Calculate parameters for Three-Phase Autotransformer with given resistances for series and common winding

8 次查看(过去 30 天)
Hello,
I want to do a simple load flow in Simscape electrical using an Autotransformer. I used the predefined block "Three-Phase Autotransformer with Tertiary Winding" However I need to input RHL, RHT and RLT for the resistances. However I have only given the Resistance of the series and common winding (0.04Ω and 0.06Ω, tertiary winding will be assumed to 1Ω). The documentation of this block gives a formula:
R1 = (R12*(1+k)/(1-k) + R13 - R23_pu)/(1-k)/2 R2 = (R12 - R13 + R23_pu)/(1-k)/2 R3 = (-R12 + R13 + R23*(1-2*k))/(1-k)/2
I then assumed, to just do a matrix calculation out of this and calculate R12, R13 and R23 which correspond to RHL, RHT and RLT. However I need to convert these values in pu and I don't understand how this is done. The Block always gives errors that some of the resistances will be negative when I try to input my results.
Here is my code for the calculation:
k = 500/345;
Pnom = 300e6;
Vhv = 500e3;
Vlv = 345e3;
Vte = 12e3;
Zb = Vhv^2/Pnom;
Zb2 = Vlv^2/Pnom;
Zb3 = Vte^2/Pnom;
% R1 resistance in Ohm series winding
% R2 resistance in Ohm common winding
% R3 resistance in Ohm tertiary winding
R1 = 0.04/Zb;
R2 = 0.06/Zb;
R3 = 0.1/Zb;
WR = [R1; R2; R3];
M = [(1+k)/(1-k) 1 -1;
1 -1 1;
-1 1 (1-2*k)]*(1/((1-k)*2));
R = M\WR;
I have a screenshot for the remaining parameters as well:
I tried with different Zb's but nothing will work and give me the correct behaviour.
I hope someone can help me

回答(1 个)

MULI
MULI 2025-6-11
To avoid the limitations of the standard model that may result in negative resistances or very uneven sharing of losses between winding 1 and winding 2, you may choose an alternate model for computing winding resistances as descibed in this documentation:
This alternate model assumes that the Joules losses corresponding to R12 are equally shared between winding w1 and winding w2 (R2 = R1), which is close to real world behaviour.

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