Quote from: fletcher on September 05, 2012, 06:50:08 PM
Hi Marcel .. what is stopping you from posting a few photo's with explanations & arrows showing things etc ?

A few posts back you made the comment that a single layer ZED showed unity [100% I think you said] i.e. work input equaled work output, & you expected that.

Then immediately you said the unit still had potential [energy] that could be used to pre-charge another unit so an estimated reduction in input cost of estimated 50% & back & forth & so on.

This unity comment would now seem a contradiction - if the unit has additional potential after one complete cycle it can't be 100% because you haven't accounted for the raised potential in pressures or whatever.

Webby1 & Mondrasek are happy to post photo's with explanations of their finding as they progress forward - when you get over you euphoria perhaps you could find the time to post a pic or two ?

A complete logical cycle sequence would be good including pre-charge etc if there is any - thanks in advance.

I'm quite busy last days, I'll see what can be done later.

For the start please read my post #1860 with 2D model there are three steps.

Then calculate whats is potential energy of water after each step in risers. Initial, precharge and stroke.
This is your input. Hope you will see work done equals energy needed to do precharge and lift step.
But still after work is done water has still potential to do some work, risers are full!

Marcel

Do not forget that not only are the risers still full, but the total head pressure is the SUM of the number of risers, giving more potential energy than you might think.

Hi,
I showed this to my colleagues.

Marcel

@MT. OK, let us be clear what your diagram is saying.


In Zed one on the left , water is pumped in under pressure . Energy consumed is A


Zed One rises, lifting a load. output energy is B


A=B, so Zed, in this ideal case is 100% efficient.At the top of the stroke there is still potential energy left in the system, which is exported to Zed Two.Call this D . This partially charges Zed two.


To complete the input of Zed Two, we need more energy,C. This energy is part of the output of Zed one, which is fed back.


So the input to a Zed is in two parts, the exhaust from its partner [D] plus extra energy from the output [lift] of its partner.


Is that a reasonable summary of what the diagram shows?

Quote from: neptune on September 06, 2012, 08:30:52 AM
So the input to a Zed is in two parts, the exhaust from its partner [D] plus extra energy from the output [lift] of its partner.
Is that a reasonable summary of what the diagram shows?

Exactly. And C < A, always. Usable energy remainder is in fact whats left of B after feeding C. Maybe somebody have a better diagram, this is how I see it.
thank you,
Marcel