1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Modeling with First Order Differential Equation

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    A tank contains 70 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

    Find the amount of salt in the tank after 2 hours.

    I'm failing to figure out how to manipulate the rate-in to include it in a differential equation that models this process.
    2. Relevant equations
    Q= amount of salt in tank
    [tex]\frac{dQ}{dt}= Rate in - Rate out[/tex]

    Rate out: [tex]\frac{Q}{2000} \cdot 6(Liters/min)[/tex]

    The rate in is 12L/min but of pure water, so I can't just have a simple rate in - rate out as I had envisioned.

    3. The attempt at a solution
    Realizing the 12L/min is a rate that affects the volume of water, I solved this simple differential equation and put it in my equation for rate out: rather than dividing Q by 2000, I'm dividing it by 12t +2000.

    Problem is that this doesn't seem to work. So I need help on figuring out how or where this rate of pure water falls into my differential equation.
  2. jcsd
  3. Feb 7, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Your aren't accounting for the 6 liters of solution volume leaving every second. The volume in tank at time t isn't 2000+12t.
  4. Feb 7, 2010 #3
    Q = amount of salt in tank.

    dQ/dt has units of mass/time right?
    So the "rate in" and "rate out" must also have these same units.

    Now, pure water enters the tank at the rate of 12L/min. That rate in contains no salt, therefore rate in = 0.

    The rate out is tricky. The rate of salt leaving is the concentration of salt times the the rate solution leaves. Notice you have (kg/L)*(L/min), which gives kg/min--exactly the units. The concentration is the amount of salt( which is Q at all times) over the volume of the solution (which is initially 2000L, but then changes by 6L/min - 12L/min).

    You should see it from here.
  5. Feb 7, 2010 #4
    I don't get it, how do I account for it? The incoming rate is pure water, the outgoing rate is solution, I can't mix those two rates can I? It wouldn't make sense unit-wise.

    Is this problem meant to be reduced to a rate-in - rate-out business or should I let go of that idea?
  6. Feb 7, 2010 #5
    Try to model a diff eq for the volume of the tank at all times. You will figure out the volume from there, and then you can use that in dQ/dt
  7. Feb 7, 2010 #6


    User Avatar
    Science Advisor
    Homework Helper

    You can mix those two because the problem says the the solution is 'mixed' before it goes out. There is no salt going in but there are 6 liters of solution going out. The concentration going out depends only on the amount of solution in the tank. How much is that?
  8. Feb 7, 2010 #7
    I can't get it man, by now I'm just randomly subtracting and adding rates and integrating the resulting equation with no real thought behind it. I need more help please.
  9. Feb 7, 2010 #8
    It's clear the rate of salt entering the tank is 0, correct?

    The rate leaving depends on the concentration and the rate of solution leaving, correct?

    The concentration depends on the amount of salt and the amount of solution.

    What is the amount of solution (aka the volume of the tank) at all times? You need this in order to know the concentration at all times, t.

    Here's a start.
    dQ/dt = rate salt enters - rate salt leaves

    That's for the salt. Now for the volume aka amount of solution:
    dV/dt = rate solution enters - rate solution leaves.
    We know solution enters at 6L/min (this is pure water, but it's still affecting the volume of tank), and solution leaves at 12L/min (this is a concentration, but it's still affecting the volume of the tank.
    dV/dt = 6 - 12 = -6

    Use that (and a given initial condition) to find an expression for V at all times. Where does V go into your dQ/dt diff eq?

    Now, we can figure out the rate salt leaves, correct?
  10. Feb 7, 2010 #9
    Yes I finally got it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

    Thank you Thank you Thank you Thank you Thank you Thank you Thank you Thank you Thank you Kizaru and Dick.
  11. Feb 7, 2010 #10


    User Avatar
    Science Advisor
    Homework Helper

    dV/dt=+6. That's PLUS SIX. 12 in. 6 out. PLUS SIX.
  12. Feb 7, 2010 #11
    Er yes, I got them mixed up. He probably caught it though.
  13. Feb 7, 2010 #12


    User Avatar
    Science Advisor
    Homework Helper

    I would guess he did. Thanks for helping!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook