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Another Hydrodynamics word problem

  1. Jun 29, 2014 #1

    HP8

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    1. The problem statement, all variables and given/known data
    The pressure difference between the main pipeline and the throat of a venturri meter is 10.5 Pascals (Pa). The areas of the pipe and the constriction are 0.1 m^2 and 0.05 m^2 respectively. How many cubic meters per second are flowing in the pipe? The liquid in the pipe is water.


    2. Relevant equations
    [itex]Q = A_1 V_1 = A_2 V_2 [/itex]
    where a = area
    v = velocity


    3. The attempt at a solution
    I don't know if I should put 15 Pa on Q, so my solution goes like this:

    [itex]10.5 = (0.1) V_1 = (0.05) V_2 [/itex]

    I don't know what the next steps are, and by the way I tried to look for the forces using F = PA, which I got F1 = 1.05 N and F2 = 0.525 N. But I still don't know yet how to continue solving it. Any suggestions or solutions if available?
     
    Last edited: Jun 29, 2014
  2. jcsd
  3. Jun 29, 2014 #2

    BvU

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    Hi HP, and a late welcome to PF.

    [STRIKE]I think you have the wrong mental picture of a venturi meter. Do some checking.[/STRIKE].

    Turns out MY picture was wrong. You need some more relevant equation material. Something with pressure. Any idea ?

    On the other hand, I have difficulty understanding where the 300 comes from (because that would be the answer for this exercise!) and I also wonder if the 10.5 Pa is really 105 Pa ?
     
  4. Jun 29, 2014 #3

    HP8

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    Oh I'm very sorry! I just realized it right after you replied! I was really wondering why my question was somewhat getting irrelevant to others, thanks and I'll edit my question.

    By the way, the 10.5 Pa you are mentioning is really 10.5, in its decimal form.
     
  5. Jun 29, 2014 #4

    HP8

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    Well there are some other formulas mentioned like this one:

    [itex] P_1 + \frac{1}{2} pv_1^2 + pgh = P_ + \frac{1}{2} pv_2^2 + pgh [/itex]

    That's it. The one I mentioned earlier was a theorem on continuity.
     
  6. Jun 29, 2014 #5
    By your equation of continuity find ratio of v1:v2...

    Make another equation as : The Bernoulli's equation!
    Note that there should be no ρgh term..
     
  7. Jun 29, 2014 #6
    It's very hard to solve a problem if you plug the pressure difference into the flow rate. Try plugging the pressure difference into the pressure difference instead. In other words, try making sure you know the meaning of the variables before solving a problem.
     
  8. Jun 29, 2014 #7

    BvU

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    Is it clear to HP that 10.5=(0.1)V1=(0.05)V2 is nonsense ? On the left you have 10.5 Pa = 10.5 N/m2, so a pressure, on the right the dimension is m2 * m/s = m3/s, a volume flowrate. That's what dauto means and I agree completely.

    Your post #4 was what I meant. The Bernouilli equation. Make life easy and take h1 = h2 ( a horizontal setup).

    The given 10.5 Pa is p1-p2. Write it out in terms of one of the v and the A's. Solve for the v and your Q is vA. Any problems ?
     
  9. Jun 29, 2014 #8

    HP8

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    What do you mean by h1 = h2? Equating them?

    And I didn't understand much this phrase "write it out in terms of one of the v and the A's".
     
  10. Jun 29, 2014 #9

    BvU

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    h1 = h2 if the pipe is horizontal. Subtract ##\rho \, gh## on both sides and forget about the h.

    What exactly did you not understand in this writing out hint? You have an expression for delta p:

    ##p_1 - p_2 = {1\over 2}\rho(v_2^2-v_1^2)## nothing magical, savvy? Your first relevant continuity equation alows you to rewrite in terms of, e.g., ##v_1##. Solve for that (i.e. get something that looks like ##v_1 = ...## and then remember that ##Q = v_1\,A_1\,##. Done !
     
