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Refrigeration Load Sizing For Walk-In Coolers, Freezers, & Other Boxes

In this Info-Tec, we will show how to calculate the heat load for any walk-in box.  First, we’ll cover the “long” way.  The long way allows you to calculate the load for any size box, in any location, with any product load.  All the data asked for in Figure 1 must be gathered and be accurate.

Figure 1.

This discussion will then take up the “short” way, or quick selection way, for small walk-in boxes.  Small will be defined as coolers up to 2,400 Cu. Ft. in size, freezers up to 1,500 Cu. Ft. in size, and no unusual loads.  These small boxes comprise over 90% of the market.  Shortcut estimating methods can be used for these small boxes.

The heat gain through walls, floors, and ceilings, will vary with the type of construction, the area exposed to a different temperature, the type of insulation, the thickness of insulation, and the temperature difference between the refrigerated space and the ambient air.

The basic formula for heat transfer through some heat transfer barrier is:  Q = U x A x TD

 Q          =          Heat transfer, BTU/Hr

U          =          Overall heat transfer coefficient BTU/(hour)(Sq.Ft.)(°F TD)

 A          =          Area in square feet 

  TD        =          Temperature differential between sides of thermal barrier; for example, between outside design                                     

                                     temperature and the refrigerated space temperature.

Q is the rate of heat flows through a medium.  It can be determined after finding the thermal resistances of the materials through which the heat flows.

Certain letter symbols are used to denote heat transfer factors.  “K” stands for Thermal Conductivity.  It is the rate of heat transfer that occurs through one inch of a material.  Different materials offer different resistances to the flow of heat.  K is expressed in units of BTU/(Hr.)(Sq.Ft. of area)(°F TD).

For example, the heat transfer in 24 hours through 2 Sq. Ft. of material, 3" thick, having a thermal conductivity factor of .25, with an average temperature difference across the material of 70°F would be calculated as follows:

                             Q = .25(k) x 2 Sq.Ft. x 24 hours x 70° TD = 280 BTU

                                                     3" thickness

Since the total heat transferred by conduction varies directly with time, area, and temperature difference, and varies inversely with the thickness of the material, it is readily apparent that in order to reduce heat transfer, the K factor should be as small as possible, and the material as thick as possible.

Thermal Resistivity, or “r” is the reciprocal of K, or 1/K.  Resistance values can now be added numerically.  R total = r1 + r2 +r3, where r1, r2, and r3 are individual resistances.  Individual r-values are used in calculating overall heat transfer coefficients.

“C” or Conductance is similar to K, except it is an overall heat transfer factor for a given thickness of material, where K is a factor per inch.

Thermal Resistance “R” is the reciprocal of Conductance, 1/C, the same way thermal resistivity was the reciprocal of conductivity.

The “U” factor is the overall coefficient of heat transfer.  It is defined as the rate of heat transfer through a material or compound structural member with parallel walls.  The U factor is the resulting heat transfer coefficient after giving effect to thermal conductivity and conductance, and is expressed in terms of BTU (Hour)(Sq. Ft. of area)( °F TD).  It is usually applied to compound structures, such as walls, ceilings, and roofs.

The formula for calculating the U factor is complicated by the fact that the total resistance to heat flow through a substance of several layers is the sum of the resistance of the various layers.  The resistance to heat flow is the reciprocal of the conductivity.  Therefore, in order to calculate the overall heat transfer factor, it is necessary to first find the overall resistance to heat flow, and then find the reciprocal of the overall resistance to calculate the U factor.

The basic relation between the U factor and the various conductivity factors is as follows:

R total  =   1   +   X1  +   X2

                 C         k1       k2

U  =  1  total


Where      C         is the conductance

    X1        is the thickness of material one

    X2        is the thickness of material two

                 k1        is the thermal conductivity of material one

And          k2        is the thermal conductivity of material two

For example, to calculate the U factor of a wall composed of 2" of material having a k factor of .80, and 2" of insulation having a conductance of .16, the U value is found as follows:

R total  =  1  +  X1

                C      K1

=   1   +     2

   .16      .80

=  6.25  +  2.5  =  8.75

U  =  1   total  =      1

         R               8.75

= .114 BTU/(Hour)(Sq.Ft.)(°F TD)

Once the U factor is known, the heat gain by transmission through a given wall can be calculated by the basic heat transfer equation.

Assume a wall with a U factor of .114 as calculated in the previous example.  Given an area of 90 square feet with an inside temperature of 0°F and an outside temperature of 80°F, the heat transmission would be:

Q  =  U x A x TD

 =  .114 x 90 Sq. Ft. x 80° TD

 =  812 BTU/Hr.

The entire heat gain into a given refrigerated space can be found in a similar manner by determining the U factor for each part of the structure surrounding the refrigerated space, and calculating as above.  Most good insulating materials have a thermal conductivity (k) factor of approximately .25 or less, and rigid foam insulations have been developed with thermal conductivity (k) factors as low as .12 to .15.

