1 00:00:23,136 --> 00:00:23,205 C 2 00:00:23,206 --> 00:00:23,275 CT 3 00:00:23,276 --> 00:00:23,345 CT 4 00:00:23,346 --> 00:00:23,415 CT E 5 00:00:23,416 --> 00:00:23,485 CT Ex 6 00:00:23,486 --> 00:00:23,556 CT Exc 7 00:00:23,557 --> 00:00:23,626 CT Exci 8 00:00:23,627 --> 00:00:23,696 CT Excit 9 00:00:23,697 --> 00:00:23,766 CT Excita 10 00:00:23,767 --> 00:00:23,836 CT Excitat 11 00:00:23,837 --> 00:00:23,906 CT Excitati 12 00:00:23,907 --> 00:00:23,976 CT Excitatio 13 00:00:23,977 --> 00:00:24,046 CT Excitation 14 00:00:24,047 --> 00:00:24,116 CT Excitation 15 00:00:24,117 --> 00:00:24,187 CT Excitation T 16 00:00:24,188 --> 00:00:24,257 CT Excitation Te 17 00:00:24,258 --> 00:00:24,327 CT Excitation Tes 18 00:00:24,328 --> 00:00:24,397 CT Excitation Test 19 00:00:24,398 --> 00:00:24,467 CT Excitation Test, 20 00:00:24,468 --> 00:00:24,537 CT Excitation Test, 21 00:00:24,538 --> 00:00:24,607 CT Excitation Test, P 22 00:00:24,608 --> 00:00:24,677 CT Excitation Test, Pa 23 00:00:24,678 --> 00:00:24,747 CT Excitation Test, Par 24 00:00:24,748 --> 00:00:24,818 CT Excitation Test, Part 25 00:00:24,819 --> 00:00:24,888 CT Excitation Test, Part 26 00:00:24,889 --> 00:00:24,958 CT Excitation Test, Part I 27 00:00:24,959 --> 00:00:25,028 CT Excitation Test, Part IV 28 00:00:25,029 --> 00:00:25,098 CT Excitation Test, Part IV: 29 00:00:25,099 --> 00:00:25,599 CT Excitation Test, Part IV: 30 00:00:28,045 --> 00:00:33,924 As mentioned earlier, the core losses include Foucault and hysteresis and its model is as you can see here: 31 00:00:33,949 --> 00:00:35,325 By using this relation, 32 00:00:35,341 --> 00:00:46,594 we can obtain the Foucault and hysteresis resistance models which Foucault resistance model is linear and independent from the voltage while hysteresis resistance model is non-linear and dependent on the voltage. 33 00:00:46,619 --> 00:00:50,434 If the CT Primary is open and voltage is fed from the secondary side, 34 00:00:50,459 --> 00:00:57,608 since Rct and the current drawn from the source is clear, the voltage of the parallel branch is calculated using: 35 00:00:57,648 --> 00:01:01,627 E= Vterminal –RCT * I 36 00:01:01,914 --> 00:01:05,531 After calculating E, the core casualties are calculated using: 37 00:01:05,556 --> 00:01:07,533 Pc= E*I *Cos ( 00:01:12,045 By having the core losses value and its extended for formula 39 00:01:12,070 --> 00:01:19,098 It is possible to calculate Ke, Kh and X by solving the three equations in the three variables system. 40 00:01:19,083 --> 00:01:23,168 To solve these equations, it is possible to apply three different voltages. 41 00:01:23,193 --> 00:01:29,649 But if the frequency is the same, its determinant turns zero and no answer can be obtained for the variables. 42 00:01:29,674 --> 00:01:33,666 Therefore, it is better to apply one of the voltages with a different frequency. 43 00:01:33,712 --> 00:01:39,421 In the first step of solving these equations with V1 and f1 as voltage and frequency, 44 00:01:39,446 --> 00:01:49,778 Ke and (Kh*(f1-x) parameters can be obtained using a two equations in the two variables system. 45 00:01:50,618 --> 00:02:06,379 Now, if one of the voltages is applied along with f2 as the frequency, “x” is obtained and by substituting it in Kh*(f1-x), Kh obtained achieved as well.By specifying these parameters, the resistance of Foucault and Hysteresis obtained. 46 00:02:06,512 --> 00:02:11,664 Due to voltage application limit for CTs with a very high saturation point, 47 00:02:11,689 --> 00:02:20,375 it is possible to find the saturation point in the nominal frequency by decreasing the voltage and frequency with the same proportion. 