Old Wiring Good or Bad

[Jim Katen]

. . .

P=(1.5)(squared)R

1.5 (squared) = 2.25

P=2.25R

So the heat developed is now 2.25 times greater than before.

That shows that resistance increaes by 2.25 times. Do resistance and heat have a linear relationship in copper?

Also, resistance increases with temperature. There's no variable for temperature in there.

The Ohm's law method would yield a straight line on a graph, but when you take heat into account, doesn't that line turn into a curve?

- Jim Katen, Oregon

[hausdok]

Ugh,

Numbers. My head began to hurt when Marc put up the equation; now Jim's making it worse.

I can't figure that kind of stuff out using just my fingers and toes.

ONE TEAM - ONE FIGHT!!!

Mike (Math moron extraordinaire)

[Terence McCann]

BTW, could you share your math on the 2-1/4 times more heat thing? It isn't that I don't believe you, I'm just curious about your method of calculation.

- Jim Katen, Oregon

Sure Jim,

Most of us know that developed power is the product of the voltage and current, but voltage is equal to the current multiplied by the resistance, Hence:

P=VI

V=IR

Substituting:

P=(IR)I

P=I(squared)R

The resistance for the conductor hasn't changed but the current has changed from 20A to 30A. 30 is 20(1.5):

P=(1.5)(squared)R

1.5 (squared) = 2.25

P=2.25R

So the heat developed is now 2.25 times greater than before.

Pardon the lack of suitable notation. I don't know how to do it on this word processor.

Marc

I think you're reaching a bit Marc. As Jim stated there are variables that you're not taking into account.

Take a 1' run of 12 ga pulling 20 amps, for 1 hour and measure temperature. Take a 70' run of 12 ga pulling 20 amps, for 1hr, and measure temperature. Do the same experiment when it is 20F and 90F.

Is the wire in free air or buried in insulation?

Last, but not least, I'm not sure you can make a direct correlation between current and heat generated as being linear (as you're eluding too). Any documentation to substantiate that?

[Marc]

That shows that resistance increaes by 2.25 times. Do resistance and heat have a linear relationship in copper?

Also, resistance increases with temperature. There's no variable for temperature in there.

The Ohm's law method would yield a straight line on a graph, but when you take heat into account, doesn't that line turn into a curve?

What we're trying to learn here is the effect of current through a conductor on the heat developed in that conductor. If we wanted to assess the effects of temperature on conductivity for a certain material, that would be an entirely different equation.

Resistance in this equation is a given length of copper wire and does not exhibit significant change in resistance with a change in current, all other variables held constant. This assumes 60 hz AC current. So the value of resistance, for all practical reasons, is constant in this situation. That leaves only power as a variable. Whereas the power developed for 20 amps was 1R, the power developed on 30 amps is 2.25R where R is the resistance of the conductor.

P=I (squared)*R is a derivation of Ohm's Law and is not a linear equation.

It does not yield a straight line.

Marc

[Terence McCann]

An interesting read:

Heat Transfer

[Marc]

Maybe stick with Ohm's Law for now. When you graduate from that, you can move up.[]

Marc

[Erby]

Y'all are making my head hurt. I'm gonna go read something else.

-

[fnorman]

The Electrical Safety Authority (ESA) in Ontario studied 160 electrically caused fires spanning five years, and identified only three instances where defective knob and tube wiring were involved - and in all cases the homeowner had been playing with the wiring. This, and inspector etc. experience, supports the statement that knob and tube wiring itself is not inherently hazardous. If it has been unsafely modified, or if the insulation is brittle or cracked (sometimes caused by high temperatures) then it can be a different story. If the wiring is in original condition ideally it should be possible to rewire only as necessary to correct deficiencies such as lack of grounds on outlets, overloaded circuits, poor splices, .. and leave the original knob and tube ceiling lighting circuits that are not causing a problem. But even when the wiring is not dangerous, the homeowner may have difficulty if home insurance companies refuse coverage for houses with knob and tube. In BC, an electrical inspection company partnered with BCAA Insurance to offer an electrical inspection service for $250 - if identified deficiencies are fixed house insurance can be obtained with a premium surcharge of 10% for knob and tube wiring. If deficiences have not been remedied or the house not inspected the surcharge is 50%. Assuming that deficiencies are fixed and safety is not an issue, the homeowner could decide to compare the costs of leaving the good wiring in (premium surcharges, inspection charges) versus the costs of replacing it, or they may just wish to replace all the knob and tube in order to eliminate difficulties with insurance.

[Garet]

... knob and tube wiring itself is not inherently hazardous. If it has been unsafely modified, or if the insulation is brittle or cracked (sometimes caused by high temperatures) then it can be a different story.

