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Energy efficiency Gain or Loss? What do you think?

  1. Sep 16, 2015 #1
    Energy efficiency related question: please help with an anwer to the following:
    Given a flat electrical coil of known dimentions, this coil is used as an induction coil, laid flat, on top of the cooking surface of an induction cooker. Basically, the induction cooker has a copper braided coil, and this second (external) coil is simply laid on top, made to work by inductive coupling generated by the induction cooker.

    Sorry about the redundant use of "induction", but I am afraid it will get worse.

    The two leads of this second flat coil are then connected to a set of incandescent lamps. The power dissipated at this "load" is 920W at 115V and 8 Amps.

    Plugged into an ordinary electrical outlet (115V, 20A max), is a Watts meter, which itself is connected to the induction cooker. The induction cooker "draws" its power from the watts meter. The watts meter registers a usage of 1,280.6W (115V and 11.13A).That is a measure of what the entire circuit (induction cooker plus the watts meter, the flat coil and the load) draws from the outlet.

    This means that the induction cooker (under load) has a power efficiency rating of 920W/1,280.6W = 71.8%.
    It is not that great, and the power loss (in eddie currents and internal resistences) justifies using an internal fan, which itself adds to the losses. We want to improve this efficiency rating, but we are limited (the finance department holding us on a tight leash) to a minimal number of changes in design.

    In order to test a few things which may increase the system's efficiency, we add another flat coil, laid on top of the aforementioned flat coil, (like stacked pankakes), which lays right on top of main induction coil of the cooker.

    Yes, there are three flat coils, stacked on top of eachother, two of them solely powered by Faraday's induction, while the first coil of the induction cooker is powered straight by AC.

    This last coill's terminals (ends) are simply connected back to the terminals (+/-) of the set of incandescent lamps, the same way the first flat coil had been connected to them. It makes a parallel connection if you will.

    By the way, the set of bulbs are, in their own layout, connected in parallel, enabling each bulb to benefit from the same voltage input from the coil(s).

    Now, we are interested in finding out if the addition of that second flat coil to join the first flat coil, in powering the bulbs, will increase, decrease or not affect, the power efficiency of the system.

    To recap, we have:

    1) outlet -> 2) Watt meter -> 3) Induction cooker -> a flat coil on top of it, connected to a set of bulbs.
    To this, we add -> an additional flat coil, which is also connected to the bulbs (hopefully to add power to it, without draining more power from the outlet).

    My quizz question to you is what do you think will happen in terms of efficiency:
    a) Power drawn by the entire system will increase, as the added top coil is subject to termodynamics losses, even though it might increase power output (fed to the bulbs) a bit. Basically, efficiency will DROP, because we add another coil, which has its own losses, added to the total losses (related to impedence) of the system.
    b) Same input and same output. Overall efficiency remains just about the same (to the third digit or so).
    c) we will GAIN efficiency, because we add a coil which is powered by the inductive coupling of the flat coil laying underneath it, which itself is powered by the induction of the cooker. A good use of inductive coupling.

    Looking forward to your input.
  2. jcsd
  3. Sep 16, 2015 #2


    Staff: Mentor

    A diagram would help us visualize what you are talking about.

    Remember that coupled coils are merely transformers.
  4. Sep 16, 2015 #3
    The entire system is 100% efficient. The use of incandescent lights indicates you are trying to convert electricity into heat. Every component you've described does this. Even the cooling fan stirs the air ultimately causing the eddy currents (of air) to heat the surroundings.

    More pedantically, efficiency is measured as a percent. So whether the third coil adds or subtracts depends on whether it is more or less efficient than the second coil. Typically it would be less efficient, but it could easily go either way.

    BTW, lots of Tesla's work had to do with resonant energy transfer, like the induction stove. Induction furnaces operate over a narrow frequency range. Matching frequencies is an important consideration in the circuit you described.
  5. Sep 16, 2015 #4


    User Avatar

    Staff: Mentor

    I'm not convinced you have measured the power input correctly: an induction cooker is unlikely to have a unity power factor, so you likely cannot get watts by multiplying volts and amps.

