Power System Assignment - 02| POWER SYSTEM ASSIGNMENT-02 (odd) |
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Submitted by : Reza Ashraf (Rabbi), B.Sc. in EEE, 30 | B | Day, SUB.
Book – Principle of power system (V.K. Mehta)/Chapter 10 – Performance of transmission lines/Page – 229/Q: 1 [A single phase overhead transmission line delivers 4000 kw at 11 kv at 0.8 p.f. lagging. If resistance & reactance are 0.15 Ω & 0.02 Ω respectively, calculate: – (i) percentage regulation (ii) sending end power factor (iii) line losses] Given that – PR = 4000 kw VR = 11 kv R = 0.15 XL = 0.02 Ω Cos фR = 0.8 Lagging Sin фR = 0.6 Lagging Z = 0.15 + j0.02 = 0.15 < 7.6 I = 4000 × 103 / 11 × 103 × 0.8 = 454 A VR = 1100 + j0 = 11000 < 0 I = 44 (0.8 – j0.6) = 363 – j272.4 = 453.8 < –36.89 VS = 11000 + 453.8 < –36.89 × 0.15 < 7.6 = 11000 + 68 < –29.3 = 11000 + 59.3 – j33 = 11059.3 – j33 = 11059.34 < – 0.17 →% Regulation = 11059.3 – 11000 / 11000 ×100 = 0.53% Cos фS = 11000 × 0.8 + 453.8 × 0.15 / 11059.34 = 0.8 lagging →Line loss = (453.8)2 × 0.15 = 30 kw Answer. Book – Principle of power system (V.K. Mehta)/Chapter 10 – Performance of transmission lines/Page – 240/Q: 2 [A 3 phase, 50 Hz overhead transmission line has the following constants - Resistance / phase = 9.6 Ω Inductance / phase = 0.097 mH Capacitance / phase = 0.765 цF If the line is supplying a balanced load of 24000 kvA & 0.8 p.f. lagging at 66 kv, calculate: – (i) sending end current (ii) line value of sending end voltage (iii) sending end power factor (iv) percentage regulation (v) transmission efficiency ] Given that – f = 50 Hz R/ ф = 9.6 Ω L = 0.097 mH C = 0.765 цF PR = 24000 kvA VR = 66 kv Cos фR = 0.8 Lagging Sin фR = 0.6 Lagging XL = 2 × 3.14 × 50 × 0.097 × 10-3 H = 0.03 H Y = 2 × 3.14 × 50 × 0.765 × 10-6 = 2.4 × 10-4 F Z = 9.6 + j0.03 = 9.6 < 0.18 VR = 66 × 103 / √3 = 38105 v IR = 24000 × 103 / √3 × 66 × 103 = 209.9 A VR = 38105 + j0 = 38105 < 0 IR = 209.9 (0.8 – j0.6) = 167.92 – j125.9 = 209.9 < – 36.86 VI = 38105 + 209.9 < – 36.86 × 4.8 + j0.015 = 38105 + 209.9 < – 36.86 × 4.8 < 0.178 = 38105 + 1007.5 < – 36.7 = 39195 +807.8 – j602 = 38912.8 – j602 = 38917.5 < –0.89 IC = j × 2 × 10-4 × (38912.8 – j602) = 0.144 + j9.33 = 9.33 < 89.12 (i) IS = 167.92–j125.9 + 0.144 + j9.33 = 168 – j115.67 = 203.96 < – 34.54 (ii) VS = 38912.8 – j602 + 203.96 < – 34.5 × 4.8 < 0.178 = 38912 – j602 + 979 < – 34.33 = 38912 – j602 + 808.5 – j552 = 39721.3 – J1154 = 39738 < – 1.66 →Line to line voltage = √3 × 39742.9 = 68826 = 68.82 kv (iii) Angle between VR and VS (Ѳ1) = 1.65 & angle between VR and IS (Ѳ2) = 34.54 ФS = 1.65 + 34.54 = 36.2 Cos ФS = Cos 36.2 = 0.8 lagging (iv) % Regulation = (39738 – 38105 / 38105) × 100 = 42.8 % (v) PS = 3 × 39738 × 203.96 × 0.8 = 19451909.95 w PR = 24000 × 103 × 0.8 = 192 × 105 →Efficiency (ηT) = (1920000 / 19451909.95) × 100 = 98.7% Answer.
