Post on 02-Jun-2018
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MESFIN INDUSTRIAL ENGINEERING
2008
Refrigeration system for food storages
(selection guide)
Commercial coolers and freezersFISEHA M.
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Table of Contents
Contents Page No
1. Introduction . 3
2. Panel thickness calculation.. 10
3. Refrigeration load calculation. 14
4. Refrigeration equipment selection.. 21
4.1.Packaged equipment selection 22
4.2.Packaged equipment selection form 28
4.3.Evaporator selection 32
4.4.Compressor selection . 36
4.5.Condenser selection. 39
4.6.Expansion valve selection 40
4.7.System balance 40
5. Appendix . 41
5.1.Table1: Influence of RH on system temperature difference (TD). 41
5.2.Table2: Frost and dust (fin spacing) reduction factor as function of fin spacing. 41
5.3.Table3: Number of air changes per 24 hours 41
5.4.Table4: Air change factor . 42
5.5.Table5: Specific heat for various products 43
5.6.Table6: Heat of respiration of variousproducts 44
5.7.Table7: Heat equivalent of electric motors 45
5.8.Table8: Heat equivalence of occupancy.. 45
5.9.Table9: Operating condition for storage.. 46
5.10. Table10: Insulation Requirements for Storage Rooms.. 465.11. Table11:Suggested Freezer Temperatures 46
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Refrigeration system for food storages
(Commercial coolers and freezers)
1. Introduction1.1.Refr igerationRefrigerationis defined as the process of extracting heat from a lower temperature heat source,
substance, or cooling medium and transferring it to a higher temperature heat sink. Refrigeration
maintains the temperature of the heat source below that of its surroundings while transferring theextracted heat, and any required energy input, to a heat sink - atmospheric air, or surface water.
1.2. Refrigeration system (Equipment)
A refrigeration system is a combination of components and equipments connected in a sequentialorder to produce the refrigeration effect.
1.2.1. Classification of refrigeration systemsA. Classificationsby the type of input energy and the refrigeration process:a) Vapor compression systemsIn vapor compression systems, compressors activate the refrigerant by compressing it to a
higher pressure and higher temperature level after it has produced its refrigeration effect. Thecompressed refrigerant transfers its heat to the sink and is condensed to liquid form. This liquid
refrigerant is then throttled to a low-pressure, low temperature vapor to produce refrigerating
effect during evaporation. Vapor compression systems are the most widely adopted refrigeration
systems in both comfort and process air conditioning.
b) Absorption systemsIn an absorption system, the refrigeration effect is produced by thermal energy input. After
absorbing heat from the cooling medium during evaporation, the vapor refrigerant is absorbed byan absorbent medium. This solution is then heated by direct-fired furnace, waste heat, hot water,
or steam. The refrigerant is again vaporized and then condensed to liquid to begin the
refrigeration cycle again.
c) Air or gas expansion systemsIn an air or gas expansion system, air or gas is compressed to a high pressure by mechanical
energy. It is then cooled and expanded to a low pressure. Because the temperature of air or gas
drops during expansion, a refrigeration effect is produced.
B. Classification according to size (tonnage in capacity) of the refrigeration system:
Small (2.5 and = 75tons)
C. Classification according to their evaporating temperatures (Te):
Low temperature systems - 40 Fo
eT 0 Fo
(-40 Co eT -18 Co )
Medium temperature systems 0 Fo
eT 32 Fo
(- 18 Co eT 0 Co )
High temperature systems eT >32 Fo ( eT >0 C
o )
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D. Classification according to application1) Domestic refrigerationfor domestic application2) Commercial refrigerationfor commercial application3) Industrial refrigerationfor industrial application
E. Classification based on the method employed for extracting or delivering heat1) Direct systemsA direct refrigeration system is one in which the evaporator or condenser of the refrigeration
system is in direct contact with air or other media to be cooled or heated.
2) Indirect systemsAn indirect refrigeration system uses a secondary coolant to cool the air or other substance
after the coolant is cooled by the refrigeration system.
F. According to the probability that leakage of refrigerant will enter the occupied space High-probability system- Any refrigeration system in which a leakage of refrigerant from a
failed connection, seal, or component could enter the occupied space (the space is frequently
occupied by people) is a high-probability system. A direct system is a high-probabilitysystem.
Low-probability system -This is a refrigeration system in which leakage of refrigerant froma failed connection, seal, or component cannot enter the occupied space. An indirect system
that uses secondary coolants, such as chilled water, brines, or glycols, is a low-probabilitysystem.
1.2.2. Applications of refrigeration systemsAir or gas expansion refrigeration systems are limited in aircraft and cryogenics
Vapor compression systems dominate in both comfort and processing air conditioning
Absorption systems are
1.3.Refrigeration system selectionMost commercial coolers and freezers, refrigeration systems used for commercial application,
are categorized under small and medium size refrigeration systems.
For small and medium-size refrigeration systems, vapor compression refrigeration systems are
widely adopted.
1.4.Classification of vapor compression refrigeration systems
1.4.1. According to the type of compressor used, such as:
Reciprocating
Rotary
Scroll
Screw or
Centrifugal systems
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1.4.2. According to the type of evaporator, such as:
Direct-expansion (DX) coolerIn a DX cooler system space air is directly cooled and dehumidified by the expanded and
evaporated refrigerant in a DX coil.
Liquid chillers
A liquid chiller is a factory-assembled refrigeration package in which chilled water is producedin its evaporator. Air is then cooled and dehumidified by chilled water in air-handling units
(AHUs) or in fan coils.
For a refrigeration system, the type of compressor, system size, and whether a DX coilor a
water chi ll eris used are often interrelated. A centrifugal chiller is often a large-tonnagerefrigeration system. Small and medium refrigeration systems are often an air-cooled DX cooler
using a reciprocating, rotary, or scroll compressor.
For commercial coolers and freezers, air cooled direct-expansion systems using reciprocating
compressors are more economical.
1.4.3. Air cooled DX, Reciprocating systemIn a DX system, a direct-expansion coil is used as the evaporator, and R-22, R-134A, R- 404A
and R-502A are the primary refrigerants used in new and retrofit projects. The air-cooledcondenser is often combined with the reciprocating compressor(s) to form a unit called the
condensing uni t.The condensing unit is generally located outdoors. Either the DX coil is
installed in a rooftop packaged unit consisting of fan(s), filters, condensing unit, and controls,
mounted on the roof of the building; or it is a system component in a split or indoor packagedunit located indoors. In supermarkets and many industrial applications, DX coils are also
installed in refrigerated display cases and food processing and food storage facilities. Air isforced through the DX coil by a fan for better heat transfer. DX systems are most widely used inreciprocating and scroll refrigerating systems.
compressor
condenser
evaporator
expansion
valve condenser fan
evaporator fan
Fig a) schematic diagram of single stage vapor compression
1
23
4
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Refrigeration CyclesThe refrigerants undergo a series of evaporation, compression, condensation, throttling, and
expansion processes, absorbing heat from a lower-temperature reservoir and releasing it to ahigher temperature reservoir in such a way that the final state is equal in all respects to the initial
state. It is said to have undergone a closed refrigeration cycle.
1.7.Refrigeration Effect, Refrigeration Load, and Refrigerating CapacityRefrigeration Load(or cooling load)Refrigeration load is the total heat load of the storage to be refrigerated.
Refrigeration effectThe refrigeration effect is the heat extracted by a unit mass of refrigerant during the evaporating
process in the evaporator.
