Measuring velocity in a pitot-static tube and manometer

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Homework Help Overview

The problem involves measuring the velocity of air using a Pitot-static probe connected to a water manometer, with a specific deflection measurement provided. The context includes the application of Bernoulli's equation and the relationship between static and dynamic pressures.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the height change in the manometer to dynamic pressure and applies Bernoulli's equation to derive an expression for air velocity. Some participants question the applicability of Bernoulli's equation in this context, particularly regarding the streamline considerations of the Pitot and static pressure sensors.

Discussion Status

Participants are exploring the relationship between the manometer readings and Bernoulli's equation. There is a mix of interpretations regarding the assumptions made, particularly about the pressure changes and the conditions under which Bernoulli's equation can be applied. Some guidance has been offered regarding the correctness of the original assumption about pressure relationships.

Contextual Notes

Assumptions about the fluid properties and the setup of the manometer are under discussion, including the density of the fluids involved and the conditions of the airflow being measured.

JJBladester
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Homework Statement



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.

Homework Equations



http://composmentisconsulting.com/12-36.jpg

Pressure in a manometer = ρfluidgh

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

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.
 
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But, how can I relate this to the Bernoulli's equation?
 
Adam Nur said:
But, how can I relate this to the Bernoulli's equation?
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!
 
Assume your manometer is water. If you look at the manometer on both ends, then you'll see that
P01 = P02
Ps1 + 1/2*rhowater*V12 + rhowater*g*h1 = Ps2 + 1/2*rhowater*V12 + rhowater*g*h2
assuming that the change in height is only dependent on the static pressure,
dPs,water = Ps2 - Ps1 = rhowater*g*(h1 - h2)

Now if you assume that the wind streamline starts at zero (1), and compare to the wind at the inlet to the manometer (2)
P01 = P02
Ps1 + 1/2*rhoair*V12 + rho*g*h1 = Ps2 + 1/2*rhoair*V12 + rho*g*h2
assume that the height is negligible
dPs,air = Ps2 - Ps1 = 1/2*rhoair*(V12 - V22)

Since the change is static pressure of the wind is directly affecting the change in static pressure at the manometer,
dPs,air = dPs,water
rhowater*g*(h1 - h2) = 1/2*rhoair*V22
If you measure the manometer's change in height, then you can find the wind velocity. So yes, your assumption is correct.
 

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