Water Heater Circuit

[dvcraven0522]

I have a water heater that has Simultaneous Dual Element Operation (Unbalanced Three Phase connection) both element are 5000watts. Terminals - L2 is listed at 41.6 amps and L1 @ L3 are listed at 24 amps.

41.6 x 120V = 4,992watts phase B
24 x 120V = 2880watts phases A & C

This is 120/208V service. Total 3-phase power - 10,752

Does this look correct?

 

[ptonsparky]

I have a water heater that has Simultaneous Dual Element Operation (Unbalanced Three Phase connection) both element are 5000watts. Terminals - L2 is listed at 41.6 amps and L1 @ L3 are listed at 24 amps.

41.6 x 120V = 4,992watts phase B
24 x 120V = 2880watts phases A & C

This is 120/208V service. Total 3-phase power - 10,752

Does this look correct?

IDK why you used 120v. I suspect 208.

 

[kwired]

Sort of. guessing there is no neutral used here so 120 volts doesn't exactly play in there. A and C connect to one end of different elementws, the B line is connected to both elements opposing A and C - presuming both elements are same resistance that would make the B line carry 1.73 times the current of the other two lines.

 

[kwired]

IDK why you used 120v. I suspect 208.

Kind of coincidental that it comes out to approximately same watts but then the 1.73 factor times 120 does give you 208 so that is why it comes out about the same, difference really is in rounding off that square root of three and what stage did you round it at.

What OP really has is two independent loads that are operating at 208 volts and 5000 watts each connected across three phases. Current will be 1.73 times higher on the common line regardless which one you make the common. If he connected each element from line to neutral then the watts would go down by nearly 75% but current on the two lines and the neutral conductor would all be near identical.

 

[ptonsparky]

Still a 10KW load. Maybe only 5KW at times. Dependent on the control. I suspect some additional internal fusing or protection as well.

 

[synchro]

There would be 4992 VA and also 4992 watts drawn from phase B, because the line current would be in-phase with the L2-N voltage on a 208/120V system as a result of the two balanced loads.

2880 VA would be drawn from both phase A and phase C. Because each of the currents is 30° from its L-N voltage, the real power drawn = VA x cos(30°) = 2880VA x 0.866 = 2494 watts (which is half of the 5000W element rating within significant figures).

You can add the VA or kVA all together to get a total, but I'd be careful what you do with that because you'll be loosing information about the loading on each line conductor.

 

[ptonsparky]

There would be 4992 VA and also 4992 watts drawn from phase B, because the line current would be in-phase with the L2-N voltage on a 208/120V system as a result of the two balanced loads.

2880 VA would be drawn from both phase A and phase C. Because each of the currents is 30° from its L-N voltage, the real power drawn = VA x cos(30°) = 2880VA x 0.866 = 2494 watts (which is half of the 5000W element rating within significant figures).

You can add the VA or kVA all together to get a total, but I'd be careful what you do with that because you'll be loosing information about the loading on each line conductor.

Where does the Neutral come into play on 5kw 208volt elements?

 

[jim dungar]

Where does the Neutral come into play on 5kw 208volt elements?

It doesn't. But your first post infers you might have one because you talk of a VA rating on a single hot conductor.

You have one heater connected L1-L2 which is 24A @ 208V.
You have a second heater connected L3-L2 which is also 24A @ 208V.

To combine these into a single current of 41.6A on L2 you need to consider the phase angle between the two heater elements.

 

[synchro]

Where does the Neutral come into play on 5kw 208volt elements?

It's just a common reference point for all of the line voltages in a system that has a neutral like 120/240V, 208/120V, etc. It doesn't effect the current on loads to which a neutral is not connected, as Jim has just mentioned.

 

[ptonsparky]

It doesn't. But your first post infers you might have one because you talk of a VA rating on a single hot conductor.

You have one heater connected L1-L2 which is 24A @ 208V.
You have a second heater connected L3-L2 which is also 24A @ 208V.

To combine these into a single current of 41.6A on L2 you need to consider the phase angle between the two heater elements.

