refrigeration system for food storages

8/10/2019 Refrigeration System for Food Storages

1/46

MESFIN INDUSTRIAL ENGINEERING

2008

Refrigeration system for food storages

(selection guide)

Commercial coolers and freezersFISEHA M.


2/46

Page 2of 46

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


3/46

Page 3of 46

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 )


4/46

Page 4of 46

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


5/46

Page 5of 46

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


6/46


7/46

Page 7of 46

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.


8/46

Page 8of 46

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


9/46

Page 9of 46

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.


10/46

Page 10of 46

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


11/46

Page 11of 46

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


12/46

Page 12of 46

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


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


13/46

Page 13of 46

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


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.


14/46

Page 14of 46

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


15/46

Page 15of 46

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


is thermal conductivity of insulation material

ih is inside 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)


16/46

Page 16of 46

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:


17/46

Page 17of 46

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.


18/46

Page 18of 46

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.


19/46

Page 19of 46

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 = ____________


20/46

Page 20of 46

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


21/46

Page 21of 46

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


22/46

Page 22of 46

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


23/46

Page 23of 46

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.


24/46

Page 24of 46

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


25/46

Page 25of 46

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


26/46

Page 26of 46

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


27/46

Page 27of 46

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



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)


28/46

Page 28of 46

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


29/46

Page 29of 46

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)


30/46

Page 30of 46

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


ETD = () or specified by the manufacturer

7. Refrigerant flow rate (.

m ).

m =41 hh

CapacityCooling

Where


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


31/46


32/46

Page 32of 46

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.


33/46

Page 33of 46

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


34/46

Page 34of 46

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)


35/46

Page 35of 46

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.


36/46

Page 36of 46

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


37/46

Page 37of 46

Where


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


38/46

Page 38of 46

L is stroke length

Compressor capacity ( compQ ) can be calculated by:

compQ = compm.

( 1h - 4h )

Where


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


39/46

Page 39of 46

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




40/46

Page 40of 46

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


41/46

Page 41of 46

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.


42/46

Page 42of 46

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


43/46

Page 43of 46

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


44/46

Page 44of 46

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


45/46

Page 45of 46

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


46/46

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

refrigeration system for food storages

Documents