    What kind of watt meter is it?
  6. Sep 17, 2015 #5
    Hello!Thanks for your reply. You're right a diagram is better. I did one by hand - the attached JPEG. Hope it is clear enough.
    Looking forward to what you say.Best regards, Teslafanclub.

    Click below here:

    Attached Files:

  7. Sep 17, 2015 #6


    User Avatar
    Gold Member

    Not being very knowledgeable about induction cookers, I don't know what is cooker and what is measurement devices. My initial impression was that the incandescent loads were a measurement method and the coils are part of the cooker, but then again......?.

    All the designs I looked at were a single induction cooking coil, and then a simulated load coil.

    Why are you adding the second/third coil where you would normally put a cooking load?
  8. Sep 17, 2015 #7
    Hello Russ!
    Lovely photo of you with that gorgeous telescope!
    Okay, I am happy to confirm we have a 1.00 power factor with the cooker ON, the first coil resting nicely on top, and that coil connected to the load of about 2800W of light bulbs. On the next trial, we also got a 1.00 power factor with the addition of the second coil, connected in parallel to the first coil. My watts meter is Hampton Type 46-0110 125v and 15A resistive at 60Hz. As you probably know, transformers connected to a load can express a power factor, whereas you will not have a value approaching 1 when the transformers are not under load. You just get gibberish. When it's all powered up nicely, the Hampton says 1. So no need for cos phi. You had a good point. So, what do you think?

    Attached Files:

  9. Sep 17, 2015 #8
    Hello MeBigGuy!
    I should have joined a diagram. I added one for you to this reply. The cooker is the source of power. The flat coil No1 and No2 are powered by induction coupling performed by the internal (also flat) coil of the cooker. The idea was to find out if we could increase the power efficiency of a simple induction cooker, by adding a flat coil on top, and then adding yet another one on top of it (like mama's saturday morning's pancakes). The measure of efficiency is power out divided by power in. You "draw" 2000Watts from the mains, you use 1000W at the load, and you have a nightmare because it's only 50% efficient - and it's embarrassing because you cannot account for the lost sheeps! They usually turn up somewhere in your system as unwanted heat. So, the experiment can only go three ways: Better efficiency, just abour the same, worse. What do you think would happen? Thanks for you input.

    Attached Files:

  10. Sep 17, 2015 #9


    Staff: Mentor

    I didn't fully realize what you did until I saw your diagram. It is novel and it made me think. But it also sounds dangerous. You sound like you know what your doing, but you also reported a 1.0 PF which is not believable so I have to question if you are fully aware of what you're doing. More important, other people in the future might be tempted to try what you did, and it could lead to injury or death. Therefore, as soon as I post this reply, I will report this tread to the moderator as breaking PF rules about dangerous topics. Hopefully, he will close the thread.

    Now to your question. At first I was going to direct you to the equivalent circuit for a three-winding transformer. But then I noticed that you have the two secondary windings connected in parallel on the load side. That is highly unusual; what does it represent. I had to scratch my head a bit.

    Consider taking a twisted pair of wires, shorted to each other on each end. We then have electrically a single conductor. Wind that twisted pair wire to make the secondary of a transformer. Is there any difference between that an a secondary wound with ordinary wire? (Ignore internal resistance). The answer is no. You have an ordinary two-winding transformer.

    Now start untwisting the pair and separating the halves by a bit leaving the ends shorted. Use them to make transformer secondaries. Now the amound of flux linking the two halves will be unequal. In the most extreme case, all the flux will link with the second coil and none with the third coil. In that case you have a two-winding transformer with an inductor connected across the secondary terminals in parallel with the load.

    Your acual case is somewhere between these two extremes. Exactly where would require detailed 3D modelling. Neither I nor anyone else can tell you the answer based on the info provided.

    Congratulations on posing a novel question. But, I urge you to stop this experiment before you get hurt.
  11. Sep 17, 2015 #10


    Staff: Mentor

    Thread closed for moderation.
  12. Sep 17, 2015 #11


    User Avatar

    Staff: Mentor

    Because of the dangers involved for the OP and others, this thread will remain closed.
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