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Math Help - I have a question

  1. #1
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    Question I have a question

    When crude oil flows from a well, water is frequently mixed with it in an emulsion. To remove the water the crude oil is piped to a device called a heater-treater, which is simply a large tank in which the oil is warmed and the water is allowed to settle out. Operating experience in a particular oil field indicates that the concentration C of water in the treater's output can be modeled by the following equation if the temperature is close to 134 F DEGREE and the holding time is close to 2 hours:C=0.04 - 0.0032h^2 - 10^-6 t^2 + 4.1.10^-5 ht where h is the holding time in hours and t is the temperature in degrees F.
    1. Assuming that t = 134 and h = 2, find C to four significant digits.
    2. Sketch or print out a graph of the curve near the point (2,134). (This means that h is on the horizontal axis and t is on the vertical axis.)
    3. Because of random fluctuations in the well's flow rate the holding time actually varies slightly around 2 hours. Suppose you are given a simple control device that can change the temperature proportionally to the measured change in holding time. What constant of proportionality would best compensate for small holding time fluctuations and keep the water concentration C as constant as possible. Hint: a constant of proportionality is really a slope.
    4. Using your answers to part 3, what will your device change the temperature to be if the holding time is decreased by 2 minutes (1/30 of an hour)?
    Last edited by hadi; October 22nd 2006 at 06:36 PM. Reason: error in q
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  2. #2
    dan
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    Quote Originally Posted by hadi View Post
    C=0.04 - 0.0032h^2 - 10^-6 t^2 + 4.1.10^-5 ht where h is the holding time in hours and t is the temperature in degrees
    I did not under stand this equation...
     C=0.04 - 0.0032h^2 - 10^{-6} t^2 + 4.1.10^{-5}
    What does  4.1.10^{-5} mean?

    sorry I didn't get it...

    `dan
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  3. #3
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    Exclamation sorry


    I am sorry i mean

    C=0.04 - 0.0032h^2 - 10^-6 t^2 + 4.1*10^-5 ht
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  4. #4
    dan
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    that helps...(:
     .04 - (.0032(2^2)) - (10^{-6}*134^{2}) + (4.1X10^{-5}) = .009285


    but don't quote me on that...I aways make mistakes...

    ~dan
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