Refrigeration capacity (or cooling capacity)
Refrigerating capacity is the actual rate of heat extracted by the refrigerant in the evaporator. Itcan be expressed as follows:
The cubic feet per minute (CFM) suction vapor of refrigerant required to produce 1 ton of
refrigeration (liters per second to produce 1 kW of refrigeration) depends mainly on the latentheat of vaporization of the refrigerant and the specific volume at the suction pressure. It directly
affects the size and compactness of the compressor and is one of the criteria for refrigerant
selection.
1.8.Sub cooling and Superheating
Sub-coolingCondensed liquid refrigerant is usually sub-cooled to a temperature lower than the saturated
temperature corresponding to the condensing pressure of the refrigerant. This is done to increase
the refrigerating effect. Sub-cooling is shown in Fig. 1b. The degree of sub-cooling depends
mainly on the temperature of the coolant (atmospheric air) during condensation, and theconstruction and capacity of the condenser.
Superheating
Superheating isheating the saturated vapor slightly above its saturation temperature. This is doneto avoid compressor slugging damage. Superheatingis shown in Fig. 1b. The degree of superheat
depends mainly on the type of refrigerantfeed and compressor as well as the construction of the
evaporator. Generally, the degrees of super heating and sub cooling have the following ranges:
Super heat temperature = 5 to 7 Co
temperature of a gas above the saturation point
Sub cooling temperature = 3 to 8 Co
- temperature of a liquid below the saturation point
1.9.COMPRESSORS
The four main types of compressors used in commercial refrigeration today are: Open - belt driven (low speed, 500-1750 rpm)
Open - direct driven (medium speed, 1160 or 1750 rpm)
Semi-hermetic (1750 rpm)
Hermetic (welded, 3500 rpm)The compressor type used is often a matter of personal preference but it is important to be aware
that compressor life decreases with increased speed and increased condensing temperature.
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On commercial refrigeration applications, compressors are most commonly used with air-cooled
condensers. They are also used with water-cooled condensers and occasionally with evaporative
condensers. Water restrictions in recent years and simpler maintenance are the reasons for thepopularity of air-cooled systems.
The air-cooled condenser may be an integral part of the compressor unit (air-cooled condensing
unit) or it may be remotely located (on the roof, for example).Compressor/condensing units are generally classified as high, medium or low temperature.Approximate evaporating temperatures are:
High +30F to +50F
Medium -10F to +30F Low 40F to -10F
In a reciprocating compressor a piston (single or double acting) in a cylinder is driven by a
crankshaft via a connecting rod. At the top of the cylinder are a suction valve and a discharge
valve. There are usually two, three, four, or six cylinders in a reciprocating compressor.Vapor refrigerant is drawn through the suction valve into the cylinder until the piston reaches its
lowest position. As the piston is forced upward by the crankshaft, it compresses the vapor
refrigerant to a pressure slightly higher than the discharge pressure. Hot gas opens the dischargevalve and discharges from the cylinder. The gaseous refrigerant in a reciprocating compressor is
compressed by the change of internal volume of the compression chamber caused by the
reciprocating motion of the piston in the cylinder. The cooling capacity of a reciprocating
compressor ranges from a fraction of a ton to 200 tons (1 to 700 kW). Refrigerant HCFC-22,HFC-134a, HFC- 404A, HFC- 407A, and HFC- 407C in comfort and process air conditioning
and R-717 (ammonia) in industrial applications are the currently used refrigerants in
reciprocating compressors because of their zero ozone depletion factors, except HCFC-22 whichwill be restricted from the year 2004 due to its ozone depletion.
The maximum compression ratio ( dP / sP ) of a single-stage reciprocating compressor is about 7.
Volumetric efficiency of a typical reciprocating compressor v decreases from 0.92 to 0.65 when
dP / sP increases from 1 to 6, and the isentropic or compressor efficiency ise decreases from 0.83
to 0.75 when dP / sP increases from 4 to 6. Methods of capacity control during part-load operation
include on/off, cylinder unloader, and hot-gas bypass controls.Reciprocating compressor design is now in its mature stage. There is little room for significant
improvement. Reciprocating compressors are still widely used in small and medium-size
refrigeration systems. However, they will be gradually replaced by rotary, scroll, and screw
compressors.
Hermetic, Semi hermetic, and Open Compressors
In a hermetic compressor, the motor and the compressor are sealed or welded in the same
housing. Hermetic compressors have two advantages: They minimize leakage of refrigerant, andthe motor can be cooled by the suction vapor flowing through the motor windings, which results
in a small and cheaper compressor-motor assembly.
Motor windings in hermetic compressors must be compatible with the refrigerant and lubricationoil, resist the abrasive effect of the suction vapor, and have high dielectric strength. Welded
compressors are usually used for small installations from
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repair during a compressor failure or for regular maintenance. Other features are similar to those
of the hermetic compressor. Most of the medium compressors are semi hermetic.
In an open compressor, the compressor and the motor are enclosed in two separate housings. Anopen compressor needs shaft seals to minimize refrigerant leakage. In most cases, an enclosed
fan is used to cool the motor windings by using ambient air. An open compressor does not need
to evaporate the liquid refrigerant to cool the hermetic motor windings. Compared with hermeticcompressors, open compressors may save 2 to 4 percent of the total power input.Many very large refrigeration compressors are open compressors.
Direct Drive, Belt Drive, and Gear DriveHermetic compressors are driven by motor directly or driven by gear trains. Both semi hermetic
and open compressors can be driven directly, driven by gear trains, or driven by motor through V
belts. The purpose of a gear train is to increase the speed of the compressor. Gear drive is
compact in size and rotates without slippage. Like belt drive, gear drive needs about 3 percentmore power input than direct- drive compressors. Some large open compressors may be driven
by steam turbine, gas turbine, or diesel engine instead of electric motor.
1.10. Power absorption of refrigeration systems
Power absorption of a refrigeration system is the power input at the compressor shaft, brake
horse power. Power rating of refrigeration equipments is the brake horse powerwhich doesn'tinclude mechanical and motor losses. But the actual power requirement of the system is higher
because of mechanical loss at the coupling of compressor shaft and electric motor and electric
motor inefficiency.
1.11. Coefficient of performance(COP) of refrigeration systems
The coefficient of performance is an index of performance of a thermodynamic cycle or athermal system. COP is defined as the ratio of the refrigeration effect to the work input.
1.12. Energy efficiency ratio (EER) of refrigeration systems
EER is energy use index defined as the ratio of the net cooling capacity to the electric power
input under designated operating conditions.
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2. Panel thickness calculation
Sandwich panel thickness is the sum of insulation thickness plus laminate thickness. Butlaminate thickness is so small that for calculation purpose it can be neglected. The main factors
which affect insulation thickness calculation are:
- Cost- Condensation of vapor on wall surfaces
2.1.Economic analysis of insulation thicknessIncrease in insulation thickness will decrease the operational costs but increase the investment
for the insulation. However, by increasing the insulation thickness a little investment for thecooling unit can be saved. Therefore, all these factors are to be taken in to account to determine
the optimal thicknessthickness which yields minimum annual total cost. Therefore, total
annual cost as function of insulation thickness can be expressed as follows:
C (x ) = I (x ) + C op (x ) + Constants
Where
C (x ) is total annual cost
I (x ) is investment cost per year
C OP(x ) is operational cost per year
x is insulation thickness
Investment cost is the sum of:
- Material cost:is the sum of all costs of insulation, hook, laminates, labor etc- Equipment cost:is cost of refrigerating machinery
Life expectancy of the machinery and insulation is assumed to be same. For example 10 years.
Therefore, the total costs of insulation and machinery are to be multiplied by 0.1 to convert them
to annual costs.
