Unbalanced water heater overcurrent protection

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hhsting

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See attached data sheet 3 of 4 water heater which has 6 kw simultaneous AO Smith unbalanced water heater 480V three phase. Water heater terminal full load amps are L1 has 21.6A and L2 has 12.5A and L3 has 12.5A. How do you size over current protection? 3 pole breaker with one pole 30A, second pole 20A and third pole 20A?
 

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You apparently should wire it as a straight 30a circuit.

Otherwise, yes, you'd need to assemble a hybrid breaker with a 3-ph handle tie.
 
hhsting said:
See attached data sheet 3 of 4 water heater which has 6 kw simultaneous AO Smith unbalanced water heater 480V three phase. Water heater terminal full load amps are L1 has 21.6A and L2 has 12.5A and L3 has 12.5A. How do you size over current protection? 3 pole breaker with one pole 30A, second pole 20A and third pole 20A?
Click to expand...
I think I would wire it and protect it all to the maximum current draw on any phase, i.e., three phases at 21.6A.
 
Is appliance marked with minimum circuit or max overcurrent? If so go with whatever that says and it will comply with 422.11(E)(1)

Otherwise 422.11(E)(3) says: "Not exceed 150 percent of the appliance rated current if the overcurrent protection rating is not marked and the appliance is rated over 13.3 amperes. Where 150 percent of the appliance rating does not correspond to a standard overcurrent device ampere rating, the next higher standard rating shall be permitted"

Which is what lets us put typical 4500 watt 240 volt single phase water heaters on 30 amp breakers when otherwise they would be required to be on a 25 amp breaker. Thing that is different with your application though is you do have two conductors with load under 13.3 amps.

Reason you have this unbalance is it only has two elements connected in open delta configuration. Same appliance is likely intended to be connected to single phase supply by connecting the two elements parallel to one another, and in that case you would have 25 amps on each line and would require 31.25 amp conductors and 35 amp overcurrent protection - still supplying the same elements.

My guess is 30 amp three pole breaker is acceptable, if anything because max load on any one conductor of the circuit demands it, you still must use 30 amp conductor on those other two conductors though.
 
hhsting said:
Don’t follow if it’s unbalanced load then why wouldn’t neutral be brought to the water heater?
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Because each element runs at 208v, not 120v. The phase that connects to both elements is wired as you would a neutral on 120/240v, but no actual neutral is required.
 
See attached. At 480V, 6kw, simultaneous dual elements L1 has 21.6A and L2 & L3 has 12.5A. If both elements are single phase 480V then where is 21.6A coming from?
 

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hhsting said:
Don’t follow if it’s unbalanced load then why wouldn’t neutral be brought to the water heater?
Click to expand...

It is two 480 volt elements of same rating connected in open delta fashion. The most balanced you can get is with one supply leg carrying 1.732 times the current as the other two. If you bring a neutral there is nothing to connect it to, if you replace one leg with neutral you only get 277 volts across which ever element(s) it connects to and that will result in less wattage.

Same unit can be made to operate off single phase input by simply putting the two elements in parallel.
 
hhsting said:
See attached. At 480V, 6kw, simultaneous dual elements L1 has 21.6A and L2 & L3 has 12.5A. If both elements are single phase 480V then where is 21.6A coming from?
Click to expand...
Each element draws 12.5A when supplied with 480V, two-wire. Element 1 is connected from L1 to L2. Element 2 is connected from L1 to L3.

So the line currents on L2 and L3 are both 12.5A. The line current on L1 is sum of the two load currents, taking into account the phase angle. That sum is 21.6.A.

No neutral conductor is involved.

Cheers, Wayne
 
Just expanding on what Wayne said, the component (i.e., portion) of the current flowing through one heating element that's in-phase with an equal component flowing in the other element is cos(30) x 12.5A = 0.866 x 12.5A = 10.8A. So both in-phase components added together is 2 x 10.8A = 21.6A on L1.
So what happened to the rest of the 12.5A element current? It didn't disappear. Instead it flows from one heating element on through the other. It is the same amount of current that would flow if L1 was left open and the 480V across L2 and L3 would be split evenly as 240V on each element. Therefore this current would be 12.5A/2 = 6.25A. It can also be viewed as the component of the 12.5A element current that is 90 degrees to the L1 output current, or sin (30) x 12.5A = 6.25A.
All this can be double-checked by taking the RMS sum of the in-phase and 90 degree (quadrature) components of the element current:

sqrt ( (10.8)**2 + (6.25)**2 ) = 12.5 A, which is back to the total RMS current flowing in one element.
 
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