faramarzi_2

8/11/2019 faramarzi_2

1/7

4 6 PR ACTICAL G UIDE N ovem b e r 199 9

Efficient Display Case RefrigerationBy Ramin Faramarzi, P.E.

Associate Member ASHRAE

PRACTICAL GUIDE

r e f r i g e r a t i o n

T

his article discusses the cooling load compo-

nents of an open refrigerated display case.

Refrigeration system energy use is importantbecause it accounts for approximately half the

electricity use of a typical large supermarket. Re-

frigerated display cases account for most of this

usage.

This article presents a series of simplified cooling loadestimation equations that provide basic guidelines fordetermining the individual loads. It further describes theresults of a recent test conducted by Southern CaliforniaEdisons Refrigeration Technology and Test Center toquantify the main cooling load components of a typicalopen, multi-deck, meat display case. Using those test re-sults, this article can provide:

A sense of the magnitude of the cooling load com-ponents and their relative ranking. The test results indi-cate that infiltration constitutes the largest cooling loadcomponent in an open vertical display case.

The basis for more accuracy in simulating supermar-ket energy systems.

Better forecast of refrigeration effects on human com-fort (intrinsic value).

Better understanding of the main loads on displaycases leading to more intelligent end user display casepurchasing decisions.

Additional insight to help display case manufactur-ers design more efficient cases.

Display Case Heat GainRefrigerated display cases are used to merchandise

perishable food and provide desirable storage tempera-tures. The refrigeration system controls product storagetemperatures by removing all the heat gain componentsof the display case.

The heat transfer in a display case involves interac-tions between the product and the internal environmentof the case as well as heat from the surroundings that

enters the case. The constituents of heat from the sur-rounding environment include transmission (or conduc-tion), infiltration, and radiation.

The product exchanges heat with the environmentinside the display case through convection and radia-tion. Convection is typically the main mode of heat transferthat cools the product. It takes place when the refriger-ated air comes in contact with the product. The tem-perature difference between the surface of the productand the surfaces inside the fixture governs the rate ofheat transfer into and from the product through radia-tion.

Equation 1expresses Newtons Law of Cooling thatcan be used to determine the convection heat transfer atthe surface of the product.

Qc

= Ap

hconv

(Tsp

Tcase

) (1)

where,Q

c= Convection load, Btu/h

Ap

= Surface area of the product, ft2

hconv

= Overall coefficient of convective heattransfer, Btu/hft2F

Tsp

= Surface temperature of product, FT

case= Air temperature inside the case passing over

the product surface, F

Conduction, radiation, and infiltration loads from thesurroundings into the case as well as heat exchangesbetween the product and the inside components of the

case depend on the temperatures of the ambient air andthe air within the case. Open vertical display cases relyon their air curtains to keep warm ambient air from pen-etrating into the cold environment inside the case. Aircurtains play a significant role in the thermal interactionof the display case with the surrounding air.

The total refrigeration load is normally calculated overa 24-hour period (i.e., Btu/24 hours). The required

The following article was published in ASHRAE Journal, November 1999. Copyright 1999 American Society of Heating, Refrigerating and Air-

Conditioning Engineers, Inc. It is presented for educational purposes only. This article may not be copied and/or distributed electronically or in paper

form without permission of ASHRAE.

Ramin Faramarzi, P.E., is the project manager at South-ern California Edisons Refrigeration Technology and TestCenter in Irwindale, Calif.


2/7

Practical Guide

N ov em b er 19 9 9 P R A C T I C A L G U I D E 4 7

cooling capacity of the refrigeration system in Btu/h canbe determined by dividing the total cooling load ofdisplay cases and walk-in coolers over a 24-hour periodby the compressor run time. Compressor run time is afunction of the defrost period and cooling load fluc-

tuations. Defrost is necessary for any evaporator thatoperates below the freezing point of water. The needfor defrost makes refrigeration systems different fromair-conditioning systems. It is not realistic to design arefrigeration system that operates continuously. There-fore, refrigeration compressor capacity should be de-termined as a function of total daily cooling loads andcompressor run time.