The total annual cost can be written as:
C (x ) = IC (x ) + MC (x ) + OPC (x ) + Constants (1)
Where
IC (x ) is annual cost of insulation as function of thickness
MC (x ) is annual cost of refrigerating machinery as function of thickness
OPC (x ) is annual operating cost as function of thickness
Annual cost of insulation can be approximated by:
IC (x ) = 0.1 AxPI + Constants 1a
Where
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IP is price of insulation material per m3of insulation
A is area of insulation0.1 is a factor which converts total cost to annual, assuming 10 years of life expectancy of the
cold room
Annual cost of refrigerating machinery can be approximated by:
MC (x ) = 0.1
QPM + Constants 1b
Where
MP is price increase of refrigerating machinery per watt increase of capacity
Q is refrigerating capacity in watts
Refrigerating capacity is approximated by:
Q = TUA + Constants
Where
U is over all heat transfer coefficientA is envelope area
T is temperature difference
Annual operating cost of the refrigerating machinery can be approximated by:
OPC (x ) = elP (ave
ave
COP
Q
) 310
Where
elP is price of operating energy (electricity) per kWh
is annual operating time ( in hours/year )
aveQ
is annual average refrigerating capacity in Watt
aveCOP is annual average coefficient of performance of the machinery
Annual average refrigerating capacity can be approximated by:
aveQ
= aveTAU + Constants
Where
aveT is annual average temperature difference approximated by:
aveT = aveT - roomT
Where
av eT is annual average temperature of the surrounding
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roomT is room temperature
By substituting all the values in equation (1) and setting the derivative of equation (1) with
respect to insulation thickness (x ) equal to zero and solving for U gives:
ecoU =3101.0
1.0
ave
el
aveM
I
COP
PTPT
P
Where
ecoU is U value for which total annual cost is minimum
Over all heat transfer coefficient of an envelope neglecting laminate effect can be calculated by:
U=oi hxh /1//1
1
Where
x is insulation thickness
is thermal conductivity of insulation material
ih is inside film or surface conductance
oh is outside film or surface conductance
For still air, ih = oh = 9.37 W/ Cm
o
.
2
. If the outer surface is exposed to 6.1m/s wind, oh isincreased to 34.1 W/ Cm o.2
Simplifying and solving the above equation for thickness gives:
x =
OR
U
1
Where
OR =oi
oi
hh
hh
, assuming still air then, OR = 0.21 ( Cm
o.2 ) / W
Therefore, the economical thickness will be:
ec ox =
O
eco
RU
1
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2.2.Condensation of water vapor on wall surfaceIf an envelope is poorly insulated, the out side surface temperature of the envelope may be below
the dew point temperature in the ambient air. For such conditions moisture in the ambient air willcondense on the wall surface which is undesirable because it may cause:
Loss of thermal insulationCorrosion of sheet metal laminatesDamage the cold room etc.
Therefore, to avoid condensation on the out side wall surface, the surface temperature should behigher than the dew point temperature of the out side air. This consideration results in the
following in equality:
roomatm
dewatmo
TT
TThU
WhereU is over all heat transfer coefficient
oh is outside film or surface conductance
atmT is ambient air temperature
dewT is dew point temperature of the ambient air
Note:
First economic thickness (U value) is to be calculated and then to be checked for condensation of
vapor on wall surfaces.
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3. Refrigeration load calculation
Refrigeration system design/selection is based on a heat load (refrigeration load) calculation
conducted for the environment we are aiming to cool and dehumidify. Therefore, first
refrigeration load calculation is to be conducted then equipment selection based on the calculatedload is to be done.
3.1. Refrigeration load segments
The primary function of refrigeration is to maintain conditions of temperature and humidity that
are required by a product or process within a given space. To perform this function, equipment of
the proper capacity must be installed and controlled on a 24-hour basis. The equipment capacityis determined by the actual instantaneous peak load requirements. Generally, it is impossible to
measure the actual peak load within a refrigerated space. These loads must be estimated. The
total refrigeration load is the total of the following load segments:
Transmission Load - heat gain through walls, floors and ceilings.
Air Change Load - heat gain associated with air entering the refrigerated space, either byinfiltration or ventilation.
Product Load - heat removed from and produced by products brought into and stored in therefrigerated space.
Internal Load - heat produced by internal sources such as lights, electric motors, and peopleworking in the space.
1) Transmission LoadThe transmission load is sensible heat load caused by the refrigerated space being located
adjacent to a space at a higher temperature. Heat always travels from the warmer to the cooleratmosphere. The sensible heat gain through walls, floors and ceilings will vary with the
following factors:
Type and thickness of the insulation
Type of construction
Area
Temperature difference (TD) between the refrigerated space and the outside ambient,adjusted to allow for solar heat load on any surface exposed to the sun.
The following explains the formulae used to calculate the transmission load:
Q = A x U x TD
Where
Q is transmission heat loadA is area of the wall, roof, floor, etc
U is over all heat transfer coefficient
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TD is temperature difference between the refrigerated space and the outside ambient,
adjusted to allow for solar heat load on any surface exposed to the sun.
Over all heat transfer coefficient of an envelope neglecting laminate effect can be calculated by:
U=oi hxh /1//1
1
Where
x is insulation thickness
is thermal conductivity of insulation material
ih is inside film or surface conductance
oh is outside film or surface conductance
For still air, ih = oh = 9.37 W/ Cm o.2 . If the outer surface is exposed to 6.1m/s wind, oh is
increased to 34.1 W/ Cm o.2
Note:Latent heat gain due to moisture transmission through walls, floors, and ceilings of modern
construction refrigerated facilities is negligible and can be ignored.
2) Air Change or Ventilation Load
Each time a door is opened to a refrigerated room from an adjacent unrefrigerated space, someoutside air will enter the room. This untreated warm moist air will impose an additional
refrigeration load and must be taken into account in the heat load calculation. Usually, the
infiltration air's moisture content is more than that of the refrigerated space. As this air is cooled
to the space temperature, the moisture will condense out of the air. This imposes both a sensible
and latent heat load in the space which must be removed by the refrigeration equipment.Infiltration air quantities are difficult to determine accurately. Usually, a number of air changes
per day are estimated. Table3indicates the number of air changes that may be expected in agiven size of room over a 24-hour period. The data contained in this table have been determined
by experience and may be used with confidence. Please note that the air change factors vary for
rooms above and below 32F. For rooms below 0F, some further reduction of the air changes
may be considered. There is usually less traffic involved in a 0F room, with less air movementresulting. Having determined the number of air changes to be expected, the room volume is then
multiplied by the number of air changes. To obtain the infiltration load, a factor is then obtained
from Table 4and is applied to the total volume of air. The Btu/cubic foot factors in Table 4 are
based on the dry bulb temperature and the relative humidity of the infiltration air.
Infiltration load formula
Infiltration load=No. of air changes per 24hr(Table3) x Room Volume x Air Change Factor
(Table 4)
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3) The Product Load
The heat gain caused by the product must be considered in the total refrigeration loadcalculation. The product heat gain will include some or all of the following:
1. The load due to the product being placed in the refrigerated space at a higher temperature
than the design refrigerated space.2. The heat removed by freezing or chilling the product.3. The heat of respiration caused by chemical reactions occurring in the product.
Specific Heat
A product cooling from its initial temperature requires the removal of sensible heat. Sensibleheat is heat that can be detected and recorded on a dry bulb thermometer. The sensible heat to be
removed is known as the specific heat which is the amount of heat that must be removed to
reduce the temperature of the product. The specific heat will vary with the type of product and is
different above and below 32F. Specific heat figures are listed in Table 5for various producttypes.
Latent HeatWhen the product is cooled to a temperature of 32F or lower, the latent heat load is also a partof the product load. This process is called the latent heat of fusion. The latent heat load is the
quantity of heat involved in changing the state of a substance without changing its temperature.
