Lecture 13 - Symmetrical Faults_2ba.pptx
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Oct 08, 2024
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Lecture 13 - Symmetrical Faults Outcome-Analyze symmetrical faults in power systems 1
Electrical faults - Causes- Types - Symmetrical faults Unsymmetrical faults 2
Fault Scenario The currents which flow in different parts of a power system immediately after the occurrence of a fault differ from those flowing a few cycles later just before circuit breakers are called upon to open the line on both sides of the fault. And all of these currents differ widely from the currents which would flow under steady-state conditions if the fault were not isolated from the rest of the system by the operation of circuit breakers . 3
Fault Scenario... Two of the factors on which the proper selection of circuit breakers depends are the current flowing immediately after the fault occurs and the current which the breaker must interrupt. In fault analysis values of these currents are calculated for the different types of faults at various locations in the system . The data obtained from fault calculations also serve to determine the settings of relays which control the circuit breakers. 4
Purpose of fault Analysis It is also known as short-circuit study or short circuit analysis. To determine the values of voltages and currents at different points of the system during the fault Determination of the ratings of the required circuit breakers Selection of appropriate relaying 5
Transients in RL series circuits The selection of a circuit breaker for a power system depends not only on the current the breaker is to carry under normal operating conditions, but also on the maximum current it may have to carry momentarily and the current it may have to interrupt at the voltage of the line in which it is placed. In order to approach the problem of calculating the initial current when a system is short-circuited, consider what happens when an ac voltage is applied to a circuit containing constant values of resistance and inductance. 6
Transients in RL series circuits Let the applied voltage be Vmax sin(ꞷt+ɑ), where t is zero at the time of applying the voltage. Then, ɑ determines the magnitude of the voltage when the circuit is closed. If the instantaneous voltage is zero and increasing in a positive direction when it is applied by closing a switch, a is zero. If the voltage is at its positive maximum instantaneous value, ɑ is π/2. The differential equation is 7
Here first term varies sinusoidally with time.The second term is nonperiodic term is called the dc component of the current 8
Current as a function of time in an RL circuit for ɑ -ፀ = 0 Note that DC term does not exist if the circuit is closed at a point on the voltage wave such that ɑ -ፀ = 0 9
Current as a function of time in an RL circuit for ɑ -ፀ = π/2 When switch is closed at a point on the voltage wave such that ɑ -ፀ = π/2, the dc component has its maximum initial value, which is equal to the maximum value of sinusoidal component 10
AC component of the symmetrical short-circuit current in an alternator In a synchronous machine the flux across the air gap is not the same at the instant the short circuit occurs as it is a few cycles later. The change of flux is determined by the combined action of the field, the armature, and the damper windings or iron parts of the round rotor. After a fault occurs, the subtransient, transient, and steady-state periods are characterized by the subtransient reactance X d ” , the transient reactance X d ’, and the steady-state reactance X d , respectively. These reactances have increasing values ( that is ,X d ” < X d ’ < X d ) and the corresponding components of the short-circuit current have decreasing magnitudes ( |1"| > |1'| > |I| ). With the dc component removed , the initial symmetrical rms current is the rms value of the ac component of the fault current immediately after the fault occurs. 11
Example 1 A synchronous generator and a synchronous motor each rated 20 MVA, 12.66 KV having 15% subtransient reactance are connected through transformers and a line as shown in Fig. The transformers are rated 20 MVA, 12.66/66 KV and 66/12.66 KV with leakage reactance of 10% each. The line has a reactance of 8% on a base of 20 MVA, 66 KV. The motor is drawing 10 MW at 0.80 leading power factor and a terminal voltage 11 KV when a symmetrical three-phase fault occurs at the motor terminals. Determine the generator and motor currents. Also determine the fault current 12
Answer All reactances are given on a base of 20 MVA and appropriate voltages 13
14 Equivalent circuit during fault
voltage behind subtransient reactance (generator) 15
Fault current 16
Example 2 A three phase transmission line operating at 33 kV and having a resistance and reactance of 5Ω and 20Ω respectively is connected to a generating station busbar through a 15 MVA step up transformer which has a reactance of 0.06 per unit. Connected to the busbars are two generators, one 10 MVA having 0.1 per unit reactance and another 5 MVA having 0.075 per unit reactance. Calculate the short circuit MVA and the fault current when a three phase short circuit occurs (a) at high voltage terminals of the transformer (b) at the load end of the transmission line 17
Answer G1 = 10 MVA, 0.1 pu G2 = 5 MVA , 0.075 pu Tr= 15 MVA , 0.06 pu Let 15 MVA be taken as base then G1 = j0.15 pu, G2 = j0.225 pu Zpu = ( 5 + j 20) 15/(33) 2 = (0.06887+ j0.27548) 18
Three ph fault at Fa Ze pu = (j0.15)|| (j0.225) + j0.06 = j0.15 pu Short ckt MVA fed at fault point Ssc = 15/0.15 = 100 MVA Fault current= If= 100X 10 6 / √3 x 33x10 3 = 1749.5 A 19
Three ph fault at Fb Ze pu = (j0.15)|| (j0.225) + j0.06 +0.06887 +j 0.275448 = 0.431 pu Short ckt MVA fed at fault point Ssc = 15/0.431 = 34.8 MVA Fault current= If= 34.8X 10 6 / √3 x 33x10 3 = 608.8 A 20
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