2. Transformer Calculations
X
XF3
F4
X
X
F1
F2
51
50 51
51G
87T
63
SPR
ANSI / IEEE C37.91
F1: Fault cleared by
transformer primary side
relays
F2: Fault cleared by
transformer primary
side devices or neutral
protection relays
F3: Fault cleared by
transformer primary
side devices or by
secondary side relays
F4: Fault cleared by feeder
protection relays
Transformer Calculations
4. Transformer Calculations
1. Instantaneous units should be set so they do not
trip for fault levels equal or lower to those at
busbars or elements protected by downstream
instantaneous relays.
2. Time delay units should be set to clear faults in a
selective and reliable way, assuring the proper
coverage of the thermal limits of the equipment
protected.
Protection Coordinate Principles
6. Transformer Calculations
A margin between two successive devices in the
order of 0.2 to 0.4 seconds should be used to avoid
losing selectivity due to one or more of the following
reasons:
Instantaneous units should be set so they do not
trip for fault levels equal or lower to those at
busbars or elements protected by downstream
instantaneous relays.
Time delay units should be set to clear faults in a
selective and reliable way, assuring the proper
coverage of the thermal limits of the equipment
protected.
Coordination Time Interval (CTI)
7. Transformer Calculations
Operating time defined by IEC and ANSI/IEEE:
t = Relay operating time in
seconds
TD = Time dial, or time multiplier
setting
I = Fault current level in
secondary amps
IP = Tap or pick up current
selected
B = Constant
p = Slope constant
A = Slope constant
P
M =
I
I
* IEEE equation constants are defined at TD of 5
=
A
TDt
− 1M
p
IEEE equation *
IEC equation
=t B+
A
− 1M
p
TD
5
Expression for Time Delay Setting
8. Transformer Calculations
Setting Time Delay on Overcurrent Relays
IDMT Curve Description Standard p A B
Moderately Inverse IEEE 0.02 0.0515 0.114
Very Inverse IEEE 2 19.61 0.491
Extremely Inverse IEEE 2 28.2 0.1217
Standard Inverse IEC 0.02 0.14
Very inverse IEC 1.0 13.5
Extremely inverse IEC 2.0 80.0
ANSI/IEEE and IEC constants for overcurrent relays
* IEEE equation constants are defined at TD of 5
9. Transformer Calculations
Operatingtime(s)
Current (Multiples of Is)
IEC EI
IEC VI
IEC SI
UK LTI
Current (Multiples of Is)
Operatingtime(s)
IEEE MI
IEEE VI
IEEE EI
US C08
US C02
IEC/UK
overcurrent
relay curves
IEEE/US
overcurrent
relay curves
IEC curves are IMDT
(Inverse Minimum Definite Time)
Standards of Time/Current Characteristics
10. Transformer Calculations
Dac Y
Dy11
X
X
X
1.0
1.0
1.0 1.0
1.0
1.0
1.0
1.0
1.0
0.58
0.58
0.58
Three phase fault on secondary
I
X
VI LG
F
==
31
2
I
N
N
IIdelta
==
III deltaprimary
== 3
Coordination across D-y Transformer
11. Transformer Calculations
Dac Y
Dy11
X
X
0.5
1.0
0.5 0.87
0.87
0
0.87
0.87
0
0.5
0.5
0
Phase-to-Phase Fault
I
x
V
X
V
I LG
F
2
3
2
3
2
===
22
3
1
2 I
N
N
IIdelta
=××=
III deltaprimary == 2
LL
Coordination across D-y Transformer
13. Transformer Calculations
Fault Iprimary Isecondary
Three phase I I
Phase-to-phase
I 0.87I
Phase-to-earth 0.58I I
For SLGF on wye side, current withstand capability is
58% of that utilized for Delta-Delta
For Phase-Phase Faults on wye side, only 87% fault
current flows in secondary while 100% flows in primary
protector. 16% margin must be taken into account to
maintain primary to secondary protection selectivity.
Coordination across D-y Transformer
17. Transformer Calculations
Category II:
Thermal Limit: I ² t = 1250
Mechanical Limit: I ² t = K = (1/ZT) ² x t
K is determined at t = 2 seconds and at maximum I
in terms of multiples of self-cooled full load current.
The transition from thermal to mechanical plot
occurs at 70% of max possible multiplies of I, or
0.70x(1/ZT)
Category II Transformers
19. Transformer Calculations
Thermal Limit: I ² t = 1250
ANSI C57.12 Appendix
Short-time thermal loading capability oil-immersed transformers
Time Multiples of I RATED
2 s 25.0 x
10 s 11.3 x
30 s 6.3 x
60 s 4.75 x
300 s 3.0 x
1800 s 2.0 x
All Category Transformers
20. Transformer Calculations
Times Normal Base Current
I TN = 16.67 p.u.
