Cadman’s Hydrostatic Displacement Engine

[Cadman]

Hi Cadman,

Don't be discouraged by the naysayers.  If you have done the experiments and proven to yourself that the idea has merit that is really all that matters.  I am very involved in a couple of large projects right at the moment but intend to pursue your idea when I have time.  I am sure there are others that also believe you might have a workable idea.

Respectfully,
Carroll

Hi Carroll,

I think Floor is sincere. Not a naysayer, just healthy skepticism.
I repeated my experiment yesterday. It's confirmed.

Regards,
Cadman

[citfta]

Hi Carroll,

I think Floor is sincere. Not a naysayer, just healthy skepticism.
I repeated my experiment yesterday. It's confirmed.

Regards,
Cadman

I'm sorry.  My post was misunderstood.  It wasn't directed toward Floor.  I am aware of some others that have been very negative about your idea.  It was those I was referring to.  My apologies to Floor.  I also greatly respect his work and in fact am trying to replicate some of it for my own testing.

Glad to hear you have successfully verified your results.   

Sincerely,
Carroll

[Cadman]

Sorry, my mistake. I too respect his work, although I have trouble understanding it sometimes.

Well, the math behind my idea doesn't lie.

To anyone who is negative about my idea please post your rebuttal and supply the math to back it up. Prove your contention with the math.

There, the gauntlet is thrown.

Cadman

[Cadman]

Dear Floor,
This is for you. I unexpectedly find myself at a stop on my build so here is the analysis you asked for. I hope it brings understanding. If I have made any mistake in the math please point it out.

The water weighs ~8.345 lbs per gallon. 231 cubic inches per gallon.
The area of a circle is Pi times the radius squared.
The area of the 6" piston is 28.2743 sq. inches. Call it 28.27.
The area of the 3/4" diameter rod is 0.4417 sq. inches. Call it 0.45.
The area available is the piston area minus the rod area. 28.27−.45 = 27.82 sq. in. This area is for both pistons. Actually the area of the bottom piston increases back to 28.27 as soon as the rod loses contact with the valve, but I'm going to ignore that and use the lesser area.
The stroke is 6".
The operational sequence begins with the piston assembly at the bottom of the stroke, the feed line open and the bypass line closed.

The height of the water column in the feed pipe is 63" to the bottom of the 6" piston. That gives a working pressure of 2.2759 psi.
The force available from the piston is the area of the piston times the psi applied to it. 27.82 × 2.2759 = 63.315538.
Call it 63.3 lbs of force.

The weight of water in a vertical tube is the area of the tube diameter times the tube height, divided by 231, times the water weight per gallon.
The displacer piston is 40" tall. There is 1.5 in dia passage bored through the piston from top to bottom with a check valve on the top. The fluid in this passage weighs 2.5535 lbs.
Call it 2.6 lbs.

The displacer piston itself is of light weight semi-hollow construction and weight is 5 lbs.

There are 3 other water columns above the pressure piston, 6" 0.5" and 7". 13.5" total height. Their combined weight is 13.7892 lbs.
Call it 13.78 lbs.

The 0.75" chromed rod weighs 1.5 lbs/ft and is 72 " or 6 ft long. Weight is 9 lbs.

The pressure piston is 1" thick plastic and steel assembly. Piston weight is 2 lbs.
Lets toss in another 1 lb for seals, collars and misc hardware.
Weight is 3 lbs.

Total weight rod, pistons, hardware and water is  2.6 + 5 + 13.78 + 9 + 3 = 33.38 lbs.

At the bottom of the stroke we have 63.3 lbs of force lifting 33.38 lbs.
The surplus is 63.3 – 33.38 = 29.92 lbs. This is at the beginning of the up stroke.

As the piston rises through it's 6" stroke the height of the water column in the feed pipe relative to the pressure piston decreases from 63" to 57". That means there is a gradual decrease in the force applied to the pressure piston, it drops to 2.0591 psi. However the water volume above the displacer piston also drops to 0 as it spills into the tank. That lowers the combined water column height to 7.5". Total water weight decreases to 7.6606 lbs, a difference of 13.7892 – 7.6606 = 6.1286 lbs

So we now have 2.0591 x 27.82 = 57.2841 lbs of force.
A total lift weight of 33.38 – 6.1286 = 27.2514 lbs
Surplus is 57.2841 – 27.2514 = 30.0327 lbs.

So with this engine configuration there is an almost constant surplus force throughout the up stroke of ~30 lbs. This force shifts the valve which blocks the feed pipe and opens the pressure area at the bottom of the piston to the top of the piston which removes the pressure holding up the piston assembly, and the down stroke begins.

The down stroke action is almost a no-brainer. We have a piston assembly dead weight of 5 + 9 + 3 = 17 lbs. A constant force supplied by gravity. With the valve shifted the pressure on the top and bottom of the pressure piston equalizes and the water flows from below the piston to the top of the piston, it sinks. Gravity is pulling it down.
The water below the displacer can not reverse flow back into the lower section because of the check valve in the line. The weight of the piston assembly causes the displacer to sink in the upper cylinder section also. The fluid trapped below the displacer piston finds a path upward through the displacer and it's check valve, which is now free to open, and ends up residing above the displacer when the piston assembly has sunk to the bottom of the stroke. The water has been displaced to a higher elevation by gravity. The speed which this occurs can be slowed by flow restriction but the force of the piston assembly supplied by gravity is always available up to a maximum of 17 lbs.

The cylinder is now at the bottom of it's stroke and it's 17 lbs of weight will shift the valve and the whole process can repeat.

Respectfully,
Cadman

PS. To any naysayer. I have been designing and building industrial hydraulic machines for a living for over 30 years.
You have an idea, you work it out on the cad, study the motion, apply the forces, and if it looks good you build a prototype. The prototype is to test the machine and discover any unforeseen problems. If the problems can be corrected you continue testing and improving until it's ready for production.

This one looks good.

This is how it's done in the real world.

[tak22]

Nice. One could consider magnetic coupling to avoid the internal shaft gland seal between the pistons. Magnets embedded in the piston edges, magnets external to the shell, with guide rods and linear bearings. Probably many ways to achieve this.