Refrigeration Load Calculations

18. REFRIGERATION LOAD CALCULATIONS

18.1. MULTIPLE PURPOSES OF LOAD CALCULATIONS

It is mandatory that refrigeration load calculations be performed during thedesign stage in order to properly size the components. Load calculations areuseful on other occasions as well, such as to evaluate the effectiveness inconserving energy of a proposed plant modification. Still another situationwhere a load calculation may be useful or necessary is in troubleshooting,when, for example, a refrigerated space cannot be maintained at the designtemperature or a production rate is lower than design.

Many designers use computer programs and/or checklists as aids inperforming load calculations so that all the major sources are included. Themost significant error in load calculations probably results from the completeomission of a contributor, in contrast to a 20 to 30% error in any onecalculation. Using a computer program for the load calculation will inherentlyincorporate all the items if the program has been written properly in the firstplace. The major contributors to refrigeration loads are: (1) the heattransmission through the roof, floor, and walls (2) infiltration through opendoorways (3) internal loads from lights, people, motors, and lift trucks (4)defrost heat and (5) the product load—cooling, freezing, and maintaining thetemperature of products. All of these categories will be addressed in thischapter.

REFRIGERATION LOAD CALCULATIONS

18.2. HEAT TRANSFER THROUGH SINGLE MATERIALS

When heat flows by conduction through a wall, as in Fig. 18.1, the equationthat expresses the rate of heat transfer is as follows:

(18.1)

where: q =rate of heat transfer, W (Btu/hr) k =conductivity of material, W/m·K (Btu/hr·ft· °F) A =cross-sectional area of wall, m (ft ) x =thickness of wall, m (ft)t and t =temperatures on opposite surfaces of wall, °C (F)

Equation 18.1 applies when the temperatures are constant and theconductivity is uniform throughout the wall. The equation indicates that therate of heat transfer is proportional to the conductivity, the cross-sectionalarea of the wall, and the temperature difference. The rate of heat transfer isinversely proportional to the thickness of the material. The thermalconductivities of some insulating materials used in low-temperatureapplications are shown in Table 18.1. Three columns of conductivities areshown—one in SI units and two in I-P units. Most heat-transfer books, whenpresenting conductivities in I-P units, use Btu/hr·ft·°F. In many catalogs,however, conductivity is expressed in Btu·in/hr·ft ·°F. The user ofconductivity values must express the thickness of the wall in feet when usingk in Btu/hr·ft·°F and express the thickness in inches when using k inBtu·in/hr·ft ·°F.

Table 18.1. Thermal conductivities of some insulating materials.

Figure 18.1. Heat transfer by conduction through a single wall.

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1 2

2

2

Conductivity

Table 18.2. Conductance and resistance of some materials.

18.3. CONDUCTANCE

The conductivity is one thermal characteristic of materials, but an alternatemethod of presenting thermal transfer data for materials is to use theconductance, C , which has units of W/m ·K or Btu/hr·ft ·°F. The conductancefits into the heat-transfer equation representing the process of Fig. 18.1 asfollows:

Material W/m·K Btu/hr·ft·°F Btu·in/hr·ft ·°F

Cellular foamglass 0.050 0.029 0.35

Cellular polyurethane 0.023 0.013 0.16

Expanded polystyrene 0.035 0.020 0.24

Extruded polystyrene 0.027 0.015 0.18

Glass fiber 0.036 0.021 0.25

Polyisocyanurate 0.020 0.012 0.14

2

Conductance Resistance

Material W/ m·K

Btu/hr·ft ·°F

m ·K/W

hr·ft ·°F/Btu

8-in Cinder concrete block 3.3 0.58 0.30 1.72

Hollow clay tile, 2 cells 3.75 0.66 0.27 1.51

Built-up roofing, ... 10mm(.394 in)

17.0 3.0 0.06 0.33

Asphalt roll roofing 36.9 6.50 0.027 0.154

2 2

2 2

2 2

(18.2)

Comparison of Eqs. 18.1 and 18.2 shows that C = k/x for slabs of uniformconstruction. Sometimes, it is more convenient to use the conductance inpreference to the conductivity. A concrete block with hollow cores, forexample, poses ambiguities as to what k values and x values to use, but canbe expressed by the conductance such that the heat-transfer rate can becalculated by Eq. 18.2. Table 18.2 shows examples of several materials thatare characterized better by conductance than by conductivity.

18.4. CONVECTION AND THE CONVECTION COEFFICIENT

The calculation of the rate of heat transfer through a single material, asillustrated in Fig. 18.1, is a building block in the procedure for calculating therate of heat transfer in walls and roofs of both refrigerated andnonrefrigerated structures. An additional concept that should be introducednow is that of convection. Equation 18.1 permits computation of the heat-transfer rate when the opposite surface temperatures are known. A moreprevalent situation occurs when the air temperatures on either side of thewall are known, which then requires that the heat-transfer process betweenthe air and surface be quantified. In Fig. 18.2, heat flows from the warm wallto the colder air, the surface temperature is t , and the air temperature is t. The rate of heat transfer is expressed by the following equation:

(18.3)

where: h =convection coefficient, W/m ·K (Btu/hr·ft ·°F)

Equation 18.3 can be thought of as the definition of the convection

Figure 18.2. The convection coefficient.

1

i

c 2 2

coefficient, which is designated by the symbol h . The letter h is also thesymbol in refrigeration for enthalpy, but usually the context will provide theclue to the meaning of h. As a further means of avoiding confusion in thisbook, the following conventions will be employed:

h with no subscript refers to an enthalpy.

hf or h refer to enthalpies of saturated liquid and vapor.

h refers to an enthalpy of air.

h , h , and h express convection coefficients, with the subscripts i ando indicating inside and outside, respectively.