The problem is, in 10 years I have yet to see a k&t installation that hasn't been monkeyed with by some fool who had no business mucking around in an electrical system. They're all screwed up. Every last one of them.

[Jim Katen]

... knob and tube wiring itself is not inherently hazardous. If it has been unsafely modified, or if the insulation is brittle or cracked (sometimes caused by high temperatures) then it can be a different story.

The problem is, in 10 years I have yet to see a k&t installation that hasn't been monkeyed with by some fool who had no business mucking around in an electrical system. They're all screwed up. Every last one of them.

Yes, the fallacy of the impossible contingency.

If the Trojan Horse had born foals, then horses today would be much cheaper to feed. . .

- Jim Katen, Oregon

[Charlie R]

I write in the report "I strongly recommend you hire a licensed electrical contractor to inspect the knob and tube wiring as seen in the (location) to verify that the wiring is still safe to be in use and reduce the potential for fire and personal injury. Many home insurance companies will not provide coverage to homes with this obsolete wiring system. Budget to replace the wiring as soon as possible."

[John Kogel]

BTW, could you share your math on the 2-1/4 times more heat thing? It isn't that I don't believe you, I'm just curious about your method of calculation.

- Jim Katen, Oregon

Sure Jim,

Most of us know that developed power is the product of the voltage and current, but voltage is equal to the current multiplied by the resistance, Hence:

P=VI

V=IR

Substituting:

P=(IR)I

P=I(squared)R

The resistance for the conductor hasn't changed but the current has changed from 20A to 30A. 30 is 20(1.5):

P=(1.5)(squared)R

1.5 (squared) = 2.25

P=2.25R

So the heat developed is now 2.25 times greater than before.

Pardon the lack of suitable notation. I don't know how to do it on this word processor.

Marc

Thanks, Marc. I'll try to say it a different way that doesn't hurt peoples heads. []

Power = I squared times R, current squared times the resistance.

Resistance ® thru Cu does not change.

At 20 amps, the current (I) squared would be 40, for 30 amps I squared is 90, 2 and a half times more, times R = 2.5 times the power.

Now we need convincing that that produces heat. [:0]

[Marc]

Thanks, Marc. I'll try to say it a different way that doesn't hurt peoples heads. []

Power = I squared times R, current squared times the resistance.

Resistance ® thru Cu does not change.

At 20 amps, the current (I) squared would be 40, for 30 amps I squared is 90, 2 and a half times more, times R = 2.5 times the power.

Now we need convincing that that produces heat. [:0]

Power is a time rate of energy, just as acceleration is a time rate of speed and speed is a time rate of displacement. The energy developed by an electric current flowing through the resistance of a copper conductor is in the form of heat energy. Halogen or tungsten flashlight bulbs would be a little different. A small portion of the electric energy is in the form of visible light (radiation) due to the high temperature of the filament. Flashlights with LED light sources are much more efficient in changing electric energy into visible light. That's why they don't get as hot. Much more of the electric energy is transformed directly into radiation. More light, less heat, smaller and longer lasting battery.

Did I get across to anyone?

Marc

[John Kogel]

[]

Now we need convincing that that produces heat. [:0]

Power is a time rate of energy, just as acceleration is a time rate of speed and speed is a time rate of displacement. The energy developed by an electric current flowing through the resistance of a copper conductor is in the form of heat energy. Halogen or tungsten flashlight bulbs would be a little different. A small portion of the electric energy is in the form of visible light (radiation) due to the high temperature of the filament. Flashlights with LED light sources are much more efficient in changing electric energy into visible light. That's why they don't get as hot. Much more of the electric energy is transformed directly into radiation. More light, less heat, smaller and longer lasting battery.

Did I get across to anyone?

Marc

We know that a thin filament in a light bulb or space heater gets hot, but we are discussing a #14 copper wire in an old house, maybe supplying a couple of lights in the ceiling of an old house.

Let's say 6 or 7 60 watt bulbs at the most. Supposing the connections have all been cleaned up, will the wire even get slightly warm?

Why I ask, typical K+T left in the ceiling for light fixtures on the main floor, hidden up there behind the lath and plaster. Should we worry about it?

[Marc]

We know that a thin filament in a light bulb or space heater gets hot, but we are discussing a #14 copper wire in an old house, maybe supplying a couple of lights in the ceiling of an old house.

Let's say 6 or 7 60 watt bulbs at the most. Supposing the connections have all been cleaned up, will the wire even get slightly warm?

Why I ask, typical K+T left in the ceiling for light fixtures on the main floor, hidden up there behind the lath and plaster. Should we worry about it?