This calculation is applied to all products that must be frozen. The latent heat of fusion of anyproduct is that of water -144Btu/lb - multiplied by the percentage of water content of the product.
Actual corrected latent heat figuresfor various products are shown in Table 5.
Heat of Respiration
Certain food products experience chemical changes after storage. This is true of most fruits andvegetables, and some dairy products. This chemical change results in heat production which
must be considered in the load calculation. The heat of respiration occurs at temperatures over32F and varies depending on the product and the storage temperature. Table 6indicates the heatof respiration for various products at common storage temperatures. Please note that this heat
load increases considerably at higher temperatures.
Product Load Formulae
Sensible load (Btu/24 Hours) = Specific heat of products (Table 5) x temperature reduction of
products x mass of product
Latent heat of fusion (Btu/24 Hours) = Latent heat of product (Table 5) x mass of product
Heat of respiration (Btu/lbs/24 Hours) = Heat of respiration of product (Table 6) x mass ofproduct
4) Miscellaneous LoadsAll electrical energy used by lights, motors, heaters, etc., located in the refrigerated area, must be
included in the heat load. These loads are calculated as follows:
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a) Lights
Lights = Total lighting wattage x hours in use x 3.41Btu/Watt for incandescent or 4.2 for
fluorescent lights.
Note:
Coolers and freezers = 1 to 1-1/2 Watts per square foot of floor area. Allow up to double thisamount for work areas
b) MotorsThe heat input from motors varies with the motor size, BHP output, efficiency and whether it islocated within, or outside of, the refrigerated space. The heat equivalent of one BHP is 2545
Btu/hr. Motor efficiencies vary from 40% and less for small fan motors to 80% or more for
integral horsepower motors. The motor output will be its BHP x 2545 Btu/hr x hours of
operation.If the motor is located inside the refrigerated area, divide its output by its efficiency. If the motor
is located outside the refrigerated area, its inefficiency will be dissipated outside, then only itsoutput will figure in the room load. If the motor is located in the room and the load is outside,
only the inefficiency will be added to the room load. Multiply the output by (1 - efficiency).For motors rated in Watts output, divide by 746 to obtain the heat equivalent horsepower rating.
See Table 7, Heat Equivalent of Electric Motors.
c) Occupancy LoadPeople working in a refrigerated storage area dissipate heat at a rate determined by the room
temperature. The heat load added to the room equals the number of people, the hours of
occupancy and heat equivalent per person. Multiple occupancies of short duration will carryadditional heat into the room. See Table 8, Heat Equivalent of Occupancy.
d) Equipment related loadsHeat gain associated with the operation of the refrigeration equipment consists essentially of the
following:
Fan motor heat where forced air circulation is used Reheat where humidity control is part of the cooling Heat from defrosting where the refrigeration coil operates at a temperature below freezing
and must be defrosted periodically, regardless of the room temperature
Equipment heat gain is usually small at space temperatures above approximately -1C. Where
reheat or other artificial loads are not imposed, total equipment heat gain is about 5% or less of
the total load. However, equipment heat gain becomes a major portion of the total load at freezertemperatures. For example, at-30C the theoretical contribution to total refrigeration load due tofan power and coil defrosting alone can exceed, for many cases, 15% of the total load. This
percentage assumes proper control of defrosting so that the space is not heat excessively.
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5) Safety FactorA minimum 10% safety factor is normally added to the total refrigeration load to allow for minor
omissions and inaccuracies.
6) Total refrigeration Load
To arrive at the total Btu/24 hr load, sum all four main sources of heat gain and add a 10% safetyfactor as recommended.
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3.2.Load calculation form
i) Site conditions1. Describe the application
2. What are the outside room dimensions (ft.)? (w).x (l) .. x (h)
3. Describe the insulation Type Thickness ..inches4. What is the overall wall thickness? .. inches
5. What is the outside or surrounding air temperature? ..F
6. What is the storage room temperature? ...F7. What is the temperature reduction (TD)? (Subtract line 6 from line 5) ..F
8. What is the electrical load watts including lights and motors? ...watts
9. How many people occupy this space? ..................................................................................
10. What is the total product weight? .. 11. Product load information: ..
ii) Load calculation
A. Transmission (Wall) Load ( TQ ) per 24 hrsTQ = U x A x TD= _______________Btu / 24 hr
a. Interior wall surface (A)sum of the following areas:2 x (W) x (L) = __________
2 x (L) x (H) = __________2 x (W) x (H) = __________
+ __________
b. Over all heat transfer coefficient (U)
U=oi
hxh /1//1
1
c. Temperature difference (TD)TD = outside or surrounding air temperature - storage room temperature
B. Air Change (Infiltration Load) ( iQ )
iQ = a x b x c = _______________Btu / 24 hr
a. Interior room volume L x W x H = ________ cubic ft (Inside room dimensions)
b. Table 3 air changes per 24 hours = ____________
c. Table 4 Btu/cu. ft = ____________
C. Product Load ( pQ )
pQ
= apQ
, + HFQ + bpQ
, = ____________________Btu / 24 hr1. Product temperature reduction load above freezing ( apQ , )
apQ , = a x b x c =_________________ Btu / 24 hr
a. Total product weight = _____________ lbsb. Product temperature reduction to freezing = ______________ Fc. Table 5 specific heat above freezing = ____________
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2. Latent Heat of Fusion Load ( HFQ )
HFQ = a x b = _________________ Btu / 24 hr
a. Total product weight = _____________lbs
b. Table 5 latent heat of fusion = _____________Btu/lb3. Product temperature reduction load below freezing
a. Total product weight = __________ lbsb. Product temperature below freezing = ______________ F
c. Table 5 specific heat below freezing = ______________
bpQ , = a x b x c =________________ Btu / 24 hr
D. Miscellaneous Load
a. Electrical load (Watts) __________ x 3.42 x 24 =___________________Btu / 24 hr
b. Number of occupants________ x (Table 8) ________ x 24 = __________ Btu / 24 hr
E. Total Load without safety factor.....................................................= __________ Btu / 24 hr
F. Safety Factor (add 10% of Btu load per 24 hours) .......................=__________ Btu / 24 hr
G. Total Load with safety factor (Add E and F) ............................... = _________ Btu / 24 hr
REFRIGERATION LOAD (G) = _______________________BTU / 24hr
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4. Refrigeration equipment selection
Refrigeration equipment selection is done based on operating condi tions of the systemand aheat load calculation conducted for the environment we are aiming to cool and dehumidify.Equipment selection means relating product rating and specifications with cooling load and
prevailing working conditions. Inputs or information from parti cular applications(cooling loadand operating conditions) together with product performance dataand specificationsfrommanufacturers' literature will result in an informed selection decision. Different manufacturers'
may use different rating and specification parameters; therefore, all the possible rating and
specification parameters and their relation with refrigeration load and working conditionparameters are to be seen.
Operating conditions & Cooling load
To design or select a system the heat loadand operating conditionsof the system must beknown. The operating conditions of the system are determined from the application and working
environment of the system.
Operating condition parametersOperating (working) condition parameters are determined from application and prevailing
environment of the storage. These are: Room air dry bulb temperature and relative humidity or wet bulb temperature Ambient air dry bulb temperature and relative humidity or wet bulb temperature Atmospheric pressure or altitude of the site Power supply
Cooling load (refrigeration load)
Refrigeration load is the sum of all heat load elements, per day, of the space to be refrigerated.
This load is to be extracted by the refrigerating equipment during run time of the compressoroperation time of compressor per day.
