Tommey Reed's Implosion motor experiments

This is a simulation on output due to centrifugal for using real mathematics.

https://www.youtube.com/watch?v=pYL3yVd6-2s

Tom

This first experiment was 4ft diameter, but needed too much hp to test run.

https://www.youtube.com/watch?v=4Q7vfucSq_8

I have a interesting experiment pump design

At 2000 rpm's, this is producing 1,116 lbf form centrifugal force.

Calculating mass of water in rotating tube.

((pi x r^2) x 24") =((169.6460033cu/in /1728)*8.33)= 0.8177958376/2.2

m = 0.3717253807 kg

1/3.28 radius of rotating mass

r=0.3048780488 meters

Centrifugal force

[((2000/60)*(2pi))^2] =

43820.4444 x (r*m)=

4966.210791/4.45 = 1116.002425 lbf

I would like some input on this design.

What is this device supposed to do? What is its purpose?

Hello SolarEagle,

That is a great question.

If centrifugal force is able to create large amounts of pressure force due to rotation, then could it not also raise water to a higher altitude with a simple discharge at the top.

Most pumps have a enclosed housing where the impeller is submerged in fluids, this becomes fluid drag.

First lets look at head pressure using say 500 psi. If centrifugal force is able to create with very little amount of power like spinning a flywheel.

Then would it not also raise water 500*.433=1154.73ft ?

The interesting part about the simple build is that the discharge is 360deg at the top.

The fluids never collides with the rotating mass.

Tom

In a centrifugal pump the water flows inside the impeller....

Without the force of the water being slung to the exterior raceway, no force can be attained... also I don't understand your equation....500psi times .433 what's this?

Hello SolarEagle,

If you want to raise water to a height say 10ft, you would need 4.33 psi.

Each foot of height creates .433 psi, or 2.31ft=1 psi

So if you can produce 500 psi, then you would be able to raise water at a height of 500psi /.433=1154.734 ft.

Tom

The psi needed would depend on the quantity of water, which is determined by the inner diameter of the pipe....The bigger the pipe the more water volume and weight, so the more force(psi) is needed....

100 gallons of water stored at a height of 10 ft will only produce 4.33 psi on the bottom. Even if the mass of water is 833lb, the pressure is still 4.33 psi.

If you are using the weight of water, then that's a totally different load like on a water wheel.

Tom

100 gallons release in 1 minute=

(100*833)*10ft=8330lb of mechanical work

8330/33000=0.25hp

Tom

So wouldn't that be the static pressure that had to be overcome to pump water in?

...and wouldn't that increase with height, so at 100 ft it would be 43.3 psi static pressure...and 500 times .433 = 216.5

You are correct, but I'm taking about 500 psi, not 500ft.

500ft x 0.433=216.5 psi.

216.5 x 2 = 433 psi

67 psi / 0.433= 154.73 ft

500ft +500ft+ 155.73ft=1154.73ft

My point is, if centrifugal force can be harness by creating fluid pressure, could it be used in a efficient means to move fluid upward as a pump station.

I would say that the load would be equal to a flywheel of the same mass if it was a cylinder shape.

I don't really know without testing my implosion cylinder without the pump.

Tom

So you're talking about a hydraulic pump? rather than an electric one....?...but you still need a separate pump to supply the hydraulic fluid...

That becomes a pump when rotating it with a motor, as the fluid is pulled upward due to centrifugal force pushing fluid outward at the top.

Maybe you don't understand that the unit is attached to a motor that runs it.

So this is a different type of pump? It's run by electricity and works like a centrifugal pump....so what is different?

That's something that needs testing, before I can make any claims.

OK I'm guessing you are trying to improve the efficiency...and I'm guessing it's by impeller design not motor efficiency....

https://www.sciencedirect.com/science/article/pii/S0029549318305491

Yeah, experiments always override theories.

Even those with <...real mathematics...>?

"... The psi needed would depend on the quantity of water, which is determined by the inner diameter of the pipe...The bigger the pipe the more water volume and weight, so the more force(psi) is needed.... ...."

.

This absolutely wrong. The pressure at the base of a stand pipe has nothing to do with the diameter. What do you imagine might happen with two stand pipes, one of half the diameter of the other that had a connection at the bottom....would the level in the smaller diameter pipe br four times as high at the bigger pipe because the bigger pipe would have four times the water per length? No, of course not, the level would be the same aside from capillary action.

Also, "(psi)" is not a unit of "force". Pressure and Force as distinct, related by the equation P = F/A ...with A being Area.

