3 phase load to "2 phase" load conversion 3

[bandy2000]

HI!

I'm looking for the formulas to calculate the equivalent 2 phase Load (only two impedances between phases 1,2, and 2,3) out of a "usual" 3-phase Delta connected Load and vice versa. What are the formulas or where can i find them?
(You may be used to name the phases R,S,T or R,Y,B which i name 1,2,3)

 

[infomaniac]

bandy 2000,

Unless I am crazy, if you look at the impedance from any two phase points into a delta circuit and assume a balanced load (i.e., Z1 = Z2 = Z3 from each delta leg), then the total impedance will be equal to the following assuming we reference 1 & 2:

Z12 equivalent = [(Z23+Z13)*Z12]/(Z12+Z13+Z23)

between 2 & 3:

Z23 equivalent = [(Z12+Z13)*Z23]/(Z12+Z13+Z23)

and finally, between 1 & 3:

Z13 equivalent = [(Z12+Z23)*Z13]/(Z12+Z13+Z23)

Then to calculate the current just use the phase-phase voltage (i.e., V12, V13, or V23) divided by the equivalent Z.

 

[bandy2000]

thanks for your answer, but this was not exactly that what i'm looking for or maybe i don't get it
right.If you assume a 3phase delta connected load. (balanced or unbalanced) with impedanzes
connected as Z12 Z23 and Z31. Then if you take out say Z31. Of what value are now my
remaining Impedances Z12 and Z23 to keep exactly the same conditions as if there would be all
three Impedances. If i assume a balanced, purely resistive Load then (as i think) Imaginary parts
have to be added to the remaining two Impedances to keep the 3 currents equal. Over all this
Impedances will cancel out so that the Load as seen from the supply is still purely resistive.
I hope this describes my prblem in a better way and that somebody can help me!!

 

[Bung]

Assume the loads are named L12 and L23 and areconnected phases 1-2 and 2-3 respectively. The current in L12 is I12=V12/Z12 where Z12 is the load impedance (assuming it is not voltage dependent). The current in L23 is I23=V23/Z23. The total load one the three phase system is V12*I12+V23*I23.

Basically, ohm's law still applies, irrespective of the number of phases. (Try it out on a "3-phase" load - it is really just 3 single pahse loads connected to 3 voltage sources - the voltaes just depend o a Y or D connection of the loads).

Bung
Life is non-linear...

 

[bandy2000]

Ok this was still not what i wanted to know. It tried to do it on my own.
If you assume a 3phase delta connected load. (balanced or unbalanced) with impedanzes
connected as Z12 Z23 and Z31. Then if you take out say Z31.

In order not to change the currents there have to be values added to the impedance of the remaining Impedances Z12res and Z23res.
The formula looked like following:
Z12res=1/(-(U31/(Z31*U12))+1/Z12)
Z23res=1/(-(U31/(Z31*U23))+1/Z23)
Assuming balanced Voltage and positive sequence
V1=V1 V2=a^2*V2 V3=a*V3
a=-1/2 +sqrt(3)/2 equals 120 degrees
a^2=-1/2 -sqrt(3)/2 eqals -120 degrees or +240
Z12res=1/(-(a/(Z31))+1/Z12)
Z12res=1/(-(a^2/(Z31))+1/Z12)

If i connect Impedances with the calculated values the condition for the Supply hasn't changed, but i have only two impedances!
Cheers!