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Volume flow rate DE - Help needed

  1. Feb 9, 2012 #1
    Hi all,

    So I've got a fairly straight forward problem to solve here regarding flow rate out of a tank with uniform cross-sectional area. I am treating this is a volumetric flow problem since there is assumed to be volume flow out of the tank, and volume flow into the tank.

    I have two Qout terms (one out of a hole in the bottom of the tank, and one from fluid consumed by an engine), as well as one Qin term (via a pump feeding fluid into the tank at a constant rate).

    Qouthole (Qh) = a*C*sqrt(2*g*z)
    Qoutengine (Qe) = constant
    Qin (Qi) = constant

    NOTE: a = exit hole area
    C = energy loss coefficient
    g = gravity
    z = head height in tank

    My DE looks like:

    -A*(dh/dt) = Qh - Qi + Qe

    Separating the terms and solving for time by integrating h from Hi to Hf and t from 0 to T, I get this...

    ((a*C)/A)*sqrt(2*g)*(2/3)*((Hf^(3/2))-(Hi^(3/2))) = (1/A)q*t

    I combined the Qi and Qe terms early on since they are constants (easier to integrate), therefore, q in the above equation equals Qi-Qf.

    The problem is, solving for t eliminates the tank cross-sectional area (A) since it is divided out.

    I feel confident with my general equation, but may have made a mistake somewhere in solving for t. Any help would be GREATLY appreciated! Thank you
  2. jcsd
  3. Feb 11, 2012 #2
    Could you outline how did you arrive at the expression you found after integrating?
  4. Feb 12, 2012 #3
    I found one (stupid) mistake I made in my original equation (divided out the Qleak term rather than adding it to the other side). So ignore that equation... Below is how I arrived at the new equation, and this one still has a problem.

    Qleak = a*C*sqrt(2*g*h)
    Qi = constant
    Qe = constant

    integrate dH from Hi to Hf
    integrate dt from 0 to T

    1. -A * (dH/dt) = Qleak - Qi + Qe

    2. ... simplify by combining constants... Qi + Qe = Qtot

    3. dH = (-Qleak/A + Qtot/A)dt

    4. dH = 1/A*(-Qleak + Qtot)dt

    5. (A*dH) / (-Qleak + Qtot) = dt

    6. int((A*dH) / (-Qleak + Qtot)) = int(dt)

    7. ... integrate dt from 0 to T

    8. int((A*dH) / (-Qleak + Qtot)) = T


    This is where I stopped since I could not find an easy way to integrate the left side. The basic form of the integral would be: dH/(h^1/2 + constant)

    Any tips on where I may have gone wrong, or how to solve that integral would be appreciated! Thanks
  5. Feb 13, 2012 #4
    Are H and h the same variable, or related in some way?
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