INTRODUCTION

The previous chapter introduced you to some of thesources of heat that refrigeration equipment mustovercome. It also presented basic mathematical pro-cedures for determining the heat gain from eachsource, and showed you how to use these data toestimate the total heat load. This chapter broadensthe scope of information on the subject of heat loadcalculations. Let’s begin by reviewing some importantterms and concepts.

Heat gain through walls, floors, and ceilings of refrig-erated enclosures varies with the type of construc-tion, the temperature difference between therefrigerated space and the ambient air, the areaexposed to ambient temperatures, and the type andthickness of insulation. Letter symbols are used toidentify the various heat transfer factors. A workingknowledge of these symbols is very important ininterpreting heat load data.

THERMAL CONDUCTIVITY

The letter “K ” represents thermal conductivity, whichis the rate of heat transfer through a material as mea-sured in Btuh per square foot of area per degreeFahrenheit of temperature difference per inch ofthickness. The total heat transferred by conduction,therefore, varies directly with time, area, and temper-ature difference, and inversely with the thickness ofthe material through which it passes.

Different materials offer varying resistances to heat.Assume, for example, that you want to calculate theheat transfer through a material with a thermal con-ductivity (K-value) of 0.25. The material in questionhas an area of 2 ft2 and is 3 in. thick. If the averagetemperature difference across the material is 70°F,you will be able to calculate the total quantity of heat

transferred (Q ) over a 24-hr period with this simpleequation:

It should be apparent from this example that in orderto reduce heat transfer, the thermal conductivity mustbe as low as possible and the material as thick aspossible. Note: In some technical literature, K-valuesare based on thickness per foot instead of per inch.Unless otherwise specified, all references to K-valuesin this chapter are based on heat transfer per inch ofthickness.

CONDUCTANCE

The letter “C ” represents thermal conductance,which, like conductivity, is a measure of the rate ofheat transfer through a material or heat barrier. Con-ductance is used frequently with specific buildingmaterials, air spaces, etc., and differs from conduc-tivity in one significant way. Thermal conductance isa specific factor for a given thickness of material orstructural member. Thermal conductivity is a heattransfer factor per inch of thickness.

SURFACE FILM CONDUCTANCE

Heat transfer through any material is affected by theresistance to heat flow offered by its surface. Thedegree of resistance depends on the type of surface,its relative roughness or smoothness, its vertical orhorizontal position, its reflective properties, and therate of air flow over it. Surface film conductance isnormally denoted by the symbols “F i” (for inside sur-faces) and “Fo” (for outside surfaces). It is similar to

QK

=0.25 ( ) 2 (ft ) 24 (hr) 70 F ( T)

3 (in. of thickness)

=840

3= 280 Btu

2× × × ° ∆

1

Refrigeration ServiceEngineers Society1666 Rand RoadDes Plaines, Illinois 60016

ESTIMATING THE REFRIGERATION HEAT LOADPart 2

© 2004 by the Refrigeration Service Engineers Society, Des Plaines, ILSupplement to the Refrigeration Service Engineers Society.

630-145Section 10

This document, together with 630-144, replaces prior publication 630-11.

conductance in that its value is expressed in Btuh persquare foot of area per degree Fahrenheit of temper-ature difference. Use 1.65 as the F i factor for insidewall surfaces where the air is relatively “still” (that is,not exposed to outdoor conditions). Use 6.0 as the Fo factor for outdoor walls exposed to winds of up to15 mph. These values are sufficiently accurate formost calculations.

THERMAL RESISTANCE

The letter “R ” represents thermal resistance. By def-inition, the resistance of a material to the flow of heatis the reciprocal of its heat transfer coefficient. Inother words, the R-value is the reciprocal of eitherthe K-value or the C-value. For example, if a materialhas a conductance (C-value) of 0.25, then resistanceis:

You can see, then, that for this material it would takea 4°F temperature difference across a 1-ft2 area, or a1°F temperature difference across a 4-ft2 area, toresult in a heat flow of 1 Btuh. Thermal resistance isimportant primarily because individual resistance values can be added numerically:

Rtotal = R1 + R2 + R3 + …Rn

This makes the use of R-values very convenient incalculating overall heat transfer coefficients.

OVERALL COEFFICIENT OF HEAT TRANSFER

The letter “U ” represents the overall coefficient ofheat transfer, which is the overall rate of heat transferthrough a material. U-values are usually applied tocompound structures, such as walls, ceilings, androofs, and must include the additional insulatingeffect of the air film that exists on either side of thesurface.

