chapter 6a pdf

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Heating Ventilating and Air Conditioning

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Heating, Ventilating, and Air

Conditioning

Heating,Ventilating,and

AirCondi

tioning

Department of Mechanical Engineering Technology

Yanbu Industrial CollegeMET

D E

P A R T M E

N T

M E C H A N

I C A

L

E N G INEE R I N G

T E C

H N O L O G Y


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Chapter6

Cooling and Heating Load Calculations

4.1- Introduction

Before the cooling and heating load determination process starts, the customerhas to be contacted effectively in order to define all technical specification andconditions under which the plant must run. This information then is used to designthe plant. The determination of the cooling load is an early step in the design.The cooling load will be used in later calculations to determine the selection of themechanical refrigeration or AC plant.

Most customers carry out a cooling load calculation to see weather their ownestimate of the load matches with those in the various tenders. Thus it isimportant to both customers and contractors to be able to evaluate cooling loadsaccurately. If the estimate made is too conservative an excessively large and

expensive plant will result. If the load is underestimated the plant may not performto specification.

In this section the technical calculations for cooling load determination will bedescribed. Most of these follow the traditional approaches available in a numberof textbooks. There are also some update methods not found in textbooks, butrecommended by Professor Donald Cleland (Vice President of the InternationalInstitute of Refrigeration) as being more accurate.

4.2- Cooling/Heating Load Definition

The rate at which heat must be removed from the refrigerated or AC space ormaterial in order to produce and maintain the desired temperature conditions iscalled cooling load, the refrigeration load or the heat load.

4.3- Cooling /Heating Load main Sources

Cooling/heating Load is the summation of the heat, which usually evolves fromseveral different sources. Some of the more common sources of heat that supplythe load on refrigerating/AC equipment are:

1. Heat that leaks into the refrigerated or AC space from the outside byconduction through the insulated walls (The Wall Gain Load).

2. Heat that is brought/ taken out into/from the space by warm/cold outside airentering the space through open doors or through cracks around windows anddoors (Air Infiltration/Interchange/Ventilation Load).

3. Heat that enters the space by direct radiation through glass or other transparentmaterials (Effect of Sun Radiation).

4. Heat given off by warm products (The Product Load,).

5. Heat given off by the people occupying the space and by equipment locatedinside the space such as lights, electronic equipment (TV, VCR, computers,….), fork lift, …etc. (The Miscellaneous Load)


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4.4- Equipment Running Time

In any refrigeration application using air, frost will accumulate if the evaporatorsurface is below 0 oC. Periodic defrosting of the evaporator(s) is necessary tomaintain the heat transfer and air flow performance of the evaporator(s) and fans.Therefore it is not practical to design the refrigerating system in such a way thatthe equipment must operate continuously in order to handle the load in mostcases.

Experience has shown that when defrosting is required, the maximum allowablerunning time is usually between 16-20 hrs out of each 24 hours period. Then thetotal cooling load should be multiplied by (24hrs/maximum allowable running time(RT )) in order to find the required system capacity (Qs.c.).

Load Cooling Total RT

hrsQ c s

24..

If maximum allowable running time=18 hrs, then the cooling load should bemultiplied by 24/18, which in fact increases the required system capacity of 33.3%so that the selected equipment handle the 24 hrs cooling load in the required 16hrs of operation time.

4.5- Cooling and Heating Load Calculations

To simplify cooling load calculations, the total cooling load is divided into a numberof individual loads according to the sources of heat supplying the load. Thesummation of these individual loads is the total cooling load on the equipment.

4.5.1-The Wall Gain Load

The wall gain load, sometimes called the wall leakage load, is a measure of theheating flow rate by conduction through the walls of the refrigerated space fromoutside to the inside or from inside to outside of AC space in winter. The wall gainload is common to all refrigeration and AC applications and is ordinarily aconsiderable part of the total cooling load. Commercial storage coolers andresidential A/C applications are both example of application wherein the wall gainload often accounts for the greater portion of the total load. Some exceptions tothis is liquid chilling applications, where the outside area of the chiller is small andthe walls to the chiller are well insulated.

Factors determining the wall gain

The quantity of heat transmitted through the walls of refrigerated space per unit oftime is the function of three factors whose relationship can be expressed in thefollowing equation.

wQ = A U T

Where wQ : the rate of heat transfer in Watts (W).

A: the outside surface area of the wall (m

2

).


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U : the overall coefficient of heat transmission in W/m2.K.

T : the temperature differential across the wall

The value of U factor depends on the thickness of the wall and on the materials

used in the wall construction.

Determining of U factors

The U factor for any type of wall construction can be calculated and provided foreach of the material used in the wall construction by knowing the thermalconductivity (k ) of the thermal conductance and the heat transfer coefficients ofoutside and inside air, ho and hi . The thermal conductivity of most homogenousmaterials used in wall construction and heat transfer coefficients can be tabulatedin Tables6.1 and6.2 of Appendix A. For any Cold store insulation material, thethermal overall heat transfer coefficient is tabulated in Table6.3 of Appendix A.

The overall U factors can be expressed as,

i

n

1 j j

j

o

th

h

1

k

x

h

1

1

R

1U

where

ho outside air heat transfer coefficient, W/m2 K, see Table6.2. of Appendix A

hi inside air heat transfer coefficient, W/m2 K , see Table6. 2. of Appendix A

x insulation material thickness, m

Temperature Differential across Cold Storage Walls

The design temperature differential across cold store walls is usually taken as thedifference between the inside and outside design temperatures.

The inside design temperature is that which is to be maintained inside therefrigerated space and usually depends upon the type of product to be stored andthe length of time the product is to be kept in the space. The recommendedstorage temperature for various products is given in Tables6.5 –6.8 of Appendix A.

The outside design temperature depends on the location of the cooler.

