Feeder circuit for electric heaters

[PhaseShift]

I am calculating the size for a feeder circuit to feed (4) 5kW heaters at 240V. Each heater has a rated current of 21A.

We will have a feeder circuit fed from a disconnect with the feeder circuit feeding a 120/240V panel. The panel will then have individual branch circuit to feed each of the (4) heaters.

Section 424.3(B) states that heaters must be treated as a continuous load and thefore each branch circuit must be sized at 125%. So in my case the branch circuit for each heater will be 21A*1.25= 26.25A. Therefore each heater will have a #10AWG fed from a 30A OCPD.

For the feeder circuit to calculate the toal load I add up all (4) heaters and then take 125% of this value because they are treated as continous loads. Doing this I get 21A *4 * 1.25 = 105A. This tells me that the feeder circuit must have an ampacity of at least 105A. Is this correct? Is there anything in the code that allows this to be rounded down to 100A or must it be 105A at a minimum.

Assuming feeder circuit must have an ampacity of 105A I was going to use a disconnect rated larger than 100A. I'm not sure what the next size up is but I'll assume there is a disconnected rated for 125A. So I will use a 125A rated disconnect with 110A fuses since feeder cable will be existing #2AWG cable. Fuses in disconnect must be 100A in order to meet size of load as well a protect #2 cable rated at 115A. The 120/240V panel has a rating of 200A with a 200A MCB.

Does my above design sound correct?

 

[Dennis Alwon]

I believe the feeder would calculate as 4 units at 5000 watts and then take 125% of just one unit. Thus you would have 4*5000 watts= 20000.
Then add 25% of 5000 watts= 1250. Thus you have 21250/240==88 amps. You can use a 90 amp breaker.

 

[PhaseShift]

I believe the feeder would calculate as 4 units at 5000 watts and then take 125% of just one unit. Thus you would have 4*5000 watts= 20000.
Then add 25% of 5000 watts= 1250. Thus you have 21250/240==88 amps. You can use a 90 amp breaker.

Why would you only take 125% of one of the heaters as opposed to adding up the heaters and then taking 125% of the continuous load as you would with any other continuous load?

 

[ActionDave]

NEC allows for "demand factors." While they love there 1.25%, the code makers are reasonable enough to know that not every heater is going to be at max current draw at the same time.

 

[suemarkp]

Is this for a dwelling? If so, can you use the Optional calculation sections 220.82(C)(4) or (5)? Your feeder conductors must still have an ampacity of 100 to use this method. What I am unsure of is whether this method can be used for feeders that don't serve the whole dwelling. It doesn't have the full restriction wording of 310.15(B)(6), but it seems to me as long as the dwelling unit as a whole meets the rule, I don't know why you can't apply it to sub feeders with an ampacity of 100 or more.

 

[PhaseShift]

Is this for a dwelling? If so, can you use the Optional calculation sections 220.82(C)(4) or (5)? Your feeder conductors must still have an ampacity of 100 to use this method. What I am unsure of is whether this method can be used for feeders that don't serve the whole dwelling. It doesn't have the full restriction wording of 310.15(B)(6), but it seems to me as long as the dwelling unit as a whole meets the rule, I don't know why you can't apply it to sub feeders with an ampacity of 100 or more.

Each of these heaters will be located in a storage conex, and each heater will be running at the same time.

Even if we know that all heaters will be running at the same time and will be continuous loads can we still apply divirsity factors in this situation? What makes these heaters different from any other type of continuous load?

 

[ActionDave]

While you are required to treat them as a continuous load for the calculations, in reality they are not. T-stats cycle the load on and off and the resistance changes as the heating elements warms up.

 

[hurk27]

don't think you can apply a demand factor to space heating equipment"

220.15 Fixed Electric Space Heating.
Fixed electric space heating loads shall be computed at 100 percent of the total connected load; however, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.
Exception: Where reduced loading of the conductors results from units operating on duty-cycle, intermittently, or from all units not operating at the same time, the authority having jurisdiction may grant permission for feeder and service conductors to have an ampacity less than 100 percent, provided the conductors have an ampacity for the load so determined.

