Homework Help: Choosing repeating variable in pi Buckingham theorem

1. Apr 10, 2016

foo9008

1. The problem statement, all variables and given/known data
why we cant form the pi group by using repeating variable of (μ, ρ , v) or (D, v , μ ) ?
http://www-mdp.eng.cam.ac.uk/web/li...mal_dvd_only/aero/fprops/dimension/node9.html
http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section5/dimensional_analysis.htm

2. Relevant equations

3. The attempt at a solution
according to the principle , when we choose the repeating variable , it cant form the dimensionless group , right ? for (μ, ρ , v) , i have (M/LT)(M/(L^3))(L/T) , i got (M^2)(L^-3)(T^-2)
for (D, v , μ ) , i have (L)(L/T)(M/LT) , i got ML(T^-2) , both group are not dimensionless, why is it a must to form group by using (D, ρ , v) ??

2. Apr 11, 2016

Orodruin

Staff Emeritus
In general, it is perfectly fine to choose any group which is dimensionally independent. But be aware that it is not sufficient to check the product of the variables to conclude this as you seem to be doing.

Edit: Just one additional thing. Although you are free to pick any dimensionally independent group, some might work better or worse for your purposes. The theorem does not tell you which is the best choice.

Last edited: Apr 11, 2016
3. Apr 11, 2016

haruspex

4. Apr 11, 2016

foo9008

so , we have to try and error to get the correct ans ?

5. Apr 11, 2016

foo9008

6. Apr 11, 2016

haruspex

As I wrote in the final post on the thread, any three of D, ρ, v or μ could have been used.

7. Apr 11, 2016

foo9008

do you mean no matter which combination of D, ρ, v or μ , i would get the same answer too?

8. Apr 11, 2016

haruspex

Try it.

9. Apr 11, 2016

foo9008

taking ρ, v and μ as repeating variables , i form 2 pi group , but i cant get the ans .... why ?

Attached Files:

• 1420002.jpg
File size:
16.6 KB
Views:
95
10. Apr 11, 2016

Orodruin

Staff Emeritus
You will not get the same $\pi$s, but the relations will be equivalent.

11. Apr 11, 2016

haruspex

You left something out of the L equation in the left hand column.

12. Apr 11, 2016

foo9008

deleted

Last edited: Apr 11, 2016
13. Apr 11, 2016

foo9008

ok , i have done the correction , and my ans is Fρ / (μ^2) = f(ρvD / μ) , but my ans is not consistent with the author's working which is
μ/ ρvD = f ( F / ρ(v^2)(D^2) )

Attached Files:

• l1.jpg
File size:
16.1 KB
Views:
103
14. Apr 11, 2016

foo9008

i have also tried out forming pi group by taking the repeating variable as ρ , μ , D , the ans that i got is same as post #13 .
however , when i choose repeating variable as v , μ , D , my ans is different ( i got (FD/ vμ ) = f(vDρ/μ) instead of (Fρ/ μ^2 ) = f( vDρ/μ) ) . why is it so ?

15. Apr 11, 2016

haruspex

It is consistent.
Let's write x=μ/ ρvD and y=Fρ / (μ^2). You got y=f(x), the author got x=f(y/x2) for some other function f. These are effectively the same.

16. Apr 11, 2016

foo9008

sorry , i dont understand . do you mean from the author's working , i can remove the f( ) , and make μ/ ρvD = F / ρ(D^2)(V^2) , so eventually , i will get μDV = F ? for pi group with (ρ , V , μ , F) and (ρ , V , μ , D) , i will get Fρ/ (μ^2) = f( ρvD / μ) , after rearranging , i will also get μDV = F

17. Apr 11, 2016

haruspex

No, I mean that saying y is a function of x is, generally speaking, equivalent to saying x is a function of y/x2.
Examples:
Y=2x is the same as x= 2/(y/x2)
Y=x4 is the same as x=(y/x2)1/2.

But it doesn't always work. There is no way to express y=x2 in the form x=f(y/x2)
Translating that back, your form can express F=AρV2D2 but the author's cannot. Likewise, the author's can express ρvD=Aμ, but yours cannot. That is a limitation of writing things in the form y=f(x); it cannot express a vertical line, i.e. x is constant. That problem would be solved by writing instead f(x,y)=0.

18. Apr 11, 2016

foo9008

So, my working in post 13 is correct??

19. Apr 11, 2016

foo9008

in my working , i only show that F(ρ)/ (μ^2) = f(ρvD / μ) , where is F=AρV2D2 ?
can you point out which part of the author's working show ρvD=Aμ ? i didnt see it

20. Apr 11, 2016

haruspex

No, you misunderstand. F=AρV2D2 is merely an example of Fρ/ (μ^2) = f(ρvD / μ) . Specifically, it is if the function f is f(x)=Ax2.

21. Apr 11, 2016

haruspex

Yes.

22. Apr 11, 2016

foo9008

oh , no ... i am getting more confused now , can you explain in another way round ?

23. Apr 11, 2016

Orodruin

Staff Emeritus
The theorem only states that the relation must be in the form of a function of dimensionless variables. It does not tell you what this function is and depending on your choice of repeating variables, the function describing your relationship will be different (because you will have different dimensionless variables). What haruspex gave you was an example.

24. Apr 11, 2016

foo9008

for repeating factor ( v, μ , D ) , i can form pi group with ( v, μ , D , F ) and ( v, μ , D , ρ) , i got (F/ v μ D) = f(v D ρ / μ) , after rearranging , i got F = ρ (D^2)(v^2) , is my working correct ?

Attached Files:

• 143.jpg
File size:
28.3 KB
Views:
71
25. Apr 12, 2016

Orodruin

Staff Emeritus
The Buckingam pi theorem tells you nothing about what the function f is. Again, haruspex's Ax^2 was an example.