## A problem of fluid flowing through a hole

I’m sure you’re all already as bored of Christmas as I am so I thought I’d do you all a favour by giving you something interested to do to distract you from the yuletide tedium,
The cute problem of the water tank I posted a while ago seemed to provide a diversion for many – although only about 10% of respondents go it right – so here’s a similar one. It’s not multiple choice so you will have to write your answers to the two parts in the comments box. As a hint, I’ll  say that this is from some notes on dimensional analysis, and it’s one of the harder problems I have in that file!

An incompressible fluid flows through a small hole of diameter d in a thin plane metal sheet. The volume flow rate R depends on d, on the fluid viscosity η and density ρ, and on the pressure difference p between the two sides of the she

(a) Find the most general possible relationship between the quantities  R, d, η,  ρ, and p.

(b) Measurement of the flow rate R1  through this the hole for a pressure difference p1 is made using a particular fluid. What can be predicted for a fluid of twice the density and one-third the viscosity?

### 4 Responses to “A problem of fluid flowing through a hole”

1. andyinkuwait Says:

R = d^4√(ρP^3)/η^2

9√2 change in R

• James Fry Says:

Well, since no one else has tried –
(a) With three units (length, mass, and time), assuming powers of the four variables, \eta, rho, p, and d, leaves some degeneracy in the exponents,
R ~ [\sqrt{\rho p} d / \eta]^\alpha \sqrt{p/\rho} d^2
where the factor in brackets is dimensionless and \alpha is arbitrary. \sqrt{p/\rho} is dimensionally a velocity, and the bracketed quantity is dimensionally density times velocity times diameter divided by viscosity, a Reynolds-number sort of thing. In full generality, any function of the bracketed factor works.
(b) For laminar flow, the choice \alpha = 1 or R ~ p d^3/\eta seems right, leaves no dependence on density. One-third the viscosity, three times the flow rate.

2. telescoper Says:

Nobody’s quite got this right yet. There are in fact two dimensionless combinations that can be formed from the relevant variables: one excludes R and the other excludes \rho. If you call these N and M then the most general expression is f(N,M)=0.

3. A Problem of Dimensions | In the Dark Says:

[…] some things about units and dimensional analysis. Thinking about this reminded me that I posted a dimensional analysis problem (too hard for first-year students) on here a while ago which seemed to pose a challenge so I […]