The differential across ceilings and floors

Ceiling

The design temperature of the ceiling depends on the cooler ceiling location.When the cooler is located inside a building and there is an adequate clearancebetween the top of the cooler and the ceiling of the building to allow freecirculation of air over the top of the cooler, the ceiling is treated the same as aninside wall. Likewise when the top of the cooler is exposed to the outdoors, theceiling of the cooler is treated as an outdoor wall.

Floor

The design floor temperature depends on the floor design configuration. When

the floor of a cooler is laid directly on a slab on the ground, the ground


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temperature under the slab varies only slightly all year round and is alwaysconsiderably less than the outdoor design dry bulb temperature for the region insummer.

When the floor of a freezer is laid directly on a slab on the ground, some provisionshould be made to prevent any damage to the floor slab and creation any frozenwater layer on the freezer floor which will be dangers to manpower and themachine. Preventive measures usually include warm air ducts, electric heatingcables, and pipe coils for the circulation of brine or antifreeze solutions (Glycol).

4.5.2-The Air Infiltration/Interchange Load

Air infiltration/interchange is the term used to describe the replacement of cold airin refrigerated space by the warm air from the outside. Whilst there usually isleakage through door seals, some infiltration due to pressure equalized systemand sometimes deliberate infiltration when cool storage ventilation is required,

generally, the bulk of infiltration occurs during door openings.The door entering the refrigerated space will lose both sensible heat in cooling,and latent heat which is released on condensation or frosting of some of the watervapour in the incoming air. Air infiltration is generally a greater problem in coldand cool stores than in chillers and freezers.

The space heat load gain resulting from air changes in the refrigerated space isdifficult to determine with any real accuracy except in those few cases where aknown quantity of air is introduced into the space for ventilating purposes. Whenthe mass flow rate of the outside air entering the space is known, the space heatgain resulting from air infiltration can be determined by applying the following

equation:

iaQ , = am (ho – hi ) x 10

3 = am c pa T

Where

iaQ , air infiltration load (W)

ho: enthalpy of outside air (kJ/kg)

hi: enthalpy of inside air (kJ/kg)

am : mass flow rate of air (kg/s)

However, since air quantities are usually given in units of volume rather than inunits of mass, to facilitate calculations the heat gain per liter of outside air enteringthe space is listed in Tables6.12A and6.12B of Appendix A for various inside andoutside air conditions. To determine the air infiltration rate in liters per second bythe appropriate enthalpy change factor from Tables6.6A or 6.6b of Appendix A.

Example 6.1

The rate of air infiltration into a refrigerated space is 10 /s. If the inside of the

cooler is maintained at 5 oC and the outside dry bulb temperature and relativehumidity are 30 oC and 60% respectively. Determine the air change load in kW.


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Solution

From Table 6.12A of Appendix A, the enthalpy change factor = 0.061 kJ/liter

Then, the Air infiltration Load =10*0.061 =0.61 kW

Average air Infiltration Rate

The quantity of outside air entering a refrigerated space through door openings ina 24 hrs periods depends upon the size, and location of the door(s) and upon thefrequency and duration of the door openings. Since the combination effect of allthese factors vanes with the individual installation and is difficult to predict withaccuracy, it is general practice to estimate the air change quantity on the basis ofexperience with similar application. Experience has shown that, as a general rule,

the frequency and duration of door openings and , hence the air change quantity,depends on the inside volume of the cooler and the type of usage. Table 6.11 of Appendix A lists the approximate infiltration rate s for various cooler sizes. Thevalues given are for average usage

Example 6.2

A storage cooler has outside dimensions of 6.5m X 6m X 3.45m. The outsidetemperature is 30 oC and RH 60%. The inside of the cooler is maintained at 5 oC.The walls of the cooler are approximately 150mm thick. Calculate the airinfiltration Load.

Solution

Since the walls thickness are 150 mm, the inside dimensions of the cooler are0.3m less then the outside dimensions, so the inside volume is (4.2 X 5.7 X 3.15)75.411 m3 say 75 m3. From Table 6.11 of Appendix A, the infiltration rate is 9l/sec .

Air infiltration Load = Infiltration rate * Enthalpy Change

= 9 * 0.061 = 0.549 kW

Air Interchange load

The instantaneous cooling load due to air interchange

)W ( 10v

1 )hh(

606024

V nQ 3

o

io s

ic

Where

n Number of air changes per 24 hours (see Table 6. 13 of Appendix A)

V s Store volume (m3)

vo Specific volume of fresh air m3/kg (from psychrometric chart at outside

conditions)

ho Enthalpy at the outdoor conditions, see the psychometric chart


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hi Enthalpy at the indoor conditions, see the psychometric chart

Ventilation load

(W)10v

1hh10

Person

snQ 3

o

io

3-

pv

where

n p Number of people and

Person

s from Table 6. 14 of Appendix A.

4.5.3-Effect of Sun Radiation

Whenever the walls of a refrigerator are so situated that they receive an excessiveamount of heat by radiation, either from the sun or from any hot body, the outsidesurface temperature of the wall will usually be considerably above the temperatureof the ambient air. The amount by which the surface temperature exceeds thesurrounding air temperature depends upon the amount of radiant energy strikingthe surface and upon the reflectivity of the surface. The energy waves are eitherreflected by or absorbed by any opaque material that they strike. Light colored,smooth surfaces will tend to reflect more and absorb less radiant energy thandark, rough textured surfaces. Hence, the surface temperature of smooth, lightcolored walls will be somewhat lower than that of dark, rough-textured walls underthe same conditions of solar radiation. Since any increase in the outsidetemperature will increase the temperature differential across sunlight walls must

be corrected to compensate for the sun effect. Correction factors for sunlight wallsfor cold store the temperature difference due to sunlight, sT , can be calculated

directly from Table 6. 16 A of Appendix A, while for air conditioning can becalculated as,

o

sh

I T

15.1

Where is the surface absorption factor, see Table 6. 15 of Appendix A.

