Most Efficient Way To Waste Power?

Discussion in 'General Electronics Chat' started by Manfred Mans Earth Band, Oct 26, 2014.

Amazing MS-Paint skills demonstrated here

  1. Yes

    100.0%
  2. Yes

    0 vote(s)
    0.0%
Multiple votes are allowed.
  1. Manfred Mans Earth Band

    Thread Starter New Member

    Oct 5, 2014
    17
    1
    Hi,

    I have been wondering what the most efficient way of wasting power is. High power resistors and delta fans for cooling? Magnetic fields causing causing mechanical tension? An electromagnetic spring stretching/compression device?

    I want to build a LiPo discharger with variable power draw, preferably 1-12s (3.3V - 51V) at 3-5000W - With as much of that watt range at any Voltage. I'm gonna use a 12-position rotary switch to switch between 1-12s and a "coarse and fine" system to adjust power. The power draw must be monitored with a 100+Hz refresh rate and preferably a system to engage a mechanical relay to shut the power draw off to the main discharging mechanism. Just how I'm going to achieve the latter function - I'm unsure, I don't know of any specific component (Except user programmed microcontrollers, but I don't know any language) that can provide this function. I'll figure it out eventually.

    Oh - And I intend to make it 100% powered by the LiPo connected and as small and light as possible - Field convenience.

    Any input is appreciated!
    (The poll is a joke btw.)

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    proj1333333333.jpg
     
  2. nerdegutta

    Moderator

    Dec 15, 2009
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    I have noe idea about this but your MS Paint skills are from another world!
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    Wasting energy efficiently is an oxymoron.
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    LOL MrChips, good one :)

    The 'simplest' method to waste power is to waste it using the type of energy for which you intend to waste.
    For example, to waste electrical energy the simplest way is to use a resistor. This includes resistive devices like regular power resistors and light bulbs, although light bulbs are harder to deal with because they are more fragile and also have a changing resistance over their voltage range.
    For mechanical rotation, probably a generator is the simplest because then you have options for what is connected to the generator.

    To actually 'waste' power efficiently though would mean to 'convert' that energy into another form that is usable. For example, if instead of resistors you use high power LED's then you can gain some useful work light while wasting that energy. If it is really a significant amount of power and it will be continuous over many years, then you might look into line tied converters that can take your energy and pump it back into the power line and that will reduce your electric bill or you could sell the energy back to the power company. If your converter is 80 percent efficient then you'll only be wasting 20 percent of the energy while still having a true load on whatever it is you are loading.
     
  5. BR-549

    Well-Known Member

    Sep 22, 2013
    1,987
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    Cern.
     
    killivolt and Alec_t like this.
  6. Manfred Mans Earth Band

    Thread Starter New Member

    Oct 5, 2014
    17
    1
    Aha. Well I've been considering LEDs too, but getting enough for a decent power draw is relatively expensive. Even with cheap ones like these - I would need 10 to get at least 1000W dissipated - $60.

    Regarding the first concept - the first drawing - Do you think this would be possible? After having a look at some electromagnets myself I don't think so... This one for example draws just 10W and puts out 400N ("40Kg")... :eek::eek::eek:
     
  7. Manfred Mans Earth Band

    Thread Starter New Member

    Oct 5, 2014
    17
    1
    ?o_O
     
  8. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,797
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    Give it to a politician? :)
    Seriously though, why not dissipate it in the heating element of a coffe-maker/kettle to make a refreshing brew?
     
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  9. Manfred Mans Earth Band

    Thread Starter New Member

    Oct 5, 2014
    17
    1
    I only drink coffee in the morning :D
    But yeah, seems like the best solution. I would however need to make the device large enough so that the electronics are separated from the heating element(s), assuming no cooling. Alternatively, I could make a case with a large heatsink and some extremely high power delta fans to cool it off enough for longer periodic use. (Generally I'll be using it to discharge around 1500-2400W in 15-60min.)

