Flow Rates & Equilibrium

Starting from an empty tank (with dimension L" x W" x D"), what is the
system of equations necessary to determine the diameter of a drain
hole (the drain is always open) necessary to fill the tank and
maintain a x" water level for an indefinite period of time in a L" x
W" x D" cube with a constant inflow rate of r gph? Naturally, part of
this solution is to account for the differential pressure across the
orifice caused by the height of the water.

The answer should include an abstract of the solution; include what
steps were taken to get to the solution.

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Request for Question Clarification by hedgie-ga on 16 Mar 2003 20:42 PST

Pygovian

  Are  you  (still) interested in (more numerical)
 answer, which would elaborate my comment below?

  Formula would include L  - the Length of the drain.

I would need  clarification of that 'filter in the inlet'.

Word 'inlet' can be used for pipe bringing the water to the tank
but also for the drain.
If there is a free surface, under the inlet,
then filter does not influence the water level reached
in steady state.

Some initial thoughts:
The tank begins empty and must retain some water in order to fill to
the required constant depth.  Therefore, the outflow rate must
initially be lower than the inflow rate.  As the depth increases, the
increasing pressure will increase the outflow rate until at some point
it exactly matches the inflow rate, and this point must occur at the
required depth.
The question therefore reduces to finding the diameter of a drain that
will pass r gph with a pressure head of x".

Some worries:  turbulence and vortex effects, and believe it or not -
latitude!
I also wonder if the required depth is approached asymptotically, and
in fact, never "really" achieved.

I would agree with your initial assessment, which would appear to be
simple enough s.t. the answer would be given by the equation
k*sqrt(2*g)*A*sqrt(h)=gph, where k a constant of the medium (water is
assumed to be 1), A is the area of the orifice and h is the height of
the water.  Solving for A, then in turn solving for D, would give the
answer for the diameter necessary to balance the inflow and outflow
rates.

However, part of my solution is also on the fact that time is of the
essence.  Hence, my problem.  I am having difficulty setting up a
system of equations (in which I am sure a lot would be, ughh, DE) to
determine the exact level of water with relation to an arbitrary time
t.

As for turbulence, vortex effects, capillary effects (because of a
medium in the tank), and other physics/fluid dynamic variables that I
cannot think of (I’m an econ type, not an engineer!) there are many
variables that I would like added to the model.  Thus, I have just
been looking for a more idealized solution, modeling what I can
explain, as I cannot speak intelligently upon the minutiae of the
system.  Any help in modeling these variables in would be greatly
appreciated.

Good point on approaching the level asymptotically.  Inclusion would
be good within the model, such that a given epsilon is the maximum
deviation allowed from the asymptote.

In reality, this should not be a big issue.  The error on the water
level can be +-5% of the height.

Another factor involving the minutaie, there will be a carbon filter
diffusing the stream at inflow.

This is actually pretty easy, because you are only concerned with the
steady state situation.  That is, the situation that holds when the
volume of fluid in the container (and thus the fluid level) does not
change (assuming the width and length of the container are fixed --
the container must not deform as it fills), which means that the flow
of fluid into the container is exactly balanced by the flow out. 
There's no need to resort to differential equations to obtain the
relationship you are seeking.

Outline of solution:

1. Volumetric flow rate out of container (Q.out) = orifice area *
fluid velocity through exit orifice (A*v)
2. Volumetric flow rate into container (Q.in) = a constant (r)
3. Use Bernoulli's equation to find fluid exit velocity as a function
of fluid height above orifice
4. For the steady state case (i.e. not change in fluid depth as a
function of time) Q.out = Q.in = r
5. The equation resulting from step 4 provides the desired
relationship between the steady-state fluid depth and orifice area.

Complications:

The velocity of the exit stream across the diameter of the orifice
will not be exactly what is predicted by the Bernoulli equation
because of the non-zero viscosity of any real fluid and frictional
effects that depend on the details of the geometry of the exit orifice
(e.g., shape, thickness of the vessel walls, surface roughness, etc.),
the fluid properties (density, viscosity), and the fluid velocity.  
In practice, the equation for volumetric flow out of the container
(Q.out) must include a "fudge factor", usually called the "discharge
coefficient" (k; k <= 1) that accounts for this: Q.out = k * A * v. 
In general you will need to determine k experimentally.  If you assume
k=1, then the calculations given below will tell you the *minimum*
steady-state fluid height that an orifice of a given diameter will
sustain.

This calculation further assumes that any pressure variations in the
fluid due to movement of the fluid within the container will average
out to zero along all streamlines from the surface to the orifice.  I
practice, this means that we don't account for any persistent currents
induced by the water stream entering the container.  An obvious
consequence of this assumption is that the solution will also be
grossly incorrect if a vortex (whirlpool) forms at the drain.

Solution:

Solving Bernoulli's equation for an incompressible, low-viscosity
fluid along a streamline that starts at the upper surface of the fluid
and ends at the orifice (which assumed to be on the bottom of the
container at fluid height h = 0) yields the following equation:

	rho * g * h(t) + [p.atm(h=0)) - p.atm(h(t))] = 0.5 * rho * v(t)^2 

where, 
v(t) is the ideal fluid velocity out of the exit orifice (which at
this stage, we allow to be a function of time)
rho is the fluid density
h(t) is the fluid height above the orifice (also allowed to be a
function of time)
g is the acceleration of gravity
p.atm(x) is the atmospheric pressure at elevation x

If the container is relatively small, we can assume that the
atmospheric pressure at the exit orifice and the fluid surface is the
same (p.atm(h=0)) = p.atm(h(t))) , so the second term on the left hand
side of the above equation vanishes, resulting in:

	g * h(t) = 0.5 * rho * v(t)^2
	v(t) = sqrt[ 2 * g * h(t)]

Obviously, this also assumes that the discharge orifice is open to the
air, and is not submerged outside the container.

