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Differential Equation- Height of water in tank

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A water tank is filled by an inflow x(t), the tank is emptied by the outflow y(t)

    The outflow if controlled by a resistance R

    The water depth in the tank is represented by d(t)

    The surface area of the water is A, independent of depth

    The tank is 1.5m high with a 1m diameter and a valve resistance of 10s/m^2

    We are being asked to write a differential equation in terms of tank dimensions and valve resistance


    2. Relevant equations

    The outflow is related by y(t) = d(t)/R

    d/dt(Volume of tank) = x(t)-y(t)

    dh/dt = -a*e^(10t) --> I pulled this from another example online
    3. The attempt at a solution

    I don't understand how to express this without using the terms in regards to outflow and inflow. From the given equation we can calculate...
    d(t) = y(t)R --> From here I assume the following
    d(t ) = x(t) - y(t)R , can we assume that x(t) is the initial volume of the tank?
    I honestly don't know where to go from here, the only times I have seen a problem like this has been in regards to inflow and outflow, not tank dimensions and valve resistance.

    d(t) = pi(.5^2)(1.5) - y(t)*10

    I feel like I am overlooking something very simple here, I haven't slept in about a day and a half so that might be it :p
     
  2. jcsd
  3. Sep 28, 2012 #2
    The net volumetric flow (m^3/sec) which is expressed as

    dv/dt = x(t) - y(t) = x(t) - h/R where h is the height of water in the tank

    But dv/dt = A * dh/dt where A is cross sectional area.

    So write your ODE. The independent variable is time, and the dependent variable is h.
     
  4. Sep 28, 2012 #3
    So then after some algebraic manipulation I have

    dh/dt= x(t)/A - h/RA

    This still uses output flow as a parameter, which isnt what the question wants. I'm sorry that I'm so hung up on this problem
     
  5. Sep 29, 2012 #4
    The output flow is y(t). It's not in the ODE. The input flow is present.
     
  6. Sep 29, 2012 #5
    Right but that isn't in the terms requested right? We can't have it in terms of output flow..or does y(t) somehow relate to valve resistance?
     
  7. Sep 30, 2012 #6
    dh/dt= x(t)/A - h/RA

    A is the area and it is based on the tank dimension diameter. R is the resistance. The overall height of the tank is 1.5 m but that is extraneous information. You could change the A to pi*D^2/4.
     
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