Heat transmission coefficients for many commonly used building materials are shown in Table 1.

Table 1.

Extensive studies have been made of weather bureau records for many years to arrive at suitable outdoor design temperatures.  Naturally, the maximum load occurs during the hottest weather.  However, it is neither economical nor practical to design equipment for the hottest temperature, which might ever occur, since the peak temperature might occur for only a few hours over the span of several years.  Therefore, the design temperature normally is selected as a temperature that will not be exceeded more than 1% of the hours during the four month summer season.

For Wisconsin, the outdoor design temperatures are 90°F dry bulb, 74°F wet bulb.  For Minneapolis, Minnesota, they are 92°F dry bulb, 75°F wet bulb.  For Illinois, they are 94°F dry bulb, 75°F wet bulb.

If the box has any exterior walls or ceiling, an allowance for radiation from the sun must be taken into consideration.  If the walls of the refrigerated space are exposed to the sun, additional heat will be added to the heat load.  For ease in calculation, an allowance can be made for the sun load in refrigeration calculations by increasing the temperature differential by the factors listed in Table 2.

Table 2.

Insulation thickness must be increased as the storage temperature decreases.  Table 3 lists recommended insulation thickness from the 1981 ASHRAE Handbook of Fundamentals.  The recommendations are based on expanded polyurethane, which has a conductivity factor of .16.  If other insulations are used, the recommended thickness should be adjusted based on relative k factors.

Table 3.

Table 4 lists the approximate heat gain in BTU per 1°F temperature difference per Sq. Ft. of surface per 24 hours for various thicknesses of commonly used insulations.  The thickness of insulation referred to is the actual thickness of insulation, and not the overall wall thickness.  For example, to find the heat transfer for 24 hours through a 6' x 8' wall insulated with 4 inches of glass fiber, when the outside is exposed to 95°F ambient temperature and the box temperature is 0°F.  Calculate as follows:  1.9 factor x 48 Sq. Ft. x 95o TD = 8664 BTU 

Table 4.

Any outside air entering the refrigerated space must be reduced to the storage temperature, thus increasing the refrigeration load.  In addition, if the moisture content of the entering air is above that of the refrigerated space, the excess moisture will condense out of the air, and the latent heat of condensation will add to the refrigeration load.

Because of the many variables involved, it is difficult to calculate the additional heat gain due to air infiltration.  Various means of estimating this portion of the refrigeration load have been developed based primarily on experience, but all of these estimating methods are subject to the possibility of error, and specific applications may vary in the actual heat gain encountered.

The traffic in and out of a refrigerator usually varies with its size or volume.  Therefore, the number of times doors are opened will be related to the volume rather than the number of doors.

Table 5 lists estimated average air changes per 24 hours for various size refrigerators due to door openings and infiltration for a refrigerated storage room.  Note that these values are subject to major modification if it is definitely determined that the usage of the storage room is either heavy or light.  Heavy usage is usually defined as four or more door openings per hour.

Table 5.

Another more accurate means of computing infiltration into a refrigerated space is by means of the velocity of airflow through an open door.  When the door of a refrigerated storage space is opened, the difference in density between cold and warm air will create a pressure differential, causing cold air to flow out the bottom of the doorway and warm air to flow in the top.  Velocities will vary from maximum at the top and bottom to zero in the center. The estimated average velocity in either half of the door is 100 feet per minute for a doorway 7' high at a 60oF TD.  The velocity will vary as the square root of the height of the doorway and as the square root of the temperature difference.

For example, the rate of infiltration through a door 8' high and 4' wide, with a 100oF TD between the storage room and the ambient can be estimated as follows:

Velocity = 100 FPM  x     8    x    100

                                        7    x    60

= 100  x  2.83  x   10  

               2.65      7.74

= 138 FPM

Estimated rate of infiltration   138 FPM   x   8 Ft.  x  4 Ft.  =   2210 Cu. Ft. per min.


Infiltration velocities for various door heights and TDs are plotted in Table 6.

Table 6.

If the average time the door is opened each hour can be determined, the average hourly infiltration can be calculated, and the heat gain can be determined as above.  (If positive ventilation is provided for a space by means of supply or exhaust fans, the ventilation load will replace the infiltration load if greater) and the heat gain may be calculated on the basis of the ventilating air volume.  This situation is very unusual (most walk-ins are not ventilated).

Once the rate of infiltration has been determined, the heat load can then be calculated from the heat gain per Cu. Ft. of infiltration as given in Table 7. The product load is composed of all the heat gain occurring due to the product in the refrigerated space.  The load may arise from a product placed in the refrigerator at a temperature higher than the storage temperature, from a chilling or freezing process, or from the heat of respiration of perishable products.  The total product load is the sum of the various types of product load, which may apply to the particular application.

Table 7.

Fruits and vegetables are living organisms.  Their life processes continue for some time after being harvested, and as a result they give off heat.  Certain other food products also undergo continuing chemical reactions, which produce heat.  This is called heat of respiration.  Meats and fish have no further life processes and do not generate any heat.  The amount of heat given off is dependent on the specific product and its storage temperature.  The heat of respiration varies with the storage temperature.

