Hi, I'm wiring four new electric baseboards in my basement. All four are 220Volt units: 500W, 500W, 1000W, 1500W, for a total of 3500W. From the main panel to the farthest unit is less than 100ft. If I understood the rules, putting all of this on one circuit will require a 20Amp breaker (1.25*3500W/220V), and the wire can be #12AWG. Did I get that right, and am I missing any other design factor?
Many thanks, especially if you can reference the NEC so I know I'm doing it right. Theodore
OK first things first, heaters are rated at 240v not 220 so your total load is 14.6a.
OK lets look at the code (Fixed electric space heating art 424)
************** 424.3 Branch Circuits. (A) Branch-Circuit Requirements. Individual branch circuits shall be permitted to supply any size fixed electric space-heating equipment. Branch circuits supplying two or more outlets for fixed electric space-heating equipment shall be rated 15, 20, 25, or 30 amperes. ************* OK there, 20a is a listed size*******************
424.3(B) Branch-Circuit Sizing. Fixed electric space-heating equipment shall be considered continuous load. *******************"Continuous load" means you size the circuit to 125% of total load (14.6*1.25=18.25a) so 18.25a is OK on a 12 ga wire with a 20a breaker
Modern space heating equipment should meet the "disconnecting means" requirement with a switch marked "off" that disconnects both hot wires. (424.19(C))
I think you are good to go.
Fantastic. Thank you soooo much for the detailed reply. Very much appreciated.
You're going over the 80% trade standard load (16amps on a 20amp circuit) at that point, though...If you're going over 50ft or so, I'd probably use 10/2wg for the main run and come off that with a 12/2wg tail for each of the heater units. As each tail is only going to be asked to carry whatever heater it's connected to will ask for, not the combined load for all of them.
While I wouldn't expect to see an overheat fire risk situation using
12/2wg for the main run (If it's not for a long distance run...), I'd feel better knowing the main line I ran to feed them was more than upto the task and wasn't near full capacity if I ran one or all of the heaters at the same time.
Where is the over 16 amps coming from? All I see are 14.6 and it meets NEC.
Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so the
80% has been accounted for.
The load is 14.6A under perfect conditions of a wire 100ft or less in total length, yes. However, if the last heater happens to be a bit further away from the panel than 100ft as the OP suggested, there's a real risk of voltage drop to the last heater with the others running because the 12/2 wire just can't provide the full load that far out to the last device. A voltage drop will increase the amount of amps required by that heater to do it's job. The 12/2 isn't going to appreciate that and neither will the breaker. It will tolerate being over 20amps until it heats up enough to trip.
It's why I suggested 10/2. So, the OP won't even have to worry about problems down the road if their estimates are off. It certainly won't hurt the situation.
The end of run outlet is where the starvation is the worst and any inductive load will suffer the worst at that point so you must consider this as you overall design limit at 100feet.
A 20amp circuit fully loaded at this outlet will drop 5.6% where 5% is the RECOMMENDED maximum and we must install 10gage wiring if we still insist on a 20amp rated circuit. The other choice is to derate this long run to a 15 amp circuit on this 12g wire and you could possibly expect to be at 5% drop at 140-150feet. Tthe full 15 amp dribbles off at around 120 feet!
20amp circuit loaded at 18amp is 5%16amp will result in 4.5% missing at that far outlet.
The loss will be accumulative along the run as other 'stuff' is plugged in.
Most equipment that uses electricity is designed to function well at plus or minus 10%,whetner it is a resitive load like light bulb or an inductive load like a lighting ballast or motor.
This all becomes even more crucial when the power company is experiencing voltage sags in the summer high loads and the volatage that arrives at the pole down the street is already considerably LOW!!!---and if you designed your circuit lengths at the extra long end ---you are causing motor loads to labor,and lights to be dim!
Go with the far outlet at 100feet MAX...and you will be with the majority!!!
from the OPS original post: From the main panel to the farthest unit is less than 100ft.
So, the last heater on the circuit, depending on how far away it actually is from the panel and if the others are all running or not, could exceed the rated capacity for the wire and/or the breaker. IE: it's too far and the wire is too skinny to get the needed amps to the device.
I understand there is a cost difference going with 10/2 and some individuals consider that to be more important than the total amount of amps the circuit might pull when fully energized (all heaters on
- high) So, for short distance, and if the OP is very close with his figure on distance to last device on the circuit, your suggestion will save him money and work fine. If he's wrong or a bit off and it's a bit further, he might have to break up the heaters into seperate fed circuits to keep the amps low enough. Which will use additional wire and another breaker or two in his panel, assuming there's room for that.
