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Is this practical?

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

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  1. Eric R Snow

    Eric R Snow Guest

    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?
  2. Doug Miller

    Doug Miller Guest

    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. 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)

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

    Fleetie Guest

    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.

  6. Eric R Snow

    Eric R Snow Guest

    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.
  7. Doug Miller

    Doug Miller Guest

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

    Doug Miller Guest

    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. Eeyore

    Eeyore Guest

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

  10. Eeyore

    Eeyore Guest

    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.

  11. Eric R Snow

    Eric R Snow Guest

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

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

    CptDondo Guest

    They do make heat tape.....

    And it's cheap and legal.....
  15. Also you just happen to pull numbers out your ass huh? Fucking morons in
    this NG...
  16. No, you need to fucking learn to read instead of looking for ways to feed
    your ego.
  17. Eeyore

    Eeyore Guest

    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 !

  18. Eeyore

    Eeyore Guest

    Your explanation was very poor.

  19. default

    default Guest

    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 Fields

    John Fields Guest

    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,
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