The U-value of a wall is the reciprocal of its totalresistance. Resistance, in turn, is the reciprocal ofconductivity. But calculating the U-value is compli-cated by the fact that the total resistance to the flowof heat through a wall made of several layers is thesum of the resistances of the individual layers. In

order to calculate the overall heat transfer coefficient,you first must find the overall resistance to heat flow.The various heat transfer factors must be taken intoconsideration when you calculate the total resistance:

In the equation above, F i represents the heat trans-fer through the inside air film, and Fo represents theheat transfer through the outside air film. C1, C2 (and,if necessary, C3, C4, etc.), represent the C-values of individual materials (if the heat transfer through the material is expressed as a conductance). If conductivities apply rather than conductances, thenK1, K2, …Kn represent the K-values of the individualmaterials involved, and X1, X2, …Xn represent thethicknesses of the materials. Once you have used theequation above to calculate the sum of the thermalresistance values, you simply find the reciprocal ofthe total resistance in order to determine the U-valueof the wall. For example, assume that you want to cal-culate the U-value of a wall made of a material with aconductivity of 0.25. If the wall is 2 in. thick, with oneside exposed to the outside air, proceed as follows:

Now use the reciprocal of your answer to solve for theoverall coefficient of heat transmission:

As you can see, the U-value for the wall in this exam-ple is 0.114 Btuh per square foot of area per degreeFahrenheit of temperature difference. For typical heat

UR

=1

=1

8.773

= 0.114

total

RF F

XKtotal

i o

=1

+1

+

=1

1.65 +

1

6.0 +

2

0.25

= 0.606 + 0.167 + 8

= 8.773

RF F C C

X

K

X

Ktotal

i o 1 2

1

1

2

2

=1

+1

+1

+1

++

R =1

C=

1

0.25 = 4

2

transfer coefficients, see Table 1 at the end of Part 1of “Estimating the Heat Load.”

CALCULATING HEAT LOADS

For load calculation purposes, the heat transferredthrough a wall can be determined by using the basicheat transfer equation:

Q = U × A × TD

where: Q = the total heat transfer (Btuh)U = the overall heat transfer coefficientA = the area (ft2)

TD = the temperature differential (sometimesalso signified by the symbol ∆T) betweenthe outside design temperature and thedesign temperature of the refrigeratedspace (°F).

Consider the wall in the previous example (with aU-value of 0.114). The heat transmission per hourthrough a 90-ft2 area with an inside temperature of0°F and an outside temperature of 80°F is calculatedas follows:

Q = U × A × TD

= 0.114 × 90 × 80 = 821 Btuh

You can determine the amount of heat gained bytransmission into a refrigerated space in a similarmanner. Calculate the U-value for each part of thestructure around the refrigerated space, then proceedwith the equation shown above.

Extensive tests have been conducted to determineaccurate values for heat transfer through all commonbuilding and structural materials. Certain materialshave a high resistance to heat flow (a low thermalconductivity). They are used as insulation to reducethe level of heat transfer into a refrigerated space.Many good insulation materials have K-values of0.25 or less, and some foam insulations have thermalconductivities of 0.12 to 0.15.

OUTDOOR DESIGN TEMPERATURES

In refrigeration and air conditioning applications,maximum load conditions generally occur during the

warmest weather. But it is neither economical norpractical to design equipment for the highest possibletemperature that might occur in a given area. Afterall, the temperature may be at that record level foronly a few hours over a period of several years.National Weather Service records spanning manyyears have been used to establish suitable outdoordesign temperatures for specific geographical areas.The outdoor design temperature is a temperaturethat, according to past data, will not be exceededmore than a given percentage of the time during thesummer season in that area.Table 5 at the end of thischapter lists typical summer design temperatures.Actual temperatures should equal or exceed thoselisted only during 1% of the hours of the four-monthsummer season.

MAKING ALLOWANCES FOR SOLAR HEATRADIATION

Radiant heat from the sun contributes significantly toheat gain. The heat load is much greater if the wallsof a refrigerated space are exposed to the sun’s rays,either directly or indirectly. In the northern hemi-sphere, the sun’s ray’s fall on east, south, and westwalls. To allow for solar radiation in heat load calcula-tions, simply increase the temperature differential bythe applicable figure shown in Table 6 on the nextpage (or in a similar table). Note that the figures givenin Table 6 are for the summer season, and are to beused for refrigeration load calculations only. They arenot accurate for estimating air conditioning loads.

MAKING QUICK HEAT LOAD CALCULATIONS

There may be times when you need to make a non-critical heat load calculation, either for basic cost-estimating purposes or for some other reason.Table 7at the bottom of the next page will help you make aquick calculation of heat transmission through insu-lated walls. It lists the heat gain for various thick-nesses of common insulations. The values given inTable 7 are in Btu per degree Fahrenheit of tempera-ture difference per square foot of surface area per 24 hours. Note that the “thickness of insulation” refersto the insulation itself, not the thickness of the overallwall.