Display Case Cooling Load ComponentsThe cooling load of a typical display case has both sen-

sible and latent components. They can be determined by:

Qcase = [QC + QR + Qis + QL + QF + QD + QASH

+ Qpulldown

]sensible

+[Qil+ Q

pr]

latent(2)

where,Q

case= Display case cooling capacity, Btu/h

QC

= Transmission heat gain, Btu/hQ

R= Radiation heat gain, Btu/h

Qis

= Infiltration sensible heat gain, Btu/hQ

L= Heat gain from lights, Btu/h

QF

= Heat gain from fan motor, Btu/hQ

D= Heat gain from defrost heater, Btu/h

QASH

= Heat gain from anti-sweat heater, Btu/hQ

pulldown= Product pull-down load, Btu/h

Qil

= Infiltration latent heat gain, Btu/hQ

pr= Product latent heat of respiration, Btu/h

For medium temperature display cases used to mer-chandise meat, deli, dairy, produce, poultry, and fish,the main cooling load components are infiltration, lights,fan motors, radiation, and conduction.Figure 1depictsthe comparative load contribution for an 8 ft (2.4 m)open vertical meat case tested at ambient room condi-tions of 75F(24C) dry-bulb (DB) and 55% relative hu-midity (RH). As shown, infiltration is the largest constitu-ent of the case cooling load.

Conduction

The conduction load refers to the transmission ofheat through the display case envelope. The tempera-ture difference between the air in the room and theair inside the case is the driving force for this heattransfer. Transmission load makes up the smallest por-tion of the display case total load. The first task indetermining the transmission load is to determine theoverall heat transfer coefficient of the case walls. Thisinvolves the outside and inside air film convectivecoefficients, the thermal conductivity of the outer andinner surfaces of the case, and the thermal conductiv-ity of the insulation between the inner and outer walls.

The inside film conductance of the case is a functionof forced convection, whereas the exterior film con-ductance depends on natural convection heat trans-fer.Equation 3describes how this overall heat transfer

coefficient can be determined.

U = 1/[(1/hi) + (L

1/k

1) + (L

2/k

2)

+ (L3/k

3) + (1/h

o)] (3)

where,U = Overall heat transfer coefficient of the case

walls, Btu/hft2Fh

i= Convective coefficient for inside case air film

against case inner wall, Btu/hft2FL

1= Thickness of outer shell of the case, in.

k1

= Thermal conductivity of outer shell of case,Btuin./hft2oF

L2

= Thickness of insulation within the case walls,in.

k2

= Thermal conductivity of insulation within thecase walls, Btu-in./hft2F

L3

= Thickness of inner shell of the case, in.k

3= Thermal conductivity of inner shell of case,

Btuin./hft2Fh

o= Convective coefficient for outside/room air

film against case outer shell, Btu/hft2F

Once the display case overall heat transfer coefficientis determined, the transmission load can be quantifiedusingEquation 4.

Qc = U A (Troom Tcase) (4)

where,Q

c= Transmission, or conduction, load of the

case, Btu/h

A = Total surface area of case walls that areconducting heat, ft2

Troom

= Dry-bulb temperature of the air in the room,F

Tcase

= Dry-bulb temperature of the air inside thedisplay case, F

Figure 1: Contribution of individual components to the re-frigeration load of the display case.


3/7

Refrigeration

4 8 P R A C T I C A L G U I D E N ov e mb e r 19 9 9

Pull-Down LoadThe pull-down load has two com-

ponents: product shelving and postdefrost.

Product shelving load.Pull-down

load due to product shelving comesfrom delivering products into the caseat a temperature higher than the des-ignated storage temperature. It is theamount of cooling required to lowerthe product temperature to a desiredtarget point (Equation 5).

Qps

= m Cp(T

s T

f)/dT (5)

where,Q

ps= Pull-down load due to

product shelving, Btu/h

m = Product mass, lbCp

= Product specific heatcapacity, Btu/lbF

Ts

= Temperature at the timeof shelving, F

Tf

= Final temperature, FdT = Time required to lower

product temperature, h

Post-defrost pull-down load.

