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I Pitot tube

  1. Nov 12, 2017 at 5:55 AM #1
    fig_16.3.gif

    I have a problem here. I dont get why the stagnation pressure equal the height of the water in the pitot tube.

    The liquid manometer is easy. We analyse the forces on the horizontal contact surface of the tube and the liquid manometer so P1 = Density * g * h

    However I fail to analyse the forces on the curved section of the pitot tube. How can I prove that P2 = density* g H
    so I can use P2 - P1 = density *g (H-h)
     
  2. jcsd
  3. Nov 12, 2017 at 8:29 AM #2
    Have you learned about Bernoulli's equation yet?
     
  4. Nov 12, 2017 at 10:05 AM #3
    Yes I have.
    If we choose a stream line that passes through the tube then change in height is equal to zero. Also since the fluid will have zero velocity at the opening of the tube we get that P1 +0.5 density V^2 = P2

    What I want you to clarify please is how can I relate P2 to the weight of the fluid in the tube using free body diagram as we did in the simple liquid manometer. What I cant analyse is the pressure at the curve. Also wouldn't there be a pressure due to the weight of the water above the stagnation point (Hydrostatic pressure)?
     
  5. Nov 12, 2017 at 11:13 AM #4
    Well, if it were above P1 location, height would be ##P_1/\rho g##. The additional height ##\Delta h## over the stagnation tube would be ##\rho g \Delta h =\rho v^2/{2}##.
     
  6. Nov 13, 2017 at 11:41 AM #5
    I think I have a problem in understanding something in the derivation of brenouli's equation

    In the derivation of brenouli's equation we assume that we have a constant pressure P that is pushing the fluid and it changes according to the changes in kinetic and potential energy, and I know the derivation and understand it

    But when I try for example find the pressure at a point in the fluid, I assumed it will be P + density * g * h where h is the height of the fluid above it. I thought I could perhaps not use that the change in potential energy is mg delta h and say that gravity will cause pressure on each slice of the fluid and will do work as same way that P does. So I integrated the pressure resulting from gravity over the area to find the net force then added it to PA and continued to derive brenouli's equation as ##W_{ext} = dKE## but I was not able to arrive to the same equation using this method.

    What am I doing wrong? If it is right and I have done something wrong with the calculation, Could you post a proof of that because it takes a quite long time between replies and this problem caused me quite a mess in my schedule :C
     
    Last edited: Nov 13, 2017 at 11:52 AM
  7. Nov 13, 2017 at 1:28 PM #6
    Sorry. I don't understand what you are saying in your discussion.
     
  8. Nov 13, 2017 at 1:43 PM #7
    Oh sorry, let me make it in the form of questions.

    First we assume in a tube we have a static pressure as though we are pushing it with a piston from the other side is that right? Then if I want to find a pressure at a certain point, Shouldnt the total pressure be P + density * g * h? where pgh is the pressure resulting from the weight of the fluid above it? If not, Why please?
     
  9. Nov 13, 2017 at 2:55 PM #8
    if it is in one of the tubes rising above the pipe, then it is just ##\rho gh## at the bottom of the tube.
     
  10. Nov 13, 2017 at 3:04 PM #9
    https://i.imgur.com/oVVX7Tk.png

    I mean like this, Just a tube and you have a fluid flowing in it. Do we have a change in pressure in the direction perpendicular to the direction of the flow due to the weight of the fluid above it? So for example at the red point, The total pressure Pt = P +pgh
     
  11. Nov 13, 2017 at 6:36 PM #10
    I don't understand your diagram. But, I do understand your original pitot tube diagram. What would you say the fluid velocity is at the horizontal entrance to the pitot tube? What would you say the pressure is at this location in terms of P1, rho and the upstream velocity v? What would you say the pressure is at this location in terms of rho, g, and h, where h is the height the fluid rises in the vertical section of the pitot tube?
     
  12. Nov 13, 2017 at 11:39 PM #11
    1)zero
    2)P1 +0.5 p V^2 = P2
    3)P2 = pgh

    Let's forget a bit about brenoulli's equation and pitot tube. Maybe this will make you understand what I am thinking.

    If we have a cup of water, the pressure at the surface is atmospheric pressure and The pressure at a point underneath the surface with distance h is Patm + pgh
    So what we see here is as if the atmospheric presssure is there all over the water but the changes in the vertical direction is caused by the weight of water. While in the horizontal we have the same pressure.
    https://i.imgur.com/9g6MR3u.png

    Now rotate that cup 90 degress, Break the base of it and we have a tube. Patm would be a static pressure that is lets say applied by a piston and we will still have variation in pressure on different stream lines because of the weight of the water above each point
    https://i.imgur.com/obLjyq7.png

    In brenoulli's equation, We say the force is equal to PA as if the pressure is constant on every point on the surface but shouldn't the pressure change according to pgh as in the static water cup? (Pressure variation normal to stream lines)
     
    Last edited: Nov 14, 2017 at 1:41 AM
  13. Nov 14, 2017 at 4:54 AM #12

    Nidum

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    You are saying that you think that pressure should vary with height in the pipe carrying the main flow ?