OK I have a 6" diameter pipe 50' high and 2" diameter pipe also 50' high, both filled with water, what is the psi gauge reading at the bottom of the pipes? Now start the pumps and what does the psi peak to? Is it the same on both pipes?

Unanswerable without knowing the pump curves. However, if the flow happens to be the same, the discharge pressure of the 2" pump will be greater because of greater flow resistance of the smaller pipe.

So if the pipe resistance is the same, which pump draws the most amperage? Wouldn't you agree that to move a larger quantity of water requires more power? and that the impeller creates pressure that exerts force to move the water...?

Hp = (Q x H) ÷ (3,960 gallons per minute per foot x eff)

I agree!

No. My post was premised on equal flow, not moving a larger quantity of water.

The power needed by a pump solely for the purpose of moving a liquid is the product of the rise in pressure across it and the flowrate through it. Real world pumps consume a bit more than that due to fluid friction within it, fluid slippage, shaft bearing, drive and motor heating losses.

Pipe resistance is a function of fluid viscosity and density, and pipe characteristics such as diameter, length and smoothness.

The two situations need to be looked at using the characteristics of the pump as well as the characteristics of the pipes in order to determine the answer, because the two pumps will be operating at slightly different points on their pump performance curves and one does not immediately know the gradient between those two points on the power consumption curves without seeing them.

100ft/2.31 =43.29 psi.

A++

Same

6" 50' 21.65 psi

2" 43.29 psi

Not looking to go back to school

This depend on flow of the pump to calculate output pressure.

<...Not looking to go back to school...>

It would be worthwhile, though. This thread is GCSE-level physics.

Centrifugal pumps do not create pressure, they create flow. Pressure is the resistance to flow.

You need to convert the driver energy to kinetic energy by accelerating the fluid.

To convert some of that kinetic energy into pressure energy you need to slow down the fluid with a change of section, as is done in the volute or diffuser of a centrifugal pump.

Another way to do this in your (I think) plane spinning tube is to introduce a Pitot tube into the flow as below. Without this, the fluid will just go round and round.

Picture courtesy of Weir Pumps

I agree about pressure, but maybe I can clarify this pump design.

https://www.youtube.com/watch?v=4Q7vfucSq_8

In this video I did some testing rotating this 4ft diameter drum with (16) 2.25 dia/id tubes around 400 rpm's.

This was not a good design due to loss in mass at the tip of the discharge, but it did have back pressure because of the 0.25 jets.

Because is hydraulics and the pump was moving fluid upward into the 3" rotary union to the rotating mass.

Any pressure would create mechanical load as the calculation hp=gpm*psi/1714 and could not get above 500 rpm due to 2hp motor.

This is why I did some changes on the design, cutting off the tubes allow more fluid to rotate to get the centrifugal forces needed to get greater flow that the pump can deliver.

The very fact of moving more fluid outward then the pump can deliver is the main purpose of this experiment.

If centrifugal force can cancel out the pump with the exception of not losing prime and create cavitation. I believe the centrifugal force would by theory cause the pump to accelerate due to the increase of gpm.

Anyhow this is my theory and that's why I build this to test in the first place.

"Experiments always has more merit then a idea or theory!"

Tom.

Good on you for experimenting, but for the life of me I cannot understand where in your design (as I understand it) you change kinetic energy to pressure energy and thus flow. Have you tried to construct a PV diagram for your system similar to the one below for an impeller. All I can make out is a liquid filled flywheel.

Good point, let me explain.

When we talk about psi it is related to amount of force, in other words 100 psi is 100lb/per inch, 12 inches 100 ft/lb was created.

When centrifugal/centripetal force is created due to rotation, on the circumference of the inner walls of a spinning drum.

A great example is a cd in a cd player. What happens when the cd has a limit of amount of rpm, it explodes into pieces.

Take titanium disk spinning at a maximum speed where the centrifugal force is so great that it also explodes with very little energy.

Unlike spinning disks, I'm working with forcing fluids outward through ports where these forces are created due to rotation.

Maybe this simple drawing will help.

In this picture you have 3 hydraulic cylinder that is attached to each other, when rotated to a maximum rpm, the outer weight will produce greater force to lift the 1000lb mass.

I do understand that the expanding upper cylinder will change in load, that was not my point.

It is a fact, by rotating mass you can create work.

Instead of a fixed mass, I'm mover fluid outward in a 360deg rotation that will create a fixed vacuum to drive a hydraulic vane motor at any load.

