Hi all,

A couple of neighbors were bickering about this.

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity of a
normal household garage.

How would one calculate the minuscule energy loss from such a static
conductor just sitting and radiating (and/or whatever).

I tend to agree with the one neighbor that there's bound 'some' loss...
but probably on the order of, oh maybe a penny's worth every few
centuries; but haven't a clue as how one would calculate something like
this, or even what loses would be involved.

Inquiring minds and all that... Thanks in advance!

Erik

On Tue, 13 Aug 2013 19:02:53 -0700, Erik wrote:

Hi all,

A couple of neighbors were bickering about this.

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity of a
normal household garage.

How would one calculate the minuscule energy loss from such a static
conductor just sitting and radiating (and/or whatever).

I tend to agree with the one neighbor that there's bound 'some' loss...
but probably on the order of, oh maybe a penny's worth every few
centuries; but haven't a clue as how one would calculate something like
this, or even what loses would be involved.

The main loss will be through the AC current driving the cable
capacitance. There's also radiation losses but at 60Hz they are
negligible. A typical capacitance might be, oh, 100 pF/m, so the total
under 5nF, or about half a megaohm at 60Hz. The current in the cable
itself is of course non-dissipative, but the 0.2 mA will flow through the
house wiring and cause some power loss. You are right that it will be
minuscule: 1 Watt of electrical energy costs on average around 1$/year,
and this is far less than a Watt.

"Erik" wrote in message
...
Hi all,

A couple of neighbors were bickering about this.

Lets say there is a 100' extension cord (with metal copper
conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged
into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity
of a
normal household garage.

How would one calculate the minuscule energy loss from such a static
conductor just sitting and radiating (and/or whatever).

I tend to agree with the one neighbor that there's bound 'some'
loss...
but probably on the order of, oh maybe a penny's worth every few
centuries; but haven't a clue as how one would calculate something
like
this, or even what loses would be involved.

Inquiring minds and all that... Thanks in advance!

Erik

Loss by antenna radiation
http://en.wikipedia.org/wiki/Radiation_resistance

Loss from dielectric heating would have to be measured. It's probably
insignificant at 60Hz, but it's how microwave ovens heat food.
http://en.wikipedia.org/wiki/Loss_tangent
"Note that the ESR is not simply the resistance that would be measured
across a capacitor by an ohmmeter. The ESR is a derived quantity
representing the loss due to both the dielectric's conduction
electrons and the bound dipole relaxation phenomena mentioned above."

This type of instrument can measure the loss:
http://en.wikipedia.org/wiki/Network...er_(electrical)
The losses can be very large at radio and microwave frequencies.

The cord is an open-circuited "stub".
http://en.wikipedia.org/wiki/Stub_(electronics)

jsw

On Tue, 13 Aug 2013 19:02:53 -0700, Erik wrote:

Hi all,

A couple of neighbors were bickering about this.

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity of a
normal household garage.

How would one calculate the minuscule energy loss from such a static
conductor just sitting and radiating (and/or whatever).

I tend to agree with the one neighbor that there's bound 'some' loss...
but probably on the order of, oh maybe a penny's worth every few
centuries; but haven't a clue as how one would calculate something like
this, or even what loses would be involved.

Inquiring minds and all that... Thanks in advance!

There is loss in the cord.

It is teeny.

It is subject to so many manufacturing and environmental variables
(moisture, how many elephants are standing on the cord, what it's made
of, temperature, etc.) that calculating the losses would be as inaccurate
as they would be time consuming.

Measuring the losses could be done, but you'd have to do much more than
just measure the current into the cord, because most of that current is
reactive (meaning, it's from the capacitance, and doesn't actually
dissipate power).

Possibly the easiest way to measure the losses in the cord would be with
a calorimeter -- pile the cord up into a styrofoam cooler with a
thermometer (electronic, probably), let everything settle for days to
reach an even temperature, then measure the temperature rise in the
cooler with the cord plugged in, then estimate the thermal
characteristics of cord and cooler to figure out the power dissipated.

It'd be a good senior project for a techno-dweeb with a sense of humor.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

"Tim Wescott" wrote in message
...

Measuring the losses could be done, but you'd have to do much more
than
just measure the current into the cord, because most of that current
is
reactive (meaning, it's from the capacitance, and doesn't actually
dissipate power).

200' of 12AWG cord didn't register on a Kill-A-Watt, i.e. it drew less
than 0.01A. Watts and VA stayed at zero too.
jsw

On Wed, 14 Aug 2013 18:40:24 -0400, Jim Wilkins wrote:

"Tim Wescott" wrote in message
...

Measuring the losses could be done, but you'd have to do much more than
just measure the current into the cord, because most of that current is
reactive (meaning, it's from the capacitance, and doesn't actually
dissipate power).

200' of 12AWG cord didn't register on a Kill-A-Watt, i.e. it drew less
than 0.01A. Watts and VA stayed at zero too.
jsw

I would expect that the actual power loss in an extension cord is about
as close to "practically zero" as you can get, and still be burning some
minuscule amount of real power.

The amount of power lost in the cord may well be significantly less than
the amount of power lost in the business end on a humid day vs. a dry one.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On 2013-08-14, Erik wrote:
Hi all,

A couple of neighbors were bickering about this.

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity of a
normal household garage.

How would one calculate the minuscule energy loss from such a static
conductor just sitting and radiating (and/or whatever).

I tend to agree with the one neighbor that there's bound 'some' loss...
but probably on the order of, oh maybe a penny's worth every few
centuries; but haven't a clue as how one would calculate something like
this, or even what loses would be involved.

What you need is to determine the capacitance between the hot
lead and the combination of the ground and neutral. It will be lower
with a heavier gauge wire (because it will typically have thicker
insulation, and thus greater separation between conductors.

