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Why does reactive power use transmission line capacity?

Does reactive power take part of the cables power transfer capacity and if yes why?

I know reactive power cannot possibly heat up cables because that would mean that reactive power produces energy in the form of heat for the cables and then the cables will use energy from reactive power but reactive power cannot give away real energy like active power.

Then I have two questions:

1) How does it work exactly that a certain amount of reactive power limits the capacity for active power of a transformer ?

2) How does it work exactly that a certain amount of reactive power limits the capacity of a cable or transmission line to transmit active power?

As far as I know reactive power cannot do Work (J) and cannot give away or take Energy (kWh).


Okay. I received some nice explanations.

Basically you said that reactive current can still heat up conductor because i^2.R losses. Please tell me:

Does this mean that part of the reactive power is lost (dissipated in heat in the conductor) on its way to and back from source to sink ?

That would mean that while reactive power is circulating .. the power going out of source is not equal to the power going back to source, because part of the power went in the direction of heating the conductors ?

Does this mean that reactive power can produce real work (J) in the process of heating the conductor? I am lost.

Or does it mean that reactive current is no different than active current when it comes to facing the resistance?

2 Answers

  • 6 years ago
    Favorite Answer

    Generators and transformers are rated in both KW and KVA. The KVA indicates the maximum current each is capable of producing, while the KW indicates the maximum 'work' (real power) each can supply.

    An example to answer your question would be as follows.

    A generator (or transformer) rated at 1000 KVA at 0.8 power factor is capable of supplying only 800 KW at the stated power factor,

    Now suppose you have a load of 500KW with a power factor of 0.5, the KVA requirement for this load will be KVA = KW / PF or KVA = 500 / 0.5 = 1000. This indicates the 500 KW load uses all the KVA the generator or transformer can safely produce. because of the poor power factor (caused by excessive reactive power). The complete system is thus limited to 500 KW on a theoretical 800 KW system. Incidentally, reactive power is not registered on your watt-hr meter.

    In the above example system, the capacity of the system can be recovered by installing capacitors in the grid at either (or both) the supply end or the user end of the grid.

    I assume your additional questions are directed to David. I will not answer these questions.


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  • David
    Lv 6
    6 years ago

    Reactive power is a result of the current and voltage being out of phase. But the current is still there. You can still have i²R losses in the transmission lines and other wires (which will heat them up) regarless of the phase of the current.

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