Air Conditioner Sizing and SelectionClimate Control

Cooling Electronic ControlCabinetsMost electronic control systems generate a substantial amount of heatduring operation. This heat factor isintensified as electronic controls aremade more compact, perform morefunctions and are placed in confinedareas. Additional problems areencountered when the electronicprocess control system is located on-line in an industrial setting, rather thanin a clean computer room. The factoryenvironment can be hostile to thepoint that performance and effectivelife of the electronic components are materially reduced or the control system fails completely. Ambient temperature might be excessivelyhigh, as that found in a steel mill.Moisture-laden air and airborneparticulate matter might be presentto adversely affect the electroniccomponents, as in the papermanufacturing industry. Airconditioners are designed to performreliably under many of these harshconditions and to provide the coolingand environmental protection requiredby sensitive electronic production control systems.

Factors Affecting ModelSelectionThis section is presented as abasic outline or checklist of thevarious application conditions to beconsidered when choosing a coolingunit. These are the factors which mustbe considered when selecting acooling unit:

Internal Heat LoadThis is the heat dissipated by the electronic controls. It is expressed inwatts. One WATT equals 3.413BTU/HR. Thus, to obtain the approximate cooling capacity requiredto remove a specific heat load, the following formula can be used: Watts x 3.413 - BTU/HR.For example, a heat load of 800 watts,require an air conditioner capable ofremoving at least 2,730 BTU/HR.

Resistance to Air Flow in theEnclosureAir flow is measured in cubic feet perminute (CFM). To create an air flow ofany desired velocity requires that pressure be produced by the blower.Resistance to this blower-produced airflow is created by obstructions withinthe cabinet in the air flow path. Theresistance itself is called staticpressure (SP) and is measured ininches of water column.The effect ofsignificant restrictions in the cabinetair flow path are as follows:• the obstructions cause static

pressure.• static pressure results in a pressure

drop, or differential, from the airvelocity produced by the blower.

• this reduction in cool air flow will decrease the effective capacity of the cooling unit. Allowance must be made for static pressure.

Heat Load from theSurroundingsAmbient conditions can cause a heatgain in the enclosure. The rated capacity of the cooling unit must besufficient to handle this heat gain.When evaluating the additional heatload gained from the surroundingsthere are two possible conditions:• the cabinet is insulated and well

sealed• the cabinet is not insulated.

Cabinet Insulated - Normally, well-insulated cabinets do not gainsufficient heat from the surroundingsto affect the air conditioner operation.BTU/HR ratings for Kooltronic airconditioners have been established atthe maximum ambient operating temperature of 125˚F. A substantialimprovement in heat removal occurswhen operating in ambienttemperatures below 125˚F.

Cabinet Not Insulated - Obviously,this design places more of a burdenon the cooling unit. Heat is conductedto the cool side. Thus, high ambientheat will be readily transmitted into the

cooler enclosure. To determine theadditional capacity required of an airconditioner installed in an uninsulatedcabinet, the surface area of the enclosure must be calculated to obtainthe total effective heat transfer area.For this calculation, use the surfacearea of the sides, plus the area of thetop and omit the bottom area of thecabinet. Air movement outside the uninsulatedcabinet will increase the heatconducted from the ambient into theenclosure. When there is little or no aircirculation outside the cabinet, thelayer of air immediately adjacent tothe exterior cabinet walls acts as aninsulating film. Exterior air movementdissipates this insulating layer of air inproportion to the velocity of the airflow. Substantial ambient aircirculation will increase the transmittedheat load imposed on the cooling unit.If the cabinet being cooled is not airtight, high ambient relative humiditywill adversely affect the cooling effectiveness of the air conditioner.When humid air infiltrates a poorlysealed enclosure, the air conditioneris required to use up valuable capacityjust to condense the moisture from theinternal air.Conversely, if the cabinet is wellsealed, high ambient relative humidityhas very little effect on the ratedcapacity of the air conditioner.

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Notes: Dimensions are in inches. Millimeters shown are for reference only. Data subject to change without notice.

The proper selection of an air conditioner is determined by the following criteria:

1) Required cooling capacity BTU/HR.

2) Mounting requirements (top, side or internal mounting)

3) Dimensions of air conditioner & enclosure

Air Conditioners Sizing & SelectionClimate Control

Fourth: (Watts x 3.413) + [1.25 x Area ft.2 x ∆ T (˚F)] = BTU/HR.

