The Carnot cycle uses only two thermal reservoirs – one at high temperature T₁ and the other at two temperature T₂.

If the process undergone by the working fluid during the cycle is to be reversible, the heat transfer must take place with no temperature difference, i.e. it should be isothermal.

The Carnot cycle consists of a reversible isothermal expansion from state 1 to 2, reversible adiabatic expansion from state 2 to 3, a reversible isothermal compression from state 3 to 4 followed by a reversible adiabatic compression to state 1.

The thermal efficiency, h is given by

h = Net work done / Energy absorbed as heat

During processes 2-3 and 4-1, there is no heat interaction as they are adiabatic.

Similarly for the process 3-4,

Net heat interaction = Net work done

= RT₁ln(v₂/v₁) + RT₂ln(v₄/v₃)

= RT₁ln(v₂/v₁) - RT₂ln(v₃/v₄)

The processes 2-3 and 4-1 are reversible, adiabatic and hence

T₁v₂^g-1 = T₂v₃^g-1

Or, v₂/v₃ = (T₂/T₁)^1/(g-1)

And T₂v₄^g-1 = T₁v₁^g-1

Or, v₁/v₄ = (T₂/T₁)^1/(g-1)

v₂/v₃ = v₁/v₄ or v₂/v₁ = v₃/v₄

h = {RT₁ln(v₂/v₁) - RT₂ln(v₃/v₄)} / RT₁ln(v₂/v₁)

h = (T₁ – T₂)/T₁

= 1- T₂/T₁

The Carnot Principles

1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between same two thermal reservoirs.

2. The efficiencies of all reversible heat engines operating between the same two thermal reservoirs are the same.

Lets us assume it is possible for an engine I to have an efficiency greater than the efficiency of a reversible heat engine R.

h_I > h_R

Let both the engines absorb same quantity of energy Q₁. Let Q and Q₂ represent the energy rejected as heat by the engines R, and I respectively.

W_I = Q₁ - Q

W_R= Q₁ – Q₂

h_I = W_I / Q₁ = (Q₁ - Q)/Q₁ = 1-Q/Q₁

h_R = W_R/Q₁ = (Q₁ - Q₂)/Q₁ = 1-Q₂/Q₁

Since h_I > h_R,

1-Q/Q₁ > 1-Q₂/Q₁

or, Q < Q₂

Therefore, W_I (= Q₁-Q) > W_R (=Q₁ – Q₂)

Since the engine R is reversible, it can be made to execute in the reverse order. Then, it will absorb energy Q₂ from the reservoir at T₂ and reject energy Q₁ to the reservoir at T₁ when work W_R is done on it.

If now engines I and R are combined, the net work delivered by the combined device is given by

W_I – W_R = Q₁ – Q – (Q₁ – Q₂) = Q₂ – Q

The combined device absorbs energy (Q₂ – Q) as heat from a single thermal reservoir and delivers an equivalent amount of work, which violates the second law of thermodynamics.

Hence, h_R ³ h_I

Carnot principle 2

Consider two reversible heat engines R₁ and R₂ , operating between the two given thermal reservoirs at temperatures T₁ and T₂.

Let h_R1 > h_R2

Q₁= energy absorbed as heat from the reservoir at T1 by the engines R₁ and R₂, separately.

Q = energy rejected by reversible engine R₁ to the reservoir at T₂

Q₂ = energy rejected by reversible engine R₂ to the reservoir at T₂.

W_R1 = Q₁ - Q = work done by a reversible engine R₁ .

W_R2 = Q₁ –Q₂ = work done by a reversible engine R₂

According to assumption,

h_R1 > h_R2

Or, 1 – Q/Q₁ > 1- Q₂/Q₁

Q₁ –Q >Q₁-Q₂ or W_R1 >W_R2

W_R1 – W_R2 = (Q₁ –Q) – (Q₁- Q₂) = Q₂ – Q

Since the engine R₂ is reversible, it can be made to execute the cycle in the reverse by supplying W_R2.

Since W_R1 > W_R2 the reversible engine R₂ can be run as a heat pump by utilizing part of the work delivered by R₁.

For the combined device,

W_R1 – W_R2 = Q₂ – Q, by absorbing energy Q₂ – Q from a single thermal reservoir which violates the second law of thermodynamics.

Hence h_R1 > h_R2 is incorrect.

By similar arguments, if we assume that h_R2 > h_R1 then,

h_R1 ³ h_R2

Therefore, based on these two equations,

h_R1 = h_R2

The efficiency of a reversible heat engine is also independent of the working fluid and depends only on the temperatures of the reservoirs between which it operates.