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Insane Dimensional Analysis Problem

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data
    • The American Petroleum Institute has published a correlation for determining the hydrocarbon emissions from fixed-roof storage tanks


    Ly = (24/1000) * (p/(14.7-p))^0.68 * D^1.73 * H^0.51 * T^0.5 * Fp * C

    where: Ly is breathing emissions, gallons/yr; p is the true vapor pressure at the bulk temperature, psi; D is the tank diameter, ft; H is the height in ft; T is the average tank outage corrected for roof volume, ft; Fp is the dimensionless paint factor; and C is the dimensionless adjustment factor.

    Is this equation dimensionally consistent? Is it possible, based on a unit analysis, that the equation is correct? If not, what units must be added to the term 24/1000 to make it dimensionally consistent?



    2. Relevant equations

    Dimensional analysis and conversions



    3. The attempt at a solution

    I simply rewrote the equation but I am totally stuck on what to do with the (p/(14.7-p))^0.68 term... how do you have a unit to a power like 0.68?
     
  2. jcsd
  3. Jun 4, 2013 #2
    It is not really different from a integer power like 2, which you find in acceleration, pressure, etc.
     
  4. Jun 4, 2013 #3

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    The term (p/(14.7-p)) is dimensionless, as you have pressure divided by pressure. Raising a dimensionless number to a power does not change the fact that it is dimensionless.
     
  5. Feb 25, 2017 #4
    Does anyone have the solution to this problem? I do not even know how to start to solve it!

    Thank you :)
     
  6. Feb 25, 2017 #5

    gneill

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    Staff: Mentor

    Hi popoff. No one here will give you the solution to a homework question. That would be against the forum rules.

    If you have the same homework question and you need help with it you'll have to show your own attempt at solution (show what your understanding is of the problem and what approaches you have already tried) before help can be offered. The best way to do this when a thread is as old as this is to start your own, new thread.
     
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