Fan Manufacturer: 145W to Move 1kg of Air 1m

[Tsar]

Homework Statement

Hi, I am trying to work out the energy needed to move a certain quantity of (ideal) gas using a fan but am finding a great discrepancy between my calculations and those supplied by the fan manufacturers.

let's say i'd like to move 1kg of air (1kg/m3) a distance of 1m. I get the energy required fs to be 10Joules. The fan manufacturers are saying that their fan to do this is 145W. I get that fans aren't 100% efficient but still, 145J to do 10J work? I'm sure I'm missing something!

Thanks!

Homework Equations

e=fs

The Attempt at a Solution

e=fs,
e=(1kg x 10ms-1) x 1m
e=10J

 

[Bystander]

You are comparing energy to power for a first source of confusion. You are not really specifying where, by what path, and how you want to move air for the second source; your energy calculation is meaningless.
Could you restate your air movement requirements?

 

[Tsar]

...Thanks Bystander.

In essence, I am trying to use a fan as an extractor. The air will simply be moved from one side of the fan to the other. I assumed that the energy required to move 1kg mass of air a distance of 1m would need 10J. The fan manufacturers state that their fan will use 145J to move 1kg of air in 1 second, hence the power of 145W.
I hope that's a bit clearer :)

 

[rcgldr]

I'm not sure what you mean by moving 1 kg of air 1 meter. It would make more sense to state the desired result in terms of power, such a 1kg of air accelerated to some velocity, say V = 10 meters / second, every second, so that the power is (1/2 1kg (10 meters / second)^2) / second = 50 joules / second = 50 watts.

If instead the goal is to create a pressure differential across some cross sectional area, then the force = pressure differential times the cross sectional area, and power = force time speed of the flow.

 

[Tsar]

...this is beginning to make sense now!
I am actually trying to use the airflow to cool down a large area. The area being cooled is 1m long and I estimated the amount of air that I needed to transfer across the area being cooled should be 1m3 of air every second (i.e. 1kg of air). I thought then to just use that as part of (force x distance) to calculate the work done and thus energy required. I'm guessing that's not the correct way to go, what am I doing wrong?

 

[Bystander]

It's not a matter of lifting the air against gravity, it's pushing the air you want to move hard enough that it displaces the air that occupies the space where you want it to go, so that it displaces air further down the line, ...

 

[Tsar]

I get it now, thanks!
I think i'd better brush up on my theory!

 

[Tsar]

Hi Guys,

I'm sort of getting there but still am a bit confused. After thinking about what I am trying to do I have managed to define the overall problem as follows:

Beginning with a large tube of diameter 1m and length of 100m that is formed into a circle so that both ends meet at a fan placed between them. Forming in effect a hollow ring so that the fan is just circulating the air inside around continuously. If the air has a density of 1kg/m3 and the total air inside is 78.5m3 or 78.5kg. With negligible friction, how would I work out the energy needed for one complete rotation of air. In other words how much energy is needed to push the air from the fan output around the circuit once ending at the input of the fan.

I hope this makes things a wee bit clearer :)

 

[rcgldr]

If friction is negligible, then power required to maintain a steady flow would be zero.

 

[billy_joule]

And the initial energy can be calculated from the change in kinetic energy of the air mass. Divide that by time to get power.
Of course frictional losses will be significant so it won't be at all realistic.

 

[Tsar]

Excellent, that's kinda what I thought!

Is there a simple way/equation I can use to estimate the frictional forces and thus how much energy would be needed to keep the air going round?

Thanks!

 

[CWatters]

Google found this..

http://www.engineeringtoolbox.com/air-duct-friction-loss-diagram-d_328.html

Looks like you can use that to estimate the pressure drop down a length of duct. The fan will have to match (eg provide) that pressure drop.

Then if I remember correctly:

power = pressure * flow rate = pressure * area * velocity

That should give you the output power. To calculate the input power you need to know the efficiency of the fan.

Edit: Interesting comments on that here...
http://www.nmbtc.com/fans/white-papers/fan_efficiency_important_selection_criteria/