I have a problem and it breaks down like this: I have a 15 ft high
tower/cylinder of water with a 10 ft. radius. The tank is sealed so
that no water will blow by when pressurized. On top of the tank I have
100 ton weight. The 100 ton weight presses down on the water in the
tower/cylinder forcing a plunger to push the water out the hole at the
bottom of the tank/cylinder. The egress pipe is 8 inches in diameter
with a nozzle the drops to  4  inches in diameter.  Question 1. What
is the exit velocity when traveling though a nozzle measured in ft
pounds.[Energy] Question 2. How quickly does the tower/cylinder drain?

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Clarification of Question by hollisterinc-ga on 27 Jul 2004 11:48 PDT

Question: A large syringe is filled with and incompressible fluid
[water]. In the first model, the water is allowed to drain from the
syringe freely with no force applied. The water drips slowly from the
end of the syringe. In the second model, a force of 10 pounds is
applied to the end of the syringe forcing the water to jet out over 12
ft. If the force applied was doubled to 20 pounds, would that double
the distance of the water jet? What is the equation that would
demonstrate the relation of force to work=energy? The volume of water
remains the same, the pressure simply increases thus increasing the
energy output.   If I had a 100 ton weight, I should be able to shoot
a jet of water 1200 hundred feet? If the syringe was 15 ft tall with a
10 ft radius and the egress pipe was 12 inches, what would be the
pressure  energy  of the water exiting the pipe? If the hole was the
size of a pin then I could stop it with my finger[Brian] ,but the size
of the egress hole is 12 inches, so there should be enough pressure to
turn a water turbine? That is 650 pounds per square foot. Should that
be divided by 144. I think that the nozzel changes the dynamics of the
equation?

 1204.8790000000001  kg/s

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Clarification of Question by hollisterinc-ga on 09 Aug 2004 11:29 PDT

Our goal is to validate the output power is stong enough to push a
water turbine. We know that there is a relationship between the weight
of the bouyant device and the force of the water striking the water
turbine. So, if one were to take a water hose and allow for the water
to "free flow," the water would only shoot out approximatly 12-16". If
one places a nozzel at the end of the hose the water shoots out 12-15
feet. The water volume remains the same. What we are trying to solve
for is the effect that weight (several tons)has on the output. There
is a direct correlation. The question is what fluid formula is used to
solve for the relationship of weight to pressure. In this example, we
have [to keep it simple] 50 tons of steel sitting on a sealed cylinder
of water. The full force [weight] of the steel pressurizes the water.
[Water is a incompressible fluid] Thus, the water is now pressurized.
The main question is the amount of force coming out of the nozzel? Am
I closer? I hope that this helps. Let me know if there is any other
information that you would like?

The surface area is 3.1416 times 10 squared = 314.16 square feet
divided into 200,000 pounds = about 650 pounds per square foot = 4.2
psi (pounds per square inch) at the the face of the 100 ton weight.
There will be additional pressure due to the column of water in the
tank up to 15 feet and up to about 7 PSI. This is best caculated by
calculus, but will average about 3 psi, making a total of 7.2 psi
pushing the water through the 4 inch nozzel. Someone else will have to
tell you how to calulate the cubic feet per minute which will decrease
some as the pressure falls; again best done by calculus or a graph or
table. The volumn of water to pass though the nozzel is the area times
the heigth = about 4700 cubic feet.   Neil

I'm quite sure foot-pounds is not the same as pounds per square feet,
and I don't see how "velocity" can be expressed in foot pounds, so
that is another calculation for which I have no solution. 550 foot
pounds per second = one housepower = 746 watts which are units of
power.
The weight of the water is 4700 times 62.5 = about 300,000 pounds and
it decends 7.5 feet on the average = 2,250,000 foot pounds plus the
weight decends 15 feet times 200,000 = 3,000,000 foot pounds.Adding
gives 5,250,000 foot pounds total energy. If you divide by the total
seconds, you will get the average power of the stream of water coming
from the nozzel, including turbulance and other friction losses which
requires a table, graph or more calculations.  Neil

ft-pounds is not a measure of velocity.  I think you must want... ft/second.

Using the ideal energy equation and the information that you have
provided I have found with google calculator

sqrt((2 * ((999 * 9.81 * 4.57200) + (88 644 / 29.18))) / 999) = 9.78694999 m/s
of course this does not include any losses, which are going to be
present as pointed out by neilzero.

which is 32.109416 ft / s (speed of free stream of water from nozzle)
note that the speed of sound in water is 1440 m/s SO we don't have to
worry about a 'choked' nozzle.  YAY less calculations. :)

From here we can find the mass flow rate as (density * area* velocity)
which comes to 122863.454 kg/s (THAT IS SO MUCH!)  But hey, we have a
100 ton weight =)

The total mass of water in the tank is about 133277.549 kg... so as
you can see the water would be flushed out by this massive weight in
just over a second.

Of course this is a rough calculation without losses, and your problem
- to be solved - required many assumptions.

OOPS I calculated the area wrong.  You can redo my calc by replacing
29.18 with 7.25833567 m^2, then recalc the mass flow rate.  then you
can have your answers again.

Also note that my answers are in SI.

Best,
SAE Miller.

hollisterinc - I'm not sure exactly what you are trying to say in your
first clarification.  Perhaps you can restate your question in detail
with clear variables.

The equation I believe your looking for is:

((1/2)*density*velocity^2+density*gravity*height+Pressure) at the
surface of the heavy steel weight =
((1/2)*density*velocity^2+density*gravity*height+Pressure) at the free
location of the nozzle.

If i reduce this for you, you get.
(Pressure) at the surface of the heavy steel weight =
((1/2)*density*velocity^2+density*gravity*height+Pressure) at the free
location of the nozzle.

This assumes no losses, if you want to account for head loss then just
divide the whole thing by (density*gravity) then you can account for
head loss.

So how do you find the pressure at the surface of the heavy steel
weight?  You take the pressure you gave me easier and then add it to
(weight*area steel weight acts on water).