CHAPTER 10 Cooling Load January 2012

Chapter 10 COOLING LOADCALCULATIONS 10-1. The Cooling Load

The cooling load on refrigerating equipment seldom results from any one single source of heat. Rather, it is the summation of the heat which usually evolves from several different sources. Some of the more common sources of heat that supply the load on refrigerating equipment are:

1.

Heat that leaks into the refrigerated space from the outside by conduction through the insulated walls.

2.

Heat that enters the space by direct radiation through glass, or other transparent materials.

3.

Heat that is brought into the space by warm outside air entering the space through open doors or through cracks around windows and doors.

4.

Heat given off by a warm product as its temperature is lowered to the desired level.

5.

Heat given off by people occupying the refrigerated space.

6.

Heat given off by any heat-producing equipment^ located inside the space, such as electric motors, lights, electronic equipment, steam tables, coffee urns, and hair driers.

It will be shown later that all of these sources of heat are not present in every application and that the Importance of any one heat source with relation to the total cooling load will vary considerably with each application. 10-2. Equipment Running Time Because of the necessity for defrosting the evaporator at frequent intervals, it is not practical to design the refrigerating system in such a way that the equipment must operate continuously in order to handle the load. In most cases, the air passing over the cooling coil is chilled to a temperature below its dew point and moisture is condensed out of the air onto the surface of the cooling coif. When the temperature of the coil surface is above the freezing temperature of water, the moisture condensed out of the air drains off the coil into the condensate pan and leaves the space through the condensate drain. However, when the temperature of the cooling coil is below the freezing temperature of water, the moisture condensed out of the air freezes into ice and adheres to the surface of the coil, thereby causing "frost" to accumulate on the coil surface. Since frost accumulation on the coil surface tends to insulate the coil and reduce the coil's capacity, the frost must be melted off periodically by raising the surface temperature of the Coil above the freezing point of water and maintaining it at this level until the frost has melted off the coil and left the space through the condensate drain. No matter how the defrosting is accomplished, the defrosting requires a certain amount of time, during which the refrigerating effect must be stopped in the coil being defrosted.*

One method of defrosting the coil is to stop the compressor and allow the evaporator to warm up the space temperature and remain at this temperature for a sufficient

length of time to allow the frost accumulation to melt off the coil. This method of defrosting is called ''off-cycle” defrosting. Since the heat required to melt the frost in off-cycle defrosting must come from the air in the refrigerated space, defrosting occurs rather slowly and a considerable length of time is required to complete the process. Experience has shown that when off-cycle defrosting is used, the maximum allowable running time for the equipment is 16 h out of each 24 h period, the other 8h being allowed for the defrosting. This means, of course, that, the refrigerating equipment must have sufficient capacity to accomplish equivalent of 24 h of cooling in 16 h of actual running time. Hence, when off-cycle defrosting is used, the equipment running time used in Equation 10-1 is approximately 16 h. When the refrigerated space is to be maintained at a temperature below 1.5°C, offcycle defrosting is not practical. The variation in space temperature which would be required in order to allow the cooling coil to attain a temperature sufficiently high to melt off the * an exception to this is where a continuous brine tray is employed to keep the coil free of frost, in which case the equipment can be selected for continuous operation. frost during every off cycle would be detrimental to the stored product. Therefore, where the space temperature is maintained below 1.5°C, some type of supplementary heat defrosting is ordinarily used. In such cases the surface of the coil is heated artificially either with electric heating elements, with water, or with hot gas from the discharge of the compressor (see Chapter 20). Defrosting by any of these means is accomplished much more quickly than when offcycle defrosting is used. Hence, the shut-down time required is less for supplementary heat defrosting and the maximum allowable running time for the equipment can be greater than for the aforementioned off- cycle defrosting. For systems using supplementary heat defrosting maximum allowable running time is usually from 18 to 20 h out of each 24-h period, depending upon how often defrosting is necessary for the

application in question. It is of interest to note that since the temperature of the cooling coil in comfort air conditioning applications is normally around 4.5°C, no frost accumulates on the coil surface and, therefore, no down time is required for defrosting. For this reason, air conditioning systems ate usually designed for continuous run and cooling loads for air conditioning applications are determined directly in kilowatts. This holds true also for other applications where there is no frost accumulation on the cooling surface. In cases where it is inconvenient to calculate the refrigeration load on a 24-h basis, the load may be determined directly in kilowatts provided that the result is multiplied by an appropriate factor which makes an allowance for the desired equipment operating time. Where the desired equipment operating time is 16 h out of each 24, the refrigeration load in kilowatts should be multiplied by (24/16) 1.5, which in effect increases the required equipment capacity of 50% so that the equipment selected will have the capacity to handle the 24-h cooling load in the required 16 h of operating time. Similar multipliers can be determined for other desired operating times.

Q

24h (qt) RT

(10- 1)

Where: Q – required equipment capacity in kilowatts RT – running times in hours (h) qt – total cooling in kilowatts (the sum of the heat loads) 10-3. Cooling Load Calculations To simplify cooling load calculations, the total cooling load is divided into a number of individual loads according to the sources of heat supplying the load. The summation of these individual loads is the total cooling l load on the equipment.

In commercial refrigeration, the total cooling load is divided into four separate loads viz: (1) the wall gain load, (2) the air change I load, (3) the product load, and (4) the miscellaneous or supplementary load.

10-4. The Wall Gain Load The wall gain load, sometimes called the wall leakage load, is a measure of the heat flow rate by conduction through the walls of the refrigerated space from the outside to the inside. Since there is no perfect insulation I there is always a certain amount of heat passing from the outside to the inside whenever the inside temperature is below that of the outside. The wall gain load is common to all refrigeration applications and is ordinarily a considerable part of the total cooling load. Some exceptions to this are liquid chilling applications, where the outside area of the chiller is small and the walls of the chiller are insulated. In such cases, the leakage of heat through the walls of the chiller is so small in relation to the total cooling load that is effect is negligible and it is usually neglected. On the other hand, commercial storage coolers and residential air conditioning applications are both examples of application wherein the wall gain load often accounts for the greater portion of the total load. 10-5. The Air Change Load When the door of a refrigerated space is opened, warm outside air enters the space to replace the more dense cold air which is lost from the refrigerated space through the open door. The heat which must be removed from this warm outside air to reduce its temperature and moisture content to the space design conditions becomes a part of the total cooling load on the equipment. This part of the total load is called the air change load.

