Voltage Regulation and Loading

[redwan hasan]

Good evening everyone.

I have two questions to ask you guys.

1. Why voltage regulation of a synchronous motor is negative with a purely capacitive load? I mean why?

2. Why Reactive loss is higher than resistive loss in a power distribution line?

thank you and have a nice day! Really appreciate any help.

 

[psparky]

Good evening everyone.

I have two questions to ask you guys.

1. Why voltage regulation of a synchronous motor is negative with a purely capacitive load? I mean why?

2. Why Reactive loss is higher than resistive loss in a power distribution line?

thank you and have a nice day! Really appreciate any help.

Not sure about your 1st question, but I'll take a shot at number 2.

The majority of your reactive losses come from inductive motors. A lot of it gets power factor corrected, a lot of it does not get power factor corrected.

There are certainly resistive losses in the power lines, but since copper or aluminum conductors are very good conductors, the total loss here does not add up the the reactive loss. (The I^2*R losses are not as big as the reactive losses). Reactive losses can be as much as 30% or in extreme cases 50% or more in motors. Just think of the millions of air conditioners running...or the fans running in furnaces in the winter, not to mention the tons of giant motors running in factories.

In factories they are likely to get power factor up to .9. They can't do perfect power factor of 1 because of the harmful harmonics it introduces. In residential, there is very little power factor correction, which also adds a lot of reactive losses.

KVA is a combination of both real power (watts) and reactive power (vars).

If you measure the amps thru a line, multiply this by the volts ( X 1.73 if three phase)... you get the KVA of your motor. If you measure the actual horsepower at the shaft of the motor, this can be converted into watts. Take these watts and divide by the KVA. This is your power factor.

In the power triangle, the bottom horizontal part is the watts. The vertical Y part is the VARS. And the hypotenuse between the two is KVA.

This looks like a pretty good article on the subject. Hopefully it agrees with what I said above!
http://electrical-engineering-portal.com/total-losses-in-power-distribution-and-transmission-lines-1

 

[donpacino]

to answer #1. Motors typically have a mostly inductive quality.This is countered with the capacitor.

 

[redwan hasan]

Not sure about your 1st question, but I'll take a shot at number 2.

The majority of your reactive losses come from inductive motors. A lot of it gets power factor corrected, a lot of it does not get power factor corrected.

There are certainly resistive losses in the power lines, but since copper or aluminum conductors are very good conductors, the total loss here does not add up the the reactive loss. (The I^2*R losses are not as big as the reactive losses). Reactive losses can be as much as 30% or in extreme cases 50% or more in motors. Just think of the millions of air conditioners running...or the fans running in furnaces in the winter, not to mention the tons of giant motors running in factories.

In factories they are likely to get power factor up to .9. They can't do perfect power factor of 1 because of the harmful harmonics it introduces. In residential, there is very little power factor correction, which also adds a lot of reactive losses.

KVA is a combination of both real power (watts) and reactive power (vars).

If you measure the amps thru a line, multiply this by the volts ( X 1.73 if three phase)... you get the KVA of your motor. If you measure the actual horsepower at the shaft of the motor, this can be converted into watts. Take these watts and divide by the KVA. This is your power factor.

In the power triangle, the bottom horizontal part is the watts. The vertical Y part is the VARS. And the hypotenuse between the two is KVA.

This looks like a pretty good article on the subject. Hopefully it agrees with what I said above!
http://electrical-engineering-portal.com/total-losses-in-power-distribution-and-transmission-lines-1

1st of all thanks for your input.
I also thought of the BOLD part but is that all?
so the main reason is the Rs is much much lower than Xs that's why the resistive loss is much lower than inductive loss?

well during experiment it was seen that with transformer and alternator, if we connect purely capacitive load we can see the no load voltage is lower than loaded voltage. why is that?

anyway appreciate ur writing. and hey we do correct power factor in our home. at least my PC power supply has active power factor correction system that keeps it at .99 all the time.

 

[redwan hasan]

to answer #1. Motors typically have a mostly inductive quality.This is countered with the capacitor.

thank u too. never thought in this way. so generators and transformers both are inductive in nature that's why with capacitive load their voltage goes higher?

can u explain e little more?
it would be really helpful!

 

[psparky]

1st of all thanks for your input.
I also thought of the BOLD part but is that all?
so the main reason is the Rs is much much lower than Xs that's why the resistive loss is much lower than inductive loss?