The magnitude of the convection coefficient, hc, is primarily a function of theair velocity along the surface. On outside surfaces of walls and roofs, thewind controls the velocity. On inside surfaces, even where there is no forcedmotion of the air, there is movement of the air due to natural convection. Inthis case, the orientation of the surface (whether the surface is horizontal orvertical) has a significant influence. Some typical values of h are shown inTable 18.3.

Table 18.3. Convection coefficients at some surfaces.

18.5. THE RESISTANCE CONCEPT

The resistance concept transfers over from electrical terminology to heat

c

g

a

c i o

1 c

Convection coefficient

Orientation and condition W/m ·K Btu/hr·ft ·°F

6.7 m/s (15mph) winter, any orientation 34 6.0

3.4m/s (7.5 mph) summer, any orient. Still air 23 4.0

..vertical surface, heat flow horizontal 8.3 1.46

..horizontal surface, heat flow upward 9.3 1.63

..horizontal surface, heat flow downward 6.1 1.08

2 2

transfer and is sometimes helpful in visualizing heat-transfer processes aswell as to facilitate calculations. Ohms’s law in electricity is as follows:

(18.4)

In general terms, the current, I, is the flow rate and the driving force is thevoltage difference. The symbols applicable to electricity are shown in Fig.18.3. Figure 18.3a represents the terms in Eq. 18.4. When severalresistances are connected in series, as in Fig. 18.3b, the total resistance isthe sum of the individual resistances, so that:

The use of the resistance concept to express rates of heat transfer hasachieved wide acceptance. For the heat-transfer process of Fig. 18.1, theequation corresponding to Eq. 18.4 is as follows:

(18.5)

where: R in m ·K/W (hr·ft ·°F/Btu)=x/k

For the convective process in Fig. 18.2, the resistance is as follows:

The resistance and conductance are reciprocals of one another, and Table18.2 shows resistance values as well as the conductance values.

18.6. OVERALL HEAT-TRANSFER COEFFICIENTS FOR WALLS,

Figure 18.3. (a) The electrical analogy for a single resistance, and (b)the total resistance as the sum of individual resistances in series.

2 2

ROOFS, AND FLOORS.

The tools are now available to compute the rate of heat transfer through thewalls, roofs, and floors that are encountered in practice. Furthermore, theuseful form of the equation expresses the rate of heat transfer in terms ofthe air temperatures on opposite sides of the solid sections, namely:

(18.6)

where: U =overall heat-transfer coefficient, which is also called the U -value, W/m ·K (Btu/hr·ft ·°F)

Comparison with Eq. 18.5 shows that the U -value is the reciprocal of theresistance:

For a single material with convection coefficients on each side, as shown inFig. 18.4, the expressions for the rate of heat transfer and the U -value are asfollows, respectively:

where:

2 2

Figure 18.4. Heat transfer through a wall of single material withconvection coefficients on both sides.

The typical roof, floor, or wall consists of a combination of several materials,as illustrated in Fig. 18.5. The principal purpose of one of the materialsmight be to provide structural strength and that of another layer to providethe thermal insulation. There might even be another layer to provide a toughsurface that is resistant to abrasion.

In the building member of Fig. 18.5, the total resistance is as follows:

Example 18.1. The vertical wall shown in Fig. 18.5 is composed of 200-mm (8-in) cinder concrete block and 75-mm (3-in) polyisocyanurate insulation. In thesummer condition, when the exterior temperature is 35°C (95°F) and theinterior temperature is −18°C (0°F), what is the rate of heat transfer throughan area of 6 m (64.6 ft )?Solution. The following resistances apply:exterior convection (from Table 18.3) 1/23=0.0435 m ·K/W (0.25 hr·ft ·°F/Btu)cinder concrete block (from Table 18.2) 0.3m ·K/W (1.72hr·ft ·°F/Btu)polyisocyanurate (from Table 18.1) (0.075m)/0.020 =3.75m ·K/W (21.3hr·ft·°F/Btu)

interior convection (from Table 18.3) 1/8.3=0.1205 m ·K/W (0.684 hr·ft·°F/Btu)

The total resistance of the wall is: 0.044+0.300+3.750+0.12 =4.214 m ·K/W

Figure 18.5. Heat transfer through a composite wall.

2 2

2 2

2 2

2

2

2

2

2

2

(23.94 hr·ft ·°F/Btu)The rate of heat transfer through the entire wall is as follows: q=(6m)[35−(−18)]/4.214=75.5 W (257 Btu/hr)

18.7. WALL AND ROOF PANELS

Combining a structural material with insulation, as in Fig. 18.5, and applyingan abrasion-resistant coating to protect the insulation is a traditional methodof constructing walls for refrigerated structures. Perhaps even surpassingthe popularity of this method is now the use of wall and roof panels, asshown in Fig. 18.6. The panel is constructed as a sandwich of insulationfoamed in place and tightly bonded to metal skins. Polyurethane,isocyanurate, or other insulating materials are protected by steel, aluminum,or plastic sheets. The edges of the panels are provided with some form oftongue-and-groove arrangement which facilitates sealing the joints.

To provide additional structural rigidity, the metal skin is usually corrugatedin some fashion, which permits wall panels to be tilted in position duringconstruction and ceiling panels to be laid on joists. Several panel thicknessesare available, providing a variety of thermal resistances, as shown in Table18.4. Panel manufacturers normally provide a standard width of panel butproduce lengths according to the customer’s order. Thick panels arerequired for long spans, and these spans sometimes extend to lengths ofperhaps 9 m (30 ft).

18.8. DESIGN TEMPERATURES

After the U -value or the R -value is determined, another set of data needed to

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2

Figure 18.6. A wall or roof panel.

calculate the rate of heat transfer from Eq. 18.6 are the interior and exteriorair temperatures. The temperature inside the refrigerated space isdetermined by the activity within the space or the product being stored. Forthe refrigeration of food, the recommended storage temperatures for variousproducts have been addressed in Chap. 17. The exterior temperature is afunction of the weather in the particular locality, and since the maximumrefrigeration load will usually occur at high outdoor temperatures, thesummer design temperatures apply.

