Name: Carnot Cycle : Third and Fourth Stage
Uploaded: 2019-07-19
Duration: 7 min 49 s
Description: Work is done on the system during the third and fourth stages of the Carnot cycle. The video teaches us how to calculate the total work done in a Carnot cycl...

Question

Describe Carnot's reversible heat engine and obtain an expression for its efficiency. Can you realise it in practice?

Accepted Answer

The Carnot engine is an ideal heat engine. The designed engine is a theoretical engine which is free from all the defects of a practical engine.

This engine cannot be realised in actual practice, however, this can be taken as a standard against which the performance of an actual engine can be judged.

The main parts of Carnot engine are $:$

Cylinder

The cylinder has conducting base and insulating walls. It consists the gas as a working substance, insulating and frictionless piston is attached with it as shown in the figure.

Source

It is a hot reservoir maintained at a constant temperature with conducting walls. It has infinite thermal capacity. Any amount of heat can be taken from it without changing the temperature of the source.

Sink

It is a cold reservoir at temperature $T_{2}$ bed. It has infinite thermal capacity, so any amount of heat can be rejected without changing the temperature of the sink.

Working substance

We use ideal gas as a working substance in the cylinder.

Insulating stand

The base of the cylinder could be placed on the insulating stand, to isolate it completely from the surroundings.

Carnot cycle

As the engine works, the working substance of the engine undergoes a cycle known as Carnot cycle.

The Carnot cycle consists of the following four strokes.

First stroke (Isothermal Expansion) (Curve AB)

The cylinder containing one mole of an ideal gas as working substance allowed to expand slowly at the constant temperature $T_{1}$ by putting it on the source.

Work done $=$ heat absorbed by the system

$W_{1} = Q_{1} = \int_{V_{1}}^{V_{2}} p d V$

$= R T_{1} \log_{e} (\frac{V_{2}}{V_{1}}) =$ area of $A B G E$

Second stroke (Adiabatic Expansion) (Curve BC)

The cylinder is then placed on the non-conducting stand and the gas is allowed to expand adiabatically till the temperature falls from $T_{1}$ to $T_{2}$ .

$W_{2} = \int_{V_{2}}^{V_{3}} p d V = \frac{R}{(γ - 1)} (T_{1} - T_{2})$

$=$ area $B C H G$ .

Third Stroke (Isothermal Compression) (Curve $C D$ )

The cylinder is placed on the sink and the gas is compressed at constant temperature $T_{2}$

Work done $=$ heat released by the system

$W_{3} = Q_{2} = - \int_{V_{3}}^{V_{4}} p d V = - R T_{2} \log_{e} \frac{V_{4}}{V_{3}}$

$= R T_{2} \log_{e} \frac{V_{3}}{V_{4}} =$ area of $C D F H$

Fourth Stroke (Adiabatic Compression) (Curve $D A$ )

Finally, the cylinder is again placed on non-conducting stand and the compression is continued so that gas returns to its initial stage.

$W_{4} = - \int_{V_{4}}^{V_{1}} p d V = - \frac{R}{γ - 1} (T_{2} - T_{1})$

$= \frac{R}{γ - 1} (T_{1} - T_{2}) =$ area of $A D F E$

Efficiency of Carnot Cycle

The efficiency of engine is defined as the ratio of work done to the heat supplied i.e.,

$η = \frac{Work done}{Heat input} = \frac{W_{net}}{Q_{1}}$

Net work done during the complete cycle

$W = W_{1} + W_{2} + (- W_{3}) + (- W_{4})$

$= W_{1} - W_{3} =$ area $A B C D$ [as $W_{2} = W_{4}$ ]

$∴ η = \frac{W}{Q_{1}} = \frac{W_{1} - W_{3}}{W_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$

or

$η = 1 - \frac{R T_{2} \log_{e} (\frac{V_{3}}{V_{4}})}{R T_{1} \log_{e} (\frac{V_{2}}{V_{1}})}$

since, points $B$ and $C$ lie on same adiabatic curve

$∴ T_{1} V_{2}^{γ - 1} = T_{2} V_{3}^{γ - 1}$

Or

$\frac{T_{1}}{T_{2}} = {(\frac{V_{3}}{V_{2}})}^{γ - 1}$ $— e q u a t i o n (I)$

Also points $D$ and $A$ lie on the same adiabatic curve

$∴ T_{1} V_{1}^{γ - 1} = T_{2} V_{4}^{γ - 1}$

Or $\frac{T_{1}}{T_{2}} = {(\frac{V_{4}}{V_{1}})}^{γ - 1}$ $— e q u a t i o n (I I)$

From Equations $(I)$ and $(I I)$ , we get $\frac{V_{3}}{V_{2}} = \frac{V_{4}}{V_{1}}$

or $\frac{V_{3}}{V_{4}} = \frac{V_{2}}{V_{1}} \Rightarrow \log_{e} (\frac{V_{3}}{V_{4}}) = \log_{e} (\frac{V_{2}}{V_{1}})$

So, Efficiency of Carnot engine, $η = 1 - \frac{T_{2}}{T_{1}}$

The efficiency of the Carnot engine depends only on the temperatures of source and sink.

The efficiency of Carnot engine does not depend on the nature of the working substance.

The efficiency of Carnot engine will be $100 %$ , if the temperature of the sink is $0 K$ . As practically, we cannot attain a sink at $0 K$ so it is not possible to have $100 %$ efficiency.

Describe Carnot's reversible heat engine and obtain an expression for its efficiency. Can you realise it in practice?

Important Questions on Carnot's Heat Engine : Second Law of Thermodynamics

Learn and Prepare for any exam you want !