Heat Engines and the Carnot Cycle

A heat engine is a cyclic process that absorbs heat and does work on the surroundings. "Cyclic" means that the system returns to its initial state at the end of each cycle so that there is no permanent change in the system. The only cycle we will work with in this course is the Carnot Cycle which is shown below on a *p-V* diagram.

The curves labeled *T _{U}* and

We can analyze the Carnot cycle as follows: The heat, *q*, for the whole cycle is

(1) *q = q _{AB}* +

and the work for the entire cycle is,

(2) *w = w _{AB} *+

Since the initial state of the cycle is the same as the final state we know that the change in *U*, the internal energy, is zero. (That's part of the first law, because the first law says that *U* is a state function, which means that the value of *U* for any state of the system does not depend on how the system got to that state, only on what the state actually is.) The first law of thermodynamics then tells us that

(3) ,

from which we see that

(4) .

Notice that when the Carnot cycle is operating as a heat engine (going around clockwise), *w* < 0, so that -*w* > 0 and *q* > 0.

As we have said, the Carnot cycle is a heat engine. That means that we want it to absorb heat and convert that heat energy into work. We certainly should care about how much work we get out for the heat we absorb. Thus we will define the efficiency, *e*, of a cycle as

(5) ,

which, when combined with Equations 1 and 4 gives

(6)

In order to conform to the usual notation we note that the A→
B segment of the cycle is at the temperature of the upper heat bath, *T _{U}*, and the segment C→
D is at the temperature of the lower heat bath,

(7)

The A→
B and C→
D segments are an isothermal expansion of an ideal gas at *T _{U}* and an isothermal compression at

(8)

and

(9)

Let's plug these heats into Equation 7, (right hand side) to get,

(10) .

The *nR* cancels leaving,

(11) .

If we are looking for an expression for efficiency in terms of temperatures Equation 11 won't do the job because of all the volumes in the equation. However, there is another set of relations between the volumes from the fact that the paths B→ C and D→ A are adiabatic and we know the volume relationships for adiabatic processes in an ideal gas. Recall that for an adiabatic expansion of an ideal gas we had the expression (using the temperatures, pressures, and volumes appropriate to our Carnot cycle here)

(12)

for the B→
C segment of the cycle and a similar equation for the D→
A segment. (Recall that *γ*
is the "heat capacity ratio," *C _{p}/C_{V}*.) Let us eliminate the pressures from Equation 12 by inserting the value of pressure from the ideal gas equation of state, remembering that at points A and B the temperature is

(13) .

Cancel *nR* from both sides and combine the volumes to get,

(14)

From which we obtain

(15)

Note that there is an equivalent expression for the D→ A leg of the cycle,

(16) .

From these two last equations we conclude that

(17) ,

or

(18)

or, on rearranging,

(19) .

We now apply this result to Equation 11 to get,

(20) .

The two logarithms are the negatives of each other so we conclude that,

(21)

.Notice that the highest efficiencies are obtained with a low temperature for the heat bath at *T _{L}* and that you could only obtain unit efficiency if the lower heat bath were at absolute zero. Since, in most cases,

You may have noticed that all of our temperatures are ideal gas temperatures. That is,they are measured on the ideal gas temperature scale. It is possible, and desirable to use the Carnot cycle to define a "thermodynamic temperature scale." We will not do this here. Suffice it to say that by choosing a suitable reference temperature you can make the two scales identical.

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