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9.A-1 Dedicated subcooling is a novel modification for frozen food refrigeration in supermarkets. With
the dedicated subcooling modification, liquid refrigerant leaving the condenser is further cooled
at constant pressure to an intermediate temperature, T4, as shown in Figure 9.A-1. The cooling
needed for this purpose is provided by another, smaller refrigeration cycle. The overall
coefficient of performance is defined as the ratio of the heat removal from the food cases to the
total work input for both refrigeration cycles.
TC= -20F
condenser
TH
= 90F
evaporator
evapQ
compressorthrottle
valve
1
234
condenser
TH
= 90F
sub-cooler
compressor
6
7throttle
valve
8
9
5
Figure 9.A-1: Dedicated subcooling system for a supermarket application.
Consider a refrigeration cycle designed to maintain food products at TC = 20F. Air-cooled
condensers in the refrigeration cycles reject heat to outdoors at a design temperature of TH= 90F.
At these design conditions, all heat exchange equipment is sized such that there is a Thx= 12F
difference between the entering cold stream temperature and the exiting hot stream temperature
(e.g., the saturation temperature in the evaporator of the refrigerated cases is -32F and the
temperature at state 6 is 12F lower than the temperature at state 4). The isentropic efficiency of
the compressors is c= 0.7. Assume that the refrigerant leaving the evaporator is saturated vapor
and the refrigerant leaving the condensers is saturated liquid. Neglect pressure losses in this
analysis. The refrigerant is R22.
a.) Determine the value of T4that maximizes the COPof the combined cycle.
b.) Compare the COP of the optimized dedicated subcooling cycle to that of a simple vapor
compression cycle that does not use subcooling.c.) Calculate the Second-Law efficiencies of the optimized dedicated subcooling cycle and a
cycle that do not use subcooling.
d.) What do you see as the advantages and disadvantages of the subcooling modification?
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9.A-2 Refrigeration equipment is often installed with more cooling capacity than is needed.
The best way to operate the equipment when the cooling load is lower than the capacity isto reduce the compressor speed with a variable speed motor. However, variable speed
motors and controllers are expensive. Shown in Figures 9.A-2(a) and (b) are two
alternative methods for reducing cooling capacity.
Figure 9.4-2 (a) and (b): Alternative methods for reducing cooling capacity
Both systems are based on a standard vapor compression refrigeration cycle with a bypass
modification for capacity control. System (a) recycles a fraction (y) of the outlet of the
compressor back to the compressor inlet whereas System (b) directs a fraction (y) of the
compressor outlet to the evaporator inlet. Both systems use ammonia as the refrigerant with
a compressor that has an isentropic efficiency of 0.68. The volumetric flow rate of the
refrigerant entering the compressor is given by
11
2
1 1d
vV V R
v
wheredV
is the compressor displacement rate = 0.20 m3/s
Ris the clearance volume ratio = 0.025
1v and 2v are the specific volumes of the refrigerant at the compressor inlet and outlet.
Assume state 5 to be saturated vapor at -15C and state 3 to be saturated liquid at 40Cfor all cases. Pressures losses in the heat exchangers and piping may be neglected.
a.) Calculate the all state points, the cooling capacity, and the COP when the bypass is
closed (i.e. y=0). Systems (a) and (b) should be identical in this case.
b.) Determine the value of yat which the cooling capacity is 80% of the full-load valuefound in part a) for both System (a) and System (b). Calculate the quantities
indicated in part a) for this operating state for both systems.
c.) Which capacity control system do you recommend and why?
d.) Determine the Second-Law efficiency of System (a) and System (b) when operatingat a capacity that is 80% of full load.
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9.A-3 Carbon dioxide has received a lot of attention as a refrigerant because it is non-toxic,
non-flammable, inexpensive and not harmful to the environment. Unfortunately, carbon
dioxide has a critical temperature of about 31C and its vapor pressure at 25C is 6.4MPa. One alternative is to employ carbon dioxide in a cascade refrigeration cycle, asshown in Figure 9.A-3, using ammonia (another natural refrigerant) in the high
temperature cycle and carbon dioxide in the low temperature cycle.
TL=225 K
WL
WHWH
50 kW
TL=320 K
1
2
3
4
5
67
8
carbon
dioxide
cycle
ammonia
cycle
Figure 9.A-3: Cascade refrigeration cycle
The purpose of this problem is to provide a thermodynamic analysis of a two-stage
cascade ammonia - carbon dioxide refrigeration cycle designed to provide 50 kW of
cooling at 225 K with heat rejection to the surroundings at 320 K. The isentropic
compressor efficiency can be assumed to be 0.76 for both compressors. The heat transfer
rates in the three heat exchangers can each be described by relations of the form:
Q UA T
In the low temperature evaporator, the T is the difference between the evaporationtemperature (T4) and TL (225 K). In the condenser the Tis difference between T7and 320
K. In the heat exchanger between the two cycles, T is difference between saturation
temperatures T8and T3. The UAvalues for all three heat exchangers are 25 kW/K. Make the
usual assumptions associated with vapor compression system analysis, i.e., saturated vapor
entering the compressors, saturated liquid exiting the condensers, adiabatic compressor
operation and negligible pressure losses in the heat exchangers and lines.
a.) Assume that T3is 250 K. Calculate the thermodynamic properties at all 8 points in the
cycle. Determine the COP at these conditions.
b.) Investigate the effect of T3by calculating the COP for values between 230 K and 280 K.
Is there an optimum value for T3? If so, what is it and why does it occur?c.) Determine the Second-Law efficiency of the cycle and the rate of exergy destruction in
each system component for the value of T3 determined in part b. Which component
results in the largest rate of exergy destruction? What, if anything, could be done to
reduce this irreversibility rate in this component?
d.) Although this cycles using natural refrigerants, it still poses several major safety hazards.
What do you see as potential safety problems with this refrigeration system?
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9.A-4 A refrigeration cycle is being designed for a small freezer room that is to be maintained at
10C in a 35C environment. The refrigeration load is estimated to be 10 kW. A
refrigeration cycle employing a liquid-to-suction heat exchanger is being considered with the
configuration shown in Figure 9.A-4. At design conditions, the refrigerant exits the
evaporator as saturated vapor and enters the cool end of the heat exchanger emerging in a
superheated state. The isentropic efficiency of the compressor is 0.60. The condenser heattransfer rate is (approximately) described by
C C C H Q UA T T
where
CUA = 3.50 kW/K
TC is the saturation temperature in the condenser
TH is the design temperature for the surroundings = 35C
Similarly, the evaporator heat transfer rate is
E E F EQ UA T T
where
EUA = 1.50 kW/K
TF is the desired freezer temperature = -10C
TE is the saturation temperature in the evaporator
REFRIGERATION CYCLE WITH HX
5
6
1 2
3
4
V
ALVE
EVAP
HTEX
COMP
COND
Figure 9.A-4: Refrigeration cycle with a liquid-to-suction heat exchanger
a.) Plot the COP of this cycle for suction to discharge heat exchanger effectivenessesbetween 0 and 1 for refrigerant R134a.
b.) Calculate and plot the COP for refrigerants propane, and ammonia on the same plot
used for R134a. What is your conclusion regarding the benefits of the liquid-to-suction heat exchanger?
c.) Calculate the exergy destruction per unit mass in each of system component for the
conditions of part (a). Can the trends evident in your plots for part (a) be explained in
terms of the exergy destruction terms?
d.) List three modifications you would suggest to improve the cycle performance in order
of priority.
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9.A-5 Many homes rely on electric water heaters. Consider a typical situation in which water is
available at 10C and heated to 55C in a 300 liter tank. The tank is equipped with two
4500 W heaters. The overall heat loss coefficient from the heated water to the
surrounding 20C environment is 15 W/K. Please answer the following questions
concerning conventional electric water heating.
a.) If the tank is initially filled with water at 10C, how much time is required to heat thewater to 55C with the electric resistance heaters?
b.) What is the First Law efficiency of the process described in part a?
c.) What is the Second Law efficiency of the process described in part a?
Heat pump water heaters have been proposed as an energy-saving alternative to
conventional electric water heaters. A heat pump water heater is a vapor compression
refrigeration cycle in which the condensing unit is located within the water storage tank.
The refrigerant for this specific unit considered here is R134a. The isentropic efficiency
of the compressor and electric motor is 0.68. The rate of heat transfer between the
R134a and the water in the condensing unit can be described by:
,( )sat c wcQ UA T T
where
c
UA = 1000 W/K
,sat cT is the condenser saturation temperature
wT is the average temperature of the water in the tank at the current time.
The volumetric flow rate of the refrigerant entering the compressor is given by
22
3
1 1dv
V V Rv
where
dV is the compressor displacement rate = 0.0020 m
3/s
R is the clearance volume ratio = 0.025
2v and 3v are the specific volumes of the refrigerant at the compressor inlet and outlet,
respectively.
Assume the compressor to operate adiabatically, with saturated vapor at the compressor
inlet and saturated liquid at the condenser outlet, and negligible pressure losses in the
evaporator and condenser.
d.) If the tank is initially filled with water at 10C, how much time is required to heat thewater to 55C with the heat pump water heater?
e.) Plot the temperature of water in the tank, the saturation temperature in the condenser
and the refrigerant flow rate as a function of time for the time period identified in part
(d).
f.) What is the First Law efficiency of the process described in part d?
g.) What is the Second Law efficiency of the process described in part d?
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h.) Compare the exergy destroyed in the throttle with the work needed to power the
compressor for the process in part (d).
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9.A-6 AevapQ
= 3-ton single-state vapor compression refrigeration system is being designed for an
application in which the condensing temperature is Tcond= 90F and the evaporator temperature is
Tevap= 0F. Refrigerants R134a, R22, R717 (ammonia), R290 (propane) and R410A are being
considered for the system. The compressor efficiency may be assumed to be c = 0.8 for all
refrigerants. Assume that the pressure losses in the heat exchangers and piping are negligible.
Assume steady-state operation.a.) Calculate and compare the refrigerant mass flow rate, the compressor power, the cycle COP,
and the required volumetric flow rate at the compressor inlet for each refrigerant. Which
refrigerant would you recommend based on these comparisons?
b.) If a reversible expansion device were used in place of a throttle valve, some power could be
produced which would offset the power needed to operate the compressor. In addition, a
greater refrigeration effect would occur. Calculate the lost power and the reduction in the
refrigeration effect resulting from throttling rather than isentropic expansion for each of the
refrigerants.
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9.A-7 You are considering installation of the conventional vapor compression refrigeration system
shown in Figure 9.A-7(a). The system must provide cooling to a space at TC= -10F and reject
heat to ambient air at TH= 85F.
TH
condenser
compressorexpansion valve
2
3
4
1
condQ
evaporator
cW
evapQ
TC
Figure 9.A-7(a): Conventional vapor compression system.
The efficiency of the compressor is c= 0.60, the volumetric efficiency of the compressor is vol
= 0.65, and the displacement rate of the compressor isdispV = 50 cfm. The condenser has an
approach temperature difference Tcond= 5 K and the pressure drop associated with the flow of
refrigerant through the condenser is Pcond= 50 kPa. The refrigerant leaves the condenser with a
subcool of Tsc= 2 K. The evaporator has an approach temperature difference of Tevap= 6 K
and the pressure drop associated with the flow of refrigerant through the evaporator is Pevap= 15
kPa. The refrigerant leaves the evaporator with a superheat of Tsh = 5 K. The refrigerant is
R134a.a.) Analyze the cycle. Print out the Arrays table containing the state point information and
generate a temperature-entropy diagram that shows all states. What is the rate of refrigeration
provided and rate of power consumed by the cycle?
b.) Check your answer by carrying out an overall energy balance.
c.) Determine the rate of entropy generation in each of the components. Check your answer bycarrying out an overall entropy balance.
d.) Determine the minimum power required to provide the same amount of refrigeration at TC
while rejecting heat to TH (i.e., the power required by a reversible refrigeration system
operating between the same two temperatures). The difference between the actual
compressor power and this minimum power is the lost work associated with entropy
generation in the cycle and therefore should be equal to the product of the rate of entropy
generation and the ambient temperature. Show that this is so.
e.) What is the Coefficient of Performance of the cycle?
The liquid overfed system shown in Figure 9.A-7(b) may provide some advantages relative to theconventional system shown in Figure 9.A-7(a). The refrigerant leaving the condenser is
expanded into a tank. Saturated liquid is pumped from the tank into the evaporator at state 1. By
controlling the flow rate provided by the pump (which is independent of the flow rate provided by
the compressor) it is possible to keep the quality of the refrigerant everywhere in the evaporator
low; this dramatically improves the thermal performance of the evaporator. Saturated vapor is
pulled from the tank by the compressor at state 4.
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TH
condenser
compressor
expansion valve
2
4
6
1
condQ
evaporator
cW
evapQ
TC
3
5
7
p
W
tank
Figure 9.A-7(b): Liquid overfed vapor compression system.
The characteristics of the compressor are the same. The efficiency of the compressor c= 0.60,
the volumetric efficiency of the compressor is vol= 0.65, and the displacement of the compressor
isdispV = 50 cfm. The condenser characteristics are also unchanged. The condenser has an
approach temperature difference Tcond= 5 K and the pressure drop associated with the flow of
refrigerant through the condenser is Pcond= 50 kPa. The refrigerant leaves the condenser with a
subcool temperature difference of Tsc= 2 K. The pump has an efficiency of p= 0.5 and the
pump is controlled so that the quality of the refrigerant leaving the evaporator is xevap,out= 0.5.
The performance of the evaporator improves with the liquid overfed modification; therefore, the
evaporator approach temperature difference is reduced to Tevap = 3 K. The pressure drop
associated with the flow of refrigerant through the evaporator is Pevap= 15 kPa. The refrigerant
is R134a. The system must provide cooling at TC= -10F and reject heat to ambient air at TH=
85F. There is no pressure loss associated with the tank and the tank is insulated.
f.) Analyze the liquid overfed system. Print out the Arrays table containing the state point
information and generate a temperature-entropy diagram that shows all states. What is the
rate of refrigeration provided and rate of power consumed by the cycle?
g.) What is the COP of the liquid overfed cycle? Compare this value to the COP of the
conventional system calculated in part (e).
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9.A-8 The compressor in an ammonia refrigeration system must be liquid-cooled because the
refrigerant would otherwise exit at very high temperature, which would decompose the
oil. In a particular case, ammonia steadily enters a compressor at -20F and 10 psia with
a mass flow rate of 126 lbm/hr. The ammonia exits the compressor at 355F, 215 psia.
Cooling water is supplied to the compressor at 70F and 850 lbm/hr. The cooling water
exits at 78F. The ammonia is condensed to a liquid at 105F. Neglecting pressurelosses in the heat exchangers and other heat losses, estimate the COP and cooling
capacity of this refrigeration cycle.
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9.A-9 A heat-powered refrigeration cycle is shown in Figure 9.A-9. In this cycle a Rankine
cycle is used to produce the power that is used to drive the compressor in the vapor
compression refrigeration cycle. The system operates at steady-state with Refrigerant
R134a as the working fluid in both cycles. The isentropic efficiencies of the turbine,
compressor, and pump are 0.78, 0.72, and 0.48, respectively. At the design condition, the
evaporator must provide 54 kW of cooling at -10C with saturated vapor at theevaporator exit and 40C saturated liquid at the condenser exit. The maximum
temperature for the saturated vapor at the boiler exit (state 1) is 95C. The turbine,
compressor, pump, and valve operate adiabatically. Neglect pressure losses in the heat
exchangers and piping.
1
2
34 5
6
Boiler
Condenser Evapora
tor
Turbine Compressor
Pump Valve
7
QB
QC QE
a.) Determine the mass flow rate of R134a through the evaporator.
b.) Determine the mass flow rate of R134a through the boiler.
c.) The coefficient of performance (COP) for this cycle is the ratio of the refrigeration heat inputto boiler heat input. What is the COP at design conditions?
d.).Compare the COP determined in part c with the maximum possible COP for a heat-poweredrefrigeration system operating within the same temperature limits.
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9.A-10 An ice-making system uses R134a as the working fluid. The cooling load at design
conditions is 200 kW. The saturation temperature in the evaporator is -20C. The
compressor exit pressure is 12 bar. Pressure losses in the lines and heat exchangers are
negligible. At design conditions, saturated vapor enters the compressor and the saturated
liquid exits the compressor. The compressor isentropic efficiency is 0.80. The external
fluid circulated through the evaporator is a 40% propylene glycol-water solution thatenters the evaporator at -6C and exits at -14C. After leaving the evaporator, the glycol
solution circulates through many small tubes in an ice storage unit and ice forms on the
outside surface of these tubes. Initially, the ice storage tank contains water at 0C. After
12 hours of operation, all of the water has frozen to become ice at 0C.
a.) What is the required mass flow rate of the glycol solution? (Note that property data
for propylene glycol and other brines are available in the BrineProp2 external routine.
