Thursday, October 31, 2019

Today's Electrical Engineering Question

For a sheet of 12 gauge steel one foot wide and 10 feet long, how much heat will be produced by passing 100 W at 12 V through it?  This page says "The resistivity of steel is on the order of 10−7 Ωm."  That seems to imply that the total length of the sheet of steel, not area or mass is what matters.  That is counter-intuitive.

Question answered and a better solution offered.  My objective is melt enough snow at the edge of the driveway to expose a little asphalt to sunlight.  Exposed asphalt absorbs heat and rapidly clears the rest of the snow.  Instead of building my own heater, deicing cable was suggested.  This puts out 5W per foot and nestled up to the edge of the driveway will likely do the job.  Two 100W PV panels and an inverter should be enough for a 30 foot section.  The 200 foot cable will require six cables to do all 1200 feet of driveway, but even doing some of it should get the rest largely clear on the sunny days that follow most serious snowfalls.  A 200 foot cable will gobble at least 1000 watts (with losses from the inverter), so that might be a big investment in PV panels.


  1. Clayton,

    Those units are correct. Resistance of a piece of metal is resistivity times length divided by cross-sectional area. All of the length units cancel, leaving only ohms.

    Based on this resistance calculator, you are looking at about 0.4 milli-ohms for a piece of steel that size. You cannot put 12V into that sheet of steel and get 100W. Ohm's law states that for a fixed voltage, the power is P=V^2/R. Based on this, you would get a power of 360 kW for a 12V source providing 30,000 amps (think about welding for an example).

    To get 100W from that piece of steel, you need a power supply to deliver 500 amps at 0.2 volts (again, think of welding). At that resistance, the quality of your connections becomes an ENORMOUS problem. If each of four connections has only .1 milli-ohm of resistance, ONE HALF of your power is going into your connections.

    Honestly, this doesn't sound like a winning plan.

  2. So nichrome is used for heating because steel does not produce enough heat until you get to very high amperage. To produce a high heat with durability (like cars passing over it) I would be better using nichrome under an electrically insulating sheet with the steel on top.

  3. Nichrome has a fairly high resistivity. It can also tolerate very high temperatures. That is also why heating elements have long, thin wires. Resistance increases linearly with the length and decreases linearly with cross-sectional area.

    Best of luck!

  4. The point is that you need high resistance (and thus high bulk resistance) if you don't want to have to use insane amperage for a large conductor. Steel is way too good a conductor.

    To do what you want, you'd have to cut that steel down to a small wire - maybe a millimeter diameter (I'm too lazy to do the math).

    Your power in is P = E*I ( watts = amps * volts). Your amps is I = E / R ( amps = volts / ohms ).

    For most electrical power calculations, those two formulas will let you figure out anything you want, perhaps with a bit of trivial algebra. You can combine those and get P = E * E / R;

    E means electromotive force, which is measured in volts - i.e. the voltage.

  5. Short and wide has low resistance. Long and skinny has high resistance. While a simple rectangular sheet will have low resistance, you can get high resistance by cutting channels so the sheet becomes a long skinny trace that zig zags back and forth