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Differential Equation: Mixing Problem

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data

    The air in a small room 12x8x8 ft (768 ft^3 is the volume) is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is blown into the root at rate of 100 ft^3/min. If the air in the room flows out at same rate (100 ft^3/min), when will air in room be 0.01%?

    2. Relevant equations

    s'(t) = (rate entering)(concentration entering) - (rate exiting)(concentration exiting)

    concentration = amount/volume

    3. The attempt at a solution

    I don't need you to solve the problem for me, but I am having trouble getting started and figuring out what the concentration is.

    s'(t) = (rate entering)(concentration entering) - (rate exiting)(concentration exiting)
    s'(t) = (100)(?) - (100)(?)

    concentration = amount/volume
    Since it said at time t=0, fresh air containing *NO* carbon monoxide..., will the concentration entering be 0?
    And for the concentration exiting, will it be s(t)/768?

    so the s'(t) is: 0 - 100s(t)/768

    Is this correct?

    I am contemplating about the 3% carbon monoxide, could this be the concentration entering??
     
  2. jcsd
  3. Jun 16, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes. But notice that it does NOT say "at time t= 0"- the problem says "starting at time t= 0". There is fresh air containing NO carbon monoxide entering the room for all positive t.

    Well, that would depend upon what "s(t)" means- and you didn't tell us that!

    If s(t) is the amount of carbon monoxide in the room at time t, then the concentration will be s(t)/768 per cubic foot. But s'= ds/dt is amount of carbon monoxide per minute. Since air is flowing out at 100 ft^3/min, it carries with it (s(t)/768 carbon monoxide/ft^3)(100 ft^2/min) carbon monoxide per minute.
    so the s'(t) is: 0 - 100s(t)/768

    No, it is the initial value: If s(t) is the amount of carbon monoxide in the room then s(0)= .03(768).

    I cannot comment on what
     
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