# Cyclic process

A cyclic process is a thermodynamic process which begins from and finishes at the same thermostatic state. It is a closed loop on a P-V diagram. The area enclosed by the loop is the work done by the process:

[itex] W = \oint P \ dV [itex].

This work is equal to the balance of heat transferred into the system:

Equation (1) makes a cyclic process similar to an isothermal process: even though the internal energy changes during the course of the cyclic process, when the cyclic process finishes the system's energy is the same as the energy it had when the process began.

If the cyclic process moves clockwise around the loop, then it represents a heat engine. If it moves counterclockwise then it represents a refrigerator.

 Contents

## Types of Cycles

A cyclic process can be (ideally) constructed on a P-V diagram out of 3 or 4 of these processes: isochoric, isobaric, isothermal, adiabatic. For example, a Carnot cycle is constructed out of a pair of quasi-parallel isothermal processes on the top and bottom of the loop, and a pair of quasi-parallel adiabatic processes left and right of the loop. The adiabatic processes are described by Q=0 so they are impermeable to heat. Heat flows into the loop through the top isotherm and some of that heat leaves the loop through the bottom isotherm. The heat which remains is equal to the work done by the process, which equals the area enclosed by the loop.

An Otto cycle is constructed out of a pair of quasi-parallel adiabatic lines on the top and bottom of the loop, and a pair of parallel isochoric processes on the left and right sides of the loop. The adiabatic processes are impermeable to heat: heat flows into the loop through the left pressurizing process and some of it flows back out through the right depressurizing process, and the heat which remains does the work.

A Stirling cycle is like an Otto cycle, except that the adiabats are replaced by isotherms. Heat flows into the loop through the top isotherm and the left isochore, and some of this heat flows back out through the bottom isotherm and the right isochore, but most of the heat flow is through the pair of isotherms. This makes sense since all the work done by the cycle is done by the pair of isothermal processes, which are described by Q=W. This suggests that all the net heat comes in through the top isotherm. In fact, all of the heat which comes in through the left isochore comes out through the right isochore: since the top isotherm is all at the same warmer temperature [itex] T_H [itex] and the bottom isotherm is all at the same cooler temperature [itex] T_C [itex], and since change in energy for an isochore is proportional to change in temperature, then all of the heat coming in through the left isochore is cancelled out exactly by the heat going out the right isochore.

### Reference

• Halliday, Resnick & Walker. Fundamentals of Physics, 5th edition. John Wiley & Sons, 1997.
• Chapter 21, Entropy and the Second Law of Thermodynamics.

## State Functions and Entropy

If Z is a state function then the balance of Z remains unchanged during a cyclic process:

[itex] \oint dZ = 0 [itex].

If entropy is defined as

[itex] S = {Q \over T} [itex]

so that

[itex] \Delta S = {\Delta Q \over T} [itex],

then it can be proven that for any cyclic process,

[itex] \oint dS = \oint {dQ \over T} = 0. [itex]

### Demonstration

#### Part 1

Draw a rectangle on a P-V diagram, such that the top and bottom are horizontal isobaric processes and the left and right are vertical isochoric processes. Such a rectangle should be made really small, so that change in temperature can be averaged out, and so that the cycle will enclose an area [itex] \Delta \mbox{area} [itex].

Let the top left corner be labeled A, then label the rest of the corners clockwise starting from A as ABCD.

Missing image
Cyclic_process.PNG
Image:Cyclic_process.PNG

Assume that the system is a monatomic gas. Then

[itex] W_{AB} = P_A (V_B - V_A) [itex]
[itex] Q_{AB} = {5 \over 2} n R (T_B - T_A) [itex]
[itex] \Delta S_{AB} = {Q_{AB} \over T_{AB, avg}} = 5 n R \left( { T_B - T_A \over T_B + T_A } \right) [itex]

Process BC:

[itex] W_{BC} = 0 [itex]
[itex] Q_{BC} = {3 \over 2} n R (T_C - T_B) [itex]
[itex] \Delta S_{BC} = {Q_{BC} \over T_{BC, avg}} = 3 n R \left( { T_C - T_B \over T_C + T_B } \right) [itex]

Process CD:

[itex] W_{CD} = P_C (V_A - V_C) [itex]
[itex] Q_{CD} = {5 \over 2} n R (T_D - T_C) [itex]
[itex] \Delta S_{CD} = {Q_{CD} \over T_{CD, avg}} = 5 n R \left( {T_D - T_C \over T_D + T_C } \right) [itex]

Process DA:

[itex] W_{DA} = 0 [itex]
[itex] Q_{DA} = {3 \over 2} n R (T_A - T_D) [itex]
[itex] \Delta S = {Q_{DA} \over T_{DA, avg}} = 3 n R \left( {T_A - T_D \over T_A + T_D } \right) [itex]

Process ABCDA (cyclic):

[itex] \Delta S_{cyc} = 5 n R \left( {T_B - T_A \over T_B + T_A } \right) + 3 n R \left( {T_C - T_B \over T_C + T_B } \right) + 5 n R \left( {T_D - T_C \over T_D + T_C } \right) + 3 n R \left( { T_A - T_D \over T_A + T_D } \right) [itex]
[itex] \Delta S_{cyc} = 5 n R \left( {V_C - V_A \over V_C + V_A } \right) + 3 n R \left( {P_C - P_A \over P_C + P_A } \right) + 5 n R \left( {V_A - V_C \over V_A + V_C } \right) + 3 n R \left( { P_A - P_C \over P_A + P_C } \right) [itex]
[itex] \Delta S_{cyc} = 5 n R \left( {V_C - V_A \over V_C + V_A } \right) + 3 n R \left( {P_C - P_A \over P_C + P_A } \right) - 5 n R \left( {V_C - V_A \over V_C + V_A } \right) - 3 n R \left( { P_C - P_A \over P_C + P_A } \right) [itex]
[itex] \Delta S_{cyc} = 0 \qquad [itex]

#### Part 2

Any loop can be broken up into a rectangular grid of differential areas. The line integral of the entire loop is equal to the sum of the line integrals of each of the constituent differential areas. Let all these integrals be done clockwise. Then any pair of adjacent differential areas will be sharing a process as a common border, but one area will add that process in one direction while the adjacent area adds that process in the reverse direction, so that process is cancelled out. Therefore all processes internal to the loop cancel each other out (see Green's theorem), and the result of the summation is equal to the line integral of the contour of the loop:

[itex] \Delta S ( \mbox{contour} ) = \sum_\mbox{area} \Delta S ( \Delta \mbox{area} ) = \sum_{\mbox{area}} 0 = 0 [itex]

Q.E.D.

### Conclusion

The fact that entropy is a state function is what puts entropy on the map (the P-V diagram).

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