Tuesday, June 19, 2012

Estimating the Cooling Requirements of an Electronic System

Estimating the Cooling Requirements of an Electronic System:

Any electronic system generates waste heat during normal operation. This heat must be removed – otherwise, it might damage the system components and cause malfunctions. Whether you are designing a system or diagnosing a system’s cooling requirements, you need to know which parameters to look at and how to estimate the airflow that will maintain a safe temperature within the system.
In a typical cabinet-mounted system, there are usually one or two power supplies, electronic circuits and displays, all of which can be assumed to generate heat within the cabinet. From the system’s power requirements, a fair idea of the power input may be estimated. If the system is cooled by simple convection, the thermal capacity of air can be taken to be 0.569W-minute/°C/ft³.
That means, every cubic foot of moving air can remove 0.569 W of dissipated heat every minute when its temperature changes by 1°C. To express it reciprocally, to dissipate 1W of heat, and maintain a 1°C change in temperature, an air stream of 1.757cfm (1.757 cubic feet per minute) will be required. Therefore, once you have estimated the heat dissipation within the system, estimating a cooling fan’s rating in cfm will depend on the internal temperature rise you allow. However, until you have completed your measurements and fitted the right size of fan, there will always be the risk of failure of system components. Therefore, for experimentation, what you need is a representative model.

Such a representative model is readily made by replacing the internals of the system with a power resistor and feeding DC current into the resistor from a variable power supply. The wattage of the resistor must be about 125% of the estimated power dissipation of the system electronics. Feed power to the resistor from a DC power supply outside the cabinet and vary the voltage and current settings until the power input matches with the estimated power dissipation of the system electronics. Using this arrangement you can dissipate the same amount of heat as the system electronics will, but now there is no risk of the circuit electronics frying while you experiment with the size of the fan.
To estimate whether the chosen fan is doing its job, you will need to measure the temperature of the exhaust air from the cabinet. To do this, you can use a multimeter with a thermocouple temperature probe. Place the temperature probe tip at the exhaust port of the cabinet and set the multimeter to measure temperature.

You may have to wait a while before the temperature inside the cabinet settles down to a steady value. You can now switch the fan on, and note the drop in temperature of the exhaust air.
Assume your system operates from a 400W AC/DC supply, and you load the supply to only 70% of its rating. Furthermore, the efficiency of the power supply is 75% – therefore, the supply is dissipating 25% of its power as heat. The total wasted heat power is therefore:
Pdiss = 70% x 125% x 400W = 350W.
This heat has to be removed by the fan.
It is typical to design the system to operate in an ambient temperature of 35°C, and let the temperature inside rise by about 15°C (TD). This will make the exhaust air about 50°C, as seen on the multimeter, and the fan must operate to maintain these conditions.
The effective airflow required in cfm will be n = k x Pdiss/TD. Here k is a constant with a value 1.757cfm.C/W. This gives n = 1.757 x350/15 = 41 cfm. Since the fan will be used in a closed cabinet, it will experience a considerable backpressure. Consequently, it may provide only 50% of its rated cfm. Therefore for our example we would select a fan with around an 80 cfm rating.

No comments:

Post a Comment