Baseboard heat, continuous load?

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mark32

mark32

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Currently in NJ
Hi, I will be running a feeder for two 8' baseboard heaters, both on the same t-stat. The total draw will be 16.6a, just wondering if baseboard heaters would be considered a continuous load. I assume they would cycle during a three hour period, just want to make sure.
 
Wow you are right on the cusp. Article 220.5(B) could have given you some leeway

220.5(B) Fractions of an Ampere. Where calculations result in a fraction of an ampere that is less than 0.5, such fractions shall be permitted to be dropped.
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Looks like you just missed it. I bet in reality your voltage is higher than 240 so your amp. would be lower.
 
"I bet in reality your voltage will be higher than 240 so you amp will be lower"

Not true IMHO, a simple resistance such as an electric heater will draw current in proportion to the voltage applied.
Therefore if the voltage is 5% high at 252 volts, the current will also increase by 5% to 17.5 amps.
 
090224-1013 EST

Dennis:

Broadgage is correct. Increase the voltage and the current goes up.

At 30 deg F outside the on-to-off time might be 1 to 3. But at -10 outside it is almost certainly on continuously. I would design for continuous. What does NEC say?

If the heater is working in an on-off cycle, and you increase the voltage, then averaged over several on-off cycles the average current goes down. This is because the average power remains constant. Assuming constant temperature control of the on-off cycle.

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090224-1221 EST

I should clarify that the average power dissipation in the wiring will remain constant if the average power dissipation of the heater is held constant by the thermostat as the voltage varies. The heater is a fixed resistance and so is the wiring. So if the average power is constant in one it will be in the other.

.
 
Gar, I don't have my codebook with me but 424.3(B) states something like: "Fastened in place electrical heating equipment shall be considered a continous load".

Thanks guys
 
LarryFine said:
Oops. Looks like someone needs a refresher course in Ohm's Law. :cool:
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Obviously I do--- I was just plugging in numbers in the formula
I= Watts/voltage--- my bad. Hey it wasn't even 7 am on that post. Larry you were probably just getting to bed. :D

As always thanks for making me look like an ass :D
 
Dennis Alwon said:
Obviously I do--- I was just plugging in numbers in the formula
I= Watts/voltage--- my bad.
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The one that would require an equipment change to work? :D

Hey it wasn't even 7 am on that post. Larry you were probably just getting to bed. :D
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Sometimes. :roll:

As always thanks for making me look like an ass :D
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Well, if the foo . . . :cool:
 
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