I is the sunlight intensity (W/m2), see Table 6. 16 B of Appendix A for airconditioning.

ho is the heat transfer coefficient (W/ m2 K), see Table 6. 2 of Appendix A.

The sT values should be added to the normal wall temperature differential.

4.5.4-Windows Load

Conduction load

The heat transfer across the windows glass can be expressed as,

T U AQ g g

Where


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A g is the window glass area m2.

U is the window glass overall heat transfer coefficient, see Table 6.17of Appendix A.

T : the temperature differential across the window which normally isequals to the temperature differential across the wall.

Direct sun load

The windows glass direct sun load is

SC Q sun ggs A,Q

Where

sunQ is the heat gain per unit area ( W/m2), see Table 6. 18 A of Appendix A

SC is the Shading Coefficient, see Table 6. 18 B of Appendix A.4.5.5-Calculating the product Load

The heat that transfers from a product when it enter a storage space can becomputed by the following equation:

Q p = ( m ) (C ) T

WhereQ p the quantity of heat in kJ per kg.

m mass of the product (kg)

C the specific heat before freezing kJ/kg K(Can be obtained from

Tables 6.5-6.8 of Appendix A).T the change in the product temperature (K).

Example 6.3

Five ton of fresh beef enter a chilling cooler at 40 oC and chilled to 6 oC each day .Calculate the product land in kJ.

Solution

From Table 6.7 of Appendix A the specific heat (C) of beef above (before) freezingis 3.14 kJ/kg K.

Then

Q p

= 5000 X 3.14 X (40-6) = 533,800 kJ

As we can see from the above equation has no time element. Since time isalways considered in determining the cooling rate, then the above equation can bewritten as follows:

ondsintimecooling desired

T C mQ p

sec

Then if we assumed the beef described in above example is chilled in 20 hrs.


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pQ = [5000 X 3.14 X (40-6)] /(20 X 3600) = 7.414 kW

Chilling Rate Factor

During the early part of the chilling period, the product load on the equipment isconsiderably greater than the average hourly product load. Because of the hightemperature difference which exists between the product and the space air at thestart of the chilling period, the chilling rate is higher and the product load tends toconcentrate in the early part of the chilling period. Therefore, where theequipment selection is based on the assumption that the product load is evenlydistributed over the entire chilling period, the equipment selected will usually haveinsufficient capacity to carry the load during the initial stages of chilling when theproduct load is at a peak. Consequently, a significant rise in the spacetemperature can be expected during the early part of the chilling period.

When such a rise is undesirable, a chilling rate factor is sometimes introduced intothe chilling load calculation to compensate for the uneven distribution of thechilling load.

The effect of the chilling rate factor is to increase the product load calculation byan amount sufficient to make the average hourly cooling rate approximately equalto the hourly load at the peak condition. This results in the selection of largerequipment. Then the products cooling rate can be expressed as follows:

Chilling rate factors for various products are listed in Tables 6.5 through 6.8 of Appendix A.

)()sec( factor rateChilling ondsintimecooling desired T C mQ p

Example 6.4

Recalculated the product load described in the previous example employing theappropriate chilling rate factor.

Solution

From Table 6. 7 of Appendix A, the chilling rate factor for beef is 0.67. TheProduct load

pQ = [5000 X 3.14 X (40-6)] /(20 X 3600 X 0.67) = 11.065 kW

Product Freezing and Storage

When a product is to be frozen and stored at some temperature below its freezingtemperature, the heat involved is calculated in three parts:

1. The quantity of heat given off by the product in cooling from the enteringtemperature to it freezing temperature.

2. The quantity of heat given off by the product in cooling from its freezing

temperature to its final storage temperature


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3. The quantity of heat given off by the product in solidifying or freezing orfreezing.

The method of determining the quantity of heat resulting from temperaturereduction (parts 1 and 2) has already been established. The quantity of heatresulting from freezing (part 3) can be calculated from the following equation:

Q f = ( m ) ( h )

Where

Q f the quantity of heat given off by product solidification or freezing kJ /kg.

m mass of the product (kg)

h the product latent heat in kJ/kg.

The latent heat for various products are listed in Tables 6.5 through 6.8 of

Appendix A.Example 6.5

Five thousand kilograms of poultry enter a chiller at 5 oC and are frozen andchilled to a final temperature of –15 oC for storage in 12 h. Determine the productload.

Solution

From Table 6. 7 of Appendix A

Specific heat above (before) freezing = 3.18 kJ/kg.K

Specific heat below (after) freezing = 1.55 kJ/kg.K Latent heat = 246 kJ/kg

Freezing temperature = -2.75 oC

Chilling rate factor =1

1. To cool poultry from entering temperature to freezingtemperature, applying

Q p,1 = ( m ) (C ) T

= (5000)(3.18)[5-(-2.75)] = 123225 kJ

2. To freeze, applying

Q f = ( m ) (hl )

= (5000)(246) = 1230000 kJ

3. To cool from freezing temperature to final storagetemperature, applying

Q p,2 = ( m ) (C ) T

= (5000)(3.18)[-2.75-(-15)] = 194775 kJ


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Where

Q : Heat load due to light (W)

N l : Number of light

w : Wattage of each light (W)

b- People

peQ = t N pe x Percent of present time (i.e 8/24)

Where t is the total human heat (sensible + Latent), see Tables 6.20 and 6.21of Appendix A

peQ : Heat load due to people (W)

N pe: Number of people

c- Mechanical devices

md Q = Z N m m x used time/24

Where N m: Number of motors of equal rating

m Nominal motor rating (W)

Z Motor rating correction factors which is often about 0.9.

The total miscellaneous load is equal to the summation peQ + peQ

+ md Q

4.5.7- Defrost heat Load

There are two load-related effects of defrost-extra heat added to the refrigerated

space that is not removed with the melt water, and the loss of refrigeration effectwhile a coil is being defrosted.