During the defrost cycle, the tempera-ture of the product inside the caserises. Once the defrost is complete,the refrigeration system must haveenough capacity to remove the accu-mulated defrost heat and lower theproduct temperature to a desirablesetpoint in short time. The time re-quired to remove post-defrost pull-down load depends on:

Defrost types and heat intensity. Defrost termination controls. Specific heat of the product. The desired target storage tem-

perature.During the defrost cycle, the com-

pressor serving the display case does

not operate (Figure 2). Therefore, therefrigeration system needs enoughpost-defrost cooling capacity to main-tain desirable product temperature.

Figure 3 illustrates the effect ofdefrost on the product temperatureand cooling load of an open verticalmeat case over four defrost cycles.The product temperature rises dur-ing defrost cycles resulting in a highercooling load following each defrost.In addition, the auxiliary defrost heat

Figure 2: Compressor power over 24 hours, including defrost period (actual testdata).

Figure 3: Cooling load, compressor kW and product temperature over 24 hours,including defrost period (actual test data).

added to the evaporator must be re-moved once compressors operate atthe end of the cycle.

Qpd

= m Cp(T

pd T

i)/ dT (6)

where,Q

pd= Pull-down load due to

defrost, Btu/hm = Product mass, lbC

p= Product specific heat

capacity, Btu/lbFT

pd= Post-defrost temperature, F

Ti

= Final desired temperature,F

dT = Time interval to lowerproduct temperature, h

Defrost LoadAs the evaporator of a display case

operates at temperatures below thefreezing point of water, ice or frost

Figure 4: Effects of coil frost accumulation on cfm, UA, air pressure drop and airtemperature leaving coil.


4/7

Practical Guide


forms on the surface of the evaporator coil. Moisture inthe air that circulates over the evaporator is the mainsource of frost formation. As water molecules contact thecold surface of the coil at temperatures below their dewpoint, they condense and lose their latent heat of vapor-

ization. If the surface temperature is below freezing, thewater further gives up its heat of fusion and converts toice. A buildup of ice on the heat transfer surface de-creases the evaporator heat transfer coefficient (U-value)and also increases the resistance to airflow across thecoil. The reduction in airflow is more detrimental thanthe decrease in U-value. Hence, refrigeration applica-tions where frost can accumulate should have some typeof defrost mechanism.

Figure 4depicts the effects of coil frost accumulationon the case air circulation rate (cfm), effective heat trans-fer coefficient (UA), air pressure drop across the coil,and coil leaving air temperature.

Defrost mechanisms typically cycle according to pre-determined time intervals. Some controls maintain thedefrost cycle for a set time period. Others initiate on atime cycle and terminate when the evaporator surfacereaches a preset temperature. Depending on the refrig-eration application, defrost mechanisms may vary fromsimple off-cycle and reverse airflow, which rely on thetemperature difference between the store and the case,to electric and hot gas defrost, which introduce supple-mentary heat.

Medium Temperature Defrost.If the temperature ofthe air returning to the evaporator is above high 20s to30s (F) an off-cycle defrost usually will be used. Off-cycle defrost can be accomplished by allowing the fan tocontinue to run while the compressor is shut down, ei-ther for a preset time interval or until the coil tempera-ture rises a few degrees above 32F (0C).

Reverse air defrost is also used for applications wherethe evaporator temperature is not too low. Under thisdefrost mechanism, display case fan(s) induce air fromthe store into the case. It is similar to off cycle exceptwarm store air is forced into the case. Both defrost mecha-nisms rely on the heat content of store air to melt the icewithout auxiliary heat.

Low Temperature Defrost.Low temperature appli-cations need a source of auxiliary heat to melt the ice/frost on the coil. Electric defrost applies heat externally

to the evaporator. The heating elements are typically situ-ated near the evaporator. When defrost is initiated, thefans shut off after a slight delay. Like off-cycle defrost,the refrigeration is off for the entire defrost cycle.