Then, it would draw less current in the UK than in the US,
because the frequency comes into the calculation.

O.K. I've got a reel with perhaps 100' of extension cord on it,
and a capacitance meter, so let's see:

6.142 nF (It would be less with the cord stretched out on the
ground. On the reel it is close to itself many times over.)

Now, assuming US power (60 Hz), that would be the equivalent of
431 KOhms, so you get 277.86 uA, or 33.343 mVA. (Now, if this
were resitance,it would be the same number of Watts (33
thousandths of a Watt), and compute what that would cost for a
year using your local power rates.

However -- since this is purely reactive, your power company
would not charge you for it -- *unless* you were on industrial
billing.

And -- if you have a motor running, it will cancel out a tiny
bit of the inductive reactance, so your power factor is improved
very slightly. (Motors are inductive -- computer power supplies
are mostly capacitive, so it won't help with them.

And -- in reality, there will be some resistance in the
insulation, but so high that I can't measure it, so calculating
it is not worth while.

Enjoy,
DoN.

--
Remove oil spill source from e-mail
Email: | (KV4PH) Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

"Tim Wescott" wrote in message
...
On Wed, 14 Aug 2013 18:40:24 -0400, Jim Wilkins wrote:

I would expect that the actual power loss in an extension cord is
about
as close to "practically zero" as you can get, and still be burning
some
minuscule amount of real power.

The amount of power lost in the cord may well be significantly less
than
the amount of power lost in the business end on a humid day vs. a
dry one.

Tim Wescott

Here is an example of a similar open-ended wire pair that can cause
significant problems.
http://en.wikipedia.org/wiki/Bridge_tap

jsw

Przemek Klosowski writes:

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged into
the cord, and it's one without any indicator lights or ground fault
gimmicks. Said cord is dry, other than the typical indoor humidity of a
normal household garage.
....
The main loss will be through the AC current driving the cable
capacitance. There's also radiation losses but at 60Hz they are
negligible.

A capacitive load consumes no watts.


--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

On Fri, 16 Aug 2013 05:58:27 +0000, David Lesher wrote:

Przemek Klosowski writes:

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged
into the cord, and it's one without any indicator lights or ground
fault gimmicks. Said cord is dry, other than the typical indoor
humidity of a normal household garage.
...
The main loss will be through the AC current driving the cable
capacitance. There's also radiation losses but at 60Hz they are
negligible.

A capacitive load consumes no watts.

We agree violently---I said "The current in the cable itself is of course
non-dissipative". At the same time, driving that capacitance through the
power company and house wiring does carry losses. That's why we care
about the power factor: the wiring/distribution system losses cannot be
avoided but at least at Pf=1 something is doing work at the end of the
line, whereas at Pf=0 (the case of our extension cord) it's all for
naught.

On Mon, 19 Aug 2013 03:51:20 +0000 (UTC), Przemek Klosowski
wrote:

On Fri, 16 Aug 2013 05:58:27 +0000, David Lesher wrote:

Przemek Klosowski writes:

Lets say there is a 100' extension cord (with metal copper conductors)
energized by a live household 120V 60Hz outlet. Nothing is plugged
into the cord, and it's one without any indicator lights or ground
fault gimmicks. Said cord is dry, other than the typical indoor
humidity of a normal household garage.
...
The main loss will be through the AC current driving the cable
capacitance. There's also radiation losses but at 60Hz they are
negligible.

A capacitive load consumes no watts.

We agree violently---I said "The current in the cable itself is of course
non-dissipative". At the same time, driving that capacitance through the
power company and house wiring does carry losses. That's why we care
about the power factor: the wiring/distribution system losses cannot be
avoided but at least at Pf=1 something is doing work at the end of the
line, whereas at Pf=0 (the case of our extension cord) it's all for
naught.

There are going to be capacitive losses, but I'm betting they fall
squarely below the "Too low to measure" line.

Certainly too low to measure at the power company "Demand Meter" where
they try to measure it to surcharge you for excess power factor and
sufficient transformer capacity. You need laboratory grade testing
gear (very fragile and expensive) to see something that small.

And there will be a tiny bit of resistive losses between the Hot and
Neutral & Ground conductors of the cable, no such thing as a perfect
insulator. Plus a tiny bit by conductive dirt and salts on the cord
cap and connector on the ends.

Again, there is a loss, but when the resistance is up in the
multi-megohms it will be too low to measure, so when the numbers are
0.000001A (micro-amps) of load neighborhood it's safe to just say "No
for all practical purposes."

"Bruce L. Bergman (munged human readable)"
wrote in message
...

There are going to be capacitive losses, but I'm betting they fall
squarely below the "Too low to measure" line.

This describes how the losses in a wire pair vary with frequency:
http://www.audioholics.com/audio-vid...ables-debunked
"Typical PVC insulation has a measurable dielectric loss at 10MHz
which is almost 3 decades past the limits of human hearing (~20kHz)!
Since Dielectric losses increase with frequency, they are often lumped
with skin effect losses into an overall dB loss model."

The calculation of losses may be academic for 120V 60Hz or audio but
it's very practical if you want to run high speed signals through the
power cord.
http://en.wikipedia.org/wiki/X10_(industry_standard)
"One problem with X10 is excessive attenuation of signals between the
two live conductors in the 3-wire 120/240 volt system used in typical
North American residential construction."

http://en.wikipedia.org/wiki/Power_line_communication

http://www.smartgrid.gov/the_smart_grid#smart_grid
"In short, the digital technology that allows for two-way
communication between the utility and its customers, and the sensing
along the transmission lines is what makes the grid smart."

jsw