For example, assume that you are asked to find theheat transfer over a 24-hr period through a 6-ft × 8-ft

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wall insulated with 4 in. of fiberglass.The wall’s outsidesurface is exposed to an ambient temperature of95°F. The temperature inside the refrigerated spaceis 0°F. Refer to Table 7 and calculate as follows:

Q = heat transfer value × area × ∆T

= 1.5 × 48 ft2 × 95°F = 6,840 Btu

AIR INFILTRATION

Outside air enters a refrigerator orother refrigerated space whenevera door is opened. Reducing thetemperature of this outside air tothe level inside the enclosureincreases the refrigeration load. Ifthe moisture content of the enter-ing air is above that of the refriger-ated space, the excess moisturewill condense out of the air. Thelatent heat of condensation alsoadds to the overall heat load.

Because of the many variablesinvolved, it is difficult to calculatethe additional heat gain due to airinfiltration. Traffic in and out of arefrigerated area usually varieswith the area’s size or volume. Thenumber of doors is not important,but the number of times the doorsare opened must be considered.

Table 8 lists estimated average air changes per 24 hours for refrigerated areas of various sizes.These figures assume that the storage temperatureis above 32°F. “Average air changes” are due to dooropenings and infiltration. As noted previously, how-ever, it is difficult to estimate these figures accurately.You may have to modify the values as shown if youfind that the use of the area is very heavy or very light.

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East South West FlatType of surface wall wall wall roof

Dark surfaces

Slate roofingTar roofing 8 5 8 20Black paints

Medium surfaces

Unpainted woodBrickRed tile 6 4 6 15Dark cementRed, gray, green paint

Light surfaces

White stoneLight-colored cement 4 2 4 9White paint

Figures are to be added to the normal temperature differentials for heat leakage calculations inorder to compensate for sun effect. Not to be used for air conditioning design.

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Table 6. Allowance for sun effect, in °F

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Thickness of insulation, in.

Insulation 2 3 4 5 6 7 8 9 10 11 12

K-value approximately 0.17

Expanded polyurethane 2.04 1.36 1.02 0.815 0.68 0.58 0.51 0.45 0.41 0.37 0.34

K-value approximately 0.25

Glass fiber, cork,mineral wool fill and board, 3.0 2.0 1.5 1.2 1.0 0.86 0.75 0.67 0.60 0.55 0.50expanded polystyrene

Table 7. Quick-estimate values for heat transmission through insulated walls(Btu per 1°F temperature difference per ft2 of area per 24 hours)

Another method of calculating air infil-tration into a refrigerated space is todetermine the velocity of air flowthrough an open door. When the doorto a refrigerated space is opened, thedifference in density between the coldair inside and the warm air outside cre-ates a pressure differential. This pres-sure differential causes cold air to flowout at the bottom of the doorway, andwarm air to flow in at the top. Velocitiesvary from a maximum value at the topand bottom of the doorway to zero atthe vertical center of the opening. Theaverage velocity in either half of a 7-ftdoorway, for example, is 100 ft/minwhen the temperature difference is60°F. The velocity varies with thesquare root of the height of the door-way, and with the square root of thetemperature difference.

Here’s a sample calculation. Assumethat you wish to estimate the rate ofinfiltration through a door 8 ft high and4 ft wide. There is a 100°F temperaturedifference between the refrigeratedspace and the outside air. Use the 7-ft doorway andthe 60°F temperature difference as reference figures,and find the velocity as follows:

Now, using this value for velocity, you can calculatethe estimated rate of infiltration in cubic feet perminute (cfm):

Figure 1 on page 6 plots infiltration velocities for var-ious door heights and temperature differentials. For

refrigerated spaces in which ventilation is providedby supply and/or exhaust fans, the ventilation loadreplaces the infiltration load (if the ventilation load isgreater). You can calculate heat gain on the basis ofthe ventilating air volume.

Once you know the rate of infiltration, you can calcu-late the heat load from the heat gain per cubic foot ofinfiltration. Use data such as those found in Table 9on page 7. For conditions not covered by availabledata, you can find the infiltration heat load by deter-mining the difference in enthalpy between the enter-ing air and the air inside the refrigerated space. Youcan do this most easily by using a psychrometricchart.

