Use dimensional analysis to derive Poiseuille's Law

[devon]

Homework Statement

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 ̇ .

Homework Equations

Δp/l = 8ηV ̇/(pi)r^4

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

 

[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.

 

[rude man]

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

Thus, PL^-1 = R^a η^b (dV/dt)^c
PL^-1 = L^a (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.

 

[Chestermiller]

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

Chet

 

[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?

 

[fishes]

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

Chet

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 into find c: -2=-6*-1*c which gives me -1/3. Am I doing something really obvious wrong?

 

[rude man]

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?

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.

 

[rude man]

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 into find c: -2=-6*-1*c which gives me -1/3. Am I doing something really obvious wrong?

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.

 

[fishes]

b = 1 is right. Where did you dig up -2 = -a*-1? BTW are you sticking to your original a, b and c or mine? Sorry, I should not have changed those around.

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

 

[fishes]

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.

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.

 

[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?

 

[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!

 

[rude man]

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!

All is forgiven!