Correct way to write pi buckingham theorem

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The discussion revolves around the application of the Buckingham Pi theorem in fluid dynamics, specifically regarding the representation of dimensionless groups. Participants debate whether to define π1 as the reciprocal of the Reynolds number (μ/ρVD) or as the Reynolds number itself (ρVD/μ), questioning the implications of such choices on the function relationships between the dimensionless groups. It is emphasized that either choice is valid, as one can always derive the reciprocal relationship through appropriate functions. The importance of clearly defining the function f, which relates π1 to π2 and π3, is highlighted, as it dictates how the dimensionless groups are interrelated. Ultimately, the flexibility in defining dimensionless groups is affirmed, as long as the relationships are consistently maintained.
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Homework Statement


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 ?

Homework Equations

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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wow , thi is considered as advanced physics question ?
 
I don't see any difference between taking the reciprocal of Reynold's number and taking actually Reynold's number as ##\pi_1##.
 
MathematicalPhysicist said:
I don't see any difference between taking the reciprocal of Reynold's number and taking actually Reynold's number as ##\pi_1##.
Why?
 
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?
 
MathematicalPhysicist said:
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?
It's pi Buckingham theorem, can I still do so?
You didn't state what is your precise problem here?
 
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 ?
 
Do you know what is f here?
 
MathematicalPhysicist said:
Do you know what is f here?
Ff means function, where pi1 is the function of pi2 and pi3...
 
  • #10
I mean is f given explicitly?
 
  • #11
MathematicalPhysicist said:
I mean is f given explicitly?
? What do you mean?
 
  • #12
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?
 
  • #13
MathematicalPhysicist said:
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?
Dun know
 
  • #14
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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