Chapter 12.7 - The Carnot Engine

In the previous section, we saw the basics about refrigerator. In this section, we will see the Carnot engine

Carnot Engine

The basic processes involved in the working of a Carnot engine can be written in 14 steps:
1. A Carnot engine is an ideal engine
• To make it ideal, two restrictions are imposed:
(i) Every process that occur during the working of a Carnot engine must be reversible
(ii) The working gas inside the engine must be an ideal gas
2. We have seen the details about reversible process in the first section of this chapter (see fig.12.4 of the first section)
• Based on that, we can write:
    ♦ At any instant during the working of a Carnot engine, the gas inside the engine will have the same temperature as the surroundings
    ♦ We can change the direction of the reversible process any time we want
    ♦ A reversible process can proceed only very slowly
   ♦ So the Carnot engine will be very slow. But it gives us a base to study other real irreversible engines
3. Initially, the gas is at state A
• For the engine to start operating, we have to supply heat
   ♦ So the cylinder is placed in contact with a hot reservoir
   ♦ This is shown in fig. 12.19(a) below
   ♦ The temperature of this hot reservoir is maintained at T₁ K
• So at this state, the V, P, T values of the gas will be: (V_A, P_A, T_A)
    ♦ Where T_A = T₁

Fig.12.19

4. The gas absorbs heat and expands. Thus work is done on the piston
• This expansion process should be isothermal
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason can be seen here
• At the end of this isothermal process, the gas reaches state B
    ♦ Pressure P_A decreases to a lower value P_B
    ♦ Volume V_A increases to a higher value V_B
    ♦ Temperature remains the same because it is an isothermal process
• So at this state, the V, P, T values of the gas will be: (V_B, P_B, T_B)
    ♦ Where T_B = T₁
• This process is represented by the red curve AB in fig.12.19(b)
• The first segment of the cycle is complete
5. Next we will see not the second segment, but another important segment
• Some quantity of heat is to be dumped into a cold reservoir
• Only then, can the gas return to the initial state and become ready for the next cycle
   ♦ For that, we place the cylinder in contact with a cold reservoir
   ♦ This is shown in fig.12.20(a) below
   ♦ The temperature of the cold reservoir is maintained at T₂ K
• So at this state, the V, P, T values of the gas will be: (V_C, P_C, T_C)
    ♦ Where T_C = T₂

Fig.12.20

6. The gas is compressed isothermally. Thus work is done on the gas
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason is same (but in reverse order)as that we saw for the first segment
    ♦ The discussion can be seen here
• At the end of this isothermal process, the gas reaches state D
    ♦ Pressure P_C increases to a higher value P_D
    ♦ Volume V_C decreases to a lower value V_D
    ♦ Temperature remains the same because it is an isothermal process
• So at this stage, the V, P, T values of the gas will be: (P_D, V_D, T_D)
    ♦ Where T_D = T₂
• This process is represented by the red curve CD in fig.12.20(b)
• This segment of the cycle is complete
7. The two segments mentioned above are important and inevitable segments
• During AB, the gas absorbs heat and does useful work
• But the gas has to return to it's initial state to start the next cycle
   ♦ For that, it has to dump the heat
   ♦ This dumping is done during CD
8. But we see a problem. It can be explained in 2 steps:
(i) The curve AB is separated away from curve CD
• This is obvious because, they are isotherms
    ♦ Every point in AB will be at temperature T₁
    ♦ Every point in CD will be at temperature T₂
    ♦ Two isotherms will never meet
(ii) So our problem is this:
How to reduce the temperature of the gas from T₁ (at state B) to T₂ (at state C)
9. Just after state B, we must subject the gas to a suitable process so that, the temperature falls to T₂
◼ Which process is suitable?
• An adiabatic process is the only suitable process
• Processes like isobaric, isochoric or isothermal cannot be used
    ♦ The reason can be seen here
10. An adiabatic process is most suitable for achieving our goal
• Just after state B, the cylinder is placed in a thermal insulator
This is shown in fig.6.21 below:

Fig.6.21

• The gas is then allowed to expand adiabatically
   ♦ So no heat will be lost or gained
• The gas will use it's own internal energy to expand
   ♦ This results in the decrease in temperature of the gas from T_B to T_C (T₁ to T₂)
• Thus we successfully established the connection between the first and third segments
11. Next we want to establish the connection between the third and first segments
• That is., between points D and A
• During the process from D to A,
    ♦ The volume decreases from V_D to V_A
    ♦ The pressure increases from P_D to P_A
    ♦ Temperature increases from T₂ to T₁
• So our problem is this:
How to increase the temperature of the gas from T₂ (at state D) to T₁ (at state A)
12. Just after state D, we must subject the gas to a suitable process so that, the temperature rises to T₁
◼ Which process is suitable?
• An adiabatic process is the only suitable process
• Processes like isobaric, isochoric or adiabatic cannot be used
    ♦ The reason is same (but in reverse order)as that we saw for the second segment
    ♦ The discussion can be seen here
13. An adiabatic process is most suitable for achieving our goal
• Just after state C, the cylinder is placed in a thermal insulator
This is shown in fig.6.22 below:

Fig.6.22

• The gas is then compressed adiabatically
   ♦ This results in the increase in temperature of the gas from T_D to T_A (T₂ to T₁)
• Thus we successfully established the connection between the third and first segments
• The cycle is now complete
14. Let us write a summary. It can be written in 4 steps:
(i) We supply energy to the gas during the first segment
    ♦ During this segment, the gas expands and does useful work
    ♦ The supply of energy is stopped at the end of the first segment
(ii) During the second segment also, useful work is done by the gas
    ♦ But this time, the gas expands using it's internal energy
    ♦ The temperature falls to T₂ during this segment
(iii) The gas has to return to it's original state so that another cycle can begin
    ♦ So in the third and fourth segments, the gas undergoes compression
    ♦ In the third segment, the compression is carried out at a lower temperature T₂
        ✰ So it is easier to achieve compression
(iv) In the fourth segment, the temperature rises back to T₁
    ♦ This is because, no heat is allowed to escape during the compression

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• So now we know the four processes involved in the Carnot cycle
• Next we want to calculate the work done in each of the four segments in the cycle
• We will see it in the next section

Discussion on the notes can be started here

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