240 volt heaters

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dm9289

Industrial Maintenance Electrician
Location
Pennsylvania
Occupation
Industrial process repair/ maintenance Electrician
I have worked in industry for years hooked up 3phase control circuits, water heaters,electric heaters and so forth (no neutral), and while teaching a young coworker how to hook things up he asked when you hook up an electric 240v heater why doesnt it just overcurrent. So basically he needed a good explanation of how that works and I could not give him one could someone help me with some reference material or info.

Thanks alot

Dave
 
dm9289 said:
when you hook up an electric 240v heater why doesnt it just overcurrent.
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It will if your OCPD is too small.:grin:

Use Ohms law. You know wattage and voltage. Lets say your hooking up a 10kw heater at 240V.

P?E=I

10,000?240=41.7

Due to the inherent resistance of the heater it will draw around 42 amps.
 
A tungsten filament is a similar thing: why doesn't it just overheat? It
provides a defined level of resistance (as your resistive heater does),
which limits the current, acc. the constant ratios of Ohm's law.

If resistance is greater, current will be less; if resistance is less,
current is greater. This demonstrates that these two values in a
resistive circuit are inversely proportioned. If you can convey this
idea of inverse proportionality (one goes up; the other goes down),
your student will get a long way toward understanding that heater.
 
chris kennedy said:
It will if your OCPD is too small.:grin:

Use Ohms law. You know wattage and voltage. Lets say your hooking up a 10kw heater at 240V.

P?E=I

10,000?240=41.7

Due to the inherent resistance of the heater it will draw around 42 amps.
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What if this same heater was powered with 120V? Since the resistance is fixed, I would expect less current to be drawn at 120V.
 
PhaseShift said:
What if this same heater was powered with 120V? Since the resistance is fixed, I would expect less current to be drawn at 120V.
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I'm not sure what your thinking? If you used 120V on a 1oooo watt heater it would be 83.33 amps.

It comes down to a ratio problem, if you increase or decrease one value with respects to the other implied constants across an equal sign, things have to change.

If you'd look at the wattage of a real base board heater, it would not be that high, even half that in most cases. Amps and or wattage.

Granted there is big requirements in commerical but it'd be cut in half with residental requiremnent.

If you keep these numbers in mind and bring them down a little bit to get the correct numbers that really can be readily appliable to us. :D
 
Last edited:
091202-1000 EST

I have no idea what the question is in the original post.

cadpoint:

PhaseShift's statement was correct. If you consider a specific heater its resistance can be considered approximately constant over a fairly large voltage range. My heater that I use as a test load has a room temperature resistance of 8.6 ohms (this means no voltage applied and 68 deg F). At 118 V and the same ambient room temperature its resistance is 10.6 ohms. In turn I approximate this as a 12 A load at 120 V. More accurately it is 11.32 A.

If you have a 10,000 W heater at 240 V, then at its steady state on temperature its resistance is 5.76 ohms. As a first estimate its current at 120 V will be 20.8 A, just 1/2 of its current at 240 V. And the power dissipated would be 1/4 that at 240. However, the resistance will go up slightly at 120 V because the wire in the heater is cooler. So at 120 V current and power will be somewhat less than predicted by assuming a constant resistance.

.
 
gar said:
091202-1000 EST

I have no idea what the question is in the original post.

cadpoint:

PhaseShift's statement was correct. If you consider a specific heater its resistance can be considered approximately constant over a fairly large voltage range. My heater that I use as a test load has a room temperature resistance of 8.6 ohms (this means no voltage applied and 68 deg F). At 118 V and the same ambient room temperature its resistance is 10.6 ohms. In turn I approximate this as a 12 A load at 120 V. More accurately it is 11.32 A.

If you have a 10,000 W heater at 240 V, then at its steady state on temperature its resistance is 5.76 ohms. As a first estimate its current at 120 V will be 20.8 A, just 1/2 of its current at 240 V. And the power dissipated would be 1/4 that at 240. However, the resistance will go up slightly at 120 V because the wire in the heater is cooler. So at 120 V current and power will be somewhat less than predicted by assuming a constant resistance.

.
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THANKS GAR:

You are right, I missed the response of the EXACT question, "again"!

I thought I was talking about P/V= I

I didn't talk about or address "R" or a stead state. I surely didn't have test subjects or data available to pull in gee I just did some math.

If my math is incorrect please say so, my post can be taken out if so
inept, one more won't matter... go ahead and hit the panic "!" and qualify it!

I did say that the numbers where smaller, as to what is more real or realistic.

In thinking about again these values of the available "R" or available stead state just aren't worried about. What a person on the line usually only has is available wattage required by the item and available voltage that can be used, I hope they figure it out.
 
gar said:
However, the resistance will go up slightly at 120 V because the wire in the heater is cooler. So at 120 V current and power will be somewhat less than predicted by assuming a constant resistance.
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My problem with 'real-world' interjections like this is that the question is one of theory; nobody is actually going to connect a 240v heater to 120v and then wonder why the measured current is slightly different from the calculated current.

They ask why, and we answer these questions using Ohm's Law to explain the calculations, but then add the resistance-changes-with-temperature caveat, which can only hinder the grasp of the theory by the ones trying to learn.

I think it's better to learn the theory and math using linear suppositions, and worry about the details much later, after it becomes second nature. It's hard enough to remember which are the real-world variables and which are the constants.
 
The resistance can be calculted based on the heater voltage and wattage, A 1000watt heater rated for 240 volts has a resistance of 57.6 ohms. The resistance doesn't change because it is based on the design and construction of the heater. Using ohms law if 120 vots were applied to the heater it would only produce 250 watts. The formula is P(power)=E (voltage) squared divided by R (resistance).
 
gar said:
091202-1000 EST

However, the resistance will go up slightly at 120 V because the wire in the heater is cooler. So at 120 V current and power will be somewhat less than predicted by assuming a constant resistance.

.
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I believe resistance increases linerally with increasing temperature. So a cooler wire would have a lesser resistance, and therefore would expect the current and power to be slightly greater than theory.
 
091202-1455 EST

philly:

You are absolutely correct the resistance will be smaller at 120 V. I need to do more double checking of my comments.

I will get back to you sometime on your motor question. I started an answer, but I want to try to make it clear.

.
 
gar said:
091202-1455 EST

philly:

You are absolutely correct the resistance will be smaller at 120 V. I need to do more double checking of my comments.

I will get back to you sometime on your motor question. I started an answer, but I want to try to make it clear.

.
Click to expand...

Thanks gar! I appreciate your help and your knowledge :grin:
 
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