Aerospace Velocity from Static and Total pressures

1. Jan 4, 2010

ptptaylor

Hi,
I am becoming a bit stuck with a fairly cynical question.

I want to work out the velocity of the fluid from the static and total pressures obtained in a wind tunnel experiment.
The manometor was open to air (atmospheric) at one end and measuring static or total pressure at the other. Both readings were taken.

To calculate the velocity p0=p+0.5*rho*u2

Which all seems fine.
Rearrange to find u

However, the values I have are not Ptotal or Pstatic correct?
I need to add atmospheric pressure to these?
I have an equation which is:
P0=patm-rho*g*h

So for an example,
patm=101081.39 Pa
So:
101081.39-(1.2*9.81*0.002)=P0
Would this indeed be the total pressure?

Paul

2. Jan 4, 2010

Brian_C

The manometer gives you the static pressure, which will be equal to the total pressure if it's placed at a stagnation point.

3. Jan 4, 2010

ptptaylor

Sorry, might not have been making it quite clear.
The static pressure measurement was staking perpendicular to the flow and the total (stagnation) press was taken normal to the flow.

4. Jan 4, 2010

Cyrus

Just use the bernoulli equation to back out the velocity, as per your textbook.

5. Jan 4, 2010

ptptaylor

Yes but I'm getting confused as to use as what in the bernoulli equation.
The equation is stated in my original post to find the velocity...

6. Jan 4, 2010

Cyrus

I believe your analysis is correct, but its been a few years since I've done that calculation so someone should double check my statement. I've become spoiled in getting data files of pressure measurements from digital instruments that I'm now rusty.

7. Jan 4, 2010

FredGarvin

If both of your pressure readings are referenced to atm, then you simply use the delta between the dynamic and static pressure in your calculation.

$$V =\sqrt{\frac{2 \Delta P}{\rho}}$$

That is assuming compressibility affects are negligible. If so you need to take that into account and is a function of Mach number.

8. Jan 6, 2010

mugaliens

That's 14.66 psi, off just 0.0000034 psi from atmospheric, so that's not the total pressure.

Fred's correct: use the differential pressure. Given your example, if it's representative, you'll not be encountering compressibility effects.