The relationship of the air change load to the total cooling load varies with the applica tion. Whereas in some applications the air change load is not a factor at all, in others it represents a considerable portion of the total load. For example, with liquid chillers, there are no doors or other openings through which air can pass and therefore the air change load is nonexistent. On the other hand, the reverse is true for air conditioning applications, where, in addition to the air changes brought about by door openings, there is also considerable leakage of air into the conditioned space through cracks around windows and doors and in other parts of the structure Too, in many air conditioning applications outside air is purposely introduced into the conditioned space to meet ventilating requirements. When large numbers of people are in the conditioned space, the quantity of fresh air which must be brought in from the outside is quite large and the cooling load resulting from cooling and dehumidifying this air to the space design conditions is often a large part of the total cooling load in such applications. In air conditioning applications, the air change load is called either the ventilating load or the infiltration load. The term ventilating load is used when the air changes in the conditioned space are the result of deliberate introduction of outside air into the space for ventilating purposes. The term infiltration load is used when the air changes are the result of the natural infiltration of air into the space through cracks around doors and other openings. Every air conditioning application will involve either an infiltration load or a ventilating load, but seldom both in the same application. Since the doors on commercial refrigerators are equipped with, well-fitted gaskets, the cracks around, the doors are tightly sealed and there should be little, if any, leakage of air around the doors of a commercial fixture in good condition. Hence, in commercial refrigeration, the air changes are usually limited to those which are brought about by .actual opening and closing of the door of doors. The importance of minimizing or eliminating the leakage of air from the outside to the' inside of coolers and freezers through the cracks around doors and through other openings cannot be overemphasized. Although such air leakage may or may not have an appreciable effect on the refrigeration load, the water vapor that condenses from

the warm air in the affected crack spaces and that frequently freezes into ice in these openings can be Very troublesome and should be prevented. In addition to well-aligned and well-aligned doors and, gaskets and the careful sealing of other wall openings to reduce air leakage, good design practice prescribes the use of a heater wire around the perimeter of the door to prevent condensation on these surfaces by maintaining their temperature above the DP temperature of the entering warm air. In addition the installation of a small heated air vents to equalize the pressure between the inside and outside of the cooler or freezer is also important. Without such controlled venting, negative pressures can develop inside the cooler or freezer as a result of the air pressure drop that occurs in accordance with Charles’ Law when the temperature of the warm air, which enters the space as the doors are opened and closed, is subsequently reduced to the space design temperature. Naturally, any air pressure differential between the inside and outside of the fixture greatly increases the tendency for air leakage around the door seals and through other wall openings, such as those through which refrigerant and water piping, drain lines, and electrical conduits pass to the outside. 10-6. The Product Load The product load is made up of the heat that

1

must be removed from the

refrigerated product in order to reduce the temperature of the product to the desired level. The term "product" as used here is taken to mean any material whose temperature is reduced by the refrigerating equipment arid includes not only perishable commodities, such as foodstuff, but also such .items as welding electrodes masses of concrete, plastic, rubber, and liquids of all kinds. In some instances, the product is frozen; in which case the latent heat removed is also a part of the product load.

The importance of the product load in relation to the total cooling load, like all others, varies with the application. Although it is nonexistent in some applications, in others it represents practically the entire cooling load.

Where the refrigerated cooler is designed for product storage, the product is usually chilled to the storage temperature before being placed in the cooler and no product load need be considered since the product is already at the storage temperature.* However, in any instance where the product enters a storage cooler at a temperature above the storage temperature, the quantity of heat which must be removed from the product in order to reduce its temperature to the storage temperature must be considered as a part of the total load on the cooling equipment.

* In the design of short-term storage coolers, general practice is to allow for 5°C of product cooling.

In some few instances, the product enters IV the storage fixture at a temperature below the normal storage temperature for the product. A case in point is ice cream, which is frequently chilled to a temperature of -18°C or -23°C during the hardening process, but is usually stored at about -12°C, which is the ideal dipping temperature. When such a product enters storage at a temperature below the space temperature, it will absorb heat from the storage space as it warms up to the storage temperature and thereby produce a certain amount of refrigerating effect of its own. In other words, it provides what might be termed a negative product load which could theoretically be subtracted from the total cooling load. This is never done, however, since the refrigerating effect produced is small and is not continuous in nature.

The cooling load on the refrigerating equipment resulting from product cooling may be .either intermittent or continuous, depending on the application. The product load is a part of the total cooling load only while

the temperature of the product is being reduced to the storage temperature, or while freezing is taking place. Once the product is cooled to the storage temperature, it is no longer a source of heat and the product load ceases to be a part of the load on the equipment. An exception to this is in the storage of fruit and vegetables which give off respiration heat for the entire time they are in storage at temperatures above the freezing temperature, even though there is no further decrease in their temperature (see Section 10-18).

There are, of course, a number of refrigeration applications where product cooling is more or less continuous in which case the prophet load is a continuous load on the equipment. This is true, for instance, in chilling coolers where the primary functions is to chill the warm product to the desired storage temperature.* When the product has been cooled to the storage temperature, it is usually moved out of the chilling room into a storage room and the chilling room is then reloaded with warm product. In such cases the product load is continuous and is usually a large part of the total load on the equipment.

Liquid chilling is another application wherein the product provides a continuous load on the refrigerating equipment. The flow of the liquid being chilled through the chiller is continuous with warm liquid entering the chiller and cold liquid leaving. In this instance, the product load is practically the only load on the equipment since there is no air change load and the wall gain .load is negligible, as is the miscellaneous load. In air conditioning applications there is no product load as such, although there is sometimes a "pull-down load" which tends to disappear as steady-state design conditions are attained.

10-7. The Miscellaneous Load

The miscellaneous load, sometimes referred to as the supplementary load, takes into account all miscellaneous sources of heat. Chief among these are people working in or otherwise occupying the refrigerated space, along with lights or other electrical equipment operating inside the space.

In most commercial refrigeration applications the miscellaneous load is relatively small, usually consisting only of the heat given off by lights and fan motors used inside the space.

In air conditioning applications, there is no miscellaneous load as such. This is not to say that human occupancy and equipment

*In some instances, the product is not completely chilled to the final storage temperature in the chilling cooler, and some additional chilling is accomplished in the holding storage cooler.

are not a part of the cooling load in air conditioning applications. On the contrary, people and equipment are often such large factors in the air conditioning load that they are considered as separate loads and are calculated as such. For example, in those air conditioning applications where large numbers of people occupy the conditioned space, such as churches, theaters, restaurants, etc., the cooling load resulting from human occupancy is frequently the largest single factor in the total load. Also, many air conditioning systems are installed for the sole purpose of cooling electrical, electronic, and other types of heat-producing equipment. In such cases, the equipment usually supplies the greater, portion of the cooling load.

10-8. Factors Determining the Wall Gain Load

The quantity of heat transmitted through the walls of a refrigerated space per unit of time is the function of three factors whose relationship is expressed in the following equation:

Q = (A) (U) (TD)

(10-2)

Where: Q = the rate of heat transferred in watts (W) A = the outside surface area of the wall (square metres) U = the overall -coefficient of heat transmission in watts per square metre per kelvin.

TD = the temperature differential across the wall in kelvins (K)

The coefficient of transmission or “U" factor is a measure of the rate at which heat will pass through a 1 m2 area Of wall surface from the air on one side to the air on the other side for each 1 K of temperature difference across the wall. The value of the U factor depends on the thickness of the wall and on the materials used in the wall construction. Since it is desirable to prevent as much heat as possible from entering the space and becoming a load on the cooling equipment, the materials used in the construction of cold storage walls should be good thermal insulators so that the value of U is kept as low as is practical.

According to Equation 10-2, once the U factor is established for a wall, the rate on heat flow through the wall varies directly with the surface area of the wall and with the temperature differential across the wall. Since the value of U is given in watts per square metre per kelvin, the heat flow rate through any given wall can be determined by multiplying the U factor by the wall area in square metres and by the temperature difference across the wall in kelvin or degrees Celsius, that is, by application of Equation 102.

