Unbalanced Delta Transformer Loading
Unbalanced Delta Transformer Loading
(OP)
Hello. Looking for educational assistance understanding unbalanced loading. Connected to the output of a 415V 3Ph Delta transformer are the following kW values and p.f.s: Ph AB 3kW,0.8pf; Ph CA 5kW, 0.7pf; and Ph BC 7kW, 0.6pf.
Looking for assistance calculating the line currents IA, IB, and IC.
I started breaking the components out for each phase and arrived at (I think)
Ph AB: Iab = 9A at 53deg (7.2+5.4i)
Ph CA: Ica = 17A at 164deg (12.0+12.3i)
Ph BC: Ibc = 28A at -83deg (16.8+22.5i)
IA = Iab – Ica
IB = Ibc – Iab
IC = Ica – Ibc
Looking for assistance calculating the line currents IA, IB, and IC.
I started breaking the components out for each phase and arrived at (I think)
Ph AB: Iab = 9A at 53deg (7.2+5.4i)
Ph CA: Ica = 17A at 164deg (12.0+12.3i)
Ph BC: Ibc = 28A at -83deg (16.8+22.5i)
IA = Iab – Ica
IB = Ibc – Iab
IC = Ica – Ibc






RE: Unbalanced Delta Transformer Loading
9 at 53deg = 5.4 + 7.2i
17 at 164deg = -16.3 + 4.7i
28 at -83deg = 3.4 - 27.8i
All you need to do is finish the math.
IA = (5.4 + 16.3) + (7.2 - 4.7)i = 21.7 + 2.5i = 21.8 at 6.6deg
IB = (3.4 - 5.4) + (-27.8 - 7.2)i = -2 - 35i = 35.1 at -93.3deg
IC = (-16.3 - 3.4) + (4.7 + 27.8)i = -19.7 + 32.5i = 38 at 121.2deg
RE: Unbalanced Delta Transformer Loading
3kW/0.8pf = 3.75kVA
3kW*1000/415V = 7.2A real; 3.75kVA*1000*sqrt(1-0.8^2)/(415V) = 5.4A reactive
7.2 + 5.4i : Sqrt(7.2^2 + 5.4^2) = 9 ; Cos-1(7.2/9) = 53deg
How are you developing the components for phases BC and CA?
RE: Unbalanced Delta Transformer Loading
Once determining the magnitude of the phase amps then the following is clear
17 cos(164) = -16.3 : 17 sin(164) = 4.7
28 cos(277) = 3.4 : 28 sin(277) = -27.8
RE: Unbalanced Delta Transformer Loading
IAB=7.22-5.43j .But in any case the angle has to be +/- 36.87 dgr.[if the entire circle of 2*pi=360dgr.].
ACOS(0.8)=ATAN(5.42/7.22)=0.6435 radians=0.6435*180/pi()=36.87 dgr.
RE: Unbalanced Delta Transformer Loading
If we assume three phase transformer phasor group Yd1 or Yd11, ABC phase sequence,
Voltage reference is VBC
PhAB=[3 0.8];% P pf; [kW cosfi]
PhBC=[7 0.6];
PhCA=[5 0.7];
VAB= 0.415;% kV
VBC=VAB,VCA=VBC;
a=cos(2*pi/3)+j*sin(2*pi/3);
disp('***************** EXIT')
tetaBC= acos(PhBC(2)); %rad
QBC= PhBC(1)* tan(tetaBC);%kVAr
NBC=sqrt(PhBC(1)^2 + QBC^2); %kVA
IBC = (PhBC(1) + QBC*j)/ VBC %A /0
%IBC = 16.8675 +22.4900i ; % A
% or
IBC = (PhBC(1) + QBC*j)/ j*VBC %A /90
%IBC = 3.8733 - 2.9050i; % A
RE: Unbalanced Delta Transformer Loading
Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle.
Capacitive loads are leading (current leads voltage), and inductive loads are lagging (current lags voltage).
RE: Unbalanced Delta Transformer Loading
where did I make mistake?
a=cos(2*pi/3)+j*sin(2*pi/3);
disp('***************** EXIT')
tetaBC= acos(PhBC(2)); %rad
QBC= PhBC(1)* tan(tetaBC);%kVAr
NBC=sqrt(PhBC(1)^2 + QBC^2); %kVA
IBC = (PhBC(1) + QBC*j)/ VBC %A /0
%IBC = 16.8675 +22.4900i ; % A
tetaAB= acos(PhAB(2)); %rad
QAB= PhAB(1)* tan(tetaAB);%kVAr
NAB=sqrt(PhAB(1)^2 + QAB^2); %kVA
IAB = (PhAB(1) + QAB*j)/ a*VBC %A +120
IAB = 0.1862 - 1.5451i; %A
tetaCA= acos(PhCA(2)); %rad
QCA= PhCA(1)* tan(tetaCA);%kVAr
NCA = sqrt(PhCA(1)^2 + QCA^2); %kVA
ICA = (PhCA(1) + QCA*j)/ -a*VBC %A /-120
ICA = -0.7958 + 2.8555i; %A
RE: Unbalanced Delta Transformer Loading
However if your reference voltage is VBC [VCA=a*VBC,VAB=a^2*VBC] for inductive reactive power:
IBC=16.87-22.49*j ; ICA=-16.67-4.29*j ; IAB=1.08+8.97*j
if VAB will be the reference and the phase order will be AB,BC,CA then:
IAB=7.23-5.42*j ; IBC=-27.9-3.36*j ; ICA=4.62+16.58*j
If reference will be -90 dgr[-pi()/2] and the phase order will be AB,BC,CA then:
IAB=+5.42+7.23*j; IBC=3.36-27.91*j ;ICA=-16.58+4.62*j –as jghrist said.
