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Tide, Further 2 my post yesterday on tough integration

  1. Oct 18, 2004 #1
    Tide, Further 2 my post yesterday on tough integration....

    Hey Tide! Thanks 4 answering my question, ur an absolute legend. I now know how 2 do parts one and 2 of the question. But I'm a bit stumped on part 3. I'm not sure wat 2 do when the contaminant is overflowing into the lake.


    Just 2 refresh ur memory, here is the original question.
    Thanks again!!!!!

    A holding tank of 7000 litres is full of water which has been contaminated by a pollutant. The tank contains 0.01 percent contaminant by volume. Water with a contaminant concentration of 0.001 percent now flows from a river into the tank at a rate of 5 litres per minute. Since the tank is full, there is an overflow of water into a nearby lake.

    1) What is the concentration of contaminant in the tank after 5 hours?

    2) The farmer who owns the tank claims that the contents of the tank is under the legislated upper concentration limitation of 0.002 percent for the contaminant. If the inflow has been running for 4 hours, is the farmer's claim correct?

    3) What is the volume of contaminant that has overflowed into the nearby lake after four hours?

    :smile: :smile:
     
  2. jcsd
  3. Oct 18, 2004 #2

    Tide

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    To find the volume of the contaminant multiply the concentration in the tank by the effluent flow rate (I think I called it [itex]R_{out}[/itex] in my posting) then integrate from t = 0 to the desired time.
     
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