At this point, whatever cost savings the OP had by not running 10/2 have disappeared.
Why even have to be concerned with it? Use the thicker wire, you KNOW each device will get all the amps it needs, without any possibility of an overload condition.
Like I said though, whatever choice the OP makes, it looks fine from here.
The heater is not going to decrease it's resistance and draw more current to compensate for the voltage drop. It will just be a slightly smaller heater. Heaters are rated at 120, 208 or 240v. If the voltage is lower, the heater will just have a lower output. An example is an oven element. At 240v it is rated 3600w, at 208 it is only 2700w
Like Gfre said, a resistance heater isn't going to increase it's current draw because there is some additional resistance in the circuit. It's like saying a light bulb will burn just as bright with a rheostat in the circuit to dim it, because it's going to pull additional current to compensate for the lower voltage. Put additional resistance in the circuit and the total current goes down, not up. And the difference here, the resistance in that wire is negligible.
There won't be problems if he follows the code. I can see using the heavier gauge wire if he thinks he might need to go to a larger load in the future, ie adding a heater or using larger ones.
A random internet posting is your cited source of information?
There are no inductive loads.
You don't show the calculations that is based on, and IDK how you came up with it. The resistance of 12 gauge wire is .0016 ohms per foot.
100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V . There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V drop from 240V is just 2.7%, not 5.6%.IKD where any of that is coming from, the relevant calcs are above.
What plugged in stuff? He's wiring a dedicated circuit for baseboard heaters, there are no receptacles.
And he's within 2.7%.
There is no motor load. If there were, we'd use the appropriate NEC to size for that.
The worse case is with all heaters on and that is what we sized for, 12 gauge is sufficient.
Good grief. We have used the rated capacity of the heaters. And there is already margin in the NEC, they don't say do the calcs, then upsize it more. You can if you want to, but there is no operational or safety issue here.
So, for short distance, and if the OP is very close with his
Nonsense as the NEC and above calcs show.
That's right, just use 12 gauge which NEC says is perfectly fine. You're the one throwing all kinds of FUD in and using incorrect calculations.
That isn't what I'm saying at all. I'm saying that the further out you go with 12/2 (or any wire really, it just depends on the thickness of the wire, the power you started with, and how far you're actually going), the less available voltage you'll have at the end of the run.
The less volts you have, the more amps you'll need to make up the difference to get the desired wattage.
Also, the 10/2 isn't going to be near capacity at any time, either. even if he has them all going on high at the same time. The 12/2 however, will be depending on the amount of heaters going and their setting(s).
going by Ugly:
7.9 volts lost 12/2 at 100ft 4.97 volts lost on 10/2 at 100ft240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*And that's due entirely to distance with no loads present on the line yet.
100ft on 12/2 according to my 2014 edition of Ugly's electrical reference is a net loss of 7.9 volts at the end of the run, with no load present, yet. At 125 ft out, the loss increases to 9.8 volts. And, obviously gets worse from there.OTH, the 10/2 wire loses 4.97 volts at 100ft and 6.21 volts at 125ft out. I'd rather get as many volts to the device (heaters) as is realistically possible. The closer I can get to their expected input voltage, the less amps they'll require to do their jobs. The more heat I'll get (which is obviously the point here) and the less power I'll use doing it. A win win win.
To do anything less is only costing me more money and time in the future. The 12/2 is going to heat up a bit more under various conditions than the 10/2 ever thought about doing, even if all heaters were on at the same time, on HIGH. Over time, the 12/2 wire will degrade due to heating/cooling cycles that the 10/2 won't have experienced. As it degrades, it's own resistance will increase.
Another side effect of running the wire warm/possibly hot to the touch at near full load over a period of time is that the connection points and terminals in the panel and the heaters will also become a little 'warmer' than they would if they'd been powered from the 10/2 line and pigtailed to it with a short 12/2 run.
Uhh, no. I remembered it was 100ft or so off hand as the general rule of thumb, but didn't remember how much voltage was lost as a result. I've also got various NEC books and my 2014 edition of the yellow Ugly electricians reference book. It's where the voltage drop figures I used today came from, actually.