Example 10-1 Determine the heat flow rates in watts through a wall 3 m by 6 m, if the (M factor for the wall is.0.37 W/m 2 K and the| temperature on one side of the wall is 4°cr while the temperature on the other side is! 35°C.

Solution:

Total wall area:

= (3 m)(6 m) = 18 m2

Temperature differential across wall, K=35°C-4°C = 31 K Applying Equation 10-2, the heat gain through the wall = (18 m 2)(0.37 )W/m2 K(31 K) = 206 W or 0.206 kW 10-9. Determination of the U Factor

Should it be necessary, the U factor for any type of wall construction can be readily

calculated

provided

that

either

the

conductivity

I

or

the

conductance of each of the materials used in the wall construction is known. The conductivity or conductance of most of the materials used in wall construction can be found in tables. Also, this information is usually

available from the manufacturer or producer of the material. Table 10-1 lists the thermal conductivity or the conductance of some materials frequently used in the construction of cold storage walls. The thermal conductivity or k factor of a material is the rate in watts at which heat "passes through a 1 m2 cross section of the material 1m thick for each kelvin of .temperature difference across the material and is given in watts per metre per kelvin.

Whereas the thermal conductivity or K factor is available only for homogeneous materials and the Value given is always for aim thickness of the material, the thermal conductance or. C factor is available for both homogeneous and nonhomogeneous materials and the value given in watts per square metre per kelvin is for the specified thickness of the material.

For any homogeneous material, the thermal conductance, can be determined for any given thickness of the material by dividing the k factor by the thickness in metres. Hence, for a homogeneous material,

C

K X

(10-3)

where x = the thickness of material in metres.

Example 10-2 Determine the thermal conductance for a 125 mm thickness of polyu- rethane.

Solution: From Table 10-1, k factor of

polyurethane = 0.025 W/m K

Applying Equation 10-3,

c

w mk 0.125m

0.025

c= 0.2

Since the rate of heat transmission through nonhomogeneous materials, such as the concrete building block in Fig. 10-1, will vary the several parts of the material, the C factor for nonhomogeneous materials must be determined by experiment. The resistance that a wall or a material.

Air spaces

Fig. 10-1 Concrete aggregate building block offers to the flow of heat is inversely proportional to the ability of the wall or material to I transmit heat. Hence, the overall thermal resistance of a wall can be expressed as the reciprocal of the overall coefficient of transmission, whereas the thermal resistance of an individual material can be expressed as the reciprocal of its conductivity or conductance that is

overall thermal, resistance R = Thermal resistance of an individual material = or

or

The terms 1/k and 1/C express the resistance to heat flow through a single material from surface to surface only and do not take into account the thermal resistance of the thin film of air which adheres to all exposed surfaces. In determining the overall thermal resistance to the flow of heat through a wall I from the air on one side to the air on the other | side, the resistance of the air on both sides of the wall should be considered. Air film coefficients or surface conductance for average wind velocities are given in Table 10-1.

When a wall is constructed of several layers of different materials the total thermal resistance of the wall is the sum of the resistances of the individual materials in the wall construction, including the air films, that is

=

Therefore:

+

+

+

+

(10-4)

Where = convection coefficient (surface conductance) of inside wall, floor, or ceiling = convection coefficient (surface conductance) of outside wall, floor, or roof Note When nonhomogeneous materials are used, 1/C is substituted for x/k. Example 10-3 Assuming a wind velocity of 3.35 m/s, calculate the value of 1/ for a wall constructed of 200 mm sand aggregate building blocks, insulated with 75 mm of polyurethane, and finished on the inside with 13 mm of cement plaster. Solution: From Table 10-1, 200 mm sand aggregate block Polyurethane Cement plaster Inside convection coefficient Outside convection coefficient G =5.11

=9.37

k =0.025=22.70 k =0.72 Applying Equation 10-4, the over-all thermal resistance, 

1 1 0.075 0.013 1     22.70 5.11 0.025 0.72 9.37

=0.044 + 0.196 + 3.0 + 0.018 + 0.107 = 0.37 Therefore, U = 1/3.37 =0.297 W/m2 K For the most part is the insulating material used in the wall construction that determines the value of U for cold storage walls. The surface conductance and the conductance of the other materials in the wall have very little effect on the value of U because the thermal resistance of the insulating material is so large with relation to that of the air films and other materials. Therefore, for

small coolers, it is sufficiently accurate to use the conductance of the insulating material alone as the wall U factor. As a matter of convenience, overall coefficients of transmission or U factors have been determined for various types of wall construction and these Values are available in tabular form. Table 10-2 lists U factors for typical cold storage walls, roofs, ceilings, and floors. The values listed in the table vary with the type and thickness of the wall insulation. Example 10-4. From Table 10-2, determine the U factor for a wall constructed of 150 mm clay tile with 150 mm of corkboard insulation. Solution. In Table 10-1, the k factor of cork- board is listed as 0.043 W/m K. In Table 10-2, locate the insulation thickness (150 mm) in the column at the left side of the table. Move across the table under the column headed 0.045 mm and read the wall U factor as 0.267 W/m2 K.

A realization of the full insulating values listed for the walls and materials described in the tables is dependent on proper installation and good workmanship. For example, many insulating materials are permeable to water vapor, and unless an adequate and unbroken vapor barrier is installed on the warm side of the insulating material, the higher vapor pressure on the warm side of the insulation will cause water vapor to migrate through the insulation toward the lower vapor pressure on the colder side. At some point in the travel of the vapor through the insulation, the temperature of the vapor will fall below the saturation temperature corresponding to its pressure, thereby causing the vapor to condense into water in the insulation, with the result that the insulation eventually becomes waterlogged. Moreover, in those cases where the temperature in the insulation is below the freezing temperature of the water, the condense vapor

Fig. 10-2 Temperature gradiant through a typical cold storage (see Example 10-4) will subsequently freeze into ice. In either case, since both water and ice are good conductors, the insulating value of the insulating material will be completely destroyed. The temperature gradient through a typical cold storage wall is illustrated in Fig. 10-2. Notice that the temperature drop (difference) across the individual wall components is proportional to the thermal resistance offered by that component and that the overall temperature drop (differential) across the wall is the summation of the individual temperature drops across the several wall components 10-10. Temperature Differential across Cold Storage Walls The design temperature differential across cold storage walls is usually taken as the difference between the inside and outside design temperatures. The inside design temperature is that which is to be maintained inside the refrigerated space and usually depends upon the type of product to be stored and the length of time the product is to be kept in the space.

The recommended storage temperatures for various products are given in Tables 10-8 through 10-11. The outside design temperature depends on the location of the cooler. For cold storage walls located inside a building, the outside design temperature for the cooler wall is taken as the inside temperature of the building. When cold storage walls are exposed to the outdoors, the outdoor design temperature for the region (Tables 10-3A and 10-3B) is used as the outside design temperature. 10-11. Temperature Differential across Ceilings and Floors When a cooler is located inside a building and there is adequate clearance between the top of the cooler and the ceiling of the building to allow free circulation of air over the top of the cooler, the ceiling of the cooler is treat ed the same, as an inside wall. Likewise, when the top of the cooler is exposed to the outdoors, the ceiling is treated as an outdoor wall. The same holds true for floors except when the cooler floor is laid directly on a slab on the ground. As a general rule, the ground temperature under a slab varies only slightly the year round and is always considerably less than the outdoor design dry bulb temperature for the region in summer. Ground temperatures used in determining the temperature differential across the floor of cold storage rooms are given in Table 10-4 and are based on the regional outdoor design dry bulb temperature for winter.

