Heat exchangers for cooling electronic/electrical enclosures should be sized to provide adequate cooling for the anticipated worst case conditions. This is usually when the ambient is the highest, and also when the electrical loads through the enclosure are at the maximum. The outward portion of the heat exchanger uses either water or ambient air in the cooling process. The heat exchanger cannot cool the cabinet below the temperature of that water (or air). The greater the temperature differential between 1) the hot internal enclosure air and 2) the cooling water (or air), then, the higher the capacity of any heat exchanger will be. Conversely the smaller the temperature differential available in an application, the larger the heat exchanger size needs to be to achieve the goal.

The total cooling capacity required of the heat exchanger includes the:

(A) INTERNAL HEAT LOAD, (B) SOLAR HEAT LOAD and (C) HEAT LOAD TRANSFER.

(A)

The INTERNAL HEAT LOAD is the heat generated by the components within the enclosure.

(B)

The SOLAR HEAT LOAD is the additional heat due to the sun’s rays. NOTE: Unfortunately the calculation required to properly identify the true Solar Heat Load
is too extensive to provide here. Therefore we recommend you call the Kooltronic
Sales Department. They have access to a computer program that will provide an accurate
answer after a few simple questions. If you have an outdoor application, do not ignore
the solar heat load. It can be substantial.

(C)

The HEAT LOAD TRANSFER is the heat that is lost through the walls of the enclosure. (This statement assumes the outside ambient is cooler than the air inside the enclosure.) NOTE: Refer to the Glossary/Technical Comments section for explanations of technical terms and more information about Engineering issues.

STEP 1: Calculate the Internal Heat Load by using the Incoming / Outgoing Power Test Method

The Internal Heat Load can be determined by measuring the electrical energy that stays inside the enclosure. It is assumed that this energy is eventually transformed into waste heat. To measure this electricity, the current going In and Out must be measured in amps. The voltage of this current is also important. It is critical that all wires entering and leaving the enclosure must be included. Typically, a voltmeter and a clamp-on type ammeter should be used. The data must be recorded during the time when the current flow is the highest. A qualified technician is recommended for safety and accuracy reasons. The Internal Heat Load = 3.413 x Voltage (Current IN - Current OUT.) NOTE: This equation is derived from: 3.413 BTU = 1 Watt and watts = volts x amps. For example, if you measured 220 volts, 40 amps IN, 35 amps OUT, the Internal Heat Load = 3.413 x 220 x (40 - 35) = 3754 BTU/H. Consult with an Electrical Engineer if 3 phase power or a very complicated circuit is involved.

STEP 2: Calculate the Heat Load Transfer

The heat load transfer is the additional heat lost through the enclosure walls to the surrounding ambient. This is identified by the formula: Heat Load Transfer = (Max. Outside Ambient - Max. Allowable Internal Enclosure Temperature) x Surface Area x 1.25 HLT = (MOA - MAIET) x SA X 1.25

NOTE: 1.25 is a constant for metal enclosures. Use 0.8 for a plastic enclosure or 0.6 for an insulated enclosure.

The Maximum Outside Ambient (MOA) is the warmest room temperature surrounding the enclosure that might happen all year long. The MOA might be as high as 110ºF. The Maximum Allowable Internal Enclosure Temperature (MAIET) should not exceed the heat tolerance specification of the most sensitive component in your system. The MAIET might not be allowed to go over 120ºF per the enclosure’s component specifications.
The Surface Area (SA) is calculated as follows: Surface Area = (H x W) + (H x W) + (H x D) + (H x D) + (W x D). H = height in feet, W = width in feet, D = depth in feet. For example: H = 4, W = 2, D = 3: Surface area = (4 x 2) + (4 x 2) + (4 x 3) + (4 x 3) + (2 x 3) = 46sq. ft. Therefore in our example, the HLT = (110ºF - 120ºF) x 46 X 1.25 = -575 BTU/H.

STEP 3: Calculate the Total Cooling Capacity Required

The total cooling capacity required to cool your equipment is equal to: Internal Heat Load + Heat Load Transfer. The example: 3754 + (-575) = 3179 BTU/H. The performance of heat exchangers is expressed in WATTS/ºF. Therefore you will need to convert the BTUs to WATTS, so multiply by .29 The example; 3179 BTUs x .29 = 922 watts = Total Cooling Capacity Required.

STEP 4: Selecting the Heat Exchanger Performance Rating

Calculate the Temperature Differential: MAIET - MOA. Use the numbers select in Step 2: 120ºF - 110ºF = 10ºF.
Divide the Total Cooling Capacity Required from Step 3 by the Temperature Differential to reach the required Watts/ºF for this application. Example: 922 ÷ 10ºF = 92.2 Watts/ºF. NOTE: If the Temperature Differential in step 3 can be increased to 15ºF (by changing the Maximum Outside Ambient and/or the Maximum Allowable Internal Enclosure Temperature) then a smaller heat exchanger (rated at 61 Watts/ºF) can be used. NOTE: If Heat Exchanger is determined to be inadequate for your application see Air Conditioner Selection Guide or contact Kooltronic at (609) 466-3400.

HEAT EXCHANGER CAPACITY IN WATTS:

TEMPERATURE DIFFERENCE

CABINET HOT AIR minus COOLER OUTSIDE AMBIENT AIR (or WATER)

HEAT
EXCHANGER
PERFORMANCE
RATING
(WATTS/ºF )

TO CONVERT WATTS TO BTU’S, MULTIPLY BY 3.413 1 WATT = 3.413 BTU’S

The boxes above highlighted in yellow demonstrate the benefit of having cooler ambient air available. For example, if you need to eliminate 500 Watts and have 5 degrees of temperature difference, you will need a heat exchanger rated at 118 Watts/º F. But, if ambient air can be reduced by 5º, or 5º warmer cabinet temperature is acceptable, then the temperature difference changes to 10ºF. This will allow you to select a smaller heat exchanger rated at 55 Watts/º F.

SELECT THE APPROPRIATE HEAT EXCHANGER PRODUCT LINE: