Calculating unbalanced Delta loads

[oldsparky52]

Theoretically, if you have a 120/240 Delta supply and load it to 10 amps on L1/L3 and 20 amps on L2/L3 (nothing between L1/L2), what would the amperage be on the L3 conductor (I'm assuming L1 will have a 10-amp load and L2 will have a 20 amp load)?

My understanding is you take the common current (10-amps) and multiply by 0.866 then add in the difference in the currents which is 10-amps. The would be 8.66+10 for 18.66 amps on L3.

I've been told I'm incorrect. What is wrong with my thinking?

 

[synchro]

The current on L3 would be √ ( 10² + 20² + 10 x 20 ) = 26.46A

Using this formula with equal 20A L-L loads across L1/L3 and L2/L3 we have:
√ ( 20² + 20² + 20 x 20 ) = √ ( 3 x 20² ) = √3 x 20 = 1.732 x 20 = 34.64A which is the familiar result for balanced currents.

 

[ptonsparky]

The current on L3 would be √ ( 10² + 20² + 10 x 20 ) = 26.46A

Using this formula with equal 20A L-L loads across L1/L3 and L2/L3 we have:
√ ( 20² + 20² + 20 x 20 ) = √ ( 3 x 20² ) = √3 x 20 = 1.732 x 20 = 34.64A which is the familiar result for balanced currents.

Ok. That is now in a SS.
Now what is the formula for unbalanced loads of 5, 10, 15?

 

[synchro]

For a delta with terminals A, B, and C, let I_A, I_B, and I_C be their respective line currents.
Let I_AB be the L-L load current on A and B, I_BC the current on B and C, and I_AC the current on A and C.

Then:
I_A = √ ( I_AB² + I_AC² + I_AB + I_AC )
I_B = √ (I_AB² + I_BC² + I_AB x I_BC )
I_C = √ ( I_AC² + I_BC² + I_AC x I_BC )

Now what is the formula for unbalanced loads of 5, 10, 15?

For I_AB = 5, I_BC = 10, I_AC = 15 :

I_A = √ ( 5² + 15² + 5 x 15 ) = 18.0
I_B = √ ( 5² + 10² + 5 x 10 ) = 13.2
I_C = √ (15² + 10² + 15 x 10 ) = 21.8

 

[wwhitney]

For a delta with terminals A, B, and C, let I_A, I_B, and I_C be their respective line currents.
Let I_AB be the L-L load current on A and B, I_BC the current on B and C, and I_AC the current on A and C.

Then:
I_A = √ ( I_AB² + I_AC² + I_AB + I_AC )
I_B = √ (I_AB² + I_BC² + I_AB x I_BC )
I_C = √ ( I_AC² + I_BC² + I_AC x I_BC )

It is worth noting that this assumes that the line currents I_A, I_B, and I_C are pairwise 120 degrees out of phase (for a consistent sense of positive.) I.e. all the loads are resistive, or they all have the same leading or lagging power factor.

(Also, the I_A formula has an obvious typo, the last plus sign should be a times sign.)

Cheers, Wayne

 

[synchro]

A typo in my post above:

(Also, the I_A formula has an obvious typo, the last plus sign should be a times sign.)

Cheers, Wayne

You beat me to it Wayne. I was just starting to post a correction for that. Mea culpa.

 

[wwhitney]

It is worth noting that this assumes that the line currents I_A, I_B, and I_C are pairwise 120 degrees out of phase (for a consistent sense of positive.) I.e. all the loads are resistive, or they all have the same leading or lagging power factor.

Typo in the above, I meant the load currents I_AB, I_BC, and I_CA, as my subsequent comment implies.

Cheers, Wayne

 

[ptonsparky]

Not sure what I'll do with this information @ 75 yrs, but I've got it now.

Thanks, good stuff.
I should have researched it forty years ago.

 

[Julius Right]

In complex I23=20*[cos(-120)+sin(-120)i ];I31=10*[cos(-240)+sin(-240)i]

I12=0; I23=-10-17.32i; I31=-5+8.6603i

So, I1=I12-I31=5-8.66i; I2=I23-I12=-10-17.32; I3=I31-I23 I3=-5+8.66i+10+17.32i=5+25.98i

I3=sqrt(5^2+25.98^2)=26.457