Table 18.4. Thermal resistances and recommended applications ofprefabricated wall and ceiling panels.

Table 18.5. Design summer dry- and wet-bulb temperatures for fiveU.S. cities.

Thicknessmm (inch)

Thermal resistance m ·K/W (hr·ft ·°F/Btu)

Recommendedtemperature application

50 (2) 2.92 (16.6) Ambient

75 (3) 3.66 (20.8) Down to 0°C (32°F)

100 (4) 5.87 (33.3) Down to −29°C (−20°F)

125 (5) 7.33 (41.6) Down to −45°C (−50°F)

150 (6) 8.81 (50.0) Down to −57°C (−70°F)

2

2

City Dry-bulbtemperatureequaled orexceeded 1% ofsummer hours

Wet-bulbtemperaturecoincident with1% dry-bulb

Wet-bulbtemperatureequaled orexceeded 1% ofsummer hours

Atlanta 34°C (94°F) 23°C (74°F) 26°C (78°F)

Chicago 34°C (94°F) 24°C (75°F) 26°C (78°F)

Dallas 39°C (102°F) 24°C (75°F) 26°C (78°F)

Los Angeles 34°C (94°F) 21°C (70°F) 22°C (72°F)

Table 18.5 extracts temperatures for a few U.S. cities from more-extensivetables in the ASHRAE Handbook —Fundamentals .

The design temperature that is chosen is not the highest temperature likelyto occur during the summer, because the peak temperatures occur for only afew hours at a time, and this peak is damped out because of the thermalcapacity of the wall and roof structure. The first column of data in Table 18.5indicates the outdoor dry-bulb temperature that is equaled or exceeded 1%of the hours during the summer.

The temperatures in Table 18.5 are used directly in design calculations forselecting condensers and computing the refrigeration load attributable toinfiltration. For computing the rate of heat transfer by conduction throughwalls, roofs, and floors, the outdoor design temperatures are often not useddirectly. In the first place, the temperature applicable to the floor calculationis the soil temperature and not the air temperature. Chapter 19 explains theneed for heating the soil beneath a low-temperature space, in order tomaintain the soil temperature at about 10°C (50°F). For most refrigeratedbuildings, the temperature on the bottom side of the floor will be in therange of 5 to 10°C (41 to 50°F).

New York 33°C (92°F) 23°C (74°F) 24°C (76°F)

1

Figure 18.7. Roof surface temperatures and rate of heat enteringthe space in (a) an air-conditioning application, and (b) arefrigeration application.

In the case of walls and roofs, however, the outdoor air is indeed in contactwith the outer surface of the building members, but another considerationarises, and that is the solar effect. The temperature of the outside surface ofa roof may rise to 50 to 55°C (120 to 130°F) during the day, but even thistemperature is not the outside temperature to be used in Eq. 18.6. Anotherfactor that enters into the calculation is the thermal lag in the heat transferthrough the wall or roof. Thus, if the peak temperature difference occurs atnoon, for example, the peak rate of flow into the building might not occuruntil four or five hours later. Designers performing load calculations for air-conditioning systems must cope with more complications than do designersof refrigerated structures for the reason illustrated in Fig. 18.7. Thedesigner calculating an air-conditioning load, in the situation reflected byFig. 18.7a, must perform the calculations at several different times in order toidentify the peak load. In the case of the large temperature differencesencountered in refrigeration loads (Fig. 18.7b), the percentage changethroughout the day is moderate. Many designers use a value of t in Eq.18.6 higher than the design air temperature indicated by Table 18.5, but theinfluence of the adjustment of t on the total calculated load is small.

Table 18.5 also shows design wet-bulb temperatures that are used incomputing the moisture load due to infiltration (Sec. 18.9) as well as in theselection of evaporative condensers or cooling towers. Column 3 shows thedesign wet-bulb temperature, but this value is not likely to occursimultaneously with the design dry-bulb temperature. Instead, column 2shows the mean value of the wet-bulb temperature that coincides with thedesign dry-bulb temperature. It is this coincident design wet-bulbtemperature that is normally used in conjunction with the design dry-bulbtemperature.

o

o

18.9. INFILTRATION THROUGH OPEN DOORWAYS

When an opening exists between a cold room and a warm exterior, there is anatural exchange of air between the two regions, as illustrated in Fig. 18.8.The air pressures of the cold and warm spaces equalize at the neutralpressure elevation. Above and below that elevation the air pressures differinside and outside the room due to the static force of the column of air, whichis a function of the air density. The air pressure increases at a greater ratewith respect to elevation inside the cold room, because the air density isgreater on the inside than on the outside. This means that the air pressureincreases at a more rapid rate inside the room below the neutral elevationand decreases at a more rapid rate above the neutral elevation in comparisonto pressure changes outside the room. These differences in air pressurecause the cold air to flow out from the cold room at the bottom and the warmair to flow into the room at the top of the opening. In the steady-statesituation, the mass rate of flow into the room equals that flowing out.

Because of this air exchange, there is a combination of sensible load (due tothe temperature differences of air) as well as a latent load, because theexternal air is likely to have a higher humidity ratio than that inside the coldroom. The water vapor transfer has a double effect in that it not onlyconstitutes a refrigeration load, but in the case of a room operating atsubfreezing temperatures, the moisture will ultimately deposit on the coils,requiring defrost with the attendant load and loss in efficiency.

In the practical operation of a refrigerated facility, doors must be opened attimes in order to move the product in and out. The infiltration load is one ofthe major loads in a refrigerated facility, varying from about one-fourth toone-third of the total refrigeration load, depending on such conditions as theinternal and external temperatures and humidities, the number and size ofdoors, and the amount of time the doors are open. There could be twodifferent approaches to addressing the existence of infiltration. An operatorof a plant may be satisfied to realize that infiltration constitutes a signficantload and try to reduce it by controlling door openings and/or installation ofinfiltration inhibitors (Sec. 18.10). The other approach is to quantify the load

Figure 18.8. Infiltration of air due to exchange of air between arefrigerated room and the warm environment.

