Working with closed systems.

Discussion in 'Physics & Math' started by Quantum Quack, Jun 19, 2005.

  1. Quantum Quack Life's a tease... Valued Senior Member

    If i am correct it is conventional thought to state that the universe as a whole is a closed system. The laws of thermodynamics, to me, clearly suggests that this is the case.

    But i wonder if the universe can be both closed and yet open or should I say infinite and yet finite simultaneously.

    The reasoning that occured to me and wished to share and explore could be best described using a simple mathematical model as follows:

    1+[10000-9990]+1 = 12

    The notion is that within a simple equation we can add infinite complexity yet retain the same outcome in this case being the value 12.

    so maybe it is worth asking what is the difference between
    10+2 = 12 and
    12 = 12.

    Possibly you will think this is more about philosophy and possibly you are essentially correct. However when working with "energy over time" equations the need to keep the universe a closed one yet allow for seemingly infinite values to be a part of a finite value appears to be very important.

    This extends to the example of a hydro electric station and how teh energy created by the flow of water generates energy.

    The questioned occured to me that the water has a potential energy and even though a hydro station capitalises on that potential the potential of the water remains constant regardless of how it's potential is used by the station.

    To clarify what I mean we can assume that the water is going to go to the sea regardless of whether it is diverted for a small time or not. The water retains it's potential regardless of the energy taken from it's decent to sea level.

    so in a stream of water we take a certain amount of energy from that flow but the flow retains it's potential.

    so the equation is like

    100 + [10000-9900]= 100
    meaning that energy has been generated with out any impact on the overall value of the original value.

    When thinking of a tidal generator using the out and inflow of sea water it is worth considering that the energy is happening whether we use it or not however energy can be achieved without diminishing the source of that energy as that energy is being expended regarless of whether we utilise it or not.....

    I do appologise if I have made a confusing mess of what I am attempting to describe here, and I look forward to any contributions that may help clarify the terminology needed.
    Last edited: Jun 19, 2005
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  3. kevinalm Registered Senior Member

    You're forgetting that the water that flows through the turbine loses potential energy between the top of the lake and outlet. The outlet must be lower.

    I do understand what you're getting at though. Look at it this way. The energy of all the water is going to be expended either way. Undiverted the energy will dissapate as turbulence, erosion and the like. Diverted the energy extracted by the turbine will cause that much less turbulence, etc. Energy is conserved.
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  5. Quantum Quack Life's a tease... Valued Senior Member

    The point is, I guess, that the system of water heated and then rained on a catchment area is in no way diminished by the drawing of energy from the fall of the water to the sea.

    So in effect we are taking energy from a system with out any net change to that system.

    zero net change to the system but many units of energy created and utilised.

    If I have a river that is 100 miles long and place a turbine every mile I would have theoretically 100 turbines each delivering the same amount of energy yet not diminishing the system I am drawing from.

    This I find interesting........

    so mathematically we have a system:

    3+3+3 = 9
    but with the turbines added we have

    3+ [1000000-999997] + 3 = 9

    The figure in brackets could be any figure you choose as long as it still equates in this case with the value 3.

    obviously generating electricity eventually ends up as heat in the system and the dynamics would eventually change.
    But say we stored the electricity and not use it.
    we have created an energy potential in excess of the original system potential of the water going up and down due to thermal effects of the sun.

    so in this way 3+3+3 can equal anything you want it to as long as it is in excess of 9

    Do you see the idea I am attempting to put forward.......?

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  7. superluminal I am MalcomR Valued Senior Member

    Nope. Can't happen. There is potential energy in the headwaters, gained originally by the heating and evaporation of the water from the sea. As it falls, the water causes the turbine to turn, generating electrical power. The water now has less energy, by mgh (plus loss in heat and turbulence), to give up.
  8. Quantum Quack Life's a tease... Valued Senior Member

    fair comment, however I would question how waterthat has generated electricity stred in a dam 1000 meters above sea level has lost any of it'ss potential.

    catchment area to dam, > dam to turbine > dam, dam to sea level.

    the weight of the water at sea level for all intents and purposes is the same as it is in the catchment dam, so how does turning a turbine reduce the potential of the water. [It is true it may effect the speed of the water but it certainly canot effect the weight of the water.]

    or maybe I should ask how does damming the water effect it's potential?
  9. superluminal I am MalcomR Valued Senior Member

    gravitational potential energy = mgh. m = mass, g = 9.8m/s^2, h = height.

