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Generator's under load?

  1. Aug 27, 2012 #1
    I know that in general generators under-load will eventually slow down because of the load...

    Question is... When I have a small generator that produces 100W and a load that is 80 W and mechanical Input rated in (W) about 110W. Due to losses the generator can only generate 100W of electricity from the mechanical input.
    Would that load that is rated at 80W slow the generator significantly?

    I'm confused about this point... When I run a generator with mechanical energy rated at 110W and the generator's generating 100W... And the load is 80W does that mean I have to increase mechanical energy?

    When should I consider to increase mechanical energy so that the generator would generate more?

    The reason why I'm limiting the load to 80W is because... I think that the generator can handle that load and would not need more mechanical input?

    Based on my understanding of the laws of conservation my system set up should be fine...


  2. jcsd
  3. Aug 29, 2012 #2


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    They do? I don't understand this statement. :confused: Why do you think it can't be kept spinning at the right rotational speed?
    The generator is producing 100W of electrical output? The load is taking 80W, so where is the other 20W of electrical energy goinghttps://www.physicsforums.com/images/icons/icon5.gif [Broken] It can't vanish without trace!

    The way a generator works is that the mechanical energy needed to drive it equals the electrical energy it is outputting + the losses in the machine. If it is not powering any electrical load, then the generator is easy to turn and the prime mover has to overcome only mechanical losses.

    Does that clear up a misunderstanding? :wink:
    Last edited by a moderator: May 6, 2017
  4. Sep 1, 2012 #3
    Pretty much.

    As you stated "If it is not powering any electrical load, then the generator is easy to turn" That indicates that if load is connected it would cause rotational difficulty... In that case do I have to increase the mechanical input?
    Last edited by a moderator: May 6, 2017
  5. Sep 1, 2012 #4


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    Yes: mechanical input must always be greater than electrical output. NascentOxygen said that.
  6. Sep 4, 2012 #5
    Not my question.

    What Im trying to say is that under-load do I need to increase mechanical input?
    The electrical output is generated by the mechanical input and there is a load connected to it... Would that slow the generator? Or even make it difficult to rotate?

    That question is not answered yet. NascentOxygen mentioned difficulty...
  7. Sep 4, 2012 #6
    Why is it that without load the generator spins easily and with load it's more difficult to spin? Does that mean when its becoming more difficult to spin the mechanical input has to be increased?
  8. Sep 4, 2012 #7


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    When a generator has electrical current flowing in its windings this generates a magnetic field which reacts with the field coil's magnetic field to make it harder to turn the generator

    To maintain constant speed, it is necessary to use more force to rotate the generator. So, yes, you need more mechanical power input.

    However, the generator does not always produce 100 watts (in your case) just because it is capable of producing this power. If there were no load, it would not produce any output power at all, although it would still take some mechanical power to rotate it.

    The power it produces depends on the output voltage and the resistance of the load. (Power = E2 / R. )
    If the load was 80 watts, it might take 95 watts of input mechanical power to maintain a constant speed.

    The remaining 15 watts (95 watts - 80 watts) would be lost due to the power lost in the generator due to friction, the resistance of the wires and brushes and the power needed to power the field coils.
  9. Sep 4, 2012 #8
    Regardless of whatever system, when things get difficult, you have to increase effort, otherwise things would stop working.
  10. Sep 4, 2012 #9


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    Increase it from what? From what it is when not under load? Yes: mechanical input must always be equal (ideal) or greater (real) than electrical output. So when it is not under load, mechanical input is zero and when it is under load, the mechanical input increases to remain equal to the electrical output.

    Yes, it really was the same question.
    Yes, it really is the same question. I'm not sure why it isn't sinking in.

    Lets try with me asking the questions:

    1. If the electrical load on an ideal generator is 0kW, what is the mechanical input in kW?
    2. If the electrical load on an ideal generator is 1kW, what is the mechanical input in kW?
    3. If the electrical load on an ideal generator is 2kW, what is the mechanical input in kW?

    Then back to the original question:
    If the load is 80W then the generator is generating 80W, not 100W. That's what "load" means. So the load is 80W and the input is 80+ inefficiency or in this case about 90W.
    Last edited: Sep 4, 2012
  11. Sep 4, 2012 #10
    I'll just ask simple questions to avoid all of this complication:

    1. Is it because of len'z law that causes the difficulty in mechanical rotation?

    2. Since there is a mechanical difficulty and you're saying: "mechanical input must always be equal (ideal) or greater (real) than electrical output." The truth is, its not = to the electrical output due to losses and when a load is applied I have to increase the mechanical input "EXAMPLE" 1000J Is applied to a generator, its generating 1000J of electrical power(Obviously this is theoretical in reality we would have to include losses) there is no load consuming that 1000J.

