Enclosure airflow calculations: a simple guide
01Sep
To some, enclosure air conditioning is an additional cost on top of the expensive equipment already installed in a control cabinet. However what is often ignored is the long term cost of NOT sufficiently cooling one’s kit. Over heating accounts for some 60% of faults in enclosure mounted equipment, making it the largest single cause of failure.
Whilst installing an effective cabinet cooling system may seem expensive, when compared to the cost of downtime and purchase of new inverter drives or motors, the amounts involved are insignificant. Proper air conditioning of a cabinet counters this problem, as well as two other common causes of failure - dust and humidity.
Aside from the costs of unscheduled shut downs and new equipment purchase, running motors and drives under sensible operating temperatures adds significantly to their life expectancies. Operating an inverter at 35ºC instead of 70ºC can lead to an increase in operation life-span of a factor of ten. Working out exactly what you need for effective cooling is not hard either as Jonathan Brindley, Product Manager for Rittal’s Systems Climate Control Solutions division, shows us.
Calculating the airflow required to cool an enclosure can be simple. First, calculate the amount of heat
energy being generated within the enclosure that needs to be dissipated. Next calculate the required air
temperature and establish the altitude. Then from this data the correct airflow for the application can be calculated.
In many applications users are aware of their equipment’s ability to generate heat and can provide the figure from their supplier’s manual. In other instances it may be that users can provide the specification of the equipment installed or being installed. Where the heat generation figure for equipment is unknown, the user can take an estimate by using an equipment efficiency of 97%. It is also usual to build in some allowance for unforeseen factors.
There are several software programs available on the market that can be used for the calculation of heat output. Therm 4.1 from Rittal is an example that will calculate the heat output for installed equipment and allow the user to build their own equipment library.
Whichever way you choose to calculate your heat loss, the figure is usually referred to as Qv[W]. Before
moving on the calculate airflow, it is important to calculate the difference between the outside and the required internal temperatures (ssT). This figure should always be positive (ie. the internal greater than the external temperature).
The calculation required from these figures is as follows:
V = f(Qv / ssT)
Where:
V = Required volumetric air flow (m³/h)
F = Volume (m³) of air required at given altitude per ºC change to facilitate a Watt-hour of heat transfer
Qv = Heat generated by equipment
ssT = Difference between internal requirement and external ambient
At first glance this might seem quite a simple piece of arithmetic, however, f and ssT warrant a brief
explanation. Air pressure, and as a result density, increases at lower altitudes due to the compressing effect of the air above it.
There are more air molecules per m³ at sea level than at say 1000 metres and therefore more are available for heat transfer. A fan working at sea level consequently has a greater cooling effect than the same fan operating at the same speed at a higher altitude as it is pumping more air molecules per unit volume of air.
The greater the altitude of a site location, the larger the volume of air required to achieve an equal effect at sea level. The affects of air temperature on the required airflow are very pronounced as this too affects air
density.
Air density increases as temperature decreases, and as a result, a greater cooling effect from air is achieved per m³ from a given fan flow. The best conditions for cooling are therefore at sea level on cold days under high atmospheric pressures. In order to demonstrate how simple the calculation can be, consider the following:
Two 45kW inverter drives installed in an example enclosure system would generate a heat figure in the region of 2 x 900W or 1800W in total. Allowing for unforeseen factors, the figure could be adjusted to 2100W. Therefore 2100W is the amount of heat produced by the installed equipment within the enclosure system.
Now assume that the outside temperature reaches a maximum of 25ºC, and that the installed equipment inside needs to be kept below 40ºC. In this situation, the difference (ssT) will be 15ºC. Also the example enclosure system is located near the coast at only 34 metres above sea level so f = 3.1m³.
Now all the elements required to calculate the airflow (m³/h) needed for the example enclosure system are
available and the calculation and result are shown below:
3.1m³ x (2100W/15 º C ) = 434 m³/h (the required air flow)
In summary, calculating fan flow is not a difficult exercise as long as time is taken to work through each step in sequence.
There are several ways of calculating installed heat loss within an enclosure, including self-estimation using equipment ratings or software. Calculations should include an element considering altitude, its
effect on air density, and the resultant efficiency of air at dissipating heat.
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