# Case Study: Carnot Cycle

The Carnot cycle is a thermodynamic cycle that gives the best efficiency possible. Given an amount of energy in form of heat, the Carnot cycle transforms heat put in to produce useful work by a series of reversible adiabatic (isentropic) and isothermal processes.  The efficiency of a Carnot engine is one minus the ratio of the temperature of the hot thermal reservoir to  the temperature of the cold reservoir. The Carnot cycle is an idealization and is useful in setting the limit of highest efficiency any cycle or engine can hope to achieve.

### Historical Background

The Carnot cycle is named after the French engineer, N.L. Sadi Carnot, who proposed it in 1824. Sadi Carnot is often referred to as the founder of the field of thermodynamics for discovering the relationship between heat and work. Carnot was one of the first to realize that heat is essentially work in a different form. From the Carnot cycle other aspects of thermodynamics were derived (e.g. entropy).

### Reversibility

It is important to keep in mind that the Carnot cycle is reversible. That is, at every point defined in the PV diagram the system is at equilibrium and there is no heat dissipation (the system can return exactly to its previous state). In other words, the process is quasistatic, meaning the system changes by infinitesimal amounts and is frictionless. One way to understand a quasistatic process is to imagine that the external pressure on a piston resulted from a 200L tank of water on top and the external pressure changes by 1 mL at a time. Realistically, there are no such process in the world where changes are infinitely small and that there is no lost of heat.

### The Carnot Cycle

The Carnot Cycle can be broken down into a series of steps as demonstrated below through the classic example, the piston:

#### Step 1-2

Starting from state 1 to state 2, the piston is expanded isothermally (Recall that an isothermal process is one in which the temperature stays constant between the system and the surroundings).

Because the system does work on the surrounding (the piston is pushed up), the internal energy (energy of motion of the molecules and potential energy stored in chemical bonds and position) and temperature of the system decrease. However, the temperature and change in internal energy remain constant because heat from the thermal reservoir replenishes the lost of heat due to work. Knowing that the energy lost due to work is equal to the heat gained, it is possible to calculate the heat that is gained by the system if we know how much work is done (this is just the area under the curve from $$V_1$$ to $$V_­2$$). From the first law of thermodynamics, we have

$$\Delta U = q+w$$

$$0=q+w$$

$$q=-w=P\Delta V$$

$$q=-w=\int_{V_1}^{V_2}PdV=nRT\int_{V_1}^{V_2}\frac{1}{V}dV=nRT \left(\frac{V_2}{V_1} \right)$$

#### Step 2-3

From state 2 to state 3, the piston is expanded adiabatically, where no heat is exchanged between the system and the environment. The system is thermally insulated. As the piston expands, the molecules push against the piston with less force and the average kinetic energy of the molecules decreases.

The change in the internal energy of the piston is equal to the work done during the adiabatic expansion. So we have

$$\Delta U=q+w=0+w=-P\Delta V$$

$$w=\int_{V_2}^{V_3}P\Delta V$$

Here, because we have an adiabatic condition we know that $$PV^\gamma$$ is a constant C.

$$C\int_{V_2}^{V_3}\frac{dV}{V^\gamma}=C\frac{\left (V_3^{1-\gamma}-V_2^{1-\gamma} \right)}{(1-\gamma)}$$ where $$\gamma \; is \; \frac{C_p}{C_v}$$

#### Step 3-4

From state 3 to state 4, the piston undergoes an isothermal compression. In this isothermal process (as opposed to the step1-2 one) heat is being rejected from the system.

As the piston compresses, work is being done on the system and therefore is increasing the internal energy of system. An increase in internal energy is closely associated with an increase in temperature. However, the temperature of the system stays the same because heat is flowing out of the system to the thermal reservoir which is infinitely large (so it does not change temperature). This heat, again, is equivalent to the work done to the system by an external pressure pushing on the piston. As in step 1-2, the work done in this step is

$$-q=w=-\int_{V_3}^{V_4}PdV=-nRTln\left (\frac{V_4}{V_3}\right)$$

Notice the negative sign which denotes that heat is escaping from the system.