Many engineering manuals are available with the entire product data needed to calculate product load.  (These manuals have been well circulated throughout Climatic Control Company.  The Heatcraft Engineering Manual H-ENG-1 dated June 1990 is used for this Info-Tec.)

Most products are at a higher temperature than the storage temperature when placed in a refrigerator.  Since many foods have a high percentage of water content, their reaction to a loss of heat is quite different above and below the freezing point.  Above the freezing point, the water exists in liquid form, while below the freezing point; the water has changed to ice.  The specific heat of a product is defined as the BTU’s required to raise the temperature of one pound of the substance 1°F.

Specific heats of products are listed in the manuals.  Note that the specific heat above freezing is different than the specific heat below freezing.  Also note that freezing points vary.

The heat to be removed from a product to reduce its temperature above freezing may be calculated as follows:

Q     =   W x c x (T1 - T2)

Where      Q         is the number of BTUs to be removed

                 W        is the weight of the product in pounds

                 c          is the specific heat above freezing

                 T1        is the initial temperature, °F

And          T2        is the final temperature, °F (freezing or above)

For example, the heat to be removed in order to cool 1,000 pounds of veal (whose freezing point is 29°F) from 42°F to 29°F can be calculated as follows:

Q  =   W  x  c  x  (T1 - T2)

     =   1000 pounds  x  .71 specific heat  x (42 - 29)

     =   1000  x  .71  x  13

     =   9,230 BTU

Food products have a high percentage of water content.  In order to calculate the heat removal required to freeze the product, only the water need be considered.  Since the latent heat of fusion or freezing of water is 144 BTU/Lb., the latent heat of fusion for the product can be calculated by multiplying 144 BTU/Lb. by the percentage of water content.  To illustrate, veal has a water percentage of 63%, and the latent heat of fusion for veal is 91 BTU/Lb.   63% x 144 BTU/Lb. = 91 BTU/Lb.

The heat to be removed from a product for the latent heat of freezing may be calculated as follows:

Q    =  W x h if

Q        is the number of BTUs to be removed

W       is the weight of product in pounds

h if      is the latent heat of fusion, BTU/lb.

The latent heat of freezing of 1,000 pounds of veal at 29°F is:

Q   =  W  x  h if

      =  1000 lbs.  x  91 BTU/lb.

      =  91,000 BTU

Once the water content of a product has been frozen, sensible cooling again can occur in the same manner as that above freezing, with the exception that the ice in the product causes the specific heat to change.  The specific heat of veal above freezing is .71, while the specific heat below freezing is .39.

The heat to be removed from a product to reduce its temperature below freezing may be calculated as follows:

Q   =  W x ci x (Tf - T3)

Q        is the number of BTUs to be removed

W       is weight of product in pounds

ci        is specific heat below freezing

Tf       is freezing temperature

T3      is final temperature

For example, the heat to be removed in order to cool 1,000 pounds of veal from 29°F to 0°F can be calculated as follows:

Q  =  W x ci x (Tf - T3)

     =  1,000 lbs. x .39 specific heat x (29-0)

     =  1,000 x .39 x .29

     =  11,310 BTU

The total product load is the sum of the individual calculations for the sensible heat above freezing, the latent heat of freezing, and the sensible heat below freezing.

From the foregoing example, if 1,000 pounds of veal were to be cooled from 42°F to 0°F, the total would be:

Sensible Heat above freezing            9,230 BTU

Latent Heat of freezing                     91,000 BTU

Sensible Heat below freezing          11,310 BTU

Total product load                          111,540 BTU

Note that as of now, no time limit has been imposed.  The product load of 111,540 BTU is not BTU/Hr!  Eventually a time factor must be taken into consideration.  As an example, if the 1,000 Lbs. of veal was the only load and the specifications are to have the veal get to 0°F in 8 hours, the hourly load is the total load divided by the number of hours to get to 0°F.  111,540 ÷ 8 = 13,942.5 BTU/Hr.

In addition to the heat transmitted into the refrigerated space through the walls, air infiltration, product load, and any heat gain from other sources must be included in the total cooling load estimate.

Any electric energy directly dissipated in the refrigerated space such as lights, heaters, etc. is converted to heat and must be included in the heat load.  One-watt hour equals 3.41 BTU, and this conversion ratio is accurate for any amount of electric power.

Any electrical energy transmitted to motors inside a refrigerated space must undergo a transformation.  Any motor losses due to friction and inefficiency are immediately changed to heat energy.  That portion of the electrical energy converted into useful work, for example in driving a fan or pump, exists only briefly as mechanical energy, is transferred to the fluid medium in the form of increased velocity, and as the fluid loses its velocity due to friction, eventually becomes entirely converted into heat energy.

A common misunderstanding is the belief that no heat is transmitted into the refrigerated space if an electric motor is located outside the space, and a fan inside the space is driven by means of a shaft.  All of the electrical energy converted to mechanical energy actually becomes a part of the load in the refrigerated space.