The amount of heat required to melt the frost can be directly related to the latentheat load component if a typical melt temperature of about 5 oC is assumed. Theaverage heat load to defrost can be calculated using:

)1(13.0

d

d lat d QQ

Where d Q Average heat load due to defrost (W)

lat Q Total average latent heat component (W)

d Defrost efficiency. It is suggested that defrosting efficiencies of 20%for hot gas and 30% for water

4.5.8-Fan heat load

Fan energy is a major source of energy input for refrigerated facilities using air asthe cooling medium. For facilities using liquids or refrigerants as the coolingmedium pumps are the equivalent sources.

The best estimate of fan energy use is given by


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f

f

fan N P V

Q

where fanQ fan heat load (W)

V Volumetric flow rate (m3/s)

P Pressure drop in facility or pressure boost by fan (Pa)

f Combined fan and fan motor efficiency

f Number of fans of equal ratings

4.6-Short Method Load Calculations

The cooling load can be determined by using the procedures se forth in thepreceding sections. However, when coolers are used for general-purpose

storage, the product load is frequently unknown and/ or varies somewhat from dayto day so that it is not possible to compute the product load with any real accuracy.In such cases, a short method load calculation can be employed which involvesthe use of load factors (Table 6. 22 of Appendix A) which have been determinedby experience. When the short method of calculation is employed, the entirecooling load is divided into two parts.(1) the wall gain load and (2) the usage orservice load.

The wall gain load is calculated as outlined in previous section. The usage load iscomputed by the following equation.

Usage Load (W) = Interior Volume (m3) x usage factor x T

4.7- Safety Factor

The total cooling load is the summation of the heat gains as calculated in theforegoing sections. It is a common practice to add (5%-15%) to this value as asafety factor. As a general rule 10% is used.

Home Work

1- Each day a chiller cools 16 tonne batch of cheese blocks (C=3.25 kJ/kg K)from 25 C to 7 C. Calculate the chilling load.


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2-Each day 24 tonne of peas at –12oC are loaded into a cold store –18oC. Thefrozen specific capacity is 2.1 kJ/kg. K . What is the product cooling load.

3- A cold store operating at –20 C has 8 fans with an overall efficiency of 60%.Each fan produces 10,000 cfm against a combined room and evaporator coilpressure drop of 2 inches water gauge. What is the annual cost of operatingthe fans with a unit cost of SAR0.08/kW h and a peak charge of SAR 120/kVA/quarter? Take power factor to be 0.93.


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Assignment # 1.

A cold store , 15m, 10m and 5.5m (L,W and H), maintained at 85% RH and –18oCand designed to cool 200 units of mass 40 kg of “ Beef–fresh, Average” from 23 oCto –18. The allotted and freezing hours are 8 hrs and 12 hrs respectively. Thereare 20 lights with 150 W each and 4 forklifts used by 5 occupants for 8 hours. Theaverage heat gain of forklifts is 500 W and there are 6 evaporator fans inside thecold store.. Outside air conditions are 30 oC , and 65% RH. The store walls androof are fabricated from 20cm thick expanded polystyrene (smooth).

Calculate the required system capacity if the equipment running time is 22hrs,12% S.F. and 5 oC solar radiation allowance.

Results

Roof

Floor

Walls

Glass

Solar

Lights

People

Appliances

Product: Above freezing

Product: Below freezing

Packing Load

Freezing Load

Air Infiltration

Air Interchange

Ventilation

Heat Load (W)

Safety Factor (W)

Total heat Gain (W)

Required Equipment Capacity (W)

Tons of Refrigeration


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Example 6.7

Four thousand and five hundred lug boxes of apples are stored at 2.5 oC in astorage cooler 15 m x 12 m x 6 m high. The apples enter the cooler at atemperature of 30 oC and at the rate of 250 lugs per days each day for the 18 dayharvesting period. The walls including floor and ceiling are constructed of 25 mmboards on both sides of 50 mm x 100mm studs and are insulated with 100mm ofmineral wool. All of the walls are shaded and the ambient conditions are 35 oCand RH 50%.The average weight of apples per lug box is 25 kg. The lug boxeshave an average weight of 2 kg and a specific heat value of 2.3 kJ/kg.K. Thelighting load is 500 W for 6 hrs per day. Two people and one battery operatedforklift (4.17 kW) are in the space for 4 hrs per day. Determine the average load inkilowatts on the equipment based on a 16 hrs per day equipment operating time.(Assume the insulation k factor (0.045 W/m K and S.F. 10%).

Solution


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Example 6.8

Twenty three thousand liters of partially frozen ice cream at –4oC are entering ahardening room 9 m x 6 m x 5 m high each day. Hardening is completed and thetemperature of the ice cream is lowered to -28 oC in 10 hrs. The walls, includingfloor and ceiling, are insulated with 150mm of polyurethane and the overallthickness of the walls is 250mm. The ambient temperature is 35 oC and the RH is50%. The lighting load is 500 W for 6 hrs per day. Two people and one batteryoperated forklift (4.5 kW) are in the space for 6 hrs per day Assume the averagedensity of the ice cream is 0.6 kg/L, the average specific heat below freezing is 2.1kJ/kg K, and the average latent heat per kg is 233 kJ. Determine the averagehourly load based on 18 hrs.

Solution


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Assignment # 2.

Three thousand kilograms of dressed poultry are blast- frozen on hand truckseach day (24 h) in a freezing tunnel 4 m x 3m x 3.5 m high. The poultry isprecooled to 7 oC before entering the freezer where it is frozen and its temperaturelowered to -20 oC and 90% RH for storage. The lighting load is 200 watts and thelights are on 16 hrs per day. The north and east partitions adjacent to theequipment room and vestibule are constructed of 150 mm clay tile insulated with150mm polyurethane. The south and west partitions adjacent to storage coolerare 100 mm clay tile with 50mm polyurethane insulation. The roof is a 150 mmconcrete slab insulated with 150 mm polyurethane and covered with tar, felt, andgravel. The floor is a 150mm concrete slab insulated with 150mm polyurethaneand insulated 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 thetemperature inside is approximately the outdoor design temperature for the region

(33

o

C and 60% RH). The inside design temperature for both the storage roomand the freezer is –20 oC. The vestibule temperature and relative humidity are 10oC and 70 %, respectively. Determine the average hourly refrigeration load basedon 20 hrs per day operating time for the equipment and S.F. 10%.