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through calculations, which is certainly necessary for the designer of thefacility, as well as for the operator in certain cases in order to economicallyevaluate methods of controlling infiltration.

The most widely used formula for computing infiltration is probably that ofGosney and Olama . This equation is expressed as follows:

(18.7)

where: Q =volume rate of flow, m /s (ft /s) C =infiltration coefficient = 0.692 (1.254 ) A =area of doorway, m (ft ) H=height of doorway, m (ft)ρ and ρ =air densities, kg/m (lb/ft ) of the cold and warm air,respectively

The mass rate of flow, m, is the product of the volume rate of flow and themean density of air:

where: = mass flow rate, kg/s (lb/s)

Finally, the refrigeration load (sensible plus latent) is the mass rate of flowmultiplied by the difference in enthalpies:

where: q =refrigeration load, kW (Btu/s)

Example 18.2. The conditions in a refrigerated room are −18°C (-0.4 F) and 90% relative humidity, while the ambient conditions are26° C (79°F) and 70% relative humidity. Compute the rate of heattransfer due to infiltration for 1 m (ft ) of door area if the opendoor is 3 m (9.84 ft) high.

3

3 3

inf2 2

i o 3 3

2 2

Solution. Air properties may be obtained from the psychrometric charts:

ρ =1/(0.724 m /kg)=1.381 kg/m (0.0864 lb/ft )

ρ =1/(0.868 m /kg)=1.152 kg/m (0.0719 lb/ft )

h =−16.6 kJ/kg (0.7 Btu/lb)

h =63.8 kJ/kg (35.2 Btu/lb)

Substituting into Eq. 18.7 gives

Q=0.466 m /s through an area of 1 m or (1.529 ft /s through 1 ft )

The mass rate of flow is

=0.590 kg/s through 1 m (0.121 lb/s through 1 ft )

The refrigeration load is:

q =(0.590 kg/s)[63.8 -(-16.6) kJ/kg]

q =47.4 kW per m of door area (4.19 Btu/s per ft )

The experimental facilities used to verify Eq. 18.7 were small scale, so therehas been some concern about the applicability to commercial size doors andrefrigerated warehouses. To check the validity for large-scale facilities,ASHRAE sponsored a research project in which Hendrix acquired field data.The results showed that the Gosney-Olama equation, Eq. 18.7, indicatedslightly greater infiltration rates (by 10%) than measured, but thus gaveuseful conservative predictions. Hendrix also explored two questionssurrounding the transient behavior of the air flow through a doorway, andthese were: how long does it require for the full velocities to develop afterthe door is opened, and does the flow rate subside during long durations ofdoor openings? The field tests indicated that full velocities establishthemselves within 2 or 3 seconds, so there is no basis to reduce themagnitudes of Eq. 18.7. With respect to potential reduction in the infiltration

i 3 3 3

o 3 3 3

a,i

a,o

3 2 3 2

2 2

2 2

7

load after a long door opening, because of the air inside the room close to thedoor becoming warm, the 1989 ASHRAE study of large rooms showed no suchphenomenon. These tests thus support the use of Eq. 18.7 in its basic form.

To facilitate the determination of the infiltration load of an open door, Figs.18.9 and 18.10 present graphs of the infiltration load through open door-ways for various temperatures of the infiltration air and the refrigeratedspace at four different ambient relative humidities. The relative humidityinside the space applicable to all the curves is 90%, which would be fairlytypical of facilities storing both frozen and unfrozen foods. A deviation fromthe 90% relative humidity inside the space results in little error, because achange in percent relative humidity at low temperatures results in -only asmall change in humidity ratio. The curves in Figs. 18.9 and 18.10 areapplicable to a door height of 3 m

(9.84 ft). For other door heights, the refrigeration load should be multipliedby a correction factor of , as is apparent from Eq. 18.7. Forcomparison, Fig. 18.10 shows the same heat gain as calculated for theconditions applicable to Example 18.2.

A point to emphasize is that the refrigeration load contributed by an opendoor to the refrigerated space can be considerable. For example, supposethat a door is left standing open between a −20°C (−4°F) frozen food storagespace and the dock, which is at a temperature of 10°C (50°F) and a humidityof 70%. If the door area is 9 m (97 ft ), for example, the rate of heat gain is194 kW (55 tons of refrigeration). Since such a load can hardly be tolerated,the doors should be permitted to be open only when traffic is moving throughthem.

While Eq. 18.7 or the curves in Figs. 18.9 and 18.10 offer a reasonablyaccurate estimate of the refrigeration load due to infiltration, a much moreillusive number is the fraction of time that a door is likely to be open. Thisquantity is primarily a function of the type of operation, and the selection ofthe fraction must be based heavily on past experience. A quantification of thedoorway traffic can be derived from a standard industry term, turnover,which is the number of times the complete storage capacity is depleted andrestocked. The number of turnovers per year is dependent upon the functionof the space. For a refrigerated warehouse storing mostly seasonal products,the number may be 8 to 12, while for a grocery distribution warehouse it may

2 2

be 30. With the latest concepts in material handling, annual turnovers mayreach 100. Each turnover is associated with a lift truck moving in and out,causing an opening of the door of perhaps 10 to 15 seconds.

18.10. INHIBITING INFILTRATION LOADS

Because of the considerable magnitude of infiltration loads (perhaps 25 to30% of the total refrigeration load), means of reducing these loads should beconsidered. Some possibilities are: (1) air curtains, (2) vestibules, (3) plasticstrips, and (4) refrigerated loading docks.