Because the motor efficiency varies with size, the heat load per horsepower as shown in Table 8 has different values for varying size motors.  While the values in the table represent useful approximations, the actual electric power input in watts is the only accurate measure of the energy input.

Table 8.

People give off heat and moisture, and the resulting refrigeration load will vary depending on the duration of occupancy of the refrigerated space, temperature, type of work, and other factors.  Table 9 lists the average heat load due to occupancy, but for stays of short duration, the heat gain will be somewhat higher.

Table 9.

The total supplementary load is the sum of the individual factors contributing to it.  For example, the total supplementary load in a refrigerated storeroom maintained at 0°F in which there are 400 watts of electric lights, a 3 H.P. motor driving a fan, and 2 people working continuously would be as follows:

400 Watts x 3.41 BTU/Hr. = 1364 BTU/Hr.

3 HP motor x 2,950 BTU/Hr. = 8,850 BTU/Hr.

2 People x 1300 BTU/Hr.                =      2,600 BTU/Hr.

Total supplementary load                =      12,814 BTU/Hr. 

The most accurate means of estimating a refrigeration load is by considering each factor separately.  The following example will illustrate a typical selection procedure.  The load has been chosen to demonstrate calculations and does not represent a normal loading.


   Walk-in cooler with 4" of glass fiber insulation, outside location, but shaded

   Outside Dimensions:  height 8', width 10', length 40', inside volume 3,000 Cu. Ft.

   Floor area (outside dimensions) 400 Sq. Ft. on insulated slab in contact with ground

   Ambient temperature 100°F, 50% relative humidity

   Ground temperature 55°F

   Refrigerator temperature 40°F

   1/2 H.P. fan motor running continuously

   Two 100 watt lights, in use 12 hours per day

   Occupancy, 2 men for 2 hours per day

   In storage:

     500 Lbs. of bacon at 50°F

     1000 Lbs. of string beans

   Entering product:

     500 Lbs. of bacon at 50°F

     15,000 pounds of beer at 80°F

     To be reduced to storage temperature in 24 hours

   Heavy door usage

It makes no difference in what sequence each factor that comprises the load is calculated, just so all factors are considered.  It is best to keep the same sequence so one will remember all of the factors.  In this example, we’ll do the box itself, or heat transmission load, air infiltration, product load, and then supplementary loads.


As1 = 40'  x  8'  x  2  =  640 Ft2


Qs1 = 640 Ft2  x 60°TD  x  1.9 (Table 4)                     =  72,960 BTU   


As2 = 10'  x  8'  x 2  =  160 Ft2  


Qs2 = 160 Ft x 60°TD  x  1.9                                        = 18,240 BTU



Ac = 40'  x  10'  =  400 Ft2 


Qc = 400 Ft2  x  60°TD x 1.9                                         =  45,600 BTU



Af = 40'  x  10'  =  400 Ft2 


Qf = 400 Ft2  x  15°TD x 1.9                                         =  11,400 BTU


Q total for a 24 hour transmission load                      = 148,200 BTU (cooler alone, no other loads)


Air infiltration:


3000 Ft3  x  9.5 air changes (Table 5)  x  2 usage factor  x 2.11 factor (Table 7) = 120,270 BTU



Product load factors were taken from the Heatcraft Engineering Manual H-ENG-1.

Many products have a range shown for their specific heats, such as bacon or pork.  The specific heat is shown as .46 to .55 BTU/LB/°F above freezing.  When a range for specific heat is given, average the low and high of the range to calculate the load.  The average is (.46 + .55) ÷ 2 = .505.  There is no reason to use three decimal places or more in calculating, so round off to two decimal places. 

Q bacon = 500 Lbs. bacon x .50 sp. ht. x 10°TD = 2500 BTU. 


Q beer = 15,000 lbs. beer x .92 sp. ht. x 40°TD = 552,000 BTU. 


Q beans = 1,000 Lbs. beans x .5 = 5000 BTU.



Note we had all 24 hours to reduce the temperature of the bacon from 50°F to 40°F.

Note that we have to deal only with the heat of respiration for the beans.  They were specified in long-term storage and no temperature reduction was required.  (In the real world, there probably would have been a temperature reduction, but remember we are demonstrating calculations.)  The heat of respiration for beans is shown as 4.6 to 5.7 BTU/Lb./24 hrs. at 40°F.  Again, averaging and rounding off .5 BTU/Lb./24 hrs.

The total product load is: 

500 lbs. bacon                                                   =        2500 BTU

15,000 lbs. beer                                                =   552,000 BTU

1000 lbs. beans                                                 =       5,000 BTU

Total 24 Hr. product load                                   =   559,500 BTU

The supplementary loads are: 

200 watts x 12 hours x 3.41 BTU/Hr.                =       8,184 BTU

1/2 H.P. x 4250 BTU/Hr-Hr (Table 8) x 24        =     51,000 BTU

2 People x 2 hrs/day x 840 BTU/Hr. (Table 9)  =       3,360 BTU

Total 24 hour supplementary Load                   =     62,544 BTU

We can now total all the loads to arrive at a 24-hour load. 