Where the floor of a freezer is laid directly on a slab on the ground, some provision should be made to prevent the migration and eventual freezing of ground water under the freezer floor, since the expansive forces created by such freezing will eventually cause ground heaving and severe damage to the freezer structure. Preventive measures usual include some means of

monitoring the ground temperature under the freezer, along with some means of supplying heat to the surface when ground temperatures under the freezer approach the freezing point. Warm air ducts, electric heating cables, and pipe coils for the circulation of brine or antifreeze solutions have all been employed to supply the heat necessary to maintain the ground, temperature above the freezing point. 10-12. Effect of Solar Radiation Whenever the walls of a refrigerator are so situated that they receive an excessive amount of heat by radiation, either from the sun or from some other hot body, the outside surface temperature of the wall will usually be considerably above the temperature of the ambient air. A familiar example of this phenomenon is the excessive surface temperature of an automobile parked in the sun. The temperature of the metal-surface is much higher than that of the surrounding air. The amount by which the surface temperature exceeds the surrounding air temperature depends upon the amount of radiant energy striking the surface and upon the reflectivity of the surface. Recall (Section 2-18) that radiant energy waves are either reflected by or absorbed by any opaque material that they strike. Light-colored, smooth surfaces will tend to reflect more and absorb less radiant energy than dark, rough-textured surface. Hence, the surface temperature of smooth, light-colored walls will be somewhat lower than that of dark, rough-textured waifs under the same conditions of solar radiation. Since any increase in the outside surface temperature will increase the temperature, differential across the wall, the temperature differential across sunlit walls must be corrected to compensate for the sun effect. Correction factors for sunlit walls are given Table 10-5. These values are added to the normal temperature differential. For walls facing at angles to the directions listed in Table 10-5, average values can be used.

TABLE 10-5 Allowance for Solar Radiation (Degrees Celsius to be added to the normal temperature difference for heat leakage calculations to compensate for sun effect—not to be used for air-conditioning design. 10-13. Calculating the Wall Gain Load In determining the wall gain load, the heat gain through all the walls, including the floor and ceiling, should be taken into account, when the several walls or parts of walls are of different construction and have different U factors, the heat leakage through the different parts is computed separately. Walls having identical U factors may be considered together, provided that the temperature differential across the walls is the same. Too, where the difference in the value of U is slight and/or the wait area involved is small, the difference in U factor can be ignored and the wails or parts of walls can be grouped together for computation. Example 10-5 A walk-in cooler 5 m x 7 m x 3 m high is located in the southeast corner of a store building in an area where the outdoor design DB temperatures in summer and winter are 35°C and -6°C, respectively (Fig. 10-3).

The south and east walls of the cooler are adjacent to and a part of the south and east walls of the store building. The store has a 4 m ceiling so that there

is a 1, m clearance between the top of the cooler and .the ceiling of the store. The store is air conditioned and the temperature inside the store is maintained at approximately 26°C. The inside design temperature for the cooler is 2°C. The north and west (inside) walls, floor and ceiling are insulated with 75 mm of closed-cell (smooth surface) polystyrene, and the south and east walls are insulated with 100 mm of closed-cell polystyrene. Determine the wall, gain load in kilowatts. Solution From Table 10-1, the k factor for smooth surface polystyrene is 0.029 W/m K. From Table 10-2, the U factor for the north and west walls, floor and ceiling (75 mm insulation thickness) is 0.346 W/m 2 K, and the U factor for the south and east walls (100 mm Insulation thickness) is 0.267 W/m 2 K. From Table 10-4, the design ground temperature, based on a winter, outdoor design DB of. -6°C is 25°C. From Table 10-5, the allowance for solar radiation on the south and east walls (to be added to the indoor-outdoor temperature difference) are 2°C and 3°C, respectively. Applying Equation 10-2, North and west walls, and ceiling = (71 m2) (0.346 W/m2 K)(26°C - 2°C) = 589.6 W South wall = 15 m2) (0.267 W/m2 K)(37°C - 2°C) = 140.2 W East wall = (21 m2) (0.267 W/m2.K)(38°C~2"C) = 201.9 W Floor = (35 m2) (0.267 W/m2 K)(25°C - 2°C) = 215.0 W Total wall gain load = 1147 W or 1.147 kW

10-14. Calculating the Air Change Load

The space heat gain resulting from air Ranges in the refrigerated space is difficult to determine with any real accuracy except in pose few cases where a known quantity of air introduced into the space for ventilating purposes. When the mass flow rate of the outside air entering the space is known, the space heat gain resulting from air changes can be determined by applying the following equation: Q = m(h0 - ht)

(10-5)

where: Q=air change load (kilowatts) m=mass of air entering space(kilograms per second) ho=enthalpy of outside air (Kj per kg) hi= enthalpy of inside air (KJ per kg)

However, since air quantities are usually given in units of volume rather than in units of mass, to facilitate calculations, the heat gain per liter of outside air entering the space is listed in Tables 10-6A and 10-6B for various inside and outside air conditions. To determine the air change load in kilowatts, multiply the air infiltration rate in litres per second by the appropriate enthalpy change factor from Table 10-6A or 10-6B.

Example 10-6: The rate of air infiltration into I a refrigerated space is 8 Us. If the inside of the cooler is maintained at 2°C and the out- ; side dry bulb temperature and humidity are 30°C and 50%, respectively, determine the ; air change load in kilowatts.

Solution:

By interpolation in Table 10-6A, the enthalpy change factor is 0.0598 kJ/L.

Air change load

= (8 L/s) (0.0598 kJ/L) =0.478 kw

Except in those few cases where air is purposely introduced into the refrigerated space for ventilation, the air changes occurring in the space are brought about solely by infiltration through door openings. The quantity of outside air entering a space through door openings in a 24-h period depends upon the size, and location of the door or doors, and upon the frequency and duration of the door openings. Since the combined effect of all these factors varies with the individual installation and is difficult to predict with accuracy, it is general practice to estimate the air change quantity on the basis of experience with similar applications. Experience has shown that, as a general rule, the frequency and duration of door openings and, hence, the change quantity, depend on the inside volume of the cooler and the type of usage. Table 10-7 lists the approximate infiltration rate for various cooler sizes. The values given for average usage (see table foot- The ASRE Data Book defines average and heavy usage as follows:

Average usage includes installations not subject, to extreme temperatures and where the quantity of food handled in the refrigerator is not abnormal. Refrigerators in delicatessens and clubs may generally be classified under this type of usage. Heavy usage includes installations such as those in busy markets, restaurant and hotel kitchens where the room temperatures are likely to be high, where rush periods place heavy loads on the refrigerator, and where large quantities of warm foods are often placed in it.

