DU 5 2088 J This is an example where the change in enthalpy does not equal the change in internal
3.3 The Carnot Cycle and Efficiency
In 1824, a French military engineer named Nicolas Leonard Sadi Carnot (his third name is borrowed from a Persian poet, and his surname is pronounced kar-NO) published an article that ultimately played a major—though roundabout—role in the development of thermodynamics. It was ignored at the time. The first law of thermodynamics had not even been established yet, and heat was still thought of as
“caloric.” It was not until 1848 that Lord Kelvin brought the attention of the scien- tific world to the work, 16 years after Carnot’s early death at age 36. However, the article introduced a lasting concept, the definition of the Carnot cycle.
58437_ch03_ptg01_p075-100.indd 76 27/08/13 1:03 PM
www.pdfgrip.com
3.3 | The Carnot Cycle and Efficiency 77
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Carnot was interested in understanding the ability of steam engines—known for almost a century by that time—to perform work. He was apparently the first to understand that there was a relationship between the efficiency of a steam en- gine and the temperatures involved in the process. Figure 3.1 shows a diagram of Carnot’s idealized engine. Carnot realized that every engine could be defined as get- ting heat, qin, from some high-temperature reservoir. The engine converted some heat to work, w, which it performed on the surroundings. The engine then disposed of the leftover heat in a reservoir that has some lower temperature. The engine is therefore emitting some heat, qout, into the low-temperature reservoir. Although not all engines operate exactly like this, every device we have for performing work can be modeled in this fashion.
Carnot proceeded to define the steps for the operation of an engine. These steps, collectively called the Carnot cycle, can be used to define how efficient the engine was at converting heat to work. The engine itself is defined as the system, and a schematic of the cycle is shown in Figure 3.2. The steps of a Carnot cycle are:
1. Reversible isothermal expansion. In order for this to occur, heat must be absorbed from the high-temperature reservoir. We shall define this amount of heat as q1 (labeled as qin in Figure 3.1) and the amount of work performed by the system as w1.
2. Reversible adiabatic expansion. In this step, q50, but because it is expansion, work is done by the engine. The work is defined as w2.
3. Reversible isothermal compression. In order for this step to be isothermal, heat must leave the system. It goes into the low-temperature reservoir and will be labeled q3 (this is labeled as qout in Figure 3.1). The amount of work in this step will be called w3.
4. Reversible adiabatic compression. The system (that is, the engine) is returned to its original conditions. In this step, q is 0 again, and work is done on the system. This amount of work is termed w4.
Because the system has returned to the original conditions, by definition of a state function, DU50 for the overall process. By the first law of thermodynamics,
DU505q11w11w21q31w31w4 (3.1)
Figure 3.1 A diagram of the type of engine that Carnot considered for his cycle. The high- temperature reservoir supplies the energy to run the engine, which produces some work and emits the remainder of the energy into a low-temperature reservoir. The values of qin, w1, and qout are greater or less than zero with respect to the system.
Engine System
High-temperature reservoir, T1
Surroundings Supplies heat, qin, . 0
Does work, w1, , 0
Low-temperature reservoir, T2
Emits heat, qout, , 0
Figure 3.2 A representation of the Carnot cycle performed on a gaseous system. The steps are: (1) Reversible iso- thermal expansion. (2) Reversible adia- batic expansion. (3) Reversible isothermal compression. (4) Reversible adiabatic compression. The system ends up at the same conditions it started at; the area inside the four-sided figure is represen- tative of the p — V work performed by the cycle.
Step 4 Step 1 A
D B
C Step 2 Step 3
Volume
Pressure
58437_ch03_ptg01_p075-100.indd 77 27/08/13 1:03 PM
www.pdfgrip.com
78 Chapter 3 | The Second and Third Laws of Thermodynamics
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Another way of writing this is to consider the entire work performed by the cycle, as well as the entire heat flow of the cycle:
wcycle5w11w21w31w4 (3.2)
qcycle5q11q3 (3.3)
so that
05qcycle1wcycle
qcycle5 2wcycle (3.4)
We now define efficiency e as the negative ratio of the work of the cycle to the heat that comes from the high-temperature reservoir:
e 5 2wcycle
q1 (3.5)
Efficiency is thus a measure of how much heat going into the engine has been con- verted into work. The negative sign makes efficiency positive, because work done by the system has a negative value but heat coming into the system has a positive value.
We can eliminate the negative sign by substituting for wcycle from equation 3.4:
e5 qcycle
q1 5 q11q3
q1 511q3
q1 (3.6)
Because q1 is heat going into the system, it is positive. Because q3 is heat going out of the system (into the low-temperature reservoir of Figure 3.1), it is negative.
Therefore, the fraction q3/q1 will be negative. Further, it can be argued that the heat leaving the engine will never be greater than the heat entering the engine. That would violate the first law of thermodynamics that energy cannot be created. Therefore, the magnitude 0q3/q10 will never be greater than 1, but it will always be less than or (if no work is done) equal to 1. Combining all these statements, we conclude that:
The efficiency of an engine will always be between 0 and 1.