Transmission load                                            =   148,200 BTU

Air infiltration                                                      =   120,270 BTU

Product                                                              =   559,500 BTU

Supplementary                                                  =     62,544 BTU

Total 24 hour load                                             =   890,514 BTU

Often a 5% to 10% safety factor is added to the load calculations, “to be sure the equipment is big enough”.

If, as in the example, all the data concerning the load is gathered, no safety factor need be added.  In fact, additional safety factor load may be detrimental.  The equipment may end up very oversized and cause low load problems.  Also, safety factors result in higher equipment costs.

In general, the fact that the compressor is sized on the basis of 16 to 18 hours operation in itself provides a large safety factor.  The load is calculated on the basis of the peak demand at design conditions, and the design conditions are selected on the basis that they will occur no more than 1% of the hours during the summer months. The design criterion provides plenty of safety factors if the data is complete and accurate.  More and more designers are picking 18 hour run times rather than 16 hours, due to equipment cost.  Even 20 to 22 hours are used when the temperature difference between box temperature and refrigerant temperature results in evaporator coils running close to or above freezing.  For freezers, 18 hour run time is usually acceptable. 

Tech Tip:      Hourly load for compressor sizing is simply the 24-hour load divided by the selected run time. 

For example, if 18 hours run time is selected:  890,514 BTU ÷ 18 hr = 49,473 BTU/hr.

In order to complete the equipment selection, a “split” must now be determined.  Split is the temperature difference (TD) between the refrigerant and air.  The split is selected based on the relative humidity to be maintained in the refrigerated space.  The smaller the split, the higher the humidity.

Relative humidity in a storage space is affected by many variables, such as system running time, moisture infiltration, condition and amount of product surface exposed, air motion, outside air conditions, type of system control, etc.  Perishable products differ in their requirements for an optimum relative humidity for storage, and recommended storage conditions for various products are shown in the product tables.  Satisfactory control of relative humidity in a given application can be achieved by selecting the compressor and evaporator for the proper operating temperature difference or TD between the desired room temperature and the refrigerant evaporating temperature.

Recommended splits for the most common applications are:


                          Type                     Split oF       RH %

Wet Produce Walk-In                         10              90

Dry Produce Walk-In                          13              85

Service Meat Case                             15              80

Meat Walk-In                                      18              75

All Freezers                                        10              --



For a cooler with mixed product in storage, the product requiring the highest humidity will determine the split.  To pick a split for our example consider the storage requirements for each product.  Beer would be in cans, bottles, or kegs, and is unaffected by humidity.  Pork requires 85% to 90% R.H. and beans 90% to 95% RH.  In order to prevent moisture loss and consequently ruining the bacon and beans, the split selected should be 10°F to maintain high humidity.  Once the split has been determined, major equipment selections can proceed.

The load divided by the split will equal the evaporator size at 1°F TD.  Using our example: 

49,473 ÷ 10 = 4947 BTU at 1°TD. 

To find the right condensing unit:  Box temperature minus split equals suction temperature.  In our example, 40°F - 10°F = 30°F.  The unit we want would be shown in a catalog rated at 49,473 BTU/Hr. at 30°F suction temperature.

Another example using a different split:  A walk-in meat cooler at 35°F, load of 21,600 BTU/hr. 

The calculation is as follows: 

21,600 ÷ 18 = 1200 BTU at 1° TD or 12,000 BTU @ 10° TD

The right evaporator for this job would be listed in a catalog as 12,000 BTU at 10 °TD.  Box temperature minus split equals suction temperature.  35°F - 18°F = 17°F.  The right condensing unit would be rated 21,600 BTU/hr. at 17°F suction temperature.

If the two components match, they should drop the suction to just the right pressure every time it runs.  Then, if the load calculation was accurate, the compressor will run 45 minutes and be off 15 minutes (18 hour run time).  If the load is heavier than calculated, it will run more and be off less.  If the load is less than calculated, the compressor will stay off longer.  As long as the load stays somewhere close to calculated load, the RH will be close to the requirement.

Selecting a condensing unit depends on the type of condensing medium to be used, air or water, the design ambient temperature or water temperature, and the capacity of the condenser selected.  Air-cooled condensers are commonly selected to operate on temperature differences (TD) from 10°F to 30°F, the lower to be used for low temperature applications where the compression ratio is less critical, and high TDs for high temperature applications.  Most manufacturers’ catalogs list air-cooled condensing units by ambient temperature, 90°F or 95°F being the most common.  The condenser TD has therefore been taken into consideration and condensing unit selection can be based on the ambient temperature.

Commercially available components seldom will exactly match the design requirements of a given system, and since system design is normally based on estimated peak loads, the system may often have to operate at conditions other than design conditions.  More than one combination of components may meet the performance requirements, the efficiency of the system normally being dependent on the point at which the system reaches stabilized conditions or balances under operating conditions.

Most manufacturers of commercial and low temperature coils publish only ratings based on the temperature difference between entering dry bulb temperature and the evaporating refrigerant temperature.  Although frost accumulation involving latent heat will occur, unless the latent load is unusually large, the dry bulb ratings may be used without appreciable error.

The most accurate means of determining the refrigeration load is by calculating each of the factors contributing to the load as was done in the previous example.  However, for small walk-in coolers, various types of shortcut estimating methods are frequently used.

As was mentioned at the beginning of this Info-Tec, 90% of walk-in coolers and freezers can be defined as “small”.  By limiting small coolers to a maximum of 2,500 Cu. Ft., and freezers to 1,500 Cu. Ft. in size, errors made in calculations, or assumptions made due to incomplete data, will be small.

Every attempt should be made to gather all the data listed in Figure 1.  Even if a small box, calculating the load the “long” way will produce the best results.  The “short” way or quick selection process should be used only when time restraints prevent accurate calculations, and/or data is incomplete, forcing certain assumptions to be made regarding load calculations.

There is certain minimum data required to be able to use the quick selection tables.  See Figure 1.  You need to know the answers to 1, 2, 3, 7, and 13 to arrive at a load figure.  (Of course, you’ll need to know answers to 10, 11, and 12 to properly complete equipment selection, but not to find the load.) 

Tables are attached for walk-in coolers and walk-in freezers.  The capacities given are for average applications.  If the load is unusual, these tables should not be used.  The low temperature tables do not include any allowance for a freezing load, and if a product is to be frozen, additional capacity will be required.  You would need to know how much of what is going in at what temperature and how long you have to pull down to box temperature.

Note all the small print.  The BTU’s shown are the per hour load, including the 16 - 18 hour run time safety factor. 

Some typical examples:  A customer calls for a quote on the equipment for a walk-in cooler, 14' long by 10' wide by 8' high.  The box will be located in an air-conditioned store.  He wants the box temperature about 36°F to 38°F.  He doesn’t have the slightest idea of how much of what is going to be in the box, but does know it is all long-term storage.  There is nothing else unusual about the box, like beer shafts or glass doors.  It’s a “good” box with an insulated floor.

This is a very typical example of all the information you will get to size the load for a small box.  While not perfect, it will suffice, and here’s why:

The 14' x 10' x 8' dimensions can be inside or outside dimensions; it won’t matter.  The difference between inside and outside dimensions in load will be next to nothing in the small box.  The “normal” temperatures for walk-in coolers will range from 34°F to 40°F.  While the chart was prepared for 34°F, again, because the box is a small box, no significant difference will exist in load between a 34°F box and a 40°F box.  Since you will probably never be able to exactly match the load to the equipment selected, if the box temperature is toward the high end of the range, 38°F to 40°F, select the equipment capacity closest to the load, less than the load.  If box temperature is near the low end of range, 34°- 36°F select equipment closest to the actual load.  This box is in an air-conditioned store, so the ambient temperature selected will be 80°F.  Had the box been in an un-air-conditioned building, 90°F would be selected as the ambient.  An outside box, forming its own structure, or three walls, ceiling, and floor outside of a main structure should be considered an “unusual” condition.  Which way the box faces, the color of the walls and ceiling, etc. (see previous section on sun load calculation) have to be considered and calculations made.  Avoid using the 100°F ambient temperature column for outside boxes, except very small boxes, up to 1,200 Cu. Ft.

The customer stated it was a “good” box, meaning it is well built and not leaking excessively.  Our chart specification of 3" of fiberglass insulation needs no adjustment.  No amount for any product is given, but there is always some product loading.  Product loading cannot be ignored!  Long-term storage is defined as at least 24 hours to get whatever product is put into the box down to box temperature.  The average product load column is based on extensive studies of small boxes to determine the BTU/Hr. loads.  Besides, how much product can be crammed into a small box that would significantly affect the average load figure?

The customer has told us there is nothing unusual about the load, so no further load calculations are necessary or adjustments made to the quick selection chart.

Taking the loads from the chart, we find:

Wall/Infiltration          =       6880 BTU/Hr.

Ave. Product Load    =       2790 BTU/Hr.

Total                          =       9670 BTU/Hr.

Let’s completely finish the equipment selection for this box.  We will have to select a split.  Not knowing anything about product in the box, it is best to use a TD of 10°F as the split.  This will result in high humidity, beneficial for most products and can’t harm any packaged goods.  A 15°F split could also be used.  As we will see when selecting the equipment, we can determine which split is more economical to use.  A higher split will result in a smaller evaporator, but may necessitate a larger condensing unit negating any dollar savings of the smaller, less expensive evaporator with a larger, more expensive condensing unit.

Using a 10°F split, our suction temperature will be 26°F.  Box temperature was given as “36° - 38°F”.  If a range of desired box temperature is given, always use the lowest temperature.

