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Velocity of water out of reservoir.

  1. Aug 12, 2011 #1
    I have a cylindrical tank and I know that the efflux speed is proportional to the square root of the depth of the hole from the surface. So u=k sqrt(w). I need to algebraically determine the constant or k in that situation. Has anyone got any ideas as to how I should approach this? I was thinking that I could try and find the acceleration in i and j components and integrate it for velocity but didn't get far.

    Thanks for any help.
     
  2. jcsd
  3. Aug 12, 2011 #2

    cjl

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    If you're ignoring the viscous effects, the efflux speed will be such that the dynamic pressure is equal to the static pressure just inside the hole. Dynamic pressure is 1/2*rho*v2, so rearranging for v, we can get that v = sqrt(2*p/rho). Since the pressure in a tank is simply from hydrostatic equilibrium (P = rho*g*h), we can plug in for P:

    v = sqrt(2*rho*g*h/rho) = sqrt(2*g*h).

    So, your constant is sqrt(2g).
     
  4. Aug 12, 2011 #3
    Hm I don't think taking pressure into account is necessary as it is not part of our coursework.
     
  5. Aug 14, 2011 #4
    If you want to know the velocity in fluid dynamics you need to know 2 things. Volume and pressure to find velocity. Unless you can invent some kinda new math cjl is right.
     
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