Here under, two ways of equipment selection are to be covered. These are:
1) Packaged equipment selectionPackaged equipment selection means simply selecting factory assembled equipment which canyield the required refrigerating effect economically with in the prevailing working conditions.
2) Component based selectionComponent based selection means selecting components of refrigeration system based on theprevailing cooling load and working conditions and then system balance is to be conducted. In
this paper selection of the main components is to be seen i.e.:
Evaporator selection Compressor selection Condenser selection Expansion valve selection
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4.1. Packaged refrigeration equipment selection
Possible packaged unit performance data and specification parameters are:
- Nominal electric power - Suction pressure (or SST)
- Refrigerant - Sub cooling- Nominal absorption - Superheat- Rated voltage - Condenser air flow- Cooling capacity - Evaporator air flow- Defrost type - Air throw- Discharge pressure (or SDT) - mass flow rate
Possible load and operating condition parameters are:
Power supplyPower supply is the available power input (Volt/Phase/Hz)
Required condition of room air -determined from its application, these are:
Room air dry bulb temperature (DB) Room air relative humidity (RH) or wet bulb temperature (WB)
Conditions of ambient airdetermined from site metrological data. These are: Ambient air temperature (DB) Ambient air relative humidity (RH) or WB Atmospheric pressure or altitude of the place
Cooling load
Heat load per day of the space to be refrigerated, see cooling load calculation above.
To select packaged equipment all are some of the following parameters are important. These are:
1) RefrigerantRefrigerant selection depends on many factors, but the most factors which influence refrigerant
selection are heat of vaporization of the refrigerant, ozone depilating potential and global
warming effect. Currently, for commercial refrigeration the choice of the refrigerants is asfollows: R404A and R134Athe preference is according to the order.
2) Cooling capacityCooling capacity is heat extraction capacity of the evaporator (cooling unit) per day based onrunning time of the compressor. It depends on saturated suction (evaporation) temperature (SST)
which can be calculated by:
SST = Room TemperatureTD
WhereTD is temperature difference which depends on RH, see table 1
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3) Run time and defrost operations
Run time- is the maximum time that the compressor is in operation per 24 hours
Defrosting operationis a method of removing the frost formed on the evaporator coils
Frost formation on the evaporator coils depends on saturated suction temperature (SST) which
implies SST determines defrost operation needed and consequently run time of the compressor,which influences the required cooling capacity of the evaporator.
When the design suction temperature is over 30F, a defrost cycle is not normally required, and itis common practice to select equipment on a 20 to 22 hour compressor operation.
a. Air defrostFor suction temperatures below 30F and room temperatures over 35F, off-cycle (air defrost)
can generally be used. This involves cycling the compressor off with a time clock while the
evaporator fans remain in operation and room air melts the ice on the coil. For every two hours
of compressor operation, one hour of air defrost time is needed. Therefore, compressor selectionis based on 16 hours per day.
For suction temperatures below 30F and rooms below 35F, electric defrost, hot gas defrost orwater defrost is required. With these positive methods of defrost, equipment selection can be
based on longer compressor operation, with 18 to 20 hours most common. However, thisdepends on the type of equipment used and the latent load in the storage. A modern unit cooler
or product cooler in a tight room with average latent load can be selected on 20 hour operation.
The type of defrost used is generally a matter of either contractor or owner preference. Differentgeographic regions tend to use one particular type of defrost more frequently.
As a rule, electric defrost is more common than hot gas, and hot gas more common than water
defrost.
b. Electric Defrost
Electric defrost is the most common method in use today. Equipment cost is about the same aswith hot gas but installed cost can be lower. Operating cost is about 15% higher with electricdefrost than with hot gas and a fair amount of heat and moisture is released in the room during
defrost.
c. Hot Gas DefrostHot gas defrosting is still the most efficient method of defrosting regardless of storage
temperature but, unfortunately, most contractors are reluctant to use it. Defrost is very quick withminimum room temperature rise. Hot gas defrost, however, requires care to ensure that the
compressor is protected against liquid slugging.
d. Water DefrostWhile not very common, water defrost can be used on both medium and low temperaturestorages. Water must be at least 50F and is sprayed on the coil at a rate of about 3gpm/square
foot of coil for five to 15 minutes, depending on severity of frosting. Water defrost is fast and
efficient but some moisture is re-released into the room. These systems also require moremaintenance than electric or hot gas systems.
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e. General Defrost ConsiderationsBecause of high suction pressure (and high load) after defrost, compressor selection must be
checked to see that it can operate in a higher range than the actual design point. If not, acrankcase pressure regulator may be required to keep suction pressure down to acceptable
values. If this is the case, an accumulator should also be used. This is very important for a blast
freezer. On large air defrost systems (gravity coil, for example) it is a good idea to havesolenoids in the liquid and suction lines so refrigerant will not migrate during defrost. Inaddition, large fin coil installations are often split into sections with a thermostat for each section
to compensate for uneven room loading.
It is also recommended that a pump down system be used for both off- cycle and all defrostperiods.
f. Frost and dust (fin spacing) reduction factor (table 2)While high fin density gives increased coil capacity, it also increases the problem of dirt andfrost collection, which results reduction in cooling capacity.
Cooling capacity formula
Required cooling capacity=FTimeRun
hoursperLoadCooling
24
4) Evaporator fan air quantity (or Evaporator coil face velocity)In air cooled direct expansion (DX) systems refrigeration is achieved by passing air over an
evaporator coil surface which is directly cooled by an evaporating refrigerant flowing inside atube, or tubes, over which the air passes. Therefore; the quantity of air per KW of cooling
capacity that has to pass over the evaporator coil surface to achieve the required cooling can be
approximated by:
KWV
.
=
in
exit
pT
TCP
R
1
Where
KWV
.
is evaporator air quantity (m3/s) per KW of cooling capacity
R is gas constant, for air = 277.7kJ/kg
P is pressure of entering air (atmospheric pressure) in KPapC is average specific heat capacity of air = 1.005kJ/kg. K
inT is absolute temperature of entering air
exitT is absolute temperature of leaving air
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5) Evaporator coil face velocityEvaporator coil face velocity ( v ) can be calculated by calculated by:
v =A
V.
WhereA is evaporator coil face area = Fin height x Finned length
6) Suction pressure ( sP )
Suction pressure of compressor is given by:
sP= satP - PD
Where
satP is saturation pressure @ SST
PD is evaporator pressure drop
7) Discharge pressure ( gP )
Discharge pressure is the pressure at the exit of the compressor. Discharge pressure for a given
refrigerant is the saturation pressure corresponding to condensing temperature of the refrigerant.The condenser is cooled by the ambient air; therefore, condensing temperature depends on the
ambient air temperature.
Entering temperature difference (ETD) of CondensersETD is the temperature difference of entering (ambient) air dry bulb temperature and condensing
temperature. ETD for air condensers is in the range ( ).
Condensing temperature ( cT ) = ETD + atmT
From refrigerant properties table of the selected refrigerant:
Discharge pressure = saturation pressure @ cT
8) Power consumptionPower consumption is the power to be absorbed by the system to extract the cooling load. The
power consumption of the system is calculated as follows:
Power transferred to the refrigerant ( inW ) can be calculated by:
inW =COP
CapcityCooling
WhereCOP is coefficient of performance of a system with sub cooling & superheating
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i. Power in put at the compressor shaft ( shftW ) can be calculated by:
shftW =com
inW
Where
com is compressor efficiency
ii. Required electric power supply (W) will be:
W =emmec
shftW
Where
mec is mechanical efficiency
em is electric motor efficiency
9) Condenser heat rejection (THR) and condenser fan air quantityThe heat load absorbed at the evaporator and heat of compression has to be rejected at the
condenser. To absorb this heat sufficient air should pass over the air cooled condenser coil.
a. Total heat rejection (THR) at a condenser can be calculated by:
THR = Q + inW
Where
Q is cooling capacity
inW is power transferred to the refrigerant
b. Condenser fan air quantity (m3/s) per KW of heat rejection can be calculated by:
KWV
.
