Electrical heating - Formulas

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How does electrical resistance heat work? - Example GrayFurnaceMan

  • There is a paradox: Electricity is converted to heat through resistance, but increasing resistance, actually reduces heat [1]- GrayFurnaceMan, the maker of this video nicely illustrates this. The comments are really helpful
  • Despite what the maker of the video says, it's really called resistance heating and it is an apt name, aka. Joule heating (one of the last comments)

Input

  • This is about electrical heating through solar panels. It's DC and the voltage is constant at e.g., 25V. Available max. current is depending on the sun conditions
  • Ohm's Law: R=V/I <=> I=V/R <=> V=RI
  • Power: P=V*I = R*I^2 = V^2/R
  • In general: Current creates resistance heat. That's why long-distance power lines have such a high voltage, so the current can be low
  • I is variable, and a function of V and R. It can be up to 100A, before wiring gets too hot.

Calculations - Max. heat

  • I want to have maximum power and maximum current ⇒ I=100A
  • Since V is fixed, R results from R=V/I = 25V/100A = 0,25 Ohm
  • Power: 25V * 100A = 2500W or 2.5kW

If this was a home appliance

  • The sticker would say something lik 25V - 100A - 2.5kW. This would be maximum numbers
  • If you connect it to a socket that could deliver 200A, it would still take only 100A, because it's resistance is static
  • If you connect it to a socket that could deliver only 50A, it would deliver 1.25kW, but it would still work just fine. I think it's because of the simple laws governing resistance heating, that power scales linearly with current.

Can you burn out something?

There are such things as burning out things by connecting a device that takes too much current. How could that occur here?

  • If connected to a power supply with 50V, the heater could consume double as much current. That might blow a fuse. But it's weird to connect a device to a socket with a higher voltage than usual. That's not even possible with three-phase electricity
  • So just check the label of the device.

Doubling the resistance

Now up to the paradox mentioned above and adressed in Grayfurnaceman video:

  • Doubling the resistance: R=0.25 Ohm → R=0.5 Ohm
  • I now becomes V/R = 25V/0,5 Ohm = 50A - Current halved, as in Grayfurnaceman's video
  • Power = V*I = 25V * 50A = 1,250W = 1.25kW

Doubling the resistance → Doubling the voltage

Some folks mentioned in the comments of Grayfurnaceman's video: When you double the resistance, you should also double the voltage. Seems intuitive, but let's check:

  • R=0.5 Ohm
  • V=50V
  • → I=100A
  • P=VI = 5,000W = 5kW

Indeed: Double the power. Current is the same and voltage is doubled.

Doubling the resistance - But parallel

Another frequent comment to Grayfurnaceman's video: Don't put those resistors in series, but parallel. Let's see how that works out: I suspect it might be a bit more tricky.

Input:

  • V is the same over both resistors, still 25V
  • R1=0.25 Ohm
  • R2=0.25 Ohm.

Calculations:

  • I1=V/R1 = 25V/0.25 Ohm = 100A
  • I2=V/R2 = 25V/0.25 Ohm = 100A
  • P1=V*I1 = 2500W = 2.5kW
  • P2=V*I2 = 2500W = 2.5kW.

So yes, doubling works, but you're also doubling the current. This is actually an example of the question from a couple of paragraphs back, about 'burning out something': When these two resistors are on the same socket or within the same group, changes are that a fuse will blow (or a circuit breaker will flip or however you call it).

One more thing: I remember that in electrical diagrams, you could lumb parallel resistors together as 1/R=1/R1+1/R2. Let's see how that would work out here:

  • 1/R1 + 1/R2 = 1/0.25 Ohm + 1/0.25 Ohm = 2/0.25 Ohm = 8/Ohm ⇒ R = 1/8 Ohm = 0.125 Ohm
  • V=25V
  • I=V/R = 25V/0.125 Ohm = 200A

Yes: Same outcome. Despite the 'less linearly mathematics' behind it. This stuff keeps suprising me.

Electrical heating = resistance heating?

Can you state that basically all electrical heating is resistance heating or Joule heating? That would make life quite easy.

100% efficiency in conversion

Something else that makes this stuff potentialy easy: Electrical heating basically has 100% efficiency: Heat is one of the forms of energy with the highest entropy and in the case of electrical energy, all energy converts to warmth.

Note that this not the same for e.g., firewood: Part of the energy goes into enthalpy: Gaseous burning products pushing away the air around it. Another part of the energy goes literally up in smoke.

What efficiency in warming a place?

Although you can create electrical heat with 100% efficiency, does this mean that the transfer of warmth is 100% efficient? That would mean, that the resistor has the same temperature as the medium that it is heating (note to self: Check out tegenstroom-principe).

Let's start again with an example from firewood: We have tile stoves where the hot air has to navigate a bunch of turns in pipes before exiting through the chimney. By the time it leaves the chimney, it likely hasn't completely cooled down to the environmental temperature.

But how about electical heating? I don't know yet.

AC, DC, voltage?

Note to self: Check out https://www.oemheaters.com/topic/dc-powered

Sources

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