# Wind Tunnel Test!

1. Nov 23, 2005

### rc flyer uk

I completed a wind tunnel test on a NACA 23012 aerofoil last week and am now writing a report on my findings! I need some help with working out the lift Coefficient as i seem to be going wrong somewhere.

Cl= L/.5 ρ Vw² S

To find the lift force i am subtracting the lift force which was necessary to zero the wind tunnel balance by the measured lift force
so: L=Lm-Lz

For ρ(air density) i am using 1.2kgm as this was not measured at the time of the tests.

This is where i belive i could be wrong for Vw(air speed within the working section) i am using the wind tunnel speed .075M water or 75mm water which was the speed we performed the test at. Would this be the speed in the working area if the projection manometer is connected differentially to static pressure tappings in the wall of the wind tunnel upstream of the contraction zone and within the working section. I am gusing it is but my calculations are well out of what i would believe to be normal Lift coeffients at the angles used!

Help and advice with many thanks

Rob

Ps My lift force Lz is 685 lb does this sound normall considering the lift force Lm was 76.3 lb at 0 degrees angle of attack??

2. Nov 23, 2005

### rc flyer uk

Ok i have a feeling i have identifed the error! I beleve that the problem is that Vw needs to be in m/s.

Does any body know how it is possible to convert 75mm/water in m/s??

Rob

3. Nov 23, 2005

### mezarashi

I'm not familiar with drag testing, so I don't think I can offer assistance there. As for unit conversion, can you clarify?

"75mm per water" wouldn't make sense to most people.

"75 mm of water per second" would. If that is the case, you would need the surface area, since $$\frac{\Delta V}{t} = A\frac{dx}{dt}$$

4. Nov 23, 2005

### rc flyer uk

Not quite sure to be honest! I would imagine it probally is per mm of water per second! As for surface area would this be the surface area of the fan ie where the 75mm water/second is coming from??

Is it possible to do it this way as well!

1/2 ρ Vw²= Pe-Pw/1-(Aw/Ae)² (the dynamic pressure)

So:

1/2 1.225 Vw²= 785.78 N/m²
Cheers

Rob

Last edited: Nov 23, 2005
5. Nov 23, 2005

### FredGarvin

In our end of the business, notation like .075M is usually a reference to Mach number when the M is capitalized like it is. You'll most likely have to calculate the speed of sound in the medium and then calculate the velocity from there.

6. Nov 23, 2005

### brewnog

I took millimetres/water to be an incorrectly written pressure reading, - millimetres of water (mmH2O), especially since this is how you'd likely measure air flow in a wind tunnel setup (in a U-tube manometer, with water as your medium). It's like mmHg as with many medical pressure readings, but with water instead of mercury. The capital M is merely metres.

1 mmH2O = 9.79706 Pa, at 15 Celcius.

Last edited: Nov 23, 2005
7. Nov 23, 2005

### FredGarvin

The capital M is a definite no-no. I thought the same thing though. The OP can check both pretty easy though.

8. Nov 23, 2005

### brewnog

I'm positive that the M refers to metres in this case. Shouldn't be a capital letter, but the scale is right (compare it with the millimetres), and metres of water is a standard measurement of a pressure differential for flow measurements.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?