    Now as soon as the load is attached there is a MAJOR resistance to the mechanical input. It would slow down the generator sure, but would the generator STILL GENERATE 1000J? Or will it gradually decrease? If it would decrease(Which I believe it does.) so more mechanical input has to be added, if it was 1000J maybe it has to be increased by 100,200,300,500J, etc... To rotate smoothly.

    I do apologize for not clearing out things. But now, I believe its clear enough.

    Another analogy:

    A steam engine rotates at 3000RPM without load, as soon as the load is connected it drops so more steam has to be applied to make it reach 3000RPM once more.

  12. Sep 4, 2012 #11


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    There's a little experiment my college physics prof did to start off a lesson on Newton's laws: He took the largest guy and smallest girl in the class and put them on skateboards, facing each other. He told them both to push on the other as hard as they could, then asked the class which one pushed on the other harder.

    The answer? They both pushed on each other with the same force. They have to. Forces come in pairs.

    So too with conservation of energy. You cannot apply 1000J to an ideal generator and not have a load absorb it. You can't say 1000J is being generated, but say that it isn't going anywhere. The energy has to go somewhere. One will command the other, but both will end up equal.

    What happens with a generator not connected to a load is you find there is very little torque required to spin it. It will therefore neither generate nor absorb much energy.

    Take, for example, a car engine in neutral. If you push the gas pedal down and there is no rev limiter, with the car in neutral, there is nothing applying a torque against the motor. So the engine just keeps spinning faster and faster until it destroys itself unless internal losses end up absorbing the power.

    Take, for example, a windmill: If a windmill generator has a problem and stops generating power, the windmill may just start spinning faster and faster until it destroys itself because there is no torque applied against the rotation to keep it spinning at a constant rate.
    Due to the above error, this is just a wrongly worded question. You can't say a generator is producing 1000J (btw, it would really have to be Watts, but anyway....) that aren't going anywhere and then you apply a 1000J load to it. Either it was producing 1000J before and is now producing 2000J or it was producing nothing and now 1000J or it was and still is producing 1000J.

    The question has to be fixed before the answer can be meaningful.
    That's all fine, but you aren't saying anything clear about what is being generated versus the load in that example. You haven't fully developed the example.
    Last edited: Sep 4, 2012
  13. Sep 4, 2012 #12


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    BTW, you referred to "the generator" in the OP in a way that implied you have an actual generator you're experimenting with. If you actually have a real generator, you can just read the specs off the nameplate or manufacturer's spec sheet to get clearer and more relevant answers to this!
  14. Sep 5, 2012 #13
    Interesting! I thought in some cases if the load was 80W I might need to increase the mechanical load significantly. But based on you're post if the resistance was decreased to the lowest extent. Its possibile to have 95W mechanical input and 80W load and the system would work fine! Thank you!

    True, but to what extent do you have to increase effort. You could increase a little bit and you could increase it significantly thats what I wanted to know, thanks though!
  15. Sep 5, 2012 #14
    I'd rather understand the physics of it perfectly. If you'd like to just read and follow go ahead.
    But I'd like to read,test,see, etc... To understand how things work.
  16. Sep 5, 2012 #15
    Fair enough. Thanks anyway,
  17. Sep 5, 2012 #16


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    Do you understand now?
  18. Sep 5, 2012 #17


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    just a little note - you were correct when you said does lenz' law bring about the difficulty to rotate a generator under load, the answer is yes, it does.

    Imagine a generator being rotated by some prime mover. Assume for simplification, a rotating field, stationary armature. So that dc is applied to the field to prime it, then with the rotation, that dc static field cuts the armature and induced voltage across it.

    Now, there is no current flow with no load (even though the potential exists) and the generator is easy to rotate. Imagine now connecting a resistance across the generator. Now, current flows in the armature windings to the external load. What happens physically for the generator to slow down, and for the prime mover to have to work output more, is---> as the resistance was added, a current flowed in the armature winding. As you already guesses, the current flows in the opposite direction of the field current, meaning the field oppose one another. And the more resistances we add (in parallel) the lower the overall resistance, and the more current flows. meaning, the stronger the magnetic field in the armature.

    For your interest, this is also the same effect as why dynamic braking works. We use the motor as a generator, utilizing it's counter EMF, and connecting load resistors across it. This opposition of the two magnetic fields induces a braking effect. So you can see it is true for all electrical rotating machines
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