## Heat Dissipation in Electrical Enclosures

### Selection Procedure:

1. Determine input power in watts per square feet by dividing the heat dissipated in the enclosure (in watts) by the enclosure surface area (in square feet).
2. Locate on the graph the appropriate input power on the horizontal axis and draw a line vertically until it intersects the temperature rise curve.
3. Read horizontally to determine the enclosure temperature rise
Example: What is the temperature rise that can be expected from a 48″ x 36″ x 16″ enclosure with 300 watts of heat dissipated within it?

Calculating BTU/hr. or Watts:

1. Determine the heat generated inside the enclosure. Approximations may be necessary. For example, if you know the power generated inside the unit, assume 10% of the energy is dissipated as heat.
2. For heat transfer from the outside, calculate the area exposed to the atmosphere except for the top of the control panel.
3. Choose the internal temperature you wish to have, and choose the temperature difference between it and the maximum external temperature expected.
4. From the conversion table that follows, determine the BTU/hr. per square foot (or watts per square meter) for the temperature difference.
5. Multiply the panel surface area times the BTU/hr. per square foot (or watts per square meter) to get the external heat transfer in BTU/hr or in watts.
6. Sum the internal and external heat loads calculated.
7. If you do not know the power used in the enclosure but you can measure temperatures, then measure the temperature difference between the outside at current temperature, and the present internal cabinet temperature.
8. Note the size and number of any external fans. Provide this information to Nex FlowT to assist in sizing the appropriate cooling system.

### Solution:

Surface Area => 2[(48×36) + (48×16) + (36×16)] divided by 144 = 42 square feet

Input Power => 300 / 42 = approx. 7.1 Watts/SqFt.

### From Curve: Temp. Rise = 30°F (16.7°C)

Here is the situation, some electrical equipment will be in low ambient temperatures and I need to make it warmer so that there is no damage to it. I want to calculate how much power in watts is needed to heat up equipment inside a box made of aluminum.

The box is 5 inches X 5 inches X 11.5 inches (surface area is then 1.727 ft^2) it is made of aluminum and it is .1 inches thick and insulated.

from previous tests I recorded that when the ambient temperature was at -20 Celsius for a long time (at least 1hr) the surface temperature of equipment was steady at -10 Celsius and the air between the box and equipment was -11 Celsius.

Two questions: 1) How much power in watts is being dissipated by the electronics? 2) How much power in watts is needed to bring the temperature of the surface of equipment to 0 degrees Celsius? to 10 C?

I did some research and found this: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatcond.html

heat conduction formula

Q/t = kA(Thot – Tcold)/d

where k = thermal conductivity, A = surface area, Thot – Tcold = 10, d = thickness,

The issue with this is that when I plug in k for aluminum = 205 or .5 by adjusting the units in the formula I get a large value for Watts either way… am I doing this completely wrong is there another formula that would better model this problem?