Problem Chapter 9

7/22/2019 Problem Chapter 9

1/48

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?


2/48

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.


3/48

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?


4/48

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.


5/48

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?


6/48

h.) Compare the exergy destroyed in the throttle with the work needed to power the

compressor for the process in part (d).


7/48

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.


8/48

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.


9/48

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).


10/48

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.


11/48

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.


12/48

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.


13/48

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.


14/48

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?


15/48


16/48

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?


17/48

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%.


18/48

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.


19/48

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.


20/48

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.


21/48

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.


22/48

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?


23/48

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?


24/48

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.


25/48


26/48

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.


27/48

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?


28/48

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.


29/48

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


30/48

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.


31/48

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


32/48

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:


33/48

$-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.


34/48

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


35/48

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


36/48

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


37/48

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


38/48

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.


39/48

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


40/48

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


41/48

submitted to the company as the final design specification. Based on your analysis,which system would you recommend?


42/48

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


43/48

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?


44/48

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.


45/48

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.


46/48

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?


47/48

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


48/48

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.

Problem Chapter 9

Documents

Transcript of Problem Chapter 9