(AP) 6 Application Notes Application of Individual Protection Functions 2.1.8.2 Example 2: Autotransformer (P645) - Load Tap Changer Figure 24 shows the application of a P645 to protect an autotransformer. The power transformer data has been given: 175/175/30 MVA Autotransformer, YNyn0d1, 230/115/13.8 kV. The current transformer ratios are as follows: HV CT ratio - 800/5, LV CT ratio - 1200/5 and TV CT ratio 2000/5. Since the transformer has an on load tap changer on the HV side, the nominal voltage of the HV winding must be set to the mid tap voltage level. According to the nameplate data, the mid tap voltage is 218.5 kV. The mid tap voltage can also be calculated as follows: 100 Mid tap position (5 15 ) 2 230 100 218 .5kV 175/ 175/ 30 MVA 230/ 115/13. 8 kV 230kV +5%/-15% 19 taps a b c YNynd1 A a B b C c Grounding transformer inside the protection zone a a a B b b b B c c a a b b c c B B c B B Yd 11 software interposing CT Ydy0 software interposing CT a a a b b b c c c B B B D D D Ydy0 software interposing CT P645 P4304ENb Figure 24 - P645 used to protect an autotransformer with load tap changer P64x/EN AP/A62 Page (AP) 6-41 (AP) 6 Application Notes Application of Individual Protection Functions Ratio correction: The relay calculates the ratio correction factors as follows: K amp ,T1CT Inom ,T1CT 800 S ref 175 10 6 3 Vnom ,HV K amp ,T5CT K amp ,T3CT 1.730 218.5 103 3 Inom ,T5CT 1200 S ref 175 10 6 1.366 3 Vnom ,LV 3 115 103 Inom ,T 3CT 2000 S ref 175 10 6 3 Vnom ,TV 0.273 3 13.8 10 3 To check that the differential protection does not misoperate due to errors introduced by the on load tap changer, the user may perform the following calculations. Transformer nominal rating Calculate HV full load current at both tap extremities and LV and TV full load current. 175 106 3 241500 HV full load current on tap 1 (5%) 418.37 160 HV full load current on tap 1 (5%) 418.37 A primary 2.615 A secondary HV corrected current on tap 1 1.730 2.615 4.524 A secondary HV full load current on tap 19 (-15%) HV full load current on tap 19 (-15%) 175 10 6 3 195 .510 3 516.810 160 HV corrected current on tap 19 1.730 3.230 LV full load current LV full load current TV full load current TV full load current 145 10 6 3 115 10 3 727 .963 240 3.230 A secondary 5.588 A secondary 727 .963 A primary 3.033 A secondary 30 10 6 3 13.8 10 3 1255 .109 400 516 .810 A primary 1255 .109 A primary 3.138 A secondary Determine Idiff at both tap extremities (with mid tap correction). LV correctedcurrent 1.366 3.033 Page (AP) 6-42 4.143 P64x/EN AP/A62 (AP) 6 Application Notes Application of Individual Protection Functions TV correctedcurrent 0.273 3.138 0.857 Idiff at tap 1 0.476 A Idiff at tap 19 4.524 4.143 0.857 5.588 4.143 0.857 0.588 A 0.476 5 0.588 5 0.095 pu 0.118 pu Determine Ibias at both tap extremities (with mid tap correction). The currents used in the Ibias calculation are the currents after ratio and vector correction. Ibias at tap 1 Ibias at tap 19 4.524 4.143 0.857 2 4.762 A 5.588 4.143 0.857 2 5.294 A 4.762 5 5.294 5 0.9524 pu 1.059 pu Determine relay differential current. Iop = Is1, (Ibias Is1/K1) Iop = K1 Ibias, (Is1/K1 Iop = K1 Is2 + K2 Ibias Is2) (Ibias - Is2), (Ibias Is2) Ibias at tap 1 is less than 5 A (1 pu) and greater than 3.33 A (0.667 pu); since Is2 is set to the rated current (5 A), Iop is calculated as follows: Iop = 0.3 4.762 = 1.429 A Ibias at tap 19 is greater than 5 A; since Is2 is set to the rated current (5 A), Iop is calculated as follows: Is2 = 1 pu = 1 Iop = 0.3 5=5A 5 + 0.8 (5.294 - 5) = 1.735 A Check Idiff < Iop by a 10% margin for each tap extremity and adjust Is1 and/or K1 as necessary. Tap 1: Since Idiff = 0.476A and 0.9Iop at tap 1 = 0.9 x 