Employing factors from Tables 10-6 and 7, the space heat gain resulting from air infiltration can be determined by the following equation:

Air Change Load = (infiltration rate L/s) (enthalpy change kJ/L)

(10-6)

Example 10-7 A storage cooler has outside dimensions of 4mx5mx3m high. The outside temperature is 25°C and the RH is 50%. The inside of the cooler is maintained at 2°C and usage is average. The walls of the cooler are approximately 150 mm thick. Calculate the air change load in kilowatts. Solution: Since the walls of the cooler are 150 mm thick the inside dimensions of the cooler are 0.3 m less than the outside dimensions, so that the inside volume is (3.7 m x 4.7 m x 2.7 m) 47 m 3. From Table 10-7 by interpolation, the infiltration rate for an inside volume of 47 m 3 is 7.3 L/s. By interpolation in Table 10-6A the air change factor is found to be 0.0451 kJ/L. Equation 10-6. Air change load = (7.3 L/s) (0.0451 kJ/L) 0.329 kW

Applying

10-15. calculating the Product Load

When a product enters a storage space at a temperature above the temperature of the space, the product will give off heat to the space until it cools to the space temperature. When the temperature of the storage space is maintained above the freezing temperature of the product, the amount of heat given off by the product in cooling to the space temperature depends upon the temperature of the space and upon the mass, specific heat, and entering temperature of the product. In such cases, the space heat gain from the product is computed by the following equation:

Q = (m) (G) (AT)

(10-7)

Where: Q = the quantity of heat in kilo joules per kilogram m = mass of the product (kilo grams) c = the specific heat above freezing, kilojoules per kilogram per kelvin. A T = the change in the product temperature (K)

Example 10-8 Thirty-five hundred kilograms of fresh beef enter a chilling cooler at 39°C and are chilled to 7°C each day. Compute the product load in kilojoules. Solution:

From Table 10-8, the specific heat of beef above freezing is 3.14 kJ/kg Applying Equation 10-7, Q = (3500)(3.14)(39°C - 7°C) = 351,680 kJ Notice that no time element is inherent in Equation 10-7 and that the result obtained is merely the quantity of heat the product will give off in cooling to the space temperature. In Example 10-8 the product-is to be cooled over a 24-h period. However, in many instances the desired cooling time may be less than 24 h. Since time is always a consideration in determining the cooling rate, in all cases the product load or product cooling rate is determined by following equation:

Q

(m)(c)(T ) (desired cooling time in seconds)

Example 10-9 Determine the product load in kilowatts assuming that the beef described in Example 10-8 is chilled in 20 h. Solution: Applying Equation 10-8, product load (3500 kg)(3.14 kJ/kg K)(39°C - 7°C) 20(3600 s) = 4.88 kW

10.16. Chilling Rate Factor

During the early part of the chilling period, the product load on the equipment is considerably greater than the average hourly product load as calculated in

the previous examples. Because of the high temperature difference which exists between the product and space air at the start of the chilling period, the chilling rate is higher and the Product load tends to concentrate in the early part of the chilling period (Section 9-23). Therefore, where the equipment selection is based on the assumption that the product load is evenly distributed over the entire chilling period, the equipment selected will usually have insufficient capacity to carry the load during the initial stages of chilling when the product load is at a peak. Consequently, a significant rise in the space temperature can be expected during the early part of the chilling period.

When such a rise is undesirable, a chilling rate factor is sometimes introduced into the chilling load, calculation to compensate for the uneven distribution, of the chilling load. The effect of the chilling rate factor is to increase the product calculation by an amount sufficient to make the average hourly cooling rate approximately equal to the hourly load at the peak condition. This results in the selection of larger equipment, having sufficient capacity to carry the load during the initial stages of chilling.

Chilling rate factors for various products are listed in tables 10-8 through 1011. The factors given in the tables are based on actual tests and on calculations and will vary with the ratio of the loading time to total chilling time. As an example, test results show that in typical beef and hog chilling operations the chilling rate is 50% greater during the first half of the chilling period than the average chilling rate for the entire period. The calculation without the chilling rate factor will, of course, show the average chilling rate for the entire period. To obtain this rate during the initial chilling period, it must be multiplied by 1.5. For convenience, the chilling rate factors are given in the tables in reciprocal form and are used in the denominator of the equation. Thus the chilling rate factor for beef as shown in the table is 0.67 (1/1.5),

Where a chilling rate factor is used, Equation 10-8 is written

Q

(m)(c)(T ) (desired cooling time in seconds)

(10-9)

(chilling rate factor)

Example 10-10 Recalculate the product load described in Example 10-9 employing the appropriate chilling rate factor. Solution: From Table 10-8, the chilling rate factor for beef chilling is 0.67. Applying Equation 10-9, product load = (3500 kg)(3.14 kj/frg K)(39°C - 7°C) 20(3600 s)(0.67) = 7.3 kW Chilling rate factors are usually applied to chilling rooms only and are not normally used in calculation of the product load for storage rooms. Since the product load for storage rooms usually represents only a small percentage of the total load, the uneven distribution of the product load over the cooling period will not ordinarily cause overloading of the equipment and/or unacceptable fluctuations in the space temperature and therefore no allowance need to be made for this condition. 10-17. Product Freezing and Storage When a product is to be frozen, and stored at. some temperature below, its freezing temperature, the heat involved is calculated, in three parts:

1. The quantity of heat given off by the product in cooling from the entering temperature to its freezing temperature. 2. The quantity of heat given off by the product in solidifying or freezing. 3. The quantity' of heat given off by the product in cooling from its freezing temperature to the final storage temperature. The method of determining the quantity of resulting from temperature reduction (1 and 3) has already been established. The quantity of heat resulting from freezing (part2) is calculated from the following equation: Q = (m)(hlf)

(10-10)

where : m = the mass of the product in kilograms hjf = the product latent heat in kJ/kg

The summation of the three parts is divided by the desired processing time in seconds to determine the equivalent product load in kilowatts. Example 10-11: Three hundred kilograms of poultry enter a chiller at 5°C and are frozen and chilled to a final temperature of -15°C for storage in 12 h. compute the product load in kilowatts. Solution: From Table 10-8, Specific heat above freezing = 3.18 kJ/kg K Specific heat below freezing = 1.55 kJ/kg K Latent heat = 246 kJ/kg

Freezing temperature = -3°C

To cool poultry from entering temperature to freezing temperature, applying Equation 10-7 = (300 kg)(3.18 kJ/kg K)[5°C-(-30C)] =7630 kJ To freeze, applying Equation 10-10 = (300 kg)(246 kJ/kg) = 73,800 kJ To cool from freezing temperature to final storage temperature, applying Equation 10-7 = (300 kg)(1.55 kJ/kg K)[(-3°C) - (-15°C) = 5580 kJ Total heat given up by product (summation of 1, 2, and 3) = 87,010 kJ Equivalent product load =87010Q kJ 12(3600 s) = 2.01 kW

10-18. Respiration Heat

Fruits and vegetables are still alive after harvesting and continue to undergo changes while in storage. The more important of these changes are produced by respiration, a process during which oxygen from the air combines with the carbohydrates in the plant tissue and results in the release of carbon dioxide and heat. The heat released is called respiration heat and must be considered as a part of the product load where considerable quantities of fruit and/or vegetables are held in storage at temperatures above the freezing temperature. The amount of heat evolving from the respiration process depends upon the type and temperature of the product. Respiration heat for various fruits and vegetables is listed in Table 10-12.