I don't see where a reduction can be given since anytime they are first turned on, they will all come on, lets say you have a power failure, and the temperature lowers below the thermostat setting, when power is restored they will all come on.


I don't see the requirement for using the 125%, but if the feeder breaker is not rated for continuous use then you will have to derate the circuit, or again apply the 125% most 100% breakers will be bolt in.

 

[hurk27]

edit the above to change the article number to :220.51, didn't relized this change in the 2008

 

[ActionDave]

Oops. Demand factor was a poor choice of words since there are specific percentages for those.

Still, 125% for the first heater, 100% for the remaining is all that is required, no need for 125% of all the heaters.

 

[PhaseShift]

don't think you can apply a demand factor to space heating equipment"




I don't see where a reduction can be given since anytime they are first turned on, they will all come on, lets say you have a power failure, and the temperature lowers below the thermostat setting, when power is restored they will all come on.


I don't see the requirement for using the 125%, but if the feeder breaker is not rated for continuous use then you will have to derate the circuit, or again apply the 125% most 100% breakers will be bolt in.

This makes sense considering this is a feeder. So since its a feeder we can use 100% of total load as referenced above.

However if the case were that all (4) heaters were on a branch circuit in parallel then would the branch circuit have to be calculated according to 424.3(B) and use 125% of total heater load to size circuit and OCPD?

 

[brother]

This makes sense considering this is a feeder. So since its a feeder we can use 100% of total load as referenced above.

However if the case were that all (4) heaters were on a branch circuit in parallel then would the branch circuit have to be calculated according to 424.3(B) and use 125% of total heater load to size circuit and OCPD?

Yes, if its a 'branch circuit' it must be calculated at 125%.

 

[david luchini]

This makes sense considering this is a feeder. So since its a feeder we can use 100% of total load as referenced above.

I disagree with this. As you point out in the OP, article 424 says the heater loads should be treated as "continuous load."

Article 220.51 says that heater load shall be calculated at 100% of connected, that is to say, no demand factors apply. So your heater load on the feeder is 84Amps, continuous load.

Article 215.2(A)(1) says that the feeder shall have an ampacity not less the required to supply the load as calculated in Art. 220 (see 220.51 above.) It also says the feeder shall have an ampacity not less than the noncontinuous load plus 125% of the continuous load. The feeder, per 215.2, would have to have an ampacity not less than 105A.

 

[brother]

I disagree with this. As you point out in the OP, article 424 says the heater loads should be treated as "continuous load."

Article 220.51 says that heater load shall be calculated at 100% of connected, that is to say, no demand factors apply. So your heater load on the feeder is 84Amps, continuous load.

Article 215.2(A)(1) says that the feeder shall have an ampacity not less the required to supply the load as calculated in Art. 220 (see 220.51 above.) It also says the feeder shall have an ampacity not less than the noncontinuous load plus 125% of the continuous load. The feeder, per 215.2, would have to have an ampacity not less than 105A.

I think you have over 'thought' this process. If its a 'feeder' supplying heaters, you are alowed to calculated only the 100% of the rated load. the 125% calculation does not apply with heaters only on a feeder.

This is allowed because of the reduction of risk of the ALL heaters NOT LIKELY TO BE running all the time at the same time in the same place. This also tells why the CMP says the 'feeder in no case can be smaller than the largest 'branch' circuit of the largest heater'.

 

[david luchini]

I think you have over 'thought' this process. If its a 'feeder' supplying heaters, you are alowed to calculated only the 100% of the rated load. the 125% calculation does not apply with heaters only on a feeder.

What code section says that? 215.2 says that the minimum feeder conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125% of the continuous load. I don't see where it says "125% of continuous load, except for heaters."

424.3(B) says that fixed electric space heating equipment shall be considered continuous load.