Instead of head pressure from a pump, centrifugal force is creating just by rotating the drum.

If any hydraulic pump is moving fluid forward for output, then then the input is pulling fluid in.

This is what I'm trying to harness and calling it a implosion motor created from centrifugal force.

In the <...simple drawing...>, it is atmospheric pressure that would be raising the <...1000lb mass...>. The force is the difference between the outer and inner pressures multiplied by the area of the piston.

Once the device is spinning the lift action will stop when the reducing pressure in the liquid on the upper side of the lift piston reaches the boiling pressure of the liquid above it at whatever the temperature is. The cause is the pistons in the upper cylinders moving outwards in what Newtonian Mechanics is called centrifugal force.

Once boiling starts then the two side plungers will travel outwards unrestrained. Once they are ejected from the cylinders in which they travel there will be a loud bang from the atmosphere rushing in, which is rightly labelled an <...implosion...>, at which point the central lift piston will fall unrestrained until either:

• something mechanical stops the fall, or
• the lift piston leaves the lower tube still attached to the load, at which point the fluid will drain from the central cylinder

whichever is sooner.

All that has been made by this device is some combination of mechanical damage and a fluidic mess, the reason for which remains abstruse.

What is shown in the <...simple drawing...> is actually a vacuum hoist, not a pump. The same lift action on the <...1000lb mass...> could be caused in principle by discarding the motor and attaching one of these to the top of the central cylinder:

In order to lift the <...1000lb mass...> at sea level on this planet and 0degC the lift piston needs to be something over 19in in diameter, rising in diameter with both altitude above sea level, which affects atmospheric pressure, and temperature, which affects the boiling pressure of the fluid.

If it doesn't work, that might be why.

You still can't grasp what I'm working on, maybe this is beyond your line of thinking.

Please explain what I'm working on, I'm not here to attack you.

Read up on my post, so we can have a logical discussion. I have a open mind...

Moving fluids around with equipment is something that has been practised over many centuries up until the current day, with many successful solutions being applied. The physics behind it is well-established and adopted in those solutions.

<...can't grasp...> is not likely to be a solo shortcoming. In such circumstances it behoves the originator of the concept to explain it in ways that can be understood by a varied audience.

Current thinking is based upon trying to avoid an over-unity device being designed and built, and to save the heartache of repeated failure. A computer-generated simulation video like the one in the links is not the same as something that has been built, tried, tested and works. That it works would validate the design activity. The claim of <...-100psi...> raises doubts on the design process as it falls outside current scientific principles based on practical and repeatable measurement.

Of course, if this thread is in pursuit of either a sales process or a patent application then unsubscription from it will follow immediately, naturally.

<...can't grasp...>

Peer review is an important principle in Science and Engineering circles, so that others may repeat experiments and verify the claims. Any invitation to do that needs to be thrown to a wider audience, which probably justifies posting on CR4, and certainly justifies subscription to a thread as a learning opportunity.

Why is this called an "implosion" motor; where is any implosion?

Maybe I can explain it in layman's term.

Hydraulics have two types of pressure from output flow and intake flow.

Both are equal.

If you have a 100psi output pressure then won't you also have a -100psi (vacuum)?

I call vacuum a "implosion effect", due to the fluid static flow.

<......a -100psi (vacuum)?...>

That's nonsense.

Atmospheric pressure is only around 14.7psig.

...at sea level.

PWslack

Listen I give you some slack, you clearly don't know what you're talking about. Your mind is set in water pumps.

Hydraulics is totally different when pulling fluid and pushing fluids.

It show you are over reacting with your emotions and not realizing air pressure/vacuum is another subject that is not related to static pressure.

Fluids can't be compressed, what makes you think they can create a vacuum to boil?

A vacuum will boil at 30"mercury if it has a pocked of air above the fluid, but not if its in a solid compartment like a hydraulic cylinder.

Clearly this is beyond your understanding, study it more before looking foolish like the last few posting.

Studies completed. Due diligence applied. No change to the claim. Emotions discarded.

In this new illustration of a piston in a closed cylinder, the force on the down-rod needed to move the piston will increase until the material above it is the vapour phase. In no case will it exceed atmospheric pressure multiplied by the area of the piston. It will drop until the difference in pressure both sides of the piston multiplied by the area of the piston equals the force applied by the weight of the piston and the down rod, at which point things will stop.

