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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 432092 times)

Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #645 on: March 06, 2014, 05:28:21 PM »
This suffers from the tyranny of the:  N*X/N2 = X/N problem.
The other problem that you have is that it requires head to support the payload weight.  You have to pick a payload that is less than the initial uplift force, because the uplift force declines steadily towards zero with lift.  If you pick a payload weight that is the same as the uplift force then the thing never moves.  If you pick one slightly less as you propose then it moves only slightly.

MarkE, the initial payload is slightly below the maximum buoyancy Force in State 2.  And the payload also declines linearly as the system lifts up.  The payload "slug" of water does not sit on the piston at the end of State 3.  It has been allowed to run off unimpeded across the top surface shown that includes the bore.  The correct Energy equation reduces to .5*Fmax*S.

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Re: Mathematical Analysis of an Ideal ZED
« Reply #645 on: March 06, 2014, 05:28:21 PM »

Offline MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #646 on: March 06, 2014, 05:36:05 PM »
MarkE, the initial payload is slightly below the maximum buoyancy Force in State 2.  And the payload also declines linearly as the system lifts up.  The payload "slug" of water does not sit on the piston at the end of State 3.  It has been allowed to run off unimpeded across the top surface shown that includes the bore.  The correct Energy equation reduces to .5*Fmax*S.
You are still subject to the N*(X/N)2 tyranny.  Let's say that you set-up your payload such that it runs off at the rate of 64.649N/m to match the rate at which the riser lift force runs down, then the  Wpayload = 0.30078N - 64.649N/m.  Now perform the integral math on the work you impart doing the lifting:

E = integral( F*ds )
F = 0.30078N - 64.649N/m
integral from 0 to 4.653mm is:
0.30078N * 0.004653m - 0.5 * 64.649N/m * (0.004653m)2 = 0.7mJ

The internal change in energy from State 2 to State 3 was 1.024mJ.  So you are still stuck with a 30% loss.


Offline TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #647 on: March 06, 2014, 05:37:36 PM »
TK, regarding your "peanut oil" comment earlier (sorry, but I'm too lazy to go back and find it).  I was wondering if you, or anyone else, had experience with Pentane?  I did a little bit of research on a low and high density fluid and had settled on Pentane and an Aqueous Sodium Polytungstate solution for candidates to use in a production level ZED system.  Only downside to Pentane though (that I can tell) is it is flammable.  Any thoughts?

http://en.wikipedia.org/wiki/Pentane

http://en.wikipedia.org/wiki/Sodium_metatungstate
Pentane is flammable, volatile, toxic and carcinogenic probably. Sodium polytungstate? Man, you are weird. How about using galinstan and kerosene?





Or just a big spring and a rock.

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Re: Mathematical Analysis of an Ideal ZED
« Reply #647 on: March 06, 2014, 05:37:36 PM »
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Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #648 on: March 06, 2014, 05:43:23 PM »
No, the payload is a fixed mass.  It does not decrease as it moves up.  A payload just less than the State 2 lift force barely moves.

Incorrect.  It is the case I explained and that you address below.

If you create some arrangement where you lift the payload slightly and then it falls off of a cliff, then you need to account for the loss in PE that occurs when you do that.  You cannot pretend that you lift what you drop, or that you lift what runs off.  No matter how you slice you have to perform the integral math.  And no matter how you slice it, when you convert potential energy stored in some device into potential stored in multiple devices, you get screwed by the N*X/N2 problem.

There is no need to be concerned with where the exiting Energy goes off to once it is properly calculated as leaving the system.  The water could "fall off a cliff" or sit on top of an infinitely large surface.  It does not matter.  The only thing that matters is that this method allows for a correct calculation of the Energy that leaves the system during the transition from State 2 to State 3.

Offline TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #649 on: March 06, 2014, 05:43:39 PM »

   Question,
                 how much does it cost to pump a little bit of "virtual water"?
                                                      John.

Thanks for reminding me. I wuz gonna post this last night but my uplink went down and I fell asleep. Here's a Little demonstration of "virtual water" that I made back in 2012. Note that there is _real data_ being displayed here.

The term "virtual water" has been used for a long time, but in a different context. It refers to the water involved in producing and processing and delivering food to the table. Like how much water do the cows drink, how much water did it take to grow the grain to feed the cows, etc , per hamburger patty served up to your kids. It is a concept very familiar to those who actually work for a living, growing our food.   ;)

However, the term "virtual water" seemed like a very obvious way to describe what happens with these nested buoyancy things. It's a result of displacement, it is buoyancy, it is gravity, it is looking at the inside of the pod chamber instead of the outside. It is the key to the "inverted Travis effect". It's still a Red Herring, though.

My virtual water demonstration. Take note of the scale readings at each step, and the water levels. Uploaded August 13, 2012.

http://www.youtube.com/watch?v=qFjqBaH_NWU

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #649 on: March 06, 2014, 05:43:39 PM »
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Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #650 on: March 06, 2014, 05:53:03 PM »
How about using galinstan and kerosene?

I assumed galinstan would be much more costly and probably have a high viscosity.  Plus it didn't come up in my searches.  They were not extensive.

Kerosene is interesting.  Is it safer than Pentane?  Any idea on the least dense liquid that is non-toxic, and non-flammable?