This is allowed because of the reduction of risk of the ALL heaters NOT LIKELY TO BE running all the time at the same time in the same place. This also tells why the CMP says the 'feeder in no case can be smaller than the largest 'branch' circuit of the largest heater'.

The original poster mentioned that all of the heaters WILL be running at the same time:

Each of these heaters will be located in a storage conex, and each heater will be running at the same time.

So you have four 5kW, 240V heaters (21A each) considered continuous load, all running at the same time for a total of 84Amps of continuous load. How would a feeder with an ampacity of less than 105A comply with 215(A)(1)?

 

[Dennis Alwon]

I had the impression these were 4 different units. If this is one unit fed with 4 branch circuits of 5kw each then we have a different animal in calculating the feeder--

Wayne is absolutely correct however if the situation were as above then I would run a feeder large enough for a 20kw units because that is what you have. Most central electric heater have one or more elements but are figured as one, I am curious as to this setup.

 

[david luchini]

I had the impression these were 4 different units. If this is one unit fed with 4 branch circuits of 5kw each then we have a different animal in calculating the feeder--

Dennis, I read this as four separate heaters, each with its own branch circuit, being fed from a panel. The question is what size must the feeder be for the panel.

The feeder load would be 84Amps, continuous. Therefore the feeder should have an ampacity of at least 105Amps, per Art. 215.

 

[brother]

What code section says that? 215.2 says that the minimum feeder conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125% of the continuous load. I don't see where it says "125% of continuous load, except for heaters.

The feeder calculation is found in 220.51 for the Fixed Electric space Heating. The thing I think you are missing is that you do NOT apply any rules of the branch circuits calculations to the feeder calculations unless it specifically tells you to.

424.3(B) says that fixed electric space heating equipment shall be considered continuous load.)

You have to read that 'whole' section, NEC 2008 424.3(B) deals with BRANCH CIRCUITS not FEEDERS. thats where 125% will apply, not to FEEDERS. you go to section 220 to calculate feeders, in which it tells you to use only the 100% of the heaters. To go even futher, the exception allows the AHJ to let you go even lower than the 100% of the rated load on FEEDERS for heaters if you can show it can handle the load. Thats where the 'unspoken' demand factor is applied.

The original poster mentioned that all of the heaters WILL be running at the same time:.)

Even then, this is still dealing with FEEDERS not branch circuits. and in reality all the heaters I know of have a 'thermo cuttout'. So even then the odds of the all heaters running to overload a feeder is very small. Thats why its allowed to use the reduction in 220 CALCULATIONS.

So you have four 5kW, 240V heaters (21A each) considered continuous load, all running at the same time for a total of 84Amps of continuous load. How would a feeder with an ampacity of less than 105A comply with 215(A)(1)?

Again you are misapplying these sections. I think you meant NEC 2008 215.2(A)(1) It even tells you to use NEC section 220 for the FEEDER calculations NOT branch circuits. Therefore the 125% of continous load for the BRANCH CIRCUITS of section 424.3(B) for heaters does NOT apply to the sizing/ampacity of the FEEDERS.

You can almost think of this as the sizing of 'service' , this is why you can use smaller wire when feeding the buildings. I hope I explain this better.

 

[david luchini]

The feeder calculation is found in 220.51 for the Fixed Electric space Heating. The thing I think you are missing is that you do NOT apply any rules of the branch circuits calculations to the feeder calculations unless it specifically tells you to.

I think the LOAD calculation is found in 220.51. It says that four heaters at 5kw shall be calculated as a load of 20kW (ie, 100% of connected) even if the heaters may not run at the same time, UNLESS, the AHJ grants you permission to use a lower load.

Article 215.2(A)(1) tells you the minimum feeder size for the LOAD calculated in Article 220.