PRODUCT LOADS

If the product to be stored in the refrigerated space isdelivered pre-cooled, the product load is usually neg-ligible, especially in small walk-in cooler installations.For large-volume applications, however, the productload may be substantial enough to require separate

138 ft/min 8 ft 4 ft

2= 2,208 cfm×

×

velocity = 100 ft/min 8

7

100

60

= 100 2.83

2.65

10

7.75

= 100 1.07 1.29

= 138 ft/min

× ×

× ×

× ×

5

Volume, Air changesft3 per 24 hours

200 44.0

300 34.5

400 29.5

500 26.0

600 23.0

800 20.0

1,000 17.5

1,500 14.0

2,000 12.0

3,000 9.5

4,000 8.2

5,000 7.2

Volume, Air changesft3 per 24 hours

6,000 6.5

8,000 5.5

10,000 4.9

15,000 3.9

20,000 3.5

25,000 3.0

30,000 2.7

40,000 2.3

50,000 2.0

75,000 1.6

100,000 1.4

Table 8. Average air changes per 24 hours due to door openings and infiltration, for storage rooms above 32°F

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Note: For heavy usage, multiply the above values by 2. For long periods of storage, multiply the above values by 0.6.

calculations for each product. This requires knowingthe freezing point, specific heat, percent water, etc. ofeach product.

The product load, whether it is heavy or light, isdefined as any heat gain resulting from the productsthat are being stored in the refrigerated space. Theheat given off depends on the specific product and itsstorage temperature.

Product heat gain consists of some or all of:

ª the heat released from products that were initiallystored and/or transported at a higher temperaturethan that maintained in the refrigerated space

ª the heat removed in the process of freezing orchilling the products

ª the heat of respiration from chemical reactionsthat take place in some products.

Note that the values resulting from the last item varywith the storage temperature. Refer to Table 2 at theend of the previous chapter for storage requirementsand properties of a wide variety of products.

CALCULATING SENSIBLE HEAT

Above-freezing temperatures

Most products are at a higher temperature than thestorage temperature when they are first placed in therefrigerated space. Many food products have a highpercentage of water content. Their reaction to a lossof heat is quite different above the freezing point thanit is below the freezing point. Above the freezing

6

Figure 1. Infiltration air velocity through open doors

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OR

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RAT

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point, the water continues to exist as liquid. Below thefreezing point, it changes to ice.

The specific heat of a product is defined as the amountof heat (Btu) required to raise the temperature of 1 lb

of the substance 1°F. Look back at Table 2 at the endof the previous chapter, which shows the specificheats of various food products. Note that the specificheat of a product differs depending on whether thetemperature is above freezing or below freezing. The

7

Temperature of outside air, °F

40 50 90 100

Storage room Relative humidity, %

temperature, °F 70 80 70 80 50 60 50 60

30 0.24 0.29 0.58 0.66 2.26 2.53 2.95 3.35

25 0.41 0.45 0.75 0.83 2.44 2.71 3.14 3.54

20 0.56 0.61 0.91 0.99 2.62 2.90 3.33 3.73

15 0.71 0.75 1.06 1.14 2.80 3.07 3.51 3.92

10 0.85 0.89 1.19 1.27 2.93 3.20 3.64 4.04

5 0.98 1.03 1.34 1.42 3.12 3.40 3.84 4.27

0 1.12 1.17 1.48 1.56 3.28 3.56 4.01 4.43

–5 1.23 1.28 1.59 1.67 3.41 3.69 4.15 4.57

–10 1.35 1.41 1.73 1.81 3.56 3.85 4.31 4.74

–15 1.50 1.53 1.85 1.92 3.67 3.96 4.42 4.86

–20 1.63 1.68 2.01 2.09 3.88 4.18 4.66 5.10

–25 1.77 1.80 2.12 2.21 4.00 4.30 4.78 5.21

–30 1.90 1.95 2.29 2.38 4.21 4.51 4.90 5.44

Temperature of outside air, °F

85 90 95 100

Storage room Relative humidity, %

temperature, °F 50 60 50 60 50 60 50 60

65 0.65 0.85 0.93 1.17 1.24 1.54 1.58 1.95

60 0.85 1.03 1.13 1.37 1.44 1.74 1.78 2.15

55 1.12 1.34 1.41 1.66 1.72 2.01 2.06 2.44

50 1.32 1.54 1.62 1.87 1.93 2.22 2.28 2.65

45 1.50 1.73 1.80 2.06 2.12 2.42 2.47 2.85

40 1.69 1.92 2.00 2.26 2.31 2.62 2.67 3.06

35 1.86 2.09 2.17 2.43 2.49 2.79 2.85 3.24

30 2.00 2.24 2.26 2.53 2.64 2.94 2.95 3.35

Table 9. Heat removed in cooling air to storage room conditions (Btu/ft3)

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freezing point, which in most cases is below 32°F,also varies among products.

The heat to be removed from a stored product tolower its temperature to a point above freezing canbe calculated as follows:

Q = W × c × (T1 – T2)

where: Q = the quantity of heat (Btu) to be removedW = the weight of the product (lb)c = the specific heat of the product above

freezingT1 = the initial temperature (°F)T2 = the final temperature, at or above the

freezing point (°F).

