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Problem with energy in a cycle

  1. Jan 16, 2014 #1
    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 theorical 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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    Last edited: Jan 16, 2014
  2. jcsd
  3. Jan 16, 2014 #2

    Dale

    Staff: Mentor

    You are really doing it the hard way. A much easier way is to use conservation of energy to begin with. I assume that you want to neglect the density of R2, so then all you need to do is to calculate the potential energy of the two segments of R1.

    The formula for the area of a trapezoid is ##A=h(a+b)/2## where a is the length of the top, b is the length of the bottom, and h is the height of the trapezoid. The formula for the height of the centroid is ##y=h(2a+b)/(3a+3b)+y_0## where ##y_0## is the height of the bottom of the trapezoid. So the the PE of each segment of R1 is given by ##A\rho g y##. Also, since you have fixed the slope you have ##a=b-2h\tan(\alpha)##.
     
  4. Jan 16, 2014 #3
    You right, it's because I would like to verify the conservation of energy with integrales. It's a good exercice because I know the solution to correct myself, but not here I don't find. I'm agree with your method but I would like to find my error.

    Don't forget R2 has no slopes (slopes of the trapezoid) because slopes are R1, not R2.

    For compute I look at inside forces and outside forces.

    http://imagizer.imageshack.us/v2/800x600q90/163/trq7.png [Broken]

    Fe1 and Fe2 are forces from water
    Fg1 and Fg2 are forces from gas

    H = 5 m
    P = 1 bar (pressure inside R2)
    Pl = 0.1 bar (pressure everywhere except inside R2)
    p = 1 m (thickness of the system)
    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 = 0.189 m3. So e' is e'(S2'-2*e'*tan(α))=V=0.189 so e'=0.053 m.

    1/ Inside R2 (small container), there is gas under pressure at P = 1 bar, like slopes are not R2 but R1, there is always a difference of surface: S2-S1 > 0 all the time. In this case gas in R2 works, not ? I take this in integrales.

    2/ Outside R2, there are forces from pressure of water, this pressure come from gravity and fixed pressure of 0.1 bar. So water works in negative value, not ?

    The sum of integrales is 64000 J, what's wrong ?
     
    Last edited by a moderator: May 6, 2017
  5. Jan 16, 2014 #4

    Simon Bridge

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    Has also posed this problem in other forums. i.e.
    http://physics.stackexchange.com/questions/93323/problem-with-sum-of-energy?newreg=63be9baf16694771bc6bbc0bab8f7662 [Broken]

    If I read this correctly:
    The immediate fatal problem that springs to mind is the assertion of the same pressure and volume for the gas section no matter where it sits in the liquid. This cannot be the case - the gas section at the bottom must have a smaller volume and a higher pressure since it has to support the weight of the liquid on top of it.

    This and probably many other problems will result in non-zero energy for the entire cycle.
     
    Last edited by a moderator: May 6, 2017
  6. Jan 16, 2014 #5

    Simon Bridge

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    Oh dear - looking closer I see a repeat of the hoop of water problems.

    Fe1 and Fe2 are from the pressures - Fe1 > Fe2 always - that's what gives buoyancy.

    Fg1 and Fg2 look like they are supposed to be gravity somewhere ... are they supposed to be the gravitational attraction of the masses of liquid above and below the gas volume?

    Both of these should be lots smaller than the pressure-force arrows.
    So small that, of this scale, they should be invisible.
    The dominant gravity force is the weight of the gas due to the Earth.

    Already stated - there is also work done to prevent the gas expanding or contracting due to changes in the pressure due to it's location.

    I don't know why bother making the lower surface bigger than the upper one - buoyancy already points upwards.
    I think OP needs to reread D Simaneks buoyancy page - and follow past advise to learn about how buoyancy works before doing any more of these setups.
    It's just another pmm by the back door.
     
  7. Jan 17, 2014 #6
    No, I control position of S1 and S2 for have the same volume of R2 when it moves down. It's possible with hydraulic cylinder for example.

    Not at all, I said Fg1 and Fg2 are forces from gas, not from gravity. Inside R2, there is gas under pressure 1 bar, so there is a force on S1 and on S2. Fg2 > Fg1 because S2 > S1.

    Because there is gas in R2. And like slopes surfaces are supported by R1 not R2, R2 has only 4 walls not 6. Vertical walls don't works because force is perpendicular to movement. There are only 2 surfaces that works S1 and S2. For you gas don't works ?

    Edit: in R2 there is 1 bar, but water is under pressure from gravity (buoyancy forces) and fixed pressure of 0.1 bar.

    For me, gas inside R2 lost temperature, gas accelerate S2 and gas is accelerated by S1, but S2>S1. This is not possible, 2° principle of thermodynamics. I can repeat cycle several times, at final gas in R2 will be at 0.
     
    Last edited: Jan 17, 2014
  8. Jan 17, 2014 #7

    Simon Bridge

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    If you used, say, a hydraulic cylinder - you would need to apply greater pressure to the fluid at the lower level than at the upper level to support the weight of the liquid.

    How can the 1bar have any impact on the liquid?

    The rest of what you wrote makes no sense to me.
     
  9. Jan 17, 2014 #8
    Sorry if I don't use good physics words.

    Hydraulic cylinders is used only for control the position of 4 walls of R2. They don't give additionnal pressure. Because it's not stable for each wall. Inside R2 there is 1 bar (gas), so the pressure push 4 walls, S1, S2 and 2 vertical walls (side wiew for show vertical walls). Don't forget R2 has no slope walls. And volume of R2 is always the same. Outside R2 what is the pressure ? it's the pressure from buoyancy and fixed pressure of 0.1 bar from gas, pressure is different inside and outside R2. Do you understand this point ?

    http://imagizer.imageshack.us/v2/800x600q90/547/4qz1.png [Broken]


    http://imagizer.imageshack.us/v2/800x600q90/513/rka2.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. Jan 17, 2014 #9

    Dale

    Staff: Mentor

    Your primary error is that you forgot to include the forces at the gaskets as well as the work required to change the shape of S2 and keep it stable against the difference between the external and internal forces. These forces are very complicated to calculate and will make a big difference in your integrals.

    In the end, the simple conclusion is that it is always possible to make a scenario which is too complicated for you to analyze correctly, particularly if you insist on using the more difficult analysis technique.

    Unfortunately, I agree with Simon Bridge here. This has all of the hallmarks of a traditional buoyancy PMM, and we don't discuss PMM's even to debunk them.

    Gh778, please, since you are learning, stick to the simplest scenario possible to convince yourself that the easy analysis gives you the same answer as the complicated analysis. Then, when you do a more complicated scenario, use the simple analysis. Any other approach is doomed to failure.
     
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