A thermodynamic cycle describes a set of processes through which a system returns to the same state that it was in initially. Typically cycles are used to produce power or provide refrigeration. Since the system returns to its initial state after the cycle has been completed, all the properties have the same values they had originally. The advantage of executing a thermodynamic cycle is that by having the system return to its initial state, we can repeat the cycle continuously. There are many different examples of ther- modynamic cycles; in this section, we examine one such cycle—the Carnot cycle.12 In Chapter 3, we will learn that a Carnot cycle represents the most effi cient type of cycle we can possibly have.
11 In general, the energy balance of throttling processes reduces to this simple form.
12 This cycle was conceived in 1824 by Sadi Carnot, a French engineer, to explore the maximum possible effi ciency that could be obtained by the steam engines of his time.
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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 103
Figure 2.17 shows an ideal gas in a piston–cylinder assembly undergoing a Carnot cycle.
In this cycle, the gas goes through four reversible processes through which it returns to its initial state. Two processes occur isothermally, alternating with two adiabatic processes.
These processes were analyzed, individually, in Section 2.7. Consider a gas that is initially in state 1 at a pressure P1 and a temperature T1 as shown at the top of Figure 2.17. The fi rst step of a Carnot cycle is a reversible isothermal expansion, in which the gas is exposed to a hot reservoir at temperature, TH; it gains energy via heat, QH, as indicated on the diagram.
During this process, which takes the system from state 1 to state 2 (at P2 and T2 ), the pres- sure decreases while the temperature stays the same. The piston–cylinder assembly is then transferred into an adiabatic (well-insulated) environment and expanded further to state 3.
In this step, both T and P decrease. In both expansion processes, work is done by the system on the surroundings; that is, we get useful work out. The system then undergoes two revers- ible compression processes. First, it is isothermally compressed by being placed in contact with a cold thermal reservoir at temperature TC. The gas loses an amount of energy via heat, QC to the cold reservoir. This process takes the system to state 4 1P4, T42. The system returns to its initial state (state 1) through an adiabatic compression.
The net work obtained in a Carnot cycle is given by the sum of the work obtained in all four processes:
2Wnet5 0W1201 0W230 2 0W340 2 0W410 (2.51)
Since the overall effect of the power cycle is to deliver work from the system to the surroundings, the sign of Wnet is negative. The subscript “ij” on the terms for work in Figure 2.17 An ideal gas undergoing a Carnot cycle. The Carnot cycle consists of four reversible processes by which the gas is returned to its original state.
Constant TH
Constant TC State 1 State 2
State 2 State 3 State 1 State 4
State 4 State 3
Isothermal compression Adiabatic
compression Adiabatic
expansion Isothermal
expansion
Well insulated QH
T1
P1
P4 P3 P2 T2 = T1
T1
P1 P4
T4 T2
P2 P3
T3
T4 = T3 T3
QC
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104 ► Chapter 2. The First Law of Thermodynamics
Equation (2.51) refers to the work obtained in going from state i to state j. Absolute val- ues are used to explicitly distinguish the steps where we get work out from those where we must put work in.
The net work obtained from a Carnot cycle can also be calculated by applying the fi rst law to the entire cycle. Since the cycle returns the system to its original state, its internal energy must have the same value as at the start of the cycle. Thus,
DUcycle505Wnet1Qnet (2.52)
Comparing Equations (2.51) and (2.52), we see that:
2Wnet5Qnet5Q121Q231Q341Q415 0QH0 2 0QC0
We see that the net work obtained is the difference in heat absorbed from the hot reservoir, QH, and expelled to the cold reservoir, QC. An alternative way of schematically representing a Carnot cycle is shown in Figure 2.18a. This schematic gives an overview of the energy transferred between the Carnot engine and the surroundings. Inside the circle labeled “Carnot engine” are the four processes depicted in Figure 2.17.
0 0
Figure 2.18 Alternative representation for the Carnot cycle. (a) Carnot “engine”; (b) Carnot refrigerator.
