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Bernoulli's equation and water pipe

  1. Jun 4, 2013 #1
    Hi all i'm trying to teach myself a bit of bernoulli's principle but i am struggling.
    I'm pretty sure I need to use a combination of bernies equation and continuity equation but I get lost when trying to rearrange to find velocity.

    Water moves through a constricted pipe in steady, ideal
    flow. At the lower point, the pressure
    is P1 17500 Pa and the pipe diameter is 6.00 cm.
    At another point y 0.250 m higher, the pressure is P2 12000 Pa and the pipe diameter is 3.00 cm. Find the speed of flow (a) in the lower section and (b) in the upper
    section. (c) Find the volume flow rate through the pipe.


    2. Relevant equations

    P1/ρg+v^2/2g+y1 = P2/ρg+v^2/2g+y2
    A1v1=A2v2

    3. The attempt at a solution

    Information I have is
    P1= 17500 Pa
    P2= 12000 Pa
    Y1= 0
    Y2= 0.25m
    A1= 0.0028 m2
    A2= 0.00071 m2

    I need to find v1 and v2 and volume flow rate

    I know buy using the the continuity equation I can substitute the valve of v1 for v2,
    but when I come to rearranging the formula it becomes a mess. I have searched online but lots of examples tend to skip important steps and just show the final equation.
    I would like to know step by step how to do it. (please note that it has been nearly 20 years since I last did maths :redface:)
     
    Last edited: Jun 4, 2013
  2. jcsd
  3. Jun 4, 2013 #2

    SteamKing

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    Homework Helper

    First step: plug in the known quantities in the first equation. Where possible, make any arithmetic simplifications.

    Second step: collect all unknown quantities on one side of the equal side and all constants on the other side.

    Note: Steps 1 and 2 will rely somewhat on the use of algebra. If this subject is a bit hazy, now would be the time for a refresher.

    For your problem, you are trying to find v1 and v2. Use the continuity equation to get v1 or v2 in terms of the other. Substitute back into your original equation and solve for one of the unknown velocities. Once v1 or v2 is known, the other can be readily calculated.
     
  4. Jun 5, 2013 #3
    So I've had a go and I must be making a simple mistake somewhere but can't see it. see attached sheet.
     

    Attached Files:

  5. Jun 5, 2013 #4

    SteamKing

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    Your algebra is OK when you reach this line:

    6.1 + ((A2/A1)*V2)^2 = V2^2

    You can't take the square root of both sides because of the constant factor of 6.1 added to the LHS.

    Instead, you must bring the expression for V2 to one side first:

    A2/A1 = 0.25, because d2 = (d1)/2

    so,

    6.1 = (1 - 0.25)*V2^2 = 0.75*V2^2

    V2^2 = 6.1/0.75 = 8.13

    V2 = SQRT (8.13) = 2.852 m/s

    then V1 = 0.25*V2 = 0.713 m/s
     
  6. Jun 5, 2013 #5
    Thanks for taking the time to help, but i'm still a bit confused,

    these parts i understand
    6.1 + ((A2/A1)*V2)^2 = V2^2
    A2/A1 = 0.25, because d2 = (d1)/2

    but

    6.1 = (1 - 0.25)*V2^2 = 0.75*V2^2
    on this line where did the 1 come from and where has the other v2^2 gone because when I look at it there is a v2^2 on both lhs and rhs
    6.1 + ((A2/A1)*V2)^2 = V2^2

    I see this as
    6.1 + (0.25*V2)^2 = V2^2
     
  7. Jun 5, 2013 #6

    SteamKing

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    I got in a hurry and made a small mistake:

    6.1 + (0.25*V2)^2 = V2^2

    which should be

    6.1 + 0.0625V2^2 = V2^2

    collecting V2 on one side and constants on the other:

    6.1 = V2^2 - 0.0625V^2 = (1 - 0.0625)*V2^2 = 0.9375*V2^2

    V2^2 = 6.1/0.9375 = 6.507 (m/s)^2

    V2 = SQRT (6.507) = 2.551 m/s

    V1 = 0.25*V2 = 0.638 m/s
     
  8. Jun 5, 2013 #7
    Thank you, i'll be able to sleep well tonight :smile:.
     
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