In this section, we will examine the basic elements of common industrial power and refrigeration systems. These systems employ a thermodynamic cycle in which a working fl uid is alternatively vaporized and condensed as it fl ows through a set of four processes.
Recall from Section 2.9 that, after completing a cycle, the working fl uid returns to the same state it was in initially so that the cycle can be repeated. We will use the principles of energy conservation and entropy to analyze the performance of power and refrig- eration cycles. We will fi rst examine the Rankine cycle, which is used to convert a fuel source to electrical power. We will then look at a vapor-compression cycle operated in
“reverse” to expel heat from a cold reservoir and produce refrigeration.
The Rankine Cycle
Say we wish to convert a fossil-fuel, nuclear, or solar energy source into net electrical power. To accomplish this task, we can use a Rankine cycle. The Rankine cycle is an ide- alized vapor power system that contains the major components found in more detailed, practical steam power plants. While hydroelectric and wind are possible alternative sources, the steam power plant is presently the dominant producer of electrical power.
A schematic of the Rankine cycle is shown in Figure 3.7. The left-hand side shows the four unit processes in order: a turbine, a condenser, a compressor, and a boiler.
States 1, 2, 3, and 4 are labeled. The right-hand side identifi es each of these states on a Ts diagram. Each of the four individual processes operates as an open system at steady- state, such as those modeled in Section 2.8. Moreover, these processes are assumed to be reversible; hence, the effi ciency we calculate will be the best possible for a given design scenario.
The working fl uid that fl ows through these processes is usually water. We will for- mulate our analysis on a mass basis, in anticipation of using the steam tables for ther- modynamic data. Electrical power is generated by the turbine, while the energy from combustion of the fuel is input via heat transfer in a boiler. Energy transfer between the surroundings and the system is further needed to return the system to its initial state and complete the cycle. This energy transfer occurs via heat expulsion in the condenser and via work input to the compressor. A more detailed analysis of the four processes in the Rankine cycle follows.
Fuel air
Turbine Rankine cycle
1
4
3 Condenser
Compressor
Boiler Cooling
water 2
Ws
Wc
QC
QC
QH
Wc
Ws
QH
1
2 3
4
s T
Figure 3.7 The ideal Rankine cycle used to convert fuel into electrical power. The four unit processes are sketched on the left, while the path on a Ts diagram is shown on the right. The working fluid is typically water.
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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 165 We start from state 1 in the diagram in Figure 3.7, where the working fl uid enters the turbine as a superheated vapor. As it goes through the turbine, it expands and cools while producing work. It exits the turbine at state 2. The rate of work produced can be determined by applying the fi rst law [Equation (2.50)]. Assuming that bulk kinetic and potential energy and heat transfer are negligible, the power produced by the turbine becomes:
W#
s5m# 1h^22h^12 where m# is the fl ow rate of the working fl uid.
Since the process is reversible with negligible heat transfer, the entropy remains constant, as depicted by the vertical line in the Ts diagram:
s^15s^2
The steam enters as a superheated vapor, and does not condense signifi cantly in the turbine. If the steam were saturated when it entered the turbine, a signifi cant fraction of liquid would be formed when the temperature dropped isentropically. The dashed line on the Ts diagram illustrates this case. This option is impractical, since too much liquid causes erosion and wear of the turbine blades.
The steam next enters a condenser; it exits in state 3 as saturated liquid water. The change of phase occurs at constant pressure and requires that energy be removed from the fl owing stream via heat. Thus, a low-temperature reservoir is needed. A fi rst-law bal- ance around the condenser gives:
Q#
C5m# 1h^32h^22
Next it is desired to raise the pressure of the liquid, which is accomplished using a compressor. High-pressure water exits the compressor in state 4. The work required to compress the liquid is given by:
W#
c5m# 1h^42h^32 5m#v^11P42P32 (3.33)
where Equation (3.32) was integrated assuming v^l is constant. Since the molar volume of the liquid is signifi cantly less than that of the vapor, the work required by the compressor is much less than that produced by the turbine. Typically, a small fraction of the power produced by the turbine is used to compress the liquid, and the remaining power is the net power obtained by the cycle. The liquid that enters the compressor is saturated, by design, since most compressors cannot handle a two-phase mixture.
Finally, the high pressure liquid is brought back to a superheated vapor state in the boiler. It is in this step that energy released from the combustion of fuel is transferred to the working fl uid. The fuel provides the high-temperature reservoir for the boiler. The boiler isobarically heats the liquid to saturation, vaporizes it, and then superheats the vapor. The rate of heat transfer in the boiler is given by:
Q#
H5m# 1h^12h^42
The vapor exits the boiler in state 1 and the cycle is repeated.