Cold reservoir, TC
Hot reservoir, TH Hot reservoir, TH
Carnot engine
Carnot refrigeration QH
QC QC
QH
Wnet
Cold reservoir, TC
Wnet
(a) (b)
QH QC
Efficiency
The effi ciency, h, of the cycle is defi ned as the net work obtained divided by the heat absorbed from the hot reservoir:
h; net work
heat absorbed from the hot reservoir5 Wnet
QH (2.53)
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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 105 For a given amount of energy available from the hot reservoir via QH, the greater the effi - ciency, the more work we obtain. For example, say the high temperature in the hot reser- voir is obtained from the combustion of coal. A high effi ciency means we can reduce the amount of coal we need to combust to produce a given amount of work.13
A refrigeration cycle allows us to cool a system so that we can store some ice cream and so on. In this case we want to expel heat from a cold reservoir. It takes work from the surroundings to accomplish this task. For example, your freezer at home needs electricity to keep the ice cream cold. Figure 2.18b shows a schematic way of represent- ing the energy transferred in a refrigeration cycle. We supply work to the cycle in order to absorb energy via heat QC from the cold reservoir. We then expel the energy via heat QH to the hot reservoir. Thus, the direction of heat transfer is opposite that of the power cycle depicted in Figure 2.18a. The effectiveness of a refrigeration cycle is measured by its coeffi cient of performance, COP, which is defi ned as follows:
COP5 QC
Wnet
(2.54) We can see from Equation (2.54) that the higher the COP, the less work it takes to produce a desired level of cooling.
Can you draw the analogous cycle to Figure 2.17 that goes in the circle labeled
“Carnot refrigerator”?
13 The steam engine was patented by James Watt in 1765. These fi rst steam engines had effi ciencies of only about 1%. Indeed, we can see there was much engineering that remained to be done!
Consider 1 mole of an ideal gas in a piston–cylinder assembly. This gas undergoes a Carnot cycle, which is described below. The heat capacity is constant, cv5 13/22R.
(i) A reversible, isothermal expansion from 10 bar to 0.1 bar.
(ii) A reversible, adiabatic expansion from 0.1 bar and 1000 K to 300 K.
(iii) A reversible, isothermal compression at 300 K.
(iv) A reversible, adiabatic compression from 300 K to 1000 K and 10 bar.
Perform the following analysis:
(a) Calculate Q, W, and DU for each of the steps in the Carnot cycle.
(b) Draw the cycle on a Pv diagram.
(e) Calculate the effi ciency of the cycle.
(d) Compare h to 12 1Tc/TH2.
(e) If what is found in part (d) is true, in general, suggest two ways to make the above process more effi cient.
SOLUTION (a) We will analyze each of the steps separately, with a little help from the results of Section 2.7. We label each state in a manner consistent with Figure 2.17.
