QUICK CALCULATION AND TESTING METHOD FOR TRANSFORMER DIFFERENTIAL WITH AN EXAMPLE

FULL LOAD CURRENT:

P=1.732VI
I = p/(1.732*V)

Ihv = 393.65A in primary.
Ihv = 0.874 in secondary value.

Ilv = 1312.1A in primary.
Ilv = 0.820 in secondary value.

MATCHING FACTOR:
K = 1/flc

Matching factor Khv = 1/0.874 = 1.14
Matching factor Klv = 1/0.820 = 1.21

PICKUP CAL:
Ipickup = Idiff/(matching factor),
  Ipickup = Idiff/(matching factor*Vector compesation factor) if 3I0 to be eliminated on Wye side of transformers

Idiff lowset:0.15

Pickup HV 3Ph = 0.15/(1.14) = 0.131A
Pickup HV 1Ph = 0.131 * 1.732 = 0.226A

Pickup LV 3Ph = 1Ph = 0.15/(1.21) = 0.131A 
  Pickup LV 1Ph = 0.15/(1.21*0.577) = 0.123A if zero sequence element to be eliminated.


The vector compensation factor for manual calculation is mostly used on Wye connected side in case Zero sequence components are to be eliminated. The consideration of zero sequence component in differential function is mandatory and that depends the system scheme, if not considered, the protected zone system may get malfunction. We know that, almost the load connected to the transformer could be unbalanced. There the unbalance current 3I0 flows on neutral line(@Wye) but on other side which is delta that I0 appears in the phases. So, we can say that the protected zone system current on both side is different in unbalance condition, this may create a differential current in relay current coils and results in differential tr


It is possible to manage this kind of issue by two ways easily:


1.     Physically changing the CT connection to the relay current coil junction. Connecting either side of ct wiring in delta structure instead of star to eliminate the zero seq. comp.


2.     Selecting mode to eliminate the zero sequence component in modern numerical relays..



Vector compensation factor for manual calculation:


Kvect = 0.577 for the Vect.Grp wich consist clock number oddly 1,3,5,7,9,11.
Kvect = 0.666 for the Vect.Grp wich consist clock number evenly 0,2,4,6,8,10.
        

Note: Here concepts and formulas are preferred from RET, P63, p64 relays.

HV FLC: 0.874A

LV FLC: 0.820A

Idiff=Iw1+Iw2

Ibias= (Iw1-Iw2)/2

SIMPLE CALCULATION METHOD:

1. STABILITY:

Iw1=HV FLC*Matching factor=0.874*1.14=0.996A

Iw2=LV FLC*Matching factor=0.820*1.21=0.992A

IA diff=(0.996<0°)+(0.992<180°)= 0.004A

IA bias= ((0.996<0°)-(0.992<180°))/2 = 0.994A

IB diff=(0.996<240°)+(0.992<60°) = 0.004A

IB bias= ((0.996<240°)-(0.992<60°))/2 = 0.994 A

IC diff=(0.996<120°)+(0.992<300°) = 0.004 A

IC bias= ((0.996<120°)-(0.992<300°))/2 = 0.994 A

Note: Here, why? we were used 180, 60, 300 on LV side current followed by transformer displacement principle instead of 150, 30, 270. This is simple calculation method we are avoiding here use of matrix calculation, we’ll see detailed in second method with all its original values on Next Paper.

2. UNSTABLE: Generously we may change angle to make the zone unbalanced or otherwise ramp the current magnitude down/up.

IA diff=(0.996<0°)+(0.992<30°) = 1.92A                                      

IA bias= ((0.996<0°)-(0.992<30°))/2 = 0.257A                                   

IB diff=(0.996<240°)+(0.992<270°) = 1.92A                                    

IB bias= ((0.996<240°)-(0.992<270°))/2 = 0.257 A                            

IC diff=(0.996<120°)+(0.992<150°) = 1.92 A                                        

IC bias= ((0.996<120°)-(0.992<150°))/2 = 0.257 A    

Note: 30, 270, 150 are appeared instead of 0, 240, 120.

3. SLOPE:

Slope2 = 15% = 0.15

End section2 = 300% = 3

Slope3 = 60% = 0.6

a. Slope 1: For manual calculation we need 2 points to get the slope region by applying those values in below formula;

M2=((Id2-Id1)/(Ib2-Ib1))*100

i. Point1: Bias on both side refer from 1.5*FLC

HV: 1.31A*1.14 = 1.49

LV: 1.23A Ramped to 1.51A*1.21 = 1.82A

IA diff=(1.49<0°)+(1.82<180°)= 0.33A

IA bias= ((1.49<0°)-(1.82<180°))/2 = 1.65 A

IB diff=(1.49<240°)+(1.82<60°) = 0.33 A

IB bias= ((1.49<240°)-(1.82<60°))/2 = 1.65 A

IC diff=(1.49<120°)+(1.82<300°) = 0.33 A

IC bias= ((1.49<120°)-(1.82<300°))/2 = 1.65 A

ii. Point2: Bias on both side refer from 2.5*FLC

HV: 2.18A*1.14 = 2.48A

LV: 2.05A Ramped to 2.46A*1.21 = 2.97A

IA diff=(2.48<0°)+(2.97<180°) = 0.49A

IA bias=((2.48<0°)-(2.97<180°))/2 = 2.72 A

IB diff=(2.48<240°)+(2.97<60°) = 0.49 A

IB bias=((2.48<240°)-(2.97<60°))/2 = 2.72 A

IC diff=(2.48<120°)+(2.97<300°) = 0.49 A

IC bias=((2.48<120°)-(2.97<300°))/2 = 2.72 A

M2 = (Id2-Id1)/(Ib2-Ib1)

M2=((0.49-0.33)/(2.72-1.65))*100 = 0.149*100 = 14.9%

Similarly we can achieve slope3 by taking any 2points randomly but equal or above 300% in base point reference. Ex, Point1 = 4 & point2 = 5 then find its Slope3 region by noting those Id & Ib currents respectively.

4. 2nd HARMONIC BLOCK Set Percentage = 15% i. Note the level of blocking diff trip. ii. Note the level of unblocking diff trip. We can able to operate Harmonic block by applying double frequency current more than given setting.

Ex, commonly followed 0.15*0.874 or 0.15*0.820 current or more than that with double frequency on either side of applying sources. 

...................Paper 2 will be updated soon...........

Thanks for your learning along with me