3 Replies Latest reply on Apr 25, 2013 3:57 PM by Mike Pogue

# Fans use power, does this input heat to the system?

I have some pc fans that are around 1 amps, 12 volts, 12 watts, times 6, that's 72 watts of power. Does this heat end up in the output of the fan?

how would i simulate this, just change the temperature coming out of the fan if it is an external inlet fan? or should i make a body and make it 12 watts, then blow the fan through it?

im assuming it will not make much of a difference, but i am doing a simulation now that is very close to the limit i want it at, and any added heat would make me depressed.

-Shane

• ###### Re: Fans use power, does this input heat to the system?

Technically yes there is some heat addition.  But most cooling fans are very efficient and most of that energy is converted into motion.

If you are concerned you would calculate something along these lines...

12W = 12 J/sec

cp (air) = ~1J/gK

density (air) = ~1.2kg/m^3

assuming fan flow = 1cubicmeter / minute = 1200g/min

heat input = 12J/sec=720J/min

so 720J/1200g = 0.6J/g

0.6J/g / 1J/gK => 0.6deg K temperature rise of incoming air

(multiply or divide according to the airflow of your actual fan)

This is worst case assuming all of the electrical power is converted to heat.

• ###### Re: Fans use power, does this input heat to the system?

Fans only add heat into the system if you have a temperature in their definition. I think this is the best way to include that effect vs the other methods that you describe. They are valid, but I don't think are the best fit. Regarding how much power, it isn't likely 12w. Consider that most of it probably goes to turning the fan. Go with a worst case input temp plus a couple degs maybe for the energy that is converted to heat.

• ###### Re: Fans use power, does this input heat to the system?

You can bound the heat that the fan dissipates. Moving some mass rate x [kg/s] of air from a dead stop to some velocity v can be converted to a power P (x*v^2/2, I think.). Then the heat dissipated will be volts*amps - P. This heat will be an upper bound, because getting the air started through a turbulent fan actually involves a lot of non-conservative work. I'd be curious whether this is useful or whether the bounds would be so conservative it doesn't really help.