(i) The fi rst process is a reversible, isothermal expansion at 1000 K from state 1 at 10 bar to state 2 at 0.1 bar. By defi nition, the internal energy change for an ideal gas at constant temperature is:
DU50 We can calculate the work using a result from Section 2.7:
W53nRT
P dP5nRT ln P2
P1
5 238,287 3J4 EXAMPLE 2.23
Carnot Cycle Effi ciency
(Continued)
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106 ► Chapter 2. The First Law of Thermodynamics
The negative sign indicates that the system is performing work on the surroundings (we are getting useful work out). To fi nd the heat, we apply the fi rst law:
QH5 DU2W538,287 3J4
(ii) The second process is a reversible, adiabatic expansion from 0.1 bar and 1000 K to state 3 at 300 K. The pressure decreases during this process. By the defi nition of an adiabatic process:
Q50
At constant heat capacity, the change in internal energy becomes:
DU5ncv1T32T22 5 28730 3J4 Applying the fi rst law gives:
W5 DU5 28730 3J4
(iii) The third process is a reversible, isothermal compression at 300 K. Again:
DU50 and,
W5 23PdV53nRT
P dP5nRT lnP4
P3
(E2.23A) However, we now need to fi nd P3 and P4. From Section 2,7, we know PVk5const for the polytropic, adiabatic processes (ii) and (iv). We fi rst fi nd k:
k5cP
cv 5cv1R cv 51.67 Setting PVk equal for states 2 and 3 gives:
PVk5P2V21.675 1nRT221.67
P20.67 573475 1nRT321.67 P30.67
Solving for P3, we get:
P35 B1nRT321.67 7347 R
1.5
50.0049 bar Similarly for P4:
PVk5P1V11.675 1nRT121.67
P10.67 53415 1nRT421.67 P40.67 and,
P45 B1nRT421.67
341 R
1.5
50.49 bar Thus, the work given by Equation (E2.23A) is:
W5nRT ln 0.49 bar
0.0049 bar 511,486 3J4
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2.9 Thermodynamic Cycles and the Carnot Cycle ◄ 107
The work is positive for this compression process. From the fi rst law, QC5 DU2W5 211,486 3J4
(iv) The fourth process is a reversible, adiabatic compression from state 4 at 300 K and 0.52 bar back to state 1 at 1000 K and 10 bar (process 4S1). After this process, the gas can repeat steps (i), (ii) . . . Again, for this adiabatic compression:
Q50
At constant heat capacity, the change in internal energy becomes:
DU5ncv1T12T42 58730 3J4 Applying the fi rst law gives:
W5 DU58730 3J4
TABLE E2.23A Results of Calculations for Carnot Cycle in Example 2.23
Process DU3J4 W [J] Q [J]
(i) State 1 to 2 0 238,287 38,287
(ii) State 2 to 3 28,730 28,730 0
(iii) State 3 to 4 0 11,486 211,486
(iv) State 4 to 1 8,730 8,730 0
Total 0 226,800 26,800
TABLE E2.23B T, P, and v for Carnot Cycle in Example 2.23
State T [K] P [bar] v 3m3/mol4
1 1000 10 0.0083
2 1000 0.1 0.8314
3 300 0.0049 5.10
4 300 0.49 0.051
A summary of DU, W , and Q for the four processes and the totals for the cycle are presented in Table E2.23A. We get a net work of 26.8 kJ after one cycle.
(b) To sketch this process on a Pv diagram, we fi rst calculate the molar volume at each state using the ideal gas law. The results are presented in Table E2.23B. A sketch (not to scale) of the Pv diagram is presented in Figure E2.23. The work for a reversible process is given by the area under the Pv curve; hence, the net work is given by the shaded area in the box in Figure E2.23.
Isotherms, TH and TC, for processes (i) and (iii) are also labeled.
(c) The effi ciency is given by Equation (2.53)
h; net work
heat absorbed from the hot reservoir526,800
38,28750.70 (E2.23B) In practice, electrical power plants have effi ciencies around 40%.
(Continued)
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108 ► Chapter 2. The First Law of Thermodynamics
(d) Applying the relation in the problem statement, we get:
12TC
TH512 300
100050.7 (E2.23C)
Comparing the values of Equations (E2.23B) and (E2.23C), we get:
h 512TC
TH
(E2.23D) (e) If Equation (E2.23D) holds, the process can be made more effi cient by raising TH or lowering TC. Note that these options will push the isotherms depicted in Figure E2.23 up and down, respectively. Thus either raising TH or lowering TC will serve to make the shaded box, which represents net work, larger. We will learn in Chapter 3 that, indeed, Equation (E2.23D) is true in general. However, we can also reach this conclusion by realizing that the isotherms in Figure E2.23 are fi xed on the Pv plane.
Figure E2.23 Pv diagram of a Carnot cycle. The shaded area represents the net work obtained from one cycle.
3 2
P
TH
Isotherm
wnet
TC
Isotherm Q= 0
Q= 0
−Q
C =W QH
=−W
v 4
1