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Correct way to write pi buckingham theorem

  1. Jun 16, 2016 #1
    1. The problem statement, all variables and given/known data
    in this problem , the author make π1 = D(dp/ dx) / ρ( V^2) , and make π3 as μ/ ρVD , how if i want to make μ/ ρVD (reciprocal of reynold number ) as π1 and make D(dp/ dx) / ρ( V^2) as π3 ?



    2. Relevant equations


    3. The attempt at a solution
    since we know that π1 is function of ( π2 , π3 )
    is it necessary to change μ/ ρVD (reciprocal of reynold number ) to reynold number (ρVD / μ ) ?
    which is correct ? Re = f ( D(dp/ dx) / ρ( V^2) , Ks/ D ) or μ/ ρVD = f ( D(dp/ dx) / ρ( V^2) , Ks/ D ) ? which is correct ?
     

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  2. jcsd
  3. Jun 17, 2016 #2
    wow , thi is considered as advanced physics question ?
     
  4. Jun 18, 2016 #3

    MathematicalPhysicist

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    I don't see any difference between taking the reciprocal of Reynold's number and taking actually Reynold's number as ##\pi_1##.
     
  5. Jun 18, 2016 #4
    Why?
     
  6. Jun 18, 2016 #5

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    If you take ##\pi_1 =f (\pi_2 , \pi_3)## for some function ##f## of ##\pi_2 , \pi_3## in which case you can find a function ##g## such that ##\pi_1^{-1} = g(\pi_2 , \pi_3)##; so it doesn't matter which one you choose your ##\pi_1## to be, your function will of course be different for different cases, but you don't seem to know what is your function ##f##, right?

    You didn't state what is your precise problem here?
     
  7. Jun 18, 2016 #6
     
  8. Jun 18, 2016 #7
    Re = f ( D(dp/ dx) / ρ( V^2) , Ks/ D ) or μ/ ρVD which is 1 / Re = f ( D(dp/ dx) / ρ( V^2) , Ks/ D ) ? which is correct ?
     
  9. Jun 18, 2016 #8

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    Do you know what is f here?
     
  10. Jun 18, 2016 #9
    Ff means function, where pi1 is the function of pi2 and pi3...
     
  11. Jun 18, 2016 #10

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    I mean is f given explicitly?
     
  12. Jun 18, 2016 #11
    ? What do you mean?
     
  13. Jun 18, 2016 #12

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    I mean do you know how is f given? I mean do you know what is f(\pi_2, \pi_3) what is this function of \pi_2 and \pi_3?
     
  14. Jun 18, 2016 #13
    Dun know
     
  15. Jun 18, 2016 #14

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    As it mentioned in one of the pics you rearrange only for convenience, i.e. it doesn't matter if you take Reynold's number or the reciprocal of Reynold's number as a function of the other dimensionless variables since you can always take the reciprocal of the function if you have ##Re = f(\pi_2 , \pi_3)## then you can take ##1/Re = 1/f = g(\pi_2,\pi_3)##.

    If on the other hand ##f## were given then you'd know how to rearrange the equation.
     
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