For example, assume that you want to calculate theheat that must be removed in order to cool 1,000 lbof veal from 42°F to 29°F. The highest freezing pointof veal is 29°F. The specific heat, taken from Table 2at the end of the previous chapter, is 0.76 Btu/lb/°F.Proceed as follows:

Q = 1,000 lb × 0.76 × (42°F – 29°F)

= 1,000 × 0.76 × 13 = 9,880 Btu

Below-freezing temperatures

Once the water content of a product has been frozen,sensible cooling occurs again, just as it did when theproduct was above freezing. But now the ice in theproduct causes the specific heat to change. Forexample, the specific heat of veal above freezing is0.76. Its specific heat below freezing is 0.41.

The heat to be removed from a stored product tolower its temperature to a point below freezing canbe calculated as follows:

Q = W × c i × (Tf – T3)

where: Q = the quantity of heat (Btu) to be removedW = the weight of the product (lb)c i = the specific heat of the product below

freezingTf = the freezing temperature (°F)T3 = the final temperature, below the freezing

point (°F).

For example, to calculate the heat that must beremoved from the same 1,000 lb of veal in order tocool it from its freezing point (29°F) to 0°F, you wouldproceed as follows:

Q = 1,000 lb × 0.41 × (29°F – 0°F)

= 1,000 × 0.41 × 29 = 11,890 Btu

CALCULATING THE LATENT HEAT OF FREEZING

Most refrigerated food products contain a high per-centage of water. When calculating the amount ofheat that must be removed in order to freeze a prod-uct, you need to know its water content.Table 2 at theend of Part 1 also lists the water content percentagefor various food products.

The latent heat of freezing (also called the latent heatof fusion) is defined as the amount of heat (Btu) thatmust be removed in order to change 1 lb of a liquidto 1 lb of solid at the same temperature. You can finda product’s latent heat of freezing by multiplying thelatent heat of water (144 Btu/lb) times the percentageof water in the product. Table 2 at the end of Part 1shows that veal can have a water content of 70%. Italso shows the latent heat of fusion for veal to be 100Btu/lb, but you could have arrived at approximatelythe same figure by calculating as follows:

0.70 × 144 Btu/lb = 100 Btu/lb

To calculate the amount of heat that must beremoved from a product in order to freeze it, then,simply proceed as follows:

Q = W × h f

where: Q = the quantity of heat (Btu) to be removedW = the weight of the product (lb)h f = the product’s latent heat of fusion

(Btu/lb).

Therefore, the latent heat of freezing 1,000 lb of vealat 29°F is:

Q = W × h f

= 1,000 lb × 100 Btu/lb = 100,000 Btu

8

CALCULATING THE TOTAL PRODUCT LOAD

The total product load is the sum of the individual cal-culations just completed. It includes the sensible heatabove freezing, the latent heat of freezing, and thesensible heat below freezing. Using the previous cal-culations as an example, if 1,000 lb of veal is to becooled from 42°F to 0°F, the total product load wouldbe:

Sensible heat above freezing 9,880 BtuLatent heat of freezing 100,000 Btu

Sensible heat below freezing + 11,890 BtuTotal product load = 121,770 Btu

SUPPLEMENTARY LOADS

In addition to product load, air infiltration, and theheat transmitted through walls, the total cooling loadestimate must include any heat gain from othersources. Among these supplementary sources iselectrical energy directly dissipated in the refriger-ated space. This energy may come from lights,motors, heaters, or other appliances. It is convertedinto heat, which becomes part of the load on therefrigeration system.

Calculating and applying this heat gain value to acooling load estimate is not difficult. One equation isacceptably accurate for converting any amount ofelectrical energy:

1 watt-hour = 3.41 Btu

This conversion factor was covered in Part 1, but isworth repeating here.

An additional source of supplementary heat load ishuman occupancy. This is another considerationworth repeating, since it is often overlooked. Varioustables are available that estimate heat gain accordingto the number of occupants and the degree of activ-ity (sitting, standing, engaged in light work, heavywork, etc.).

The total supplementary heat load is the sum of allthe individual sources that contribute to it. Consider,for example, a refrigerated storeroom in which thetemperature is maintained at 0°F. The space isequipped with 300 W of electric lighting and a 3-hp

motor driving a fan. Two people work continuously inthe space. First consult Tables 3 and 4 in Part 1 toobtain the necessary values. Then you can calculatethe total supplementary load as follows:

300 W × 3.41 Btuh 1,023 Btuh3-hp motor × 2,950 Btuh 8,850 Btuh

2 people × 1,300 Btuh + 2,600 BtuhTotal supplementary load = 12,473 Btuh

HOURLY LOAD CALCULATIONS

Refrigeration equipment is designed to function con-tinuously. Normally, the compressor operating time isdetermined by the defrost requirements. Load is cal-culated on a 24-hr basis. To find the required hourlycompressor capacity, therefore, you divide the 24-hrload by the desired number of hours of compressoroperation during the 24-hr period. It is customary toadd a reasonable “safety factor,” which enables thesystem to recover rapidly after a temperature rise,and to handle any load that might exceed the originalestimate.

