Fall 2010 PQ Assignment _6 Solution

Fall 2010 PQ Assignment _6 Solution - Assignment #6 Chapter...

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Assignment #6 – Chapter 5 Fall 2010- EE 362Q /EE 394.9 Power Quality Prof. S. Santoso (instructor) Problems 5.6 – 5.14 Part [5.6] The single phase MVAR rating of the capacitor bank is given by, MVAR 4 . 38 800 8 6 1 = × × = × × = unit s p kVAR n n MVAR φ Thus, MVAR 2 . 115 4 . 38 3 3 1 3 = × = × = MVAR MVAR The single phase voltage rating of each unit is given by, kV 36 . 159 92 . 19 8 1 = × = kV Hence the capacitance is given by, 6 32 38.4 10 4.01 F 2 60 (159.36 10 ) C μ π × == ×× × × Alternatively, the rated terminal-to-terminal capacitor voltage is 02 . 276 3 36 . 159 = = LL kV kV The capacitance at the rated terminal-to-terminal capacitor voltage can then be computed as follows F C 01 . 4 02 . 276 60 2 2 . 115 2 = × × = Part [5.7] Rated per phase voltage is kV 36 . 159 1 = kV Rated per phase current, 6 3 38.4 10 240.96 A 159.36 10 I × × Part [5.8] The capacitor bank is energized when the substation voltage is 0.98 p.u. In other words, substation voltage, kV 4 . 225 98 . 0 230 = × = sub V Thus, the actual reactive power injected into the system is given by, 1
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2 225.4/ 3 3 38.4 76.82 MVAR 159.36 actual MVAR ⎛⎞ × = ⎜⎟ ⎝⎠ Part [5.9] When bank #1 is online, Voltage rise 00668 . 0 11500 82 . 76 = = Δ V p.u. Therefore, voltage at the bus is V = (0.00668 x 0.98) + 0.98 = 0.9865 p.u. or in kV it is 226.9 kV When bank #2 comes online, the MVAR delivered by bank #2 is 2 #2 226.9/ 3 3 38.4 77.84 MVAR 159.36 actual MVAR × = Voltage rise 00676 . 0 11500 84 . 77 = = Δ V p.u. Therefore, voltage at the bus is V = (0.00676 x 0.9865) + 0.9865 = 0.993 p.u. or in kV it is 228.42 kV Possible alternative calculation for voltage rise when both banks are online: Under steady state conditions, the two banks appear in parallel. Hence the combined capacitance of both the banks is 2C. (as both banks are in parallel) S X CC S V Therefore X C = 0.5X C as the capacitive reactance are in parallel. Therefore voltage rise is given by, sc cap C S C S C S MVA Q X X X X X X V φ 3 ' 2 2 5 . 0 = = = = Δ When both banks are switched online, the voltage rise is given by, Voltage rise 01336 . 0 11500 82 . 76 2 = × = Δ V p.u. or 1.336% respectively 2
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Hence voltage at bus is given by V = (0.01336 x 0.98) + 0.98 = 0.993 p.u. or in kV it is 228.42 kV Part [5.10] Worst case energizing scenario for bank #1 is 2V m (In this case the substation voltage is at 0.9865p.u. i.e. at 225.4kV before bank #1 is energized) Hence maximum transient overvoltage = 07 . 368 3 2 4 . 225 2 = × × kV Now system impedance is given by Ω = × = 6 . 4 11500 230 230 S Z Neglecting the system resistance, the system inductance can be calculated as follows: 6 . 4 = S L ω or 2 . 12 60 2 6 . 4 = × × = π S L mH
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Fall 2010 PQ Assignment _6 Solution - Assignment #6 Chapter...

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