3 phase asymmetrical load calc

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Joethemechanic

Joethemechanic

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Electro-Mechanical Technician. Industrial machinery
I think I'm having a senior moment. This is a 3 phase, two element water heater (booster). What is the actual current on leg B?

Please forgive my drawing skills

three-phase-three-wire-y-connection.jpg
 
It's a problem in vector algebra. The sum of the current vectors going into the BØ node between the two resistors has to add up to zero.

IA+IB+IC =0 IB=-IA-IC

The resistors are connected line-to-line so the voltage won't be at the same phase angles as the line-to-neutral voltages given. The current will be in-phase with the line-to-line voltage across the respective resistor. So calculate the voltage vectors first:

VAB = VA-VB = 120/0° - 120/120°
VBC = VB-VC = 120/120° - 120/240°

Then use Ohm's law to calculate the currents in A & C, which you can use to solve the first equation.

Do you know how to add and subtract the vectors?
 
The short answer is that when you add two identical currents I that are 60 degrees out of phase, the sum is not 2 * I , but sqrt(3) * I. So the answer is 20A * sqrt(3) = 34.6A.

That's what the previous answer boils down to in this special case, which arises often enough that I find it a useful shortcut to remember.

Cheers, Wayne
 
wwhitney said:
The short answer is that when you add two identical currents I that are 60 degrees out of phase, the sum is not 2 * I , but sqrt(3) * I. So the answer is 20A * sqrt(3) = 34.6A.
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Thanks Wayne. 1.73 was what came into my head at first but then for some reason I had doubt.. This is the 3rd time I've had to reengineer this whole heater relocation because I was getting bad info from the field. First they told me there was 480 3 phase available, then that changed to single phase 208, and when I went and looked for myself it was 208 three phase. It was originally powered by 480 three phase in it's old location.
JoeStillman said:
Do you know how to add and subtract the vectors?
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Thanks Joe. I need to refresh myself with the math. I did know it at one time, but when I tried to remember I realized I hadn't done it in quite a few years
 
@Joethemechanic ,
If that resistive load was supplied by 480V, 3-phase before, the load amps on the B leg would be 34.6A. But since the new source was verified by you to be 208V, 3--phase, the current on leg B will be 15.01A (Ohmic value of resistors = 480/20 = 24 ohms, Amps on 208V = 208/24 = 8.67A and leg B amps = 8.67 X 1.732 = 15.01A)
 
topgone said:
@Joethemechanic ,
If that resistive load was supplied by 480V, 3-phase before, the load amps on the B leg would be 34.6A. But since the new source was verified by you to be 208V, 3--phase, the current on leg B will be 15.01A (Ohmic value of resistors = 480/20 = 24 ohms, Amps on 208V = 208/24 = 8.67A and leg B amps = 8.67 X 1.732 = 15.01A)
Click to expand...
Had to order the correct elements this morning. The 208v ones were where the 20 amp number came from. Had to order a different control transformer and a bunch of stuff. Every time I had it all figured out, they changed the voltage or the phases on me
 
JoeStillman said:
Sorry, my guess was you had a homework problem and not a real-world application. I was trying not to give you the answer.
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No problem, Nahhh, not homework, I'm almost 62 and was having a senior moment. Too many changes in what I was designing in a short period of time. I appreciate your help
 
Actually it is not a sum but a difference.
 

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JoeStillman said:
What's the difference between a sum and a difference? Do two differences make a sum?
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Sum / difference impacts the direction of vectors. A difference is essentially a sum with one of the vectors in the opposite direction (rotated 180 degrees).

As noted in post#3, you're looking for the sum of two currents 60 degrees out of phase. Post #2 gives the context on how to get to 60 degrees. Solving the vector addition graphically using 30/60/90 triangles is probably the easiest way to get the resultant vector magnitude of 20*sqrt(3).
 
PD1972 said:
Sum / difference impacts the direction of vectors. A difference is essentially a sum with one of the vectors in the opposite direction (rotated 180 degrees).

As noted in post#3, you're looking for the sum of two currents 60 degrees out of phase. Post #2 gives the context on how to get to 60 degrees. Solving the vector addition graphically using 30/60/90 triangles is probably the easiest way to get the resultant vector magnitude of 20*sqrt(3).
Click to expand...
A vector is called a vector because it has in itself a direction aside from its magnitude. IMO, the correct term used is "summation of vectors".
 
JoeStillman said:
Do two differences make a sum?
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Sort of. Though there will still be a difference kicking around at the end of the day.

The difference between B and C is subtracted from A, to make the difference between A and (B - C):
A - (B - C)

When you distribute the negative sign, this simplifies to:
A - B + C

Which we can rearrange to make it more intuitive to understand:
A + C - B

It started as two nested differences, and now it simplifies to a sum and a difference.
 
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