    Take one gallon of the water. Pour it through the turbine. It drops 100 meters, generating one watt. Gather up that same gallon of water at the bottom. How will you ever use it to generate that one watt again?
  10. Quantum Quack Life's a tease... Valued Senior Member

    well in the example given the evaporation of water leads to rain which repeats the cycle. The point though is how the turbine removes energy from this weather system.

    say we imagine the weather system as a relatively closed system, and the turbine is a parasite on this system. How does the turbine effect the weather system?
  11. Quantum Quack Life's a tease... Valued Senior Member

    you know it is friggin near impossible to describe complex ideas on the net.....sheesh!!
  12. glaucon tending tangentially Registered Senior Member

    Interesting stuff.
    I just finished reading Russell's 'Introduction to the Philosophy of Mathematics'. I think what doesnt work here is the analogy. It's understood that for a given series of natural numbers in series, there are indeed an infinite series of rational numbers in between each of them. This is explained by the fact that all natural numbers are really just 'expressions' (for lack of a better term) for classes composed of similar elements. So, while the series may (from one perspective) seeem to be a closed system, and indeed can behave as if it is, in other ways it may not quite behave as such.
    n.b.: I'm sure most of the heavy math guys in here will go to town on my comments, but I'm still digesting Russell's 'so called' Introduction...

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  13. superluminal I am MalcomR Valued Senior Member

    Ok. Assume the turbine is 100% efficient. Assume also that the sea is at the bottom of the dam. From the headwater to the sea is 100m.

    Say it takes 10 watts of solar energy to evaporate one unit of water from the sea. It rains back into the sea. It takes another 10 watts of solar heating to loft that unit of water back into the air to a height of X.

    Now, how much energy does it take to loft a unit of water from the headwater to a height of X? Less than from the sea right? It's already 100m up. Say the difference is 1 watt, so that it takes only 9 watts of solar heating to loft the unit of water into the air.

    Ok. It takes 10w to get water into the air from the sea. When it rains into the sea, it gives up that energy by falling. If we coulld tap it all we could get 10 watts from it. Well, some of it fell into the headwater. The weather cycle naturally only has to expend 9 watts of energy to loft any unit of water from the headwater.

    By letting the unit of headwater fall the remaining 100m through the turbine, we can extract the remaining 1w it took to get it there in the first place(imaginary 100% efficient turbine remember). By doing this we have affected the weather system by now requiring a unit of water that only needed 9w to evaporate and cycle to now require 10w once again.

    Hopefully this makes some sort of sense.

    NOTE: The numbers are purely fictitious and used for illustration purposes only.
    Last edited: Jun 20, 2005
  14. kevinalm Registered Senior Member

    Ok QQ, I think I see what your getting at now. You're thinking of the earth as a closed thermodynamic system. It isn't. You have heat enterring the atmosphere at a high temp (sunlight) and leaving at a low temp (radiant cooling of cloud tops, night side ground infrared radiation, and so on). That's called a Carnot cycle. A heat engine that can do usefull work like refill a dam, or some not so usefull work, like hurricanes, tornados...

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  15. Quantum Quack Life's a tease... Valued Senior Member

    OK Me thinks to start again:
    I drew these quick animations to show my point a bit better [I hope]

    the first one is simple naked and clear:

    <img src=>

    we have a simple energy system.[system A] The ball is raised to the height by energy applied to a conveyor belt system.

    The energy to raise the ball is deemed to be 100 units.

    This system is constant. Energy use is always the same for every cycle.

    now the next animation shows how the balls fall is innterupted by a gate.The energy derived by the movement of the gate is 10 units.

    <img src=>

    Even though we have gained 10 units of energy from system A we have not increased the energy burden on that system as it is still 100 units.

    So in the first animation the energy burden is 100 units.
    and in the second the burden is also 100 units but we have an output of 10 units.

    In essence we have utilised the potential of the ball achieved by the ongoing expenditure of 100 units, and created an extra 10 units thus feeding off the potential with out increasing the burden.

    So we have no net change to system A except time even though we have gained 10 units.