1.429 = 1.286 A Therefore there is sufficient safety margin with K1 = 30% and Is1 = 0.2 pu. Tap 19: Since Idiff = 0.588A and 0.9Iop at tap 19 = 0.9 x 1.735 = 1.562 A Therefore there is sufficient safety margin with K1 = 30% and Is1 = 0.2 pu. 66.7% of transformer nominal rating Calculate HV, LV and TV load current at 66.7% of the nominal MVA rating. The 66.7% is the interception between Is1 and K1. It is determined as Is1/K1 100% = (0.2/0.3) 100 = 66.7%. HV full load current on tap 1(5%) 0.667 HV full load current on tap 1(5%) P64x/EN AP/A62 279 .05 160 175 10 6 3 241500 279.05 A primary 1.744 A secondary Page (AP) 6-43 (AP) 6 Application Notes Application of Individual Protection Functions HV corrected current on tap 1 1.730 1.744 3.017 A sec ondary 175 10 6 HV full load current on tap 19 (-15%) 0.667 3 195 .510 3 344.71 160 HV full load current on tap 19 (-15%) 2.154 A secondary HV corrected current on tap 19 1.730 2.154 LV full load current 0.667 LV full load current 485 .551 240 TV full load current 0.667 837.158 400 TV full load current 145 10 6 3 115 10 3 344 .71A primary 3.727 A sec ondary 485 .551A primary 2.023 A secondary 30 10 6 3 13.8 10 3 837.158 A primary 2.093 A secondary Determine Idiff at both tap extremities (with mid tap correction). LV corrected current 1.366 2.023 2.763 A TV corrected current 0.273 2.093 0.571A Idiff at tap 1 3.017 2.763 0.571 Idiff at tap 19 0.317 A 3.727 2.763 0.571 0.393 A 0.317 5 0.393 5 0.063 pu 0.0786 pu Determine Ibias at both tap extremities (with mid tap correction). The currents used in Ibias calculation are the currents after ratio and vector correction. Ibias at tap 1 3.017 2.763 0.571 2 Ibias at tap 19 3.727 2.763 0.571 2 3.176 A 3.531A 3.176 5 3.531 5 0.635 pu 0.706 pu Determine relay differential current. Iop = Is1, (Is1/K1 Ibias) Iop = K1 Ibias, (Is1/K1 Iop = K1 Is2 + K2 Ibias Is2) (Ibias - Is2), (Ibias Is2) Ibias at tap 1 is less than 3.33 A (0.667 pu); then Iop = 0.2 pu = 1 A Ibias at tap 1 is less than 5 A (1 pu) and greater than 3.33 A (0.667 pu); since Is2 is set to the rated current (5 A), Iop is calculated as follows: Page (AP) 6-44 P64x/EN AP/A62 (AP) 6 Application Notes Application of Individual Protection Functions Iop = 0.3 3.531 = 1.059 A Check Idiff < Iop by a 10% margin for each tap extremity and adjust Is1 and/or K1 as necessary: Tap 1: Since Idiff = 0.317 A and 0.9Iop at tap 1 = 0.9 x 1 = 0.9 A Therefore there is sufficient safety margin with K1 = 30% and Is1 = 0.2 pu. Tap 19: Since Idiff = 0.393 A and 0.9Iop at tap 19 = 0.9 x 1.059 = 0.953 A Therefore there is sufficient safety margin with K1 = 30% Is1 = 0.2 pu. Figure 25 shows the bias characteristic, the CT and tap changer errors (assumed as 20%), and the bias, differential coordinates corresponding to full load current and 66.7% of full load current. It can also be seen that it is necessary to check the safety margin at the two knee-points of the bias characteristic. 2.5 Bias characteristic CT and tap changer errors 2 nominal MVA, tap 1 nominal MVA, tap 17 66.7% of nominal MVA, tap 1 66.7% of nominal MVA, tap 17 1.5 1 0.5 0 0 0.5 1 1.5 2 Ibias/Inom 2.5 3 3.5 4 P4305ENa Figure 25 - Safety margin at the two knee-points of the bias characteristic Vector correction and zero sequence filtering Vector correction is done by setting LV Vector Group in SYSTEM CONFIG to 0 and TV Vector Group in SYSTEM CONFIG to 1. The zero sequence filtering is done by setting HV Grounding, LV Grounding and TV Grounding in SYSTEM CONFIG to Grounded. The TV winding is grounded using a grounding transformer inside the protected zone. The system configuration and differential protection settings are shown in Figure 26: P64x/EN AP/A62 Page (AP) 6-45