RL load and power factor

Discussion in 'General Electronics Chat' started by almotions, Jul 9, 2009.

  1. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    Hi all,
    i have a slight confusion here. Assuming there is a AC source supplying an RL load and drawing certain amount of current. When people say that increasing load,what do they mean besides the current drawn increases? What are they actually increasing ? R or L? Will they power factor change when load increases? PLease advice.Thanks.

    Regards,
    alvin
     
  2. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    Hi Alberto,
    i'm so sorry,i read through the link but i still can't figure out your explanations.Could you explain in more detail .Thanks.

    Regards,
    alvin
     
  3. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
    760
    When dealing with loads in AC Power supplies you need to know 2 things
    1. Resistive Load
    2. Reactive Load

    In resistive load - An increase in load means a decrease in resistance, causes an increase in current and power. Resistive load has a PF of 1 or unity
    A Reactive load also known as the Capacitive or Inductive. These introduces an opposition to the Power that is uses to drive it. Which results in the current leading or lagging creating the Power factor thingy.
    In AC there is this thing called apparent power which is virtual and is not real power.
    It mainly consists of real power and lost power during energy consumption.

    So to say Reactive loads introduces huge loss in AC and PF is used to correct this to increase efficiency

    Rifaa
     
  4. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    Hi Rifaa,
    thanks for your explanations.However,i must say i'm still confused. According to Alberto, if load increase L will decrease and power factor will increase .i know that for a RL load(Ie:an induction motor) ,the current is always lagging the voltage by theta.What i'm still confuse is if when we say load increases which means R and L decreases proportionally,why does the power factor change? Shoudn't it be still the same since the (VL/VR) ratio still remains the same? Thanks.

    Regards,
    alvin
     
  5. mindmapper

    Active Member

    Aug 17, 2008
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    The resistive resistans R is usually fixed. Depends on resistans in the wire, which doesn't alter much. It change a little with temperature. So it is the inductive (reactive) part that change the most. Because of that, R and L doesn't change proportionally.
     
  6. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    Just out of a question for a moment.Assuming a AC source supplying a RL load through a line impedence.Initially the current is lagging the terminal voltage by theta. After installing a perfect capacitor which provides a PF of 1,the current is reduced and in phase with the terminal voltage. Assuming the the source voltage is a constant supply,shoudn't the terminal voltage across the RL load increase when current decrease ? Because i see a lot of graphs from papers showing that the terminal voltage doesn't seem to change after power factor correction. And if i plot a rough phasor diagram,it seems that terminal voltage have to increase a lot in order of that after adding the line impedence drop will be equivalent to the source voltage. Please advice.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi Almotions,

    Regarding "terminal voltage" regulation.

    Depends what you mean by terminal voltage and in what context.

    In the simplest case the terminal or source voltage is often assumed to have zero internal impedance / resistance. In that situation the terminal or source voltage is fixed, irrespective of the applied load.

    In the analysis of (large) power systems, the source or terminal may often be taken as the point of connection to an "infinite bus". Voltage and frequency at an infinite bus are (by convention) assumed to be unaffected by the local load conditions. This is probably the underlying assumption for the various papers you may have seen in which terminal voltage is unchanged by load variations.
     
  8. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    If your RL load is an induction motor;

    at full speed, no load, your rotor is travelling along with the stator flux. Your reactance, based on frequency difference, is lowest. Current is lowest and pf is close to unity.

    Now you begin to add load. Your rotor begins to slip in relation to the stator flux and the difference in frequency causes your reactance to increase, which in turn increases your apparent power and current, resulting in a lower pf.

    The resistive, or inphase current of the motor will peak at about 75% full speed, due to voltage differences within. It is approx at this point where inphase and out of phase currents are equal, and the peak torque of the motor delivered. Increaed load/slippage beyond that has the out of phase current exceeding the in phase, and the motor will stall.

    At stall, or 100% slippage, you will have the max frequency difference between rotor and stator, which in turn increases your reactance to max, your current to max, your pf to it's lowest.
     
    Last edited: Jul 13, 2009
  9. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    Hi Tnk,
    my context of terminal voltage would be the voltage across the load or at the point of common coupling(PCC).I understood your explanations but there is one further question to ask.When we talk about power factor correction for distribution lines with varying loads,we are not talking about regulation voltage at PCC but just to reduce the reactive power drawn from the source right? When we talk about voltage regulation ,we are talking about solving voltage swell/sag or flickers which are not caused by PF issues right? Or is there a link between both?

    HI GetDevice Info,
    Sorry for asking a stupid question:p When you said at no load,the reactance is the lowest,how come the current is the lowest since I=V/Z and vice versa(I'm assuming R is fixed).

    Thanks both so much!
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I believe power factor correction is primarily used to reduce the losses and costs of transmitting reactive current beyond the PCC into the consumer's network. But not all consumers will have a good 'solid bus' at their point of connection. For instance, PFC could also be advantageous for a consumer sitting at the end of long feed line and having significant power factor issues within their installation - e.g. with a high use of induction motors. They may have to pay an unnecessary penalty in $ cost and possible line voltage regulation issues for an uncompensated low (<< unity PF) power factor load.

    Power system operators probably encourage consumers to adopt PFC to reduce their own network / transmission losses and infrastrsuture rating limits.

    I think you are correct about the issues of voltage swell/sag etc. - this may have more to do with power system stability and dynamics (such as high load perturbations). PFC is probably more a (quasi) steady - state consideration, but it may well have an impact on overall system dynamics. I'm not an expert in these matters.
     
  11. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    because the rotors voltage is also it's lowest. At no slip there is no lines of magnetic flux being cut, hence no induced voltage into the rotor. This is theorectical of course as some slip always takes place to overcome internal losses.
     
  12. almotions

    Thread Starter Active Member

    Feb 6, 2009
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  13. almotions

    Thread Starter Active Member

    Feb 6, 2009
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    hmm.i think i need to start reading on motors.Oh my,it seems there are just endless stuffs to read.zzzz...
    Thanks anyways guys:)
     
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