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Gh778
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Hi,
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 water inside
R2 = small volume, this volume is constant always, it has gas under pressure 1 bar inside
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/ Add water around R2, this need energy e1
4/ Destroy slopes walls of R2 (put walls in R2 and imagine very thin walls), R1 is fixed, R2 move down with volume of R2 = constant, this gives energy 64000 J and gives energy e0. In this step it's very important to destroy slopes walls of R2. Sure, this need gasket and theoretical walls, but it's poosible to build it. When R2 move down, water can pass from bot to top because the thickness of R2 is smaller than R1.
5/ Rebuild walls of R2, Dettach it, water has moved up: this give 5840 J
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 64000 +5840 - e1
But e1 can be near 0 if space between R1 and R2 is very small at step 1. Look at image please.
H = 5 m
P = 1 bar
p = 1 m
S1 at start 0.892 m (around)
S2 at start = 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.
[itex]E_{step4} = \int_e^H p(P-Pl)S2(1+2xtan(\alpha))-pg\rho xS2(1+2xtan(\alpha)) dx - [/itex]
[itex]\int_0^{H-e'} p(P-Pl)S1(1+2xtan(\alpha))-pg\rho xS1(1+2xtan(\alpha)) dx[/itex]
Integrales with WxMaxima:
float(integrate(1*(100000-10000)*1*(1+2*x*tan(15))-1*1000*10*x*1*(1+2*x*tan(15)),x ,0.2,5)-integrate(1*(100000-10000)*1*(1+2*x*tan(15))-1*1000*10*x*1*(1+2*x*tan(15)),x ,0,4.8));
When R2 move down, S2>S1 all the time. So, pressure P inside R2 give a net force to the bottom, sure water give an up force to the top with the same difference of surface. But, pressure of water come from only with pressure of gravity and a fixed term 0.1 bar, if R2 move down until 9 meters, not more, pressure P give more energy than pressure of water.
I don't know where I'm wrong. I think integrales be good. Maybe I lost energy needed in a step, where ?
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 water inside
R2 = small volume, this volume is constant always, it has gas under pressure 1 bar inside
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/ Add water around R2, this need energy e1
4/ Destroy slopes walls of R2 (put walls in R2 and imagine very thin walls), R1 is fixed, R2 move down with volume of R2 = constant, this gives energy 64000 J and gives energy e0. In this step it's very important to destroy slopes walls of R2. Sure, this need gasket and theoretical walls, but it's poosible to build it. When R2 move down, water can pass from bot to top because the thickness of R2 is smaller than R1.
5/ Rebuild walls of R2, Dettach it, water has moved up: this give 5840 J
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 64000 +5840 - e1
But e1 can be near 0 if space between R1 and R2 is very small at step 1. Look at image please.
H = 5 m
P = 1 bar
p = 1 m
S1 at start 0.892 m (around)
S2 at start = 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.
[itex]E_{step4} = \int_e^H p(P-Pl)S2(1+2xtan(\alpha))-pg\rho xS2(1+2xtan(\alpha)) dx - [/itex]
[itex]\int_0^{H-e'} p(P-Pl)S1(1+2xtan(\alpha))-pg\rho xS1(1+2xtan(\alpha)) dx[/itex]
Integrales with WxMaxima:
float(integrate(1*(100000-10000)*1*(1+2*x*tan(15))-1*1000*10*x*1*(1+2*x*tan(15)),x ,0.2,5)-integrate(1*(100000-10000)*1*(1+2*x*tan(15))-1*1000*10*x*1*(1+2*x*tan(15)),x ,0,4.8));
When R2 move down, S2>S1 all the time. So, pressure P inside R2 give a net force to the bottom, sure water give an up force to the top with the same difference of surface. But, pressure of water come from only with pressure of gravity and a fixed term 0.1 bar, if R2 move down until 9 meters, not more, pressure P give more energy than pressure of water.
I don't know where I'm wrong. I think integrales be good. Maybe I lost energy needed in a step, where ?
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