Someone can help me for english translation for an idea ?

[rc4]

I posted an english thread here:

http://physics.stackexchange.com/questions/93323/problem-with-sum-of-energy

But people don't understand, maybe a problem with english or maybe my system is not described very well. Could you help me to write good text please ? I think I will delete the thread and restart a new one if you help me to translate good message.

The question will be block if I can't explain better.

I think there is a problem with sum of energy.


The text:

**It's a theoretical study.** The system is under gravity, on Earth. Here I consider walls like very thin, so thin that I don't compute weight and volume. I consider water like not compressible for reduce computational complexity. I consider gas at the same temperature, ambiant temperature 10°C for example.



The system is composed of one container R1 with water inside. One container R2 with gas inside. R1 and R2 are trapezoid shapes like front and side views showing. Container R1 is at alitude Z0. The gas inside R2 is the same than outside (same pressure, same temperature).


I explain step by step what I do with R1 and R2:

**Step 1:** I put R2 on R1, this don't need energy (Archimedes law). Water is under pressure from outside gas, 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 behind them. Very few water for don't compute it, just for have pressure from water and allow water to move up. If it's necessary to compute it, it's possible to imagine thickness of water like 1µm. This need a little energy to move up this water, 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 for the example. The only limitation it's the capillary action that limit the thickness, but I don't compute it. Look details on image:


**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 my big problem. I don't know how evaluate the energy needed here. But if the volume of gas is 0 I think the energy lost here can be near 0. Maybe it's necessary to do step 4 before step 3, because for cancel P pressure from R1, this need a little volume where there is no water. Maybe add a little chimney with gas inside for cancel pressure. If I take in account the vapor pressure of water I need to fixed a limit for low pressure Pl. It's possible to take Pl at 10000 Pa for example, this cancel the problem of vapor pressure and allow to limit losses from pump.


**Step 5:** I move down surfaces of R2.

With α angle of slope of trapezoid shape, P the pressure of gas, p the depth of R1 and R2, g the acceleration of gravity, ρ the density of water.

Look at image:

Maybe the integral need to multiply by x each term like that :


But the result is not 0 too it's 227 000 J and this don't seems to be good if I test with the work on each mean force. So I think I don't need to multiply by x.

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. Side walls moves perpendiculary to the force. I done integrales 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 ° or 0.261 rd 
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 computed integrales for this example, **the sum is 62400 J**

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

**Step 6:** I dettach R2 and R1. I reshaped 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 integrales, or this principle is not possible to do, in this case why ? I don't know if I can consider vacuum pump to need 0 energy maybe the problem come from this step.


**Integrales with WxMaxima:**

Sum of energy = integrate(1*(100000-10000)*1*(1+2*x*tan(0.261))-1*1000*10*x*1*(1+2*x*tan(0.261)),x,0.2,5) - integrate(1*(100000-10000)*0.892*(1+2*x*tan(0.261))-1*1000*10*x*0.892*(1+2*x*tan(0.261)),x,0,4.95);

[rc4]

I wrote the message in stackechange but I prefer to save it in the case they delete question. Maybe, here, if someone want to help me ?

-------------------------------------------------------------------
FIRST CASE
-------------------------------------------------------------------

I compute sum of energy in a cycle and I don't find 0, could you help me to find the error ? I started an equivalent thread in another physics forum but nobody understand. I hope my english is clear enough for understand the problem.

The system is under gravity and everywhere outside R1 and R2 there is 0.1 bar. It's a theoretical problem.

R1 = big volume, with gas at low pressure 0.1 bar for example. 
R2 = small volume, this volume is constant always, it has gas under pressure 1 bar inside. It's very important to understand that R2 has only 4 walls when it is in R1. Slopes walls are assured by R1 and gaskets. Each wall of R2 is independant from each other. So, like pressure inside/outside each wall of R2 is different, each wall is unstable. So I need to use hydraulic cylinder not for add pressure, but for control the position of each wall.

Gaz inside R1, R2 and outside are at the same temperature T. It's only the pressure that is different inside R2.

1/ Move up R2 above R1, this cost energy e0 
2/ Change walls of R1 for have R2 in R1, this don't cost energy in theory 
3/ **Destroy slopes walls of R2** (put walls in R2 and imagine very thin walls), R1 is fixed. In this step it's very important to destroy slopes walls of R2. Sure, this need gasket and theorical walls, but it's poosible to build it.   
4/ R2 move down with **volume of R2 = constant**, this gives energy 110000 J and gives energy e0.
5/ Rebuild walls of R2, Dettach it 
6/ Rechaped R2 like need in step 1/ this don't cost energy 




Repeat the cycle.

