Electrical Calculation Question (Academic)

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
Reply to
Erik
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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.
Reply to
Przemek Klosowski
Loss by antenna radiation
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Loss from dielectric heating would have to be measured. It's probably insignificant at 60Hz, but it's how microwave ovens heat food.
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"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:
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(electrical) The losses can be very large at radio and microwave frequencies.
The cord is an open-circuited "stub".
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(electronics)
jsw
Reply to
Jim Wilkins
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.
Reply to
Tim Wescott
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
Reply to
Jim Wilkins
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.
Reply to
Tim Wescott
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.
Reply to
DoN. Nichols
Here is an example of a similar open-ended wire pair that can cause significant problems.
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jsw
Reply to
Jim Wilkins
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.
Reply to
Przemek Klosowski
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."
Reply to
Bruce L. Bergman (munged human readable)
"Bruce L. Bergman (munged human readable)"
This describes how the losses in a wire pair vary with frequency:
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"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.
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(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."
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"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
Reply to
Jim Wilkins

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