Information on BrineProp2 is available from the Function Info menu item in the
Options menu. Click the EES library routines button and then click BrineProp2.lib in
the list. Clicking the Function Info button will provide documentation.b.) Determine the heat exchanger effectiveness of the evaporator.
c.) What is the COP of this refrigeration cycle?
d.) Determine the mass of ice that is produced in the 12 hour operating period.
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9.A-11 Cooling at two different temperatures is needed in many situations, such as in
household refrigerators and in supermarket refrigeration systems. Figure 9.A-11
shows a schematic diagram for a vapor compression refrigeration system in a
supermarket that has a single compressor, a single condenser and two evaporators
providing cooling capacity at two different temperatures and operating at steady
conditions. The compressor isentropic efficiency is 0.78. Saturated liquid exits thecondenser at 105F. The low temperature evaporator provides a cooling capacity of
3 tons at a saturation temperature -5F which is used to maintain a freezer at 5F.
The higher temperature evaporator provides a cooling capacity of 2 tons at 25F,
which is used to maintain a dairy case at 35F. The refrigerant is R-134a. Assume
saturated vapor exists at the evaporator exits and neglect pressure losses in the
piping and heat exchangers.
condenser
compressor
expansion valve
23
41
high temperature
evaporator
cW
low temperature
evaporator
2 tons35F
3 tons5F
expansion valve
5 6
7 8
90F
Figure 9.A-11: Refrigeration cycle with two evaporators
a.) Determine the properties at all points in the cycle.
b.) Determine the power required to operate the cycle at steady-state.
c.) Indicate how you would define a COP for this cycle and determine its value.
d.) Determine the total rate of entropy generation for steady-state operation.
e.) Define a Second-Law efficiency and determine its value.
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9.A-12 A vapor compression refrigeration system is to be installed in a dairy for the purpose
of cooling 0.05 liters/sec of milk from 30C to 5C. (Assume milk to have the same
thermodynamic properties as water.) In addition, 1300 W must be removed from the
storage unit to maintain the stored milk at 5C. While the milking is taking place,
0.08 liter/sec of water at 60C are needed for cleaning purposes. The water supply is
available at 10C. It has been suggested that the heat rejected form the condenser ofthe cooling unit could be used to supply some or all of the energy needed to heat the
water. A proposed system is shown in Figure 9.A-12. In this system, the water that
is needed for cleaning provides the sink to which heat is rejected. The condensing
unit consists of two heat exchangers. The first, called the condenser, is where the
refrigerant changes from saturated vapor (state 4) to saturated liquid (state 5) at the
condenser saturation temperature. The second heat exchanger is called the
desuperheater. The refrigerant is cooled from state 3 to saturated vapor in this heat
exchanger. The isentropic efficiency of the condenser is 0.76. The saturation
temperature in the evaporator is -2C. Assume that saturated liquid is produced at
the condenser exit and saturated vapor exits the evaporator at design conditions.
1 2
34
5
6 7 8
Milk, 0.05 l/s, 30C Milk, 5C
Water10C0.08 l/s
x=1-2C
x=1
desuperheatercondenser
evaporator
compressor
Valve
Figure 9.A-12: Refrigeration system for a dairy application
a.) Determine the steady-state cooling load for this system
b.) Determine the required refrigerant flow rate
c.) Assume the heat exchanger effectiveness is 1 in order to calculate the lowest possible
saturation temperature in the condenser. What is this temperature?d.) Determine the saturation temperature in the condenser if the effectiveness of the
condenser is 0.9
e.) Determine the COP for this system with the heat exchanger as indicated in d.)
f.) What temperature is the water at state 8 heated to? What fraction of the energy
needed to heat the cleaning water is supplied by this system?
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9.A-13 It has been suggested that the refrigeration equipment in an old household
refrigerator, which is no longer needed for refrigeration, could be modified to serve
as a heat pump for heating water in residence. The purpose of this problem is to
evaluate this suggestion. The refrigerator uses R134a. The compressor is driven by
an electric motor that spins at 1750 revolutions per minute. The compressor has a
displacement of 4.5 in3, an adiabatic efficiency of 0.78 and a volumetric efficiencyof 0.70 when operated at condensing temperature of 110F and an evaporating
temperature of 0F.
a.) Determine the COP and cooling capacity (in tons) of the refrigerator when it is
operated as designed.
The air-cooled condenser is replaced with a water-cooled condenser in order to heat
water for domestic purposes. The evaporator temperature increases to 60F since
energy is supplied to the evaporator from indoors. The condenser must operate at
140C to ensure that the water is heated to 130F. The volumetric efficiency is
reduced to 0.60 at these conditions. Assume that the isentropic efficiency is
unaffected by the change in conditions.
b.) Determine the number of gallons per hour of water that could be heated from 50F to
130F with this system and the COP.
c.) What is your evaluation of the suggestion to use the refrigerator as a water heater?
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9.B-1 The single-stage LiBr-H2O absorption system shown in Figure 9-B-1 will be used to cool
water that enters at state 14 at 0.5 kg/s from 12C to 8C. The evaporator heat exchangeris designed such that the saturated refrigerant vapor exiting the evaporator at state 10 is at
4.5C. Cooling water at atmospheric pressure is provided to the absorber (state 11) at 0.8
kg/s and 25C to maintain the temperature of the dilute absorbent at state 1 at 32C. The
generator uses low pressure steam to concentrate the absorbent solution. Thetemperatures of the released water vapor (state 7) and the concentrated absorbent solution
(state 4) and 7 are both 95C. Saturated liquid refrigerant (water) exits the condenser at
state 8 at a temperature that is 5C higher than the cooling water exit temperature at state13. Assume the pump to operate with a 50% isentropic efficiency. Neglect pressure
losses and assume that all of the equipment operates adiabatically.
Figure 9.B-1: Schematic diagram of the basic lithium bromide-water absorption cooling cycle
a.) Determine the COP, and all energy flow rates for this absorption cooling cycle
assuming that a solution heat exchanger is not employed.b) Include a solution heat exchanger as shown in Figure 9.B-1 in the analysis and
prepare a plot of the COP as a function of the solution heat exchanger effectiveness.
c) Prepare a plot that shows how the COP varies with generator temperature between
80C and 120C for system that has a solution heat exchanger effectiveness of 40%.
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9.B-2 A solar air-conditioning system uses a lithium bromide-water absorption chiller having
the configuration shown in Figure 9.B-2. The solar collectors provide thermal energy at arate sufficient to maintain the solution in the generator at 200F (states 4 and 7). The
temperature of the strong solution entering the generator at state 3 is 160F, while the
temperature of the solution leaving the absorber (state 1) is 100F, the evaporation
temperature (states 9 and 10) is 40F, and the temperature of the saturated liquid water atstate 8 is 100F. The evaporator provides 2 tons of cooling capacity. Assume state 10 is
saturated vapor at 40F and neglect the pressure losses in the piping. The pump power
should be small so it is sufficient to assume that it operates isentropically.
1
2
3
4
5
6
7
8
9
10pump
generatorcondenser
evaporator absorber
throttle
from solarcollectors
to solarcollectors
Coolingat 40F
HeatRejection
HeatRejection
Figure 9.B-2: Solar air conditioning system
a.) Determine the system coefficient of performance at these operating conditions.
b.) Determine the required collector area if the solar radiation is 950 W/m2 and the
collector efficiency at these conditions is 0.30.
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9.B-3 Figure 9.B-3 shows a schematic diagram of a gas-based low temperature refrigeration
cycle. High pressure argon gas enters a concentric-tube counter-current heat exchanger(the recuperator) at 8.5 MPa and 300 K with mass flow rate 0.25 kg/s (state 1) where it is
cooled by argon gas at state 4 returning from the load heat exchanger. The gas returning
at state 4 has the same flow rate and is at pressure 0.1 MPa. The high pressure argon gas
leaving the recuperator at state 2 is throttled from 8.5 MPa to 0.1 MPa (state 3) in anexpansion valve. The argon at state 3 enters the load heat exchanger where the argon
picks up heat from a refrigerant that is evaporating at a saturation temperature of 200 K.