Book – Principle of power system (V.K. Mehta)/Chapter 10- Performance of transmission lines/Page – 240/Q: 3 [A 3 phases overhead transmission line delivers 10 MW at 0.8 p.f. lagging and at 66 kv. The resistance & inductive reactance of the line per phase are 10 Ω & 20 Ω respectively while capacitance admittance is 4 × 10-4 sicmen. Calculate: – (i) sending end current (ii) sending end voltage – line to line (iii) sending end power factor (iv) transmission efficiency. Using nominal T method] Given that – PR = 10 Mw VR = 66 kv R/ ф = 10 Ω XL / ф = 20 Ω Y = 4 × 10-4 δ Cos фR = 0.8 Lagging Sin фR = 0.6 Lagging Z = 10 + j10 & Z/2 = 5 + j10 = 11.18 < 63.4 Now, VR = 66 × 103 / √3 = 38105 IR = 10 × 106 / √3 × 0.8 × 66 × 103 = 109 A VR = 38105 + j0 = 38105 < 0 IR = 109 (0.8 – j0.6) = 87.2 – j65.4 = 109 < – 36.86 VI = 38105 + 109 < – 36.86 × 11.18 < 63.4 = 38105 + 1218.62 < 26.54 = 38105 + 1090 + j544.5 = 39195 + j544.5 =39198.8 < 0.8
IC = j × 4 × 10-4 × (39195 × j544.5) = – 0.217 + j15.67 = 15.67 < 90 (i) IS = – 0.217 + j15.67 + 87.2 – j65.4 = 86.98 – j49.73 = 100 < – 29.75 →Sending end current = 100 A (ii) VS = 39195 + j544.5 + 100 < – 29.75 × 11.18 < 63.4 = 39195 + j544.5 + 930.67 + j619.5 = 40125.67 + J1164 = 40142 < 1.66 →Line to line voltage = 40142 × √3 = 69527 =69.52 kv (iii) Angle between VR and VS (Ѳ1) = 1.66 & angle between VR and IS (Ѳ2) = 29.75 ФS = 1.66 + 29.75 = 31.41 Cos ФS = Cos 31.41 = 0.85 lagging (iv) PS = 3 × 40142 × 100 × 0.85 = 10236210 w →Efficiency (ηT) = (10 × 106 / 10236210) × 100 = 97% Answer.
Book – Principle of power system (V.K. Mehta)/Chapter 10 Performance of transmission lines/Page – 241/Q: 6 [A 3 phase, 50 Hz overhead transmission line, 110 kv between the lines at the receiving end has the following constants: Resistance per km per phase = 0.153 Ω Inductance per km per phase = 1.21 mH Capacitance per km per phase = 0.00958 цF The line supplies a load of 20000 kw at 0.9 power factor lagging. Calculate using nominal π representation. The sending end voltage, current, power factor, regulation and efficiency of the line neglect leakage. ] Given that – f = 50 Hz Line length = 100 km L / ф / km = 1.21 mH C = 0.00958 цF PR = 20000 kw VR = 110 kv Cos фR = 0.9 Lagging Sin фR = 0.44 Lagging Total R/ ф / km = 0.153 × 100 = 15.3 Ω XL / ф / km = 2 × 3.14 × 50 × 1.21 × 10-3 = 38Ω VR = 110 × 103 / √3 = 63508 IR = 20000 × 103 / √3 × 0.9 × 110 × 103 = 116 A VR = 63508 + j0 = 63508 < 0 IR = 116 (0.9 – j0.44) = 104.4 – j51.04 = 116.2 < – 26 IC1 = 63508 × j × 3 × 10-7 / 2 = j9.52 IL = 104.4 – j51 + 0 + j9.52 = 104.4 – j41.5 = 112.34 < – 26.67 (i) VS = 63508 + 112.34 < – 21.67 × 15.3 < j38 + 112.34 < – 26.67 × 40.96 < 68 + 4601.44 < 46.33 + 3177.31 + j3328.15 = 66685.3 + j3328.35 = 66768.3 < 2.85 →Line to line voltage = √3 × 66768.8 = 115.46 v IC2 = j66685.3 + j3328.35 × 3 × 10-4 / 2 = – 0.5 + j10 IS = 104.4 – j41.5 – 5.0 + j10 = 103.9 – j31.5 = 108.57 < –16.86 →Angle between VR and VS (Ѳ1) = 2.85 & angle between VR and IS (Ѳ2) = 16.86 ФS = 2.85 + 16.86 = 19.71 & Cos ФS = Cos 19.71 = 0.94 lagging →% Regulation = (66768.3 – 63508 / 63508) × 100 = 5.13% VS = 3 × 66768.3 × 108.57 × 0.94 = 20442276 w →Efficiency (ηT) = (20000 × 103 / 20442276) × 100 = 97.81% Answer.
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| Last Updated on Thursday, 08 October 2009 07:33 |
