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Closet cooling and fan static pressure

kaspor

Electrical
Aug 12, 2021
41
Hello

I have a closet which is approx 2.7m high by 1m deep by 1.2m wide. I have l equipment in there that needs to sit at max 30degC. There are two double doors on the closet.

In order to keep it at 30degC I’ve installed 4 extraction fans on a thermostat. The fans draw air in through gaps in the doors and from underneath and exhaust it at the top. The fans are similar to these


The problem is the noise. They’re generation 30dba each. I’ve insulated them as much as I can but they’re still loud.

So I then started looking at quieter fans such as these


They have a much lower db but looking at the performance curve it looks to have a much lower static pressure rating


Does this mean they will inherently take longer to get the temp in the closet down as they are not able to draw the same amount of air in?

I’m worried that if I change the fans over that the newer fans will be useless as they won’t be able to draw air in like my current ones.

Also cutting holes or vents for intake isn’t an option unfortunately.
 
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The 30dB noise is the noise of the fans, not including the sound that vibration of the wall they are attached to. 30dB is around a whisper; even 40dB is similar to a quiet library.

If the fans seem too loud, perhaps they need better isolation from the wall, perhaps with soft duct work.
 
Affinity laws applied to fans predict that for a given speed reduction the flow will vary same as speed reduction and the static pressure output will vary in accordance with the square of the speed.

So as I indicated above your existing fan outputs 8 Pa static pressure at 80 m3/hr, if you reduce the speed by 50% the new fan output at same point on curve will be 2Pa at 40 m3/hr which is very low. So you can create a new curve for 50% reduction in speed by selecting points on the original 100% speed curve and use the affinity laws to draw the new curve of pressure versus flow. Here is information on the affinity laws:


One thing though is that you really don't know the heat dissipated by the electrical equipment so you really don't know the true required ventilation rate. When designing a ventilation cooling system such as this you would normally add up all the heat from the equipment then calculate the required flowrate of ventilation air. The flow of ventilation air is determined by the following equation:

Q = 1.1(CFM)(delta T)

Q is heat input (BTU/HR)
CFM is the flowrate to be calculated (CUBIC FEET PER MINUTE)
delta T is the temperature rise of the air between inlet and outlet (Deg F)

The heat dissipated by the electrical equipment might be obtained from manufacturer's literature. For purely electronic components such as servers I would think that all electrical power input would eventually be dissipated as heat just like a light bulb? For items such as motor starters or electrical panels only the loss due to resistance of the wiring is dissipated as heat while the rest of the power is transferred to the user via the current flow.
 

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