First Second Third

1 Watt = 3.413 BTU/HRDetermine the internal Watts of heat to bedissipated.

1 M2 = 10.76 ft2

Calculate the area of the enclosure which isexposed to the ambient air:

2H” (W”+ D”) + (D” x_W” ) = Area (ft2)144

1˚C ∆ T = 1.8˚F∆ TDetermine the temperature differential ∆ T (˚ F) by subtracting the maximum allow-able internal cabinet temperature (Ti) from

the maximum ambient temperature outside ofthe enclosure (To): To - Ti = ∆ T

H

W

D

TiTo

To determine air conditioner capacity required:

Calculations

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Notes: Dimensions are in inches. Millimeters shown are for reference only. Data subject to change without notice.

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To calculate the total internal enclosure temperature using the Heat Exchanger selected, add the temperature rise (∆ T)inside the cabinet to the maximum ambient temperature. Divide the internal heat load (WATTS) by the heat exchangerperformance (WATTS/˚F + Area Ft.2 x .22) and add the maximum ambient temperature outside of the enclosure (To),as in the equation:

Internal Heat Load (WATTS)Heat Exchanger Performance (WATTS/˚F) + (Area Ft.2 x .22)

Heat Exchanger Sizing and SelectionClimate Control

Calculate the area of the enclosure whichis exposed to the ambient air.

2H” (W” + D”) + (D” x_ W” ) = Area (ft2)144

+ To

Determine the internalWatts of heat to be dissipated.

Total CabinetTemperature =

Fourth: (Watts /∆∆ T ˚F) - (0.22 x Area ft.2) = WATTS/˚F (Required Heat Exchanger Performance)

H

W

D

TiTo

First Second Third

Determine the temperature differentialby subtracting the maximum ambienttemperature outside of the enclosure(To) from the maximum allowable internal cabinet temperature (Ti).

(Ti) - (To) = ∆T ˚F

Note: The maximum allowable internalcabinet temperature (Ti) should not

exceed the heat tolerance specification ofthe most sensitive component in yoursystem.

The proper selection of a Heat Exchanger is determined by how many WATTS can bedissipated at a given temperature differential.Since a Heat Exchanger uses ambient air forcooling, the enclosure cannot be cooled below atemperature slightly above the surrounding airtemperature. Therefore, the greater thetemperature differential, the higher the capacityof the Heat Exchanger.

NOTE: This selection process applies only to indoor gasketed enclosures which are uninsulated.

To determine required Heat Exchanger performance in WATTS/˚F:

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Notes: Dimensions are in inches. Millimeters shown are for reference only. Data subject to change without notice.

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Heater Calculation SheetClimate Control

Enclosure Heaters are used to maintain minimum operating temperatures and to help prevent failure of electronic components caused by condensation and corrosion.

Follow steps 1-5 to determine the heating requirement of an enclosure (US units - left column, metric - right column)

STEP 1: Determine Surface Area (A) exposed to open air

Enclosure Dimensions: height = feet meterswidth = feet metersdepth = feet meters

Assuming a free-standing enclosure ⇒ A = 2 (Height x Width) + 2 (Height x Depth) + 2 (Width x Depth)

STEP 2: Choose the Heat Transmission Coefficient (k) for your enclosure

painted steel = 0.511 W/(ft2 • K)5.5 W/(m2 • K)stainless steel = 0.344 W/(ft2 • K)3.7 W/(m2 • K)

aluminum = 1.115 W/(ft2 • K) 12 W/(m2 • K)plastic (or insulated stainless) = 0.325 W/(ft2 • K)3.5 W/(m2 • K)

STEP 3: Determine the Temperature Differential (∆∆ΤΤ))

Desired interior temperature = °F °CLowest ambient temperature = °F °C

Temperature differential = °F °C = K⇓

Calculation requires ∆Τ to be in Kelvin (K)

Divide ∆T (°F) by 1.8 for

STEP 4: Determine Heating Power (PV), if anyPV = components in the enclosure which generates heat (i.e. transformers, power supplies, etc.)

STEP 5: Calculating the Required Heating Power (PH)

If enclosure is located inside:

If enclosure is located outside:

A = ft2 m2or

PV = W Wor

∆∆T = K Kor

k = W/(ft2 • K) W/(m2 • K)or

PH = (A x k x ∆∆T) - PV = W

PH = 2 x (A x k x ∆∆T) - PV = W

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