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166 ► Chapter 3. Entropy and the Second Law of Thermodynamics
We defi ne the effi ciency of the cycle as the ratio of the net work obtained divided by the heat absorbed from the boiler:
hRankine5 0W#
s0 2W#
c
Q#
H 5 0 1h^22h^12 0 2 1h^42h^32
1h^12h^42 (3.34) In this defi nition, it is assumed that the heat absorbed is proportional to the amount of fuel consumed.
The Ts diagram provides a useful graphical aid in interpreting the Rankine cycle.
From the defi nition of entropy,
qrev53Tds
Since we are assuming reversibility for the cycle, the heat absorbed by the water in the boiler, qH, and the heat expelled in the condenser, qC, are equal to the respective area under the Ts curve. These graphical depictions are illustrated in the fi rst two diagrams of Figure 3.8. The net work produced by the cycle is given by the difference of these two quantities:
qH2 0qC0 5 0ws0 2wc5wnet
Thus, the net work is equal to the area of the box in the third diagram. If we can make this box bigger relative to qH, we increase the effi ciency. Can you think of ways to accom- plish this?
q H − qC = wnet
2
s s s
T T T
3 4
1
2 3
4
1
2 3
4
1
wnet
qC qH
Figure 3.8 Graphical representation of the heat absorbed in the boiler, qH, the heat expelled in the condenser, qC, and the net work, wnet, in an ideal Rankine cycle.
Steam enters the turbine in a power plant at 600ºC and 10 MPa and is condensed at a pressure of 100 kPa. Assume the plant can be treated as an ideal Rankine cycle. Determine the power produced per kg of steam and the effi ciency of the cycle. How does the effi ciency compare to a Carnot cycle operating between these two temperatures?
SOLUTION We can refer to Figure 3.7 to identify the states of water as it goes through the cycle. It is useful to refer to this fi gure as we are solving the problem. Examining Equation (3.34), we see that we need to determine the enthalpies in the four states to solve for the EXAMPLE 3.14
Calculation of the Power and Effi ciency of a Rankine Power Cycle
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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 167
effi ciency. Steam enters the turbine at 600ºC and 10 MPa. Looking up values from the steam tables (Appendix B), we get:
h^153625.3 3kJ/kg4
and, s^156.9028 3kJ/1kg K2 45s^2
Since the entropy at state 1 is equal to the entropy at state 2. We also know the pressure at state 2, P25100 kPa. Thus, state 2 is completely constrained, and we can determine its enthalpy from the steam tables. Since it exists as a liquid–vapor mixture, we must determine the quality, x, of the steam as follows:
s^256.90283kJ/1kg K2 45112x2s^l1xs^v5 112x2 11.30253kJ/kg K4 21x17.35933kJ/kg K4 2 Solving for x gives:
x50.925 Therefore, the enthalpy in state 2 is given by:
h^25 112x2h^l1xh^v510.0752 1417.443kJ/kg4 210.92512675.53kJ/kg4 252505.33kJ/kg4 The power generated by the turbine is found by the enthalpy difference between state 2 and state 1:
w^s5h^22h^152505.323625.35 21120.0 3kJ/kg4 The enthalpy of state 3 is a saturated liquid at 100 kPa:
h^35h^15417.44 3kJ/kg4 The enthalpy at state 4 can be determined from Equation (3.33):
h^45h^31v^l1P42P325427.343kJ/kg4
Note that since we are increasing the pressure of water in the liquid state, the work required is only 9.9 kJ/kg. This value is less than 1% of the work produced by expansion of vapor through the turbine (1120 kJ/kg). The net work is given by:
w^net5w^s1w^c5 21120.019.95 21110.1 3kJ/kg4 Solving for the effi ciency from Equation (3.34), we get:
hRankine51 0W#
s02W#
c2/Q#
H5 3 0 1h^22h^12 02 1h^42h^32 4/1h^12h^42 50.347
The 34.7% effi ciency is the best-case scenario for this cycle, since we assumed reversible processes. In reality, we would not even achieve this value!
The Carnot effi ciency is given by Equation (3.9):
hCarnot512 TC
TH512373 87350.573
The Rankine effi ciency is lower than the Carnot effi ciency. We can see the basis if we compare the net work graphically, as we did in Figure 3.8. The box representing net work for each cycle is shown in Figure E3.14. Recall that the steam that enters the turbine of the Rankine cycle is superheated to eliminate wear and corrosion on the turbine blades. In modifying the cycle in this way, we “crop off” a signifi cant portion of the rectangle that represents the Carnot cycle.
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168 ► Chapter 3. Entropy and the Second Law of Thermodynamics
Redo the analysis of the Rankine cycle of Example 3.14 but include isentropic effi ciencies of 85% in the pump and turbine. Determine the net power and the overall effi ciency of the power cycle.
