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Is this practical?

My neighbor and I were talking the other day about the frozen pipes in his shop. This is because he didn't insulate enough the pipes in the walls and because it's only heated when he is working in it on the weekends. Now, with the walls covered and with benches permanently mounted he doesn't even want to consider trying to add more insulation. As it is just repairing the split pipes will take him at least 30 hours. He could drain the pipes but he forgot just once hence the split pipes. He could also leave the water on but that would mean pumping a lot of water out of the well. So I opened my big mouth and suggested running a low voltage through the pipes to warm them up. Like the way plumbers will do sometimes using a welder or similar power supply. The power would be controlled by a thermostat. The pipe is copper and is in an open loop. The pipe does go under a concrete slab at one point for about 30 feet. The ground under the slab is dry all year. So, could this be practical? Since the load is purely resistive and the power source is AC can I just measure the voltage and current to get watts just like a DC circuit? Thanks, Eric

Reply to
Eric R Snow
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Let me see if I have this straight.

Electrical code requires metal water pipes to be bonded to the grounding electrode of the building's electrical system, to ensure that the water piping is at zero potential with respect to ground and thus eliminate the possibility of electric shock from touching a water pipe.

And you're proposing deliberately placing a voltage on those pipes.

--
Regards,
        Doug Miller (alphageek at milmac dot com)

It\'s time to throw all their damned tea in the harbor again.
Reply to
Doug Miller

The resistance of the pipes would be very small and would actually require a very large voltage.

Yes, you use ohms law because essentially you just have a huge wire and when you do this, you have P = V^2/R = I^2*R for the power dissipation. Notice that since R is going to be so small it will require V to be fairly large.

Say you want to dissipate 1W/ft and the piping/ft is equivilent to AWG 0 which has ~ 1/10k ohms/ft.

Then V = sqrt(10k) = 100 V

But even if that was ok then think about the current required!

If its 1/10k Ohms/ft and you have, say, 100 ft, then thats 1/100 Ohms. Now V = I*R or V/R = I so we have I = 100/(1/100) = 10k A. This means it will require a power source of 1MW.

The problem is, is that the piping is not meant to dissipate heat efficiently. Its not a tranducer of heat.... unless they happen to be very very small and then there not good for piping.

Probably the easiest method might be to preheat the water before it enters the building and have a routing that can circulate the water. Another possibility, which I have no idea if it would work, would be to ultrasonically vibrate the water so that its less likely freeze.

You could possibly run a resistive wire through the piping and use that to heat the water... If you electrically isolate it then you don't have to worry about shorts.

In any case I think your method isn't practical so you'll need to find some other solution. Theres just not enough resistance in the piping use for heating. (although maybe you could heat them indirectly but I'm not sure if this would be very good or efficient)

Jon

Reply to
Jon Slaughter

um... have you ever heard of electrical isolation or transformers?

Reply to
Jon Slaughter

very large voltage.

Complete rubbish. You have it backwards. Yes, the resistance will be small, so you only need a small voltage to drive a large current. I=V/R.

you do this, you have P = V^2/R = I^2*R for the

V to be fairly large.

Rubbish. P=I^2R. R is small, so CURRENT has to be large. The voltage drop from one end of the piping to the other will be fairly small.

You need to do some exercises concering Ohm's law.

Martin

--
M.A.Poyser                                                  Tel.: 07967 110890
Manchester, U.K.          http://www.livejournal.com/userinfo.bml?user=fleetie
Reply to
Fleetie

Yes Doug, the pipe is indeed grounded with two grounding rods driven into the soil about 8 feet. Just one end. I propose applying 12 to 20 volts to both ends of the pipe. And since the pipe is still at ground potential there will be no more shock hazard than there exists now with the neutral line from the mains AC also connected to the same grounding rods as the pipe. Unless I'm missing something. If so please tell me what, that's why I posted the question in the first place. Eric

Reply to
Eric R Snow

Not relevant. Run that scheme past the local electrical inspector, and see how he likes it.

--
Regards,
        Doug Miller (alphageek at milmac dot com)

It\'s time to throw all their damned tea in the harbor again.
Reply to
Doug Miller

What you're missing is that it's almost certainly illegal. Talk to your local electrical inspector and see what he thinks of the idea.

--
Regards,
        Doug Miller (alphageek at milmac dot com)

It\'s time to throw all their damned tea in the harbor again.
Reply to
Doug Miller

It'll be more like 1 to 2 volts I suspect.

Graham

Reply to
Eeyore

I suggest you learn some basic electricity.

It will require I to be very large. Given the low value of R, I expect V will be quite small too.

Conceivably something like 5 volts and 200 amps might be required.

Graham

Reply to
Eeyore

The welder has about 20 volts open circuit and this drops to about 12 when welding. When connected to the pipes will the lower resistance cause to voltage to drop even more? Eric

Reply to
Eric R Snow

Do you jsut f****ng read what you want? I never said current was small. Fucking moron.

Reply to
Jon Slaughter

Um... f*ck you. How bout you read the whole post next time?

Reply to
Jon Slaughter

They do make heat tape.....

And it's cheap and legal.....

Reply to
CptDondo

Also you just happen to pull numbers out your ass huh? Fucking morons in this NG...

Reply to
Jon Slaughter

No, you need to f****ng learn to read instead of looking for ways to feed your ego.

Reply to
Jon Slaughter

be

It would be purely a guess if I said anything.

You need to measure the resistance of the pipes. Also, if a voltage drops that much, all the heat is warming up the wrong thing !

Graham

Reply to
Eeyore

Your explanation was very poor.

Graham

Reply to
Eeyore

Hey John. This is similar to the soldering guns of old, 100 watt iron would drive a three inch piece of something like #10 wire to soldering temperatures.

Primary was ~500 turns of wire on a rolled core and secondary was one turn of wire (brass or hard copper tube) that developed about two tenths of a volt.

Your comment about the voltage requirement to be high is just wrong. And in one of your statements you mention the power you want to dissipate per foot as a few watts, then come to a conclusion that it will take megawatts to do it - that should have been a clue.

Of course you knew that and were just having some fun . . .

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Reply to
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--- You have no clue as to the length of the pipe or its OD and ID, so you can't calculate its resistance. Not only that, you don't know the thermal conductivity of the material which surrounds it, so you can't calculate how much power will be required to keep the water in it from freezing. In short, your suspicions are just like you are, bogus.

-- JF

Reply to
John Fields

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