# Measuring velocity in a pitot-static tube and manometer

1. Apr 4, 2012

1. The problem statement, all variables and given/known data

A Pitot-static probe connected to a water manometer is used to measure the velocity of air. If the deflection (the vertical distance between the fluid levels in the two arms) is 7.3 cm, determine the air velocity. Take the density of air to be 1.25 kg/m3.

2. Relevant equations

http://composmentisconsulting.com/12-36.jpg [Broken]

Pressure in a manometer = ρfluidgh

Dynamic pressure from the Bernoulli equation = $\rho \frac{V^2}{2}$

3. The attempt at a solution

I reason that the height change in the water of the manometer is due only to the dynamic pressure entering the device. Therefore,

$$\rho_{air} \frac{V^2}{2}=\rho_{H_{2}o}gh$$

$$V=\sqrt{2\frac{\rho_{H_{2}O}}{\rho_{air}}gh}$$

$$V=\sqrt{2\left (\frac{1000}{1.25} \right )\left (9.81m/s^2 \right )(.073m)}=33.8m/s$$

I think I'm doing the math correctly. I just want to be sure my original assumption (that the pressure in the manometer = dynamic fluid pressure) is correct.

Last edited by a moderator: May 5, 2017
2. Nov 4, 2015

But, how can I relate this to the Bernoulli's equation?

3. Nov 4, 2015

### rude man

Well, if you don't get any other replies, my take is that you can't. Bernoulli applies to a streamline only, and the pitot and static pressure sensors are not in the same streamline.
Other takes solicited!