Magnetizing current consumes no power. Reactive power is not real. However, the current in the wires is real. As I understand it, the reactive power DOES actually consume energy each cycle but it is like a spring. It compresses/decompresses/compresses, etc. providing the energy for the next cycle. The reason a low PF is bad is because the cables that supply the motor, or the cables that supply your residential area, have a curernt capacity associated with them, and thus when a low PF exists, the current in the wires is higher than it needs to be... which results in heat dissipation losses. THEREFORE, the magnetizing current DOES actually result in energy consumption, just not in the intended load. Heat dissipation is real. Energy cannot be created or destroyed. It can just change forms... This is why you get charged a lot for buildings that have low PFs because the utility company doesn't care and is simply charging you for the total energy consumed, both in the distribution lines and in the loads in your building. Is this correct logic!? Please, I am going crazy trying to comprehend this.
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Power Factor - Power Consumption 2
- Thread starter jdogg05
- Start date
mikekilroy
Electrical
said another way, say u put your pf caps 10 feet away from this motor. nameplate is 100amps, no load say it shows 40 amps on clamp on ammeter, friction to rotate is 1 amp say. you will have 40 amps on those 10 feet of wires out to the pf caps, then the rest of the wires from there on back to the power company will have only the real current of 1 amp on them. this imaginary current is only circulating back and forth.
- Thread starter
- #22
Ok yes that is exactly what I was trying to get at, thank you. So, if we know that the reactive current is 40 amps and full load current is 100, and we have a capacitor at the motor, then the wiring from the the motor back to the power supply don't need to be spec'd for any larger than 60 amps... right? Anything else would be overkill.
I looked up the circuit for a PF correction capacitor and a motor. It shows the capacitor in parallel with the motor. So I guess the power supply wires will only carry 1 amp of current and the wires connecting the capacitor will carry 39 amps (for the no load case).
I looked up the circuit for a PF correction capacitor and a motor. It shows the capacitor in parallel with the motor. So I guess the power supply wires will only carry 1 amp of current and the wires connecting the capacitor will carry 39 amps (for the no load case).
electricpete
Electrical
Maybe a nitpick, but quantitatively important:
If motor was drawing FLA and you had 100% correction (you shoudln't have that much if caps are switched with motor) then the real component supplied upstream of capacitors would be sqrt(100^2 - 40^2) = 91.6A
=====================================
(2B)+(2B)' ?
- Thread starter
- #24
So in this case, at full load, the 100% PF correction only reduces the real supply current by 8.4%. BUT, this doesn't actually change the PF (it just corrects it) because the apparent current is still 100 amps and the real current is still 91.6. The power factor is 0.916 in both cases. If there were no correction, the real current would still be 91.6 and the apparent 40; however, the current in the supply lines would then be 100 amps.
Please god tell me this is right...
Please god tell me this is right...
- Moderator
- #25
A simple example of why power companies are justified in charging penalties for poor power factor:
A 1 kW motor at 100% power factor will require 1 KVA transformer capacity.
A 1 kW motor at 50% power factor will require 2 KVA transformer capacity.
If you don't like it, correct your power factor. ROIs are often less than one year.
Bill
--------------------
"Why not the best?"
Jimmy Carter
A 1 kW motor at 100% power factor will require 1 KVA transformer capacity.
A 1 kW motor at 50% power factor will require 2 KVA transformer capacity.
If you don't like it, correct your power factor. ROIs are often less than one year.
Bill
--------------------
"Why not the best?"
Jimmy Carter
mikekilroy
Electrical
Jdogg05 you about got it! couple last comments since you are ready 
remember that the reactive current is exactly 90 degrees out of phase with the real current (ELI the ICEman). so the math is like finding the lengths of the sides of a triangle not algebraic addition and subtraction; hence the sum of the squares.
when a motor is loaded, the reactive current goes up due to leakage inductance inside so the 40 amps no load in my example is probably closer to 50 or 55amps at full load; so the math shows probably sqrt[100^2-55^2] or 83amps real for a more realistic .83 pf....
remember that the reactive current is exactly 90 degrees out of phase with the real current (ELI the ICEman). so the math is like finding the lengths of the sides of a triangle not algebraic addition and subtraction; hence the sum of the squares.
when a motor is loaded, the reactive current goes up due to leakage inductance inside so the 40 amps no load in my example is probably closer to 50 or 55amps at full load; so the math shows probably sqrt[100^2-55^2] or 83amps real for a more realistic .83 pf....
LionelHutz
Electrical
jdogg05 said:
Just to be clear something up, Your local electrical code will tell you that you can never size the feeder wire to a motor by using the corrected current. There are good reasons why sizing the feeder cable to match the motor rated current is not overkill, such as the capacitor failing and being able to start the motor without an excessive voltage drop.
- Thread starter
- #28
FWIW, real world power factors are typically 80% plus. Generators are sold like that for ex., 100 KVA @ 80% pf = 80 KW. For large consumers, if the electric company is able to bill by load PF (most are not, my utility for ex., does not) they would love to make you pay more when your PF falls below 80 or 90%, whatever they can get the regulating agency to approve. So examples of 0% or 10% PF are just theoretical examples, not normally seen in the real world.
