Thank you for a (second) real life proof.
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16 years ago
Thank you for a (second) real life proof.
But that loss is independent of downstream load...
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??? Loss at a series connection resistance ("problem") depends on the downstream load.
The original statement (Mike) was "If you have a high resistance joint anywhere in your supply then total power consumption will always fall."
A thermostatically controlled resistance heater will use the same power at the heater. I agree with Bob that the power consumption will go up because the power loss at the "problem" adds to the power used at the heater.
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That's true--I knew this was going to come back when I sent it--just _after_ "Send". :(
If that really was the claim rather than a resistance heater was going to use more power at the lower voltage simply by running longer at the lower input to output the same heating...
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So where is the connection that got hot in his example????
That may be, but you haven't specified anything significant. How much resistance makes all the difference in the world.
But the original point was that a single *connection* that is corroded and offers a high resistance is a fire hazard, but will *not* cause a higher power bill. That's how it works.
For a given run of wire, as you are describing, if there is enough power lost to leakage between the conductors to raise the power bill significantly, the insulation is going to suffer serious damage and soon result in a direct short. But, until there is a short, it will not reduce the power available to other loads.
Another issue, which obnoxious John wants to discuss, is a high resistance cable loop. That's a whole different beastie, and is a very common problem. It *will* increase the power bill, and might cause damage to electric motors that require high torque for starting (motors driving compressors in refrigeration units are good example).
These are all significantly distinct problems and should not be confused with each other. The OP's is correct in saying that his investigation of high power usage lead him to discover potentially dangerous problems. His description did not make it clear enough that he knew the discovered problems were not the cause of the high power usage, and that lead to some ornery comments by Mike, who was technically correct but had misread what the OP meant.
You're welcome. Looks like Floyd the cyberstalker is still nippin' at your heals though. Sad and funny at the same time.
John
-- John De Armond See my website for my current email address
That wouldn't be unusual if the wire were wet, for example. PVC insulated wire that has overheated breaks down and one of the breakdown products is hygroscopic. Thus, moisture in the conduit would not be unusual.
That resistance wouldn't be dissipating any significant power, however. Consider
10kohms and 240 volts. (10kohms would be a very low resistance for such a circumstance so consider it an extreme example) P = Vsquared/R or about 6 watts. The bulk of the heating was simple I^2R losses from the current passing through the conductors and perhaps some conducted heat from the hot joint.John
-- John De Armond See my website for my current email address
Forget Stefan-Boltzmann and look at the volts, amps and ohms *that* is what matters.
Actually, it is a good fire, a cold one, and a suitable wench that matters.
You got it.
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