Hot gas defrost uses a portion of the heat content ofdischarge gas from an operating compressor to defrostthe evaporator(s). Basically, in this method the evapora-tor becomes a condenser. Either a timer or a tempera-ture sensor typically terminates the defrost cycle.

Both electric and hot gas defrost typically deliver moreheat than needed just to melt the ice. A large portion ofthe extra heat warms the coil metal and dissipates inside

the case. This extra heat adds to the refrigeration load.The extra heat can be as much as 85% of the total defrostheat input, which means only 15% of the defrost heat isuseful ice melting heat.

The following equations provide a simple approach

to quantify the defrost load.

Electric defrost:

QE-Defrost

= (W k1) Q

im(7)

where,W = Electric defrost heat input, wattsk

1= Conversion factor from watts to Btu/h

Qim

= Heat required to melt the ice as a function offrost mass, heat of fusion of ice and time,Btu/h

Hot gas defrost:

QHG-Defrost

= Qrefrig

Qim

(8)

where,Q

refrig= Hot gas defrost heat input as a function of

refrigerant mass flow and refrigerant enthalp-ies entering and leaving the coil, Btu/h

Anti-Sweat Heater LoadAnti-sweat heaters are used on most low temperature

open display cases as well as reach-in type cases withglass doors. These electric resistance heaters are locatedaround the case handrails and door frame/mullions toprevent condensation on metal surfaces. They also re-duce fogging of the glass door that can hurt productmerchandising. Anti-sweat heaters typically operate con-tinuously (8,760 hours per year). Their power usage andresulting cooling load often can be reduced by applyingsmart controls than can reduce heater operation depend-ing on indoor humidity. Equation 9gives the coolingload contribution of anti-sweat heaters.

QASH

= W k1k

2(9)

where,W = Connected electric load, watts

k1 = Fraction (%) of heat dissipation into the casek

2= Conversion factor from watts to Btu/h

Radiation LoadThe heat gain of the display case through radiation is

a function of the conditions inside the case, includingwall temperature, wall emissivity, wall area, view factorwith respect to the surrounding (store) walls/objects,floor, ceiling, and their corresponding emissivities andareas. For simplicity, each surface can be represented bya set of equivalent emissivity areas and temperatures.

The case load due to radiation heat transfer can be


5/7

Refrigeration


determined by modeling the system as two gray surfaces,one surface being the total surface area of the room (walls,floor, ceiling), the other being an imaginary plane thatcovers the opening of the display case. All the radiationleaving the room surface arrives at the imaginary plane.

The imaginary plane at the case opening, in turn, ex-changes all its radiation with the interior surfaces of thedisplay case.

Figure 5shows a simplified diagram that identifies thesurfaces exchanging heat through radiation. The insideof the display case (back, top, bottom, and sides) wasdesignated Surface 1. The room surfaces (floor, ceiling,and walls) were designated Surface 2, and the imaginaryplane covering the case opening was designated Sur-face 3. Using the reciprocity relation (as a function ofarea [A] and view factor [F]), it can be shown that A

1F

13

= A3F

31. In this case, F

31is 1, because the plane over the

opening of the case can see the entire inside surface of

the case. Also, F13= F12because the inside surfaces ofthe case must look through the opening of the case tosee the room surfaces. Therefore, F

12= F

13= A

3F

31/A

1

= A3/A

1.

Once this view factor is determined,Equation 10canbe used to calculate the radiation load on the case.

Qrad

= (Tw

4 Tc4)/[(1

w)

wA

w

+ 1/Aw

Fcw

+ (1 c)/

cA

c] (10)

where,Q

rad= Radiation heat transfer between room walls

and display case, Btu/h = Stefan-Boltzmann Constant, 0.1714 108

Btu/hft2R4

Tw

= Surface temperature of the room walls, RT

c= Surface temperature of the display case inner

walls, Rw

= Emissivity of the room wallsA

w= Total area of room surfaces, ft2

Fc-w

= View factor from case to surfaces of the room

c= Emissivity of the inside walls of the case

Ac

= Total area of the inside walls of the case, ft2

Internal LoadsThe display case internal load includes the heat from

the case lights and the evaporator fan motors. The lamps,

ballasts, and fan motors are typically located within thethermodynamic boundary of the case. Hence, in mostcases, their total heat dissipation should be consideredpart of the case load.

