Carnot engine

• A Carnot engine is named after Carnot scientist.
• It is a reversible heat engine operating between two temperatures.
• It has a maximum efficiency which no other engine can have.

Conditions of a reversible engine

• It should be quasi static which means the change happens gradually that we do not notice the change happening
• It should be non dissipative – No loss due to friction or any other thing

Theory

Consider a Heat engine, hot reservoir at a temperature T₁ (Source of heat) and a cold reservoir at a temperature T₂ (Sink of heat).  The heat engine consists of the perfect gas which acts as a working substance. 

The carnot cycle consists of 4 steps –

• Heat is absorbed from the hot reservoir without change in temperature - Isothermal expansion
• Increase of temperature in the heat engine from T₁ to T₂ so as to perform step 3 with same temperature gradient; There should not be any heat release during this process – Adiabatic expansion
• Heat engine releases heat Q₂ to sink at temperature T₂ without change in temperature – Isothermal compression
• Resetting the temperature back so as the repeat the cycle again. Hence, without heat release temperature changes from T₂ to T₁ – Adiabatic compression

Work done

• Isothermal expansion of gas – Q₁ is the amount of heat absorbed in this process and gas changes its volume and pressure from A(V₁ , P₁) to B(V₂ , P₂). Since it is expansion volume increases from V₁ to V₂ . There is no change in temperature and it remains at T₁ This is represented in the graph as AB where we can notice increase in volume and decrease in pressure. So for simplicity, we can say Isothermal expansion process as change from A (V₁ , P₁,T₁) to B (V₂ , P₂,T₁)

Work done by the gas W₁ = Q₁ =    = RT₁ log_e(V₂/V₁) = Area ABMKA

The above derivation is based on the standard gas equation PV = RT.  It is actually work done in an isothermal expansion and if we do that here it will take us off the topic.

• Adiabatic expansion of gas – Temperature changes from T₁ to T₂ The gas changes its volume and pressure from B(V₂ , P₂) to C(V₃ , P₃). Since it is expansion, volume increases from V₂ to V₃ and pressure decreases. This is represented in the graph as BC where we can notice increase in volume and decrease in pressure. So for simplicity, we can say adiabatic expansion process as change from B (V₂ , P₂,T₁) to C (V₃, P₃,T₂)

Work done by the gas W₂  =    = R (T₂ – T₁)/ (1 - ϒ) = Area BCNMB

The ϒ here is equal to C_p / C_v which is ratio of specific heat of gas at constant pressure to specific heat of gas at constant volume; Also, the above derivation which is actually work done in an adiabatic expansion, will take us off the topic here.

• Isothermal compression on the gas – Q₂ is the amount of heat released in this process and since it is a compression process, volume decreases here. There is no change in temperature. This is represented in the graph as CD where we can notice decrease in volume and increase in pressure. So for simplicity, we can say Isothermal compression process as change from C (V₃ , P₃,T₂) to D (V₄ , P₄, T₂)

Work done on the gas W₃ = Q₂ =    = - RT₂ log_e(V₄/V₃) = RT₂ log_e(V₃/V₄) = -  Area CDLNC

• Adiabatic compression on the gas – Here again there is only change in temperature adiabatically from T₂ to T₁. So for simplicity, we can say adiabatic expansion process as change from D (V₄, P₄,T₂) to A (V₁, P₁,T₁)

Work done on the gas W₄  =    = - R (T₂ – T₁)/ (1 - ϒ) = - Area DAKLD

Total work done by the engine per cycle

Total work done by the gas (in steps (a) and (b))  = W₁ + W₂

Total work done on the gas (in steps (c) and (d))  = W₃ + W₄

Net work done  by the gas in a complete cycle = W =  (W₁ + W₂) – (W₃ + W₄)

W₂ and W₄ are equal in magnitude and opposite in direction. Hence, they cancel out each other

Hence, W = W₁ – W_e = Q₁ – Q₂  which is area enclosed by the curve ABCDA

Efficiency η

Efficiency is defined as total work done by the heat engine in a cycle divided by the amount of heat energy absorbed per cycle from the source which is W / Q₁

Putting W = Q₁ – Q_{2 ,} we get η = (Q₁ – Q₂) / Q₁ = 1 -  Q₂ / Q₁

Points A,B and C,D  lie on the same isotherm . Also, points B, C and D, A lie on the same isotherm

Hence, P₁V₁ = P₂V₂                   -------(1)

                P₃V₃ = P₄V₄                         -------(2)                                     

P₂V₂^ϒ  = P₃V₃^ϒ                     -------(3)

P₄V₄^ϒ  = P₁V₁^ϒ                     -------(4)

Multiply all LHS on one side and RHS on other side

( P₁V₁) (P₃V₃) (P₂V₂^ϒ)( P₄V₄^ϒ) =  ( P₂V₂) (P₄V₄) (P₃V₃^ϒ)( P₁V₁^ϒ)

Taking V₂^ϒ = V₂^ϒ V₂^ϒ-1  and simplifying further we get V₂  /  V₁ = V₃ / V₄

Taking log on both sides, log (V₂  /  V₁)  = log (V₃ / V₄)

Let us consider the work done equation for isothermal expansion and isothermal compression

W₁ = Q₁  =  RT₁ log_e(V₂/V₁)

W₃ = Q₂  = RT₂ log_e(V₃/V₄)

Dividing to find Q₂ by Q₁ , we get Q₂ / Q₁ = RT₂ log_e(V₃/V₄) /  RT₁ log_e(V₂/V₁)

Q₂ / Q₁ = T₂ / T₁ as the log terms and R gets cancelled

η = (Q₁ – Q₂) / Q₁ = 1 -  Q₂ / Q₁

 Hence, η = 1 – T₂ / T₁

1. The above expression for efficiency indicates that the carnot engine depends only on the temperature of the source and sink. The efficiency does not depend on the nature of the working substance
2. The RHS of the equation shows that it is less than 1. Hence, efficiency of the engine cannot be greater than 100%
3. When η = 1, then either T₁ should be infinity or T₂ should be zero. As source at infinite temperature is not attainable, efficiency cannot be 100%
4. Carnot engine is a reversible engine and each step of the engine is reversible
5. We can conclude
   • Working between the given two temperatures of source and sink, no engine can have an efficiency more than carnot engine
   • The efficiency of carnot engine is independent of the nature of the working substance

The above two points forms Carnot therem.

Irreversible engine

Second Law of Thermodynamics: It is impossible to extract an amount of heat Q_H from a hot reservoir and use it all to do work W. Some amount of heat Q_C must be exhausted to a cold reservoir. 

But if this irreversible engine is more efficient than the reversible engine, then the second law of thermodynamics is violated. And, since the Carnot cycle represents a reversible engine, we have the first part of Carnot theorem:

No irreversible engine is more efficient than the Carnot engine operating between the same two reservoirs.