Measuring velocity in a pitot-static tube and manometer

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A Pitot-static probe connected to a water manometer measures air velocity based on the deflection of fluid levels. The calculated air velocity, using the formula derived from Bernoulli's equation and the manometer's height change, is approximately 33.8 m/s. The discussion confirms that the pressure in the manometer corresponds to dynamic fluid pressure, allowing for the relationship between static pressure and air velocity. It emphasizes that while Bernoulli's equation applies to streamlines, the assumptions made about static pressure changes are valid for this scenario. Overall, the calculations and assumptions regarding the relationship between manometer height and air velocity are deemed correct.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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