Simple Wattage & Resistance Question Regarding Heat

[Bulgdoom]

Hello, I'm looking to make handlebar grip heaters for my motorcycle, and so I read a few things online regarding the basics. The most important thing i need to figure out is just how much resistance i need from my wires for the right temperature.

Ohms Law states V = I x R and Wattage = V x I

Fine, here's the deal. I'm using standard 12V from my battery (12.7 in reality, but for simplicity let's stick with 12). If I use 4 ohms of wire, at 12V, that means the current is 3 amps. 3 x 12 = 36 Watts, sounds toasty! If I decide to double the amount of wire, the resistance is now 8 ohms, and current is 1.5 amps times voltage = 18 watts.
4 ohms, 36 watts vs 8 ohms 18 watts, which is hotter?

Now... I understand that more resistance will mean more heat given off, but the total power (wattage) decreases following the equation. I have always thought of higher wattage = more heat generated. When you buy a soldering gun, 40Watts > 15 watts, and similarly a 1000 watt microwave is more powerful than an 800 watt one. So what's the deal here?


I looked at a few kits on the market as a basis, one being Symtec with 36 watts and 3 amps on High which is supposed to be very hot, and Oxford with 1.3 amps and 18 watts on High, which one person measured to get as hot as 124F.

Thank You.

 

[cepheid]

Welcome to PF Bulgdoom!

In the higher power scenario, the wires are giving off energy in the form of heat at a faster rate, and will feel hotter than the lower power scenario. So, the higher the wattage, the hotter the heater.

I'm not sure where you're seeing a contradiction here, unless it's just in your statement that "more resistance will mean more heat given off", which is simply not true.

 

[Bulgdoom]

I'm not sure where you're seeing a contradiction here, unless it's just in your statement that "more resistance will mean more heat given off", which is simply not true.

But I thought that it was the resistance property of the wire that made it give off heat, hence why the name resistance heating wire (nichrome). What does a wire's resistance have to do with the heat it will give off then?

 

[Averagesupernova]

More resistance will give off more heat if you leave the resistance that is already there alone and add the second resistance in parallel. You are getting series and parallel resistances mixed up.

 

[sophiecentaur]

There is a bit of confusion about 'cause and effect' here, I think.
If you just use the formula
P =V²/R
you will get the power delivered to a particular resistance. (This is simply what you get when you jiggle the two formula we started off with.) It removes the necessity of calculating the current - just assumes voltage is constant.

There is always the possibility of over-heating / local charring with systems like this. You need to make sure that there is suitable thermal insulation between hands and cable - and this sort of design factor is one of the things you pay for when buying an off the shelf system. I'm not saying it can't be done but that you will need some glass wool / equivalent if any part of the element is likely to get much more than 100C. imho, you should choose a long length of low resistance wire rather than a short length of higher resistance wire (of equivalent overall resistance) in order to spread out the source of heat and avoid hotspots.

 

[mdjensen22]

The power is V*I or I^2 * R.
The resistance alters the current draw which has a larger impact on the wattage than the resistance. Consider two resistors, a 10 ohm and 20 Ohm in your 12V system.

The 10 Ohm resistor allows 1.2A of current (V/R) and 14.4W (V*I)
The 20 Ohm resistor allows 0.6A of current (V/R) and 7.2W (V*I)As you can see, lower resistance = more current = more heat.
Think of it in terms of a short across a battery - there will be a lot of heat given off since there is little resistance (short current = large). However, if you put a large value resistor (1M) across your resistor, the current is low, and thus the heat generated is low).

 

[cepheid]

But I thought that it was the resistance property of the wire that made it give off heat, hence why the name resistance heating wire (nichrome). What does a wire's resistance have to do with the heat it will give off then?

Yes, resistance causes heating. So, more resistance causes more Ohmic losses IF you can keep the current constant:

P = IV = I(IR) = I²R

So, for constant current, more R would mean more power dissipated as heat.

The trick with your situation is that you don't GET to keep the current constant. The battery maintains a constant *voltage*. At a constant voltage, the current drawn actually decreases with R. So, a more appropriate way to write the equation would be:

P = VI = V(V/R) = V²/R

So, at constant voltage, the power dissipated as heat decreases with increasing R.

EDIT: beaten to the punch several times. That's what happens when composing a post on one's phone and having to leave it aside for awhile!

 

[sophiecentaur]

It's the early worm.

 

[Bulgdoom]

Yes, resistance causes heating. So, more resistance causes more Ohmic losses IF you can keep the current constant:

P = IV = I(IR) = I²R

So, for constant current, more R would mean more power dissipated as heat.

The trick with your situation is that you don't GET to keep the current constant. The battery maintains a constant *voltage*. At a constant voltage, the current drawn actually decreases with R. So, a more appropriate way to write the equation would be:

P = VI = V(V/R) = V²/R

So, at constant voltage, the power dissipated as heat decreases with increasing R.

EDIT: beaten to the punch several times. That's what happens when composing a post on one's phone and having to leave it aside for awhile!

I get it now, thanks for clearing it up. I simply need to keep the current constant to make use of the extra resistance, otherwise it does me no good to purchase high resistance wire. So here's the next question, how do i do that?

The market ready kits have a high and low switch, which from the pictures i discovered is done by making 2 circuits on each handlebar. I had an adjusting knob in mind, kind of like the ones you see for your lightning where you can control the brightness of the lamps instead of just having and on and off position. I'm guessing that they work by altering the current and the voltage stays the same, which would be great in my case.

As for the other design factors which some of you pointed out, hot spots can be a problem which is why ribbon wire makes the most sense, and why all the market kits use them. Electrical tape or some liquid insulator on the handlebars will protect me from getting shocked and thwarting heat upward. As for heat distribution, epoxy to cover the wires will do that, and the rubber grips on top will be further insulation for my hands. Wiring to the battery is easy, although it would be even best to have it linked to the ignition cable so its only on when the bike is.

 

[cepheid]

I get it now, thanks for clearing it up. I simply need to keep the current constant to make use of the extra resistance, otherwise it does me no good to purchase high resistance wire. So here's the next question, how do i do that?

Hmm? No, I think you...missed the point entirely. At the end of the day, the POWER is what determines how much heat you get. You are working with a battery. Therefore, you are stuck with s steady voltage (more or less). Therefore, to increase the power output, you need to REDUCE the resistance (if 36 W is not enough for some reason). So much the better. You can get more power using less wire, saving costs. Of course, there is a natural limit to this. As others have said: be careful! If the resistance gets too low, you'll basically just be shorting your battery terminals, and that leads to "fire and destruction!" Don't exceed the current for which the gauge and length of wire you're using is rated. If I were you, I'd stick to a kit. Hobby projects that draw several amps should make anyone nervous

 

[Bulgdoom]

Hmm? No, I think you...missed the point entirely. At the end of the day, the POWER is what determines how much heat you get. You are working with a battery. Therefore, you are stuck with s steady voltage (more or less). Therefore, to increase the power output, you need to REDUCE the resistance (if 36 W is not enough for some reason). So much the better. You can get more power using less wire, saving costs. Of course, there is a natural limit to this. As others have said: be careful! If the resistance gets too low, you'll basically just be shorting your battery terminals, and that leads to "fire and destruction!" Don't exceed the current for which the gauge and length of wire you're using is rated. If I were you, I'd stick to a kit. Hobby projects that draw several amps should make anyone nervous

I got the power point and that its the Watts that determine the heat. I can easily buy a ready kit for $45 which I may end up doing, but I just wanted to learn how its done, that's all.