Ch 11 - Generator Protection

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Symmetrical ComponentsExample Problem• I a0 = 0, I b0 = 0, I c0 = 0• I a1 = 5.78 /-30° , I b1 = 5.78 /-150° , I c1 = 5.78 /90°• I a2 = 5.78 /30° , I b2 = 5.78 /150° , I c2 = 5.78 /-90°GENERATOR CONTROL AND PROTECTION
Example ProblemSymmetrical Components• Note: the components I c1 and I c2 havedefinite values although line c is open andcan carry no net current. As expected, thesum of these currents is zero.• The sum of the currents in line a is 10/0°• The sum of the currents in line b is 10/180°GENERATOR CONTROL AND PROTECTION
Page 1 and 2: Generator ProtectionGENERATOR CONTR
Page 3 and 4: Generator Protection• Loss of Fie
Page 5: Steam Generator Stator WindingsGENE
Page 10 and 11: Cylindrical Rotor in Need of Repair
Page 14 and 15: Symmetrical ComponentsGENERATOR CON
Page 20 and 21: Symmetrical ComponentsExample Probl
Page 24 and 25: Symmetrical ComponentsSingle Phase
Page 29 and 30: Fault Current Behavior of aSynchron
Page 35 and 36: Generator GroundingGENERATOR CONTRO
Page 37 and 38: Generator Stator Phase FaultProtect
Page 43 and 44: Generator Field Ground FaultProtect
Page 45 and 46: Generator Stator Ground FaultProtec
Page 51 and 52: Loss of Field Protection (40Q, 40Z)
Page 53 and 54: Loss of Field Protection (40Q, 40Z)
Page 55 and 56: Overexcitation and OvervoltageProte
Page 57 and 58: Out of Step Protection (78)E AE BAZ
Page 59 and 60: Negative Sequence Protection (46)
Page 61 and 62: Inadvertent Energization Protection
Page 63 and 64: Inadvertent Energization Protection
Page 65 and 66: Loss of Voltage TransformerProtecti
Page 67 and 68: System Backup Protection (51V, 21)G
Page 69 and 70: System Backup Protection (51V, 21)
Page 71: Conclusion• Generators must be pr

Ch 11 - Generator Protection

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