**I done numerical application:**

Sum of energy (energy giving in each cycle) is 110000  J



H = 5 m 
P = 1 bar 
Pl = 0.1 bar 
p = 1 m 
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'=0.05 m. 

$$E_{step4} = \int_e^H p(P-Pl)S2(1+2xtan(\alpha)) dx - $$
$$\int_0^{H-e'} p(P-Pl)S1(1+2xtan(\alpha)) dx$$

Integrales with WxMaxima:

float(integrate(1*(100000-10000)*1*(1+2*x*tan(15)),x ,0.2,5)-integrate(1*(100000-10000)*1*(1+2*x*tan(15)),x ,0,4.);

When R2 move down, S2>S1 all the time. So, pressure P inside R2 give a net force to the bottom. But, pressure outside R2 is lower at 0.1 bar, pressure P give more energy than pressure of 0.1 bar. It's possible to give oustide R1 and R2 atmospheric pressure (near 1bar), inside R1 at 1 bar too, and inside R2 2bars. The advantage to use water in R1 is to change fixed pressure more easily in R1 than with 2 gases.

Like S2 > S1 all the time when R2 move down, Gas in R2 works more for S2 than S1 works for gas, so for me gas is decelerating when R2 move down. And this is not possible in thermodynamics I think ?

-------------------------------------------------------------------
SECOND CASE
-------------------------------------------------------------------

It's possible to study this case:


All the system is on Earth, all around the system there is atmospheric air gas at temperature 10°C. 

I use the cycle :

1/ R2 is like image shows 
2/ I move down R2 with volume = constante. R2 has no slopes, there are gaskets like first case 
3/ At bottom, I build walls for R2 on slopes (imagine thin walls), like that R2 is closed 
4/ I modify the shape of R2, this don't cost energy in theory 
5/ I move up R2 this cost only energy for move up walls and gas but it is recover at step 2, or it is possible to have the gravity perpendicular to the screen 
6/ I "destroy" slope walls of R2 (in fact, destroy it and put it in R2) 

Repeat cycle


There is gas under pressure at 1 bar and temperature T = 10°C in R2.  R2 give energy from gas under pressure when it move down. Sure strictly vacuum is not possible but with iron (walls), for example, the vapor pressure can be very low. Each cycle, R2 lost temperature I think, it's possible to increase temperature inside R2 from the external temperature with a wall of R2 (side view) in contact with atmospheric temperature. Even the small gas inside vacuum container is at 10°C like the pressure is very low, the temperature of vacuum area increase very slowly, and it's possible to cool it with external gas I think.


Note, it's possible to do the same with a liquid inside R2 and nothing outside R2. A liquid can be under pressure easily and it's possible to recover energy from liquid's pressure (in theory). So, move R2 like show allow to recover energy from nothing in fact.

[rc4]

I simplify the problem with only one liquid. It's logical in this case (and mathematical with integrales).

No gravity here. It's a theoretical study. R2 has a liquid inside it. R2 is composed of 6 walls but when R2 move down it is composed only with 4 walls, 2 slopes wall are assume with fixed slopes. Walls are very thin. Dimension of walls can change, when R2 move down, the volume of R2 must be exactly the same, so walls must change their length. It's a little difficult in practise but it's possible to imagine it. 4 Walls are controlled in position with 4 hydraulic cylinders, 2 hydraulic cylinders of vertical walls don't need and don't recover energy. 1 hydraulic cylinders give energy to S1 and the last recover energy from S2.


Outside R2 there is 1 bar of pressure.

0/ I compress liquid at 11 bars, it is compress for all the time, this step don't repeat again. 
1/ R2 is put between slopes, I suppress slope walls and add gaskets, walls are put in R2, imagine very thin walls. 

2/ I move down R2 with volume = constant.   

S2 > S1 and the works give from R2 is :

With :

p : thickness 
Pl : Pressure in liquid 
Po : Pressure outside R2 
alpha : angle of slope 
e : height of R2 at start 
e' : height of R2 at final 
H : Height of slopes 
S1 : Top surface   
S2 : Bottom surface 


NA:

H = 5 m 
Pl = 11 bar 
Po = 1 bar
p = 1 m 
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'=0.05 m. 

The result is an exceed of energy of : 122000 J for each cycle.

3/ I restore slope walls of R2 this no need energy in theory. 

4/ Move up R2

It's possible to repeat cycle at step1.


For understand, that change lenght of S1 and S2 don't need energy if thickness of wall is veery small:


Gaskets can be very small in theory, so they don't use energy.

The system is unstable ? Sure, but it's only a net force in one direction. It can be controlled by a hydraulic cylinder. It's possible with technology and in theory more !

[vasik041]

Hi rc4,

It would be interesting to know what actually you trying achieve ?
Why these manipulations needed ?

Thanks.

[rc4]

Hi Vasik041,

To recover more energy than I give. Push S1 with a hydraulic cylinder and recover energy from another hydraulic cylinder. The volume is always constant so liquid is always at the same pressure. Integrales show the system recover more energy than it need.

Try to compute with Wxmaxima the formula :

float(integrate(1*(100000)*1*(1+2*x*tan(0.261)),x,0.2,5)-integrate(1*(100000)*0.892*(1+2*x*tan(0.261)),x,0,4.947));

The result is 122000 J in one cycle.