Just to clarify, the calculations I provided in this post aren't FUD, unless you're able to dismiss the yellow Ugly electrician reference book. I haven't seen any 240volt baseboard heater wired with a 12/2 in sometime, actually. I think the last time I actually observed that was with a trailer. Alas, they're built with heavy consideration on cost. IE: as cheap as you can get away with. I don't work like that, I don't offer advice with that in mind first and foremost.
If it were me, writing only for myself, I'd spend the extra money for the heavier gauge wire and use the 12/2 wire for pigtailing off of it into the heater units. I've already explained why I'd run it in this manner. No real point in doing so again.
With no load, there is no voltage drop at all.
If you are actually dropping 7.9v you do not have a 1500w heater anymore. The element is going to be still around 38.4 ohms (the element did not change, just the voltage) so at 232.1 v heater draws
6.04a and becomes a 1402.8w heater.This means nothing at this point unless you are changing the element to get 1500w at 232v.
Heaters are supposed to be intelligent enough to figure out when they need to shorten their elements :-)
I'm often surprised by how little some people know about electricity. Voltage drop depends on current flow. There's no power (wattage) without current flow either.
So you have a load which is not present. How is it drawing power?
[ship]
I sincerely do apologize for having missed that in my original reply to you. With that said, I'd still have opted for 10/2 to feed them, and come off the feed with a 12/2 for each heater. This reduces heat creation along the wire, extends the life of the chosen wire, ensures the greatest possible amount of power is available to each heater on the circuit.
Although power loss will still occur at the last one, the voltage drop won't be as bad as it would have been with a 12/2 run from the panel; and the feed wire won't run as hot supplying power to an individual heater or all of them at the same time. A bit more costly, but, a decent enough tradeoff imo, that I would personally have used
10/2.As I said though, at the end of the day, it all looks great from here.
[snip]
Where can I get one of these magic heaters, that uses 1500 watts when it isn't connected?
I sometimes feel the same way.
If you put 120volts AC on a 12/2 wire, and run that wire, 100 ft out, and take a measurement at the end of the run, the available voltage at the end of the run isn't going to match the volts going into the wire. Some is lost during transit due to the wires resistance and the length.
If you try this experiment using DC power, it's even worse. DC really doesn't travel distance well.
For example, an electric space heater may have a resistance of ten ohms, and the wires which supply it may have a resistance of 0.2 ohms, about 2% of the total circuit resistance. This means that approximately 2% of the supplied voltage is lost in the wire itself. Excessive voltage drop may result in unsatisfactory operation of, and damage to, electrical and electronic equipment.
The simplest way to reduce voltage drop is to increase the diameter of the conductor between the source and the load, which lowers the overall resistance. In power distribution systems, a given amount of power can be transmitted with less voltage drop if a higher voltage is used. More sophisticated techniques use active elements to compensate for the undesired voltage drop.
You aren't losing power due to the load, You've already lost it in route to the load via wire resistance and the distance said electricity has to travel on the wire. Thicker wire, less resistance, less voltage drop. Simple concept, really.
Like you wrote though, it does sometimes amaze me how little people know/understand about electricity.
The voltage drop is due to the wire's own resistance, and the distance the voltage must travel. The wire we're using isn't a super conductor. It has a certain amount of resistance to it. As a result, some volts are no longer available to us, we spent them getting the rest down the line. Nothing for free, you know.
Watts is watts, man. Available voltage determines how many amps it's going to take to get them, though. Lower voltage=more amps to do the same job.
Voltage drop still depends on the load. If there is no load, there is no voltage drop. If you have a wire with no load at all, there will be full circuit voltage at both ends.
If you have a lower voltage, the watts will be lower. In our example, that 1500 watt heater (at 240v) will be 1407 watts at 232v. Current will actually be less, not more.
Hmm? Oh.. ROFL, I see. That's what happens when you don't pay attention to context and choose to selectively quote.
Those are load calculations based on a 1500watt element running at two different voltages. It shows the amount of amps required to get
1500watts at those voltages.The different voltages are directly related to length and diameter of wire and have nothing to do with the load from the heater at this point.
Thanks to the wire and the wire alone, some volts have already been lost (as in they'll never reach the heater) in transit, due again, to the wires own resistance and length. In this case, the length is the same; 100ft. But, the wire size or gauge is not. 12ga is on the top,
10ga is on the bottom; for comparison.Might I suggest a simple english comprehension class or two? Muggles suffers from a similar problem. So, you're in very good company.
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