# Stagnation Point - Bernoulli Equation

1. May 6, 2013

### LauraMorrison

1. The problem statement, all variables and given/known data
Can anyone tell me why, in the figure attached, the pressure in the manometer at 1 is a stagnation pressure?
I understand that you get stagnation pressure at a stagnation point but point one is below the stagnation point, not on the stagnation point. Therefore how can it be a stagnation pressure?

2. Relevant equations

Ps = P1 + 1/2ρV12

3. The attempt at a solution

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2. May 6, 2013

### rcgldr

The stagnation pressure is the pressure within that section of pipe where the velocity of the air has been declerated to zero, and not at the point below it where the air is still moving.

3. May 6, 2013

### Staff: Mentor

At the very entrance to the "manometer", the velocity of the air is zero; otherwise, there would be a flow of air through the manometer. So the very entrance to the manometer is a stagnation point. The pressure at that location is the stagnation pressure given by your equation.

4. May 6, 2013

### LauraMorrison

Yes, that is what I thought. The solutions to the question do not agree with this though. I have attached the part of the solution relating to the manometer as the overall question asks to calculate the speed V2. I don't know if this changes anything?

I also emailed my lecturer about this, however his explanation is fairly unsatisfactory as I still don't understand how the height between 1 and the stagnation point cancels. He said:

"The fact that the lower branch of the tube is directed into the flow signifies that stagnation pressure acts there. The change in height between point 1 and the tube entry is cancelled out in when calculating the pressure acting at the level of the gauge fluid (water) in the manometer. That is you could take BE between point 1 and tube entry, then relate pressure at tube entry to gauge fluid level, which gives the same result."

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5. May 6, 2013

### Staff: Mentor

The final result in the solution relating to the manometer looks OK to me. The cancellation explanation given to you by your instructor also seems correct to me. Suppose z represents the distance between point 1 and the stagnation location of the manometer. Since the free stream velocity is constant between point 1 and the location of the manometer inlet (except, of course, immediately at the stagnation point), the pressure in the free stream at the elevation of the manometer inlet is p1airgz. So the exact stagnation pressure at the manometer inlet is

p1airgz+(1/2)ρairv12.

Now, lets go inside the manometer. Let's ask what the pressure is within the manometer at the elevation of point 1. This is at the top surface of the water. The top surface of the water is located at the elevation of the manometer inlet minus the distance z. So the pressure at the surface of the water is

p1airgz+(1/2)ρairv12airgz = p1+(1/2)ρairv12

Note that, as your instructor indicated, the term ρairgz has cancelled out, and the pressure at the top of the water surface is just equal to the stagnation pressure at point 1.

6. May 7, 2013

### LauraMorrison

Thank you very much, great help.