A bit off topic, as it has to do with house wiring. But how does power factor
affect dissipation in the conductors? I'm wondering, for example, if a 1500
watt capacity and a 30% power factor would overload #14 wire.

--
All relevant people are pertinent.
All rude people are impertinent.
Therefore, no rude people are relevant.
-- Solomon W. Golomb

clifto wrote:

A bit off topic, as it has to do with house wiring. But how does power factor
affect dissipation in the conductors? I'm wondering, for example, if a 1500
watt capacity and a 30% power factor would overload #14 wire.

To put it into perspective...NEC allows #14 for up to 15 amps when used
as a conductor in a 3-core cable. The wire alone is rated for 32 amps
for open chassis wiring. I'm not sure how its rated loose in conduit
but I suspect that NEC still treats it as 15 A.
That said, I don't think NEC takes into consideration...or actually they
do in a roundabout way by derating...the power factor variance.
I'd say you are in the clear both from a practical and a 'legal'
standpoint with 1500 watts load. However, if I were doing this from
scratch I'd bump up to #12 or #10 because 14 wire could create an
undesirable voltage drop depending on distance.

I'm assuming you're in the U.S. and by "#14" wire you mean 14 AWG.

A power factor of 30% means that only 30% of the product of V and I is
power being delivered to a load. The rest is simply being moved back and
forth in the power system. To get 1500 watts of power at a power factor
of 100%, you need P / V amperes. For V = 120 volts, that works out to
12.5 amps. But to get 1500 watts of power with a power factor of 30%,
you need P / V / 0.30 = 12.5 / 0.30 = 41.67 amps. Heating in wires is
due to the current going through them. Ordinary #14 wire in houses is
rated for 15 amps. So you'd be grossly exceeding the rating of the #14
wire in getting 1500 watts with a 30% power factor. [All voltages and
currents in this posting are in RMS.]

When the power factor is less than 100%, you need more current for the
same amount of power, because it requires current to move the "reactive
power" -- power moved back and forth without being used up in a load.

If by "1500 watt capacity" you actually mean "1500 volt-ampere
capacity", then you're ok. In this context, watts are real power, but
volt-amperes are the product of V and I, which is the sum of the real
power (watts) and "reactive power" (VARs, for "Volt Ampere Reactive").
If you limit your volt-ampere product to 1500 volt-amperes, then your
maximum current is 12.5 amps. But if your power factor is 30%, you'll
only be delivering 0.30 * 1500 = 450 watts to the load.

This is why you'll generally see transformers and other power devices
rated by volt-amperes rather than watts.

Roy Lewallen, W7EL

clifto wrote:
A bit off topic, as it has to do with house wiring. But how does power factor
affect dissipation in the conductors? I'm wondering, for example, if a 1500
watt capacity and a 30% power factor would overload #14 wire.

clifto wrote:
A bit off topic, as it has to do with house wiring. But how does power factor
affect dissipation in the conductors? I'm wondering, for example, if a 1500
watt capacity and a 30% power factor would overload #14 wire.

For heating wire, the I-squared-R losses are what matter. Same for most
small circuit breakers.

1500 watts at 120 volts would be a RMS current of 12.5 Amps, if the
load was purely resistive.

With a power factor of 30%, the RMS current might be 40 Amps. That's a
lot more than the 15 Amps that electrical codes allow on 14-gauge wire.
But you'd be tripping breakers if you really were putting 40 Amps
through that circuit and it's house wiring AND it's up to code AND the
breaker was Ok AND (yadda yadda yadda).

But I don't know what a "1500 watt capacity" is. I don't think you're
really talking about 1500 watts. Maybe you meant Volt-Amps (VA).

Tim.

To figure out your current draw:
Amps = Watts (in Kw) X 1000/(voltage X 1.732 X Pf)
Thats a very low power factor, what are you running?

Lew


"steve carey" wrote in message
...
what could you possibly be operating has a 30% power factor
"clifto" wrote in message
...
A bit off topic, as it has to do with house wiring. But how does power
factor
affect dissipation in the conductors? I'm wondering, for example, if a
1500
watt capacity and a 30% power factor would overload #14 wire.

--
All relevant people are pertinent.
All rude people are impertinent.
Therefore, no rude people are relevant.
-- Solomon W. Golomb

Lew wrote:
To figure out your current draw:
Amps = Watts (in Kw) X 1000/(voltage X 1.732 X Pf)
Thats a very low power factor, what are you running?

I'm not. It's just one of those silly theory questions my mind cooks up from
time to time. Answering the silly questions helps me understand the non-
silly ones.

--
All relevant people are pertinent.
All rude people are impertinent.
Therefore, no rude people are relevant.
-- Solomon W. Golomb

On Mon, 27 Feb 2006 12:07:53 -0500, "Lew"
wrote:

To figure out your current draw:
Amps = Watts (in Kw) X 1000/(voltage X 1.732 X Pf)
Thats a very low power factor, what are you running?

That would be the current per phase into a three phase load.

Paul