Results

Roof

Floor

Walls

Glass

SolarLights

People

Appliances

Product: Above freezing

Product: Below freezing

Packing Load

Freezing Load

Air Infiltration

Air InterchangeVentilation

Heat Load (W)

Safety Factor (W)

Total heat Gain (W)

Required Equipment Capacity (W)

Tons of Refrigeration


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Appendix A

Cooling and Heating Load Calculations Tables


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Table (6.2) Heat transfer coefficient (W/m2 K)

Surface type Heat direction h

A-Still air Horizantal Horizantal Vertical

B-Moving Air

6.7 m/s (24 km/h) 3.35 m/s (12 km/h)

UpDown

Horizantal

Any direction Any direction

9.376

34.122.7

Table (6.3) Coefficients of heat transfer (U factor) of typical cold storage walls,roofs and factors (W/m2 K)

Insulation thermal conductivity, k, (W/ m K)

Insulationthickness

(mm)

0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060

25

50

75

100

125

150

175

200

0.732

0.420

0.295

0.227

0.182

0.153

0.136

0.119

0.834

0.489

0.346

0.267

0.218

0.184

0.159

0.140

0.931

0.556

0.397

0.308

0.252

0.213

0.185

0.163

1.013

0.617

0.443

0.346

0.283

0.240

0.208

0.184

1.091

0.675

0.489

0.383

0.315

0.267

0.232

0.206

1.163

0.731

0.533

0.420

0.346

0.294

0.256

0.227

1.229

0.784

0.576

0.455

0.376

0.320

0.279

0.247

1.289

0.834

0.617

0.489

0.405

0.346

0.302

0.267

Table (6.4) Design temperatures of major cities in KSA

City Dry bulb temp o

C Wet bulb temp o

C.

Jeddah 41 29.5

Riyadh 43.5 25.5

Dhahran 44.0 29.5


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Table (6.5) Storage requirements of perishable products

Chilling

ratefactor

Water

Content %

Approx.

Storagelife

Approx.Freezing

temp.oC

Latent

heat kJ/kg

Specific Heat kJ/kg K

RHStorage temp

oCProduct Belowfreezing

Abovefreezing

Fruits and Melons

0.67843 to 8 m-1.12801.893.6590 to 95-1 to 4 Apples0.67 1 to 2 w-1.12841.903.6890 to 950 Apricots

0.67652 to 4 w-0.32171.653.0185 to 907 to 10 Avocados - Green

0.17 --0.82501.783.3585 to 95-Bananas

2 to 3 d-0.82841.903.6890 to 1000Blackberries

0.67822 w-1.62741.863.5890 to 1000Blueberries

0.9925 to 15 d-1.23071.993.92902 to 4Cantetoupe(Rock

Melon)

0.9934 to 6 w-1.131023.9585 to 957 to 10Casaba Melons

802 to 3 w-1.82671.842.4195-1 to 0Cherries

471 to 2 m-0.91571.433.5180 to 850 to 2Coconuts872 to 4 m-0.92901.933.7590 to 952 to 4Cranberries

8510 to 14 d-12841.903.6890 to 95-0.5 to 0Currents

206-12 m-16671.091.575 or less-18 or 0Dates- Cured

3 d-1.32841.903.6890 to 950Dew Berries

239 to 12 m771.121.6150 to 600 to 4Figs- Dried

787 to 10 d-2.42601.813.4585 to 90-1 to 0- Fresh-6 to 12 m---90 to 95-23 to -18Frozen Fruits

891 to 2 w-1.12971.953.8290 to 950Gooseberries

0.8894 to 6 w-1.12971.953.8285 to 9014 to 16Grapefruit

823 to 6 m-22741.863.5895 to 100-1 to 0Grapes

933 to 4 w-0.931023.95907 to 10Honeydew Melons

1.0891-6 m-1.42971.953.8285 to 9015 to 18Lemons0.9866 to 8 w-1.62871.923.7285 to 909 to 10Limes

812 w-0.92701.853.5585 to 9013Mangoes

821 to 2 w-0.92741.863.58900Mectarines754 to 6 w-1.42501.783.3585 to 907 to 10Olives - Fresh

0.7873 to 12 w-0.82901.933.7585 to 905Oranges

893 to 6 w2971.953.82-1 to 20range Juice911 to 3 w-0.83041.983.889013Papaw

0.80892 to 3 w-0.92971.953.8290 to 950Peaches

0.80832 to 6 w-1.62771.883.6190 to 95-1.6 to 0Pears

0.9932 w-0.831023.9590 to 957 to 10Persian Melons

783 to 4 m-2.22601.813.4590-1Persimmons

0.67851 to 4 w-12841.93.6885 to 9020Pineapples

0.67861 to 4 w-0.82871.923.7290 to 95-0.5 to 0Plums

822 to 4 m-32741.863.58900Pomegranates

862 to 4 m0.82871.923.7290 to 95-1 to 0Prunes - Fresh

285 to 8 m--1.191.7755 to 600 to 4- Dried

2 to 3 m-22841.93.6890-1 to 0Quinces

812 to 3 d-1.12701.853.5590 to 1000Raspberries 905 to 7 d-0.83001.973.8590 to 1000Strawberries

872 to 4 w-1.12901.933.7590 to 950Tangerines

932 to 3 w-0.431023.9585 to 905 to 10Watermelons


24/36

MET 412


23/36


Chillingrate

factor

WaterContent

%

Approx.Storage

life

Approx.Freezing

temp.