The vertical type air curtain shown in Fig. 18.11 uses ambient air. The aircurtain forms a two-dimensional free-stream jet, into which outside air isinduced into the outside surface and cold air induced into the inside surface.When the jet reaches the floor, the induced air from the ambient spills back tothe ambient, and the air induced from the refrigerated space spills back tothat space. The air spilling back to the refrigerated space has mixed withwarm humid air, so the infiltration load is that represented by the mass rateof flow of inside spill air multiplied by the difference in enthalpy of this spillair and the air in the refrigerated space.

Hayes and Stoecker conducted a combined analytical and experimentalstudy of air curtains and found that from 65 to 80% of the infiltration loadcan be avoided by using a properly applied air curtain in comparison to anopen door. The tests verified the bending of the air curtain into the coldspace at the top and away from the space at the bottom. A failure of the aircurtain occurs when the jet velocity is so low that the curtain breaks and thejet fails to reach the floor. In that case, cold air pours out at the floor leveland is replaced by warm air flowing in at the top.

5 , 6

Figure 18.9. Refrigeration loads through open doorways for 30 and50 percent outdoor relative humidities. For door heights other than3 m (9.84 ft) multiply load by H m/3 (H ft/9.84).

Figure 18.10. Refrigeration loads through open doorways for 70 and90 percent outdoor relative humidities. For door heights other than3 m (9.84 ft), multiply load by H m/3 (H ft/9.84).

Figure 18.11. An air curtain resisting infiltration through an open

Some efforts in the Hayes/Stoecker tests to measure pressure differencesbetween the cold space and the ambient suggest a fundamental rule forapplying air curtains. Air curtains are very flexible therefore, they can easilydeflect to change the rates of spill (as shown in Fig. 18.11) into the space andinto the ambient. Thus, if a gust of wind strikes the outside surface of thecurtain, the curtain deflects to allow net flow of air into the space to build upthe pressure in the space, countering the overpressure on the outside. Theair curtain breathes when one air curtain protects a door of a space thatotherwise has few leaks. The installation of air curtains on opposite sides ofa refrigerated space is to be avoided, because both air curtains will deflect toallow an unacceptable rate of infiltration due to the wind blowing directlythrough the space.

Another concept in air curtains is the horizontal curtain with recirculated air, shown schematically in Fig. 18.12a. The air is directed from the supply side

on one side of the door, and most of the air is drawn into the return plenumon the opposite side of the door, as shown in Fig. 18.12b. This conceptappears to provide a stiffer curtain than is possible in the vertical,nonrecirculated arrangement, and the conditions of the air in the activeportion of the curtain are a blend of those of the ambient and therefrigerated space.

Vestibules are effective in reducing air infiltration but are expensive and alsoslow fork-lift traffic in and out of the space.

The third approach to reducing the infiltration load is to use plastic strips atthe doorways in one of the commercially available forms. The simplest form ofplastic strip curtains is as a single fixed-in-place curtain immediately inside oroutside the sliding door. Tests conducted in a refrigerated warehouseindicate that these strips are about 90% effective in reducing infiltration incomparison to an open door when there is no traffic through the doorway.When there is fork-lift truck traffic through the doorway, the effectiveness isapproximately 85%, depending on the intensity of the traffic. While fixed-in-place plastic strip curtains are effective in reducing infiltration, they canbecome cold and stiff and are subject to breakage. Also, they are not popularwith fork-lift truck operators. To overcome these drawbacks and still achievethe benefits of plastic strips, they are available on hinged and sliding doors

door.

7

8

as well. The light weight of these doors facilitates easy movement of them.

18.11. CALCULATING THE REFRIGERATION LOAD FOR DOCKS

The fourth concept in reducing infiltration is to provide a refrigeratedloading dock, as in Fig. 18.13, which is a design feature that is almost arequirement for a frozen food storage space. The first benefit, and perhapsthe most important, is the protection of the frozen product. Rather thanrequiring any residence time on a non-refrigerated loading dock, the producthas at least a cool ambient temperature during those situations when theproduct does not move directly between the truck and the storage space.The second benefit of refrigerated docks is in the reduction of infiltrationload. During periods of heavy traffic, the infiltration air entering from theoutdoors that is possibly warm and humid first enters the loading dock, andthat infiltration load is absorbed by the evaporator coils lo cated within thedock area. These coils can be served by refrigerant evaporating at anintermediate temperature level, which offers lower-cost refrigeration than inthe frozen food space. The infiltration to the low-temperature space is ofdock air, which is maintained in the 4–8°C (39–46°F) temperature range. Stillanother benefit of the refrigerated dock is that considerable moisture iscondensed on the dock coils as liquid. Thus, the moisture is prevented fromentering the low-temperature space, where it will deposit as frost on the coilsand increase the frequency of defrost.

Figure 18.12. (a) A horizontal, recirculated air curtain, and (b) theplan view.

Example 18.3. A frozen food storage space operates at −20°C (−4°F) and90% relative humidity. When the ambient conditions are 35° C (95°F) and 50%relative humidity, compute the rate of heat gain due to infiltration (a) if thereis no refrigerated dock, and (b) if a refrigerated dock at 10°C (50°F) and 70%relative humidity protects the refrigerated space, as in Fig. 18.14.Solution. (a) Figure 18.9 indicates a heat gain rate of 65 kW/m (1.7 tons/ft) for the unprotected space.

(b) When the space is protected by a dock at 10°C (50°F) and 70% relativehumidity, Fig. 18.10 shows that the rate of infiltration heat gain is only 21kW/m (0.55 tons/ft ), which is about 32 percent of the load with no dock.Furthermore, each kg (lb) of air from the dock holds 0.0053 kg (lb) of water incomparison 0.0178 kg (lb) of ambient air.

