Fluid Flow and Hole Size
Does the size of the hole affect the water pressure?
This is from a student conducting a test. She wanted to know if you
the hole size (not the location) will the rate of water flow out of that
hole decrease or increase? But more specifically, what is the
between the hole size and water flow rate - for example, if a hole is
doubled its size does the water flow rate also double? Also, is there a
point where a hole becomes so small that water flow is non existent.
This depends largely on the source of water.
If you have a pump pressurizing the water, then the pressure it is able to
maintain will drop with larger holes, because more water will be escaping.
(More flow at less pressure)
If the water pressure behind the hole is constant, (say, a relatively small
hole at the bottom of a constant source such as a dam or water tower), then
you pressure will remain constant, and flow will increase even more.
I do not believe you will ever get a hole small enough to completely prevent
water flowing through it, (at least not unless you are drilling holes the
size of molecules), however, a small enough hole will have such a small
amount of water flowing through it that it can be disregarded for all
Yes, you will certainly get more water out of larger holes. If the water is
flowing out of a tank where the air above the water is at atmospheric
pressure as is the air outside the hole, and if the area of the surface of
the water is much larger than the area of the hole, then, by Bernoulli's
Law, the water will leave through the hole at a speed v = sqrt (2gh), where
h is the height of the surface of the water above the hole. Notice that
this is the speed an object would get by falling through a height h. So the
volume of water leaving per second will be
Q = A*v.
Incidentally, the water accelerates as it leaves the hole since the air
pressure outside the hole is less than the water pressure inside the hole,
causing the diameter of the stream of water to be less than the diameter of
the hole . So Q is actually less (by a factor of roughly 2/3) than the
value given above. It could be an interesting experiment to measure this
effect by measuring Q as a function of h. As h is increased, the speed of
the stream increases, its diameter shrinks and Q decreases by an increasing
amount from the simple calculation above.
To be sure this calculation is clear to you, I will calculate it for h = 1m
(just over 3 feet) and a hole 1 cm in diameter (A = pi*R^2 = pi/4 cm^2 =
7.8E-5 m^2 = 0.000078 m^2.) g = 9.8 m/s^2 so v = sqrt (2*9.8 m/s^2*1m) =
4.43 m/s. Finally Q = Av = 7.8E-5 m^2*4.43 m/s = 3.45E-4 m^3/s = 345
Water will continue to flow out of the hole as its size is decreased until
the hole diameter becomes comparable to the diameter of a water molecule.
That's a few angstrom units, which is a few times 1E-10 m. Any hole you
could possibly make is much larger than that. This argument ignores surface
tension, which is important as the hole gets smaller. A circular drop of
water at the outside of the hole generates a pressure holding the water
inside the tank. The pressure is given by p = s/r, where s is the surface
tension of water and r is the radius of the drop. s = 0.073 N/m. A drop
with radius 7.3E-7m (about 0.03 thousandths of an inch)would generate a
pressure of 1 atmosphere = 1E5 N./m^2 and could support the water in a tank
32 feet high!
Best, Dick Plano...
The above web site has a animation which illustrates
Bernoulli principle, which is
"Bernoulli's Principle states that as the speed of a moving fluid increases,
the pressure within the fluid decreases."
So, decreasing the size of the hole, lets less water
through the hole, thus building up the pressure, with increased pressure the
water flows faster.
If the hole decreases indefinitely, one would think that
there is an ultimate limit where the molecules, or atoms
in the water cannot get through the hole, but that case is a theoretical
Venturi followed Bernoulli and developed equations to establish flows rates
this site illustrates the equations, and you can answer
the problem about doubling the size.
From what little I know about lasers, they do not work
as one expects from Bernoulli's principle. In other words,
here the discussion deals with water, but light does not work
by the same logic, meaning a laser is not just concentrated
light rays. I think the laser works by exciting atoms.
Glad you got to refine that question.
I presume the hole is in the side of a container of water.
The water pressure behind the hole stays the same, is independent of the
size of the hole.
The flow in response to that pressure is what changes.
Flow usually varies as some power of the hole size.
The power might be 1 (proportional), 2 (square), or even 3 (cube).
The flow in a long, thin tube tends to go as the diameter cubed.
The power-law in effect for a hole might be less because of the inertia or
viscosity of the water in the immediate vicinity of the hole,
most holes in thin walls have flow effects about like a short tube whose
length is something like 1/2 the diameter.
So then the effective tube-length increases when the diameter increases,
so perhaps the flow will go as the square of the diameter.
For this square-law, doubling the diameter would give 4 times the flow.
But the power-law-number in effect will depend on major changes in scale.
If the hole is very small or the liquid is very viscous, the power-law
might be different than if the hole is large and the liquid is watery.
Flow through small holes at low pressures can completely stop,
if there is air rather than water on the outside of the hole,
and the wall-substance is water-repellant.
A tiny half water-drop forms, bulging out through the hole, and the
inside that water drop can be a larger pressure that the water-pressure
inside the container.
So that surface-tension can hold back the flow.
There are a bunch of things you can do to break this surface-tension and
allow the water to flow again,
at the rate normally predicted by the power-law for that size of hole.
They include: throwing soapy water over the hole,
continuously pouring water down the outside of the container,
hanging wet paper across the hole on the outside,
or sitting your container in a bucket of water filled to just about the
level of the hole, or higher.
If higher, then the pressure is the difference between the water level
inside and out,
not between the water level inside and the height of the hole.
I think I recommend hanging wet paper or cotton string.
Gluing a vertical toothpick just beside the hole might work too.
I think you need to measure and discover that power-law for yourself.
Your set-up might give a different power-law than I would predict.
The flow of a fluid can be very complicated, especially if the flow
becomes turbulent -- for example, when you pull the plug in a basin or
sink and the water forms a "whirl pool". But let us make some simplifying
arrangements, as follows: A large cylindrical tank with a small horizontal
exit pipe of radius R and length L near the bottom of the large tank. In
this configuration the level of the water does not change very much during
the time period in which the flow from the exit pipe is being measured.
Then the volume rate (volume / time) V = (pi / 8) * P * R^4 / n * L
where: pi = 3.14, P is the pressure difference between the ends of the small
horizontal pipe, L is the length of the small horizontal pipe, n is the
viscosity of the fluid, and R is the radius of the small horizontal pipe. So
you can see that the flow rate is VERY sensitive to the size of the exit
pipe -- the radius of the hole in the large tank. Putting in some typical
values, say L = 1 meter, R = 2 mm (pretty small), the depth of the water in
the tank = 0.5 meters, n (the viscosity of water) = 1x10^-3 kg/meter*sec.,
then the numbers fall out to be about 3x10^-5 (cubic meters / sec). Since
the flow varies as R^4 doubling the radius of the small pipe will increase
the flow rate by 2^4 = 16 times, so the flow rate is very sensitive to the
size of the hole. Note: the long, narrow exit pipe is to ensure that the
fluid flow is smooth. There is a minimum size below which the fluid (in this
case water) will no longer flow freely. This is related to the surface
tension of the water, but then once more the equations get complicated
For fluid flow, there is a relationship among flow rate, flow velocity,
and flow area, expressed by the following formula:
Q = A x V (Q = A times V)
where Q = flow rate or discharge
A = cross-sectional flow area
V = velocity of flow
In American units, V is usually expressed in terms of ft/s (feet per
second), therefore A should be in ft2 (feet squared), and the units for Q
would be ft3/s (feet cubed per second). It is important to use the
appropriate units in the above equation so that the results are
dimensionally correct. I hope that this helps.
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Update: June 2012