Since respiration rate is given in watts per kilogram, the product load accruing from respiration heat is found by the following equation.

Q (watts) = Mass of product (kg) x respiration rate (W/kg)

(10-11)

10-19. Containers and Packing Materials When a product is chilled in containers, such as milk in bottles or cartons, eggs in crates, and fruit and vegetables in baskets and lugs, the heat given off by the containers and packing materials in cooling from the entering temperature to the space temperature

must be considered as a part of the product load (see Example 1015). 10-20. Calculating the Miscellaneous Load The miscellaneous load consists primarily of the heat given off by lights and electric motors operating in the space and by people working in the space. The space heat gain from lights in watts is computed by multiplying the wattage of the lights by the number of hours per day the lights are in use divided by 24 hours. Q( watts ) 

wattsxhours 24hours

The space heat gain from motors and from people working in the space is listed in TABLES 10-13 and 10-14, respectively. The following calculations are made to determine the heat gain from motors and occupancy: Q (kW) = Motor output kW x factor (Table hours in use 10-13) x hours in use / 24hours

eq. (10-13)

Q (kW) = No. of people x heat equivalent (kW) (Table 10-14) eq. (10-14)

The heat gain from other miscellaneous heat sources, such as forklift trucks and other handling or processing equipment, should also be taken into account in determining the total rate of heat gains for the refrigerated space. 10-21. Use of Safety Factor The total cooling load is the summation of the heat gain as calculated in the foregoing section. It is common practice to add 5%

to 10% to use as value as a safety factor. The percentage used depends upon the reliability of the information used in calculating the cooling load. As general rule 10% is used. After the safety factor has been added, the cooling load is multiplied by 24 hours and divided by the desired operating time in hours for the equipment to determine the average load (see Section 10-2). The average cooling load is used as a basis for equipment selection.

10-22. Short Method Load Calculations

Whenever possible the cooling load should be determined by using the procedures set forth in the preceding sections of this chapter. However, when coolers are used for general-purpose storage, the product load is frequently unknown and/or varies somewhat from day to day so that it is not possible to compute the product load with any real accuracy. In such cases, a short method of load calculation can be employed which involves the use of load factors which have been determined by experience. When the short method of calculation is employed, the entire cooling load is divided into two parts: (1) the wall gain load and (2) the usage or service load. The wall gain load is calculated as outlined in Section 10-13. the usage load is computed by the following equation:

Usage load (W) = interior volume (m3) x usage factor x TD

(10-15)

Notice that the usage factors listed in Table 10-15 vary with the interior volume of the cooler and are given in watts per cubic metre

per kelvin. Too, an allowance is made for normal and heavy usage. Normal and heavy usage have already been defined in Section 1014. No safety factor is used when the cooling load is calculated by the short method. The calculated cooling load is multiplied by 24 hours and divided by the desired equipment operating time in hours to find the average kilowatt load used to select the equipment (see Example 10-1-3). Since the values listed in Table 10-15 are average values based on experience, some judgment should be exercised in their use.

Example 10-12 A storage cooler 6 m x 4 m x 3.4 m high is insulated with 100 mm of glass fiberboard. Overall wall thickness is approximately 200 mm. The outside temperature is 30°C and usage is average. Twelve, hundred and fifty kilograms of wet mixed vegetables are cooled 25°C to the storage temperature of 5°C each day. Compute the cooling load kilowatts based on a 16-h/day operating time for the equipment. Solution: The outside surface area of the cooler is 116 m 2 and the inside volume is (5.8 m x 3.6 x 3 m) 60.5 m3. From Table 10-1, the k factor of glass fiberboard is 0.036 W/m K. From Table 10-2, for 100 mm insulation thickness and a k factor of 0.035 W/m K, the wall U factor is 0.31 W/m2 K. By interpolation in Table 10-7, the infiltration rate for an interior volume of 60.5 m3 is 8.13 L/s. Assuming an inlet air RH of 50%, the air change factor from Table 10-6A is 0.0536 kj/L.

From Table 10-9, the specific heat of wet mixed vegetables-Is 3.77 kj/kg K. From Table 10-12, an average respiration heat for vegetables is 0.097 W/kg

Applying Equation 10-2,

Wall gain Ioad = (116 m2)(0.31 W/m2 K)(30O C - 5OC) = 899 W or 0.899 kW

Applying Equation 10-6,

Air change load = (8.13 L/s)(0,0536 KJ/L) =0.43 kW Applying Equation 10-8,

Product cooling load

=

(1250 kg)(3.77 kj/kg K)(5 K) 24 (3600 s).

=0.273 kW Applying Equation 10-11, Respiration load = (1250 kg)(0.097 W/kg) = 121.25 W or 0.121 kW

Summation of heat loads = 1.729 kW Safety factor (10%) = 0.173 kW Total cooling load = 1.902 kW Required equipment capacity (kW) = (1.902 kW)(24 h) 16h =2.85 kW

Example 10-13 Recalculate the refrigeration load for the cooler described in Example 10-12 using the short-form calculation. Solution: From Example 10-12, the wall gain load is 0.B99 kW and the interior volume is 60.5m3. By interpolation in Table 10-15, the usage factor for an interior volume of 60.5 m2 is 0.642 W/m3 K

Applying Equation 10-15, Usage load = (60.5 m3)(0.642 W/m3 K) (30°C - 5°C) = 971 W or 0.971 kW Summation heat loads = 1.87 kW Equipment load (kW) = (1.87kW)(24h) (16 h) = 2.81 kW

Example 10-14: A holding cooler 6m x 10 m x 4 m high is to be used for the short- term storage of fresh beef. Thirty four hundred kilograms of beef enter the cooler at 7°C and are cooled to the storage temperature of 2°C each day. All the walls are partitions adjacent to unconditioned spaces (30°C and 50% RH) except the east wall (6 m x 4 m), which is adjacent to a chilling room maintained at the same inside design temperature. Wall-construction is 100 mm cinder block insulated with 100 mm corkboard equivalent. The floor, located over an unconditioned space, is a 125 mm concrete slab insulated with 100 mm corkboard equivalent and finished with 75 mm of concrete. The ceiling, situated under an unconditioned space, is a 100 mm concrete slab with wood sleepers and insulated, with the equivalent of 100 mm of corkboard. Two people work in the space during the loading periods (4 h), usage is average, the lighting load is 500 W and the lights are in use 4 h/day. Determine the required equipment capacity in kW based on a 20 h operating time.

Solution: All of the walls including floor and ceiling are insulated with 100 mm of insulation having a k factor of approximately 0.043 W/m K (Table 10-1). From Table 10-2, the wall U factor, based on 100 mm insulation thickness and k factor of 0.045, is 0.383 W/m2 K. The outside surface area of the cooler, excluding the east wall, which is adjacent to a chilling room maintained at the same temperature as the storage cooler, is 224 m 2. The inside volume is (5.6 m x 9.6 m x 3.5 m)

188

ft.' From Table 10-7, the air infiltration rate is 13.56 L/s

based on an inside volume of 190 m 3. By interpolation in Table 10-6A, the air change factor is 0.0598 kJ/L From Table 10-14, the heat gain per person is 261 W. From Table 10-8, the specific heat of fresh beef is 3.14 kJ/kg K.

