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Got a real world problem here

  1. Oct 4, 2007 #1
    Hi! I'm new here and trying to get some help solving a problem I have.
    It's a real problem, but for discussion purposes I've simplified it as such:

    Problem Description:
    A large tank is initially full of pure distilled water at volume of 5,000 cubic feet (cf)
    50% pure vinegar is being poured into the tank at 2 cfm
    A faucet on the bottom of the tank is open pouring out at 2 cfm
    The tank has a mixer that circulates and mixes the fluid internally

    After how many minutes would it take for the faucet to pour fluid containing 10% vinegar?

    The mixing of the vinegar and the rest of the fluid in the tank happens instantaneous and homogeneous.

    Please be very descriptive in your solution and how you did it. I really forgot most of my DE knowledge. Thanks!

    I only know that:
    tank concentration of vinegar after t = (volume of vinegar in after t - volume vinegar out after t) / tank volume
    volume of vinegar out after t is the integral of the tank concentration x flow rate out x t

    Please help!
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 5, 2007 #2
  4. Oct 6, 2007 #3


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    I find it hard to believe this is a "real world" problem. It looks like a standard "introduction to differential equations" problem (from chapter 1).
    Oh, well. Everyone's "real world" is different!

    Let V(t) be the amount of vinegar, in cubic feet, in the tank at time t, in minutes. The concentration of vinegar in the tank, then, is V(t)/5000 (fortunately the same amount of solution is coming in as is going out so the volume of solution does not change.)

    dV/dt is the "rate of change of the amount of vinegar in the tank" and the amount of vinegar is changing for two reasons:
    Some vinegar is coming in: "50% pure vinegar is being poured into the tank at 2 cfm" so 1 cubic foot of vinegar is coming in every minute.
    Some vinegar is going out: "A faucet on the bottom of the tank is open pouring out at 2 cfm" so solution with concentration V(t)/5000 is going out at 2 cubic feet per minute: 2V(t)/5000= V(t)/2500 cubic feet of vinegar is going out every minute.
    The differential equation is dV/dt= 1- V(t)/2500 with initial condition V(0)= 0. That a simple "separable" differential equation. You can solve it by a simple integration.
    Then find t so that V(t)= 10% of 5000= 500 cubic feet of vinegar.
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