I think the Rs loss happens along the power lines. The Xs is essentially lost in the magnetic field of the motor. If you were talking exclusively the loss the the power lines, I believe the Rs loss is bigger than the Xs loss. But once u add in the Xs loss of the magnetic field, the Rx does outweigh the Rs loss. It's the actualy amps in the power lines that you get the I^2*R loss. The amps are from the KVA. (or probably better to say the other way around) The KVA can be divided into x and y components, but still the Rs is bigger than Xs thru lines.

well during experiment it was seen that with transformer and alternator, if we connect purely capacitive load we can see the no load voltage is lower than loaded voltage. why is that?

Hmmmm, well your transformer is essentially a bunch of coiled wire and your alternator is a bunch of coiled wire so you have two highly inductive devices. You then counter it with the exact opposite capacitve load. So you can clearly see that...(insert your explanation here)

anyway appreciate ur writing. and hey we do correct power factor in our home. at least my PC power supply has active power factor correction system that keeps it at .99 all the time.

There certainly are some power factor correcting devices, but certainly not all!

I'm not a 100% sure of all my explanations above, but until someone possibly corrects it, it's all we have at the moment.

 

[redwan hasan]

are you sure? cause what i think is in transmission line Xs is larger than Rs because Rs depends only on the length, material and temperature but there are a lot of magnetic loss like loss due to skin effect and capacitance,impedance etc. ok i guess. i will keep digging though.

It helped thanks!

 

[psparky]

are you sure? cause what i think is in transmission line Xs is larger than Rs because Rs depends only on the length, material and temperature but there are a lot of magnetic loss like loss due to skin effect and capacitance,impedance etc.


ok i guess. i will keep digging though.

It helped thanks!

Not sure. Consider my info food for thought.

You tell us after you figure it all out.

 

[psparky]

Here's one more piece for thought.

Let's say the entire world only had light bulbs. The only inductive machines now are transformers feeding these lights from the power plants. (which also leads to more thought above, clearly the transformers add to reactive losses.)

Would the reactive losses still outweigh the resistive losses? Why or why not?

 

[sophiecentaur]

I think it would depend upon the voltages used for transmission and the details of the generator used. The 'reactive' losses you mention are not inherent losses but will depend upon the source resistance of the generator. A reactance cannot dissipate power 'per se', can it? But it can, for instance, cause excess current to flow in a resistor, which is a loss mechanism.

 

[redwan hasan]

Here's one more piece for thought.

Let's say the entire world only had light bulbs. The only inductive machines now are transformers feeding these lights from the power plants. (which also leads to more thought above, clearly the transformers add to reactive losses.)

Would the reactive losses still outweigh the resistive losses? Why or why not?

i guess it will still be larger. transmission lines will loose due to skin effect, capacitive loss, inductive loss and all the reactive should be greater than the resistive loss.

 

[redwan hasan]

Not sure. Consider my info food for thought.

You tell us after you figure it all out.

i'm trying. :S
i will concentrate more on this after figuring out some stupid topic of quantum physx and signals.

 

[redwan hasan]

I think it would depend upon the voltages used for transmission and the details of the generator used. The 'reactive' losses you mention are not inherent losses but will depend upon the source resistance of the generator. A reactance cannot dissipate power 'per se', can it? But it can, for instance, cause excess current to flow in a resistor, which is a loss mechanism.

i do know that reactive losses do not contribute to power dissipation but what do u mean by the excess current flow? what i know is current flow through the resistor will depend on the voltage across the resistor.

 

[Babadag]

In my opinion the "loss" is only "active”. There are no reactive losses actually. The increased reactive power will increase the current and so the "active" losses will go up. The reactance depends on the distance between phase conductors. In a three phase cable the distance between phases is small then the reactance is small too. In an overhead line the distance is big [relatively] then the reactance is elevated. If it is a small conductor cross section area the reactance of a cable is too small then may be neglected with respect the resistance.
By the way both cable and overhead line present a capacitance also. It is not a series capacitance as the inductance but could influence the total reactance.
The reactive power "consumed" in a receiver depends on current position with respect the supplied voltage. If the current lag the voltage as in all receivers connected with a magnetic field as coils and induction motor windings the reactive power is considered "negative" but in a capacitor where the reactive power is connected with the electric field the current leads the voltage and this reactive power is considered "positive".
The synchronous generator supplies the current and the voltage as required by the "consumers" the active power and the reactive power too. So, nevertheless the generator reactive power is generated by means of a magnetic field the current supplied "change" the sense and leads the voltage. If a capacitor- bank -will overcompensate the induction motor reactive power then the required reactive power will be "positive" and the generator has to change the supplied “negative “reactive energy. The excitation current controls the delivered reactive power. The active power will be provided by primary driving unit-turbine or Diesel Motor, usually.