chapter 6a pdf
TRANSCRIPT
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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
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Table (6.6) Storage requirements of perishable products
Chillingrate
factor
WaterContent
%
Approx.Storage
life
Approx.Freezing
temp.
oC
Latentheat kJ/kg
Specific Heat kJ/kg K
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
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Table (6.7) Storage requirements of perishable products
Chilling
ratefactor
Water
Content %
Approx.
Storagelife
Approx.Freezing
temp.oC
Latent
heat kJ/kg
Specific Heat kJ/kg K
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
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Table (6.8) Storage requirements of perishable products
Chillingrate
factor
WaterContent
%
Approx.Storage
life
Approx.Freezing
temp.
oC
Latentheat kJ/kg
Specific Heat kJ/kg K
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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