Before equipment selection can continue, we need the answers to Figure 1, 11, 12, and 13.  The customer has told us the condensing unit will be outdoors, either 115 VAC or 230 VAC single phase is available, and to quote pump down as an option.  (All equipment selected will be equipment Climatic Control Company is handling as of this date.)

First, we select the unit cooler.  A Chandler RLC092F1 is rated 9200 BTU at 20°F suction, 10°F TD.  At 26°F suction, the RLC092F1 will be about 9300 BTU, 10°F TD, as close as we can get to the calculated load.  Next, select the condensing unit.  Some designers will take a 2° or 3°F temperature penalty for suction line loss when selecting a condensing unit.  It is deducted from the temperature found after determining the split.  With all the built-in safety factors, it isn’t really necessary.  It is convenient to use those 2° or 3°F to adjust the suction temperature to correspond to BTU ratings shown on manufacturers tables for condensing units.  In our example, using 25°F will be convenient.

A Tecumseh AJ9486EC condensing unit is rated 9500 BTU/Hr. at 25°F suction, 90°F ambient temperature.  It is single phase 208/230 Volt.  It meets all the specifications and is a good match for the RLC092F1 unit cooler.  After selecting the condensing unit, look at the detailed specification sheet(s) for the unit.  We will need to know the liquid line size to be able to select the liquid line components.  We need to find out if the unit has a low-pressure control, needed for pump down control or the primary control if no thermostat is used.  It is at this point we can compare the cost of the major components to see if using a 15° split will make any difference in price.  Using the 15° TD, the RLC067F1 will be the unit cooler.  The AJ9486EC at 20°F suction will be 8600 BTU/Hr., a little small, but the next larger unit is far too big.  It is still a close enough match to use the AJ9486EC with the RLC067F1.  In this case, there is a significant savings using a 15°F split.

If this condensing unit was going to be inside a heated building, nothing more would be added to it, but the customer has told us it is going outdoors.  This requires the addition of an appropriate RDM, outdoor enclosure, crankcase heater, and head pressure control valve (see Info-Tec 10).  A larger receiver will be needed because of the HP valve.  The rest of the liquid line components should be selected, dryer, and sight glass, based on the liquid line size and tonnage of the condensing unit.  Select the TXV based on the load in tons.  The customer asked for pump down control as an option.  Pump down should be on all refrigeration systems, so should be quoted all the time even if the customer doesn’t request it.  If not asked for, include pump down as an option.  To complete a pump down system, a low-pressure control, liquid line solenoid valve, and thermostat are needed.  A high-pressure control or dual pressure control should also always be quoted as options.  Check the condensing unit specifications sheets to see what may or may not be included with the unit.  For instance, as of this date, Tecumseh does not include low-pressure controls on units up to and including one H.P., but does include them on units with a rating of 1.5 horsepower and up.  Some units have crankcase heaters included, etc.  Chandler’s units vary considerably as to the included devices.  The point is, check the catalogs and specification sheets for what is or is not included.

Quick load selection for small walk-in freezers is very similar to walk-in coolers.  Using the “walk-in freezers” chart, find the box size closest to size you’re given, pick the ambient temperature that is appropriate, and then use the column closest to the box temperature shown, 0°, -10°, or -20°F.  Use the 0°F column for boxes up to +10°F.  Any freezer over a box temperature of +10°F will have to be calculated the long way.

To the BTU/Hr. figure for the transmission load, add the figure for “average product load BTU/Hr”. 

Example: a freezer with the following dimensions: 10’ x 10' x 9’ It is also in a building without air-conditioning. The box temperature is -20°F.

Box                   =     8,760 BTU/Hr.

Average load     =     2,520 BTU/Hr.

Total load          =   11,280 BTU/Hr.

All freezers use a split of 10°F TD, so the suction temperature will be -30°F.  Pick equipment rated for the load at -30° F suction temperature.  Don’t forget we are now dealing with unit coolers that will need some method of defrost other than air temperature during the off cycle.  Electric defrost is 99% of the method chosen to defrost for small boxes.  Hot gas defrost is used with supermarket rack systems and in some large boxes.  The 11,280 BTU/Hr. load at -30° would use a Chandler ELC 122F2.  A defrost timer needs to be added to the equipment list, an 8145-20 Paragon.  Pump down is not an option; it must be used!  An accumulator is highly recommended on any job, but is a must on low temperature jobs.  Always quote the condensing unit with an accumulator and note that to the customer.  If the defrost heater load connected to the defrost timer exceeds the timers electrical load rating, use a contactor to carry the heater’s load.  The rest of the equipment selection is the same as for coolers -- dryer, sight glass, TXV, thermostat, etc.

There are many other loads that will be encountered that have to be added to the load found through long way calculations or through using quick selection tables.