RE: Unbalanced Delta Transformer Loading
For a single phase unity power factor load on AB, the current In each winding of the supply transformer will be equal and equal to 50% of the load current. The current in the in-phase winding will be at unity power factor.
The current in one phase will be 60 degrees leading (50% PF) and the current in the third winding will be 60 degrees lagging (50% PF).
To understand this consider a three transformer bank split into an open delta (broken delta in IEC land) and a single phase transformer.
The virtual single phase transformer formed by the open delta may have its values described by the resultant vectors of the addition of the vector descriptions of the two real transformers.
These resultant vectors will be equal to the vectors of the transformer that was removed.
What I am trying to say is that for the case of a single phase load on a delta transformer bank, the transformer bank may be considered to be a real transformer in parallel with the virtual transformer formed by the open delta. The real transformer and the virtual transformer will share the load equally (assuming three identical transformers forming the delta bank).
In the case of a wye delta transformer bank with the primary neutral connected, these currents will be reflected in the three independent primary currents.
Now if the three transformers have different PU impedances and/or are rated on different KVA bases, the division of the current is a little more difficult to solve.
If a solution for one single phase load on phase A-B does not show a current on C phase the solution is incorrect.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Unbalanced Delta Transformer Loading
PhBC=[7 0.6];
PhCA=[5 0.7];
VAB= 0.415;% kV
VBC=VAB,VCA=VBC;
% an inductive load has a lagging power factor,the capacitive load
% has a leading power factor, resistive load has a unity power factor
% we assume 3phase voltage balanced, AB,BC,CA phase order, with inductive load(P-jQ) relative to voltage associated,
[
%for AB,BC,CA phase order VBC reference
a=cos(2*pi/3)+j*sin(2*pi/3);
VBC= 0.415;
VCA= a*VBC;
VAB= a^2*VBC;
clc
disp('***************** * SAIDA-VBC reference')
tetaBC= acos(PhBC(2)); %rad
QBC= PhBC(1)* tan(tetaBC);%kVAr
IBC = (PhBC(1) - QBC*j)/ VBC; %A
tetaCA= acos(PhCA(2)); %rad
QCA= PhCA(1)* tan(tetaCA);%kVAr
ICA = (PhCA(1) - QCA*j)/VCA; %A
tetaAB= acos(PhAB(2)); %rad
QAB= PhAB(1)* tan(tetaAB);%kVAr
IAB = (PhAB(1) - QAB*j)/ VAB; %A
IA = IAB - ICA ;
disp({'modIA angIA' abs(IA),angle(IA)*180/pi})
IB = IBC - IAB;
disp({'modIB angIB' abs(IB),angle(IB)*180/pi})
IC = ICA - IBC;
disp({'modIC angIC' abs(IC),angle(IC)*180/pi})
***************** * EXIT for VBA,VBC or VCA as reference
'modIA angIA' [22.1556] [36.7605]
'modIB angIB' [35.1998] [-63.3534]
'modIC angIC' [38.1575] [151.5093]
%%%%%% jghrist calculation
% IA = 21.8 at 6.6deg
% IB = 35.1 at -93.3deg
% IC = 38 at 121.2deg
RE: Unbalanced Delta Transformer Loading
RE: Unbalanced Delta Transformer Loading
Iab = 9A at 53deg
Ica = 17A at 164deg
Ibc = 28A at -83deg
were correct. The magnitudes were consistent with the stated kW and pf (with rounding). The angles were consistent with a starting Vab angle of 90°.
RE: Unbalanced Delta Transformer Loading
You're right. I calculated
'modIAB angIAB' [9.0361] [83.1301]
'modIBC angIBC' [28.1124] [-53.1301]
'modICA angICA' [17.2117] [-165.5730]
The modules are equal.
If you add 30 degrees in each angle that you calculated,
then the angles are equal.
Except ang IAC. - I do not know why they are different!
RE: Unbalanced Delta Transformer Loading
RE: Unbalanced Delta Transformer Loading
Anyone knows?
RE: Unbalanced Delta Transformer Loading
You considered VBC as reference then you get the same modules as jghrist but the angle is different. You have to take VAB=90 [PI()/2] ; VBC=a*VAB and VCA=a^2*VAB and then you'll get the jghrist's results.
RE: Unbalanced Delta Transformer Loading
That is a very interesting post. My results have different phase angles than you guys here. So, should complex apparent power equals to voltage vector multiplied by CONJUGATED current vector?
Regards
RE: Unbalanced Delta Transformer Loading
RE: Unbalanced Delta Transformer Loading
I thought that the modules should be equal and the angles between the three phasors
of reference system should be the same as another reference!
RE: Unbalanced Delta Transformer Loading
RE: Unbalanced Delta Transformer Loading
%for AB,BC,CA phase order and load , operator is 1 + a + a^2
see 2.9 connected loads page 34 - Power system analysis-Hadi Sadat
7anoter4
below is the results for 3 reference(VAB,VBC,VCA) and confirm The angle between the currents will be the same no matter the reference.(see Check)
Unbalanced Delta Transformer Loading
thread238-352772: Unbalanced Delta Transformer Loading
odlanor
current module- Ampere
IAB IBC ICA
VAB ref. 9.0361 28.1124 17.1127
VBC ref. 9.0361 28.1124 17.1127
VCA ref. 9.0361 28.1124 17.1127
current angle-Degree check
1+a+a^2 IAB IBC ICA IAB-IBC IBC-ICA ICA-IAB
VAB ref. -36.86 -173.13 74.42 136.27 112.45 111.28
VBC ref. 83.13 -53.13 -165.57 136.26 112.44 111.3
VCA ref. -156.86 66.86 -45.57 136.28 112.43 111.29