Understanding Dimensional Analysis: Simplifying Fluid Flow Equations

[Beer w/Straw]

This is an elusive subject with added order of magnitude.

If that doesn't make sense, well... It tells that I'm confused.


Say rate of fluid flow in a uniform cross-sectional area pipe |A|=L^2,|R|=L^3*T^-1. Is it customary that I get all variebles to the same order of magnitude and be able to express it as the normal equation d^2*V for fluid flow rate?

 

[Andy Resnick]

I don't follow your question- the variables don't have to be 'the same order of magnitude'- in fact, the dimensionless number that results (i.e. the Reynolds number) gives information about the relative importance of various processes.

 

[Beer w/Straw]

What I said was very similar to a question I had last term, and I still don't understand how orderr of magnitude is related to dimensional analysis:

A pipe of uniform cross-sectional area drains (fluid) into into a bucket. Use dimensional analysis to determine the rate of fluid flow to a within a multiplicative constant of order 1.

Dimesional analysis, although seems simple, is in none of my textbooks

I'm stuck on |R|= L^3T^-1 with cross-sectional area |A|=L^2

 

[Renge Ishyo]

I'm not sure if I understand the question either, but if you are talking about "why do I have to express the measurements in base units" before performing the calculation (i.e. express them as the base SI units)...well you don't *have* to, but then when you perform the calculation the units will not cancel and you will get a different number with freaky units attached to it. If you use dimensional analysis so that all length or area measurements are in meters (for example) then when you perform the calculation the "meter units" can combine together (multiplication) or cancel each other (division) and it makes the units on the final answer much more simple to interpret.

 

[Andy Resnick]

That's a strange question. [R] = L^3/T is easy: volume flow (liters/min, for example). Maybe the idea is to construct a dimensionless number: [R]/[A] is a velocity [L/T]. For your problem, you may also need to consider the viscosity [L^2/T].

 

[Renge Ishyo]

From dimensional analysis alone, if you assume a direct linear relationship you can solve for the units of the missing component (x). I.e.

R = (x)A

or R/A = (x)

Since the units of R = L^3/t and A = L^2, upon cancelling the units we get for x that the units are L/t. These are the units of velocity such that:

R = vA

Is that what it was asking for?Edit: Ah, Andy you beat me too it. I think what he is asking it why does he have to make sure the dimensions of L for both the area and the volume flow rate are the same, but we won't know until he replies.

 

[Beer w/Straw]

Thank you guys, I'll ponder the simplicity of your answer(s). My proffesor is a weirdo who I don't understand half the time. It took me quite some time to understand he wanted a scalar "projection" and not a vector one in one question (that was supposed to be simple.)