oC

Latentheat kJ/kg


RHStoragetemp oC

ProductBelow

freezing

Above

freezing

Vegetables

82 w-1.22801.893.6595 to 1000Artichokes -Globe

805 m-2.5267843.4790 to950-Jerusalem

0.9932 to 3 w-0.631023.9595 to 1000 to 2Asparagus

897 to 10 d-0.72971.953.8295 to 1007 to 10Beans - Green

1 to 2 w-0.495 to 1000Beetroot-Bunch

3 to 5 m-0.92941.943.7895 to 1000-Topped

9010 to 14 m-0.63001.973.8595 to 1000Broccoli

853 to 5 w-0.82841.903.6895 to 1000Brussels Sprouts

921 to 4 m-0.93071.993.9298 to 1000Cabbage

0.80884 to 6 m-1.42941.943.7898 to 1000Carrots-Topped, Immature

0.80884 to 5 m-1.42941.943.7898 to 1000- Topped, Mature

0.80922-4 w-0.83071.993.9295 to 1000Cauliflower1941-2 m-0.53142.023.9895 to 1000Celery

744-8 d-0.62471.763.3195 to 980Corn - Sweet

9610 to 14 d-0.53202.044.0595 to 10010Cucumbers

937 d-0.83102.03.9590 to 957 to 10Eggplant

932 to 3 w-0.131023.9590 to 1000Endive (Escarole)

6 to 12 m-23 t0 -18Frozen Vegetables

616 to 7 m-0.82031.62.8865 to 700Garlic-Dry

7510 to 12 m-1.82501.783.3595 to 1000Horseradish

0.70873 to 4 w-0.52901.933.75950Kale

0.7902 to 4 w-13001.973.8590 to 1000Kohlrabi

851-3 m-0.72841.903.68950Leeks - Green

952 to 3 w-0.23172.034.0295 to 1000Lettuce - Head

913 to 4 d-0.93041.983.88950Mushrooms

0.301 to 8 m-0.82941.943.7865 to 700Onions-Dry

851 to 2 m-1.12841.903.6895 to 1000Parsley 792 to 6 m-0.92641.833.4898 to 1000Parsnips

0.67741 to 2 w-0.62471.763.3195 to 980Peas - Green

0.67126 to 8 m0.991.247010- Dried

922 to 3 w-0.73071.993.9290 to 957 to 13Peppers - Sweet

126 m0.991.2460 to 700 to 10- Dry, Chili

78-0.72601.813.4590 to 957Potatoes - Culinary

694 to 6 m-1.32301.73.1585 to 9013 to 16- Sweet

912 to 3 m-0.83041.983.8885 to 9013Pumpkins

0.70953 to 4 w-0.73172.034.0290 to 950Radishes -Topped

952 to 4 w-0.93172.034.02950Rhubarb

892 to 4 m-1.12971.953.8290 to 950Rutabaga

0.80931 to 2 w-0.331023.9595 to 980Silver beet (Spinach)

941 to 3 w-0.53142.023.9895 to 1007Squash - Button

851 to 3 m-0.82841.93.6885 to 9013- Hard Shell

944-7 d-0.53132.023.9890 to 955 to 7Tomatoes - Firm, Ripe

931 to 2 w-0.631023.9590 to 9513- Mature, Green0.67924 to 5 m-1.13071.993.92950Turnips

743 to 6 m2471.763.3185 to 9016Yams


25/36

MET 412


24/36


Chilling

ratefactor

Water

Content %

Approx.

Storagelife

Approx.Freezing

temp.oC

Latent

heat kJ/kg


RHStoragetemp oCProduct Below

freezing Abovefreezing

Meat-Fish -Shellfish

0.6762 to 771 to 6 w-2.2 to -

2.7206 to257

1.6 to 1.82.9 to 3.488 to 920 to 1Beef - Fresh, average

0.56705 d-1.72331.713.18900-Liver

0.56661 to 7 d2201.663.05900 to 1- Veal

6 to 12 m90 to 95-23 to -18- Frozen

0.7560 to 705 to 12 d-2.2 to -

1.7200 to233

1.6 to 1.72.8 to 3.285 to 900 to 1Lamb- Fresh , average

8 to 12 m90 to 95-23 to -18-Frozen

741 w-2.82471.763.3185 to 90-2 to 0Poultry- Fresh, average

8 to 12 m90 to 95-23 to -18- Frozen

681 to 5 d2271.693.1190 to 950 to 1Rabbits -Fresh

62 to 815 to 14 d-2.2207 to270

1.61 to1.85

2.91 to1.85

95 to 100-1 to 1Fish – Fresh, average

6 to 12 m90 to 95-29 to -18- Frozen

8012 d-2.22671.843.5195 to 1000 to 1Scallops-Meat

7612 to 14 d-2.22541.793.3895 to 100-1 to 1Shrimp

875 to 8 d-2.22901.933.751000 to 2Oysters, Clams-Meat and

Liquid

805 d-2.82671.843.5195 to 1005 to 10Oysters – In shell

3 to 8 m90 to 95-29 to -18Shellfish - Frozen


26/36

MET 412


25/36


Chillingrate

factor

WaterContent

%

Approx.Storage

life

Approx.Freezing

temp.

oC

Latentheat kJ/kg


RHStoragetemp oC

ProductBelow

freezing

Above

freezing

Miscellaneous

32 to 373 to 6 m-2.2106 to123

1.271.99- 18Bread - Frozen

163 to 13 w-20 to

0.6531.041.3775 to 850 to 4Butter

12 m70 to 85-23Butter - Frozen

3712 m-131231.32.07650 to 1Cheese -Cheddar

376 m-131231.32.07654.4-Cheddar

16 to 12 m3.30.850.8740-18 to 1Chocolate Milk

10 to 152 to 4 m33 to

500.96 to1.03

1.17 to1.34

80 to 852 to 3Coffee -Green

0.85665 to 6 m-2.22201.663.0580 to 85-2 to 0Eggs - Whole

662 to 3 w-2.22201.663.0570 to 7510 to 13- Whole741 y +2471.763.31-18 or less-Frozen, Whole

Severalyears

45 to 551 to 4Furs and Fabrics

141 y +571.051.4Below 10Honey

Severalmonths

50 to 60-2 to 0Hops

0.8587-0.62901.933.750 to 1Milk-Whole, Pasteurized

3 to 68 to 12 m10 to

200.88 to0.91

0.94 to1.04

65 to 750 to 10 Nuts

161 y +531.041.3760 to 702Oleomargarine

104 to 6 w330.961.17850 to 4Popcorn- Unpopped


27/36

MET 412


26/36

Table (6.9) Heat of Respiration: Watt/Tonne

ProductStorage temperature oC

0 5 10 15 20

Fruits and Melons Apples 6-10 31-20 11-56 1-80 8-5

Apricots 36 -17 35-27 61-102 -3

Avocados-Green

1-80 3-281 360-415 35-915

Blackberries 8-68 -136 905-432 388-582

Blueberries -31 9-36 86 303-183 154-259

Cantaloupe(Rock Melon)

26-30 300-114 132-192

Cherries -Sweet

32-16 9-42 8-133 83-195

Cranberries 39-14 6-68 33-54

Figs –Fresh11

-39386

-188 169-282Gooseberries 20-26 16-40 6-56

Grapefruit 1 52

Grapes 8-6 -16 98 96-31

HoneydewMelons

1-8 59-71

Lemons -17 8 67

Limes 3-13 20-55

Mangoes 311 223-449

Olives- Fresh 6-116 114-145

Oranges 5-13 30-19 1-60 60-90

Papaw 33-16 80-60

Peaches 12-19 35-27 91-59 5-396 176-304

Pears 8-15 3-39 6-3 101-231Persimmons 3 1-42 59-71

Pineapples 4-6 9-34 1-50 65-105

Plums 6-9 39-27 9-165 1-37 53-77

Raspberries 52-74 59-114 386-281 -103 301 340-727

Strawberries 36-52 85-98 99 933-274 303-581

Watermelons 51-74


28/36

MET 412


27/36

Table (6.10) Heat of Respiration: Watt/ Tonne

ProductStorage temperature oC

0 5 10 15 20

Vegetables Artichokes-Globe 6-133 5-178 369-292 995-430 404-692

Asparagus 3-238 369-404 13-904 89-971 809-1484

Beans - Green 303-104 369-173 99-276 351-386

Beetroot - Topped 36-21 9-28 1-40 0-69

Broccoli -68 309- 8 3-1008 825-1011

Brussels Sprouts 86-71 56-144 3-251 91-317 267-564

Cabbage - White 3-40 99- 64 16-98 -170

Carrots - Topped 86 51 33 209

Cauliflower 1 63 300 31 238

Celery 90 10 300 170

Corn - Sweet 396 910 119 81 855

Cucumbers 6-86 3-98 92-143Garlic - Dry 5 -19 3-29 9-29 11-81 30-54

Horseradish 98 19 5 132

Kohlrabi 10 85 51 386

Leeks - Green 9-85 -86 35-202 98-347

Lettuce - Head 9-50 80 -59 3-119 338-121 178

Mushrooms 1-130 930 782-939

Onions-Dry 5 30 93 11 50

Parsley 5-137 356-252 15-487 89-662 582-757

Parsnips 18-46 96-52 63-78 56-127

Peas - Green 50-139 361 -227 10-600 728-1072

Peppers - Sweet 81 6 130

Potatoes- Immature 1 89-62 89-92 54-134

- Mature 3-20 90-30 90-35 20-47

Radishes- Topped 36-18 91-24 8 9-97 142-146

Rhubarb-Topped 24-39 11-54 59-135 119-169

Rutabaga 6-8 38-15 19-47

Silver beet (Spinach) 316 19 13 682

Tomatoes - Colored andRipe

36 6-75 65-115

- Mature, Green 31-22 81-75 75-110

Turnips- Roots 96 9-30 68-71 71-74


29/36

MET 412


28/36

Table (6.11) Average Air Infiltration Rates in s/ due to door openings

Room Volume(m

3)

*

s/ Infiltration

Roomsabove 0

oC

Roomsbelow 0

oC

8.510152025304050

75100150200250300400500600700800900

1000

3.13.43.74.45.05.55.96.87.5

9.010.212.213.915.316.719.021.423.624.325.927.128.9

2.32.62.83.33.84.24.65.45.8

6.97.99.4

10.911.912.914.916.818.118.620.421.923.1

* For heavy usage, add 50% to table values


30/36

MET 412


29/36

Tables (6.12 a) Kilojoules per liter removed in cooling air to storage conditions aboveoC

Storageroom temp.

oC

Inlet Air Temperature oC

25 10 15 40

Inlet air relative humidity %0 60 0 0 60 0 0 60 0 60

151050

0.01280.02660.03880.0493

0.01860.03230.04450.0550

0.02460.03820.05020.0606

0.02810.03190.05360.0639

0.03570.04910.06100.0713

0.04410.05740.06930.0794

0.05000.05910.07080.0808

0.05630.06940.08100.0910

0.06630.07920.09060.1003

0.07950.09920.10360.1141

Tables (6.12 b) Kilojoules per liter removed in cooling air to storage conditions below oC

Inlet Air Temperature oC

StorageRoom

temp oC

353025105

Inlet air relative humidity %60 0 60 0 60 0 0 0 0 0

0.0921 0.10040.10710.11370.12030.12650.13250.13820.1440

0.0820 0.09030.09700.10370.11020.11650.12250.12830.1341

0.0724 0.08090.08770.09450.10130.10770.11380.11970.1256

0.0650 0.07360.08050.08730.09410.09980.10670.11260.1185

0.0562 0.06490.07190.07880.08570.09220.09850.10450.1106

0.0505 0.05920.06620.07320.08010.08660.09290.09890.1050

0.0154 0.02470.03210.03950.04680.05370.06040.06680.0732

0.0142 0.02350.03090.03830.04560.05250.05910.06560.0720

0.0111 0.02100.02880.03670.04440.05230.05880.06570.0725

0.0092 0.01930.02710.03500.04270.05010.05710.06400.0708

0

-5-10-15-20-25-30-35-40

Table (6.13) Typical data for the number of refrigerated room volumes of air

interchange in a 24 hour period as a function of room volume (m

3

)