Fan motor heat gain is a direct function of flow workand combined motor/fan efficiencies. Flow work is alsoa function of air velocity, discharge opening dimensions,and total pressure loss across the coil. For load calcula-tions, however, simply the nameplate horsepower andthe rated wattage of lamps and ballasts could be used in

Equations 11 and12.

Qfans

= Wfans

k (11)

Qlights

= Wlights

k k

1(12)

where,Q

fans= Case load due to fan motors, Btu/h

Qlights

= Case load due to lighting, Btu/hW

fans= Wattage consumed by the fan motors, W

Wlights

= Wattage consumed by the light fixtures inthe case, W

k = Conversion factor, 3.413 Btu/h/Wk

1= Percentage of heat coming to the case as a

function of ballast and lamp location

Infiltration LoadThe infiltration load of the display case refers to the

entrainment of warm, moist air from the room, acrossthe case air curtain, into the refrigerated space. Thetotal performance of the air curtain and the amount ofheat transferred across it depends on several factors,including:

Air curtain velocity and temperature profile.

Number of jets.Air jet width and thickness.Dimensional characteristics of the discharge air hon-

eycomb.Store and display case temperatures and humidity

ratios. Rate of air curtain agitation due to shopper walking

by.Effects of turbulence and eddy viscosity in the initial

region of the jet.An air curtain consists of a stream of air discharged

from series of small nozzles within a honeycombed con-

Figure 5: Surfaces participating in radiation heat transfer.


6/7

Practical Guide


figuration at top of the display case. For vertical fixtures,

the air discharges downward toward a return grille lo-cated approximately 2 ft (0.6 m) above the floor on thefront panel of the case. The air is drawn into a circulat-ing fan and cooling coil (or evaporator) assembly, whereit gives up sensible and latent heat. The chilled air thenreturns to the discharge grille. The discharged air passesbetween the still air in the store and still air within thecase. The still air on both sides of the air curtain mixeswith the discharged air.

Figures 6 and7depict the air curtain velocity streamlines and temperature profiles of an 8 ft (2.4 m) openvertical meat case. These velocity stream lines representthe actual flow using a sophisticated visualization tech-nique. The cross sectional temperature distributions weredeveloped using infra-red photography. As shown, theentrainment of warm room air into the case takes placeat several locations along the plane of the air curtain.Based on the law of conservation of mass, an equal (andsubstantial) amount of cold air from the case spills intothe room near the return air grille of the case (Figure 6).

The temperature gradient between the cold and warmsides of the air curtain along with the velocity profile withinthe mixing (or entrained) zones causes heat from the warm(store) side to transfer to the cold (case) side of the aircurtain inside the display case.Figure 8shows this mixingon a psychometric chart using data from an actual testunder 75F (24C) DB and 55% RH indoor conditions.

This figure shows the properties of the air inside the caseand the entrained or infiltration air. The mixed airstreamis the condition entering the display case cooling coil.

The infiltration load has two componentssensibleand latent. The sensible portion refers to the directtemperature-driven heat added to the display case. Thelatent portion refers to the heat content of the mois-ture added to the case by the room air drawn into thecase through the air curtain. As air passes through theevaporator, it loses its sensible heat and is dehumidi-fied. The sensible portion of the infiltration load isgiven byEquation 13.

Qsenseinf

= V

C

p k

(T

room T

case) (13)

where,Q

senseinf= Sensible portion of the infiltration load,

Btu/h = Density of air, lb/ft3

V = Volume flow rate of air entrained into thecase, ft3/min

Cp

= Specific heat of air, Btu/lbk = Conversion factor, 60 min./h

The main source of latent load for a display case is themoisture content of the ambient air that is entrained intothe case across the air curtain. In some cases, product res-piration generates additional moisture within the displaycase.Equation 14calculates the latent load of the fixture.

