Question

Two three-phase power transmission lines with a nominal operating voltage of 220 kV and a frequency of 60 Hz are connected in series as shown below.

E.I.H.-1 R/2 Ei.H.-2 min 2 00000 R/2 2 10000) L 10000 1. IN N10 load

Per-phase constants of the first 110 km long energy transmission line (E.İ.H.-1); r = 0.01035Ω / km, L = 1.554 mH / km, and C = 7.387 ∙ 10-3 µF / km. Constants per phase of the second 100 km long energy transmission line (E.İ.H.2); Zh = 1.035 + j58.58 Ω / phase and Y = 278.47 µS / phase.
The two lines are symmetrical in themselves. When an inductive load is connected to the end of the energy transmission line, it is seen that the voltage per line is 241.4312.97 kVolt and the current is 260.369 -23.186 Amper. According to the operation of the power transmission line under these operating conditions,

a.) Find the equivalent A, B, C, D constants of the line according to the connection status of the two Energy Transmission lines above. and Examine the results you found under "b" and "c" according to the loading status of the line and explain the necessary explanations.

b.) Calculate the phase voltage and current at the end of the line. and the value of the apparent power drawn by the load and the efficiency of the line.

Answer 1

Answer :- The Given data :- > Two three phase power transmission lines with operating voltage = 220 KV 60 H3 frequency (f) 12 R 피 R12 wh 42 D. R12 42 wemo m2 니2 2/2 c E 1. Hi EiH 2 Per. phase constants of first 110km long energy (Ei H-1) transmission line 0.01035 s/km 10554 MHIkm L c = 7+3878 10-3 MF lkm -> per phase constants of second 100 km long energy transsmission line (Ei-H.2) Iphase 2 1.035 +358:58 mm y = 278.47 115 l phase M 241.443 232.970 kr I = 260.36 (-23-186° A

- hi a) 2. NINI 12 Zzi Y, V2 1 load It represents T model Assume a model It represents Ti model Assume T model as TB as TA where T ITA]* (to Here Zi = R +jxL C': f= 60H3) R = 0.010351 | Kma 0.01035 X 100 1.1385.0 XL = WL = 27152 = 2 1 X 60 x 11550x103 Xllo 64.432 z 101385 + 3 64.43 m , phase Zi = 64.44 289° Iphase y, jwc j 2T X607,387410 x 110 yi - j306-33 ms (phase Given that 32=1.035+3 58.58 n 1 phase 58.59 2890 Iphase = 42 j278.47 LU / Phase

+요. ㅜ A [1) + 다 돼 + Z) 9 2 +252 뽐 2 14 가 닦던 +1 이 셔a* 1107 T- +I (219] Zz 1+23 32 미나부 요 + Z9) (+2392 2 'z 7 [17] + z [4] (22] 사조부러사필기의가쪽 iz +! (1) 22 + z, [1+1년] [+] 를 143볼 Z292 yt 1

1-514 4 vart jau sus 7 (390-31]]((rear-tjet 13 ( 21.04.2) .sation se ng kanyang inal) 3 278.17 Jzis ut), 7 (-0352js3:38 ] [1278007]]+[1+ '1385 jour us]/4306.30]]5218104 14 (1035+j584836278 473) lit H+(1.235 +35840) (j 278.47]]1278:47 시 2 -0.35+j;58 88) j278:47 (146.2 2) 746.025 - 1385'+j6uu3) (336-28](-0.35+jst 5)+ (-1365+3 6+43)(1+(+158 taj bu kali 306-28 2 (4-138+j64-43] (j 306-33 (3 306-22) (1.035735858)+(1+(4-1387 (0035 +358.58 ) (3278-U72] I [C 1+ 2 2

10.9960.01°)(0.99K0.080)+(62-44289°) (0.9420.005)(0.99620.00)/778.41290° (-9920.1) (58.59189°) + 64.44484 10.99401)/0.99 0.08 T= (306-3390°) (0.9442.008) + (0-9960-01) (18.47440*)/0.996 0.0001) (306-33670)(58-59 (84) + (0-1920.0f) (0-994-0-006 0.963 10.34 121.16 289 ㅏ 5:77X10-4 290° 0.963 20:34

A B where T C D Therefore A= 0.963 20.034° 121.16289° = 5.77 2104 190° D=0.963 20.034' Here B shows the impedance of the transmission line - represented in 'is' c shows the admittance of the transmission where MS line , represented in b) Where Vi= a vzt biz I = (v2 + die Given vi = 241:43 L12.97 201.43 vi per phase 13 139.39 212.97 I, = 260.36 L-23.186 139. 39 412197 0.963 2004 V2 + 121.16 189 Iz 260.36 %-23.186 = 5.77X10" L90 V₂ + 0.963 20.034 Iz

Here by soking the equation and equation @ so V2 = 141 v I2 = 249.5 Therefore the phase. Voltage = 141 V current per phase - 249.5 A is 3V₂ Iz * The apparent power drawn by load 3 xiul * 249.5 105538.5 VA р = TOS: 538 KVA Therefore the apparent power load drawn by is 105.536 kVA V3x V, XI where Input power apparent -V3x 2u1.43(12.97 x 260.36 223-18 6 (107.152 + 19.315) KVA where Efficiency - Input paser 107.152 kVA apparent power 105.538x103 Input power 107.152 x102 Efficiency - 0.784 Efficiency 7. = Therefore the efficiency of the line is 98.46: 98.47.