Unbalanced load calculations.

[StanMan122]

I've been searching EVERYWHERE for a simple formula to calculate 2 unbalanced loads on a 3 phase line. This is an issue when calculating FLA on machines. i.e. If I have 3 230vac, 2300watt single phase heaters across the three phases, no problem. P/E * 1.732. If only two heaters of equal wattage, no problem. P/E = I in the uncommon legs, I * 1.732 in the common leg. The problem is when I have 2 heaters of different wattage. If one heater is between Phase A & B and the second is between Phase B & C, what is the current at Phase B? I found one formula, but it must be worng because my result on Phase B was greater that I * 1.732 of the higher wattage heater. Unfortunately, the formula is at work and I am at home. To the best of my recollection, the formula was something like :
(((Ia + Ic)^2) + (((Ia + Ic)^2)*.866))^.5
Again, I think the formula was wrong. Perhaps a typo somewhere. Anyone know what the formula is???
Any help would be greatly appreciated.

 

[waross]

You need the vector sum of the currents. You can sketch the vectors and solve the problem with Pythagoris' theorem.
basically use the theorem to resolve one current into the in phase and quadrature components relative to the other current current. Then use the theorem again to calculate the magnitude of the resultant current.
The power factors of the loads must be the same.
Generally, resolve the smaller current and any rounding errors will be less significant.
yours

 

[Birch]

Ia = V/R1
Ib = V/(2*R2)*sqrt(((R1 + 2*R2)/R1)^2 + 3)
Ic = V/R2

R = V^2/P

Example:
V = 230 V
P1 = 2300 W
P2 = 5000 W

R1 = 23 ohm
R2 = 10.58 ohm

Ia = 10 A
Ib = 28.1 A
Ic = 21.7 A