So if I have 5kw load L1-L2 and 2kw L2-L3, 24 and 9.6 A respectively, I have 29.15 A on L2

 

[kwired]

There would be 4992 VA and also 4992 watts drawn from phase B, because the line current would be in-phase with the L2-N voltage on a 208/120V system as a result of the two balanced loads.

2880 VA would be drawn from both phase A and phase C. Because each of the currents is 30° from its L-N voltage, the real power drawn = VA x cos(30°) = 2880VA x 0.866 = 2494 watts (which is half of the 5000W element rating within significant figures).

You can add the VA or kVA all together to get a total, but I'd be careful what you do with that because you'll be loosing information about the loading on each line conductor.

Where does the Neutral come into play on 5kw 208volt elements?

It doesn't. But your first post infers you might have one because you talk of a VA rating on a single hot conductor.

You have one heater connected L1-L2 which is 24A @ 208V.
You have a second heater connected L3-L2 which is also 24A @ 208V.

To combine these into a single current of 41.6A on L2 you need to consider the phase angle between the two heater elements.

You have 2 5kW loads.

The way engineering would label it on a panel schedule would be done the way synchro mentioned with 120 volts to neutral as a reference for each position in the panel. So two positions in the panel are supplying ~2880 VA each and the third is supplying 4992, that gives you 10752 total VA which is within reasonable accuracy considering rounding and the fact that actual voltage is likely a little more than the 208 rating.

But you can't view the current of 24 amps on one leg to be @ 120 volts as that would only be 2880 watts and we know that element is seeing 208 across it.

 

[synchro]

So if I have 5kw load L1-L2 and 2kw L2-L3, 24 and 9.6 A respectively, I have 29.15 A on L2

The currents on the L1-L2 and L2-L3 loads are 60° apart. If one is I_a and the other is I_b, then the resultant current on L2 would be
√ [ I_a² + I_a x I_b + I_b² ] . Note that if I_a = I_b, then the current will be the familiar √ 3 x I_a .
And so for your example above, the current on L2 would be √ [ 24² + 24 x 9.6 + 9.6² ] = 29.98A, or 30A to 2 significant figures.

 

[ptonsparky]

It doesn't. But your first post infers you might have one because you talk of a VA rating on a single hot conductor.

You have one heater connected L1-L2 which is 24A @ 208V.
You have a second heater connected L3-L2 which is also 24A @ 208V.

To combine these into a single current of 41.6A on L2 you need to consider the phase angle between the two heater elements.

I must be missing something. Usually am, but can you highlight which post, which area?

The currents on the L1-L2 and L2-L3 loads are 60° apart. If one is I_a and the other is I_b, then the resultant current on L2 would be
√ [ I_a² + I_a x I_b + I_b² ] . Note that if I_a = I_b, then the current will be the familiar √ 3 x I_a .
And so for your example above, the current on L2 would be √ [ 24² + 24 x 9.6 + 9.6² ] = 29.98A, or 30A to 2 significant figures.

Those dang details. Thank you.

 

[jim dungar]

I must be missing something. Usually am, but can you highlight which post, which area?

My error. It was the OP that mentioned VA per conductor.

 

[ptonsparky]

My error. It was the OP that mentioned VA per conductor.

Whew...

 

[winnie]

One thing to remember is that a single wire doesn't carry power. (You can even argue that the power doesn't even flow in the wires at all, but that is a different thread... )

Power requires voltage and current. You can measure the current in a single wire, but voltage always requires a reference point.

In the original post, you have only L-L connected loads, and you have 3 line currents, but for purposes of assigning 'VA per phase', the neutral was selected as the voltage reference. This was an arbitrary selection which happens to make 'VA per phase' proportional to phase current, useful if you are trying to balance the panel.

If the only thing you cared about was total power used, then you don't need to select the neutral as the reference. For a 3 phase system with only L-L loads, you could select one of the line conductors as your voltage reference. Such an unbalanced reference would give meaningless values for 'VA per phase' but is totally fine for measuring total system power.

Personally the idea of VA per phase is unnecessary. I prefer to think of the current loading on each phase. I mean what is the VA per phase on a corner grounded delta? If you use the 'natural' reference of the grounded conductor than you assign 0 VA to the grounded conductor, and the reference that gives the correct results (neutral) isn't even present in the system.

Jon