Pressure losses in both heat exchangers can be neglected. During a test the temperaturesat states 4 and 5 were measured to be 190 K and 280 K, respectively.
1
2
3
4
5
200 K
load heat exchanger
expansion
valve
recuperator
0 25 kg/s
high
in
m .
P
T
= 8.5 MPa
0.1 MPalowP = 300 K
200 K
Figure 9.B-3: Schematic of a gas-based cooling system
a.) Determine the temperatures at all points in the cycles.
b.) Determine the effectiveness of both heat exchangers.c.) The argon provided to cycle at state 1 must be compressed from 0.1 MPa. What is
the minimum power required to compress the argon from the low to the high pressureassuming that is it done isothermally at 300 K.
d.) Determine the COP for this cycle using the power determined in part c.
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9.B-4 Figure 9.B-4 illustrates a cryogenic refrigerator that is used to maintain a detector at 70
K. The refrigerator utilizes a reverse-Brayton cycle with neon as the working fluid.
Neon enters the compressor at state 1 with pressure Plow= 152 kPa and mass flow rate m
= 0.0006 kg/s. The isentropic efficiency of the compressor is 0.78. The neon leaves the
compressor at state 2 with pressure Phigh= 532 kPa. The neon leaving the compressor is
quite hot and is therefore cooled in an aftercooler. The aftercooler rejects heat at rate HQ
to the ambient temperature at TH= 20C with an approach temperature difference is Tac
= 10 K. Neon flows through the recuperative heat exchanger where it is cooled by heattransfer with the cold neon returning from the cold end of the device. The neon exits the
turbine at state 5 with pressure Plow. The turbine operates with an isentropic efficiency of
0.84. The neon leaving the turbine enters the load heat exchanger where it is warmed bya heat transfer from the detector which is at temperature TC = 70 K. The approach
temperature difference for the load heat exchanger is TLHX = 2 K. Finally, the neon
leaving the load heat exchanger at state 6 passes through the recuperator where it iswarmed by heat transfer with the high pressure neon flowing in the opposite direction.
The recuperator has an effectiveness of 0.92. Neglect pressure losses in the heat
exchangers and assume neon behaves as a real fluid for the conditions in this cycle.
Figure 9.B-4: Reverse-Brayton refrigeration cycle.
a.) Determine the temperature and pressure at each of the six numbered states in Figure
9.B-4.
b.) Determine the rate of work transfer required by the compressor ( cW ), the rate of work
transfer from the turbine ( tW ), and the rate of heat transfer to the neon in the load
heat exchanger ( CQ ).
c.) Determine the Coefficient of Performance (COP) of the cryocooler.
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d.) Conduct numerical experiments with your model to determine the importance of the
isentropic efficiencies of the compressor and turbine on the COP of this cycle.
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9.B-5. A reverse Brayton cycle using air as the working fluid has been proposed as an
alternative to the conventional automobile air-conditioning system. A major advantageof this refrigeration cycle is that it does not involve ozone depletion or global warming
issues that are associated with leakage of conventional refrigerants. The necessary
equipment consists of a belt-driven compressor and turbine mounted on the same shaft.
(Similar equipment is commonly used in modern cars for turbocharging). In a particularcase shown in Figure 9.B-5, outdoor air enters the compressor at 35C, 100 kPa, 0.135
kg/s and is adiabatically compressed with an isentropic efficiency of 0.74. The air is then
in a heat exchanger by air that is blown across the outside of the heat exchanger. Becausethe external flow rate is high, the effectiveness of the heat exchanger can be estimated to
be
1 exphx NTU
The overall heat transfer coefficient-area product (UA) for the heat exchanger is 225
W/K. (See Section 6.6.6 for a discussion of heat exchangers.) After leaving the heatexchanger, the air expands through a turbine to 100 kPa. The turbine operates
adiabatically with an isentropic efficiency of 0.82. The cold air exiting at state 4 is blown
into the cabin of the automobile to provide the air-conditioning.
Figure 9.B-5: Air-based cooling system for a vehicle
a.) Calculate and plot the temperature at state 4, the cooling capacity, the required power
and the COP of this cooling system as a function of compressor pressure ratio for
ratios between 1.5 and 10.
b.) What pressure ratio would you recommend for this application?
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9.B-6. A simplified ammonia-water absorption refrigeration cycle is shown in Figure 9.B-6.
Solution having an ammonia mass fraction of 0.48 steadily enters the generator at point 1at 80 C, 13.5 bar and 0.05 kg/s. Heat is supplied to the flash generator at a rate
sufficient to maintain its contents at 115C. Assume that the vapor and liquid streams (2
and 3) leave the generator in equilibrium at 13.5 bar, 115C. Cooling water is used to
maintain the temperatures at points 4 and 7 at 27C. The evaporator pressure is 3.0 bar.The refrigerant exits the evaporator at 5C. Pressure drops in lines and components
(except the valves) are assumed to be negligible. Assume the pump to operate ideally.
Note: Properties for ammonia-water mixtures are provided in EES with the NH3H2O
external procedure. Information relating to the use of these properties is available by
selecting Function Info from the Options menu. Click the External Routines button and
scroll to the NH3H2O entry in the list of routines. Select the NH3H2O and click the
Function Info button to view the documentation.
9
76
5
4
3
2
1
8
10
Condenser
EvaporatorAbsorber
GeneratorFlash
Pump
Solutionheat exchanger
Figure 9.B-6: Ammonia-water refrigeration system
a. What is the temperature of the refrigerant mixture entering the evaporator?b. What is the effectiveness of the solution heat exchanger?
c. What would the temperature at state have to be to allow all of the refrigerant to
evaporate?d. What is the coefficient of performance for this cycle?
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9.B-7 Figure 1 illustrates a system that is used to provide short term cooling for a cryogenic
detector on a spacecraft.
aftercooler
Tamb
1
TC
load heat exchanger
recuperator
loadQ
2
4
acQ
3
throttle
valve
5
bottle ofargon
Pbottle
Preg
Pexh
Figure 1: Joule-Thomson system for detector cooling.
A bottle of argon is stored onboard the spacecraft. The initial pressure of the bottle is Pbottle=3500 psi and the volume of the bottle is Vbottle = 10 liter. Heat transfer from the ambient
environment onboard the spacecraft maintains temperature of the contents of the bottle at Tamb=
20C. When cooling is desired, a valve on the bottle is opened and argon passes through apressure regulator that maintains an exit pressure of Preg = 2000 psi. The temperature of the
argon may change as it passes through the regulator; therefore an aftercooler is placed
downstream of the regulator so that the argon is returned to ambient temperature before it enters
the recuperative heat exchanger at state 1. The recuperator has an approach temperature
difference of Trec= 5 K and the pinch point is at the warm end. The argon leaving the high
pressure side of the recuperator at state 2 is throttled to the exhaust pressure, Pexh= 0.25 atm atstate 3 before being heated in a load heat exchanger to state 4 and passing through the low
pressure side of the recuperator where it is exhausted to space at state 5. Pressure losses in the
recuperator are negligible for both the low and high pressure streams. This problem will ignorethe initial cool down transient for the detector and consider only its steady state operation at TC=
140 K. The approach temperature difference associated with the load heat exchanger is TLHX=1 K. The total refrigeration load required to maintain the temperature of the detector at TC is
loadQ = 15 W. The system is deactivated when the bottle pressure reaches the regulator pressure.
a.) Determine the time that the system can maintain the detector at its operating temperature.b.) Plot the operating time as a function of the regulator pressure and determine the optimal
regulator pressure that maximizes the operating time of the system.
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9.C-1 Carbon dioxide is of interest as a refrigerant because it is inexpensive, non-toxic, non-
flammable, and it has excellent thermodynamic and transport properties. However, carbon
dioxide has a critical temperature of about 31C. Since the ambient temperature is often thishigh or higher, and a temperature difference is needed to drive heat transfer rates, the
carbon-dioxide refrigeration cycle must operate with the high temperature cooler at
supercritical pressures. The purpose of this problem is to design a supercritical carbon-dioxide air conditioning cycle to provide 10 kW of cooling at 5C with heat rejection toambient air at 35C. We will assume that the high temperature heat exchanger can bedesigned to cool the high pressure refrigerant to 40C. The mass flow rate of refrigerant canbe represented in terms of a volumetric efficiency with the following relation.
b
n
suction
edisch
v
RPMV
p
pCCm
1
arg.