If frost does not accumulate on the evaporator, nodefrost period is necessary. Compressors for suchapplications should be selected for 18- to 20-hr peri-ods of operation. When refrigerant temperatures arelow enough to cause frosting, it is common practice todefrost by stopping the compressor and allowing thecirculating air to melt any frost or ice from the evapo-rator coil. Compressors for such applications shouldbe selected for 16- to 18-hr periods of operation.

Low-temperature applications require some integralmeans of defrosting. With normal defrost periods, 18-hr compressor operation is usually acceptable.However, many such systems are designed for con-tinuous operation except during the defrost period.

It is tempting to allow an additional 5 to 10% safetyfactor to ensure that the equipment will not be under-sized. This practice may be desirable if there areuncertainties about the potential load calculation. Ingeneral, however, the fact that the compressor issized for 16- or 18-hr operation provides an adequatesafety factor.

You should always calculate the load on the basis ofpeak demand at design conditions. Normally, design

9

conditions are based on the premise that peakdemand will occur for no more than 1% of the hoursduring the summer months. If the overall load calcu-lations are accurate, and if the equipment is sizedproperly, an additional safety factor may result in theequipment being oversized.This may cause operatingdifficulties under light load conditions.

ESTIMATING A SAMPLE LOAD

The most accurate way to estimate a refrigerationload is to consider each factor separately. The follow-ing example presents a typical selection problem.(The load in this case has been chosen to demon-strate the types of calculations required. It does notrepresent a normal load.) The following conditionsapply:

ª The refrigerated space is a walk-in cooler con-structed of 6-in. expanded extruded polystyrene.The cooler is in a location where it receives nosunlight, either direct or indirect.

ª The cooler’s outside dimensions are 8 ft (height)× 10 ft (width) × 40 ft (length). Its floor area (out-side dimensions) is 400 ft2. The floor is on a slabfoundation in contact with the ground. Thecooler’s inside volume is 3,200 ft3.

ª The ambient temperature is 100°F. The relativehumidity is 50%.The ground temperature is 55°F.

ª The temperature inside the refrigerated space is40°F.

ª There is a 1⁄2-hp fan motor running continuouslyinside the cooler.There are also two 100-W lightsin use 12 hours per day.

ª Two men occupy the cooler for 2 hours per day.Door usage is qualified as “heavy.”

ª The products currently in storage are 500 lb oflettuce and 1,000 lb of green beans.

ª The products to be brought into the cooler are500 lb of bacon at 50°F and 15,000 lb of beer at80°F. The temperature of the entering productsmust be reduced to the storage temperature in24 hours.

Use the sample estimate form on the next page tocalculate the total heat load for the example givenabove.You should find all of the data you need in thischapter or the previous chapter. Answers may differsomewhat, depending on certain variables. See thecompleted form on page 16 for one solution.

10

11

REFRIGERATION LOAD ESTIMATE FORM

Application Room: Height Width

Room conditions °F RH Length Volume

Ambient conditions °F RH Insulation: Thickness

Temperature difference (TD) °F Type

Product load

Lights, motors

Occupancy, miscellaneous

Heat transmission load

Side walls L × H × 2 = area × TD × U-factor =

End walls W × H × 2 = area × TD × U-factor =

Ceiling L × W = area × TD × U-factor =

Floor L × W = area × TD × U-factor =

Glass area × TD × U-factor =

Subtotal =

× 24 hours =

Air infiltration

ft3 volume × air changes per 24 hours × usage factor × Btu/ft3 =

Product load

Product temperature reduction

lb × specific heat × TD =

lb × specific heat × TD =

lb × specific heat × TD =

Latent heat of freezing

lb × specific heat × TD =

lb × specific heat × TD =

Heat of respiration

lb × Btu/24-hr heat of respiration =

lb × Btu/24-hr heat of respiration =

Supplementary load

watts × hours × 3.41 Btuh =

hp × hours × Btuh =

people × hours × Btuh =

Subtotal

% safety factor

Total 24-hr refrigeration load

Required hourly compressor capacity based on 16-hr operation

12

Dry bulb, Wet bulb,Location °F °F

ALABAMABirmingham 94 75Mobile 94 77Montgomery 95 76Tuscaloosa 95 77

ALASKAAnchorage 71 59Fairbanks 81 61Juneau 74 60Nome 69 57

ARIZONAFlagstaff 85 56Phoenix 110 70Tucson 104 65

ARKANSASFayetteville 95 75Fort Smith 99 76Little Rock 97 77

CALIFORNIABakersfield 104 70Fresno 103 71Long Beach 92 67Los Angeles 85 64Sacramento 100 69San Diego 85 67San Francisco 83 63San Jose 93 67Santa Barbara 83 64

