## 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 R _{1} through this the hole for a pressure difference p_{1} is made using a particular fluid. What can be predicted for a fluid of twice the density and one-third the viscosity?*

As usual, answers through the comments box please!

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December 19, 2017 at 5:06 pm

R = d^4√(ρP^3)/η^2

9√2 change in R

December 23, 2017 at 12:28 am

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.

January 2, 2018 at 5:58 pm

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.

August 21, 2019 at 5:55 pm

[…] 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 […]