1
where Cis the clearance volume ratio, nis the polytropic index, Vis the compressor cylinder
displacement, RPMis the compressor speed and vbis the specific volume of the refrigerantat the compressor inlet. A typical value for Cis 0.025. The compressor speed is 3500 rpm.
The polytropic index can be assumed to be the isentropic index (ratio of cp/cv) at the
compressor inlet conditions. The compressor power can be represented by defining a
combined compressor and motor efficiency defined as the ratio of the isentropic to actual
power. The combined efficiency is 0.62. Compressors are often assumed to be adiabatic.
However experience has shown that small compressors have significant heat loss. In this
case, the ratio of the heat transfer rate to the compressor power input is 0.25. It will be
necessary to determine the optimum compressor discharge pressure, the associated discharge
temperature, the refrigerant mass flow rate, the compressor displacement, and the
compressor power. A schematic of the proposed refrigeration cycle is shown in Figure 9.C-1. Note that this figure includes a liquid-to-suction heat exchanger. At design conditions,
the refrigerant exits the evaporator as saturated vapor and enters the cool end of the heat
exchanger emerging in a superheated state. The advantage of the liquid-to-suction heat
exchanger should be investigated in your analysis. Neglect pressure losses in the heat
exchangers and piping.
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Suction-lineHX
Evaporator
Heatexchanger
Compressor
3
4
5
6
1
2
Air at 35 C
Figure 9.C-1: Carbon dioxide refrigeration cycle with suction-line heat exchanger
a.) Assume that the liquid-to-suction heat exchanger is not present (or alternatively, it has a
small heat exchanger effectiveness) and that the discharge pressure is 95 bar. Determine
the COP of this cycle and plot the state point information on temperature-entropy and
pressure-enthalpy diagrams.
b.) Determine optimum discharge pressure (at state 4) between 75 bar and 125 bar that
maximizes the COP of this cycle for the conditions or part a. If you find that there is an
optimum discharge pressure, please explain why it exists.
c.) Prepare a plot of the optimum COP as a function of the liquid-to-suction heat exchanger
effectiveness. (You will need to determine the optimum discharge pressure at each heat
exchanger effectiveness value.) What is your conclusion regarding the benefits of the
liquid-to-suction heat exchanger?
d.) How does carbon dioxide compare to other refrigerants under similar operating
conditions? What is your overall assessment of carbon dioxide as a refrigerant?
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9.C-2 The purpose of this problem is to determine the performance of residential vapor
compression heat pump operating in the heating mode using refrigerant R134a. The heat
pump system has a fixed speed reciprocating compressor that provides a constant
displacement rate ( dV ) when it is operating. The volumetric flow rate of the refrigerant
entering the compressor is given by:
22
3
1 1dv
V V Rv
where
dV is the compressor displacement rate = 0.0055 m
3/s
Ris the clearance volume ratio = 0.025
2v and 3v are the specific volumes of the refrigerant at the compressor inlet and outlet,
respectively.
The compressor isentropic efficiency is a function of temperatures and it can be
expressed approximately as:
0.84 0.0075c cond evapT T where
condT and evapT are the saturation temperatures (C) in the condenser and
evaporator, respectively. The heat exchange rates in the condenser and evaporator can
both be modeled with effectiveness relations of the form:
Q = CminTmax where
is the heat exchanger effectiveness which can be determined as (1-exp(-NTU))Cmin is the minimum capacitance rateTmax is the maximum temperature difference between the entering air and thesaturationtemperature of the refrigerant
Air is supplied to the condenser at 25C and 1 kg/s. The outdoor evaporator coil sees an
air supply at the outdoor temperature and 1 kg/s. The overall heat transfer coefficients
for the condenser and evaporator are both 40 W/m2-K. Assume that the evaporator outlet
is saturated vapor and the condenser outlet is saturated liquid. Pressure drops in the heat
exchangers can be neglected.
a. Assume the outdoor air temperature is -5C and the heat transfer surface area of
condenser and evaporator are each 35 m2. Calculate the heating capacity (kW) and
COP at this condition and plot the cycle on a pressure-enthalpy diagram.
b. Calculate and plot the heating capacity and COP as a function of the outdoor
temperature between -30C and 20C. Explain the behavior you observe in the plot.
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9.C-3 Determine the heating season performance factor (HSPF) for operation of the heat pump
described in 9.C-2 for a building located in Madison, WI. The HSPF is defined as a heat
pumps estimated seasonal heating output in BTU divided by the amount of electrical
energy that it consumes in watt-hours. The heating season is defined as the hours during
the year that the ambient temperature is less than 15C. Include the fan power and
auxiliary energy in your calculated HSPF value. When operating, the condenser and
evaporator fans consume a combined power of 68 W. Compare your calculated value
with recommendation of SHPF>7.7 provided by the U.S. DOE.
Table 9.C-3 provides bin temperature information for Madison WI. The first column
shows the building load, i.e., the rate of heat transfer required to maintain the building at
a comfortable temperature when the outdoor temperature is at the value indicated in the
second column in C. The third columns provides the number of hours during the
heating season that the temperature is between -2.5C and 2.5C of the temperature
shown in the second column. For example, there are 136 hours for which the ambient
temperature is between -22.5C and -17.5C. If the heat pump capacity is less than the
building load, then auxiliary electrical heating is provided with a COP=1. If the heat
pump capacity exceeds the building load, the heat pump is cycled off and on as necessary
to supply the load. In this case, the heat pump operates for a period equal to
hours*load/capacity.
Table 9.C-3: Heating load and number of hours at each bin temperature in Madison, WI
Load
[W]
T_avg
[C]
Hours
[hr]
16438 -30 12
14737 -25 55
13037 -20 136
11336 -15 341
9636 -10 364
7935 -5 784
6234 0 1487
4534 5 961
2833 10 1058
1133 15 1315
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9.C-4 You own and operate a natural gas-fired gas turbine engine that provides electrical powerto the grid. The gas turbine engine is shown in Figure 9.C-4(a).
1
compressor
turbine
combustor
2 3
4
ambient air
amb am bm,T ,P
fuel,f
m
net power
Figure 9.C-4(a): Gas turbine engine.
Both the ambient air temperature and the price that you can sell your electricity vary
dramatically. During the day, the ambient air temperature is high and the resulting air-conditioning load causes a high demand for power and therefore leads to high electricityprices. During the night, the ambient air temperature is low and the electricity demand isalso low. As a result, you cannot sell your electricity for very much money. For thisproblem, we will divide the day into two periods. The daytime period has durationtimeday= 10 hr/day and ambient temperature Tday= 96F. During the daytime period youcan sell the electricity that you produce at a rate evday= 0.25 $/kW-hr. The nighttimeperiod has duration timenight= 14 hr/day and ambient temperature Tnight= 72F. Duringthe nighttime period you can sell the electricity that you produce at a rate evnight= 0.05$/kW-hr.
The compressor takes in air with ambient conditions, Tamb (equal to Tday or Tnight,depending on the time of day) and Pamb= 1 atm. The compressor pressure ratio is PR= 5
and the compressor efficiency is c= 0.80. The compressor is a fixed volume device - itis designed to process a specific volumetric flow rate of air. The volumetric flow rate at
the inlet to the compressor isc,inV = 15500 cfm. The air leaving the compressor at state 2
enters the combustor where it is mixed with natural gas and combusted. The heat ofcombustion for natural gas is HC= 50 MJ/kg and the cost of natural gas is ngc= 0.25$/kg. The air-fuel ratio is adjusted so that the combustion products leaving the combustorare at Tt,in = 900C. The combustion products leaving the turbine are exhausted to
atmosphere. The efficiency of the turbine is t = 0.84. Net power produced by theturbine is provided to a generator and converted to electricity (assume that the generator
is 100% efficient).The combustion products can be modeled as air and air can be modeled as an ideal gas.Neglect pressure drop through the combustor.
a.) Model the cycle for the daytime period. Prepare a T-sdiagram and overlay the stateson it. Print out an arrays table that includes each of the states and their entropy,enthalpy, temperature, and pressure. Determine the efficiency, power output, and theair-fuel ratio.
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b.) Determine the profit that you make each day when you run the power plant during thedaytime period.
c.) Determine the profit that you make each day when you run the power plant during thenighttime period.