  11. Jun 29, 2014 #10

    Nathanael

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    He basically means drop the [itex]ρgh[/itex] terms from the equation (because [itex]h_1=h_2[/itex])

    This is because they both have the same gravitational potential energy (per volume) so it's irrelevant to the problem

    So you can simplify the bernouilli equation into:

    [itex]P_1+\frac{ρv_1^2}{2}=P_2+\frac{ρv_2^2}{2}[/itex]



    Verbally, this means "the difference in kinetic energy (per volume) is equal to the difference in pressure(*)"

    So you can at least solve for the difference in the kinetic energy between the two pipe sections.

    Then the areas of the pipes will give you another restriction (the equation in your OP) which enables the problem to be solved.




    (**Pressure naturally has units of energy per volume (Force per area = Joules per volume) so it is indeed consistent)
     
    Last edited: Jun 29, 2014
  12. Jun 29, 2014 #11

    HP8

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    Well, after what you suggested, here's what happened:

    [itex]2(P_1-P_2) = p(V_2^2-V_1^2) [/itex]

    after that I plugged 10.5 Pa into pressure difference, then I divided it by the density of water (1000kg/m^3), so I got an answer to the difference between two velocities.

    [itex]0.021 m^2/s^2 = p(V_2^2-V_1^2) [/itex]

    and I'm wondering if I should square them to get the correct unit of velocity in my answer.
     
  13. Jun 29, 2014 #12

    HP8

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    Hi, BVU suggested to me that I should get the difference in velocity, which is 0.14 m/s. I am still confused on how should I plug this value into my original equation when my answer is stated as difference in velocity?

    [itex]Q = A_1 V_1 = A_2 V_2 [/itex]
     
  14. Jun 29, 2014 #13

    Nathanael

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    First off, the difference in velocity is not 0.14, you actually don't know the difference in velocity.
    (I'm guessing you got 0.14 by taking the square root of 0.021? The reason that this doesn't work is because 0.021 is the difference of the squares, NOT the square of the difference)


    You can solve it just like any system of equations

    You know that:

    [itex]V_2^2-V_1^2=0.021[/itex]

    So,

    [itex]V_2=\sqrt{0.021+V_1^2}[/itex]

    Just plug that into the equation [itex]A_1V_1=A_2V_2[/itex] and you'll be able to solve for [itex]V_1[/itex] (and the rest of the problem)
     
    Last edited: Jun 29, 2014
  15. Jun 29, 2014 #14

    HP8

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    Wouldn't the two [itex]V_1[/itex] cancel each other out because I'll transpose [itex]V_1[/itex] after simplifying the radical expression. This is what I'm saying:

    [itex]V_1=(A_1\sqrt{0.021+V_1^2})/(A_2)[/itex]
     
  16. Jun 29, 2014 #15

    Nathanael

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    [itex]A_1V_1=A_2\sqrt{0.021+V_1^2}[/itex]

    [itex]A_1^2V_1^2=A_2^2(0.021+V_1^2)[/itex]

    [itex]V_1^2(A_1^2-A_2^2)=0.021A_2^2[/itex]
    [itex]V_1=\sqrt{\frac{0.021A_2^2}{A_1^2-A_2^2}}[/itex]



    I don't understand how the [itex]V_1[/itex]'s would cancel out?



    (I may have made an algebraic mistake, I wasn't being very careful)
     
  17. Jun 29, 2014 #16

    HP8

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    Oh sorry, my bad, I never thought of moving the square root of the equation on the right to the left, instead, I canceled out the square root and got confused because moving the V_1 to the left will make it equal to 0. Sorry I caused you a little confused about my work. Finally, I got the value for [itex]V_1[/itex] and I got the volumetric flow rate.

    I am really thankful for you. You had helped me solve two problems about it. I am really sorry because it took me a while to understand what others are saying, because I don't really have much knowledge about this specific topic in Physics. I really appreciate your help.
     
  18. Jun 29, 2014 #17

    Nathanael

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    That's nothing to be sorry about! No one is born with physics knowledge.

    Besides, I'd MUCH rather help someone who is actually interested in learning (like you) than someone who just wants the answer.
     
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