatic processes called Carnot Cycle. It is a theoretical heat engine with which the efficiency of Practical engines is compared.

2.0Carnot Engine Diagram

3.0Parts of Carnot Engine

1. Cylinder-The central part of the engine has a conducting base and insulating walls; it is fitted with an insulating and frictionless piston.
2. Source- It is a heat reservoir at a higher temperature T₁ from which the engine draws heat. The source is supposed to have an infinite thermal capacity, so any quantity of heat can be extracted from it without altering its temperature.
3. Sink-It is a heat reservoir at a lower temperature T₂  to which the engine can discard an arbitrary amount of heat. It also has infinite thermal capacity. Therefore, any quantity of heat can be added to it without altering its temperature.
4. Working Substance - The working substance is an ideal gas in the cylinder.
5. Insulating Stand - When the base of the cylinder is attached to the insulating stand, the working substance is isolated from its surroundings.

4.0Carnot Cycle Diagram

5.0Carnot Cycle Working and Derivation

1. Isothermal Expansion

• Place the cylinder on the source so that the gas acquires its temperature T₁. The slow outward motion of the piston allows the gas to expand, causing the gas's temperature to fall. As the gas absorbs the required heat from the source, it expands isothermally.
• If Q₁ heat is absorbed from the source and W₁ work is done by the gas in isothermal expansion which takes it state from (P₁,V₁,T₁) to (P₂,V₂,T₁) then

${\mathrm{W}}_1={\mathrm{Q}}_1=\mathrm{nR}{\mathrm{T}}_1\log \left(\frac{V_2}{V_1}\right)=\text{ area ABGEA }$

2. Adiabatic Expansion

• The gas is placed on the insulating stand and allowed to expand slowly until its temperature falls to T₂.
• IIf the gas does W₂ work in the adiabatic expansion, which takes its state from (P_2,V_2,,T₁) to (P₃,V₃,T₂), then $W=\frac{nR\left(T_1-T_2\right)}{\gamma -1}=\text{ area }\mathrm{BCHGB}$

3. Isothermal Compression

• The gas is now placed in thermal contact with the sink at temperature T_2,.It is slowly compressed so that as heat is produced, it quickly flows to the sink. The gas maintains a constant temperature(T₂) throughout. 
• If   the gas releases  to Q₂ the sink and W₃ work is done on the gas by the surroundings in the isothermal compression, which takes its state from (P₃,V₃,T₂) to (P₄,V₄,T₂), then

$\mathrm{W}=\mathrm{nR}{\mathrm{T}}_2\log \left(\frac{V_3}{V_4}\right)=\text{ area CDFHC }$

4. Adiabatic Compression

• The cylinder is positioned on the insulating stand. The gas is further compressed slowly till it returns to its initial state (P₁,V₁,T₁)
• If W₄ is the work done in the adiabatic compression from ( P₄,V₄,T₂) to (P₁,V₁,T₁) then 

$W=\frac{nR\left(T_1-T_2\right)}{\gamma -1}=\text{ area DAEFD }$

Net work done by the gas per cycle

Total work done by the gas =W₁+W₂

Total work done on the gas=W₃+W₄

Net work is done by  gas in one complete cycle.

W=W₁+W₂ - (W₃+W₄)          (∴W₂=W₄)

W=W₁ - W₃=Q₁ - Q₂

W= area ABCDA

In the Carnot Engine, The work done by the gas per cycle is numerically equivalent to the area enclosed by the Carnot cycle on a pressure-volume diagram.

6.0Efficiency of Carnot Engine

It is defined as the ratio of the engine's net work done per cycle to the amount of heat absorbed  by the working substance from the source.

$\eta =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}$

7.0Carnot Engine Efficiency Formula

• $\eta =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}$
• $\eta =1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}\times 100\%$

The Efficiency of The Carnot Engine depends on

• Temperature of the source and sink
• It Is independent of the nature of the working substance
• Directly proportional to the temperature difference
• Is the same for all reversible engines working between the same two temperatures

8.0Carnot Cycle

The Enclosed area represents work done by the engine.

• It is a hypothetical engine because the working substance is an ideal gas.
• Its Efficiency is maximum but not 100%.
• This cyclic process has two isothermal and two adiabatic processes.

9.0Non-Practicability of Carnot Engine

The Carnot Engine is ideal, but it cannot be realized in practice due to specific factors.

