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Fluid Flow Question

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    In a horizontal rectangular open channel 20 m wide the water depth is 9 m. When a
    smooth hump 1.5 m high is introduced in the channel floor, a drop of 1 m is produced in
    the water surface. What is the flow rate, neglecting energy losses? It is proposed to place
    a pier at the centre of this channel on the hump. Determine the maximum width of this
    pier if it is not to cause any backwater effects.



    2. Relevant equations
    all the equations i can think of are :
    E=h+(v^2/2g) also h+(v^2/2g)=constant
    E=h+(q^2/2gh^2)
    hc=(q^2/g)^1/3
    Emin=1.5hc
    Q=qB
    E1=y+E2
    v=q/h

    Where E=specific energy
    h=depth of flow
    hc=critical depth
    q=flow rate per unit width
    Q=flow rate



    3. The attempt at a solution

    First of i tried to find velocities at each point using :

    h+(v^2/2g)=constant

    9+v1^2/2g = 6.5+v2^2/2g

    and i got to v1-v2=7m/s, wasnt sure what to do from there, so i tried a different method...

    I assumed the critical depth would be 6.5, Emin=1.5hc given Emin=9.75

    I then subbed that into the equation E=h+(v^2/2g) to give v=7.99m/s

    i then subbed into eq v=q/h to give q=51.935

    then Q=qB
    = 51.935 x 20
    =1038.7 m^3/s

    however answer is
    832.52 m^3
    /s
    any help would be appreciated. Thanks




     
  2. jcsd
  3. Jan 20, 2015 #2

    TSny

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    I'm not sure I'm interpreting the problem correctly. I have attached a figure. I assume the hump goes across the entire width of the channel.

    When using the formula h + v2/(2g) = const. you have apparently chosen two points in the fluid flow to apply the equation. Can you describe the location of these two points? Where are you choosing h = 0? Have you measured the height h for the two points from the same reference level (h = 0)?

    Also note that ##\sqrt{v_2^2 - v_1^2} \neq v_2 - v_1 ##.
     

    Attached Files:

    Last edited: Jan 20, 2015
  4. Jan 20, 2015 #3

    haruspex

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    Think again about that 6.5m. Remember that this term represents the lost PE. How does the height of the bump come into that?
     
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