How to figure amp draw?

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Steelhead

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Hi Guys,

I have a three phase circuit feeding a bunch of heaters. Specifically, there is a 3 pole distribution block with 6 conductors terminated to each phase. I have measured 2.9A on each of the 6 conductors. This holds true for each of the other two phases as well. When I measure the total amp draw per phase I get about 14.7A, but adding up each of the 6 conductors (2.9A x 6) I get 17.4A. Why do I get 14.7A when the added amp draws equals 17.4A?
 
It sounds like this is a delta connected configuration where you have three groups of three heaters. The currents in each group are out of phase with the other groups. So the currents do not add directly. However I would expect the difference between the line current and the phase current to be by a factor of sqrt(3).

Can you provide additional details on the voltage and the wiring configuration?
 
I_line = sqrt(3) * I_phase

We take I_phase to be the current in one bank of heaters, 2.9 * 3 = 8.7 A.

Thus, I_line is 15.1 A. This is close to the 14.7 A you measured.
 
This may be important and explain it as well. Are those single phase loads off to the right? (Connected from L1 to L3.)
 
bwat said:
This may be important and explain it as well. Are those single phase loads off to the right? (Connected from L1 to L3.)
Click to expand...

Yes, basically L1-L2=4200W, L2-L3=4200W and L1-L3=4200W. My math then is: 4200W/480V=8.75A, Line current would then be 8.75A*1.73=15.1A

I was getting about 14.7A

Am I looking at this correctly?
 
The way I see it you have two 700 watt elements in series across 480 VAC. So amps per element is 700/240 = 2.9167. So 8.75 amps for each of the three single phase loads (20-20A + 22-22A + 24-24A). So 8.75 X 1.73 = 15.1375 amps.

This assumes 480 VAC on the dot between all phases and each element being perfectly matched. And yes, what Larry says what are 4-ac, 5-ac, and 6-ac?
 
Steelhead said:
When I measure the total amp draw per phase I get about 14.7A, but adding up each of the 6 conductors (2.9A x 6) I get 17.4A. Why do I get 14.7A when the added amp draws equals 17.4A?
Click to expand...

I agree with PeterOven.
The reason that the line current is √3 ≈ 1.733 times the current through each of two equal delta connected loads is that the load currents are at 60° from each other.

Just to hopefully understand this better, first consider the following:
If instead the load currents were in-phase (at zero degrees from each other) then of course they would sum directly and you'd get twice the current with two equal loads.

But if two loads were driven by opposite phases on a 3-wire split-phase system at 180° from each other, then two equal L1, L2 load currents would cancel at the neutral. All of the current would flow through each load to the other and there would be zero current through the neutral conductor.

Now in a 3-phase delta the 60° phase relationship is between these 0° and 180° cases, and the portion of the current on one load that is in-phase with an equal load on the other leg of the delta is sin(60°) = √3/2 ≈ 0.866. Then if you add the two equal in-phase portions from each leg you get √3 ≈ 1.733 times the load current on each leg to get the total line current.
Note that this does not mean any current disappeared because you didn't get 2 times the load current. It just means that some of the load current in one leg of the delta flows over via its connection to the other leg of the delta, instead of all of it going through the line conductor that's connected to the two legs. As mentioned above in the split-phase case all of the load current would pass to an equal load on the other leg instead of the neutral, and so this phenomena should not be too difficult to understand (I think).
 
Russs57 said:
The way I see it you have two 700 watt elements in series across 480 VAC. So amps per element is 700/240 = 2.9167. So 8.75 amps for each of the three single phase loads (20-20A + 22-22A + 24-24A). So 8.75 X 1.73 = 15.1375 amps.

This assumes 480 VAC on the dot between all phases and each element being perfectly matched. And yes, what Larry says what are 4-ac, 5-ac, and 6-ac?
Click to expand...

It’s actually RC. It’s an option that is not installed on the machine so those can be disregarded, sorry
 
Steelhead said:
Yes, basically L1-L2=4200W, L2-L3=4200W and L1-L3=4200W. My math then is: 4200W/480V=8.75A, Line current would then be 8.75A*1.73=15.1A

I was getting about 14.7A

Am I looking at this correctly?
Click to expand...

Don't forget that there are manufacturing tolerances to your heaters, they may not be exactly 4200W, then there is the tolerance of the supply voltage it is not likely 480.0V, as the voltage goes down so does the current draw.
 
I'll add that you are lucky to have a well designed system. You mention six leads on each phase so I'll assume total system is 6 X 12.6 KW = 75.6 KW

Imagine of you had six 4,200 watt elements in series on each phase (call them 1A thru 1F on A phase and 2A thru 2F on B phase and 3A thru 3F on C phase). Then tie all the "F" ends together. Call it a floating Wye scheme. Burn out any one element, say 1D, and you are down to 50% capacity. Burn out another on a different phase, say 3B, and you have absolutely no heating at all.
 
Russs57 said:
I'll add that you are lucky to have a well designed system. You mention six leads on each phase so I'll assume total system is 6 X 12.6 KW = 75.6 KW

Imagine of you had six 4,200 watt elements in series on each phase (call them 1A thru 1F on A phase and 2A thru 2F on B phase and 3A thru 3F on C phase). Then tie all the "F" ends together. Call it a floating Wye scheme. Burn out any one element, say 1D, and you are down to 50% capacity. Burn out another on a different phase, say 3B, and you have absolutely no heating at all.
Click to expand...

Interesting.
 
You seem to be interested in learning. That is something us old farts love to see (and I assume you are younger than me).

So see if you can figure out why a loss of any one element brings you to 50% rather than say 66%. The internet makes it so easy to cheat these days.

Being that I'm old and senile it could be that I'm wrong. And of course, sometimes us old curmudgeons like to send the new kid back to the shop for ....well never mind:)
 
Russs57 said:
So see if you can figure out why a loss of any one element brings you to 50% rather than say 66%. The internet makes it so easy to cheat these days.
Click to expand...
To be clear, you meant the loss of one of the lines/phases (L1,L2, or L3), right? And not the loss of an element. Losing an element in a delta connected heater takes you to 2/3 capacity. Losing a phase takes you to 1/2 as you stated.
 
bwat said:
To be clear, you meant the loss of one of the lines/phases (L1,L2, or L3), right? And not the loss of an element. Losing an element in a delta connected heater takes you to 2/3 capacity. Losing a phase takes you to 1/2 as you stated.
Click to expand...
I believe Russ57 was talking about an wye configuration of loads with their center connection allowed to "float"and not a delta.
The only discrepancy I see with saying the power goes down by 1/2 with a loss of one of the three element chains is that the element resistance will be less than at full power due to its variation with temperature. Therefore the power won't go down quite as much as by 1/2. But that's probably beyond the scope of the intended problem.
 
synchro said:
I believe Russ57 was talking about an wye configuration of loads with their center connection allowed to "float"and not a delta.
The only discrepancy I see with saying the power goes down by 1/2 with a loss of one of the three element chains is that the element resistance will be less than at full power due to its variation with temperature. Therefore the power won't go down quite as much as by 1/2. But that's probably beyond the scope of the intended problem.
Click to expand...
ahhh.. I believe you're right. I missed the switch to wye then. Thought we were still on the OP's original delta connection. Apologies to Russ.
 
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