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Homework Help: Pipe and Bernoulli's equation

  1. Feb 23, 2008 #1


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    1. The problem statement, all variables and given/known data

    In the figure below, fresh water flows through a horizontal drainage pipe and then out into the atmosphere at a speed v1 = 15 m / s. The diameters of the left and right sections are 5.0 cm and 3.0 cm.

    (a) Find the mass of water that flows into the atmosphere in one hour.

    (b) What is the gauge pressure in the left section of the pipe?

    http://img174.imageshack.us/img174/5987/picture1lj9.th.jpg [Broken]

    2. Relevant equations

    [tex] P_1 + 1/2 \rho v^2 + \rho gy_1 = P_2 + 1/2\rho v^2 + \rho g y_2 [/tex]

    A1v1= A2v2

    P= F/A

    [tex]A_{circle}= \pi r^2 [/tex]

    [tex] m= \rho \Delta V = v_1A_1 \Delta t [/tex]

    3. The attempt at a solution

    First of all I'm sort of confused with how you can estimate since A1>> A2 but I don't think I can do that here so would that change anything??

    I think that i know this:

    [tex]A_1= \pi r^2 = \pi (3cm/2)^2= 7.0685 cm^2 [/tex]

    [tex]A_2= \pi r^2 = \pi (5cm/2)^2= 19.6349cm^2 [/tex]

    [tex]P_1= 1.013x10^5 Pa [/tex] ====> I'm not completely sure it's atmospheric pressure

    [tex] P_1= ? [/tex]
    [tex] V_2= ? [/tex]

    1hr=> 3600s

    a) find mass of water that flow into the atmosphere in one hour

    I'm not sure how to find the mass..but I think I'd use
    [tex] m= \rho \Delta V= v_1 A_1 \Delta t [/tex]

    so would the mass be (since v1 is given...

    [tex] m= \rho \Delta V= v_1 A_1 \Delta t = (15m/s)(0.00070685 m^2)(3600s)= 38.1699 [/tex] I'm not sure about the units though...

    b) the gauge pressure in the left section of the pipe

    I think I'd use bernoulli's equation here

    [tex] P_2 + 1/2\rho v^2 + \rho g y_2= P_1 + 1/2 \rho v^2 + \rho gy_1 [/tex]

    in this case it's P1 as the atmospheric pressure so I switched the sides up.

    [tex]P_2-P_1 = 1/2 \rho (v_1^2- v_2^2) [/tex]

    gauge pressure= [tex]P_2-P_1 [/tex]

    BUT I'm confused since I don't have P2 but I also think I don't need it since P2-P1 = gauge pressure right?

    I don't have the v2 however...but I think I can find it from the continuity equation Or do I use the mass of water found then use that to find the volume (from density) then based on the time interval that the water was flowing and the mass of water to find volume..then use that as the flow rate?

    well not sure which is right so I'll use the continuity equation...

    [tex] A_2v_2= A_1v_1 [/tex]

    [tex] v_2= (A_1v_1) / A_2 [/tex]

    [tex] v_2= (7.0685 cm^2 / 19.6349cm^2)*(15m/s) [/tex]

    [tex]v_2= 5.40m/s [/tex]

    now plugging in...

    [tex]P_2-P_1 = 1/2 \rho (v_1^2- v_2^2) [/tex]

    [tex]P_2- P_1= 1/2(1000kg*m ^ {-3} )(15m/s-5.40m/s ) = 4.8 Pa [/tex]

    edit: above I went and typed 1000kg m ^-3 but for some reason it keeps showing 8722-- well that's not supposed to be there.

    Wait...did I find the gauge pressure?

    Can someone see if I did this correctly?
    and if not can you help me?

    THANKS very much. :smile:
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 23, 2008 #2

    Doc Al

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    Staff: Mentor

    That trick doesn't apply here, since it's not true that A1 >> A2 (or vice versa) like it was in that needle problem.

    Right idea, but you dropped off the density in the last term.

    Fix this.
    Yes, gauge pressure is P - Patm, which in this case is P2 - P1. So that's all you need. (But you could certainly figure out P2 if you wanted to.)

    Good. The continuity equation is the way to go.

    Two problems: (1) you forgot to do the squaring; (2) redo the arithmetic (you have to anyway).
  4. Feb 23, 2008 #3


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    okay..I was thinking that it would mean that v1 aprox = v2 right? thus the bernoulli's equation would simplify.

    oh..I did.so that's why I couldn't figure out the mass unit..ha, well fixing it.

    [tex] m= \rho \Delta V= v_1 A_1 \Delta t [/tex]

    so would the mass be (since v1 is given...

    [tex] m= \rho \Delta V= \rho v_1 A_1 \Delta t = (1000kg/m^3)(15m/s)(0.00070685 m^2)(3600s)[/tex]= 38,169.9kg

    that's quite large...

    I guess I could by first finding the gauge pressure then solving for P2 here [P2-P1= gauge pressure]

    fixed I think.

    [tex]v_2= 5.40m/s [/tex]

    [tex]P_2-P_1 = 1/2 \rho (v_1^2- v_2^2) [/tex]

    [tex]P_2- P_1= 1/2(1000kg/m^3)((15m/s)^2-(5.40m/s)^2 ) = 97,920 Pa [/tex]

    that number is quite large as well..hm...

    Thanks Doc Al :smile:
  5. Feb 24, 2008 #4

    Doc Al

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    Staff: Mentor

    Looks good.

    (When considering how large a pressure is, compare it to atmospheric pressure.)
  6. Feb 24, 2008 #5


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    Thanks for your help Doc Al :smile:
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