COLORADOColorado Springs 90 58Denver 93 60Pueblo 97 62

CONNECTICUTBridgeport 86 73Hartford 91 73

DELAWAREDover 93 76Wilmington 91 75

D.C.Washington 95 76

FLORIDACape Canaveral 92 78Daytona Beach 92 77Fort Lauderdale 92 78

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AE

Dry bulb, Wet bulb,Location °F °F

FLORIDA (continued)Fort Myers 94 77Jacksonville 94 77Key West 90 79Miami 91 77Pensacola 93 78Saint Petersburg 94 80Tampa 92 77West Palm Beach 91 78

GEORGIAAthens 94 75Atlanta 93 75Fort Benning 97 76Macon 96 76Marietta 94 74Savannah 95 77

HAWAIIHonolulu 89 73

IDAHOBoise 96 63

ILLINOISChicago 91 74Decatur 94 76Moline 93 76Peoria 92 76Rockford 91 74Springfield 93 76

INDIANAEvansville 94 77Fort Wayne 90 74Indianapolis 91 75Lafayette 93 75South Bend 90 73Terre Haute 93 76

IOWACedar Rapids 93 75Des Moines 93 76Fort Dodge 92 75Mason City 91 74Sioux City 94 75Waterloo 91 75

KANSASDodge City 100 70Topeka 96 75Wichita 100 73

Table 5. Summer outdoor design data

13

Dry bulb, Wet bulb,Location °F °F

KENTUCKYBowling Green 94 76Fort Knox 94 76Lexington 91 74Louisville 93 76

LOUISIANABaton Rouge 94 78Lafayette 94 78Lake Charles 93 78New Orleans 93 79Shreveport 97 77

MAINEAugusta 87 71Bangor 87 71Brunswick 87 71Portland 86 71

MARYLANDBaltimore 93 75

MASSACHUSETTSBoston 91 73Worcester 85 71

MICHIGANDetroit 90 73Flint 88 73Grand Rapids 89 73Lansing 89 73Marquette 86 69Saginaw 90 74Traverse City 89 71

MINNESOTABrainerd 88 70Duluth 84 69Hibbing 85 70International Falls 86 69Minneapolis-Saint Paul 91 73Rochester 88 72Saint Cloud 91 72

MISSISSIPPIBiloxi 92 79Jackson 95 77Tupelo 96 76

MISSOURIColumbia 95 75Kansas City 96 75Saint Louis 95 76

AS

HR

AE

Dry bulb, Wet bulb,Location °F °F

MONTANABillings 93 63Bozeman 91 61Butte 86 57Helena 90 60Missoula 91 62

NEBRASKALincoln 97 74Omaha 95 75

NEVADALas Vegas 108 66Reno 95 61

NEW HAMPSHIREConcord 90 71Lebanon 88 71Mount Washington 60 56

NEW JERSEYAtlantic City 91 74Newark 93 74Trenton 93 75

NEW MEXICOAlbuquerque 96 60Carlsbad 101 65Roswell 98 65Santa Fe 90 65

NEW YORKAlbany 90 71Buffalo 86 70New York 91 74Niagara Falls 87 72Poughkeepsie 92 75Rochester 89 73Syracuse 88 72White Plains 89 74

NORTH CAROLINAAsheville 88 72Charlotte 94 74Fort Bragg 96 77Greensboro 92 75Raleigh/Durham 93 76Wilmington 93 79Winston-Salem 92 74

NORTH DAKOTABismarck 93 68Fargo 91 71

Table 5. Summer outdoor design data (continued)

14

Dry bulb, Wet bulb,Location °F °F

OHIOAkron 88 72Cincinnati 93 74Cleveland 89 73Columbus 90 74Dayton 90 74Toledo 90 73Youngstown 88 72

OKLAHOMAOklahoma City 98 74Tulsa 100 76

OREGONEugene 91 67Pendleton 97 64Portland 90 67Salem 92 67

PENNSYLVANIAAllentown 90 73Altoona 89 72Erie 85 72Harrisburg 92 74Philadelphia 92 75Pittsburgh 89 72Wilkes-Barre/Scranton 88 71