Example 18.3 provides further insight into the influence of refrigerateddocks. With the dock in Example 18.3b, the rate of heat gain to the space is21 kW/m (0.55 tons/ft ), which is refrigeration load for the low-temperaturespace but refrigerating capacity for the dock. Furthermore, using Fig. 18.10,the infiltration load from the ambient to the dock is 26 kW/m (0.68 tons/ft ),which at least superficially means that hardly any refrigeration load isimposed on the dock cooling coils. The comparison between the two rates isactually not appropriate, because the area and degree of opening of thedoors from the ambient to the dock are different than those between thedock and the refrigerated space. Also, the relative humidity of 70% was anassumption, and if it is lower then 70%, the net refrigeration load in the dockarea increases.

Figure 18.13. Refrigerated loading dock to reduce refrigeration loadin the low-temperature space.

2

2

2 2

2 2

2 2

Even though the magnitudes of infiltration load given above may be subjectto question, they do emphasize the fact that .we really want the dock coils tobe heavily loaded in order to remove water vapor from air before it reachesthe refrigerated space. For this reason, some plants that experience heavyfrosting conditions in the low-temperature space, even though equipped withrefrigerated docks, may apply heat in the dock space to force the dock coilsto operate more and thus perform dehumidification.

The choice of dock temperatures in the range of 2 to 6°C (35 to 43°F)conforms with tradition and represents a logical approach. One of theimportant reasons for a refrigerated dock serving a frozen food space is toreduce the refrigeration load, including the inflow of moisture to the low-temperature coils. If the dock temperature is too high, the infiltration load tothe low-temperature space will be excessive. If the dock temperature is toolow, some of the operating problems, such as the frosting of coils, will simplybe shifted to the dock from the low-temperature space. Designers of newfacilities should be alert to the potential innovations in the handling of frozenfood that are appearing on the horizon. One possible trend is toward lowerdock temperatures for protection of the product. One recently constructedfrozen food storage facility operates its dock at the same temperature asthe storage space, −23°C (−10°F), so that the product is never subjected totemperatures higher than desired storage temperatures. The firm isconvinced of the improvement in quality that results from this mode ofoperation. Several of the purposes relative to refrigeration loads of anintermediate-temperature loading dock are compromised, however. Also, thedock is less comfortable for the workers.

Another potential trend is toward larger-area docks. The storage time of

Figure 18.14. Heat flow rates in Example 18.3.

10

products in refrigerated warehouses continues to shrink, and the ultimateextrapolation of this trend appears to be the situation now beingexperienced by one warehouse , which meets the need of their customersto move product from one vehicle to another. To repond to this need, thewarehouse has constructed a large refrigerated trans-loading dock just forthis purpose.

18.12. LOAD FROM LIGHTS, MOTORS, PEOPLE, AND OTHERINTERNAL LOADS

Heat liberated within the refrigerated space adds to the refrigeration load. Ingeneral, all the internal releases of heat convert directly and instantaneouslyinto refrigeration loads. Some of the major contributors are lights, motors,people, fork-lift trucks, and processing equipment. The proportions of theseloads are functions of the use of the refrigerated space. The temperaturemaintained in a meat-processing room, for example, may be 10 to 14°C (50 to57°F). The lighting level would be high enough to illuminate detail work, andelectrical consumption for lights could be in the range of 16 W/m (1.5 W/ft). In a frozen food warehouse where the lights need only provide general

illumination, the average refrigeration load attributable to lights could be 5W/m (0.48 W/ft ). The concentration of people in the meat preparation roomwould be much higher than in the frozen food storage space.

In the meat-processing room, there will likely be some processing equipmentthat is electrically driven, such as slicers and grinders. Estimates of thecontribution of such equipment to the refrigeration load may be computed bydetermining the electrical rating of the equipment. Since the motors areprobably not being loaded to their full rating, a load factor of perhaps 0.6 to0.8 can be applied. Except for stacker crane installations, fork-lift trucksoperate in frozen food storage rooms. The power ratings of fork-lift trucksmay vary from 5 to 10 kW, but once again the fork-lift truck does not operateat full power constantly.

People give off heat, and in heavily occupied work rooms the magnitude maybe noticeable on the refrigeration plant. The rate of heat release from aperson depends on the activity level and also on the temperature of thespace. A comfort air-conditioning system may need to remove 100 W (350Btu/hr) combined sensible and latent heat for each person seated quietly buttwice that amount per person engaged in physical activity. In the case of

11

2

2

2 2

refrigerated spaces, the rate of heat release increases as the temperaturedrops, as shown in Table 18.6. Also, when people go into refrigerated spacesfor short durations of time, they carry in an appreciable amount of heat intheir warm clothing.

The electric motors used to drive the fans on the cooling coils are inoperation in virtually all refrigerated spaces. All of the electricity supplied tothese motors eventually converts to heat and must be removed by the coil.Figure 18.15 shows fan power requirements as a function of coil capacity fora family of coils. The two different curves represent high and low fan speeds,which reflect themselves in different air flow rates and face velocities for acoil. Figure 18.15 applies to coils operating with a 5°C (9°F) temperaturedifference between entering air and refrigerant, typical of a frozen foodwarehouse or a blast freezer coil.

Table 18.6. Heat equivalence of occupants 9.

Space temperature, °C (°F) Heat rate per person, W (Btu/hr)

10 (50) 210 (720)

0 (32) 270 (920)

−10 (14) 330 (1,130)

−20 (−4) 390 (1,330)

An important observation from Figure 18.15 is that, in the case of the low-velocity coil, approximately 8 to 15% of the refrigeration capacity is usedsimply to remove the heat caused by its own fan motor. For the high- velocitycoil, the percentage of fan and motor heat is likely to be over 20% of the coilcapacity. This contribution to the refrigeration load is critical, particularly inlow-temperature applications where the temperature difference between theentering air and refrigerant is low. In the case of higher-temperaturerefrigerated spaces where the air-to-refrigerant temperature difference maybe twice that chosen for Fig. 18.15, the percent of parasitic heat attributableto the fan and motor is half that shown.