Applying Equation 10-2, Wall, gain load m =(224 m2)(0.383 W/m2 K)(30°C - 2°C) = 2402 W or 2.402 kW Applying Equation 10-6, Air change load = (13.56 L/s)(0.0598 kJ/L) =0.810 kW Applying Equation 10-8, Product load =(3400kg)(3.14kJ/kgK)(7°C-2°C) 24(3600 s) = 0.618 kW

Miscellaneous load: Light = (500 W)(4 h) (24 h) = 83 W or 0.083 kW People = (2)(261 W)(4 h) (24 h) = 87 W or 0.087 kW Summation of heat loads = 4.0 kW Safety factor (10%) = 0.4 kW Total cooling load = 4.4 kW

Required equipment capacity (kW) =(4.4 kW)(24 h) (20 h) = 5.28 kW Example 10-15 Three thousand lug boxes 1L of apples are stored at .2°C in a storage p cooler 16 m x 12 m x 3.4 m high. The apples enter the cooler at a temperature of 30°C and at the rate of 200 lugs per day each, day for the 15day harvesting period. The walls including floor and ceiling are constructed of 25 mm boards on both sides of 50 mm x- t00 mm studs and are insulated with 100mm of mineral wool. All of the walls are shaded and the ambient temperature is 30°C. The average weight of apples per lug box is 27 kg. The lug boxes have an average weight of 2 kg and a specific heat value of 2.5 kJ/kg- k. The lighting load is 500 W for 3 h per day. Two people and one battery operated lift truck (4.17 kW) are in the space 3 h per day. Determine the average load in kilowatts on the equipment based on a 16 h per day equipment operating time.

Solution: The load calculation is made for the maximum load, which occurs on the fifteenth day. The outside surface area is 574 m 2 and the inside volume is (15.7 m x 11.7 m x 3.1 m) 570 m3. The insulation k factor (Table 10-1) is 0.039 W/m K and is increased to 0.045 W/m K to compensate for the wood studs. From Table 10-2, the wall U factor, based on 100 mm insulation thickness and a k factor of 0.045 W/m K, is 0.383 W/m 2 K. From Table 10-7, the air change rate, assuming heavy usage is 34.41 Us. From Table 1.0-6A, the enthalpy change factor (assuming 50% RH) is 0,0598 kJ/L. From Table 10-10 the

specific heat of apples is 3.72 kJ/kg K and the chilling rate factor is 0.67. From Table 10-12, the respiration heat of apples at 2°C is 0.015 W/kg! From Table 10-14, the heat gain per person is 261 W. Applying Equation 10-2, Wall gain load = (574 m2)(0.045 W/m2 K)(30°C - 29C) = 723 W or 0.723 kW Applying Equation 10-6, Air change load =(34.41 Us)(0.0598 kJ/L) = 2.058 kW Applying Equation 10-9, Product cooling Apples =(2D0 x 27kg)(3.72kJ/kgK).(30°C- 2°C) 0.67 = 839,499 kJ

Boxes (200 x 2 kg)(2.5 kJ/kg K)(30°C - 2°C) 0.67 = 41,791 kJ. Total product load = 839,499 kJ + 41,791 kJ 24(3600 s) = 10.2 kw Applying Equation 10-11,

Respiration load = (3000 x 27 kg) (0.015 W/kg) = 1215 W or 1.215 kW Miscellaneous load: Lights = (500WH3h)/24 h = 62.5 W or 0.063 kW People = (2)(261 W)(3 h)/24h = 65.25 W or 0.065 kW Lift truck = (4.17kw)(3h) 24h = 0.521kw

Summation of heat gains = 14.845 kW Safety factor (10%) = 1.485 kW Total cooling load = 16.33 kW Required equipment capacity =(16.33 kW)(24h) 16h =24.5Kw

Regarding Example 10-15, if larger fluctuations in the storage temperature are acceptable, the chilling rate factor can be neglected, in which case the product cooling L load will be reduced by 33% and the required equipment capacity by 22.7%, from 24.5 kW to 18.94 kW. As another alternative, if the chilling rate factor is used, the respiration load could be neglected in that all of the products load, except the respiration load, will disappear on the sixteenth day.

Example 10-16: Ten thousand kilograms i of dressed poultry are blast-frozen on hand v trucks each day (24 h) in a freezing tunnel 4 m x 3 m x 3.5 m high

(see Fig. 10-4). The poultry is precooled to 7°C before entering the freezer where it is frozen and its temperature lowered to -20°C for storage. The lighting load is 200 watts and the lights are on h per day. The hand trucks carrying the poultry total 700 kg per day and have | specific heat of 0.47 kJ/kg K. The north and east partitions adjacent to the equipment room and vestibule are constructed of 150 mm clay tile insulated with 150mm polyurethane. The south and west partitions adjacent to storage cooler are 100 mm clay tile with 50 mm polyurethane insulation. The roof is a 150 mm concrete slab insulated with 150 mm polyurethane and covered with tar, felt, and gravel. The floor is a 150 mm concrete slab insulated with 150 mm polyurethane and finished with 100 mm of concrete. The floor is over a ventilated crawl space. Roof is exposed to the sun. The equipment room is well ventilated so that the temperature inside is approximately the outdoor design temperature for the region (33°C). The inside design temperature for both the storage room and the freezer is -20°C. The vestibule temperature and relative humidity are 10°C and 70%, respectively. Determine the average hourly refrigeration load based on 20 h per day operating time for the equipment.

Solution: The combined outside surface area of the floor and north and east partitions is 36.5 m2. The area of the roof is 12 m 2. The interior volume is 25 m3. From Table 10-1, the k factor for polyurethane insulation is 0.025 W/m K. From Table 10-2, the wall U factor is 0.153 W/m2 K. The roof sun factor is 11°C (Table 10-5). From Table 10-7, the air change rate, based on 25 m 3 interior volume and vestibule entrance, is (4.2 x 0.5). 2.1 Us. The enthalpy change factor (Table 10-6B), based on an entering air temperature of 10°C and 70% RH, is 0.0456 kJ/L. From Table 10-8, the specific heats of poultry above and

below freezing are 3.18 kJ/kg K and 1.55 kJ/kg K, respectively, the latent heat is 246 kJ/kg and the freezing temperature is -2.75°C. Applying Equation 1.0-2,

Transmission gain Walls and floor = (36.5 m2)(0.153 W/m2 K) [33°C - (-20°C)] = 296 W or 0.296 kW Roof = (12 m2)(0.153 W/m2 K) [33° + 11°C-(- 20°C)] =117.5 W or 0.118 kW Applying Equation 10-6, Air Change load = (2.1 Lys) (0.0456 kJ/L) = 0.096 kw Applying Equation 10-7, Product cooling load = (10,000 kg) (3.18 kJ/kg K)[7°C - (-3°C)] - 318,000 kJ = (10,000 kg) (1.55 kJ/kg K) [(-3°C) - (-20°)] = 263,500 kJ Trucks =(700 kg)(0.47kJ/kgK)[7°-(--20°C)] =8883 KW Applying Equation 10-10, Freezing load = (10,000kg)(246 kJ/kg) =2,460,000 kJ Total