Figure 6: (Left) Air curtains velocity streamlines of an actualdisplay case captured using DPIV technique. (Right) Aircurtains velocity streamlines using calibrated CFD modeling.

Figure 7: Air curtain temperature profile of an actual displaycase captured using infra-red photography.

Figure 8: Infiltration process for 8 ft (2.4 m) open multi-deckmeat case at 75F db/55% RH room condition.


7/7

Refrigeration


Qlatent

= Qil+ Q

pr(14)

where,Q

latent= Latent load, Btu/h

Qil

= Latent load due to infiltration, Btu/h

Qpr = Latent load due to product respiration,Btu/h

The contribution of the infiltration portion of the caselatent load can be determined by:

Qil= [

V

C

p k

(W

room W

case)]

h

fg(15)

where, = Density of air, lb/ft3

V = Volume flow rate of air into the case, ft3/min.C

p=

Specific heat of air, Btu/lb

k = Conversion factor, 60 min./h

Wroom = Room humidity ratio, lb/lbairWcase

= Case humidity ratio, lb/lbair

hfg

= Latent heat of vaporization of water, Btu/lb

Fresh fruits and vegetables lose moisture by respira-tion. This moisture then transports through the skin ofthe commodity, evaporates, and ends up in the surround-ings by convective mass transfer. Respiration is a chemi-cal process by which fruits and vegetables convert sugarsand oxygen into carbon dioxide, water and heat. Thegenerated heat increases the water vapor dissipationacross the skin of commodity into the refrigerated dis-play case. This cooling load can be estimated by the fol-lowing equation:

Qpr

= mvA

s n

h

fg(16)

where,Q

pr= Respiration heat, Btu/h

mv

= Mass transfer rate of water vapor leavingthe products skin, lb/hft2

As

= Surface area of the product, ft2

n = Number of productsh

fg= Latent heat of vaporization of water, Btu/lb

Total Cooling LoadThe total sensible load of the case can be calculated

withEquation 17.

Qsense

= Qcond

+ Qrad

+ Qfans

+ Qlights

+ Qinf-sens

+ Qdef

+ QASH

+ Qpulldown

(17)

where,Q

sense= Total sensible load on the display case, Btu/h

The latent load of the case can be calculated withEqua-tion 18, which is the same asEquation 14.

Qlatent

= Qil+ Q

pr(18)

The total cooling of the case can then be expressed

by:

Qtotal

= Qsense

+ Qlatent

(19)

Conclusions

Laboratory tests showed that infiltration constitutes thelargest cooling load component of an 8 ft (2.4 m) openvertical display case (Figure 9). Radiation and internalloads are the next largest constituents.

Infiltration, however, might not play the same crucialrole for other display case configurations. For instance,radiation is likely the most influential constituent of thecooling load for coffin-type fixtures. Determining theinfiltration load is the most challenging aspect of a dis-play case cooling load analysis. The lack of thermo-fluidperformance knowledge of air curtains has contributedsignificantly to this challenge. Primarily, the absence of arobust and simplified method to determine the amountof air entrained into the display is the missing piece ofthe puzzle. Additional scientific testing of various aircurtain and display case designs should be carried out toestablish a well-engineered and realistic foundation fordeveloping reliable methods for estimating infiltrated airquantities.

References1. Al-Mutawa, N.K., S.A. Sherif and G.D. Mathur. 1998. Determi-

nation of coil defrosting loads: Part III-testing procedures and datareduction.ASHRAE Transactions104(1).

2. Stoecker, W.F. 1957. How frost formation on coils affects refrig-

eration systems.Refrigerating EngineerFebruary.

3. Becker, B.R., B.A. Fricke. 1996. Transpiration and respiration of

fruits and vegetables.Refrigeration Science and Technology Proceed-

ingsOctober.

Figure 9: Cooling components of 8 ft (2.4 m) open vertical meatcase (actual test results).

faramarzi_2

Documents

Transcript of faramarzi_2