ExamplE 3.1
a. Determine the efficiency of a Carnot engine that takes in 855 J of heat, performs 225 J of work, and gives off the remaining energy as heat.
b. Draw a diagram like Figure 3.1 showing the exact amounts of heat and work going from place to place in the proper direction.
Solution
a. Using both definitions of efficiency, and recognizing the proper signs on the heat and work:
e5 22225 J
1855 J50.263 e5112(8552255) J
855 J 511(20.737)50.263 b. The drawing is left to the student.
There is another way to express efficiency in terms of the temperatures of the high- and low-temperature reservoirs, assuming an ideal gas. For the isothermal
58437_ch03_ptg01_p075-100.indd 78 27/08/13 1:03 PM
www.pdfgrip.com
3.3 | The Carnot Cycle and Efficiency 79
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
steps 1 and 3, the change in the internal energy is zero because ('U/'V)T50.
Therefore, q5 2w for steps 1 and 3. From equation 2.7, for an ideal gas, w5 2nRT ln Vf
Vi
For a reversible, isothermal process, the heat for steps 1 and 3 are:
q15 2w15nRThigh ln VB
VA (3.7)
q35 2w35nRTlow ln VD
VC (3.8)
The volume labels A, B, C, and D represent the initial and final points for each step, as shown in Figure 3.2. Thigh and Tlow are the temperatures of the high-temperature and low-temperature reservoirs, respectively. For the adiabatic steps 2 and 4, we can use equation 2.47 to get
aVB
VCb2/3 5 Tlow Thigh aVA
VDb2/35 Tlow
Thigh
Equating the two volume expressions, which both equal Tlow/Thigh: aVA
VDb2/35 aVB VCb2/3
Raising both sides to the power of 3/2 and rearranging, we get aVA
VBb5 aVD VCb
Substituting for VD/VC in equation 3.8, we get an expression for q3 in terms of vol- umes VA and VB:
q35nRTlow ln VA
VB 5 2nRTlow ln VB
VA (3.9)
Equations 3.7 and 3.9 can be divided to get a new expression for the ratio q3/q1: q3
q15
nRTlow ln VB
VA 2nRThigh ln VB
VA
5 2Tlow Thigh
Substituting into equation 3.6, we get an equation for efficiency in terms of the temperatures:
e512 Tlow
Thigh (3.10)
Equation 3.10 has some interesting interpretations. First, the efficiency of an engine is very simply related to the ratio of the low- and high-temperature reservoirs. The smaller this ratio is, the more efficient an engine is.* Thus, high efficiencies are favored by high Thigh values and low Tlow values. Second, equation 3.10 allows us to
*In practice, other factors (including mechanical ones) reduce the efficiency of most engines.
58437_ch03_ptg01_p075-100.indd 79 27/08/13 1:03 PM
www.pdfgrip.com
80 Chapter 3 | The Second and Third Laws of Thermodynamics
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
describe a thermodynamic scale for temperature. It is a scale for which Tlow50 when the efficiency equals 1 for the Carnot cycle. This scale is the same one used for ideal gas laws, but it is based on the efficiency of a Carnot cycle, rather than the behavior of ideal gases.
Finally, unless the temperature of the low-temperature reservoir is absolute zero, the efficiency of an engine will never be 1; it will always be less than 1. Because it can be shown that absolute zero is physically unobtainable for a macroscopic object, we have the further statement that
No engine can ever be 100% efficient.
When one generalizes by recognizing that every process can be considered an engine of some sort, the statement becomes
No process can ever be 100% efficient.
It is statements like this that preclude the existence of perpetual motion machines, devices that purportedly have an efficiency greater than 1 (that is, .100%), pro- ducing more work out than the energy coming in. Carnot’s study of steam engines helped establish such statements, and so much stock is placed in them that the U.S.
Patent and Trademark Office categorically does not consider any patent application claiming to be a perpetual motion machine (although some applications for such machines are considered because they disguise the claim). Such is the power of the laws of thermodynamics.
The two expressions for efficiency can be combined:
11 q3
q1512 Tlow Thigh q3
q15 2Tlow Thigh q3
q1 1 Tlow Thigh50 q3
Tlow1 q1
Thigh50 (3.11)
Notice that q3 is the heat that goes to the low-temperature reservoir in isothermal step 3 of the cycle, whereas q1 is the heat that comes from the high-temperature reservoir in isothermal step 1. Each fraction therefore contains heat and tempera- tures from related parts of the universe under consideration. Because the other two steps are adiabatic (that is, q25q450), equation 3.11 includes all of the heats of the Carnot cycle. The fact that these heats, divided by the absolute temperatures of the two reservoirs involved, add up to exactly zero is interesting. Recall that the cycle starts and stops at the same system conditions. But changes in state functions are dictated solely by the conditions of the system, not by the path that got the sys- tem to those conditions. If a system starts and stops at the same conditions, overall changes in state functions are exactly zero. Equation 3.11 suggests that for reversible changes, a relationship between heat and absolute temperature is a state function.