-
1
- #30
1capybara just wrote:
So examples of 0% or 10% PF are just theoretical examples, not normally seen in the real world.
To which I respond: 'It depends on your world...'
In my working world, we use capacitors to 'supply' the lagging vars drawn by the predominantly inductive loads connected to the grid. What we call 'low-voltage' caps [meaning with a phase-to-phase voltage of 14, 28 or 44 kV] typically have a rating of between 10 and 30 MX. Our directly-connected 'high-voltage' caps [either 115 or 230 kV] range from 96 up to a whopping 410 MX output.
Most of the HV caps and a number of the LV ones have smoothing reactors installed in series with their connection to the grid to dampen that initial 'kick' when they're placed in service. Alternatively, depending on the location of the installation and the electrical characteristics prevailing there, independent pole operation breakers may be utilized to very precisely time the individual contact closure to co-ordinate with the null or zero-crossing point in the waveform so as to facilitate smooth insertion.
On the opposite side of the fence, we also use reactors to absorb VARs from the system, particularly in association with 500 kV circuits which generate great amounts of 'lagging' VARs when lightly loaded...
But I digress.
Capacitors have almost a perfectly leading power factor; reactor have very close to a lagging power factor. The switching duty is severe, particularly for cap switching, and sulphur-hexafluoride-insulated [SF6] breakers are gradually replacing the older oil circuit breakers in LV applications.
A factor that often bears on power system operation is the need to transport large amounts of power from where it's generated to where it's used; and as an earlier poster alluded to, peak times are the worst. During these periods especially, proper deployment of reactive resources can be used to optimize the delivery limit of a circuit which might otherwise be constrained due to either thermal or stability limits. [We have also recently commissioned series capacitor installations to improve the ratings of some of our 500 kV circuits.]
A final note: capacitor output is all or nothing, and varies as the square of the applied voltage. Reactors have a similar characteristic, but since it's based on frequency and since there is typically far less variation in frequency than in voltage profile, VAR consumption by reactors is almost pegged...
The latest thing is to use Static VAR Compensators [SVCs] which consist of a combination of reactors and thyristor-switched capacitors; SVC's can thus have a soothly adjustable 'lagging' reactive buck / boost function, for example, from 40 MX 'in' to 60 MX 'out.' As a power system operator, I consider these the cat's miaow; they're almost as good as a synchronous condenser...
Autotransformer tertiary windings often prove a very handy place to hook these gizmos up.
Hope this was at least interesting, even if it didn't help...
CR
So examples of 0% or 10% PF are just theoretical examples, not normally seen in the real world.
To which I respond: 'It depends on your world...'
In my working world, we use capacitors to 'supply' the lagging vars drawn by the predominantly inductive loads connected to the grid. What we call 'low-voltage' caps [meaning with a phase-to-phase voltage of 14, 28 or 44 kV] typically have a rating of between 10 and 30 MX. Our directly-connected 'high-voltage' caps [either 115 or 230 kV] range from 96 up to a whopping 410 MX output.
Most of the HV caps and a number of the LV ones have smoothing reactors installed in series with their connection to the grid to dampen that initial 'kick' when they're placed in service. Alternatively, depending on the location of the installation and the electrical characteristics prevailing there, independent pole operation breakers may be utilized to very precisely time the individual contact closure to co-ordinate with the null or zero-crossing point in the waveform so as to facilitate smooth insertion.
On the opposite side of the fence, we also use reactors to absorb VARs from the system, particularly in association with 500 kV circuits which generate great amounts of 'lagging' VARs when lightly loaded...
But I digress.
Capacitors have almost a perfectly leading power factor; reactor have very close to a lagging power factor. The switching duty is severe, particularly for cap switching, and sulphur-hexafluoride-insulated [SF6] breakers are gradually replacing the older oil circuit breakers in LV applications.
A factor that often bears on power system operation is the need to transport large amounts of power from where it's generated to where it's used; and as an earlier poster alluded to, peak times are the worst. During these periods especially, proper deployment of reactive resources can be used to optimize the delivery limit of a circuit which might otherwise be constrained due to either thermal or stability limits. [We have also recently commissioned series capacitor installations to improve the ratings of some of our 500 kV circuits.]
A final note: capacitor output is all or nothing, and varies as the square of the applied voltage. Reactors have a similar characteristic, but since it's based on frequency and since there is typically far less variation in frequency than in voltage profile, VAR consumption by reactors is almost pegged...
The latest thing is to use Static VAR Compensators [SVCs] which consist of a combination of reactors and thyristor-switched capacitors; SVC's can thus have a soothly adjustable 'lagging' reactive buck / boost function, for example, from 40 MX 'in' to 60 MX 'out.' As a power system operator, I consider these the cat's miaow; they're almost as good as a synchronous condenser...
Autotransformer tertiary windings often prove a very handy place to hook these gizmos up.
Hope this was at least interesting, even if it didn't help...
CR
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