Offline TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #651 on: March 06, 2014, 06:07:50 PM »
I assumed galinstan would be much more costly and probably have a high viscosity.  Plus it didn't come up in my searches.  They were not extensive.

Kerosene is interesting.  Is it safer than Pentane?  Any idea on the least dense liquid that is non-toxic, and non-flammable?

You are making a _free energy_ machine. You could make it out of solid platinum and it would still pay off.

If only it worked, that is.

Kerosene is less volatile than pentane and more readily available. But see part one above: as long as it isn't made of unobtainium, the cost and availability of the materials for a _free energy machine_ may be neglected in the final accounting. You could use a light vegetable oil like soybean oil or canola oil.

Funny.... when I typed "galinstan viscosity" into the google search window I instantly found this:
0.0024 Pa-s (at 20 C)
Mineral oils are typically 0.02 to 0.05 Pa-s
Water is of course around 0.001 Pa-s

If I got the decimals in the right places, that is.

Galinstan's density is 6.44 g/cm3 (at 20 C).

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #651 on: March 06, 2014, 06:07:50 PM »
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Offline MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #652 on: March 06, 2014, 06:23:26 PM »
Incorrect.  It is the case I explained and that you address below.

There is no need to be concerned with where the exiting Energy goes off to once it is properly calculated as leaving the system.  The water could "fall off a cliff" or sit on top of an infinitely large surface.  It does not matter.  The only thing that matters is that this method allows for a correct calculation of the Energy that leaves the system during the transition from State 2 to State 3.
See the updated post above. You still lose more than 30%.

Offline minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #653 on: March 06, 2014, 06:25:07 PM »



   Hi,
       I tried with a bit of light oil over water and it's ok if you don't get any revs up.
   Once you try and increase the rate of doing work of course you just emulsion.
    When you put oil over water in your ZED you get a totally different beast than
    when it's got air over water.
                                        John.

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #653 on: March 06, 2014, 06:25:07 PM »
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Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #654 on: March 06, 2014, 06:31:17 PM »
See the updated post above. You still lose more than 30%.

Sorry MarkE, could you tell me which post has been updated?

Offline MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #655 on: March 06, 2014, 06:34:30 PM »
Sorry MarkE, could you tell me which post has been updated?
#667

Quote
Quote
Quote from: mondrasek on Today at 05:28:21 PM

    MarkE, the initial payload is slightly below the maximum buoyancy Force in State 2.  And the payload also declines linearly as the system lifts up.  The payload "slug" of water does not sit on the piston at the end of State 3.  It has been allowed to run off unimpeded across the top surface shown that includes the bore.  The correct Energy equation reduces to .5*Fmax*S.

You are still subject to the N*(X/N)2 tyranny.  Let's say that you set-up your payload such that it runs off at the rate of 64.649N/m to match the rate at which the riser lift force runs down, then the  Wpayload = 0.30078N - 64.649N/m.  Now perform the integral math on the work you impart doing the lifting:

E = integral( F*ds )
F = 0.30078N - 64.649N/m
integral from 0 to 4.653mm is:
0.30078N * 0.004653m - 0.5 * 64.649N/m * (0.004653m)2 = 0.7mJ

The internal change in energy from State 2 to State 3 was 1.024mJ.  So you are still stuck with a 30% loss.

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #655 on: March 06, 2014, 06:34:30 PM »
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Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #656 on: March 06, 2014, 06:34:51 PM »
   Hi,
       I tried with a bit of light oil over water and it's ok if you don't get any revs up.
   Once you try and increase the rate of doing work of course you just emulsion.
    When you put oil over water in your ZED you get a totally different beast than
    when it's got air over water.
                                        John.

Another reason to keep the difference in the Specific Gravities of the two working fluids as large as possible?  Or, is it a showstopper that means a gas, while lossy as far as Energy is concerned, has an advantage when Power is the ultimate goal?

Offline MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #657 on: March 06, 2014, 06:37:21 PM »
Another reason to keep the difference in the Specific Gravities of the two working fluids as large as possible?  Or, is it a showstopper that means a gas, while lossy as far as Energy is concerned, has an advantage when Power is the ultimate goal?
The irony is that the force multiplication goes up with the relative SGs, but the efficiency goes down.  The best ZED is not a ZED at all, but an ordinary hydraulic lift.

Offline mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #658 on: March 06, 2014, 06:37:48 PM »
#667

Thanks!

Now back to the 3-layer ZED.  Which one do you want to check?  We need to calculate the Energy that left that system due to the rise.  Then I think the Energy Balance Analysis is complete.

Offline MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #659 on: March 06, 2014, 06:40:43 PM »
Thanks!

Now back to the 3-layer ZED.  Which one do you want to check?  We need to calculate the Energy that left that system due to the rise.  Then I think the Energy Balance Analysis is complete.
Have you downloaded the spreadsheet?  Everything you want should be in it.  The ZED is a piece of useless junk.  I don't know where you obtained OU numbers, but it certainly is not reflected in the ZED.  The "ideal ZED" using incompressible fluids, once you've paid all the energy to prepare it and charge it up, just acts like a linear spring.  If you make one with  compressible "air" it acts like a variable rate spring with additional loss.

 

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