Volume flow rate DE - Help needed

  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. Could you outline how did you arrive at the expression you found after integrating?
     
  4. 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. Are H and h the same variable, or related in some way?
     
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