Calculation of coefficient of lift using a wind tunnel

[ajf]

I have done a lab experiment, to determine the coefficient of lift for an aerofoil, keeping the angle of attack constant.

The equation I have used is

Lift= (1/2) v times v times A times ρ times Coefficient of lLift

v= Velocity of air ρ=air density A= planform area
to give

coefficient of lift = (2 times Lift ) / ( v times v times ρ )

The answers i am getting are confusing me as the coefficient of lift should be constant for differing velocities ( ? or should it vary ? )

Also I think that the maximum coefficient of lift for any aerofoil is about 2.5

The planform area is 145mm by 223mm = 0.0323 square metres

v m/s 3.45, 4.5, 4.8, 5.0, 6.2, 6.5, 7.4
ρ 0.1, 0.13, 0.15, 0.18, 0.21, 0.25, 0.29

Lift ( N) 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8

Coefficient of lift 10.39, 7.05, 7.16, 6.87, 4.6, 4.10, 3.12

the term ρ is a mystery to me. I was reading it directly from an electronic readout, from a pitostatic tube in the wind tunnel.

the electronic readout had a label attached to it which said +-250 P , which i don't understand either - especially when the read out was only 0.1 to 0.29 ?

I am mixed up if it is air density or pressure or what units ρ is.

will you help me make sense of this please.

thanks very much,

ajf

 

[AlephZero]

$\rho$ is the air density, and for a simple low speed wind tunnel it is constant (though you need to find the value for the time and place you did the experiment, from the air temperature and barometric pressure). Your numbers between 0.1 and 0.29 don't look like densities. The density of air is about 1.2 kg/m^3.

A pitot tube measures the difference between static and dynamic air pressure, and you can put that into Bernouilli's equation to give you the air velocity.

+-250P might mean +-250 Pascals pressure difference, but we can't guess what your experimental setup was, you will have to ask somebody in the lab.

Possibly you are confusing $p$ (pressure) and $\rho$ (rho, density) in your equations?

 

[boneh3ad]

This is all not to mention that density shouldn't be changing at these speeds. I think the first step is to go back and make sure you understand the data you are getting from your instruments, which starts with knowing what each instrument is actually measuring.

 

[RandomGuy88]

The lift coefficient of your airfoil can change slightly with velocity but it won't be a large change especially if you are not near stall and you tested over a pretty small range of velocities.

The lift coefficient of your airfoil is not likely to be anywhere near 2.5, especially at these low speeds unless it has flaps. That being said your calculated lift coefficients are definitely off but it most likely is because of this mysterious ρ you are reading from an electronic readout. Its possible that the value you are reading is voltage and it needs to be converted to the correct value. But like Aleph and boneh3ad mentioned it is not the density because that is constant and you should ask someone familiar with the lab equipment.

What aerfoil section are you using?

 

[ajf]

Thanks for your replies -

Yes, I am far from happy with my understanding, my regular teacher is absent just now, and the cover teacher specialises in other subjects.

I have put a value of air density of 1.252kg /cubic metre into the equation and things have made a bit more sense.

(The aerofoil section is unidentified, and I have been unable to find out what it is.)

thanks again for your help.

 

[Cashlover123]

If you want to find what the airfoil is, i think the best way to go about is to go to this webpage http://www.worldofkrauss.com . Here you can input the thickness of the airfoil wrt to the chord of the wing, which you can easily measure. Also, you can derive the 2D Clmax of the wing and input it in along with the thickness.. It will definitely narrow down the no. of airfoils and by looking at the shape of the airfoil and comparing with that in the website, i think it will be easy to identify the airfoil..