48 00:02:20,367 --> 00:02:24,467 Because the relation of the according flux is V/f. 49 00:02:24,579 --> 00:02:29,411 But does not changing the frequency change the CT system? To explain this subject, 50 00:02:29,454 --> 00:02:36,239 we need to examine those components of the CT that are affected if the frequency and voltage are changed. 51 00:02:36,360 --> 00:02:45,835 According to previous analyses, Rct is dependent on the material of the wire and Reddy to the material of the coil. 52 00:02:45,947 --> 00:02:56,259 In Xm branch, due to the proportional change of V and f and the stability of the flux, this branch remains unchanged as well. 53 00:02:56,426 --> 00:03:04,363 Therefore, the only parameter affected by the voltage and frequency is hysteresis resistance which was obtained 54 00:03:04,472 --> 00:03:11,325 Now suppose that to obtain the saturation point, the fed voltage and frequency are split in two. 55 00:03:11,373 --> 00:03:18,710 Is it possible to obtain the saturation point in the nominal frequency by multiplying the voltage and current by two? 56 00:03:18,735 --> 00:03:21,417 This question will be answered in the following. 57 00:03:21,442 --> 00:03:41,860 Suppose that the are CT1.5 Kh=10 -3 x=1.8Ke=10 -4 knee point is obtained by decreasing the frequency to 25 Hz with 500 volts of voltage and 15 milliamps . 58 00:03:41,876 --> 00:03:43,990 with a -45 the grid 59 00:03:43,999 --> 00:03:50,351 EExc = Vterminal - Rct IExc 60 00:03:50,401 --> 00:04:00,663 Since now we have Ke، Kh and x, by dividing EExc by Reddy and Rh, la and lh currents are calculated. 61 00:04:00,688 --> 00:04:07,955 By subtracting these two from lExc, the self-inductor current is obtained. According to this relation, 62 00:04:07,994 --> 00:04:10,050 Phi=L* I 63 00:04:10,075 --> 00:04:14,124 when the flux is fixed, the self-inductor current is fixed as well. 64 00:04:14,149 --> 00:04:19,391 Therefore, by doubling the voltage of the parallel branch, this current remains fixed. 65 00:04:19,416 --> 00:04:24,198 After doubling the voltage and frequency, as it is shoawn in this formula 66 00:04:24,223 --> 00:04:32,058 the Foucault and hysteresis currents are calculated and added by self-inductor current to achieve lExc using this relation. 67 00:04:32,083 --> 00:04:42,847 Then, the voltage of the parallel branch which is 25Exc is added to Rct* lExc and the terminal voltage is obtained. 68 00:04:43,937 --> 00:04:43,979 D 69 00:04:43,980 --> 00:04:44,022 De 70 00:04:44,023 --> 00:04:44,065 Dec 71 00:04:44,066 --> 00:04:44,108 Decr 72 00:04:44,109 --> 00:04:44,151 Decre 73 00:04:44,152 --> 00:04:44,194 Decrea 74 00:04:44,195 --> 00:04:44,237 Decreas 75 00:04:44,238 --> 00:04:44,281 Decreasi 76 00:04:44,282 --> 00:04:44,324 Decreasin 77 00:04:44,325 --> 00:04:44,367 Decreasing 78 00:04:44,368 --> 00:04:44,410 Decreasing 79 00:04:44,411 --> 00:04:44,453 Decreasing t 80 00:04:44,454 --> 00:04:44,496 Decreasing th 81 00:04:44,497 --> 00:04:44,539 Decreasing the 82 00:04:44,540 --> 00:04:44,582 Decreasing the 83 00:04:44,583 --> 00:04:44,625 Decreasing the f 84 00:04:44,626 --> 00:04:44,668 Decreasing the fr 85 00:04:44,669 --> 00:04:44,711 Decreasing the fre 86 00:04:44,712 --> 00:04:44,754 Decreasing the freq 87 00:04:44,755 --> 00:04:44,797 Decreasing the frequ 88 00:04:44,798 --> 00:04:44,840 Decreasing the freque 89 00:04:44,841 --> 00:04:44,883 Decreasing the frequen 90 00:04:44,884 --> 00:04:44,926 Decreasing the frequenc 91 00:04:44,927 --> 00:04:44,970 Decreasing the frequency 92 00:04:44,971 --> 00:04:45,013 Decreasing the frequency 93 00:04:45,014 --> 00:04:45,056 Decreasing the frequency a 94 00:04:45,057 --> 00:04:45,099 Decreasing the frequency an 95 00:04:45,100 --> 00:04:45,142 Decreasing the frequency and 96 00:04:45,143 --> 00:04:45,185 Decreasing the frequency and 97 00:04:45,186 --> 00:04:45,228 Decreasing the frequency and p 98 00:04:45,229 --> 00:04:45,271 Decreasing the frequency and pe 99 00:04:45,272 --> 00:04:45,314 Decreasing the frequency and per 100 00:04:45,315 --> 00:04:45,357 Decreasing the frequency and perf 101 00:04:45,358 --> 00:04:45,400 Decreasing the frequency and perfo 102 00:04:45,401 --> 00:04:45,443 Decreasing the frequency and perfor 103 00:04:45,444 --> 00:04:45,486 Decreasing the frequency and perform 104 00:04:45,487 --> 00:04:45,529 Decreasing the frequency and performi 105 00:04:45,530 --> 00:04:45,572 Decreasing the frequency and performin 106 00:04:45,573 --> 00:04:45,615 Decreasing the frequency and performing 107 00:04:45,616 --> 00:04:45,658 Decreasing the frequency and performing 108 00:04:45,659 --> 00:04:45,702 Decreasing the frequency and performing t 109 00:04:45,703 --> 00:04:45,745 Decreasing the frequency and performing th 110 00:04:45,746 --> 00:04:45,788 Decreasing the frequency and performing the 111 00:04:45,789 --> 00:04:45,831 Decreasing the frequency and performing the 112 00:04:45,832 --> 00:04:45,874 Decreasing the frequency and performing the t 113 00:04:45,875 --> 00:04:45,917 Decreasing the frequency and performing the te 114 00:04:45,918 --> 00:04:45,960 Decreasing the frequency and performing the tes 115 00:04:45,961 --> 00:04:46,003 Decreasing the frequency and performing the test 116 00:04:46,004 --> 00:04:46,046 Decreasing the frequency and performing the test: 117 00:04:46,047 --> 00:04:46,647 Decreasing the frequency and performing the test: 118 00:04:46,672 --> 00:04:52,227 As you know, the maximum output voltage of a device in one-phase mode is “450” volts. 119 00:04:52,252 --> 00:04:58,654 Usually, to avoid damaging the measuring equipment, “Measuring CTs” go to saturation in low voltages. 120 00:04:58,679 --> 00:05:07,401 But the saturation voltage of some “Protection CTs” such as “TPX”, “TPY” and “TPZ” is too high and is beyond the ability of the device. 121 00:05:07,426 --> 00:05:13,043 To test these “CTs”, it is possible to increase the voltage until the “CT” goes to saturation but 122 00:05:13,068 --> 00:05:22,282 some scientific references suggest that high voltages must not be fed into “CTs” because high voltages can damage the insulation of the equipment. 123 00:05:22,307 --> 00:05:28,700 Another method for excitation test in “CTs” with too high saturation voltage is to decrease the frequency. 124 00:05:28,725 --> 00:05:36,319 This method is also mentioned in the standard CT” excitation test which can be performed with an “AMT105” device. 125 00:05:36,344 --> 00:05:44,739 As you saw, the knee point of a 5P10 protection CT in 50 Hz frequency was measured to be about 159 volts. 126 00:05:44,764 --> 00:05:52,003 If you wish to specify the value of the final voltage and injected frequency manually, you need to uncheck “Calc. 127 00:05:52,028 --> 00:06:07,621 From TestObject” option and fill “Frequency” and “V end” fields. For example, “25 Hz” and “186 volts” are entered as frequency and final voltage respectively which means “372 volts” in “50 Hz” frequency. 128 00:06:07,646 --> 00:06:12,972 Note that by changing the voltage, it is so much probable that the wirings change as well. 129 00:06:12,996 --> 00:06:23,347 In this method, “Demagnetize” option is checked by default so that the magnetic residue is removed and the aforementioned calculations are done. 130 00:06:23,371 --> 00:06:30,452 After adjusting this settings and specifying the test time, click on “Init Test” to initiate the test. 131 00:06:43,652 --> 00:06:52,392 As you can see in the “Signal View”, by decreasing the frequency and equating, the knee point is almost equal with what was measured in 50 Hz frequency. 132 00:06:52,416 --> 00:07:02,827 Therefore, in cases where the CT saturation point voltage is beyond the limit of the device, by decreasing the frequency, you can measure the knee point and its saturation.