Some of the more common loads and the loads to be added: 

Beer shafts 

1500 BTU/Hr. for one.  Any additional shafts, add 1000 BTU/Hr. per shaft.

Walk-in coolers with glass doors 

1 to 4 doors - 1,200 BTU/Hr. per door

5 to 7 doors - 1,100 BTU/Hr. per door

8 doors and up - 1,000 BTU/Hr. per door

Walk-in freezers with glass doors 

Freezers at 90°TD (Box to ambient)

1 to 4 doors - 2,000 BTU/Hr. per door

5 to 7 doors - 1,600 BTU/Hr. per door

8 doors and up - 1,500 BTU/Hr. per door

Freezers at 110° TD (Box to ambient)

1 to 4 doors - 2500 BTU/Hr. per door

5 to 7 doors - 2100 BTU/Hr. per door

8 doors and up - 2000 BTU/Hr. per door

Glass doors are assumed to be double pane for coolers, triple pane for freezers.

Back bars are usually constructed with 2" of fiberglass insulation or its equivalent.  Based on a 50°F TD and 16 hour run time, the load for back bars can be expressed in BTU/Hr. for specific lengths:    

Up to 6' long------------ 1100 BTU/Hr.

7' - 8' long -------------- 1400 BTU/Hr.

9' - 10' long ------------- 1800 BTU/Hr.

11' - 12' long ----------- 2100 BTU/Hr.

13' - 15' long ----------- 2700 BTU/Hr.

10' - 20' long ----------- 3500 BTU/Hr.

Loads for open display cases in supermarkets and convenience stores can be figured in BTU/Hr. per foot of case.  The stores are assumed to be air-conditioned.  Open freezers are 700 BTU/FT/Hr.  Open coolers are 500 BTU/FT/Hr.

Floors for all walk-in coolers and freezers should be insulated, but often the box is build around a concrete floor poured on grade.  Amateurs that build the box will usually do a good job insulating the walls and ceiling, but forget that cold air falls and neglect to insulate the floor.  Special problems are encountered in these situations.  Many studies have been done, and are still being done, on determining the load for uninsulated concrete on grade walk-in floors.  When the box is first put into operation, the transmission load through the floor will be large, but as time goes by, the transmission load will decrease.  How much and how fast is dependent on so many variables, too many to even list here, that there is no formula that can be used to accurately predict the load changes.  The load would have to be constantly monitored, and this is certainly not practical.

Many guidelines have been proposed to deal with uninsulated concrete floors on grade.  The old McQuay Co. (Now Heatcraft), after extensive research and testing over a long period of time, came up with a formula to add some load for uninsulated concrete on grade floors.  It is 5 BTU’s per Sq. Ft. per 24 Hrs. times DT between summer ground temperature and box temperature.  It does not matter if the floor is concrete on ground or concrete on cinder fill.  The concrete can be any thickness from 4" to 10".

A freezer should never be built with an uninsulated concrete on grade floor.  Eventually, the moisture in the ground under the floor would freeze and heave the entire box.

Example:  Walk-in cooler at 35°F box temperature with 6" thick uninsulated concrete on grade floor in Wisconsin 10' wide x 14' long.  (Ground temperature for Wisconsin = 55°F, Illinois = 60°F, Minnesota = 55°F)  Therefore, 5 x 140 x 20 = 14,000 BTU load for 24 hours.  Factoring in an 18 hour run time results in a 780 BTU/Hr. load to be added.  How to figure the concrete on grade floor loading controversy still goes on.  Obviously, the best way to avoid this problem is to have an insulated floor.

In larger boxes where the load requires approximately 10 horsepower or more, it is recommended the load be divided into multiple units.  While initial cost will be higher, the advantages gained will justify the higher costs.  Remember, everything has been sized for maximum load, which seldom occurs.  Multiple units can be staged, giving capacity control, and reducing electrical costs.  Staging also cuts demand charges by utilities.  Nothing lasts forever, so when one unit fails and needs service, the other unit(s) can prevent expensive product loss.  Once pull down of a product load has been accomplished, the box probably can be maintained at temperature by a fraction of the horsepower needed for full load, thereby saving operating and maintenance costs.  In the 10 horsepower range, recovery (payback) of the initial higher cost of equipment will typically occur in 2 to 3 years.  Smaller boxes would also benefit by using multiple units, but payback will take longer.

While having nothing to do with load calculating, it may be of interest to some why “cold” preserves food.

Spoiling of food is actually the growth of bacteria in it.  Cold slows the movement of molecules and makes all organisms sluggish.  Cold slows the growth of bacteria, thus foods to not spoil as fast.  By keeping bacteria from growing, food will be edible longer.

Most foods contain a lot of water.  Food, therefore, should be kept slightly above freezing.  The correct refrigerator temperature for fresh food is 34°F to 44°F.

Freezing food extends the storage time considerably.  If food is frozen slowly, large ice crystals are formed and their size breaks down the food tissues.  When defrosted for use, the food spoils rapidly and taste is ruined.  Fast freezing at 0°F to -15°F, forms small ice crystals which do not damage the food tissues.  That is why food freezers should be maintained at least at 0°F.  Food placed in them will freeze quickly, and when defrosted for use, even after long storage, will still be tasty.

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