A reasonable question is why choose the high face velocity coil if the fanpower is relatively high? The answer is that a coil with a given surface areawill transfer perhaps 30% more heat when operating at the high velocity for agiven air-to-refrigerant temperature difference. Furthermore, the throw ofthe high-velocity coil will be longer than that of the low-velocity coil. Whilethe first cost of the high-velocity coil will generally be less than that of thelow-velocity coil, the designer should evaluate the lifetime costs of theselection.

18.13. DEFROST HEAT

Defrosting with hot gas is the dominant method applied in industrialrefrigeration, and it cannot be denied that hot-gas defrost (also waterdefrost) adds heat to the refrigerated space and is thus a load. In addition,any refrigerant vapor that escapes through the pressure-regulating valve atthe outlet of the evapo rator during defrost increases the power requirementin the machine room to recompress this gas. Despite defrost constituting arefrigeration load, probably a minority of system designers incorporate it intheir load calculation. The calculations are cumbersome if done for all coils inthe facility, and the prediction is unreliable. The designers who do notinclude defrost load do not usually encounter trouble, because thecalculations of so many other components of the refrigeration load areconservative.

Figure 18.15. Fan power for a certain family of coils operating withtwo different face velocities. The temperature difference between theentering air and refrigerant is 5°C (9°F).

The efficiency of the hot-gas defrost process may be defined as follows:

The difference between the two thermal quantities in the efficiency equationare attributable to the energy used to raise the temperature of the coil andto the heat transferred to the refrigerated space during defrost. Somelaboratory tests, field tests, and analyses of the hot-gas defrostprocess suggest an efficiency approximately 20% and a hot-gas flow rate thatis between one and two times the refrigerant flow rate during refrigeration.(Plant operators try to ensure that at least two or three evaporators are intheir refrigerating mode while one evaporator is defrosting in order toprovide the vapor that will be needed for defrost.)

Example 18.4. To check the order of magnitude of the recommendedestimate of the defrost load, compute the average refrigeration loadcontributed by the defrost process if an ammonia coil with a nominalcapacity of 106 kW (30 tons) defrosts twice a day for 20 min each.Solution. If the refrigerant flow rate during refrigeration is 5.4 kg/min (12lb/min), assume that the flow rate during defrost is 8.1 kg/min (18 lb/min) andthat the latent heat of the ammonia is 1,240 kJ/kg (530 Btu/lb). The averagerate of heat gain distributed over the 24 hour period is as follows:

Refrigeration load=(8.1)(1,240)(40 min)/(24 hrs)=16,740 kJ/hr=4.65 kW(15,900 Btu/hr or 1.3 tons)The defrost load, in this case, is about 4% of the refrigeration load. Thedesigner must decide whether this amount is significant.

When a pressure-regulating valve controls the release of condensed liquidrefrigerant from the coil during hot-gas defrost, the escape of vapor throughthis valve adds to system load. This load can be reduced through the useof a liquid trap at the coil outlet and/or by discharging the refrigerant to anintermediate pressure in contrast to the low evaporating pressure.

18.14. PRODUCT LOAD

The refrigeration load attributable to reducing and maintaining thetemperature of products and/or freezing products may be the dominantcontributor to the total load. Such is likely to be the case in a food processing

12 , 13 , 14

15

and freezing plant where some loads, such as conduction through walls,roofs, and floors, and even infiltration might be insignificant. The calculationof the load due to the refrigerating and freezing of products, which was firstintroduced in Chapter 17, is in one sense straightforward but in anotherextremely difficult. It is simple to compute the quantity of heat that must beremoved from the product in a refrigeration or freezing process. It is alsoeasy to specify the desired time interval for the removal of this quantity ofheat. This almost effortless calculation has failed to consider a key part of theprocess, however, which is the heat transfer between the cooling medium(refrigerant, air, water, etc.) and the product. The calculation of therefrigeration rate for the product load must ensure that, for example, air at atemperature of −34°C (−30°F) can actually freeze the desired number ofboxes of meat in the blast freezer in the specified time interval.

The data on the product needed to make the load calculation are essentiallyspecific heats, latent heat of freezing if freezing is part of the process, andheat of respiration for produce stored at a temperature above freezing. Forcooling and immediate shipping of unfrozen products, only the informationon the specific heat above freezing is needed. To calculate the refrigerationload for a freezing operation, the specific heats above and below freezing areneeded as well as the latent heat of freezing of the product. When produce isstored in an unfrozen state, the heat of respiration must be considered. Suchproducts are living organisms, and as such they give off heat (data forvarious products were presented in Table 17.3). Some magnitudes ofproperties were given in Chapter 17 on the refrigeration and freezing offoods, but more extensive data are available in Reference 1 for most foodproducts. The refrigeration load encountered in freezing a product consistsof three components: (1) reducing the temperature to the freezing point, (2)freezing, and (3) reducing the temperature further to the storagetemperature. The refrigeration load in kJ (Btu) to convert a mass, m kg (lb),of warm product to a frozen food storage temperature is calculated by usingthe following equation:

The factors in the above equation are calculated as follows:

where:Q =heat removed above freezing, kJ (Btu) m=mass of product, kg (lb)Q =heat removed during freezing, kJ (Btu)Q =heat removed below freezing, kJ (Btu)C =specific heat above freezing, kJ/kg·K (Btu/lb·°F)C =specific heat below freezing, kJ/kg·K (Btu/lb·°F) L =latent heat of freezing, kJ/kg (Btu/lb)

