# Change in velocity of air due to fan

1. Feb 25, 2010

### danield

Ok i have the following problem, I have a fan connected to the nozzle, i am given the nozzle area and the volumetric flow rate generated by the fan. How can i find the velocity of the air at different distances from the tip of the nozzle?
any help is greatly appreciated

2. Feb 25, 2010

### minger

3. Feb 25, 2010

### danield

if it is not too much to ask can you help me understand the equation? i am completely lost just from watching it..

Im assuming Umax is the velocity at the tip of the nozzle, and Function U is the velocity at certain distance away from the tip of the nozzle... im also assuming rho is the density of the fluid (in this case air) but other than that im lost with the symbols

Last edited: Feb 25, 2010
4. Feb 25, 2010

### Brian_C

If the flow is incompressible and one-dimensional, you can use the 1-D mass flow rate equation and cancel out the density.

5. Feb 25, 2010

### minger

Not really. He wants at distances past the nozzle.

Sure, actually, I'm sorry that's actually incorrectly a plane jet; you really want an axisymmetric jet (I actually think I put a wrong equation in there).

OK, basically you can get the velocity at the nozzle area by simply using the definition of volumetric flow rate
$$Q = VA$$
So
$$V = \frac{Q}{A}$$
So, let's denote the velocity at that section $$V_{max}$$ and the axial distance as 0.

From there, it can be shown that velocity drops off as $$x^{-1}$$ If you would like the derivation, I can post it, although it involves a similarity solution which may be a little over your head.

Note that there are a lot of constraints in there, and if you are off-axis or anything like that, then the equations don't work; you'll need the entire thing.

6. Feb 25, 2010

### danield

we are designing a toy for a class, and now we are doing engineering models. I am supposed to do an engineering model on a fan that is supposed to levitate a ball. I was able to model the drag force generated at the tip of the nozzle, but i havent been able to do so for different heights above it. Basically the part of the toy i am in charge of is a bernoulli ball levitation...

7. Feb 25, 2010

### Brian_C

I misread the problem. Care to cite a reference for your axisymmetric jet equation? I would be hesitant to apply it without knowing more about the problem.

Last edited: Feb 25, 2010
8. Feb 26, 2010

### minger

Viscous Fluid Flow, Frank White, Chapter 4, Section 10.6, The Narrow Axisymmetric Jet.

The full axial jet velocity is:
$$u = \frac{3J}{8\pi\mu x}\left( 1 + \frac{C^2 \eta^2}{4}\right)^{-2}$$
Where J is the integral of momentum of the jet, and the constant C is determined from that. \eta is r/x. I can post the full similarity solution if needed.

So, on-axis, you're velocity will drop off by 1/x. There have been a couple of posts so far on this very topic. Danield, what have you found as far as drag on a ball?

9. Feb 26, 2010

### FredGarvin

A nozzle makes a wonderful flow meter. Can you simply put a throat pressure tap and an upstream tap to measure the pressure? Then it is pretty easy to calculate the actual flow. The fan will most likely not put out the same flow with a different back pressure on it.

10. Feb 27, 2010

### danield

well what i did was the following, i was given the area of the fan and the volumetric flow rate, so then i assumed the nozzle will operate at steady state so A1V1=A2V2 and A1V1 is equal to the volumetric flow rate so i divided this by the area of the nozzle tip giving me the velocity at that point.

Then to calculate the drag force i had to make a couple of assumptions such as standard air density, drag coefficient to be .4 (rough sphere), then the area is 2*Pi*R^2 (of the sphere)and with the velocity i found from the previous equation i got the drag force at that point... now what im trying to find is how to calculate the force or velocity of air at distance Y from the tip of the nozzle... if it drops by 1/x what happens if the x is less than a meter wouldnt that say it increases?

any help is greatly appreciated

11. Mar 1, 2010

### minger

This...is a good point. Looking back through the plane jet solution, that one is divided by x as well $$x^{1/3}$$ actually.

Going back through though, the axial variable, x, is merely arbitrary and used for similarity (i.e. $$\eta = r/x$$. Moreso, it only shows up when the similarity variable is resubstituted back into the equations.

Most importantly, it is linear, so I would argue that you can define your x anywhere you want. It might be helpful to say that x=1 is at the exit of the nozzle. Then the jet thickness grows linearly as desired.

12. Mar 2, 2010

### danield

i am still confused as how to do this... should i use Vat nozzle divided by height and that should be the answer?