Hydro Differential pressure exchange over unity system.

[TinselKoala]

Anyway, using the assumptions I made  in the 2-d model, I get 51.4 ounce-inches, or 3.21 pound-inches total work input. It turns out that the water in the lifted container has the same PE at the end as at the beginning so you are just lifting the container's weight here, and lifting the water in the other chamber. The water's PE in the other chamber is increased. So it might "look like" you are lifting the water that is still in the first container for free.

For the container you've lifted 8 ounces 2 inches, for 16 oz-in input work
For the water, the work is found by PE final - PE initial (neglecting the "g" term)
For the lifted chamber the PE  initial is 24 oz x 1.5 inch = 36 oz-inch and PE final is 13.3 oz x 2.7 inch = 36 oz inch, roughly, for a net change of zero
For the non-lifted chamber the PE initial is 48 oz x 1.5 inch = 72 oz-inch and PE final is 58.7 oz x 1.83 inch = 107.4 oz-inch for a net change of 107.4 - 72 = 35.4 oz-inch
and a grand total of 35.4 + 16 = 51.4 oz-inch of work....

A naive calculation might say, well the container is eight ounces, and the water in it is twentyfour ounces, so that's thirtytwo ounces, and you are raising it all 2 inches, so that's 64 ounce-inches of work.
But you aren't really raising all twenty four ounces a full two inches because of the increased surface area in the second container and the fact that the water levels will equalize.

OK, tear it up, where did I goof.

ETA: I drew a 2-d representation on graph paper and eyeballed the quantities from that, so don't expect great numerical accuracy. I might be a few percent off.

[TinselKoala]

So.... am I really the only one staying up late?

Anyway, the 35.4 value for the PE change in the second chamber probably should be 36 even, accounting for my eyeball errors. Then it's tempting to say, well, OK, counterbalance or eliminate the weight of the first container. Now the work input is zero, right? Since the PE of the water in this chamber before the lift, and after the lift, is the same, so no net change in PE so no work is done.

Yet the "output".... the water rise in the non-lifted chamber..... is that same 36 ounce-inches... which we got for free, since we didn't do any work lifting the first chamber and the water in it.

RIGHT?


Where did the missing dollar go, in the Whirling Dervish Motel?

[wildew]

@TK - yes, I think, correct so far.
@Webby1 - No layers here, just two very simple containers of water that are different diameters.

It's easy to say lifting .5 lb 2 inches is the same as 1 lb 1 inch ( I didn't include the time factor, so add in 1 second ) takes x joules.
Where it gets sticky is that the volume of water being lifted is changing during the lift because it's flowing through the tube - equalizing.
Not so much change in 1 second so in that case, most of the 8 fl oz would need to be factored into the lift.

I think I came up with an acceptable solution for now but am noting that this could end up being a bit of a deviation error. I started thinking I would measure the distance the cylinder moves up but that is a worthless number. I'm going to add a scale to the side of the cylinder and only measure the height change of the water in it for now.  A larger cylinder would help some too. Real sizes are 2.5" ID cylinder for input and a 7.75" ID "tank".

Here's a quick pic of where I'm at
Dale

[TinselKoala]

Time doesn't enter into the calculation of the work in Joules. The rate at which you do work or "expend" Joules is Power, not energy. So if you do a certain amount of work in one second, that's x Joules per second, or x Watts of power average during that second. If you do that same work in ten seconds, you are still doing the same number of Joules of work but your average power is now 0.1x Watts. Power isn't conserved, energy is. The same amount of energy or work in Joules could be performed in a second or in an hour. You need a lot more power to do it in a second instead of an hour, though. So force x distance gives you an energy or work result, and energy / time gives you the power, or the rate at which the work is done.

ETA: that looks very nice, by the way. Is it me, or the camera, or is it really tilted on the base just a bit?

[TinselKoala]

I know what you mean though... if you lifted instantly you would be "working" lifting the whole mass of the water along with the container. But imagine that the tube is the same diameter as the lifted container. Then the time lag to equalize is much shorter.

So I think that to do it right, and not get confused about inertia, acceleration and so on, we should assume a slow enough lift that the water levels stay about the same height as each other during the lifting. Not so fast that the lifted vessel's level gets ahead, and not so slow that your arms get tired, but reasonably slowly. A bigger connecting tube might help.

ETA: the fast lift "stores" some energy in the excess head in the lifted tube, and this energy can be recovered as the water drains back to level during equalization once the lift height is reached. That's why it takes more work to do a fast lift and that's where that work goes: into storage as the difference in head heights before equalization.