# Is this practical?

Discussion in 'Electronic Basics' started by Eric R Snow, Jun 28, 2007.

1. ### Eric R SnowGuest

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

2. ### Doug MillerGuest

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.

3. ### Jon SlaughterGuest

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

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

4. ### Jon SlaughterGuest

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

5. ### FleetieGuest

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.
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

6. ### Eric R SnowGuest

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

7. ### Doug MillerGuest

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

8. ### Doug MillerGuest

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.

9. ### EeyoreGuest

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

Graham

10. ### EeyoreGuest

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

11. ### Eric R SnowGuest

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

12. ### Jon SlaughterGuest

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

13. ### Jon SlaughterGuest

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

14. ### CptDondoGuest

They do make heat tape.....

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

15. ### Jon SlaughterGuest

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

16. ### Jon SlaughterGuest

No, you need to fucking learn to read instead of looking for ways to feed

17. ### EeyoreGuest

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

Graham

19. ### defaultGuest

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 . . .

20. ### John FieldsGuest

---
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.