![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](https://image.slidesharecdn.com/transformerovercurrent-080211-160323223614/85/Transformer-overcurrent-21-320.jpg)
![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](https://image.slidesharecdn.com/transformerovercurrent-080211-160323223614/85/Transformer-overcurrent-22-320.jpg)
![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](https://image.slidesharecdn.com/transformerovercurrent-080211-160323223614/85/Transformer-overcurrent-23-320.jpg)
![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](https://image.slidesharecdn.com/transformerovercurrent-080211-160323223614/85/Transformer-overcurrent-32-320.jpg)
![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](https://image.slidesharecdn.com/transformerovercurrent-080211-160323223614/85/Transformer-overcurrent-36-320.jpg)
Transformer overcurrent
- 1. Transformer Calculations Transformer Overcurrent Protection Calculations & Settings Transformer Overcurrent Protection Calculations & Settings
- 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
- 3. Transformer Calculations X XF3 F4 X X F1 F2 50 51 50G 51G 87H87T 50 51 87GD Current Function 50 – Phase Instantaneous Overcurrent 51 – Phase Time Overcurrent Transformer Calculations Typical Single Line Diagram
- 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
- 5. Transformer Calculations R1 CURRENT t R2 Coordination Time Interval Overcurrent inverse time relay curves associated with two breakers on the same feeder Coordination of TOC Delay
- 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
- 12. Transformer Calculations Dac Y Dy11 X X 0 0.58 0.58 1.0 0 0 1.0 0 0 0.58 0 0 Phase-to-Ground Fault 3 I Iprimary = 31 2 I N N IIdelta == I X VI LG F == 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
- 14. Transformer Calculations A t If 2 3 If 0.4 Sec Coordination of overcurrent relays for a Dy transformer Coordination across D-y Transformer
- 15. Transformer Calculations Minimum nameplate (kVA) Category Single-phase Three-phase I 5-500 15-500 II 501-1667 501-5000 III 1668-10,000 5001-30,000 IV above 10,000 above 30,000 IEEE Std. C57.109-1985 Curves Transformer Categories
- 16. Transformer Calculations X XF3 F4 X X F1 F2 Infrequent Fault Incident Zone Frequent Fault Incident Zone Category 2 > 10 Category 3 > 5 Mechanical Damage: Number Faults in Lifetime Category II & III Fault Frequency
- 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
- 18. Transformer Calculations 2 Thermal Limit: I t = 1250 Category II – Transformer Through Fault Curves
- 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
- 24. Transformer Calculations OFAF (OA/FA/FOA) 30/40/50 MVA 230kV / 34.5 kV X X X X F1 F2 F3 F4 Zs230kV Z= 7% Calculation Setting Example
- 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
- 35. Transformer Calculations I RELAY = 0.63 Amps TD # 8 Very Inverse Curve TOC Relay Curve
- 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
- 37. Transformer Calculations Questions ?Questions ? ©2008 Beckwith Electric Co., Inc.