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Homework Help: Flow measurement in venturi meter

  1. Aug 11, 2013 #1
    First I am trying to find the theoretical flow rate but I am not able to get the answer because since diameter of the pipe is 0.03m throughout, when I sub into the formula:

    My A1/A2 will be 1, and 1-1 = 0. And I will get maths error whatever is divided by 0.

    Am I going in the wrong direction?

    Please advice, thanks.
     

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    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2
    No, the diameter is not 0.03 m throughout. Read the problem statement again.
     
  4. Aug 11, 2013 #3
    I mean the 2 pipes at point 1 & 2 connecting to the venturi meter
     
  5. Aug 11, 2013 #4
    As per the given question, at point 1 it is 0.03 m and at point 2 (throat), its 0.012 m.
     
  6. Aug 11, 2013 #5
    I am confused. Is A1= Area of Left hand side pipe, A2= Area of Right hand side pipe in the theoretical flow rate formula?

    "A venturi-meter is installed in a pipe of 30mm diameter" the pipe the question is talking about is it the 2 vertical pipes or the horizontal pipe?



    Thanks.
     
  7. Aug 11, 2013 #6
    No. Your book defines A1 and A2, maybe on the previous page, no?

    Do you know about Bernoulli theorem?

    The 30 mm diameter pipe is the horizontal pipe (isn't that obvious? :biggrin: ).

    The vertical pipes are used for calculating pressures at point 1 and 2.
     
  8. Aug 11, 2013 #7
    Ok I think I got it for the pipes. I am having problem finding P1-P2

    Can I ask Since P1=P2

    P1=pg(0.045)
    P2=pg(0.027)

    P1=P2
    1000x9.81(0.045)=1000x9.81(0.027)
    441.45=264.87 -> Stuck...

    There is a note in my book saying "The pressure difference, (P1-P2) is obtained from a U-tube manometer."
    Since this is not a u-tube manometer, how do I go about finding P1-P2?


    Thanks.
     
  9. Aug 11, 2013 #8
    How P1=P2? :confused:

    What is the pressure at a depth in a liquid?
     
  10. Aug 11, 2013 #9
    I take the reference point at P1 and P2. Like how I did for Px and Py in the attached screenshot.


    What is the pressure at a depth in a liquid? I don't know sorry.
     

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  11. Aug 11, 2013 #10
    Looks good to me. Can't you find P1 and P2 in the problem you have posted?

    P1=P_(atm)+(density)*g*(h1).

    Similarly find P2.
     
  12. Aug 11, 2013 #11
    What is P_(atm)? Why do I need to take in account into my equation?

    So can I still take my reference point at point 1 & 2 like how I did for Px & Py (Px=Py) so P1=P2?


    Thanks.
     
    Last edited: Aug 11, 2013
  13. Aug 11, 2013 #12
    Sorry, I should have defined it. P_(atm) is the atmospheric pressure. I can't help you further until you learn how to calculate pressure at a depth. Pressure at depth, which is taught in hydro statics should be in your notes or you haven't yet started with it? But its unlikely that you begin with hydrodynamics before studying hydro statics.

    For this problem, P1-P2=(density)*g*(difference in heights of liquid in vertical tube).
     
  14. Aug 11, 2013 #13
    Is P1=P2?
     
  15. Aug 11, 2013 #14
    You should refer your textbook or search online for pressure at depth.
     
  16. Aug 11, 2013 #15
    I have search and don't understand the information online. My teacher said that I should find a reference point and the reference point should have the same liquid.

    So I tried to apply post #9 method on my post #1 question which is by taking a reference point. Post #9 I can find the pressure difference between 1 & 2. But for post #1, I am unable to form the equation to "P1-P2=xxxx"

    Can you tell me where I go wrong?


    Thanks.
     
  17. Aug 11, 2013 #16
    What you do in post #9 is same as calculating pressure at a depth.

    In post #9, you go down, in this problem you go up. :wink:

    Think where you should mark X and Y in this case.
     
  18. Aug 11, 2013 #17
    Ok. I have labeled my new Px and Py. But I think it's abit wrong because what do I do for the "?" part of water in Px.

    But if it is wrong I thought my reference point should have the same liquid:confused:
     

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  19. Aug 11, 2013 #18
    It would much easier if you place Px at the top most point on the liquid inside the vertical tube on the left.
     
  20. Aug 11, 2013 #19
    Ok. How about my Py? I thought they should be on the same level with each other and with the same liquid too?

    Because my post #9 referencing points are at the same level and having the same liquid.


    Please explain, I'm abit confused. thank you sir.
     
  21. Aug 11, 2013 #20
    Yes, take them on the same level.

    EDIT: And don't call me sir, I am a student myself. :D
     
  22. Aug 11, 2013 #21
    Ok same referencing point but Py is not having the same liquid as Px.

    My lecturer said that referencing point must be at the same level and is having the same liquid.

    I'm confused over this part.


    It's ok, it's a form of respect, if you don't like I won't call you that haha.. Thanks anyway for your help :)
     

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  23. Aug 11, 2013 #22
    Px=9810(0.045)+P1
    Py=9810(0.027)+P2+Patm

    Since Px=Py

    9810(0.045)+P1=9810(0.027)+P2+Patm
    P1-P2=264.87-441.45
    P1-P2=-176.58 -> Is the answer wrong?
     
  24. Aug 11, 2013 #23
    Py is actually the atmospheric pressure. Going up from P1,
    P1-9810(0.045)=Px

    If we go up from P2,
    P2-9810(0.027)=Py.

    Pressure above the liquid in the vertical tubes is same everywhere so Px=Py. Solve to find P1-P2.
     
  25. Aug 11, 2013 #24
    Don't understand at all. The questions that I did I will add the pressure together. :cry:
     
  26. Aug 11, 2013 #25
    The value you calculated for P1-P2 is correct but the sign is wrong. You can substitute that value if you wish and obtain the answer but the way you calculated is wrong. I don't know how can I explain it to you in simpler terms. We should wait for the experts to join.
     
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