Your analysis should have shown that you lose money when you operate at night. Thevalue of the electricity that you produce is not sufficient to pay for the fuel requiredbecause you get paid so little for electricity produced during off-peak times.
You are considering installing the system shown in Figure 9.C-4(b).
1
compressorturbine
combustor
2
3 4
ambient air
a mb ambm,T ,P
fuel,f
m
net power
5
Tnight
condenser2nd stage compressor
high pressureexpansion valve
flash chamber
low pressure
expansion valve
evaporator
1st stage compressor
cold storage system
Tics= -5C
precooler
Pint
12
mixer
14
7
6
8
9
10
11
13
2rc ,W
1rc ,W
pcQ
evapQ
condQ
Figure 9.C-4(b): Gas turbine engine with precooling.
The air entering gas turbine engine is precooled using a cold storage system. The coldstorage system is an ice system, except that the temperature of the ice can be adjusted bychanging the concentration of antifreeze in the frozen solution. The gas turbine engine isoperated only during the daytime period. Therefore, each day the ice in the cold storage
system is melted as it cools the air entering the compressor in the precooler. Because theair entering the compressor is colder, it has higher density and therefore the mass flowrate processed by the gas turbine is larger and its power output higher. The result is moredaytime profits. The cold storage system is recharged (i.e., the ice is re-solidified) duringthe nighttime period when the gas turbine is off. The flash intercooled refrigerationsystem shown in Figure 9.C-4(b) is operated at night in order to re-solidify the ice. Thisis advantageous because the cost of electricity at night is less than during the day and
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because the temperature for heat rejection is much lower at night, increasing the COP ofthe refrigeration system.
During the daytime period, ambient air at Tday= 96F and Pamb= 1 atm is drawn into theprecooler at state 1. The temperature of the ice in the cold storage system is Tics= -5C
and the approach temperature difference for the precooler is Tpc= 5 K. Air at T2= Tics+
Tpcenters the compressor. The remaining parameters for the gas turbine are the same(t= 0.84, c= 0.80, Tt,in= 900C, PR= 5, HC= 50 MJ/kg, ,c inV
= 15500 cfm). There is
no pressure loss in the precooler.
d.) Model the gas turbine portion of the cycle shown in Figure 9.C-4(b) for the daytimeperiod. Prepare a T-sdiagram and overlay the states on it. Print out an arrays tablethat includes each of the states and their entropy, enthalpy, temperature, and pressure.Determine the efficiency, power output, and the air-fuel ratio of the system.Determine the profit that you make each day when you run the power plant during thedaytime period, neglecting for now the cost associated with purchasing and runningthe cold storage system.
The flash intercooled refrigeration system is used to resolidify the ice in the cold storagesystem during the night. The condenser rejects heat to the ambient air at Tnight= 72F.
The condenser has an approach temperature difference of Tcond= 4 K and the refrigerantleaving the condenser is subcooled by Tsc = 0.5 K. The evaporator accepts heat fromthe ice at Ticsand has an approach temperature difference of Tevap= 5 K. The refrigerantleaving the evaporator is superheated by Tsh= 5 K. The isentropic efficiency of the firststage compressor is rc,1 = 0.65 and the isentropic efficiency of the second stage
compressor is rc,2= 0.70. The volumetric efficiency of both compressors is vol= 0.65.The working fluid in the refrigeration system is R134a. Neglect pressure loss in all of theheat exchangers. The intercooling pressure (Pint) is the average of the evaporating and
condensing pressures. The flash chamber is a large insulated tank. Saturated liquid ispulled from the bottom of the tank at state 12 and sent to the low pressure expansionvalve and evaporator. Saturated vapor is pulled from the top at state 14 and used toprovide intercooling by mixing with the superheated vapor leaving the first stagecompressor at state 7.
e.) Model the refrigeration cycle. Prepare a T-s diagram and overlay the states on it.Print out an arrays table that includes each of the states and their entropy, enthalpy,temperature, and pressure. Determine the refrigeration provided by the system (intons). Determine the coefficient of performance of the cycle and the power requiredby the refrigerant compressors.
The cost of electricity purchased during nighttime hours is ecnight= 0.12 $/kW-hr (yes, thecost of electricity to you is higher than what you could sell it for because the powercompany wants to strongly discourage you from producing power during the night). Youhave determined that the cost of the refrigeration system can be broken down into thecost of the compressors, condenser, and cold storage system. The cold storage systemcost is Costics= $50,000. The cost of the condenser is given by:
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$-K1.5
Wcond
cond
cond night
QCost
T T
where Tcond is the saturation temperature at the condenser pressure. The cost of thecompressors is given by:
3
$-s92000
mcompCost Displacement
where Displacement is the total displacement of the installed compressors. Thedisplacement of the compressor is the volumetric flow rate at the suction port of thecompressor divided by the volumetric efficiency of the compressor.
f.) Determine the net profit associated with purchasing the refrigeration system andoperating the precooled power plant for Timeec= 1 year.
As a system designer you have a few free parameters that you can use to optimize yoursystem. The temperature of the cold storage system, Tics, the approach temperature
difference in the condenser, Tcond, and the intercooling pressure, Pint.
g.) Plot the net profit as a function of Tics. Explain why an optimal cold storagetemperature exists.
h.) Plot the net profit as a function of Tcond. Explain why an optimal condenserapproach temperature difference exits.
i.) Optimize your system - determine the optimal values of Tics, Tcond, and Pintand theassociated maximum value of the net profit.
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9.C-5 Shown in the figure below is a schematic diagram of a two-stage refrigeration cycle that
is currently in the design phase. The cycle is designed to provide 50 kW of refrigeration
capacity at 225 K with a heat sink of 320 K. As with all cycles of this nature, the heat
transfer processes are responsible for most of the irreversible processes. In this problem,
assume that all of the irreversibilities are a result of heat transfer in the heat exchangers
i.e., we will assume that the two compressors operate reversibly between constant
temperature source and sink streams. The heat transfer rate across each heat exchanger
can be represented in the usual manner as follows:
L L L z
W L y w
H H x H
Q UA T T
Q UA T T
Q UA T T
where UAis the heat transfer area product and
temperatures are identified in Figure 9.C-5.
Figure 9.C-5 2-stage refrigeration cycle
The Uvalues (heat transfer coefficient) for all three exchangers are approximately equal.
However, the heat transfer areas can be adjusted to maximize the system performance. In
the present case, the sum of the three UA values is constrained to 40 kW/K that fixes the
total heat exchanger area for the cycle.
a.) Identify how to best distribute the area, i.e., what are the UA values for the three heat
exchangers that maximize system performance when Ty= 250 K The intermediate
cycle temperatures, Tw and Ty, have not yet determined. How does your result
depend on the value on your choice for Ty?
b.) How does the performance of the optimum cycle compare to that for a single stage
cycle with the same total UA value distributed over two heat exchangers? What
advantages does a two-stage cycle offer over a single stage cycle?
TL=225 K
TH=320 K
Tz
Ty
Tw
Tx
WL
WH
QL=50 kWQL=50 kW
QHQH
QWQW
TL=225 K
TH=320 K
Tz
Ty
Tw
Tx
WL
WH
QL=50 kWQL=50 kW
QHQH
QWQW
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9.C-6 You have been hired as a consultant by a food processing company. The companyemploys a blast freezing process in which very cold air (Tf = -40C) is blown overproduct in order to cause it to freeze very quickly. The rapid freezing process preventsthe growth of large ice crystals and therefore improves the quality of the frozen food.You have been asked to design the refrigeration system required by the blast freezer. The
typical refrigeration system used for this application is a compound ammonia cycle,shown in Figure 9.C-6(a).
condenser
evaporator
condQ
high pressure compressorexpansionvalve 1
1
5
6
8
flashchamber
low pressure compressor
expansionvalve 2
2
3
4
7
hm
htcW
ltcW
lm
Tamb= 26C
air
12,000 cfm
1 atm
a
amb
amb
V
T T
P P
air at Tf= -40C
to blast freezer
evapQ
9 10
Figure 9.C-6(a): Compound ammonia cycle.