• It is challenging  to realize the source and sink of infinite thermal capacity.
• Working substances are ideal, but no gas generally fulfills ideal gas behaviour.
• Cylinders cannot be provided with a Perfect frictionless piston.
• Achieving reversibility conditions is difficult because the compression and expansion process must be carried out very slowly.

10.0Carnot Theorem

This law states that

1. Carnot's theorem asserts that no engine operating between two heat reservoirs can achieve higher efficiency than a Carnot engine operating between those same reservoirs.
2. This theorem establishes the theoretical limit for the efficiency of any heat engine based  on the temperatures of the hot and cold reservoirs involved in the process.
3. The efficiency of the Carnot Engine is independent of the nature of the working substance.

• Consider two engines: I and R. 
• I-Irreversible engine and R-refrigerators
• Two engines are coupled that I run as forwards and it drives R backwards Hence, R is driven by I.
• The Engine I absorbs heat from Q₁ from the source  and performs work (W) and reject heat Q₂ to the sink, efficiency of engine I is given as ${\eta }_I=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}$
• Engine R absorb heat Q₂ from the sink work (W) is done on it and rejects heat Q₁ to the source, efficiency of engine R is given as ${\eta }_R=\frac{W}{Q_1^{\prime }}=\frac{Q_1^{\prime }-Q_2^{\prime }}{Q_1^{\prime }}$
• Suppose the engine I is more efficient than R,

${\eta }_I>{\eta }_R$

$\frac{W}{Q_1}>\frac{W}{Q_1^{\prime }}$

$Q_1<Q_1^{\prime }$

$Q_1-Q_1=\text{ Positive }$

• The source loses heat Q1to I and gains Q1'from R, Net heat gained by the source per cycle=$Q_1^{\prime }-Q_1$
• Sink gains heat (Q₁-W) from I and loses Q1'to R,Net heat lost by the sink per cycle= $Q_2^{\prime }-\left(Q_1-W\right)=Q_2^{\prime }-Q_1+W$

$Q_2^{\prime }-Q_1+\left(Q_1^{\prime }-Q_2^{\prime }\right)=Q_1^{\prime }-Q_1$

NOTE-Compound engine IR is a self acting machine which transfers heat $\left(Q_1^{\prime }-Q_1\right)$ from sink at lower temperature T₂ to the source at higher temperature T₁ without any work being done by any external agency,this is against the second law of thermodynamics so our assumption that I is more efficient than R is wrong.

• No engine can have efficiency greater than that of the Carnot Engine.

11.0Carnot Cycle Limitations

• Ideal Gas Assumption
• Reversibility Requirements
• Practical Efficiency Limits
• Engine Design Complexity
• Heat Transfer Constraints
• Limited Range of Operation

12.0Sample Questions on Carnot Engine

Q-1. What is the efficiency of a Carnot Engine working between ice and steam point?

Solution:

T₁=373 K (Steam point)

T₂=273 K (Ice point )

Efficiency $(\eta )=1-\frac{T_2}{T_1}=1-\frac{273}{373}=0.268=26.8\%$

Q-2. The efficiency of a Carnot cycle is $\frac{1}{6}$. If the sink’s temperature is reduced by 65⁰C, the Efficiency becomes \frac{1}{3}. Find the initial and final temperature between which the cycle is working.

Solution:                

${\eta }_1=1-\frac{T_2}{T_1}$                  

$\frac{1}{6}=1-\frac{T_2}{T_1}\dots \dots \dots (1)$                   

${\eta }_2=1-\frac{T_2}{T_1}$

$\frac{1}{3}=1-\frac{\left(T_2-65\right)}{T_1}\dots \dots .\text{ (2) }$

T₁=390K and T₂=325 K

Q-3. No real engine can have an efficiency greater than a car engine working between the same two temperatures; give reasons.

Solution:

A Carnot engine is an ideal engine satisfying two conditions

1. There is no friction between the cylinder walls and the piston.
2. Working substance is an ideal gas.        

$\Rightarrow$ Real engines cannot fulfil these conditions

Q-4. A heat engine is 20% efficient. If the engine does 500 J of work every second, how much heat does the engine exhaust every second?

Solution:      

$\eta =\frac{W}{Q_1}\Rightarrow 0.20=\frac{500}{Q_1}$            

$Q_1=\frac{500}{0.20}=2500\mathrm{J}$             

$Q_2=Q_1-W=2500-500=2000\mathrm{J}$