    In reality it does but with smaller pipes the variation is small and usually ignored in practical work .

    In any case if the Pitot tube is positioned along the central axis of the pipe then errors of measurement are going to be very small since the variations of pressure above and below the central axis cancel out .

    Just for interest there is an experiment often done in labs where a small diameter Pitot tube is traversed across a flow of water in a duct and the dynamic pressure profile for the flow recorded .

    Later on I had a project which required me to scan the boundary layer profile of a similar water flow in detail . I used a piece of hypodermic syringe with the end flattened down as the Pitot tube and traversed it with a DIY micrometer screw .
     
    Last edited: Nov 14, 2017 at 10:27 AM
  14. Nov 14, 2017 at 5:34 AM #13
    Yes that is what I am suggesting. Difference in pressure in the vertical direction due to the weight of the fluid. Also for a tube, If I assume pressure is isotropic, equal in every direction then I can never get steady flow in one direction only because in a circular tube if we take a horizontal cross section we will find that on this cross section points will have different pressure due to the different heights of fluid. I tried a lot and didnt find a way to make it laminar. Also if we assume gravity as a external force and then we integrate to find the pressure of gravity on one side of the tube and the other then find the net work done, I couldn't make it equal to the change in potential energy. How do we neglect gravity but it will also do work.

    If I try to make the approximation of of constant presssure, In venturi tube, They take into account that across the tube there is a pressure difference equal to pg(D/2) if we take the center line.
     
  15. Nov 14, 2017 at 6:00 AM #14

    Nidum

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    I've edited my post slightly to make it a little clearer .

    Let's forget about the Pitot tube for the present .

    Draw me a diagram showing all the pressures acting on a slice of fluid taken from the flow in a large diameter tube .
     
  16. Nov 14, 2017 at 7:05 AM #15
    I agree with Nidum. If you are asking whether the static pressure varies with depth within the horizontal pipe, then, yes it does. But this does not mean that pressure can't vary with horizontal position within the horizontal pipe. Is that what you are asking about?

    As you correctly indicated, the fluid pressure at the centerline of the horizontal pipe is higher than at the top of the pipe and lower than at the bottom of the pipe.
     
  17. Nov 14, 2017 at 7:14 AM #16
    I think I missed up :/

    https://imgur.com/a/6MCh7

    In the center point, We have the height of the fluid above it is (D/2) so the pressure is rho (D/2) where D is the diameter.
    on any other point on the slice, the height of the fluid above it is variable h. So each point has a different pressure.

    The fluid above and the bottom exerts a static pressure on the slice ( I thought of static pressure as pressure that exists everywhere and is equal in all directions)
    Lastly perhaps there would be a pressure on the bottom of the slice to counter act the imbalance of forces
     
  18. Nov 14, 2017 at 7:59 AM #17
    At a given spatial point, it is the same in all directions. But, it varies with spatial location.

    Also, in an enclosed tube, unless we know that pressure at some particular location, we can only talk about relative pressure changes. So the pressure at the centerline is ##(D/2)\rho g## higher than at the top and ##(D/2)\rho g## lower than at the bottom.
     
  19. Nov 14, 2017 at 10:38 AM #18

    Nidum

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    Re: Your pm .

    You say ' I would really like if you could show me how the pressure changes with elevation (force analysis)in the tube and then use the approximation to get the brenoulli's equation '

    Happy to help if I can but I'm not quite sure what you are asking about . Can you explain a little more clearly please ?

     
  20. Nov 14, 2017 at 10:55 AM #19

    russ_watters

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    What's throwing me is the weird inverted manometers in the diagrams. Can we get rid of them and just do pressure gauges? Because you aren't talking about the manometer fluid depth, you are talking about the height of the air in your vessel, right?

    The height of the air in a system (like a fan, air pump or wind tunnel) is usually ignored, but if you want to include it...

    ....the midpoint of the system has the average pressure (say, by force on a piston) and you can add ρgh from there to get variation with depth.

    ....er...you aren't trying to use water and air simultaneously in the same equation, are you?
     
  21. Nov 14, 2017 at 11:14 AM #20
    I don't really understand why my question isnt clear :c. Sorry.

    Can we forget about the pitot tube for a sec because if I solve this complication I can answer the original question..

    Assume there is no gravity and that we have a tube with water flowing in it. On one side we have a piston that is pushing the water with Presssure P. If the flow of water is laminar then the pressure at every point must be equal to P or I would have turbulent flow. Is this point correct?

    Now lets add the gravity to the situation, If we say that at each point the pressure is P + rho g h where h is the perpendicular distance from the point to the top of the tube, Then we have different pressure at each point which means it shouldn't be laminar? Can we find an equation that relates P and h to the pressure at a point? Should the pressure be constant at every point to have laminar flow?

    Hopefully it is clear now :/
     
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