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The percentage of carbon dioxide in the air

  1. Jul 20, 2005 #1
    Hello. I need a help again for my differential equations class.
    The problem is as follows:
    The air in a room whose volume is 10,000 cu ft tests 0.15% carbon dioxide. Starting at t=0, outside air testing 0.05% carbon dioxide is admitted at the rate of 5000 cu ft/min.
    (a) What is the percentage of carbon dioxide in the air in the room after 3 min?
    (b) When does the air in the room test 0.1% carbon dioxide.

    This problem is given after I studied the mathematical formulation of
    [tex]\frac{dx}{dt}=IN - OUT[/tex]

    I tried to write an equation using the above formula, but since there is no air coming out, I did not know how to write the euqation. The equation I wrote was

    [tex]\frac{dx}{dt} = 2.5 ( ? unit/cu ft)(5000 cu ft/min) - 0[/tex]

    I obtained 2.5 from multiplying 5000 by 0.0005. I don't even know if this is right. What I was trying to find was the amount of carbon dioxide in 1 cu ft.
    Since no air is coming out, 0 is applied in the OUT section.
    This does not seem to be right since the density of the air should be increased but there is no place that represent the change in my equation.

    Will someone help me?
  2. jcsd
  3. Jul 20, 2005 #2


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    I think you have the IN rate right (no OUT rate). That's 2.5 cu ft of carbon dioxide (100% pure) per minute entering a 10,000 cu ft room that already has how many cu ft of carbon dioxide in it already? That's it. Just 2.5. Now just solve the ODE like that for cu ft of carbon dioxide in room as a function of time. Pretty sure anyway. :smile:
  4. Jul 20, 2005 #3
    altho not mentioned, any practical application (or air system) would be withdrawing air at the same volumetric rate that air is entering. of course, concentration of CO2 leaving is not equal to that entering.

    let x(t) be the actual pure volume of CO2 in the room at time t.
    then x(0) = 10,000*0.0015 = 15 cu ft pure CO2 (the t=0 initial condition)

    at t=0, pure CO2 begins entering the room at the rate of 5000*0.0005 = 2.5 cu ft/min = IN RATE (pure CO2)

    at t=0, inside room air (presumably instantaneously mixed with the entering air) begins leaving the room with concentration x(t)/10,000 of CO2 at the rate of 5000 cu ft/min (= air entry volumetric rate), so the rate at which pure CO2 leaves the room = -5000*x(t)/10,000 = -x(t)/2 cu ft/min = -OUT RATE (pure CO2)

    dx(t)/dt = IN RATE (pure CO2) - OUT RATE (pure CO2)
    dx(t)/dt = 2.5 - x(t)/2
    dx/dt + x/2 = 2.5

    however, CO2 concentration c(t) = x(t)/10000 so that
    dc/dt + c/2 = 2.5/10000 with initial condition c(0) = 15/10000

    now you can determine concentration c(t) using Ordinary Differential Equation techniques given in your textbook.
    Last edited: Jul 21, 2005
  5. Jul 21, 2005 #4
    I didn't know that

    Thanks for your help again Saltydog and thanks for the detailed explanation for the problem, Geosonel!

    I did not know the air is going out at the same rate (I thought the air density would increase = this consept got me). (Not all students are as smart as you, so I really think the textbook should mentioned the fact either somewhere in the textbook or in the problem :frown: )

    Thanks to you, now I am sure I can solve the equation. If I can't, I will come back here again (hopefully not).

  6. Jul 21, 2005 #5


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    Thanks Geosonel. Didn't know that. Sorry if I messed you up Beez. :blushing:
  7. Jan 30, 2010 #6
    Hello! I have a problem: I need to calibrate a CO2 senzor and I can't do it right today. I need to know the percentage of CO2 in a 12 sqm room under normal conditions. Can you help me or give me an idea? Thanks
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