Store volumem3

Air change per24 hours

30

90

80

60

0

300

30

900

100

800

600

00

3000

300

9000

1000

8000>

11

99

3

121097

5.84.84.23.42.92.51.91.71.41.2


31/36

MET 412


30/36

Table (6.14) Outdoor air requirements for Ventilation rate

Person

sL

Application SmokingVentilation rate

Minimum Recommended

FlatBank

Hair dresser

Commercial ShopsFactoriesHospitalHotels

Meeting roomsOffices

RestaurantsCafeteria

SometimesSometimesSometimes

HeavyNONONO

Heavy

HeavySometimesSometimesSometimes

755

122.53.51212

147

7.53.5

9.57.57

153.55

1414

2412106

Table (6.15) Surface absorption factor ()

Material Absorption factor

Asphalt 0.89

Concrete 0.65Red bricks 0.77

White bricks 0.26

Cements 0.57

Gypsum 0.4

Thermal insulation 0.91


32/36

MET 412


31/36

Table (6.16 A ) Effect of Sun Radiation (Cold stores)

sT Average sunlightintensity (W/m2)

No of hoursSurface type

39

39

9

1

869

869

3

33

6

6

6

Wall

East West North South

511139Roof


33/36

MET 412


32/36

Table (6.17) Window glass overall heat transfer coefficient (W/m2 K)

Winter Summer

8 2.73.4

4.7

3.23.12.8

Single glass

Insulating double glass with5mm air gap6mm air gap

13 mm air gap

Table (6.18 A) Maximum heat gain per unit area, sunQ , ( W/m2)

20o North Latitude

Month DirectionWest South East North Horizontal

JanFeb

March AprilMayJuneJuly

AugustSep

OctNov.Dec

618

31

8

35

6

661

665

658

30

6699

50

6

85

161

31

311

311

316

30

160

1666

31

618

31

8

35

6

661

665

658

30

6699

50

53

5

30

390

38

36

33

396

338

30353

19

10

56

50

51

0

1

6

3896

6

Maximum 748 713 748 186 905

Table (6.18 B) Shading Coefficient (SC)

(Roller Shade)(Venetian Blinds)Withoutshading

Thickness

(mm)Glass type

DarkLightMediumLight

0.590.590.40

0.250.250.30

0.640.640.57

0.550.550.53

10.950.70

1 6-39 6

Single

Regularsheet

Colored Heat

Absorbing

0.60.6-

0.250.25

-

0.570.57

0.2-0.3

0.510.51

-

0.90.83

0.2-0.4

1 6

6

Double Regular

sheet Colored Reflective

coated


34/36

MET 412 Heating Ventilating and Air Conditioning

33/36

Table (6.16 B) Solar intensity (W/m2) ( AC space)

HourDirectionMonth

1817161514131211109876

150--

395

130

155--

600

18

90--

635

560

15--

555

745

--

45410

900

--

90220

1000

--

105-

1025

-22090-

1000

-41045-

900

15555

--

745

90635

--

560

155600

--

345

30

95

--

310

NESW

Hor

21June

390

--

5

190

33

--

16

330

--

680

80

--9

6

1

--9

6

1

--

380

99

550

--

155-

1020

-225140

-990

-42090-

980

-56525-

735

55640

--

540

120595

--

320

45365

--

110

NESW

Hor

23 July

10

--

6

9

8

--

98

0

--0

68

80

--

380

0

60

--

380

0

60

--

96

910

580

--

9

-5

-910

96

-580

-810

93

-1

-0

380

-60

-68

0

-80

10

6

--

3

45245

--

50

NESW

Hor

24August

--60

80

30

----

-

--

3

60

1

--

9

0

--

9

0

--

89

910

1

--

88

-

0

-610

86

-

1

-89

10

-

1

-0

9

-

-60

3

-

1

-80

60

-

30

----

-

NESW

Hor

22 Sep


35/36

MET 412


33/36

Table (6.19) Packaging material specific heat

Packaging Material Specific Heat (kJ/kgK)

Fiberite packaging 1.4Wood 2.3

Steel 0.5

Plastics 1.6

Aluminium 0.85

Table (6.20) Heat equivalent to Occupancy (AC)

TotalLatent heatSensible

heatTypical

applicationHuman

Activities

973166Theater Seated at rest

1174572Office, flat,

hotelSeated, very

light work

1325973Office, flat,

hotelMedium work

1325973Commercial

shopsStanding and

light work

1467373BankWalkingslowly

1628181RestaurantSeated,

eating

22913981FactoryModerate

work

293183110FactoryMoving

regularly

29220488FactoryMedium work

425255170factoryHeavy work

425255170GymnasiumSport

exercises

Table (6.21) Heat equivalent to Occupancy (Refrigeration)

Storage tem. oC

t Q

-30 450

-25 410

-20 390

-10 300

-5 275

0 250

5 225

10 200


36/36

MET 412


34/36

Table (6.22) Usage heat gain, W/m3 K

Long-termstorage

ServiceRoom Volume(m3) Heavy Average

-----------

--

0.60.450.310.240.190.160.14

0.140.13

3.973.572.762.241.961.721.611.521.451.441.371.301.231.16

-------

--

3.632.561.771.441.251.071.010.960.940.910.860.850.770.710.650.58

-----

--

0.60.851.52.03.06.08.5

11.014.017.023.028.034.043.057.085.0140.0200.0280.0560.0

1400.0

2100.02800.0

chapter 6a pdf

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