=
in
exit
pT
TCP
R
1
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Where
KWV
.
is condenser air quantity (m3/s) per KW of heat rejection
R is gas constant, for air = 277.7kJ/kgP is pressure of entering air in KPa
pC is specific heat capacity of air
inT is absolute temperature of entering air
exitT is absolute temperature of leaving air
10)Condenser coil face velocityCondenser coil face velocity ( v ) can be calculated by:
v =A
V.
WhereA is condenser coil face area = Fin height x Finned length
11)Refrigerant mass flow rate (.
m )
.
m =41 hh
CapacityCooling
Where
1h = h @ sP and sT = SST + Superheat
4h = fh @ LT = cT - sub cool
Where
Superheatand liquid temperature(sub cool) are to be taken from product rating
12)Required Air throw
Air throw can be estimated roughly as follows:
Air throw should be, at least maximum of (width, height or length of the storage room)
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4.2. Packaged equipment selection form
I. Inputs
i) Cooling load & Operating Conditions
- Available power supply (Volt/Phase/Hz):_____/____/____- Room air:
DB _____ Co or _____ Fo
WB _____ Co or _____ Fo , or
RH _____ %- Ambient air:
DB _____ Co or _____ Fo
WB _____ Co or _____ Fo , or
RH _____ %- Site condition:
Atmospheric pressure ( atmP ) ______Pa, or
Altitude above sea level (H) ______ m
- Cooling load per 24hrs: ___________KW = __________tons = ________BTU/hr
ii) Product specification (manufacturers catalog)
- Liquid temperature: ______ Co
- Superheat: ______ Co
- Temperature difference (TD) ______ Co
- Coil face area: ______ m2, or- Fin height: ______m & Finned length: ______m
- Leaving air: DB ____ Co & WB _____ Co - Leaving air velocity: ____m/s or _____CFM, or- Air throw: _____m
II. Fixing and relating the parameters
1. Refrigerant selection- Order of preference: R404A, R134a, according to heat of vaporization requirement and
evaporation temperature
2. Cooling capacity calculation2.1Fix SST ( saturated suction temperature)
SST = Room Air Temperature (DB)TD
TDuse table1 or product catalog
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2.2Fix defrost method
For SST > 30 Fo , normally no need of defrost cycle
For SST < -1 Co or 30 Fo and room temperatures > 2 Co or 35 Fo : Air defrostoff cycle defrost
For SST < 30 Fo and room temperatures < 35 Fo : a positive defrost method
Order of preference: Electrical,Hot gasand Water defrostrespectively
2.3Fix run time (compressor operation time per day) For no defrost cycle20 to 22 hrs For off cycle (air) defrosting - 16 hrs For a positive defrost method - 18 to 20 hrs
2.4Fix fin spacing
8fins/inch- down to 32 Fo 6fins/inch- hold freezers 4fins/inch- blast freezers
Cooling Capacity formula
Required cooling capacity=FTimeRun
hoursperLoadCooling
24
Where
F is fin spacing reduction factor (see table 2)
3. Evaporator air quantity per KW of Cooling Capacity
KWV
.
=
in
exitp
T
TCP
R
1
Where
P = atmP in KPa
pC = average specific heat capacity of air
exitT = Leaving air temperature (DB)
inT = ambient air temperature (DB)
3.1Evaporator coil face entering air velocity
v =A
V.
Where
Coil face area (A) is to be taken from product specification
4. Required Air throw (X)X should be, at least maximum of (width, height or length of the storage room)
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5. Suction pressureSuction pressure of compressor is given by:
sP= satP - PD
WheresatP is saturation pressure @ SST
PD is evaporator refrigerant pressure drop
6. Discharge pressureFrom refrigerant properties table of the selected refrigerant:
Discharge pressure = Saturation pressure @ cT
Where
Condensing temperature ( cT ) = ETD + atmT
ETD = () or specified by the manufacturer
7. Refrigerant flow rate (.
m ).
m =41 hh
CapacityCooling
Where
1h = h @ sP and sT = SST + Superheat
4h = fh @ LT = cT - sub cooling
Where
Superheatand sub cooling temperaturesare to be taken from product rating
8. Compressor power absorption ( shftW ) and Electric power supply (W)
Power input at the compressor shaft ( shftW )
shftW =com
inW
WhereinW =
COP
CapcityCooling
COP is coefficient of performance of a system with sub cooling & superheating givenby:
COP= bCOPYYY )1( 431
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4.3. Evaporator selection
To absorb the total heat loads at the given operating conditions the system must: Have refrigerant in the evaporator at a sufficiently low Temperature to enable it to
absorb heat from the storage room air and
Have a sufficient quantity of refrigerant flowing through the evaporator.Therefore let's work from the evaporator to determine system capacities.
The evaporator is the basis for capacity calculation - it is the component directly
responsible for absorbing heat energy from the storage.
The main factors which influence evaporator selection are: Room size, shape, orientation and application System temperature difference (TD)
Refrigerant type Saturated suction temperature (SST) - Evaporating temperature (Te) Air velocity (coil face velocity), evaporator air quantity or evaporator fan capacity Leaving air velocity (air throw)
Specification and rating parameters of DX coil evaporators
- Refrigerant - Refrigerant Flow- Evaporating Temperature (SST) - Refrigerant PD- System temperature difference (TD) - Leaving Air DB/WB- Fin spacing - Maximum Air Pressure Drop- Capacity - Leaving Vapor Velocity- Coil face area (Fin Height & Fined Length) - Coil (Tub) Type- Evaporator fan air quantity - No of Rows- Super heat - No of Circuits
- Liquid temperature ( LT ) - Fin material
Relation of specification and rating parameters with cooling load and operating conditions
Relation of cooling load and operating conditionswith specification and rating parameters are
shown above in packaged unit selection. Accordingly, DX fin and tube evaporator is to be
selected.
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Evaporator Selection Form
I. Inputs
i) Cooling load & Operating Conditions- Room air:
DB _____ Co or _____ Fo
WB _____ Co or _____ Fo , or
RH _____ %- Ambient air:
DB _____ Co or _____ Fo
WB _____ Co or _____ Fo , or
RH _____ %
Atmospheric pressure (atm
P ) ______Pa, or
Altitude above sea level (H) ______ m
- Cooling load per 24hrs: ___________KW = __________tons
ii) Product specification (manufacturers catalog)
- Liquid temperature: ______ Co
- Superheat: ______ Co
- Temperature difference (TD) ______ Co - Coil face area: ______ m
2, or
- Fin height: ______m & Finned length: ______m
- Leaving air: DB ____ Co & WB _____ Co - Leaving air velocity: ____m/s or _____CFM, or- Air throw: _____m
II. Fixing and relating the parameters
1. Refrigerant selection- Order of preference: R404A, R134a, according to heat of vaporization requirement and
evaporation temperature
2. Cooling capacity calculation
a. Fix SST ( saturated suction temperature)
SST = Room Air Temperature (DB)TD
TDuse table1 or product catalog
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b. Fix defrost method
For SST > 30 Fo , normally no need of defrost cycle
For SST < -1 Co or 30 Fo and room temperatures > 2 Co or 35 Fo : Air defrostoff cycle defrost
For SST < 30 Fo and room temperatures < 35 Fo : a positive defrost method
Order of preference: Electrical,Hot gasand Water defrostrespectivelyc. Fix run time (compressor operation time per day)
For no defrost cycle20 to 22 hrs For off cycle (air) defrosting - 16 hrs For a positive defrost method - 18 to 20 hrs
d. Fix fin spacing
8fins/inch- down to 32 Fo 6fins/inch- hold freezers 4fins/inch- blast freezers
Cooling Capacity formula
Required cooling capacity= FTimeRun
hoursperLoadCooling
24
Where
F is fin spacing reduction factor (see table 2)
3. Evaporator air quantity per KW of Cooling Capacity
KWV
.