Transition from the Mechanical to
Thermal curve occurs at 70% of I TN:
0.70 x (1/ZT) = 11.67 p.u.
I TN = 11.67 p.u.
2
Thermal Limit: I t = 1250
ITN = Times Normal Current
K = I TN = (16.67) x 2 = 556
2
Mechanical Limit: K = I t = (1/ZT) t
defined at t = 2 seconds.
for ZT = 6%
I TN = 1/.06 =16.67 p.u.
2 2
2
t = 556 / (11.67) = 4.0
K = I t = (11.67) x t =5562 2
Wye-Wye or
Delta-Delta
Category II – Transformer Through Fault Curves
21. Transformer Calculations
Category 3:
Thermal Limit: I 2 t = 1250
Mechanical Limit: I 2 t = K = [ 1 / (ZT + ZS) ]2 x t
K is determined at t = 2 seconds and at maximum I in
terms of multiples of self-cooled full load current. The
transition from thermal to mechanical occurs at 50% of
max possible multiples of I, or
0.50x1/(ZT + ZS)
Category III Transformers
22. Transformer Calculations
Mechanical to Thermal
occurs at 50% of I TN:
0.50 x 1/(ZT+ ZS) = 7.15 p.u.
I TN = 7.15 p.u.
I TN = 14.3 p.u.
I TN = Times Normal Current
Thermal Limit: I t = 1250
2
K = (I TN) x t = (7.15) x t = 40922
t = 409 / (7.15) = 8.0
2
2
Mechanical Limit
K = I t = [1/ (ZS + ZT)]
defined at t = 2 seconds.
for ZT = 6% & ZS = 1%
K = (I TN) x t = (14.3) x 2 = 40922
I TN = (1/.07) = 14.3 p.u.
Wye-Wye or
Delta-Delta
Category III – Transformer Through Fault Curves
23. Transformer Calculations
Mechanical to Thermal
occurs at 50% of I TN:
0.50 x 1/(ZT+ ZS) = 7.14 p.u.
Delta-wye:
7.14 p.u. x 0.58 = 4.14 p.u.
I TN = 4.1 p.u.
I TN = 8.3 p.u.
2 2
Mechanical Limit
2
t = 138 / (4.14) = 8.0
I TN = Times Normal Current
Thermal Limit: I t = 1250
2
K = I TN x t = (8.3) x 2 = 138
22
K = I t = [1/ (ZS + ZT)]
defined at t = 2 seconds.
for ZT = 6% & ZS = 1%
I TN = 1/.07 =14.3 p.u.
Delta-wye:
14.3p.u. x 0.58 = 8.3 p.u.
Delta-Wye
Category III – Transformer Through Fault Curves
25. Transformer Calculations
0.01 p.u. 0.07 p.u. 4.80 p.u.
VS = 1.0 p.u.
F1
X
F2/F3
X
F4
X
X
F1 ZT
A B
50
X
F3
Vs = 230kV
51
230kV / 34.5kV
30/40/50MVA
Z = 7%
ZLT = 25 MVAZS = 1.0%
on 230kV
ZS = 0.01 p.u.
ZT = 0.07 p.u.
F2
X
F4
X
51
51
50
Z Base =
30MVA
230kV 2
= 1763.3 ohm
ZLT =
25MVA
34.5kV 2
= 47.6 ohms
VBase = 230kV
MVA Base = 30MVA
Z LT at 230kV =
34.5kV
230kV 2
x 47.6 ohms = 2115.6 ohms
Z LT = (2115.6/1763.3) = 1.20 p.u.
ZL4 = 6.25MVA
Z L4 = 4.8 p.u.
ZL4 = 1/4 ZL4
Calculation Setting Example
26. Transformer Calculations
0.01 p.u. 0.07 p.u. 4.80 p.u.VS = 1.00 p.u. F1
X
F2/F3
X
F4
X
IBase at 230kV =
30MVA
3 x 230kV
= 75.3 A IBase at 34.5kV =
30MVA
3 x 34.5kV
= 502 A
IF1 =
IF2/F3 =
IF4 =
= 100.0 p.u.
= 12.5 p.u.
= 0.205 p.u.
0.01
(0.01 + 0.07)
(0.01 + 0.07 + 4.80)
1.00
1.00
1.00
IF1 = 100.0 x 75.3A = 7530A
12.5 x 75.3A = 941A
0.205 x 502A =103A at 34.5kV
IF4 =
IF2/F3 =
12.5 x 502 = 6275A at 34.5kV
0.205 x 75.3A = 15.4A
Short Circuit Currents
27. Transformer Calculations
Compatible with transformer overload capacity ≈ 200% of self-
cooled rating for wye CT’s and 350% (√3 x 200%) for Delta-
connected CT’s.
