# Homework Help: Use dimensional analysis to derive Poiseuille's Law

1. Sep 14, 2014

### devon

1. The problem statement, all variables and given/known data

Using dimensional analysis deduce the relationship between the pressure drop per unit length along a cylindrical pipe of radius r, and the radius of the pipe, the viscosity of the fluid in the pipe, η, and the volume flow rate, V ̇ .

2. Relevant equations

Δp/l = 8ηV ̇/(pi)r^4

3. The attempt at a solution

So I got the dimensions of all of the variables and made them proportional to one another

M L^-2 T^-2 = (L^3 T^-1)^a x (M L^-1 T^-1)^b x (L)^c
Pressure gtd Volume Flow Viscosity Radius

I manage to a = -2, b = 1, c= -1/3 but this seems wrong as volume flow rate is not inversely proportional in Poiseuille's, only radius is. Also should volume flow rate (a) = 1 and radius (c) = -4
Are this different values due to the dimensionless numbers of 8 and pi or have I done something completely wrong? Cheers

2. Sep 14, 2014

### devon

Wow 37 views and no replies, is it really that hard? Because that will make me feel a lot better about not getting the answer, I've wasted a lot of time with just this one question.

3. Sep 14, 2014

### rude man

You can make life easier by using pressure as one of your 3 dimensions.

Thus, PL-1 = Ra ηb (dV/dt)c
PL-1 = La (PT)b (L3T-1)c

etc. solve for a, b & c.

You should have gotten the right answer your way too but since you did not write your three equations in a, b and c I don't know what mistake you made. Your starting equation looks right.

Last edited: Sep 14, 2014
4. Sep 14, 2014

### Staff: Mentor

It looks toe like you have it set up correctly. Show us your intermediate steps, please.

Chet

5. Sep 14, 2014

### fishes

Thanks a lot for the help! So by using pressure as a dimension I can get b = 1 but I'm not sure on getting a or c. After finding b do I convert pressure back to its base dimensions to find them? I manage to get c=2 and a = -1/6 which doesn't seem right to me. Am I correct in saying that a should = -4 and b and c both = 1?

6. Sep 14, 2014

### fishes

Well I first find b = 1 as there is only 1 mass dimension on either side. I then sub that in and find a: -2 = -a*-1 which gives me a = -2. I then sub that in to find c: -2=-6*-1*c which gives me -1/3. Am I doing something really obvious wrong?

7. Sep 14, 2014

### rude man

As to your last question, the answer is yes but you could have just looked that up ...

No, you keep pressure as one of your three dimensions. Or you can do it your way which is slightly more mindboggling.

As chestermiller says, show us your three equations in a, b and c.

8. Sep 14, 2014

### rude man

Try this one again: -2 = -a*-1

And this : -2=-6*-1*c

Both equations are correct, neither is solved correctly! Again, sorry I changed the order of the powers a, b and c.

Last edited: Sep 14, 2014
9. Sep 14, 2014

### fishes

I'm sticking to mine, okay so here are my 3 equations.

b: M^1 = M^b therefore b = 1

a:T^-2 = T^-a x T^-b ---> -2 =ab (b=1) so -2 =a

c: L^-2 = L^3a x L^-b x L^c ----> -2=3a x -b x c (b =1, a = -2) -2=-6 x -1 xc -2 =6c so c = -1/3

10. Sep 14, 2014

### fishes

I'm really sorry but I must be having the biggest blank not now, I'm pretty sleep deprived but I can't see what's wrong with either equation that would get me a = 1 and c = -4

Edit: I put both equations in wolfram just to be sure and still got -1/3 and -2? Gosh this must be annoying for you guys, I'm really sorry.

Last edited: Sep 14, 2014
11. Sep 14, 2014

### rude man

Must be sleep deprivation on your part AND on Wolfram's!

-2 = -a*-1. Move a to the left and "-2" to the right:
a = -1 + 2
a = +1.

Try for c again?

12. Sep 14, 2014

### fishes

OMG I JUST REALISED WHAT I WAS DOING! I was solving them as if all of the values on either side were multipled together. Wow I feel like such an idiot. Have no idea where that came from. Can't believe that I forgot multiplying indices together pretty much means addition. Really sorry for all the trouble guys!

13. Sep 14, 2014

### rude man

All is forgiven!