# Carnot Engine - Explanation, Cycle, Work done, Efficiency, and Reversibility | Physics Grade 11

### Thermodynamics

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# Carnot Engine - Explanation, Cycle, Work done, Efficiency, and Reversibility

Carnot Engine
Carnot elaborated an ideal cycle for operation of a heat engine which is known as Carnot Cycle. The machine used for realization for this cycle is called an ideal heat engine or Carnot heat engine.

The parts of Carnot heat engine are:
1. Source Of heat: A hot body of infinite thermal capacity is a source of heat. The source is maintained at a fixed higher temperature T1 from which the working substance draws heat without changing its temperature.
2. Cylinder: A cylinder fitted with a perfectly non-conducting and frictionless piston enclosing an ideal gas. The bottom of the cylinder is a perfect heat conductor whereas the walls are perfect heat insulators.
3. Sink of heat: The sink should have the infinite thermal capacity and be at a fixed lower temperature, T2 to which any amount of heat can be rejected.
4. Working substance: An ideal gas acts as the working substance filled in a cylinder can be placed. It would completely isolate the working substance from the surroundings.
Carnot’s Cycle
Consider n-mole of an ideal gas enclosed in the cylinder. Let P1, V1, T1 be the initial pressure, volume and temperature of the gas in it. The initial state of the gas is represented by point A on the P-V diagram.
1. Isothermal Expansion: Firstly, the cylinder containing the working substance is placed on the source and the gas is allowed to expand by moving the piston outward slowly. Since, the gas draws heat from the source through the conducting base of the cylinder, the gas expands at a constant temperature. This expansion is an isothermal expansion which is represented by the curve AB. If the quantity of the heat absorbed during the expansion is Q1 and W1 be the work done from A (P1, V1) to B (P2, V2). Then according to first law of thermodynamics,
Q1 = W1
= ∫v1v2 P dV
= nRTv1v2 dV/V
= nRTlogV2/V1 ….… (i)

2. Adiabatic Expansion: The cylinder is removed from the source and then placed on the non-conducting (insulating) stand. The gas is allowed to expand further from B (P2, V2) to C (P3, V3). Since the working substance is completely thermally insulated no heat is exchanged between the system and the surroundings. Therefore, the nature of the expansion is adiabatic when it is allowed to expand more and the temperature gas falls to T2. If W2 is the work done, then
W2 = ∫v1v2 P dV

Since in adiabatic expansion, PVγ = constant = K,
W2 = K ∫v2v3 dV/Vγ
= (KV31-γ – KV21-γ) / 1-γ
= (P3V3 – P2V2) / 1-γ
= (nRT2 – nRT1) / 1-γ
= nR (T2 – T1) / 1-γ
= nR(T1-T2) / γ-1 ....… (ii)

3. Isothermal compression: The cylinder is now removed from the insulating stand and placed on the heat sink at a temperature T2. The gas is compressed slowly so that as the heat is developed in the gas which is flowed to the sink through the conducting base of the cylinder. Since the heat is rejected to the sink, isothermal compression acts on it. If Q2 be the amount of heat energy rejected and W3 be the work done then,
W3 = Q3 = -∫v1v2 P dV

The negative sign shows the isothermal compression
.: W3 = Q3 = -∫v1v2 P dV
= nRTv4v3 dV/V
= nRT2 loge V3/V...…. (iii)

4. Adiabatic Compression: The cylinder is again placed into insulating stand and the piston is slowly moved downward so adiabatic compression acts in the system. If W4 is the work done,
W4 = -∫v1v2
= nR(T1 – T2) / γ-1 …..... (iv)

Work done by the engine per cycle
In steps 1 and 2, total work done by the gas = W1 + W2
In steps 3 and 4, total work done by the gas = W3 + W4
.: Net work done by the gas in one complete cycle, W = W1 + W2 + (-W3) + (-W4)

From Equation (ii) and (iv), we have
W2 = W4
W = W1 – W2
W = Q1 – Q2 .....… (v)

So, net work done in one complete cycle by the engine is,
W = Area of ABCDA

Efficiency of Carnot Cycle
It is defined as the ratio of external work done W by the engine to the amount of heat energy absorbed from the source. i.e.
η = W/ Q1
= (Q1 – Q2­)­ / Q1
= 1 – (Q/ Q1)
= 1 – (RT2 loge V3/V2 / RT1 loge V4/V1)
= 1 – R (T2 loge V3/V/ Tloge V4/V1) …..... (vi)

As shown in figure, as A (P1, V1) and B (P2, V2) lie on the same isothermal condition,
P1V1 = P2V2 ......… (vii)

For BC,
P2V2γ = P3V3γ ....… (viii)

For CD,
P3V3 = P4V4 …..... (ix)

For DA,
P1V1γ = P4V4γ ….... (x)

Dividing equation (viii) by equation (x)
P2V2γ / P1V1γ = P3V3γ / P4V4γ ....… (xi)

From equation (vi) P1V1 = P2V2
or, P2/P1 = V2/V1

From the equation (ix)
Or, P4/P3 = V3/V4

Substituting these ratios in equation (xi), we get
(V2/V1)γ-1 = (V3/V4)γ-1
V2/V1 = V3/V4

Therefore, Equation (vi) can be written as
η = 1 – (T2 loge V3/V/ T1 loge V2/V1)
η = 1 – (T2 loge V3/V/ T1 loge V3/V)
η = 1- T2/T1

Reversibility of Carnot’s Engine
As described above, the Carnot cycle is perfectly reversible because
1. There is no friction between the cylinder and the piston
2. The operations on the gas should be performed very slowly
3. The loss of heat due to conduction is prevented by using an insulating piston and the insulating walls of the cylinder
4. During an isothermal change, the temperature remains constant, because the Carnot engine uses a conducting base for the cylinder and, the sink and source have large heat capacity.

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