@TK. Many thanks for that, I had forgotten that lowering the auxiliary tank would not sink the pod.


Incidentally, it is not quite true that "a pint`s a pound the world around"
Here in the UK we use the imperial gallon[4.546 L] as opposed to the USA Gallon [3.785L]
We still have 8 pints of water to the gallon , but we say "A pint of water weighs a pound and a quarter". So our gallon weighs 10 pounds. We have 20 fluid ounces to the pint.


This is a common cause of confusion when comparing the MPG figures of cars in UK and USA.

@TK, so I think we understand where we got off track understanding each other as far as the data set I took before leaving for vacation and that it should be valid for a calculation of input energy vs. output energy.  This is with the understanding that the losses in this test are only from the ZED test setup build itself (rubbing), my measurement errors and accuracy issues (abundant), and the accuracy of the measurement instruments themselves.  It does *not* account for the losses that would actually be incurred by a true real world full cycle water recirculation system as this is being also considered as "ideal" in this case (per our previous debate and clarifications).

The data that was previously supplied and relevant reference posts follow.  Please let me know if you can run an input vs. output energy check with this data or if something else is missing.

From post #1968:
Mass lifted:  ~2.5 lbs
ZED stoke: ~ 11.5 mm.
Water removed and introduced again to obtain repeatable(ish) stroke as measured by a "rain gage" from my yard: ~87.5mm.
Overall change in the head in the elevated fill tube during the cycle is ~ 190 mm.

Post #1981:  Painful detail of the trial runs from which this data set is gathered.

From post #1988:
So the CVS 5ml syringe rings out at about 50ml (yes, ten fills) to about 37 mm in the raingage (conservatively low reading).

Thanks,
M.

Hi,
I was owing an update. I have redone work calculation for 10 iterations. New variable is added - pipe diameter. I think now calculation represents reality much better.


Still OU. For the same cylinder diameter and height 1m and pod diameter 0.98 and height 0.9m  I'm getting COP 1.896
Pipe diameter 0.23m.


I'll post spreadsheet but first need to get some sleep over it and check it again in morning. I don't know how you but I usually I find more errors in morning.


Attached is picture visualizing how is work spent calculated now. New variable is diameter of an pipe opening between cylinder and pool beneath it.
First is target volume packetized. Number of packets = number of iterations. Since we know diameter of the tube we can calculate how high the fictive packet would be in pool. To insert such packet into the cylinder above we need to lift it over this height. The more wider pipe the less we need to lift. Work needs to be done of course also against the pressure of water column above the opening(which is increasing by each packet). As we pump the fictive packet into the cylinder water there is not raised by height of the fictive packet but only by pocket cylinder height see picture 3. Work spent against the water pressure equals work needed to lift weight of water column over packet cylinder height.


respect,
Marcel

Quote from: fletcher on September 11, 2012, 09:22:32 PM
1. at start the twin tank system has water PE of 70.5 J - 2. after lifting high enough to transfer all water the PE is 624.6 J - 3A. after transfer to around the POD [locked down] the system PE is 352.5 J - 3B. the last system on the right with purple water is the combined blue & green water volumes & masses after the POD is removed [to show something I going to discuss] - the PE is 168.6 J.

...

The upshot is that 50 kg rising 0.2 meters gives about 98.1 J - when you add that to the 168.6 J we get somewhere around 266.7 J, well short of the initial work input done.

Hi Fletcher,

Awesome work, and it all sounds good. However, I don't think this is a real situation for the setup involved.

Although the "lift-all-release-later" approach must be easier to calculate, I guess what would really happen is that as the water is lifted it would be free to flow out from the tank. This would mean less work done to raise the tank as it slowly gets lighter the higher it gets.

I don't really know how to calculate this, but I guess that with a quick enough flow, and/or a slow enough lift time, you could probably get a 1 to 1 relationship between height gained and weight lost.

Does this sound correct?

If so, can we then roughly say that the average weight lifted over the distance is about half the initial weight (as we're going from all to nothing), and so now the work done to lift the left tank of 50kg would be 306.5625 J. This seems much closer to the work output.

I've got some other ideas too, but for now I just wanted to check I was at least looking in the right direction.

Cheers
Amo

Edited to correct typos.

Quote from: fletcher on September 11, 2012, 09:22:32 PM
OK .. now we want to know what work the POD could do to add that energy back in ? - well, the work the buoyancy force can do when the POD is released is exactly the same as the f x d of 50 kg raising 0.2 meters IINM, except this is a constant force as opposed to a variable force so its easier to work out work done joules.

Also, a quick question on this.

I thought, thanks to wikipedia and other sites I've found to help me understand (a forum on physics is pretty good), that buoyancy was based on volume of the displaced water. I know we've emptied 50kg of water into the tank, but that's not what's displaced.... is it?

I thought that the initial, soon to be rapidly decreasing, buoyancy force would have been the water density * volume of the cylinder (as it's totally submerged at this point) * g would mean 1000 * .44426 * .44426 * .95 * 9.81 = 1839.36 N over 0.2m = 367.87 J. Even if we took an average between the start and stop point (0.2m), it would be: 329.15 J.

If we link in MT's idea of having a bit more water in the blue tank to start with, and as we release the pod lift the blue water tank a little further to empty the remaining contents and keep the keep the buoyancy constant, are we getting close to the OU tipping point?

I've been trying to do more drawings but it's frustratingly slow. They should be worth more than 1000 words for the effort it takes

Amo