Question

2. Shunt capacitors are installed at the receiving end to improve the line performance of Q1. The line delivers 70 MVA, 0.8 lagging power factor at 64 kV. Determine the total Mvar and the capacitance per phase of Y-connected capacitors when the sending end voltage is a) 69 kV b) 64 kV *Hint: Use (5.85) and (5.86) (H. Saadat] to compute the power angle and the receiving end reactive power.

Answer 1

Solution : 70 L cos'(0.8) Sload flere 56 Mwl + j42 MVAR V L P Q @ Now, if YO CL-1) = 69kv, We know, P= Vs Cos (0-5)- ² Caso Z. Z Here z is the impedance, o es impedance Angle Now, Z= 2+87= 7.28 <74.050 Here VS (LoL)- 69 kv, "R (L-L) = 64 kl 2 56= (69)(64) Cos (74.05 - ) - 64) Cos74.050 728 7.28 → cos(74.05-5)= 013471 S=4.3646 2 Now, Us)(Wo) SofCo-s)- Vo QR(supplied) sino 2. 근 (69)(64) Sen (74.05- 4.36) - Cu)? Sin Bos 7128 7.28 27.883 MVAR CS Scanned with CamScanner

Capacitor Therefore , MUAR Required by Qc = wad - O supplied = 42-27.883 -14.117 MVAR 2 2 (643) Now, -; K= v 12901147 m 14.11713 رس Pea Phase - Now, Shant Capacitance - pea phase (zat) (C) ㅗ 24(60) (290.(47) - 9.142 UF Now, UR = US GUKV b 56 = (64) (64) cos (74.056-8) - 6412cos 24.054° 728 7.28 → Cos (74.045 - $15013742 → S= 6.0320 & Suppina 164)(64) søn ( 74.056 - 6.0327) – (64)"son 74.0 7128 1128 = -19.2405 MVAR CS Scanned with CamScanner

43 - (-19,94099) Required Muap of (opuclaim a copa 61,240" MIIN "Copa 2 Noir (64153) 3-466.865 Qlapa Ś 61.240513 Ilow, = 39.650 uF Shunt Capacitance per phase ㅗ (30)(60) (66.5€) CS Scanned with CamScanner