=
in
exit
pT
TCP
R
1
Where
P = atmospheric pressure ( atmP ) in KPa
pC = average specific heat capacity of air
exitT = Leaving air temperature (DB)
inT = ambient air temperature (DB)
Evaporator coil face entering air velocity
v =A
V.
Where
Coil face area (A) is to be taken from product specification
4. Required Air throw (X)
X should be, at least maximum of (width, height or length of the storage room)
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5. Refrigerant flow rate (.
m )
.
m =41 hh
CapacityCooling
Where1
h = h @ sP and sT = SST + Superheat
4h = fh @ LT
Where
Superheatand liquid temperatureare to be taken from product rating
6. Selection
Knowing the above parameters, from manufacturers catalog DX fin and tube evaporator whichbest full fills the parameters is to be selected.
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4.4.Compressor selection
The compressor has 2 functions in respect to the circulation of refrigerant. It must:Discharge refrigerant into the condenser against the head pressure (high side pressure)
(determined in principle by the ambient temperature) and
Pull refrigerant through the evaporator (via the suction) to provide a nominatedsaturation temperature / pressure at an adequate flow rate (i.e. to pull the refrigeranttemperaturedown)
The flow ratevaries depending on the conditions in which the compressor is operating.
These conditions are often specified in the compressor ratings or performance graphs, whichmust be referred to in the selection process of the compressor. The following parameters are
important for compressor selection:
i) Compressor capacitySince the capacity and power requirement of a compressor and so of the system, vary with
changes in evaporating and condensing temperatures (pressures), liquid sub cooling and super
heating of the refrigerant, these conditions must be specified on selecting a compressor for anapplication.
ii) Operating conditions of the compressorThe maximum operating conditions of the compressor (suction & discharge pressures) can bedetermined from the ambient temperature and required evaporation temperature as follows:
iii) Suction pressure ( sP
)Suction pressure is the pressure at the inlet of the compressor.
Suction pressure of compressor is given by:
sP= satP - PD
Where
satP is saturation pressure @ SST
PD is evaporator pressure drop
If PD is not known, then:
When selecting compressor, use Suction Temperature 5F below evaporating temperature toallow for suction line Pressure Drop and 10F when suction accumulator is not used.
iv) Discharge pressure ( gP )
Discharge pressure is the pressure at the exit of the compressor. Discharge pressure for a given
refrigerant is the pressure corresponding to condensing temperature of the refrigerant.
From refrigerant properties table of the selected refrigerant:
Discharge pressure = Saturation pressure @ cT
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Where
Condensing temperature ( cT ) = ETD + atmT
ETD is the temperature difference of entering (ambient) air dry bulb temperature and condensingtemperature. ETD for air condensers is in the range ( ).
v) Compressor mass flow rate ( compm.
)
Mass flow rate of compressor is given by:
compm.
= suc .
ac tV
Where
compm.
is compressor mass flow rate
suc isdensity of suction vapor.
ac tV is actual compressor volume flow rate
Density of suction vapor is given by:
suc =sucv
1
Where
sucv is specific volume of suction vapor which can be read from refrigerant properties
table of the selected refrigerant at suction pressure and temperature i.e.
sucv = v @ suction pressure ( sP ) and suction temperature ( sT ) = SST + Superheat
Actual volume flow rate of compressor is given by:.
ac tV = v .
V Where
v is volumetric efficiency of compressor.
V is theoretical volume of compressor
Theoretical volume of compressor is given by:
.
V = N 4
2LD
Where
N is number of cylinders
D is cylinder bore diameter
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L is stroke length
Compressor capacity ( compQ ) can be calculated by:
compQ = compm.
( 1h - 4h )
Where
1h = h @ sP and sT = SST + Superheat
4h = fh @ LT = cT - sub cooling
vi) Power consumption (BHP/ton)Brake horse power, power required at the compressor shaft, per ton of compressor capacity is
given by:
TonBhp = COP72.4
WhereCOP is system coefficient of performance, see above
vii) Required electric power supply (W)
Required electric power supply (W) can be calculated by:
W =emmec
shftW
Where
mec is mechanical efficiency
em is electric motor efficiency
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4.5.Condenser selection
In commercial refrigeration forced circulation type air cooled condensers are used. An air-cooledcondenser is required to reject the heat load as well as the power drawn by the compressor. The
air-cooled condenser, installed outside, rejects the heat to outdoor ambient air.
The selected air-cooled condenser must meet the capacity requirements at the maximum outdoorambient air temperature of the region where the air-cooled condenser is installed.
Condenser Selection FactorsThere are factors that are required to enable the selection of a suitable condenser. These are:
- Application (low, medium, high temperature), approximate evaporating temperatures are: High +30F to +50F Medium -10F to +30F Low 40F to -10F
- Total heat rejection- Condenser fan air quantity
Selection Procedure
1. Maximum ambient operating temperature should be slightly greater or equal to ( atmT )
2. Total heat rejection (THR)Medium temperature refrigeration applications - use the following formula:
THR = Compressor capacity x THR factor x 1.05
Low temperature applications - use the following formula:
THR= Compressor capacity x THR factor x 1.1
3. Condenser fan air quantityCondenser fan air quantity (m
3/s) per KW of heat rejection can be calculated by:
KWV
.
=
in
exit
pT
TCP
R
1
Where
KWV
.
is condenser air quantity (m3/s) per KW of heat rejection
R is gas constant, for air = 277.7kJ/kg
P is pressure of entering air in KPa
pC is specific heat capacity of air
inT is absolute temperature of entering air
exitT is absolute temperature of leaving air
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Condenser coil face velocity ( v ) can be calculated by:
v =A
V.
Where
A is condenser coil face area = Fin height x Finned length
4.6.Expansion valve
The function of an expansion device in a refrigerating system is first to maintain the pressure
differential between the low pressure side (evaporator) and the high pressure side (condenser) for
a compressor driven refrigerating process. The second purpose is to regulate the refrigerant flowto match the heat flux in the heat exchangers. If the heat flux in the evaporator increases, the
mass flow through the evaporator should be increased accordingly.
Expansion devices can be divided into eight basic types:
1. Hand expansion valve 5. Electronic expansion valve2. Capillary tube 6. Low pressure float valve3. Automatic expansion valve 7. High pressure float valve4. Thermostatic expansion valve (TEV) 8. Constant level regulator
The first two are non-regulating expansion devices, while the other types adjust the flow based
on different means of signals.
In commercial refrigeration capillary tubeand TEVare commonly used.
SelectionUsing inputs from the selected evaporator, compressor and condenser suitable expansion valve
from manufacturer's catalog is to be selected. The inputs are:
- Refrigerant type- Compressor mass flow rate or capacity- Pressure drop ( P )
4.7. System balance
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APPEDEX:
Table 1: Influence of RH on system temperature difference (TD)
Relative Humidity % (RH) TD (F)
Over 90 8
80-90 1070-80 15
50-70 20
Table 2:Frost and dust (Fin spacing) reduction factor as function of fin spacing
Fin spacing (fins/inch) Cooling capacity reduction factor (F)
4 0.95
6 0.85
8 0.8
12 0.7
Table 3: Average air changes per 24hrs
Storage Rooms Below 32F Storage Rooms Above 32F
Volume
(cubic
feet)
Airchanges
per 24hrs.