A 60 watt bulb is about half an amp at 120 V. Seven of them would be 3.5 amps which will develop only (3.5/20)squared the heat for #12 conductors. That's only 3% of the wire's heat dissipation rating. No, I don't think you need to worry about it. I can't think of any situation where a conductor would have to be derated 97%.

For #14 (15 amp) that would be (3.5/15)squared = 5.5% of the heat dissipation rating of the #14 conductor. It corresponds to a 94.5% derating. Still, no issue.

Marc

[John Kogel]

We know that a thin filament in a light bulb or space heater gets hot, but we are discussing a #14 copper wire in an old house, maybe supplying a couple of lights in the ceiling of an old house.

Let's say 6 or 7 60 watt bulbs at the most. Supposing the connections have all been cleaned up, will the wire even get slightly warm?

Why I ask, typical K+T left in the ceiling for light fixtures on the main floor, hidden up there behind the lath and plaster. Should we worry about it?

A 60 watt bulb is about half an amp at 120 V. Seven of them would be 3.5 amps which will develop only (3.5/20)squared the heat for #12 conductors. That's only 3% of the wire's heat dissipation rating. No, I don't think you need to worry about it. I can't think of any situation where a conductor would have to be derated 97%.

For #14 (15 amp) that would be (3.5/15)squared = 5.5% of the heat dissipation rating of the #14 conductor. It corresponds to a 94.5% derating. Still, no issue.

Marc

Thanks Marc, that's the info we were missing above.

Are you using Heat dissipation tables, if so, from where?

So according to the math, if we're not loading down the circuits with gadgets and power supplies, a few K+T lighting circuits could be fine, even at 100 yrs old.

I always point out any grounded outlets upstairs, that's upgraded wiring, and that's your power supply for computers and TV's.

If they have all ungrounded on the wall receptacle, I call for a Repair.

What if they installed a ceiling fan with 4 light bulbs on their K+T light socket?

There's a hot idea []that could catch fire. []

[Marc]

I'm not using any dissipation tables. I'm just doing the math.

Marc

[John Kogel]

I'm not using any dissipation tables. I'm just doing the math.

Marc

How do you know a #14 copper wire with rubber and cloth will dissipate that, 5.5%?

OK, you just use 15 amps as a max rating for the wire. Right?

OK. How do you know the wire can safely carry 15 Amps? You don't. []

The old guy says put it on a smaller breaker or fuse. I think that's good advice for some people in certain situations.

Where would they get 10 amp breakers or Edison fuses?

[Marc]

Marc,

How do you know a #14 copper wire with rubber and cloth will dissipate that, 5.5%?

OK, you just use 15 amps as a max rating for the wire. Right?

OK. How do you know the wire can safely carry 15 Amps? You don't. []

The old guy says put it on a smaller breaker or fuse. I think that's good advice for some people in certain situations.

Where would they get 10 amp breakers or Edison fuses?

Sorry John. If you don't understand my point by now, it's your dang fault.

Marc

[John Kogel]

[:-party]

[John Kogel]

Just trying to understand the process better, Marc. I put up a [] when it's just light discussion with a bit of sarcasm. []

Here's a test case for your formula, correction, Ohm's formula:

The old 1912 house was rented out, main floor and attic apartment, K+T everywhere. One tenant had a home office set up in the living room. Copier, printers, fax, desktop PC, monitor, phones, etc, all plugged into an ungrounded power bar. Then for heat, a couple of plug-in space heaters. Plus the usual array of wall wart battery chargers. Attic tenant, same drill on a smaller scale, computer, TV, plug-in heaters. I guess they didn't want to buy fuel oil for the old furnace.

Down in the basement, an old 100 amp fuse panel from about 1945-50, and five or six branch circuits were over-fused with 20 Amp fuses.

Those circuits with the block heaters and gizmos could be carrying over 2000 watts.

If 60 watts draws half an amp,

120 watts draws 1 amp,

2000 watts draws 16.6 amps, enough to blow the fuse. Yep. [:-graduat

I've just got to remember 120 watts = 1 amp.

[plummen]

You can overload any kind of wire by oversizing the fuse or breaker,I come across 14ga romex all the time on 20a breakers and 20-30a fuses when im out running service calls [:-party]

[Denray]

Click to Enlarge

42.33 KB

My wife does home owner losses for insurance companies. I was helping her today look for possible water damage in the attic and pointing out to her that K&T can sometimes still have power to it. Which it did. And that all of the insulation was rotted off. There were galvanized plumbing vents nearby too, for easy grounding of one hand. Always have to be careful opening hatches that there isn't any electrical conductors for your hands to contact.

[Denray]

One more try at the picture.

Click to Enlarge

42.33 KB

[Bill Kibbel]

That's not knob & tube - it's antenna wire. It never had insulation.

What did you use to determine that it still has "power to it"? Did you actually measure voltage?