    A heating element like this one requires 220V. How can I up the voltage without using a transformer? I need variable voltage above that of the LiPo batteries. MOSFETs in series?
     
  10. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Do you mean 1500-2400W continuously for that time? That's a helluva lot of energy from a LiPo!
    Use an inverter ......a BIG one.
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490

    Hi,

    I am sorry i dont quite understand your drawing as what you are trying to show there.

    However, when you apply a DC voltage to a coil then the coil draws a current, and the loss is I*V so it's mainly just a resistive loss again anyway. If the coil resistance is 10 ohms and you apply 10 volts, then the current is 1 amp so that's 10 watts lost. If you couple two coils then the energy gets transferred to the second coil but only when the DC ramps up and then must be dissipated in the second coil with a load anyway. Once the DC stops ramping it's back to pure resistive losses again.
     
  12. ronv

    AAC Fanatic!

    Nov 12, 2008
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    More information is needed to answer your question. The easies way with the info provided would be to just put the batteries on the shelf and wait for them to go dead. :rolleyes:
    But that is probably not what you have in mind. To really answer the question depends on the AH rating and maximum discharge rate. So let us know what the current would be for the 50 volt ones and for the 3 volt ones.
     
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  13. Lestraveled

    Well-Known Member

    May 19, 2014
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    Well, before Manfred asked for help on the forum he probably used incandescent light bulbs as loads but he had to stop because he was blinded by the light.
    Then he used an electric motor but it revved up like a deuce and he had to shut it down.
    Then he used the power to run his go-kart Mozart but while checking out the weather chart found it wasn't safe outside.

    ....do I need to continue?? :D:D
     
  14. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    First, back to basics. All wasted energy turns into heat. There is no such thing as a small 5000 watt heat system. I think MrChips was out of line to call you an oxymoron. :rolleyes::D (MrChips will get that joke, even if you don't.)

    So...the first thing to design is how to get rid of 5000 watts. We do that with fans in the Air Conditioning/furnace business.
    3.413P = 1.08 CFM dT
    For 5000 watts, 15801 = CFM dT
    For 200 CFM, the air will heat up by 79 degrees F

    For water, one watt-hour per hour produces 3.413 B.T.U.s of heat.
    A B.T.U. is defined as the amount of energy required to raise 1 pound of water one degree F.
    5000 watt hours per hour times 3.413 BTU/W = 17065 pound degrees of water (in Fahrenheit)
    For 100 degrees change in water temperature you will be heating 170.65 pounds of water (20.5 gallons) for an hour.
    If you want to discharge the energy as steam, calculate the distance from ground water temperature to 212F and figure in 970.3 BTUs per pound of water to turn it into a gas.

    Whew! That was a lot of math and looking up constants!

    Now for the simple part: What do you want to use to burn the electricity? Car headlights? Wirewound resistors? Nathans Coney Island Corn Dogs? Fan motors that are busy cooling themselves?

    There''s a start, but I'm getting tired now. Most of the other helpers can finish this set of calculations.
     
  15. #12

    Expert

    Nov 30, 2010
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    ps, for those who don't believe you can rely on standards known all over the planet, these calculations assume atmospheric pressure (14.696 P.S.I.A.) at sea level and pure water.
     
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  16. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I'll take your word for it. All I know is I take in to many calories - but I don't feel warm. :rolleyes:
     
  17. #12

    Expert

    Nov 30, 2010
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    If, "ronv" stands for something like, "v ronica" we might be able to find a helper here. ;)
     
  18. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
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    If you converted here to horsepower-fortnights the calcution would have been soo much simpler :rolleyes:
     
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  19. tom_s

    Member

    Jun 27, 2014
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    using maxwell smart equation, its the '6" nail instead of a fuse' trick. local hardware store should have a variety of nail sizes and lengths to play with to put on the rotary switch. ;)

    now, where did i put those pills... :oops:
     
  20. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    You have a good point. When I turn my thermodynamics book loose on an electrical circuit, the answers don't look very much like electronics OR math. :D
     
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