We were discussing the Otto cycle and the Diesel cycle in our previous posts and we have seen there that these cycles are not totally reversible cycles but also these cycles are internally reversible cycles. Hence, thermal efficiency of an Otto cycle or Diesel cycle will be less than that of a Carnot cycle working between similar temperature limits.

Let us consider that we have one heat engine which is operating between heat source having at temperature TH and heat sink having at temperature TL. As we have already seen in our previous discussion that for a heat engine to be a complete reversible, difference between the working fluid temperature and heat source temperature or difference between working fluid temperature and heat sink temperature must never go beyond the differential amount of dT during any heat transfer process.

We can also say that heat addition or heat rejection process must be carried out isothermally and this criterion is fulfilled in Carnot cycle.

Now we will see here one more important cycle i.e. Stirling cycle, where heat addition process and heat rejection process will be carried out isothermally. But we must note it here that Stirling cycle will be different from Carnot cycle. Because there will be two constant volume regeneration processes in Stirling cycle instead of isentropic processes of Carnot cycle.

### So, Let us see here Stirling cycle

Stirling cycle is one ideal cycle for the operation of Stirling engine. First we will see here the PV and TS diagram for Stirling cycle, we will understand here the various processes involved and finally we will determine the thermal efficiency of the Stirling cycle.
As we can see here from PV and TS diagram, there will be two reversible isothermal processes and two reversible constant volume processes.

Process, 1-2: Isothermal expansion from state 1 to state 2. Heat energy will be added here from external source. Volume will be increased but pressure will be reduced during this process.
Î”U = 0,
Q1-2 = W1-2 = RT1 Log (V2/V1)

Process, 2-3: Constant volume process, internal heat transfer from the working fluid to regenerator and therefore this process is also termed as constant volume regeneration process.
Î”W = 0,
Q2-3 = dU2-3 = CV (T3-T2)

Process, 3-4: Isothermal compression from state 3 to state 4. Heat energy will be rejected here to the external sink. Volume will be reduced but pressure will be increased during this process.
Î”U = 0,
Q3-4 = W3-4 = RT3 Log (V4/V3)

Process, 4-1: Constant volume process, internal heat transfer from the regenerator back to the working fluid.
Î”W = 0,
Q4-1 = dU4-1 = CV (T1-T4)

During the process of regeneration, Heat will be transferred to the regenerator from the working fluid during one part of the cycle and heat will be transferred back to the working fluid during the second part of the cycle. Regenerator will be considered as reversible heat transfer device.

If we think the concept of regeneration where, area under 2-3 i.e. Q2-3 and area under 4-1 i.e. Q4-1 are equal, Regenerative Stirling cycle will become Carnot cycle because heat energy will be added from an external source at constant temperature and heat will be rejected too to an external sink at constant temperature and hence Regenerative Stirling cycle will become Carnot cycle and therefore it will have similar efficiency as of Carnot cycle.

Therefore, efficiency of the Stirling cycle will be written as
Do you have any suggestions? Please write in comment box.

### Reference:

Engineering thermodynamics, By P. K. Nag
Engineering thermodynamics, By S. K. Som
We will see another important topic i.e. Ericsson cycle in the category of thermal engineering