Question
(a) Using Bernoulli’s equation, show that the measured fluid speed v for a pitot tube, like the one in Figure 12.7(b), is given by v=(2ρghρ)12v = \left ( \dfrac{2\rho^\prime gh}{\rho}\right )^\frac{1}{2}, where hh is the height of the manometer fluid, ρ\rho^\prime is the density of the manometer fluid, ρ\rho is the density of the moving fluid, and gg is the acceleration due to gravity. (Note that vv is indeed proportional to the square root of hh , as stated in the text.) (b) Calculate vv for moving air if a mercury manometer’s hh is 0.200 m.
Question by OpenStax is licensed under CC BY 4.0
Final Answer
  1. Please see the solution video.

Solution video

OpenStax College Physics, Chapter 12, Problem 24 (Problems & Exercises)

OpenStax College Physics, Chapter 12, Problem 24 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .
Video Transcript
This is College Physics Answers with Shaun Dychko. This pitot tube shows the air speed based on this height difference between the fluid levels in this manometer. And we're going to show that this air speed V two is equal to two times the density of the fluid in the manometer times g times the manometer fluid height here h divided by the density of air, all square rooted, or raised to the power of one half in other words. So we start with Bernoulli's equation, which says that the pressure at position one plus one half times the air density times the speed of the air at position one squared, plus the density of air times acceleration due to gravity times the vertical height of this position equals the pressure at position two which is a little hole in this section of the tube, plus one half density times the speed of the air here V two squared plus rho g h two. Now h one and h two are pretty much the same. This little hole is not much higher than this hole here. And so we'll say since these are equal these terms are the same and they subtract to make zero, this hole here is connected to a tube that has a dead end in it. So that means that the air will not move here. So this is why V one equals zero. So this term disappears. And then we can subtract P two from both sides. Switching sides around and we could say that one half times air density times the speed V two squared equals the difference in pressures pressure one in here minus pressure two across this hole and therefore also in this region here. Or in other words, V two squared after you multiply both sides by two over density is two times pressure difference divided by air density. Now, this pressure difference between this region and this region. So here's P one in here and pressure two is in there. That difference is going to equal this height of this fluid. That's rho prime. So that's the density of this manometer fluid times acceleration due to gravity times its height. So we can substitute that in place of P one minus P two. And we do that here and then take the square root of both sides. And taking the square root is the same as raising to the exponent one half. So we've shown the V two is gonna be the square root of two times the manometer fluid density times g times the height of the manometer fluid divided by the air density.

Comments