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#130795 02/12/07 06:07 PM
Joined: Oct 2000
Posts: 2,723
Likes: 1
Broom Pusher and
Member
Quote

Scuttle Power?

Is that supposed to mean the "power" that is scuttling back and forth between the load and the source, that doesn't do any real work, but requires us to upsize conductors and such, in order to accommodate its journey?

Ding, Ding, Ding, Ding, Ding!!!!!!!

Yes, that's what the Slang Term is referring to - the way Reactive Power "Scuttles" between the Reactive Load and the Supply (typically the Secondary Windings of a Transformer).

Same Power does not perform any usable output work at the Load Device, but does result in Circuit Conductors doubling as low Power Space Heaters!!!
[Linked Image]

In addition to the extremely inefficient "Conduit Heater Properties" of the Conductors, due to the "higher than should be" Total Line Current carried in them
{per the total Apparent Power (VA) Package, for both the True Power (Wattage) and the Reactive Power (VARs)...},
one may also receive a "Special Meter", + a "Special Monthly Billing Rate" from the Utility Company, allowing someone to become a very generous donator of $$$ to the PoCo, for bouncing VARs between given loads and the PoCo Transformer.
(Detent Meter and penalties for bouncing too much Reactive Power off the PoCo's Transformer Secondary Windings)

Trying to describe this in a funny way, yet not succeeding.

Scott35


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
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#130796 02/13/07 10:37 PM
Joined: Sep 2005
Posts: 202
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Not to wander too far off topic, but what are you seeing for power factor requirements? Our new standards require a P.F. of 97%.

#130797 02/14/07 01:56 AM
Joined: Aug 2006
Posts: 21
S
Member
Power factor values are device specific and can not be averaged.

In My Professional Opinion:
Using an average power factor is as insane as finding 2.4 kids in your neighbor's house. It is just plain wrong and will always be wrong.

#130798 02/14/07 08:17 AM
Joined: Mar 2005
Posts: 1,213
S
Member
Since nobody has really explained it yet, and this could potentially be a difficult concept if you've never seen if before, this is why power factor happens:

Straight resistor circuits are simple. Volts = Amps x Ohms. When you put in more voltage, more current flows. If you look a a sinusoidal plot of 60Hz AC power over time, the Voltage and Ampere curves will be perfectly aligned.

Everyone has probably turned a coil of wire around a nail and created an electromagnet- that's a good example of inductance, where flowing current creates a magnetic field. In AC circuits, that field is constantly being built up and drained away, and that stores and releases energy. Capacitors behave in a similar way. So, inductors and capacitors store energy and release it, but not in phase with the voltage! Current flow through an inductor will peak before the voltage peaks. Current flow through a capacitor will peak after the voltage peaks. In both cases, all the energy stored in the inductor or capacitor is released later in the cycle- leading to a condition where you're not really adding or subtracting any power overall, but there is a lot more current flowing!

In the real world, nothing is ideal. Inductors and capacitors all have some resistance. A straight piece of wire acts as an inductor (that's how clamp ammeters work). And there is always some capacitance between "hot" wires and the ground. So even a very simple circuit will have some power factor that's not exactly 1.0


In our circuits, motors and transformers often have large inductive components, which usually leads to "leading" power factor. Now, as luck has it, inductors and capacitors are diametrically opposed- current stored in a capacitor while it's being released by an inductor. If you match the two together, they cancel out perfectly. So, devices with poor power factor usually can be "corrected" with capacitors.

[This message has been edited by SteveFehr (edited 02-14-2007).]

#130799 02/14/07 09:09 PM
Joined: Sep 2005
Posts: 202
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WFO Offline
Member
I guess I didn't phrase that properly.
I work for a utility. Our regulatory commission requires us to maintain a 97% power factor at our substations, which we do by switching capacitor banks in and out as the loads change.


My question was...What are the utilities requiring their customers to maintain in terms of a minimum power factor before they are penalized on their bill?

#130800 02/15/07 08:01 AM
Joined: Mar 2005
Posts: 1,213
S
Member
Isn't that done at the poco more to maximize line efficiency and reduce line loss? At the distribution level, it's often more economical just to oversize the copper a bit and live with a lousy PF. I have many sites with PF as low as 0.8 and have never heard anyone complaining about it.

#130801 02/15/07 10:08 AM
Joined: May 2005
Posts: 984
Likes: 1
G
Member
ComEd has no requirements (which may be part of their problems); but I've seen 0.89 to 0.92 for other utilities.
FYI; China changes their rate structure for every 0.01 change in power factor...and it's VERY punitive. If your power factor gets below 0.8, they cut your power off!


Ghost307
#130802 02/15/07 10:43 PM
Joined: Mar 2003
Posts: 29
R
Member
Our power factor is whatever it is as long as the POCO is carrying less than 1 MVAR. When it drifts over that, the tap changer in our feed transformer changes supplied voltage which sheds VARS back to us.

Thus, voltage, in this instance changes our power factor, albeit indirectly, as we co-generate.

#130803 03/21/07 10:17 AM
Joined: Mar 2007
Posts: 1
M
Junior Member
hello !
there are two wayes to calculate the PF.

the original definition is that the PF is the angle between the current through and the voltage on a consumer (electrical machine-not DC - , lamp atc...) - cos(fi).

the second way is to calculate the power factor by measuring the active power(P) and the reactive power [Q] (or the imaginary power S = V * I ) and use the formula PF = P/S. or by measuring the related enrgies.

this way of computing the PF used in calculating the consumer PF (at the main board ).

if there are harmonics in the sysyem then PF not equel cos(fi).

for en individual electrical equipment it is the same pf=cos(fi)

yours,
moti bohadana

#130804 03/22/07 01:58 AM
Joined: Jul 2002
Posts: 8,443
Likes: 3
Member
Gidday there Moti!.
Welcome to the group mate. wink

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