Electric heater question

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rjspark

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Location
Branchburg, NJ
I am presently working at a location that has existing electric heating. They are single phase 480V 3000W heaters. They have them fed from a 3 phase 20A breaker with 3- #8 awg feeding a 40A breaker mounted in a separate disconnect enclosure. Then they are wired through a 120V contactor, which is controlled by a 120V thermostat. Then there are 4 of the single phase 3000W heaters wired to the 3 phases. Now is it me or am I missing something:confused: I am going to bring my amp probe with me tomorrow to see exactly what these are drawing. I would appreciate any enlightenment to what I may or may not be missing. Thanks
 

augie47

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Location
Tennessee
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State Electrical Inspector (Retired)
agree, and with 4 heaters, two of the phases will see 9kw.
In most heating applications, NEC requires 125% for calculations so your 20 amp breaker is a bit small.
 

rjspark

Member
Location
Branchburg, NJ
So if you do the calculations, 4 x 6.25 = 25 x 1.73 = 43.25 X 1.25 = 54.0625A or 60A. Since the load is being distributed between the phases I'm not sure if you have to multiply the 1.73.
 

don_resqcapt19

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Staff member
Location
Illinois
Occupation
retired electrician
So if you do the calculations, 4 x 6.25 = 25 x 1.73 = 43.25 X 1.25 = 54.0625A or 60A. Since the load is being distributed between the phases I'm not sure if you have to multiply the 1.73.
You would divide by 1.73 for this application, but your method assumes that the load is equal across all 3 phases and that is not the case here.
Three heaters would have a load of 9000/480/1.73 or 10.84 amps. For two of the phases you would have to add 3000/480 or 6.25 amps for a total of 17.09. As Gus said, this is a continuous load and you have to size the OCPD and conductors for 125% of the total and a 20 amp OCPD is bit too small.
 

david luchini

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Location
Connecticut
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Engineer
I have seen arguments in the forum that the "continuous load" only applies to the branch circuits of electric heating equipment, and not to the feeders for electric heating equipment. (though i don't agree with that view)

If that is correct, then the 20A feeder circuit breaker would be properly sized (for non-continuous), and the 40A branch circuit breaker would be properly sized (for continuous load.)

However, the 40A branch circuit breaker would seem to violate 424.3(A)
 

rjspark

Member
Location
Branchburg, NJ
So then if the load is 17.09 x 1.25 = 21.3625A, breaker size would be 30A. Wire size would be #10 awg not #8 awg feeding another 40A breaker. But instead feeding a 30A Contactor box with thermal overloads sized for the individual loads on each phase for secondary OC protection for the conductors?
 

augie47

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Location
Tennessee
Occupation
State Electrical Inspector (Retired)
So then if the load is 17.09 x 1.25 = 21.3625A, breaker size would be 30A. Wire size would be #10 awg not #8 awg feeding another 40A breaker. But instead feeding a 30A Contactor box with thermal overloads sized for the individual loads on each phase for secondary OC protection for the conductors?
I have more questions than aswers :grin:
A 30 amp OCP would be permissible or a 25 amp woud be adequate.
#10- wire for either provided you don't enounter an high temperture problem.
I'm not sure where the 40 amp breaker now enters the picture, but if it being used to protect a #10 it is overisized (except for motors and other 240(G) applications).
I think the use of thermal overloads may well be restriced to motor applications
as I don't see the NEC addressing them in other overcurrent applications.
 

BJ Conner

Senior Member
Location
97006
IT's the Superposition Theory

IT's the Superposition Theory

The toatal load is 12,00 watts but it unequally connected.
Take the 9,000 that is evenly spread ( it may not be) .
The line current is
I = 9000/(480*1.73)= 10.83 Amps
Now take the same system and hook up one heater 3,000 accross two of the phase lines. The current in those two line will be I= 3000/480= 6.25 amps.
If you superimpose the two circuits two of the lines will have 17.08 Amps and the the third will have 10.83 Amps
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110406-2001 EDT

If did not make a mistake in my quick calculation I get 16.54 A for the two higher current legs.

This is from the
sq-root of [ {(6.25 + 6.25) + (6.25/2)}^2 + {0.866*6.25}^2 ] = 16.54 A

.
 

david luchini

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Staff member
Location
Connecticut
Occupation
Engineer
110406-2001 EDT

If did not make a mistake in my quick calculation I get 16.54 A for the two higher current legs.

This is from the
sq-root of [ {(6.25 + 6.25) + (6.25/2)}^2 + {0.866*6.25}^2 ] = 16.54 A

.

I agree with gar. If the load currents are Iab=12.5, Ibc=6.25 and Ica=6.25 and the loads are 120 degrees separated, then with Ia=Iab-Ica, Ib=Ibc-Iab, and Ic=Ica-Ibc:

I get Ia=16.54, Ib=16.54 and Ic=10.83.
 
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