Solved Problem: Measurement of Air Velocity with a Pitot Tube
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- Опубликовано: 27 июл 2024
- MEC516/BME516 Fluid Mechanics, Chapter 3 Control Volume Analysis, Part 8: The application of the Bernoulli equation to the measurement of fluid velocity. An example of using a Pitot Tube to measure the air speed in a wind tunnel is solved.
All of the videos for this course and copy (pdf) of this fluid mechanics presentation can be downloaded at: www.drdavidnaylor.net
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
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All the videos for this introductory Fluid Mechanics course are now available at: www.drdavidnaylor.net/
I may have gotten carried away, my apologies. Might I ask a shorter, simpler version instead, that won't req writing anything down or keeping track of variables?
I basically just have a drinking straw. I want to breathe in pulling as much air through as possible. So the question is, how can I edit the straw to pull more air through ? I've tried making one end larger or smaller than it was, as well as both ends larger or smaller than it was so start. The problem is I cant tell physically which one I'm breathing in the most air, and I don't how it works on paper so I really just can't tell lol
I appreciate your explanation and yes! Fluid Matters!
Good explanation - succinct :)
thank you
thank you sir
Thank you.
I’m looking for equations to determine the air velocity of an underground tunnel with one opening at ground level and the other opening at 100m above ground level with a wind blowing at 20 m/s.
I believe this video gives me everything I need.
Sounds like an interesting application....
@@FluidMatters It’s based of off the American Black-tailed prairie dog’s burrows. They use a mounded entrance and a level entrance to create a pressure differential when the wind blows. This creates natural ventilation.
I want to try and figure out how much wind power can be extracted with subterranean wind turbines using the same method.
ruclips.net/video/s-QHTWP26QI/видео.html
Sir please upload more mechanical engineering courses lecture of your own ...please sir
Great explanation sir.
Thanks and welcome
Great explanation and demonstration. I'm still struggling to get my head round the application of Bernoulli's equation to compressible fluids. In the above example you have one value for Rho (atmospheric pressure), but the fluid density in the duct will be higher because of static pressure. Surely that would invalidate the results of the calculation because Rho can't be treated as a constant?
Yes. That's true. But the pressure change is often small -- as this solved example, 166 Pa. So, the absolute pressure is 98.0 kPa+0.166 kPa at the stagnation point. That will increase the air density ever so slightly (0.17%). So, it's a reasonable approximation. In a different application, if the air pressure change is large relative to the absolute pressure, then using Bernoulli's equation becomes dubious.
Nice se
In this case you wouldn't include the pressure drop from the air in the manometer due to the density being so low. In other cases, the pressure from the liquid would have to be accounted for in the equation.
There is air on both sides of the manometer, except in the 20mm section. So, the pressure difference caused by the air in the manometer is ~1.2kg/m^3(9.8m/s^2)0.02m=0.23 Pa. For most purposes, this is negligible compared to the 166 Pa that is being measured. I hope that helps.
12:20 - why we are taking into account only atmospheric pressure during density calculation? what about pressure inside wind tunnel? It does not affect density of inside air? As we know, bigger pressure = bigger density. Would be great if you could let me know why we only take atmospheric pressure.
Yes, if the wind tunnel was pressurized you should include that effect. In practice, that would be unusual. Most low-speed wind tunnels are not sealed, and operate at (or very close to) atmospheric pressure. In this problem statement shows the pressure inside the wind tunnel as p_atm, a reasonable assumption.
@@FluidMatters Thanks for your respond! So, wind tunel example is clear for me. But just to be sure, please correct me if I am wrong - In case pitot measurment inside round ducting, when p_atm = 101 325 Pa, and pressure inside duct is p_duct = 12 000 Pa, for density calculation I should use p_atm + p_duct. Am I right?
@@tomaszgalek9788 Yes. You in an HVAC application should use the pressure in the duct, if you know it, to calculate the density of the air. (Keep in mind that the local atmo. pressure is not 101325 Pa -- it varies substantially (+/- a few percent) with the local weather conditions. So, using this value to (6 digits!) does not make sense. That's why you will usually find a barometer in a wind tunnel lab.)
@@FluidMatters You are right, in HVAC application, (depends of location - altitude) atmo pressure is different. 101325 Pa value was just an example. Many thanks for your time, I really appreciate it! :)
ruclips.net/video/s-QHTWP26QI/видео.html
How would this problem work with water and mercury in the manometer?
Same basic "Bernoulli" analysis. But the manometer pressure difference calculation would be different -- needs to account for the water on the RHS: h*g*(rho_merc-rho_water). I hope that helps.
How can determine volume flow rate of water?
You can use a venturi flow meter. See this video: ruclips.net/video/so1NXTcTirs/видео.htmlsi=Mr5x05bk1Xf0U0gL