# Explanation of Carnot’s cycle and Carnot engine

In this particular article ‘Explanation of Carnot’s cycle and Carnot engine’, we are going to discuss the Carnot cycle and Carnot engine in detail and easiest way possible.

Carnot’s Cycle and Carnot Engine is also called Ideal reversible engine. It has four parts

1. A cylinder which perfectly non-conducting walls. But perfectly conducting base and fitted with a perfectly insulating and frictionless piston.
2. A hot body of infinitely large heat capacity at high at high temp T1, working as a source of heat.
3. A cold body of infinitely large heat capacity, at the lower temperature T2 serving as a sink.
4. A perfectly insulating platform serving as a stand to the cylinder.

The working substance is subjected cycle of four operations consisting of two isothermal and two adiabatic operations such cycle is known as Carnot’s cycle.

## Process 1(Isothermal expansion)

The cylinder is put on the source, the presence on the piston is slowly decreased. The working substance (gas) thus expands infinitely slowly doing work against the piston. During this process, the substance takes Heat from source by conduction through the base. Let us suppose that the quantity of heat absorbed is Q1 then work done by the gas is

Q1 = w1 = ∫v1^v2 Pdv

PV = RT

and PV = RT1

w1 = Q1 = RT1 ∫v1^v2 dV/V

And w1 = Q1 = RT1 log (v2 – v1)

w1 = Q1 = RT1 log v2/v1  (1)

The cylinder is put now on the insulating stand. The pressure of piston is now deceased the substance (gas) is further expand. This expansion is adiabatic. This expansion is adiabatic because no heat will enter the substance through the cylinder. The gas does work at the expanse of its internal energy in raising the piston and temperature falls. So that work was done by the substance.

w2 = ∫v2^v3 PdV  (PV^γ = k)

or P = k/v^γ

w2 = k ∫v2^v3  1/v^γ dv

w2 = k (v3^1-γ – v2^1-γ)/ 1-γ

we have p3v3^γ = k = p2v2^γ

= ( Kv3^1-γ – Kv2^1-γ)/ 1-γ

= (p3v3^γ v3^1-γ – p2v2^γ v2^1-γ)/ 1-γ

w2 = p3v3 – p2v2 / 1-γ

but we have P2V2 = RT1

p3v3 = RT2

w2 = R(T1 – T2)/1-γ   (2)

#### Process 3 (Isothermal Compression)

The cylinder is now put on the sink and the load on the piston is slowly increased. So that substance is further compressed  until its pressure and volume become P4 and V4 (point o)

The heat produced due to the compressing passing into the sink through the base of the cylinder and the temperature of the substance remains constant at T2. Thus a quantity of Heat is given to the sink is

Q2 = w3 = ∫v3^v4 Pdv

w3 = ∫v3^v4 RT2/V dv

v3 = -RT2(log v3/v4)   (3)

The cylinder is put on the stand and by further placing the weight on the piston. The substance is compressed adiabatically. Therefore work is done on the substance by the piston and temperature increases. The process is continued until the temperature rises to T1 and state ‘A’ is recovered.

w4 = ∫v4^v1 Pdv

∴ Pvˆγ = K

w₄ = − R(T₁ – T₂)/γ − 1          (4)

###### Total work done in one cycle :-

w = w₁ + w₂ + w₃ + w₄

w = (RT₁ log v₂/v₁ ) – (RT log v₃/v₄)      (5)

T₁v₁ˆγ -1 = T₂v₄ˆγ -1

(for AD)   T₂/T₁ = (v₁/v₄) ˆγ -1                (6)

Similarly for BC curve :-T

T₂/T₁ = (v₂/v₃) ˆγ -1                                 (7)

Comparing 6 and 7

(v₁/v₄) ˆγ -1 = (v₂/v₃) ˆγ -1

(v₁/v₄) = (v₂/v₃)

Put in (5)

w =  (RT₁ log v₂/v₁ ) – (RT log v₂/v₁)

w = R (T₁ – T₂)log v₂/v₁                              (8)

But the net amount of heat absorbed by the substance in one cycle in, (Q₁ -Q₂) so that

(Q₁-Q₂) = w = R (T₁ – T₂)log v₂/v₁         (9)

Since the initial and final state of a substance is the same. Its internal energy remains unchanged. Hence from (1) law of thermodynamics-

w =   (Q₁ – Q₂)

###### Efficiency of engine

η = Heat converted into work/ heat taken by the source

And, η = (Q₁ – Q₂) / (Q₁ )

η = 1- (Q₂ /Q₁)                    (10)

Put the values,

η = [R (T₁ – T₂)log v₂/v₁ ] / R (T₁)log v₂/v₁

η = T ₁- T₂ /T₁ = 1 − T₂/T₁        (11)

It depends only on the temperature of source and sink.

We have,

T₂ / T₁ = (v₂/v₃)ˆγ -1

And we know,

v₂/v₃ = ρ = adiabatic expansion ratio

T₂ / T₁ = (1/ρ)ˆγ -1

η = 1 – (1/ρ)ˆγ -1

###### The conclusion of Explanation of Carnot’s Cycle and Carnot Engine:

In this above article, we learn about the Carnot cycle the and Carnot engine’s processes. We studied 4 types of process which take place in the Carnot engine.

a)Isothermal expansion