= 3,050,383 kJ

product load

24(3600 s)

=35.3kW Miscellaneous load Lights =

(200W)(16 h) 24h

= 133.3 W or 0.133 kW Summation of heat loads = 35.43 kW Safety factor (10%) = 3.54 kW Total heat gain = 38.97 kW Required equipment capacity = (38.97 kW)(24 h) (20 h) = 46.76 kW

Example 10-17 Twenty three hundred litres of partially frozen ice cream at -4°C are entering a hardening room 3 m x 5 m x 3 m high each day. Hardening is completed and the temperature of the ice cream is lowered to -28°C in 10 h. The walls, including floor

and ceiling, are insulated with 150

mm of polyurethane and the overall thickness-of the walls is 250 mm. The ambient temperature is 32°C and the RH is 50%. Assume the average density of ice cream is 0.6 kg/L, the average specific heat below freezing is 2.1 kJ/kg K,

and the average latent heat per kg is 233 kJ. Determine the average

hourly load based on 18 h operation.

Solution: The outside surface area is 78 m 2 and the interior volume is (2:5 m x 4.5 m x 2.5 m) 28 m3. The insulation k factor is 0.025 W/m K (Table 10-1). The wall U factor (Table 10-2), based on 150 mm insulation thickness and k factor of 0.025 W/m k, is 0.153 W/m 2 K. By interpolation in Table 10-7, the air change rate Ms 4.44 L/s. By interpolation in Table 10-6B, the enthalpy change factor is 0.1104 KJ/L).

Applying Equation 10-2, Wall gain load = (78 m2)(0.153 W/m2 K) [32° - (-28°)] = 716 W or 0.716 kW Applying Equation 10-6, Air change load = (4.44 L/s)(0.1104 kJ/L) = 0.490 kW

Applying Equation 10-7, Product cooling = (2300 L x 0.6 kg/L)(2.1 kJ/kg K) [(-4°) - (-28°C)] = 69,550 kJ Applying Equation 10-10 Product Freezing = (2300 L x 0.6 kg/L)(233 kJ/kg) = 321,540 kJ

Total product Load

=

391,090 kJ 24  3600 s 

= 4.53 kW Summation of heat

= 5.74 kW

Safety factor (10%)

= 0.57 kW

Total heat gain

= 6.32 kW

Required equipment capacity

=

(6.32kW)(24 h) = 8.43 kW (18h)

Example 10-18 A cooler 4m x 3m x 3m high equipped with 6 x 0.7 m x 1 m triple-glass doors is used for general purpose storage in a grocery store. The cooler is maintained at 2°C and the service load is heavy. The walls are insulated with the equivalent of 100 mm of corkboard and the ambient temperature is 25°C. Determine the cooling load in kW based on a 16 h operating time.

Solution : The glass area is 4.2 m2 (6 x 0.7 x 1) and the net wall area is 61.8 m 2 (664.2). The inside volume is approximately 40.4 m 3. From Table 10-2, the wall U factor, based on 100 mm insulation thickness and a k factor of 0.045 W/m K, is 0.383 W/m2 K. From Table 10-1, the conductance factor for triple-pane glass is 1.65 W/m2 K. Assuming heavy usage, from Table 10-15, the usage factor is approximately 1.18 W/m3 K. Applying Equation 10-2, Wall gain load = (61.8 m2)(0.383 W/m2 K)(25°C-2°C) = 544 W of 0.544 kW Glass = (4.2 m2)(1.65 W/m2 K)(25°C - 2°C) = 159 W or 0.159 W Applying Equation 10-15, Usage load = (40.4 m3)(1.18 W/m3 K(25°C -2° C) =1096 W or 1.096 kW Total heat load = 1.799 kW

( 1.799kW)(24  (16 h)

h)

 2.7kW

Problems 10-1 Determine the value of C for a 125 mm thickness of an insulating material having a k value of 0.043 W/m K. 10-2 Assuming a wind velocity of 12 km/h, compute the value of U for a wall constructed of-200 mm hollow clay tile insulated with 150 mm of polyurethane foam and finished on the inside with 13 mm of cement plaster. 10-3 A cooler wall 3 m by 8 m is insulated with the equivalent of 75 mm of expanded smooth cell polystyrene. Compute the heat gain through the wall in kilowatts if the inside temperature is 2°C and the outside temperature is 25°C. 10-4 A reach-in cooler is equipped with eight triple-pane glass doors, each measuring 0.8 m by 0.6 m. Compute the conduction heat gain-through the doors in watts if the temperature difference between the inside and out- side is 22 K. 10-5 The north wall of a cold storage warehouse is 4 m by 18 m and is constructed of 200 mm hollow clay tile insulated with 150 mm of corkboard. The outdoor design temperature is 35°C and the inside design temperature is -25°C. Determine the rate of heat gain through the wall in kW. 10-6 A cold storage warehouse, located where the outdoor design temperature is 32°C, has a 10 m by 16 m flat roof constructed of 100 mm of concrete covered with tar and gravel and insulated with the equivalent of 100 mm of corkboard. If the roof is unshaded and the inside of the warehouse is maintained at 2°C, compute the rate of heat gain through the roof in kilowatts.

10-7 The floor of the cold storage room described in Problem 10-6 consists of a 50 mm concrete slab insulated with the equivalent of 75 mm of cork- board and finished with 50 mm of concrete. If the outdoor design temperature in winter is 0°C, determine the rate of heat gain through the floor in kilowatts. 10-8 A frozen storage room has an interior volume of 75 m 3 and is maintained at a temperature of -25°G. The usage is light, and the outside design conditions (anteroom) are 10°C and 70% RH. Compute the air infiltration load in kilowatts. 10-9 Twenty five hundred kilograms of fresh lean beef enter a chilling cooler at 38°C and are chilled to 3°C in 24 h Compute the chilling load in kilowatts. 10-10 Eighteen hundred kilograms of green beans packed in baskets (15 kg per basket) enter a chilling cooler at a temperature of 27°C and are chilled to a temperature of 1°C in 20 h. The empty baskets have a mass of 1.4 kg and a specific heat of 2.5 kJ/kg K. Compute the product load in y kilowatts. 10-11 Three hundred kilograms of pre-pared, packaged beef enter a freezer at a temperature of 2°C. The beef is to be frozen and its temperature reduced to -18°C in 5 h. Compute the product load. 10-12 Fifty-five hundred crates of apples are in storage at 3°C. An additional 500 crates enter the storage cooler at a temperature of 29°C and are chilled to the storage temperature in 24 h. The average mass of apples per crate is 27 kg. The Crate has a mass of 4.5 kg and a specific heat of 2.5 kJ/kg K. Determine the product load in kilowatts. 10-13 A walk –in cooler 3 m by 5 m by 3m High equipped with twelve 0.6 m by 0.6 m triple-glass doors is used for general purpose storage in a drive-in market (heavy usage). The walls are insulated with 50 mm of expanded cut cell polystyrene, and

the cooler is to be maintained at 2°C. Compute the cooling load in kilowatts based on a 16-h operating time if the ambient temperature is 25°C.