Example 18.5. An IQF (individually-quick-frozen) freezer processing 0.8 kg/s(1.76 lb/s) of sweet corn kernels cools, freezes, and ultimately drops thetemperature to −18°C (0°F). The product enters the process at atemperature of 40°C (104°F). What is the refrigeration capacity required forthis IQF freezer?Solution. The specific heats of sweet corn are 3.53 kJ/kg·K (0.84 Btu/lb·°F)above freezing and 1.77 kJ/kg·K (0.42 Btu/lb·°F) below freezing, and the latentheat of freezing is 248 kJ/kg (106 Btu/lb). The freezing temperature is −0.6°C(31°F). The refrigeration rate is:q =(0.8)(3.53)[40−(−0.6)]+(0.8)(248)+(0.8)(1.77)[−0.6−(−18)]q =337 kW (96 tons of refrigeration)

The product load in refrigerated storage facilities is often small, but anunexpected load might arise from the fact that the incoming product is not atthe anticipated temperature. For example, product to be stored at 1°C (34°F)may come from a precooling facility also held at 1°C (34°F), so presumablythere is no product load expected. It is possible, however, that the productwas not held long enough at the initial site to bring the interior temperatureof the product down to the steady-state temperature. Heat will continue toemerge from the product in its new location causing a product load. Anotherreason that the temperature of the incoming product may be higher than theultimate storage temperature is that control of the temperature was lost intransit. Some public refrigerated warehouses routinely check thetemperature of the interior of sample pallets of incoming product, because ahigh temperature may be the source of quality problems that becomeapparent later.

Table 18.7. Order of magnitude of refrigeration load for several

above

freeze

below

above

below

1

applications.

18.15. LOAD CALCULATIONS IN PERSPECTIVE

Calculation of refrigeration loads is an inexact process, and because of thisfact, most designers simplify some of the procedures outlined in this chapter.In making their short cuts, designers usually add generous safety factorsresulting in extra refrigeration capacity for the plant. While this strategyresults in a larger and more expensive refrigeration plant than initiallyrequired, this extra capacity is usually put to good use later as the facilityexpands or an unforseen load develops due to new products or processes.

This evaluation of load calculations should not suggest that sloppycalculations are not without their peril. An observation is that the total plantcapacity provided by the load estimates of most competent designers isadequate, even generous. Where errors are more likely to occur is in thecalculation of an individual load, resulting in the inability to maintain thedesired temperature in a certain space or achieve the production rate of acertain freezing process. Experienced designers know that depending on thetype of service (long-term food storage, short-term food storage, freezingoperations, etc.), one of the components of the load will predominate. Forexample, the product load will be significant in a freezing operation,infiltration might dominate in short-term storage, while motors and lights aresignificant in processing areas.

Type of space Refrigeration load

kW per 1,000m tons per 1,000ft

Frozen food storage, single level 7.5 to 15 0.06 to 0.12

High-rise freezer storage 2.5 to 7.5 0.02 to 0.06

Produce storage 9.9 to 13.7 0.08 to 0.11

Shipping dock 13.7 to 25 0.11 to 0.2

Process area 20 to 60 0.16 to 0.5

3 3

Good designers intelligently use rules of thumb to check more-detailedcalculations. Table 18.7 shows orders of magnitude of loads used by onedesigner, and if the calculation differs widely from these rules of thumb,further investigation of the calculation is warranted.

REFERENCES

[1] American Society of Heating, Refrigerating and Air-ConditioningEngineers, Atlanta, Georgia, from the 1993 ASHRAE Handbook—Fundamentals .

[16.] Edmons, J., “Sizing and Application of Air Units for Low TemperatureStorage Freezers,” Seminar: A Rational Approach to Latent Heat in LowTemperature Rooms, Annual Meeting of ASHRAE, New York, New York,January 1987.

[3] Gosney, W.B. and H.A.L.Olama, “Heat and Enthalpy Gains Through ColdRoom Doorways,” Proceedings of the Institute of Refrigeration , December1975.

[16.] Hendrix, W.A., D.R.Henderson, and H.Z.Jackson, “Infiltration Heat GainsThrough Cold Storage Room Doorways,” ASHRAE Transactions , 95(Part 2):1155–1168,1989.

[5] Hayes F.C. and W.F.Stoecker, “Heat Transfer Characteristics of AirCurtains,” ASHRAE Transactions , 75(Part 2): 153–167, 1969.

[6] Hayes F.C. and W.F.Stoecker, “Design Data for Air Curtains,” ASHRAETransactions , 75(Part 2): 168–180, 1969.

[7] HCR, Inc., Lewistown, Montana, 1994.

[8] Downing C.C.and W.A.Meffert, “Effectiveness of Cold-Storage DoorInfiltration Protective Devices,” ASHRAE Transactions , 99(Part 2): 356–366,1993.

[9] American Society of Heating, Refrigerating and Air-ConditioningEngineers, Atlanta, Georgia, from the 1994 ASHRAE Handbook—Refrigeration.

[10] “Thrills and Chills,” Refrigerated and Frozen Foods , p. 30, February1994.

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[11] “Carving a Niche—State of Speciality Warehousing,” WarehouseManagement , p. 20, July/August 1994.

[12] Niederer, D.H., “Frosting and Defrosting Effects on Coil Heat Transfer,”ASHRAE Transactions , 82(Part 1): 467–473, 1976.

[13] Stoecker, W.F., J.J. Lux, and R.J.Kooy, “Energy Considerations in Hot-GasDefrosting of Industrial Refrigeration Coils,” ASHRAE Transactions , 89(Part2): 549–573, 1983.

[14] Cole, R.A., “Refrigeration Loads in a Freezer due to Hot Gas Defrostand their Associated Costs,” ASHRAE Transactions , 95(Part 2): 1149–1154,1989.

[15] Crombie, R.A., “Is Hot Gas Defrost Affordable?,” Meat Processing ,1973.

[16] Bonar, H.B. II, personal communication, 1994.

Citation

Wilbert F. Stoecker: Industrial Refrigeration Handbook. REFRIGERATION LOADCALCULATIONS, Chapter (McGraw-Hill Professional, 1998), AccessEngineering

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