Volume(cubic feet)
Airchanges
per 24hrs.
Volume(cubic feet)
Air changesper 24hrs.
Volume(cubic
feet)
Airchanges
per 24hrs
250 30.0 6000 5.2 250 38.0 6000 6.5
300 26.5 8000 4.5 300 34.5 8000 5.5
400 23.5 10000 4.0 400 29.5 10000 4.9
500 20.0 15000 2.8 500 26.0 15000 3.9
600 17.5 20000 2.5 600 23.0 20000 3.5
800 15.0 25000 2.2 800 20.0 25000 3.0
1000 13.5 30000 2.0 1000 17.5 30000 2.7
1500 12.3 40000 1.8 1500 14.0 40000 2.3
2000 11.9 50000 1.5 2000 12.0 50000 2.0
3000 7.8 75000 1.2 3000 9.5 75000 1.6
4000 6.0 10000 1.0 4000 8.2 10000 1.4
5000 5.6 5000 7.2
Note:
For storage rooms with anterooms reduce values by 50%; for heavy usage rooms, increase values by 2.
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Table 4: Heat Removed in Cooling Air to Storage Room Conditions (Btu per cu. ft.)
In Rooms Below 32F
Storage
room
temp.
F
Temperature of Outside Air F
40 50 70 90
Relative Humidity, Percent70 80 70 80 50 60 50 60
30 0.21 0.26 0.55 0.62 1.09 1.21 2.05 2.31
25 0.37 0.43 0.71 0.78 1.19 1.36 2.20 2.46
20 0.52 0.58 0.86 0.93 1.39 1.51 2.33 2.60
15 0.66 0.72 1.00 1.07 1.50 1.63 2.46 2.72
10 0.80 0.85 1.13 1.20 1.63 1.75 2.58 2.84
5 0.92 0.97 1.25 1.32 1.74 1.87 2.69 2.95
0 1.04 1.09 1.36 1.43 1.80 1.98 2 80 3.06
-5 1.15 1.20 1.47 1.55 1.92 2.05 2.90 3.16
-10 1.26 1.31 1.58 1.65 2.05 2.18 3.00 3.26
-15 1.37 1.42 1.69 1.76 2.15 2.28 3.10 3.36-20 1.47 1.52 1.79 1.86 2.25 2.38 3.19 3.46
-25 1.57 1.62 1.89 1.96 2.35 2.47 3.29 3.55
-30 1.67 1.72 1.99 2.06 2.44 2.56 3.38 3.64
In Rooms Above 32F
Storage
room
temp.
F
Temperature of Outside Air F
70 85 90 95
Relative Humidity, Percent
50 60 50 60 50 60 50 60
65 - - 0.32 0.52 0.58 0.81 0.85 1.12
60 .18 .18 0.58 0.78 0.83 1.06 1.10 1.37
55 .27 .27 0.80 1.00 1.05 1.28 1.32 1.5950 .39 .51 1.01 1.21 1.26 1.49 1.53 1.79
45 .59 .72 1.20 1.40 1.45 1.68 1.71 1.98
40 .76 .89 1.37 1.57 1.62 1.85 1.88 2.15
35 .93 1.06 1.54 1.74 1.78 2.01 2.04 2.31
30 1.08 1.21 1.78 2.01 2.05 2.31 2.33 2.64
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Table 5:Specific heat for various products
See
ASHRAE HAND BOOK 1981 FUNDAMENTALS
Section IV chapter 31 Thermal properties of foods
Page 31.4
Table 1
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Table 6:Approximate Heat of Respiration Rates at Temperature Indicated
Product
Btu/Pound/24Hours Btu/Pound/24Hours
32F 40F 60F Product 32F 40F 60F
Apples 0.45 0.8 2.05Melons -Honeydews -
0.5 1.4
Asparagus 4.70 9.0 18.5 Mushrooms 3.1 - -
Beans - Green 3.15 5.15 19.1 Okra - 6.0 15.8
Beans - Lima 1.35 2.6 12.2 Onions 0.45 0.5 1.2
Beets 1.35 1.75 3.6Onions -Green
1.8 4.9 9.0
Blueberries 0.85 - - Oranges 0.35 0.7 2.2
Broccoli 3.75 7.0 21.0 Peaches 0.6 0.85 4.2
BrusselsSprouts
2.9 4.4 10.1 Pears 0.4 0.85 5.4
Cabbage 0.6 0.85 2.05 Peas 4.2 7.4 21.0
Carrots 1.05 1.75 4.05 Peppers -Green
1.35 2.4 4.3
Cauliflower 1.95 2.25 5.05Peppers -Sweet
1.35 2.4 4.3
Celery 0.8 1.2 4.1 Plums 0.3 0.6 1.3
Cherries 0.75 1.4 6.0Potatoes -Immature -
1.3 2.4
Corn 4.65 6.0 19.2Potatoes -Mature -
0.8 1.0
Cranberries 0.33 0.45 - Raspberries 2.4 3.8 10.1
Cucumbers 0.28 - - Spinach 2.3 5.1 18.5
Grapefruit 0.35 0.50 1.55 Strawberries 1.6 2.7 9.0
Grapes 0.3 0.6 1.75SweetPotatoes
0.9 1.25 2.7
Lemons 0.35 0.65 1.8Tomatoes -Green
0.3 0.55 3.1
Lettuce - Head 1.15 1.35 4.0Tomatoes -Ripe
0.5 0.65 2.8
Lettuce - Leaf 2.2 3.2 7.2 Turnips 0.95 1.1 2.65
Melons -Cantaloupes
0.65 1.0 4.3
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Table 7:Heat Equivalent of Electric Motors
Motor
Hp
Btu/Hp/Hr
Connected Load in
Refrigerated Space1
Motor Losses Outside
Refrigerated Space2
Connected Load Outside
Refrigerated Space3
Btu/Hr Watts Btu/Hr Watts Btu/Hr Watts1/8 - 1/2 4,250 1,243 2,545 744 1,700 497
1/2 - 3 3,700 1,081 2,545 744 1,150 337
3 - 20 2,950 863 2,545 744 400 117
Note:
1. For use when both useful output and motor losses are dissipated within refrigerated space,motors driving fans for forced circulation unit coolers.
2. For use when motor losses are dissipated outside refrigerated space and useful work of
motor is expended within refrigerated space; pump on a circulating brine or chilled water
system,3. For use when motor heat losses are dissipated within refrigerated space and useful work
expendedoutside of refrigerated space;motor in refrigerated space driving pump or fan locatedoutside of space.
Table 8: Heat Equivalent of Occupancy
Cooler Temperature F Heat Equivalent/Person Btu/Hr.
50 720
40 840
30 950
20 1,050
10 1,200
0 1,300
-10 1,400
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Table 9: Operating condition for storage
Temp. Fo Humidity Operating T.D Fo Designed runningtime for unit
35 - 40 55% - 65% 20 - 25 18 hours
Table 10:Insulation Requirements for Storage Rooms
Storage Temperature Desirable Insulation
F U Factor
-50 to -25 0.01
-25 to 0 0.04
0 to 25 0.06
25 to 40 0.075
40 and up 0.1
Table 11: Suggested Freezer Temperatures F
Bread 0 Vegetables -10
Candy 0 Beef -10
Ice Cream -15 Lamb -10
Butter 0 to -10 Pork -10
Eggs 0 or lower Veal -10
Fish -10 Poultry -20
Shellfish -20