You have to read that 'whole' section, NEC 2008 424.3(B) deals with BRANCH CIRCUITS not FEEDERS. thats where 125% will apply, not to FEEDERS. you go to section 220 to calculate feeders, in which it tells you to use only the 100% of the heaters. To go even futher, the exception allows the AHJ to let you go even lower than the 100% of the rated load on FEEDERS for heaters if you can show it can handle the load. Thats where the 'unspoken' demand factor is applied.

Again, Art. 220 tells you how to calculate the load, not the feeder size. Article 424.3 deals with branch circuits (it doesn't mention feeders at all,) but how can a load be continuous on a branch circuit and non-continuous on the feeder feeding the branch circuit? It can't. If the load is continuous on a branch circuit, then it it continuous on the feeder, also.

Even then, this is still dealing with FEEDERS not branch circuits. and in reality all the heaters I know of have a 'thermo cuttout'. So even then the odds of the all heaters running to overload a feeder is very small. Thats why its allowed to use the reduction in 220 CALCULATIONS.

And if you have 4 heaters that run at the same time for a 6 hour period, what reduction would you use then? It doesn't matter what the "reality" of how the installation will operate is, the installation must comply with code.

It even tells you to use NEC section 220 for the FEEDER calculations NOT branch circuits. Therefore the 125% of continous load for the BRANCH CIRCUITS of section 424.3(B) for heaters does NOT apply to the sizing/ampacity of the FEEDERS.

Where does the NEC tell you that. Article 220 is called "Branch-circuit, feeder and service calculation." I would tend to use 220 for branch-circuit AND feeders load calculations. But in any event, the LOAD calculated for the electric-heating using Article 220, must have a feeder sized per 215.2(A)(1). This would require the feeder to have 125% of the continuous load.

 

[reddy kw]

I am calculating the size for a feeder circuit to feed (4) 5kW heaters at 240V. Each heater has a rated current of 21A.

We will have a feeder circuit fed from a disconnect with the feeder circuit feeding a 120/240V panel. The panel will then have individual branch circuit to feed each of the (4) heaters.

Section 424.3(B) states that heaters must be treated as a continuous load and thefore each branch circuit must be sized at 125%. So in my case the branch circuit for each heater will be 21A*1.25= 26.25A. Therefore each heater will have a #10AWG fed from a 30A OCPD.

For the feeder circuit to calculate the toal load I add up all (4) heaters and then take 125% of this value because they are treated as continous loads. Doing this I get 21A *4 * 1.25 = 105A. This tells me that the feeder circuit must have an ampacity of at least 105A. Is this correct? Is there anything in the code that allows this to be rounded down to 100A or must it be 105A at a minimum.

Assuming feeder circuit must have an ampacity of 105A I was going to use a disconnect rated larger than 100A. I'm not sure what the next size up is but I'll assume there is a disconnected rated for 125A. So I will use a 125A rated disconnect with 110A fuses since feeder cable will be existing #2AWG cable. Fuses in disconnect must be 100A in order to meet size of load as well a protect #2 cable rated at 115A. The 120/240V panel has a rating of 200A with a 200A MCB.

Does my above design sound correct?

Three questions:
(1) Voltage Drop-How long are your wire runs? Are feeder or branch runs long enough to require de-rating of conductor to prevent voltage drop?
(2) Temperature rating of insulation-What is maximum ambient temperature? NEC may require de-rating conductor ampacity. Temperature rating is relative to operating temp.
(3) What is insulation on existing 2 AWG feeders? Old RW/RHW/TW/THW are rated lower then modern THHN/TWHN, once again NEC requires de-rating.
Plus the de-rating is additive, I.E. if TW insulation requires 20% derating, higher ambient temp. requires de-rating of 20%, and length of wire run requires derating of 15% to keep voltage drop below 3%, the wire would be rated at 45% of its listed max. ampacity. Plus other considerations, like number of cables in raceway/conduit also affect rated ampacity.
Considering the 80% loading rule of thumb as good practice, your 84 amps over 2 AWG is pushing it to the limit. Take the NEC requirements for de-rating into acount and 2 AWG could be half the size needed.