#### Step 4-1

From state 4 to state 1, the system undergoes an adiabatic compression with no heat lost or gain from the environment. This process is effectively isentropic.

The piston is compressed reversibly (as for all of the steps above) by an external pressure. So work is done on the system by the surrounding. And because it is an adiabatic process, no heat is transferred out of the system. This means, then, that all of the work done by the surrounding on the system goes to increase the internal energy of the system. Likewise, we could calculate the work done by a similar integral as above.

### Work and Efficiency

To find the total work by in one cycle, we add up all of the work done by the system and subtract from the work done to the system. Or we simply subtract the heat put into the system and the heat rejected from the system. Essentially, that change in heat is converted into useful work under ideal conditions (reversible and frictionless).

Efficiency is then the work done by the system divided by the heat put into the system.

From step 1-2, we know that heat invested in the system is just the work done(the area under the integral from 1 to 2). Furthermore, the heat rejected from the system is also just the work that was done on the system. So we have

$$Efficiency=\frac{w_{net}}{q_{in}}=\frac{\int_{V_1}^{V_2}PdV-\int_{V_4}^{V_3}PdV}{\int_{V_1}^{V_2}PdV}=1-\frac{-nRT_c\int_{V_4}^{V_3}\frac{dV}{V}}{nRT_h\int_{V_1}^{V_2}\frac{dV}{V}}=1-\frac{T_cln\left (\frac{V_3}{V_4} \right )}{T_hln\left (\frac{V_2}{V_1} \right )}$$

But because of the relationship (in reversible adiabatic processes $$PV^\gamma$$ is constant and PV=nRT),

$$T_hV_1^{\gamma-1}=T_cV_4^{\gamma-1}\hspace {10mm} and \hspace{10mm} T_hV_2^{\gamma-1}=T_cV_3^{\gamma-1}$$

By susbstitution we see that

$$\frac {V_2}{V_1}=\frac {V_3}{V_4}$$

And,

$$Efficiency=1-\frac{T_cln\left(\frac{V_3}{V_4} \right )}{T_hln\left (\frac{V_2}{V_1} \right )}=1-\frac{T_cln\left(\frac{V_2}{V_1} \right )}{T_hln\left (\frac{V_2}{V_1} \right )}=1-\frac{T_c}{T_h}$$

It can be seen that the Carnot engine is most efficient when the ratio of heat put in and heat released is smallest. To get a hundred percent efficiency we would need the temperature to be either (or both) extremely cold (0oK) or infinitely hot (both of the conditions are clearly unattainable). Moreover, if we were to look at the PV diagram, our derivation of efficiency makes logical sense because the larger the area contained in the path 1-2-3-4, the more work we get out of our system. Having more heat entering and less heat leaving results in more work done.

### References

1. Serway, Raymond A. Physics for Scientists and Engineers. Orlando, FL: Sanders College Publishing, 1992. Print.
2. Moore, Thomas A. Six Ideas That Shaped Physics, Unit T:Some Processes Are Irreversible. 2nd Ed. New York, NY: McGraw-Hill, 2003. Print.
3. Oxtoby, David W., Gillis, H. P., Campion, Alan. Principles of Modern Chemistry. Belmont, CA: Thomson Brooks/Cole, 2008. Print.

### Problems

1. Name two key characteristics of a reversible process.
2. What would be the work done by the piston if there is no external pressure (vacuum)?
3. How would we find the heat absorbed in an isothermal reversible process (step 1-2)?
4. What would the delta U, change in internal energy, be after one Carnot Cycle? Why don't we have "delta work"? In terms of difference between the work and internal energy function?
5. Overall, why is the Carnot engine the most efficient engine?