Wye - Wye CT’s:
230kV: IFL = 75.3A IF1 = 7530A
IF1(ASYS) = 1.6 x 7530A = 12,048A
Select CT 600:5 CTR = 120
34.5kV: IFL = 502A IF2 = 6275A
IF2(ASYS) = 1.6 x 6275A = 10,040A
CT Selected at 2 x IFL
Select CT 1000:5 CTR = 200
CT Criteria
28. Transformer Calculations
Set above Inrush current:
8 x IFL (Transformers 500 to 2500KVA)
10 x IFL (Transformers > 2500KVA) *
Inrush point at 0.1 second
* For Digital Relay using DFT use 5-6 x IFL
Set above maximum asymmetrical secondary fault current:
1.6 x IF SYM for Voltage > 5kV
1.5 x IF SYM for Voltage < 5kV
IOC Setting Criteria
29. Transformer Calculations
IOC Setting Criteria
Secondary side of transformer
The IOC function at the LV side is not used unless there
is communication (interlocking) with relays protecting
the feeders
30. Transformer Calculations
Set above Inrush current: ≈ 10 x IFL at 0.1 second.
10 X 73.5A = 735A
Set above maximum asymmetrical secondary fault current:
≈ 1.6 x IF SYM
1.6 x 941 = 1506A = 20.0 x IFL
I PICKUP = 1506A ÷ CTR = 1526A ÷ 120 = 12.65 Amps
IFL =
30MVA
3x 230kV
= 75.3 A IF2/F3 = 954A
IOC Relay Setting
31. Transformer Calculations
For phase relays, three-phase faults and
maximum short time overload should be
considered
For ground relays, line-to-ground faults and max
3Io should be considered
TOC Setting Criteria
32. Transformer Calculations
For phase relays, the Pickup value is determined by:
Pickup = (OLF x IFL) ÷ CTR
For ground fault relays, the Pickup value is determined,
with the maximum unbalance, typically around 20%:
Pickup = [(0.2) x IFL] ÷ CTR
Typical OLF for transformers = 1.25 to 1.5
TOC Setting Criteria
33. Transformer Calculations
Transformer: ZT < 6%
Primary Setting < 6 x IFL
Secondary Setting < 3 x IFL
Transformer: 6% > ZT < 10%
Primary Setting < 4 x IFL
Secondary Setting < 2.5 x IFL
Tap is set to meet NEC 450-3 and ANSI C37.91*
* Primary & secondary >600V with circuit breakers
TOC Setting Criteria
34. Transformer Calculations
Pickup = OLF x IFL ÷ CTR
I Pickup = 1.5 x 75.3 ÷ 120 = 0.94 Amps
I Relay setting = 0.94 ÷ 1.5 = 0.63 Amps
TD # 8 Very Inverse Curve
IFL =
30MVA
3x 230kV
= 75.3 A IF2/F3 = 954A
TOC Relay Setting
36. Transformer Calculations
50% of I TN = 0.50 x 1/(ZT+ ZS)
t = 105.1 / (3.625) = 8.0
I TN = Times Normal Current
Thermal Limit: I t = 1250
2
Mechanical Limit
K = I TN = (7.6) x 2 = 105.1
2
defined at t = 2 seconds.
for ZT = 7% & ZS = 1.0%
I TN = 1/.08 =12.5 p.u.
Delta-wye:
2 2
K = I t = [1/ (ZS + ZT)] x t
58% x 12.5 p.u. = 7.25 p.u.
Mechanical to Thermal
Delta-wye:
50% of I TN = 0.50 x 12.5 = 6.25 p.u.
58% x 6.25 p.u. = 3.625 p.u.
IOC set at 12.6A pickup
1.6 x IFL/CTR = 1.6 x 941/120 = 12.6A
1.6 x IFL =1.6 x 12.5p.u. = 20.0 ITN
TOC set at 0.63A pickup
1.5 x ITN/CTR = 1.5 x 73.5A/120 = 0.94A
Minimum multiple of pickup is 1.5
0.94A/1.5 = 0.63A, TD#8
Very Inverse TOC curve
0
0
1
10
100
1000
10000
0 1 10 100
TIMES NORMAL BASE CURRENT
TIME(SECONDS)
IFLSYM
IFL
I inrush
0.1
0.01
0.1
2.5xIFL
4.0xIFL
IFLASYM
2
IFL SYM
3-phase
IFL SYM
phase-phase
Coordination
Interval
50/51
51
50/51
51
2
IOC & TOC Coordination
