Symmetrical Three Phase Faults

Symmetrical Three Phase FaultsFaults and TypesTransients in RL series circuitsInternal voltage of loaded machines under fault conditionsFault calculations using ZbusFaults and Types 1Temporary Faults (such as lighting)Permanent FaultsIf a temporary fault occurs, the goal of the protection is to re move the causeof the fault and restore the electric switches 100% before some switches are cut off permanently.If a permanent fault occurs, the goal of the protection is to isolate the fault fromthe system and keep the outage area to a minimum.

Faults and Types 2Open-circuit fault vs short-circuit faultBalanced fault vs unbalanced faultsBalanced fault: Three-phase fault (5%)Unbalanced faults: Single line to ground faults (70~80%) (SLG) Line to line faults (LL) Double line to ground faults (LLG)Transients in RL Series Circuits 1Why we discuss the transients ? Breaks have to carry maximum current momentarily and have to interfere it.

~RLi

The solution of this equation is whereand

Transients in RL Series Circuits 2

Sinusoid termdc component timeiFigure 10.1 (a)

Figure 10.1 (b)itimeShort Circuit Current 1cbaTimeioFig 3.19

Short Circuit Current 2

+_

+_

+_

+_~

Short Circuit Current 3Example 3.8Two generators are connected in parallel to the low-voltage side of a three-phase transformer , as shown in figure 1 . Generator 1 is rated 50,000kVA , 13.8kV . Generator 2 is rated 25,000kVA , 13.8kV . Each generator has a subtransient reactance of 25% on its ownbase . The transformer is rated 75,000kVA,13.8 /69 Y kV , with a reactance of 10% . Beforethe fault occurs , the voltage on the high-voltage side of the transformer is 66 kV . Thetransformer is unloaded and there is no circulating current between the generator . Find thesubtransient current in each generator when a three-phase short circuit occurs of the high-voltage side of the transformer .

Short Circuit Current 4

++--Solution :Generator 1

Short Circuit Current 5Generator 2

Transformer

Short Circuit Current 6

Internal Voltage of Loaded Machines Under Fault condition 1Steady -state generator equivalent circuit Circuit calculation of I

----Subtransient internal voltage----Subtransient reactanceFor calculation of I+-+-

+-Figure(a)

+-+-

+-Figure (b)Internal Voltage of Loaded Machines Under Fault condition 2

Figure 10.3 (a)Right Before the fault

+-+-

+-

Neutral

+-

-+

NeutralFigure 10.3 (b) After the fault

Internal Voltage of Loaded Machines Under Fault condition 3

Internal Voltage of Loaded Machines Under Fault condition 4Example 10.1 A synchronous generator and motor are rated 30,000kVA, 13.2kV, and both have subtransient reactances of 20%. The line connecting them has a reactanceof 10% on the base of the machine ratings. The motor is drawing 20,000kW at 0.8 power-factor leading and a terminal voltage of 12.8kV when a symmetrical three-phasefault occurs at the motor terminals. Find the subtransient currents in the generator , the motor, and the fault by using the internal voltage of the machines.

+-

-+

NeutralFigure 10.3 (a) Before the fault

+-

Internal Voltage of Loaded Machines Under Fault condition 5

Internal Voltage of Loaded Machines Under Fault condition 6For the generator :

Internal Voltage of Loaded Machines Under Fault condition 7For the motor :

Internal Voltage of Loaded Machines Under Fault condition 8In the fault :

+-

-+

NeutralFigure 10.3(b) After the fault

Figure 10.3(b) shows the paths of , and .

Internal Voltage of Loaded Machines Under Fault condition 9 Example 10.2. Solve Example 10.1 by the use of Thevenins theorem

From generator :From motor:

Internal Voltage of Loaded Machines Under Fault condition 10

Fault current from generatorFault current from motorNeglecting load current gives Fault current from generator = 3.23 1312 = 4240 A Fault current from motor = 4.85 1312 = 6360 A Current in fault = 8.08 1312 = 10,600 A The current in the fault is the same whether or not load current is considered. Fault Calculations Using Zbus 2+-+- j 0.25 j 0.125 j 0.2 j 0.25 j 0.4 j 0.10 j 0.10 j 0.20 j 0.20

1234

++--

Simulate the three-phase fault on bus 2Fault Calculations Using Zbus 3 Use to simulate a three-phase fault: equation (10.14)

12344321

The prefix is chosen to indicate the changes in t he voltages at the buses duce to the current injected into bus by the fault.

2Fault Calculations Using Zbus 4The Z bus here is not exactly the inverse of the Y bus , because subtrasient reactances are being used for the synchronous machines.

The fault current : substituting the expression for into :

Fault Calculations Using Zbus 5

Fault Calculations Using Zbus 6Neglecting the prefault current, the bus voltages are:

Fault Calculations Using Zbus 7 In general term, when the three-phase fault occurs on bus of a large scale netwo we have and neglecting prefault load currents, the voltage at

any bus during the fault

kjjiThe subtransient current from bus to bus In the line of impedance connecting those two buses

is the line impedance between i and j .

Fault Calculations Using Zbus 8Example 10.3 A three-phase fault occurs at bus of the network of Fig 10.5. Determine theinitial symmetrical rms current(that is, the subtransient current) in the fault; the voltages at buses , and during the fault; the current flow in the line from bus to bus and the current contributions to the fault from lines , and . Take theprefault voltage at bus equal to and neglect all prefault currents. 4321342321242---

11223344Fault Calculations Using Zbus 9

Fault Calculations Using Zbus 10

From bus From busFrom bus134