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Transient Tank Filling Problem (Related Rates?)

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data

    "Consider the tank and water supply system as shown in the figure below. The diameter
    of the supply pipe, D1 = 20 mm, and the average velocity leaving the supply pipe is
    V1 = 0.595 m/s. A shut-off valve is located at z = 0.1 m, and the exit pipe, D2 = 10 mm.
    The tank diameter is Dt = 0.3 m. The density of the water is uniform at 998 kg/m3."

    "At the instant the water level reaches H0 = 1 m the shut-off valve is opened. The
    instantaneous average velocity of the outflow depends on the water depth above z
    and can be expressed by, V2=0.85*sqrt(g*[H(t)-z]) where g is the acceleration due to
    gravity. Determine whether the tank continues to fill or begins to empty immediately
    after the valve is opened." (Found that it will drain when valve opens)

    Determine the steady-state value of the water depth.
    (found to be 0.9m)

    Determine the time required to achieve steady state after the valve is opened.
    (This is where I'm getting stuck)



    2. Relevant equations

    definition of volumetric flow
    V_dot=Area*Velocity

    continuity for incompressible flow:
    V_dot_in-V_dot_out=dV/dt

    relationship between tank volume and water level
    V=pi*(d/2)^2*H



    3. The attempt at a solution
    V_dot_in-V_dot_out=dV/dt

    V_dot_in=pi*(d1/2)^2*v1=1.869e-4 m^3/s

    V_dot_out=A2*v2=7.854e-5*v2

    substituting into continuity equation...
    (1.869e-4)-(7.854e-5)(0.85)sqrt(9.81(H(t)-0.1))=dV/dt

    now using the relationship between V and H...

    (1.869e-4)-(7.854e-5)(0.85)sqrt{9.81[(4*V/0.09*pi)-0.1]}=dV/dt

    now simplifying

    (1.869e-4)-(6.676e-5)sqrt(139V-0.98)=dV/dt

    now I believe I need to move the dt to the other side and move V terms to the RHS
    with dV then integrate and solve for t... but I'm not sure how to split everything up

    Sorry for the long post, and thanks in advance!
     
  2. jcsd
  3. Sep 20, 2010 #2
    Maybe I can help out. Hopefully I got it right though.

    I had similar results up to this equation and then we diverged:

    (1.869e-4)-(7.854e-5)(0.85)sqrt(9.81(H(t)-0.1))=dV/dt

    Here's what I did:

    dV/dt = v_t*A_t

    where v_t is the velocity of decreasing water level in the tank and A_t is the area for the tank.

    v_t = dH/dt

    dV/dt = v_t*A_t => dV/dt = (dH*A_t)/dt

    starting with your equation at the top and using the relationship immediately above:

    (1.869e-4)-(7.854e-5)(0.85)sqrt(9.81(H(t)-0.1))=(dH*A_t)/dt

    rearrange and we get:

    dt=A_t*dH/((1.869e-4)-(7.854e-5)(0.85)sqrt(9.81(H(t)-0.1)))

    Integrate both sides: the left side from t=0 to t and the right side from H=0.9 to H=1.0.

    My answer was 2930 seconds or 48.83 minutes.

    Hope I helped and didn't make things worse.
     
  4. Sep 20, 2010 #3
    Thanks for the response, that's pretty much what I got. I ended up also working this iteratively in Excel with similar results which seems to confirm my analytic solution. Though it turns out that the steady state water level is not exactly 0.9m and that the tank drains infinitely long- so it's just a matter of deciding when you want to call it good, 99% volume drained? 99.9%?...
     
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