Substituting this expression into the expression for the volumetric
flow out of the container (Q.out(t) = k * A * v(t), which includes our
"fudge factor") yields:

	
	Q.out(t) = k * A * sqrt[2 * g * h(t)]

where
k is the fudge factor
A is the area of the orifice

At steady state:

	Q.out = Q.in
	Q.out = r
	k * A * sqrt[2 * g * h(ss)] = r

where h(ss) is the steady-state water height.

Rearranging this gives:

	A = r/(k*sqrt[2 * g * h(ss)]

Your question implied that the orifice is circular with diameter d. 
For a circular aperture of diamater d:
	
	A = pi * d^2/4
	pi * d^2/4 = r/(k*sqrt[2 * g * h(ss)]
	d = sqrt[(4 * r)/(k * pi * sqrt[2 * g * h(ss)])]

which says that the diameter of the hole that results in a constant
fluid depth h varies as one over the fourth root of h.

If you use this to actually calculate something, make sure you use
consistent units of measurement and watch out for your unit
conversions!  Your original question involved flow rates in gallons/hr
and fluid heights in inches.  Make sure you convert gallons to cubic
inches, and express g in appropriate units.

Bernoulli equation relates change in velocity to change in presure
 when diameter of a tube changes:

 http://www.rz.uni-frankfurt.de/~weltner/

 That's not quite  the case we have here. Easy way to 
 see what you need  is to use electrical circuit
 analogy: You have a voltage (presure head) acting 
 on a resistance (drain + any filters in inlet).
 
  You may have to measure the resistance of the filter
(or getting that the spec). The resistance of the drain
you can aproximate based on  the geometry 
( diameter, LENGTH of the narrow pipe)  and viscosity
 under few assumptions (laminar flow --i.e.
low Reynolds number i.e. all is fairly slow and smooth... see
scienceworld.wolfram.com/physics/ReynoldsNumber.html

Let's clear up a few misconceptions:

1) Since the inflow rate is specified (r), the filter at the inlet
doesn't matter.

2) Though turbulence and vortex effects may well be important,
latitude doesn't matter--it takes very careful experiments where large
tanks of water are allowed to sit for many hours to become as still as
possible to detect the Coriolis effect in laboratory flows.

3) Bernoulli's equation is relevant to this problem.  In irrotational,
steady flows (we can take this flow as steady even though the depth
changes over time, because the change is slow, and though the actual
flow may not be irrotational, for a first stab at the problem it's
easier to assume it is, particularly since we aren't give information
about things like the position of the inflow, drain, etc.) at constant
density, Bernoulli's eq. is:

1/2 rho v^2 + p + rho g z = constant everywhere

rho is density, v is velocity, p is pressure, g is gravity, and z is
height.  To say Bernoulli's equation only applies to the very specific
case of flow in a pipe with a changing diameter is silliness.

Now to answer the question:

First we can use Bernoulli's eq. to find the velocity of the water
leaving the drain as a function of the height of the water, h:  v =
sqrt(2gh).  In reality, if the drain has sharp edges, the average flow
rate at the drain will only be about 62% of this (known from
experiment), but if the drain is a horizontal, smoothly tapering
'funnel' the result is correct.  Let r be the rate of inflow in cubic
meters per second (so the units work out--you can convert from gph;
incidentally, everything is assumed measured in SI units), and let R
be the radius of the drain.  Then we have:

dh/dt = r/A - a*v/A  (a = drain area = pi*R^2, A = tank area = L*W),
or

dh/dt = (r - pi*R^2*sqrt(2*g*h))/(L*W).

This differential equation is easily solved by separation of
variables, and will give you h as a function of time for a given drain
radius R.  You can use the simple method you already know to pick a
value for R, and then use the DE to find out how long it will take for
the water level to come within a given distance of its equilibrium
height for this value of R.

So which way *does* the water rotate when it drains on the equator?  
Just curious.  I would guess it is either a random direction that
self-reinforces, or the flow is chaotic.

The direction of rotation of the vortex that forms when water is
drained depends on the absolute vorticity of the water.  Generally in
something like a sink in your house, the relative vorticity due to the
the motion of the water in the sink, caused by the way the sink was
filled, or air currents, or whatever, is orders of magnitude greater
than the planetary vorticity.  So water in a sink at the equator will
drain just the same as water in a sink anywhere else on the
planet--the planetary vorticity is effectively zero everywhere, so the
fact that it's exactly zero on the equator is irrelevant.

Hi racecar:

So you're suggesting that the notion that water drains with opposite
rotation in the N and S hemisphere is either a myth, or a coincidence.
 Interesting indeed.

Hey xarqi,

Yep, a myth.  Check out 

http://www.straightdope.com/classics/a1_161.html

for corroboration.

One way to determine if Coriolis effects will be important in a given
situation is to consider the Rossby number, Ro = v/(fL), where v is
velocity, f is the Coriolis parameter, and L is the lengthscale
associated with the system.  As long as Ro is much larger than 1,
rotation is unimportant.  So for something with L~1m, as long as v is
large compared to .0001 m/s, you can forget about rotation.

My faith in physics as demonstrated on "The Simpsons" is completely shattered!
:-)