RHODE ISLANDProvidence 89 73

SOUTH CAROLINACharleston 94 78Greenville 93 74Myrtle Beach 93 79

SOUTH DAKOTAPierre 99 70Rapid City 95 65Sioux Falls 94 73

TENNESSEEChattanooga 94 75Knoxville 92 74Memphis 96 78Nashville 94 76

TEXASAbilene 99 71Amarillo 96 67Austin 98 74Beaumont/Port Arthur 94 79Brownsville 95 78

AS

HR

AE

Dry bulb, Wet bulb,Location °F °F

TEXAS (continued)Corpus Christi 95 78Dallas 100 74El Paso 101 64Fort Worth 100 75Galveston 91 82Houston 96 77Laredo 102 73Lubbock 97 67San Antonio 98 73Waco 101 75Wichita Falls 103 74

UTAHOgden 93 61Salt Lake City 96 62

VERMONTBurlington 87 71Montpelier 85 70

VIRGINIAHampton 94 78Lynchburg 93 74Newport News 95 78Norfolk 93 77Quantico 94 77Richmond 94 76Roanoke 92 73

WASHINGTONOlympia 87 67Seattle 85 65Spokane 92 62Takoma 86 65Walla Walla 98 66Yakima 95 65

WEST VIRGINIACharleston 91 73

WISCONSINGreen Bay 88 73La Crosse 91 74Madison 90 73Milwaukee 89 74Wausau 88 71

WYOMINGCasper 92 59Cheyenne 87 58Cody 91 59Rock Springs 86 54

Table 5. Summer outdoor design data (continued)

15

Dry bulb, Wet bulb,Location °F °F

CANADA

ALBERTACalgary 83 60Edmonton 82 63Fort McMurray 84 64Grande Prairie 81 62Medicine Hat 90 63Red Deer 82 62Vermilion 83 64

BRITISH COLUMBIAAbbotsford 85 67Fort Nelson 82 62Port Hardy 68 59Prince George 81 60Vancouver 76 65Victoria 79 63Williams Lake 83 59

MANITOBABrandon 87 67Churchill 77 62Winnipeg 87 68

NEW BRUNSWICKChatham 86 69Moncton 83 68Saint John 78 65

NEWFOUNDLANDBattle Harbour 65 58Cartwright 75 62Deer Lake 81 66Gander 79 65Saint John’s 76 65Stephenville 74 64Wabush Lake 76 60

NORTHWESTTERRITORIES

Cape Parry 58 53Fort Smith 82 63Norman Wells 80 62Yellowknife 77 60

NUNAVUTBaker Lake 69 57Cambridge Bay 60 53Chesterfield 66 54Coral Harbour 64 53Hall Beach 56 50

AS

HR

AE

Dry bulb, Wet bulb,Location °F °F

NOVA SCOTIAGreenwood 84 69Halifax 80 68Sable Island 70 67Shearwater 78 66Sydney 81 68Truro 79 69Yarmouth 73 66

ONTARIOArmstrong 81 66Earlton 85 69London 85 71North Bay 81 67Ottawa 86 70Sault Sainte Marie 83 69Thunder Bay 84 68Toronto 87 71Trenton 84 71Windsor 89 73

PRINCE EDWARDISLAND

Charlottetown 79 69

QUEBECBagotville 84 67Montreal 85 71Quebec 84 70Riviere du Loup 79 68Roberval 83 68Sherbrooke 84 70Val d’Or 83 67

SASKATCHEWANMoose Jaw 90 64North Battleford 86 64Prince Albert 84 65Regina 89 64Saskatoon 87 64Yorkton 86 65

YUKONBurwash 73 57Whitehorse 77 57

Table 5. Summer outdoor design data (continued)

16

REFRIGERATION LOAD ESTIMATE FORM

Application Room: Height Width

Room conditions °F RH Length Volume

Ambient conditions °F RH Insulation: Thickness

Temperature difference (TD) °F Type

Product load

Lights, motors

Occupancy, miscellaneous

Heat transmission load

Side walls L × H × 2 = area × TD × U-factor =

End walls W × H × 2 = area × TD × U-factor =

Ceiling L × W = area × TD × U-factor =

Floor L × W = area × TD × U-factor =

Glass area × TD × U-factor =

Subtotal =

× 24 hours =

Air infiltration

ft3 volume × air changes per 24 hours × usage factor × Btu/ft3 =

Product load

Product temperature reduction

lb × specific heat × TD =

lb × specific heat × TD =

lb × specific heat × TD =

Latent heat of freezing

lb × specific heat × TD =

lb × specific heat × TD =

Heat of respiration

lb × Btu/24-hr heat of respiration =

lb × Btu/24-hr heat of respiration =

Supplementary load

watts × hours × 3.41 Btuh =

hp × hours × Btuh =

people × hours × Btuh =

Subtotal

% safety factor

Total 24-hr refrigeration load

Required hourly compressor capacity based on 16-hr operation