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PROJECT rep

# PROJECT rep - 22 22 2 2232321 BUFFALO STATE COLLEGE 113...

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1 3 2 3 1 2 3 1 2 2 2 2 2 2 2 BUFFALO STATE COLLEGE DEPARTMENT OF TECHNOLOGY ENT 472 Power Systems II COURSE PROJECT Spring 2010 Name: Jeevitha Aruchamy The single-line diagram of a three-phase power system is shown. Equipment ratings are given as follows: Synchronous generators and motors : G1 1000 MVA 15 kV X d = X 2 = 0.18, X 0 = 0.07 pu Y-grounded G2 1000 MVA 15 kV X d = X 2 = 0.20, X 0 = 0.10 pu Y-ungrounded G3 500 MVA 13.8 kV X d = X 2 = 0.15, X 0 = 0.05 pu Y-grounded through reactance G4 750 MVA 13.8 kV X d = 0.30, X 2 = 0.40, X 0 = 0.10 pu Y-ungrounded Transformers (voltages and connections are given for primary, and then secondary sites ) T1 1000 MVA 15 kV / 765 kV X = 0.10 pu Delta -- Y-grounded T2 1000 MVA 15 kV / 765 kV X = 0.10 pu Delta -- Y-grounded T3 500 MVA 15kV / 765 kV X = 0.12 pu Y-ungrounded – Y-grounded T4 750 MVA 15 kV / 765 kV X = 0.11 pu Y-grounded – Y-grounded Transmission Lines 1 - 2 765 kV X 1 = 50 Ohm, X 0 = 150 Ohm 1 - 3 765 kV X 1 = 40 Ohm, X 0 = 100 Ohm

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2 - 3 765 kV X 1 = 40 Ohm, X 0 = 100 Ohm PROJECT 1 SYMMETRICAL COMPONENTS ENT 472 Power Systems II SOLUTION : Base MVA- 1000 MVA Base KV -765 KV (LINE) & 15KV (GENERATORS) I.CALCULATION OF PER UNIT VALUES OF GENERATORS: GENERATOR 1:- X 1 = 0.18 p.u X 2 = 0.18 p.u X 0 = 0.07 p.u GENERATOR 2:- X 1 = 0.20 p.u X 2 = 0.20 p.u X 0 = 0.10 p.u GENERATOR 3:- X 1 = O.15 * (13.8KV/15KV) 2 * (1000MVA/500MVA) = 0.2539 p.u X 2 = O.15 * (13.8KV/15KV) 2 * (1000MVA/500MVA) = 0.2539 p.u X 0 = O.05 * (13.8KV/15KV) 2 * (1000MVA/500MVA) = 0.08464 p.u X n = 3X 0 =3*0.08464 = 0.2539 p.u TRANSFORMERS:- X T1 = 0.10 p.u X T2 = 0.10 p.u X T3 = 0.12 * (15KV/15KV) 2 * (1000MVA/500MVA) = 0.24p.u X T3 = 0.11 * (15KV/15KV) 2 * (1000MVA/750MVA) = 0.1467p.u TRANSMISSION LINES:- Z base = V base 2 /S base = (765KV) 2 /(1000MVA) = 585.23Ω Transmission line from bus1 to bus2 X 1 = 50 Ω/ 585.23 Ω = 0.08544p.u X 2 = 50 Ω/ 585.23 Ω = 0.08544p.u X 0 = 150 Ω/ 585.23 Ω
= 0.2563p.u Transmission line from bus1 to bus3 X 1 = 40 Ω/ 585.23 Ω = 0.06835p.u X 2 = 40 Ω/ 585.23 Ω = 0.06835p.u X 0 = 100 Ω/ 585.23 Ω = 0.1709p.u Transmission line from bus2 to bus3 X 1 = 40 Ω/ 585.23 Ω = 0.06835p.u X 2 = 40 Ω/ 585.23 Ω = 0.068354p.u X 0 = 100 Ω/ 585.23 Ω = 0.1709p.u II.PER UNIT IMPEDANCE DIAGRAMS: POSITIVE SEQUENCE DIAGRAM:- Fig 2.1 Positive sequence

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NEGATIVE SEQUENCE DIAGRAM:- Fig 2.2 Negative sequence ZERO SEQUENCE DIAGRAM:- Fig 2.3 Zero sequence
III.CALCULATION OF THEVENIN EQUIVALENT: POSITIVE SEQUENCE THEVENIN EQUIVALENT Step1: Given that the fault is at bus 2. In the positive sequence diagram fig 2.1, the impedance of generator 1 and transformer 1 is in series , so 0.18p.u+0.1p.u=0.28p.u The impedance of generator 2 and transformer 2 is in series , so 0.2p.u+0.1p.u=0.3p.u

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• Spring '10
• Dr.IlyaGrinberg
• Fig, Electrical resistance, Electric power transmission, Transmission line, Impedance matching, Characteristic impedance

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