SOLUTION Recall the discussion of isentropic effi ciencies from Example 3.13. The isentropic effi ciency of the turbine is given by:
hturbine5 1W#
s2actual
1W#
s2rev
5 1w^s2actual
1w^s2rev
The reversible work was found in Example 3.14. Solving for actual work gives:
1w^s2actual5 hturbine1w^s2rev50.85121120.025 2952.0 3kJ/kg4
As we suspect, the work we get out of the turbines in the real, irreversible process is less than that for the reversible process.
Solving the energy balance around the turbine gives the enthalpy in state 2 as:
1h^22actual5 1w^s2actual1h^15 2952.013625.352673.3 3kJ/kg4
This value is higher than that found in Example 3.14, indicating that the temperature entering the condenser for the actual process is higher than it would be in the reversible case. State 3 remains the same:
h^35h^15417.44 3kJ/kg4
Since the reversible work represents the best we can possibly do, the actual work needed in the compressor must be greater than the reversible value. Hence, the isentropic effi ciency in the compressor is given by:
1w^c2actual5 1w^c2rev
hcompressor5 9.9
0.85511.6 3kJ/kg4 Solving for the enthalpy at the exit of the compressor gives:
1h^42actual5 1w^c2actual1h^3511.61417.445429.1 3kJ/kg4 EXAMPLE 3.15
Modifi cation of Rankine Analysis for Nonisentropic Steps
Figure E3.14 Graphical depiction of the net work in a Carnot cycle and a Rankine cycle.
2 3 3
4 1
2
4
1
Carnot Cycle Rankine Cycle
T T
s s
wnet wnet
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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 169
The net work can be found by adding the actual work generated in the turbine to that consumed in the compressor:
w^net5w^s1w^c5 2952.0111.65 2940.43kJ/kg4 Similarly, the actual effi ciency must be calculated using these values:
hRankine5 1 0W#
s02W#
c2/Q#
H5 3 0 1h^22h^12 021h^42h^32 4/1h^12h^42 50.294 Introducing irreversibilities in turbine and pump reduced effi ciency from 34.7% to 29.4%.
The Vapor-Compression Refrigeration Cycle
Refrigeration systems are important in industrial and home use when temperatures less than the ambient environment are required. Of the several types of refrigeration sys- tems, the most widely used is the vapor-compression refrigeration cycle. It is essentially a Rankine cycle operated in “reverse,” where heat is absorbed from a cold reservoir and rejected to a hot reservoir. Due to the constraints of the second law, this process can be accomplished only with a concomitant consumption of power.
A schematic of the ideal vapor-compression cycle is shown in Figure 3.9. The left- hand side shows the four unit processes in order: an evaporator, a compressor, a con- denser, and a valve. Each of the four individual processes operates as an open system at steady-state. States 1, 2, 3, and 4 are labeled. The right-hand side identifi es each of these states on a Ts diagram. Unlike in the Rankine cycle, the work required for refrigeration is not represented by the area enclosed on the Ts diagram because the expansion through the valve is irreversible.
The working fl uid is termed the refrigerant. In choosing a refrigerant, we must real- ize that both the evaporation and condensation processes contain phase transformations.
Thus, in each of these processes, T and P are not independent. Specifying the tempera- ture at which these processes occur, restricts the pressure for a given choice of refriger- ant. For example, the evaporator temperature is determined by the temperature, TC, required from our refrigeration system. For a given working fl uid, constraining TC also constrains the evaporator P. We typically want a species that boils at lower temperatures than water. Ideally, we choose a species that provides the desired refrigeration tempera- ture at a pressure slightly above atmospheric. In that way there is a positive pressure against the environment. Common choices are CCl2F2 (refrigerant 12), CCl3F (refriger- ant 11), CH2FCF3 (refrigerant 134a), and NH3. The fi rst two species, the chlorofl uoro- carbons, are very stable if released to the environment. They have mostly been phased out of use because they lead to depletion of the ozone layer and also contribute to the greenhouse effect, which leads to global climate change.
An analysis of the four processes in the vapor-compression refrigeration cycle follows. We start from state 1 in the diagram in Figure 3.9, where the working fluid enters the evaporator. In the evaporator, heat is transferred from the refrigerated unit to the working fl uid. This occurs at temperature TC. The working fl uid absorbs Q#
C as it changes phase.
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170 ► Chapter 3. Entropy and the Second Law of Thermodynamics
It emerges in state 2, where it is a vapor. The heat transferred is given by:
Q#
C5n#1h22h12 (3.35)
where n# is the molar fl ow rate of the refrigerant.
