Pb with sum of energy

neufneufneuf

Here I consider walls like very thin thickness, so thin that I don't compute weight and volume.

Image Y1: the system is composed of one container of water R1, one container of gas R2. The system is under gravity. Gas is at 10°C for example. R1 and R2 are trapezoid like front and side views showing. Container R1 is at alitude Z0. The volume of R2 will be always the same. Gas inside R2 is the same than outside.

Image Y4: I explain step by step what I do whit R1 and R2:

Step 1: I put R2 on R1, this need energy to move up gas. Water has pressure from outside, so it's the pressure of gas, 1 bar.

Step2: I extended walls of R1, like that R2 is now in R1. This need 0 energy in theory

Step3: I put very few water between R1 and R2 except on the 2 slopes of R2. But top wall and side walls have water. Very few water for don't compute it, just for have pressure from water and allow water to move up. Look details on image Y2. This need a little energy, but at top if thickness is 1µm, the mass of water to move is like 1000pS1*thickness = 0.001 kg, to move up this water of 0.2 m, this need 0.002 J.

Step4: I cancel pressure in R1, like the container is closed and it contains no gas, it's easy. Sure I need a vacuum pump and this cost energy, it's only the big problem I have.

Step 5: I move down surfaces of R2. Look at image Y3. Slopes of R2 don't have water (because there is no water). Pressure of gas is directly on R1 not R2, so I don't compute them for move down S1 and S2. I consider S1 and S2 length can increase without energy. Side walls move perpendiculary to the force, so this don't need energy. I done integrates of forces. VERY IMPORTANT: volume of R2 is always the same, it never change. So at start, the high of R2 is e at final it is e'. I done an example with values:

H = 5 m
P = 1 bar
p = 1 m
S2 au départ = 1 m
α = 15 °
e = 0.2 m
g = 10 m/s²
ρ = 1000 kg/m3
e' : volume of R2 is: h*(S2-2*e*tan(α))*p = 0.178 m3. So e' is e'(S2'-2*e'*tan(α))=V=0.178 so e'=0.05 m.

I compute integrates, the sum is :

799300 – 723380 = 75920 J

I don't find 0 and more water has move up from Z0 to Z1 and this add energy 1000*4.46*0.05*10 = 2232 J

And I recover energy to move down gas in R2, few energy in this example.

Step 6: I dettached R2 and R1. I reshape R2, this don't cost energy because the pressure of gas is always the same.

Step 7: I Put pressure of gas inside R1, this don't cost energy.

If you can help me to find my error ? Maybe it's integrates, or this principle is not possible to do ? I don't know if I can consider vacuum pump to need 0 energy.

neufneufneuf

Step4: with vapor pressure it's not possible to cancel all pressure, but it's possible to let 0.1 bar inside R1. This modify the result but not to 0.


The result is 62400 J

It's better to imagine all the system with vapor pressure everywhere, like that no need to cancel pressure inside water. So the energy is in this case, the works from S1 and S2 and the water moving up a little.

neufneufneuf

I think it's better to put the system in vapor pressure of water, like that no need to cancel pressure of 1 bar :

R1 = big volume
R2 = small volume, this volume is constant always

1/ Move up R2
2/ Change walls of R1 for have R2 in R1
3/ Add water around R2
4/ Destroy slopes walls of R2 (put in R2), R1 is fixed, R2 move down, this gives 64000 J
5/ Rebuild walls of R2, Dettach it, water has moved up: this give 5840 J
6/ Rechaped R2 like need in step 1/

Repeat the cycle.

Maybe replace water by gas under low pressure. Like that gas in R2 lost temperature and outside gas win temperature but less if pressure outside is lower.

neufneufneuf

It's easier to understand with only gas, no water. Here only R2 moves. R2 has gas inside: 1 bar, 10°C. Outside R2 there is 0.1 bar and 10°C. And Outside the system there is 1 bar, 10°C. R2 move down with volume = constant = 0.189 m3. At bottom, closed R2 with thin walls in it. Shape it like at start, this cost nothing of energy. Move up, and put between slopes. Destroy slope walls (and put inside R2) and add gaskets. Repeat cycle. Here with :

H = 5 m
P = 1 bar (inside R2)
Pl = 0.1 bar (outside R2)
p = 1 m (thickness of R2)
S1 at start 0.892 m (around)
S2 at start = 1 m
α = 15 °
e = 0.2 m
e' : volume of R2 is: h*(S2-2*e*tan(α))*p = 0.189 m3. So e' is e'(S2'-2*e'*tan(α))=V=0.189 so e'=0.053 m.

The energy won is 110000 J per cycle.









In fact, the same can be done with a liquid under pressure, like volume of R2 is constant, it's possible to control with hydraulic cylinders the position of S1 and S2. No need temperature in this case, so the energy come from where ?

neufneufneuf

With a liquid under pressure it's the same case ? But here I don't know where come from the energy. Someone can help me ? please.

neufneufneuf

Ok with a gas inside and outside, all is fine, the sum of energy is 0. I understood even I have problem with my integrales. The works to move S2 or S1 is the same because S1 move more quickly than S2. But if I move up a volume of gas R2 like that in water under gravity, with volume = constant, S1 move up faster than S2, but the pressure is not the same at S1 and S2, it's linear, so in this case if I move up R2 like that it's the same energy than move down without fixed slopes ?

I made mistake in integrales so it is with gas:

float(integrate(1*(100000)*1*(1+2*(x-0.2)*tan(0.261)),x,0.2,5));
float(-integrate(1*(100000)*1*(1-2*0.2*tan(0.261)+2*x*tan(0.261)),x,0,5-0.0533309));

The sum is 0.

But in water (R2 with air inside and R2 in water), the formulas are:

float(integrate(1*(10000x)*1*(1+2*(x-0.2)*tan(0.261)),x,0.2,5));
float(-integrate(1*(10000x)*1*(1-2*0.2*tan(0.261)+2*x*tan(0.261)),x,0,5-0.0533309));

The result is 334029 - 324807 = 9222 J

and for move down a volume of 0.189 m3 at 4.9 meters this need 9276 J

Sure, trapezoid is not square, so it's possible to take very thin e = 0.001 m

S2 = 1 m
S1 = 0.99946 m
V = 9.9973 e-4 m3

like this the formula is:

float(integrate(1*(10000*x)*1*(1+2*(x-0.001)*tan(0.261)),x,0.001,5)-integrate(1*(10000*x)*1*(1-2*0.001*tan(0.261)+2*x*tan(0.261)),x,0,5-0.000280523));

the result is 51.47 J

But if I move up/down the same volume of air I need 10000V*h = 49.98 J

And the result is not the same than put the object in water directly. If some can help me because here integrales are good.