The condenser rejects heat to the ambient air at Tamb= 26C. We will begin by assuming
that the condenser has an approach temperature difference of Tcond= 5 K although thesize of the condenser, and therefore the value of Tcondwill eventually be adjusted duringthe design process. The refrigerant leaving the condenser at state 7 is not subcooled. Theevaporator cools air from state 9 where T= Tamband P= Pamb= 1 atm to state 10 where T= Tf. The cold air leaving the evaporator is directed to the blast freezer. The volumetric
flow rate of air entering the evaporator is aV = 12,000 cfm. We will begin by assuming
that the evaporator has an approach temperature difference of Tevap= 5 K although the
size of the evaporator and therefore the value of Tevap will also be optimized. Therefrigerant leaving the evaporator is not superheated. The isentropic efficiencies of bothcompressors depend on the pressure ratio according to:
0.84 0.02 outc
in
P
P
where Pin and Pout are the pressures entering and leaving the compressor, respectively.The working fluid in the refrigeration system is ammonia. Neglect pressure loss in all of
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the heat exchangers. The refrigerant leaving the condenser is throttled to an intermediatepressure (Pint) and enters the flash chamber. The intermediate pressure is initiallyassumed to be half-way between the evaporating and condensing pressures (Pevap - thepressure of the ammonia in the evaporator and Pcond- the pressure of the ammonia in thecondenser):
int evap int cond evapP P f P P where fint = 0.5 (for now, but we can adjust the value of fint and therefore change theintermediate pressure during the design in order to optimize performance). The flashchamber is a large insulated tank. Saturated liquid is pulled from the bottom of the tankat state 1 and sent to the low pressure expansion valve and then the evaporator. Thesuperheated vapor leaving the low pressure compressor is directed back to the flashchamber at state 4. Saturated vapor at state 5 is pulled from the top of the flash chamberat state 5 and sent to the high pressure compressor. Note that the mass flow rate passing
through the high pressure compressor (h
m ) will not be equal to the mass flow rate
passing through the low pressure compressor ( lm ). There is no pressure drop associated
with the flash chamber (i.e., states 4, 5, 8, and 1 are all at the intermediate pressure, Pint).a.) Model the refrigeration cycle. Prepare a T-s diagram and overlay the states on it.Print out an arrays table that includes each of the states (numbered as in the figure)and their entropy, enthalpy, temperature, and pressure. Determine the refrigerationprovided by the system (in tons). Determine the coefficient of performance of thecycle and the power required by the refrigeration compressors.
You are going to use your model developed in part (a) to design a system. The companyhas specified that the design should be based on minimizing the total cost associated withpurchasing and operating the system for 5 years (neglecting the time value of money).Therefore, you need to determine the capital cost of the system as well as the operating
cost. The volumetric efficiency of the compressors is estimated according to:
1 0.025 1involout
v
v
where vin and vout are the specific volumes at the inlet and exit of the compressor,respectively. The cost of the compressor can be estimated according to the displacementrate:
3
$-s192000
mcomp dispCost V
The conductance of the evaporator (a measure of its size) is computed from:
, ln
evap
evap a P a
amb evap
T
UA m c T T
wherea
m is the mass flow rate of air in the evaporator, cP,a is the constant pressure
specific heat capacity of the air in the evaporator, and Tevap is the temperature of theammonia in the evaporator. The cost of the evaporator is estimated according to:
$-K0.75
Wevap evapCost UA
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The conductance of the condenser is computed from:
cond
cond
cond
QUA
T
wherecond
Q is the rate of heat transfer in the condenser. The cost of the condenser is
estimated according to: $-K0.75
Wcond cond Cost UA
Assume that the cost of the system is entirely related to the cost of the compressors andthe heat exchangers. The plant runs 16 hours per and day 340 days per year. The cost ofthe electricity used to power the compressors is ec= 0.10 $/kW-hr.b.) Estimate the capital cost of the system and the operating cost of the system over five
years. Determine the total cost of purchasing and operating the system for 5 years.
As a system designer you have a few free parameters that you can use to optimize yoursystem. The intermediate pressure (set by the parameter fint), the approach temperature
difference in the condenser (Tcond), and the approach temperature difference in theevaporator (Tevap) can each be adjusted in order to determine an optimal design.c.) Plot the total cost as a function of fint. Your plot should demonstrate that there is an
optimal value of the intermediate pressure.d.) Plot the total cost as a function of the condenser approach temperature difference.
Your plot should demonstrate that there is an optimal value of Tcond. Explain whythis is so.
e.) Simultaneously optimize the values of fint, Tcond, and Tevapin order to come up withthe optimal design of the compound refrigeration cycle for this application. Fill in thetable below with the salient features of the design; this table can be submitted to thecompany as the final design specification.
Table 1: Characteristics of the optimal compound refrigeration cycle.
Parameter Optimal value for
compound cycle
Optimal value for
cascade cycle
Refrigeration rate
Air exit temperature
Evaporator approach temp.
difference
Evaporator conductance
Evaporator costCondenser approach temp.
difference
Condenser conductance
Condenser cost
Cascade HX approach temp.
difference
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Cascade HX conductance
Cascade HX cost
Intermediate pressure
Intermediate temperature
Low P/T compressordisplacement
Low P/T compressor cost
High P/T compressor
displacement
High P/T compressor cost
Capital cost of system
Total compressor power
Coefficient of performance
Operating cost
Total cost of ownership for 5
years
The company has asked you to also evaluate an alternative refrigeration system for theirapplication. They have heard that the cascade refrigeration cycle, shown in Figure 9.C-6(b), is more attractive for low temperature systems. The cascade cycle consists of a lowtemperature cycle that is interfaced to a high temperature cycle by the cascade heatexchanger. The advantage of the cascade cycle over the compound cycle is that the lowtemperature circuit can utilize a high density working fluid, in this case carbon dioxide.The high temperature circuit utilizes ammonia. The high density of carbon dioxide
results in a much smaller low temperature compressor which is therefore much lessexpensive than the low pressure compressor required by the compound cycle. (Youranalysis from the previous part of the project should show that the cost of the lowpressure compressor is pretty large.) The disadvantage of the cascade cycle is therequirement of a cascade heat exchanger which adds cost to the system.
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condenser
evaporator
evapQ
condQ
high temperaturecompressor
upper stagethrottle valve
2
6
5
low temperaturecompressor
low stagethrottle valve
3
1
8
cascade heat exchanger
7
4
Tamb
= 26C
hm
htcW
ltcW
lm
air
12,000 cfm
1 atm
a
amb
amb
V
T T
P P
air atTf
= -40C
to blast freezer
ammonia
carbondioxide
9 10
Figure 9.C-6(b): Cascade ammonia cycle.
The condenser rejects heat to the ambient air at Tamb= 26C. Begin by assuming that the
condenser has an approach temperature difference of Tcond= 5 K, although the size of
the condenser will eventually be adjusted during the design process. The refrigerantleaving the condenser at state 7 is not subcooled. The cascade system must provide thesame refrigeration load; therefore, the evaporator cools air from state 9 where T= Tamband P= Pamb= 1 atm to state 10 where T= Tfand the volumetric flow rate of air entering
the evaporator is aV = 12,000 cfm. Begin by assuming that the evaporator has an
approach temperature difference of Tevap= 5 K; the optimization of this component willfollow. The refrigerant leaving the evaporator is not superheated. The isentropicefficiencies of both compressors depend on the pressure ratio as indicated above.
Neglect pressure loss in all of the heat exchangers. The ammonia leaving the condenseris throttled to state 8 where it enters the cascade heat exchanger. The ammonia
evaporates in the cascade heat exchanger as it absorbs heat from the condensing carbondioxide in the other side of the cascade heat exchanger. The ammonia leaves the cascadeheat exchanger as saturated vapor at state 5. The ammonia is compressed to state 6 andthen enteres the condenser. The carbon dioxide leaves the cascade heat exchanger assaturated liquid at state 1. The temperature at state 1 is the intermediate temperature inthe cycle (Tint). The intermediate temperature is initially assumed to be half-way betweenthe evaporating and condensing temperatures (Tevap - the temperature of the
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carbondioxide in the evaporator and Tcond - the temperature of the ammonia in thecondenser):
int evap int cond evapT T f T T where fint = 0.5 initially (but we can adjust the value of fint and therefore change theintermediate temperature during the design in order to optimize performance). The flash
chamber is a large insulated tank. Initially assume that the cascade heat exchangerapproach temperature difference is Tcascade = 5 K (we will optimize this eventually).The pinch point for the cascade heat exchanger is at the end where ammonia enters andcarbon dioxide leaves (the left side of the heat exchanger in the figure). There is nopressure loss associated with either the flow of ammonia or carbon dioxide in the cascadeheat exchanger.f.) Model the cascade refrigeration cycle. Print out an arrays table that includes each of
the states (numbered as in the figure) and their entropy, enthalpy, temperature, andpressure. Determine the refrigeration provided by the system (in tons). Determinethe coefficient of performance of the cycle and the power required by therefrigeration compressors.