We must then compress the refrigerant to a high enough pressure in state 3 that it will condense at the temperature of the hot reservoir available to us, TH.The choice of refrigerant determines the outlet pressure of the compressor required at state 3. Since we are performing the compression in the vapor phase, where molar volumes are large, a signifi cant amount of work is needed. The higher the pressure, the more work is required for a given refrigeration effect. The power of compression is given by:
W#
c5n#1h32h22 If the compression is assumed to be reversible,
s35s2
The high-pressure vapor is then condensed at TH, expelling heat Q#
H to the hot reservoir.
This process occurs in the condensor where the fl uid exits at state 4.
Q#
H5n#1h42h32
The high-pressure liquid is then expanded in a valve back to state 1 so the cycle can be repeated. A valve is used instead of the turbine that was used in the Rankine cycle.
The amount of work that would be produced by a turbine is small, so we replace it with a valve to reduce the complexity. This step is represented by a throttling process, where:
h45h1 (3.36)
Since the pressure decreases as the refrigerant passes through the valve, its entropy increases, as shown in Figure 3.9. Can you locate the evaporator and condenser on the refrigerator you have at home?
Figure 3.9 The ideal vapor-compression refrigeration cycle. The four unit processes are sketched on the left, while the path on a Ts diagram is shown on the right.
Refrigeration Cycle Valve
QH
QH
Qc
Wc
1
2
3 Condenser
High T reservoir
Compressor Refrigerated
unit at low T
4
1
3
Evaporator 4
T
s 2 QC
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3.9 Vapor-Compression Power and Refrigeration Cycles ◄ 171 The coeffi cient of performance, COP, of a refrigeration cycle measures its perfor- mance. It is defi ned as the ratio of the heat absorbed from the cold reservoir (the refrig- eration effect) to the work required:
COP5Q#
C
W#
c5 h22h1
h32h2
(3.37) In real refrigeration systems, a fi nite temperature difference is needed to get practical heat-transfer rates in the evaporator and the condenser. Thus, the evaporator must oper- ate at a lower temperature than the desired refrigeration temperature, while the con- denser must operate at a higher temperature than the ambient heat reservoir. Thus, more work is required to obtain a given refrigeration effect. Moreover, irreversibilities in the compressor must be considered, also adding to required work load and further decreasing the COP. COPs of well-designed real refrigeration systems typically fall between 2 and 5.
It is desired to produce 10 kW of refrigeration from a vapor-compression refrigeration cycle.
The working fl uid is refrigerant 134a. The cycle operates between 120 kPa and 900 kPa.
Assuming an ideal cycle, determine the COP and the mass fl ow rate of refrigerant needed.
Properties of refrigerant 134a can be found at http://webbook.nist.gov/chemistry/fl uid/. Data can be viewed in an HTML table.
SOLUTION A sketch of the process on a Ts diagram is shown in Figure E3.16.
The following saturated data for refrigerant 134a are obtained from the NIST site (see citation Pg. 27):
P [MPa]
T [K]
hl
3kJ/mol4 hv
3kJ/mol4 sl
3J/1mol K2 4 sv
3J/1mol K2 4
0.12 250.84 17.412 39.2955h2 90.649 177.895s2
0.90 308.68 25.4865h4 42.591 119.325s4 174.74
EXAMPLE 3.16 Estimation of the COP of a Vapor- Compression Refrigeration Cycle
Figure E3.16 A Ts diagram of the ideal vapor-compression refrigeration cycle of Example 3.16.
4
3
2 1
P = 0.9 MPa
P = 0.12 MPa 317.6
308.7
250.8
T [K]
WC
QC
QH
s
(Continued)
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172 ► Chapter 3. Entropy and the Second Law of Thermodynamics
To get the amount of refrigeration, we can combine Equations (3.35) and (3.36):
qC5 1h22h12 5 1h22h42 539.295225.486513.809 3kJ/mol4 To fi nd the state at the exit of the isentropic compressor, we recognize:
s35s2
At a pressure of 0.90 MPa, from the NIST website, we get the entropies to match state 3 at:
P [MPa]
T [K]
hv
3kJ/mol4 sv
3J/1mol K2 4
0.90 317.62 43,578 177.89
The work required by the compressor is:
wc51h32h22 543.578239.29554.283 3kJ/mol4 Solving for the coeffi cient of performance gives:
COP5Q#
C
W#
c5h22h1
h32h253.22 To get the desired 10 kW of refrigeration, we need:
n# 5Q#
C
qC 5 10 3kW4
13.809 3kJ/mol450.73 3mol/s4
The temperature of the evaporator is around 250 K, below the freezing point of water. On the other hand, the condenser operates between 308 K and 317 K. This temperature is warm enough to expel heat to the ambient environment.