    The thinking is:

    the only change to system A is the amount of time involved in each cycle. And I would propose that that time loss to system A is now our 10 units of generated energy.

    there is a principle involved in all this that has interested me enough to post this thread.
    and relates to space time increasing the time in a system with out reducing the energy we have a gain in energy outside that we have shifted time from one system to another. The lever provides change [time]to the outer system.
    Last edited: Jun 20, 2005
  16. James R Just this guy, you know? Staff Member


    The only difference between your first and second examples is that in your first example you waste all the energy supplied by the conveyor as heat, sound etc., whereas in the second example you use some of the energy to work the lever.

    You haven't "created" any more energy. You just aren't wasting as much.
  17. Quantum Quack Life's a tease... Valued Senior Member

    and when I used teh math:
    and compared it to
    3+[100000-999997]+3 = 9
    it could be suggested that all I have done is increased the time it takes to read the equation....ha...thus I have added time to the equation....hmmm.....
  18. Quantum Quack Life's a tease... Valued Senior Member

    James, the gate is activated by the weight of the ball [gravity]
    and yes the energy potential achieved by the conveyor is wasted in the first diagram. BUt of course this is not the point.the point being that 10 units is gained without increasing the energy burden on that system [A]
    however the increase in time is conserved by the outer systems gain.
  19. Quantum Quack Life's a tease... Valued Senior Member

    James, I do agree with your point.

    However I would argue whether waste is the right word.

    The system A is complete and it's energy consumption is 100 units, without changing that consumption we have taken 10 units from that system and as I have proposed the only difference between the two versions is the time it takes for the ball to complete it's cycles.

    noise and friction etc remain fairly constant and are insignificant to the principle i am exploring.....
  20. Quantum Quack Life's a tease... Valued Senior Member

    a little insight into the question:
    i have been thinking of doing experiments on a tidal generator for a local bay. the tidal flows are approx 4 knots in and out.

    i was thinking of floating a paddle system connected to a generator via a gear box just to see what sort of output I would get from this "free" energy source. Now the word 'free' prompted the think about just how "free" it is.

    I considered the notion of what would happen if I put 100 paddle generators in a row in the tidal flow and what effect that might have on the tidal flow.
    What isthe differernce between one generator and 100 to the potential energy of the tidal flow?

    Some one somewhere suggested that the earth moon system would see a slight change and I would argue that this is not so as the energy potential of the tide is unchanged even though we have taken use of it in the form of electricity. However the outflow would be slightly delayed and it is this delay that we are using to generate the electricity not the potential energy of the flow.
  21. everneo Re-searcher Registered Senior Member

    Regarding the "free" energy that the turbine draws :

    The kinetic energy gained by water at the cost of potential energy was imparted to turbine partly. In the absence of turbine, the impact on the floor (or at the base ) of the tunnel would be more. In both the cases, the potential energy difference is same - the total kinetic energy gained only is spent between the turbine and the floor by way of impacts.
  22. Quantum Quack Life's a tease... Valued Senior Member

    so you would say the potential energy of the water remains the same.?

    another easy diagram:

    in this diagram i pose the question: does the potential energy at gate 'd' suffer any loss due to the presence of gate 'a'?

    <img src=>

    given that kinetic energy is not in question. [ the gates are deliberately slowed so that kinetic energy is absorbed and wasted or should i say retained by the system A]

    the puzzle is that if the potential energy is unchanged between gates 'a' and 'd' then the potential energy remains in the system and the gates deliver energy to the out side with out diminishing that potential energy.

    so gate 'd' delivers the same energy as gate 'a' does and yet the potential energy is not used until the ball actually reaches the base line. Even then it has potential if it were able to fall further.

    so my point is the energy delivered by the gates does not diminish the potential energy of the ball . So the ball conveyor system retains it's energy potential. The kinetic energy potential is still absorbed by the system but in staggered amounts through the gate frames rather than one amount when the ball accelerates to and hits the base line
    So I would surmise that whether the gates are there or not has virtually no impact on the energy in the system.
    btw I might add it's no big deal and probably a moot point.

    also this has nothing to do with free energy as such.......
  23. everneo Re-searcher Registered Senior Member

    difference in potential energy is different from potential energy.

    Why do you think 'potential energy is unchanged'?

    PE = mgh ; depends on the height from sea level. You should rather talk about difference in PE; thats what is converted to kinetic energy that is partly spent in opening the gates/turbine/etc.

    PS : your animations are good.

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