Load connected to delta power supply system

Hi guys,

I'm facing with the following problem. I would like to define the right function for controlling line currents below dedicated values (e.g. 60 A per line). I've started with solve equation and Kirchoff's first law, however it doesn't work as I expected. Does anybody have a sollution how to cope with this problem? The known values are limited line currents (i.e. 60 Aef), limited active powers of every load (P1 - P5) and phase-2-phase voltages (i.e. 230 Vef).

The problem is that I don't know how to easy transform active powers (variables limited to 230V*32A = 7360 W) and known phase-2-phase voltages (230*sqrt(2)<0°, 230*sqrt(2)<120° and 230*sqrt(2)<240°) to phase-2-phase currents (x1 - x5)? By this the system would get 5 more equations and might be solvable.

So far I don't get any useful result, only the warning "Unable to find explicit solution For options, see help. ..."

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clc;clear all;

syms Y1 Y2 Y3 y1 y2 y3 x1 x2 x3 x4 x5

assume(Y1 >= 0);

assume(Y2 >= 0);

assume(Y3 >= 0);

assume(Y1 <= 60*sqrt(2));

assume(Y2 <= 60*sqrt(2));

assume(Y3 <= 60*sqrt(2));

assume(abs(x1) <= 32*sqrt(2));

assume(abs(x2) <= 32*sqrt(2));

assume(abs(x3) <= 32*sqrt(2));

assume(abs(x4) <= 32*sqrt(2));

assume(abs(x5) <= 32*sqrt(2));

assume(abs(x1) >= 0);

assume(abs(x2) >= 0);

assume(abs(x3) >= 0);

assume(abs(x4) >= 0);

assume(abs(x5) >= 0);

y1 = Y1*cos(0) + 1i*Y1*sin(0);

y2= Y2*cos(120*(pi/180))+1i*Y2*sin(120*(pi/180));

y3 = Y3*cos(240*(pi/180))+1i*Y3*sin(240*(pi/180));

y1 = x1 + x3;

y2 = -x1 - x2 - x4 - x5;

y3 = x2 + x4 + x5 - x3;

sol = solve([y1,y2,y3],[x1,x2,x3,x4,x5],'ReturnConditions',true);

sol1=sol.x1

sol2=sol.x2

sol3=sol.x3

sol4=sol.x4

sol5=sol.x5

sol.conditions;

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Answer 1

[MULI]

Hi Andraz,

I understand that you are trying to get the currents for the balanced delta load.

Since it is a balanced load, there is no need of variables Y1 ,Y2 and Y3 instead it can be formulate in the following way:

y1 = Y*cos(0) + 1i*Y*sin(0);

y2= Y*cos(120*(pi/180))+1i*Y*sin(120*(pi/180));

y3 = Y*cos(240*(pi/180))+1i*Y*sin(240*(pi/180));

Also the number of unkowns are more than the number of equations

Since its is a balanced load you can add the following conditions

1.Magnitude of all three currents are equal.

eg: abs(x1)=abs(x3);abs(x2+x4+x5)=abs(x3)

2.Consider the phase to line current relationship in delta load

eg:abs(y1)=abs(x1)*sqrt(3)

Adding these conditions you can get the number of equations equal to number of unknowns.

Hope this clears your query!

[Andraz]

Hi Muli,

The catch is that the load isn't balanced. P1 - P5 are variables standing for active load.