Fluid Dynamics

Water in a cylindrical bucket, whose axis is vertical, is moving with horizontal velocity components given by u=-ky and v=kx. Find the pressure at any point in the fluid at distance r from the axis and z from the bottom of the bucket.
On the free surface the pressure is constant and equal to the atmospheric pressure p_0. Calculate the difference in the surface level of the water at the axis from that time at the rim, given that the bucket has radius a.

Has anybody got any ideas to help me? Seriously, have no idea where to start!

Reply 1

zCtpts2T

Since the fluid is in hydrostatic equilibrium, we can use the equation:

pF = grad P ,(p ~ density of fluid, F ~ body force, P ~ pressure)

{Since the form of the velocity is circular flow about z axis, the body force on a volume of water is composed of both gravitational and centrifugal forces:

V = k(-y,x) = kr(-sinQ, cosQ) (r ~ radius, Q ~ angle relative to x axis)

i.e y = rsinQ, x = rcosQ.

(w ~ angular velocity)

|v| = kr, v = w.r -----> w = k; }

using hydrostatic equilibrium equation above:

p F(radial) = dP/dr, p F(axial, z direction) = dP/dz;

F(radial) = Centrifugal component = r.w^2 = r.k^2,

F(Z direction) = -g;

therefore two pde's to solve:

-p.r.k^2 = dP/dr -------> P = (1/2)p.r^2.k^2 + f(z)

-p.g = dP/dz --------> P = -p.g.z + h(r)

therefore P = (1/2)p.r^2.k^2 - p.g.z + constant.

Note (r = 0, z = 0) is the point at which the free surface is at a minimum height, hard to explain in words without a picture. i.e. in middle of barrel (obviously for symmetry purposes) and at bottom of surface described by water, not bottom of the barrel.

If you want to measure from the bottom of the bucket/barrel, switch the z for another variable to compensate for the change of coordinate system.

free surface part:

at the origin (in my coord system) Po ~ atmospheric pressure,

therefore Po = (1/2)p.r^2.k^2 - p.g.z + A ---> A = Po

P = (1/2)p.k^2.r^2 - p.g.z + Po.

On the free surface, insist that P = Po,

Therefore at radius a: z becomes h, the height difference from z = 0

(1/2)p.k^2.a^2 = p.g.h ---> h = k^2.a^2/(2g),

Hope this helps, would be better with a pen and paper and a picture but that is the trouble with internet answers i'm affraid.

Any other probs, don't hesistate to ask

Reply 2

DB9

wow - genius, thank you!
seeing as you're obviosuly very good in this type of maths, could you show me the way with this question too please?!

A long straight pipe of length l has slowly been tapering cross-section. It is inclined so that its axis makes an angle alpha to the horizontal and with its smaller cross-section downwards. The radius of the pipe at its upper end is twice that at its lower end and water is pumped at a steady rate through the pipe to emerge at atmospheric pressure. If the pumping pressure is twice atmospheric pressure, show that the fluid leaves the pipe with speed U given by

U^2 = 32/15(glsin alpha + p_a/p)
where p_a denotes atmospheric pressure.

Any help at all would be greatly appreciated! thank you

Reply 3

zCtpts2T

Hi, I'm actually a physicist but mostly theoretical based so I have done courses on fluid dynamics, elasticity etc (the more mathsy courses) in the past.

As for this question, its use of bernoullis equation and mass conservation. I always found that remembering a few facts is better than remembering every detail of a problem, like bernoulli's equation is just an energy conservation equation in fluid dynamics.

i.e. kinetic + gravitational + constant

so bernoulli's equation (valid for every point in the fluid) is:

(1/2)p.v^2 + p.g.z + P = constant

(V ~ velocity of fluid at the particular point, i think you may call it U but I've always called it v as velocity begins with a v?!? (I never did understand the mathematicians logic on that one) anyways, z ~ height of point with respect to choice of coordinate system, others have same meaning as in my last post)

we can write z as l.sinQ (Q ~ angle alpha)

The conservation of mass equation (usually called the continuity equation, because any conserved quantity can form a continuity equation of the form : dJ/dt + div p = 0 where J is the flow of p across some boundary per unit time) reduces to just p.A.v = constant, and since p is constant, A.v = constant.

A being the area of the pipe at the particular point of evaluation.

hence at the bottom end, radius r = a say, so pi.a^2.v(bottom) = constant,
hence at the top end, radius r = 2a, so pi.4.a^2.v(top) = constant,

so pi.a^2.v(bottom) = pi.4.a^2.v(top) ---> v(top) = (1/4)v(bottom) = (1/4)U

this makes sense as you would expect a smaller pipe to have a faster flow than a bigger pipe.

going back to bernoulli:

(1/2).p.v(bottom)^2 + Po = (1/2).p.v(top)^2 + p.g.l.sinQ + 2Po

simplifying gives:

(1/2).p.U^2 + Po = (1/2).p.(U/4)^2 + p.g.l.sinQ + 2Po

(1/2)p.U^2 = (1/32)p.U^2 + p.g.l.sinQ + Po
(1/2 - 1/32)U^2 = (15/32)U^2 = g.l.sinQ + Po/p

so U^2 = (32/15)(g.l.sinQ + Po/p) which is what I think you stated.

I assumed in this calculation that the water is flowing down the pipe from higher potential to lower potential.

hope this helps.

Are you doing university level maths/physics? If so, what year are you in?

Reply 4

DB9

Hey, I'm in my second year at uni, studying Pure Maths. Unfortunately though there is one unit of fluid dynamics for us this term and since we've started to do Bernoulli's equation I've been struggling abit! The thing with Maths I find, is that we don't always answer the questions in the most sensible approach that Physicits would usually adopt. We use random symbols to represent stuff but I never really see what the point is! I understand all of what you have written though which is a bonus, so thank you so much.

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