Should the piston move unconstrained out of the cylinder, owing to the force on the down rod exceeding the difference in pressure across the piston multiplied by its area, then a bang will indicate that an <...implosion...> has occurred and the piston and its down-rod will then fall unconstrained. The result is the same as in the earlier illustration: some combination of mechanical damage and spilled fluid may occur.

Please post a link to a video of a working prototype?

<passes the ball>

The claim <...air pressure/vacuum is another subject that is not related to static pressure....> is at variance with the earliest experiments to measure barometric pressure. in these, the variation of height of mercury in a closed-end glass column was used to determine it. Thenceforth atmospheric pressure has been measured in terms of the height of a mercury column. Sea level pressure is usually around 760mm of mercury, though this varies with weather conditions. The measuring principle is used in barometers, which have been used traditionally as weather predictors; generally a low barometric pressure is associated with stormy conditions and high as fine as there is a good correlation between them.

The mercury was in a seal tube with trap air was it not?

For mercury to rise it is stored at the bottom in a environment of trap air/gas. That's the only reason it's able to move upward.

Altitude gauges works just the same, where trap air is reacting to atmosphere pressure in a bladder.

Just like a helium balloon expands as it increase in altitude because the outside pressure is less.

Just to clarify your answer.

<...The mercury was in a seal tube with trap air was it not?...>

There was no air in the first working mercury barometer.

<...For mercury to rise it is stored at the bottom in a environment of trap air/gas. That's the only reason it's able to move upward...>

In the first barometer, the mercury became stationary as the atmospheric pressure outside the apparatus needed to support the column balanced the height of the mercury in the column.

The air pressure is the product of the acceleration due to gravity, the density of the mercury and the height of the column represented by the distance A to B above.

100 psi vacuum - that would be a first!

Not in a computer simulation with <...real mathematics...>, surely?

The 1" output ports are pressure check valve to avoid air to enter the rotating drum. This allow static flow and prevent cavitation to occur.

<...static flow...> is a new concept here. How does this work, please?

How does it <...prevent cavitation...>?

<passes the ball>

I'll wait for your insight of what I'm working on, it's best you understand my design before questions of static flow.

Please explain what I'm doing.

It seem you are missing information and going off topics.

I've been following along the best I can. I admire your curiosity and industry but you haven't really stated yet what you're trying to achieve with your device. At first you describe a water lifting (pump) function and later seem to imply a power generating function. I'll limit my comments to the pump funtion.

1. You cannot lift (vacuum) water or any other liquid anywhere close to 1000 ft (the actual limit is about 33 ft for water). Even if your rotor develops a perfect vacuum, this limit can't be exceeded.

2. Your device might act as a pump at lower lift, but I doubt it would be very efficient. One area of concern is the discharge from the arms. The fluid leaving the arms would have a tangential component of velocity as well as a radial (outward) component. Your device doesn't attempt to benefit from this work that the motor has expended to accelerate the fluid to this velocity, so a great deal of work is lost. This is a definition of inefficiency. In commercial pumps a diffuser of some type is used to recover as much of this energy as practical.

Incidentally, I watched your water turbine video. It reminded me of the steam turbine (aeolipile) described by Hero in Roman times. Except I assume yours might have some impulse component because of the torque converter vanes as well as the reaction component from the nozzles.

Hello Bigg,

The water wheel experiment was a interesting build, when head pressure was constant from the rim jets load was measured. As the torque converter rotated the pump decreasing head pressure from the main pump, in other words the rotating converter was changing the motor load with centrifugal force.

Because it was a water pump, I decide to try a fixed hydraulic gear pump at 200 psi 2.5gpm using a simple rotating devise like on the drawings. power by a electric motor to read watts.

As the rim jet thrust rotated the unit, head and watts started to decrease.

I know you all might think this is normal like a hydraulic motor running without a load, but this was different due to the fact I never change the static load, only when the rotation increase, did things change.

I was wondering if this could be a new type of motor that could use less energy to do work.

And that's what I've been working on for a few years now.

https://www.youtube.com/watch?v=yaMyKhY6kZw&t=61s

This was one of those test.

<...Please explain....>

It would seem all attempts so far have failed to do that.

This is a video I did a few years ago showing centrifugal force decreasing head pressure.

The rotation is created by water thrust.

https://www.youtube.com/watch?v=D2L_PBGgHZw

...and?

Hello all,

I have been busy doing testing on this motor, starting to get some good data.

This is a series of the build design.

https://www.youtube.com/watch?v=Xgxk71i9oSg&t=15s