You are going to use your model to design a cascade system. Again, the company hasspecified that the design should be based on minimizing the total cost associated withpurchasing and operating the system for 5 years (neglecting the time value of money).The capital cost of the compressors, condenser, and evaporator can be estimated as in part(b). The cost of the cascade heat exchanger must also be considered. The conductance ofthe cascade heat exchanger is computed from:
cascadecascade
cascade
QUA
T
where cascadeQ is the rate of heat transfer in the condenser. The cost of the cascade heat
exchanger is estimated according to:
$-K0.75
Wcascade cascadeCost UA
Assume that the cost of the system is entirely related to the cost of the compressors andthe heat exchangers. The plant runs 16 hours per and day 340 days per year. The cost ofthe electricity used to power the compressors is ec= 0.10 $/kW-hr.g.) Estimate the capital cost of the system and the operating cost of the system over five
years. Determine the total cost of purchasing and operating the system for 5 years.
As a system designer you have a few free parameters that you can use to optimize the
cascade system. The intermediate temperature (set by the parameter fint), the approachtemperature difference in the condenser (Tcond), the approach temperature difference inthe evaporator (Tevap), and the approach temperature difference in the cascade heatexchanger (Tcascade) can each be adjusted in order to determine an optimal design.h.) Simultaneously optimize the values of fint, Tcond, Tevap, and Tcascade in order to
come up with the optimal design of the cascade refrigeration cycle for thisapplication. Fill in the table with the salient features of the design; this table can be
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submitted to the company as the final design specification. Based on your analysis,which system would you recommend?
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9.C-7 The absorption refrigeration cycle shown in Figure 9.C-7 is an example of a heat-drivenheat pump. Heat-driven heat pumps operate between three thermal reservoirs
TL
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9.C-8 The performance of a heat pump cycle in a heating application can be expressed in terms
of its coefficient of performance (COPH), which is defined as the rate of energy transfer
to the heated space,H
Q , to the compressor power input. The maximum possible COP
for a machine operating between an outdoor temperature TL and a heated space
temperature THis given by the Carnot limit TH/(TH-TL). The point of this problem is to
investigate the operation of a heat-transfer limited Carnot heat pump cycle.
Consider a heat pump cycle having no internal irreversibilities that is designed to provide
a heating capacity of 10 kW for a building. (The ideal heat pump cycle is a Carnot cycleworking in reverse.) Air from the building is blown past the condenser at 25C and 2.0
kg/s. The condenser heat transfer rate can be expressed as:
,H H H p c H inQ m C T T where
His the effectiveness of the condenser given which is equal to (1-exp(-NTUH))
Hm is the condenser air flow rate (2.0 kg/s)
pC is the specific heat of air
,H inT is the entering temperature of the air, (25C)
cT is the refrigerant condensation temperature
NTUH= (UHAH)/( H pm C )
AH is the effective heat transfer for the condenser
Heat is supplied to the evaporator from outdoors with a fan-forced air stream at 2.0 kg/s
at -5C. The rate of heat supplied from outdoors can be represented as
, -L L L p L in eQ m C T T
where
L is the effectiveness of the evaporator given which is equal to (1-exp(-NTUL))
Lm is the evaporator air flow rate (2 kg/s)
,L inT is the entering temperature of the air, -5C
eT is the refrigerant evaporation temperature
NTUL=(ULAL)/( L pm C )
ALis the effective heat transfer for the evaporator
The overall heat transfer coefficients for the condenser, UH,and evaporator, UL, are both
about 40 W/m2-K, since the air flow rates through both heat exchangers are equal andthey have similar geometry. However, the total heat exchanger area for the evaporator
and condenser is fixed for cost reasons to be 35 m2. One concern is how to optimally
distribute this heat exchange area.
a.) What fraction of the total (condenser plus evaporator) heat exchanger area should be
allocated to the condenser in order maximize the performance of this heat pumpcycle?
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b.) How is your answer changed if the total heat exchanger area is increased to 70 m2or
decreased to 25 m2? Can you draw a general conclusion?
c.) Using the optimum heat exchanger area allocation determined in part a for 35 m2of
total heat exchanger surface area, plot the COP of this refrigeration cycle and the
evaporator and condenser temperatures as a function of heating capacity for
capacities ranging between 0 and 20 kW. Explain the trends that are demonstrated bythe plot.
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9.C-9 A large central chilled water plant uses a vapor compression cycle with an adiabatic flash
chamber and two-stage compression, as shown in Figure 9.C-9. The refrigerant is R134a.
The isentropic efficiencies of the high and low pressure compressors are 0.82 and 0.78,
respectively. Cooling water passing through the condenser maintains 35C saturated
liquid at state 5. The refrigerant is saturated vapor at 1C at the evaporate exit. Pressure
losses in the piping and heat exchangers are negligible, except of course, across thethrottle valves. The plant chills 1625 kg/s of water from 15C to 5C.
Figure 9.C-9: Two-stage refrigeration cycle with flash chamber
a.) Determine the temperature and pressure at all points in the cycle assuming that the
pressure at state 7 is the algebraic average of the pressures at states 5 and 1.b.) Determine the pressure at state 7 that results in the optimum COP for this cycle.
c.) Plot the optimum cycle on a pressure-enthalpy diagram.
d.) Indicate in a short paragraph the advantages and disadvantages of the two-stage
system in Figure 9.C-9.
e.) Determine the Second-Law efficiency of the optimum cycle.
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9.C-10 Temperatures and pressures at various places in a simple Linde-Hampson gas
liquefaction cycle are indicated in Figure 9.C-10. The uncondensed vapor is recycled
through a heat exchanger from which it emerges at 250 K, 1 bar. Gas from the inter-
cooled compressor enters the heat exchanger at 15.0 MPa and 275 K and exits at 155 K
and essentially the same pressure. Assume an environment at 298 K, 1 bar.
vapor
liquid
1 2 3
4
567
Satd liquid at 1 bar
163 K
15 MPa275 K
15 MPa
1 barMake-up N2298 K, 1 bar
intercooled
compressor
heat exchanger
valve
Figure 9.C-10: Simple nitrogen liquefaction system
a.) Determine the mass of liquid per kg of nitrogen gas passing through the compressor
during steady operation.
b.) What is the effectiveness of the heat exchanger effectiveness for this system?
c.) What is the Second-law efficiency of the heat exchanger?
d.) If the compressor work per kg of nitrogen compressed is 475 kJ/kg, what is the
Second Law efficiency of this liquefaction cycle?
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9.C-11 A schematic of a heat-powered refrigeration system is shown in Figure 9.C-11. This
system uses the exergy of thermal energy at 160C to move heat from a 5C source while
rejecting heat to the ambient at 25C. The heat transfer rates may be expressed:
Figure 9.C-11: Schematic of a heat-powered refrigeration system
where are 100, 150, and 300 W/K, respectively and temperatures are as shown inFigure 9.C-11.
a.) What is the maximum possible refrigeration COP (coefficient of performance) for
this heat-powered refrigeration system based on the heat input?
b.) What is the maximum possible refrigeration capacity ( LQ )?
1 2 3( ) ( ) ( )H H L L A ambQ T T Q T T Q T T
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9.C-12 A single-stage refrigeration system provides 10 kW of refrigeration from a space that is
maintained at TL=-10C, while rejecting heat to the surroundings at TH=35C. The COP
for this cycle is 2.2 at steady-state operating conditions. The heat transfer rates to and
from the refrigeration cycle can be assumed to be linear functions of the temperature
differences so that: 2L LQ T T and 4H HQ T T where T2 is the evaporator
temperature and T4is the condensing temperature.
a.) Determine the steady-state power required to operate this refrigeration cycle.
b) Determine the steady-state rate of exergy destruction resulting from operation of this
refrigeration cycle.
c.) T2 is measured to be -30C. Assuming that all of the exergy destruction in this
refrigeration cycle occurs in the heat exchangers, what are the values of the heat
exchanger coefficients, and ?
d.) The sum of and is proportional to the total cost of the heat exchangers.
Determine the values of andthatwill minimize this cost while maintaining a COP
of 2.2 and a refrigeration capacity of 10 kW.
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