Ee 1352 Monograph Unitiii

K.L.N. COLLEGE OF INFORMATION TECHNOLOGY Pottapalayam – 630 611, Sivagangai District.

Year : 2008 -2009,

Branch : III B.E. EEE

Semester : 6

Subject : EE 1352 Power System Analysis

Name of the Writer (Collaborator) : R. Ramesh, Professor (EEE), IT Dept. Unit : III

Date of Submission : 16.12.2008

Syllabus : IIIrd Unit FAULT ANALYSIS – BALANCED FAULT Introduction – Balanced three phase fau1t – short circuit capacity – systematic fault analysis using bus impedance matrix – algorithm for formation of the bus impedance matrix. EE 1352 - POWER SYSTEM ANALYSIS VI semester EEE Branch (2006-2010 Batch) Unit – III – Fault Analysis – Balanced Fault 1

Syllabus : Introduction – Balanced three phase fau1t – short circuit capacity – systematic fault analysis using bus impedance matrix – algorithm for formation of the bus impedance matrix. Introduction : Causes of fault : 1. Insulation failure – cable, insulator, insulation failure 2. Falling of a tree on a line. 3. Wind and ice loading on transmission line. 4. Vehicles colliding with supporting structures. 5. Overloading of underground cables. 6. Birds shorting the lines etc. Need for short-circuit study : 1. To determine the magnitude of currents flowing in the system during fault. 2. To select the ratings of fuses, circuit breakers, switchgear and relay setting. 3. To make cable joints. 4. To check the MVA ratings of existing CB. 5. To design the grounding system properly. Classification of faults : 1. Symmetrical faults. 2. Unsymmetrical faults. I Symmetrical faults : All 3 phases are shorted to each other and to ground also. a) Occurrence is rare. b) Severest of all the types of faults. c) Fault current is maximum.

II. Unsymmetrical faults :

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Only one phase or two phases are involved. 1. Line to Ground fault (LG fault) 2. Line to Line fault (LL fault) 3. Line to Line to Ground fault (LLG fault)

The computation of fault currents for unsymmetrical faults requires the knowledge of SYMMETRICAL COMPONENTS. Frequency of occurrence of various faults : 3-phase faults . . . . 5% LLG faults . . . . 10% LL faults . . . . 15% LG faults . . . . 70% Approximations in Modelling : 1. Synchronous machine :

Xd” – Immediately after fault subtransient reactance and subtransient voltage Xd’ – After about 3 cycles, transient reactance and transient voltage. Xd – Steaedy state reactance. Eg – Generated emf of alternator. 3. Transformer : Shunt elements of magnetizing current and core loss components are neglected.

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4. Transmission line : Shunt capacitances are neglected. 5. All series resistances in generators, transmission lines and transformers are neglected. 6. Flat profile is assumed in the initial stage. Initial prefault voltage is 1+ j0. 7. Load impedances are neglected – Prefault system is unloaded system. 8. Prefault currents are neglected. If it is to be considered, the prefault current is superimposed on post-fault currents. ----------------------------------------------------------------------------------------------------------Transients due to short circuit in 3 phase alternator Consider a 3 phase alternator running on no-load. If a 3 phase fault occurs at the terminals of the alternator then a heavy short circuit flows on the armature circuit. The oscillogram of the short circuit current after removing the DC off set current is shown below :

Fig. 1 Symmetrical three phase faults on synchronous machines The symmetrical short circuit current shown above, can be divided into three regions called subtransient, transient and steady state region. Under steady state short circuit conditions, the armature reaction of a syn. generator produces the demagnetizing flux. This effect is represented as a reactance called armature reaction reactance Xa. The sum of leakage reactance, Xl and Xa is called synchronous reactance and denoted by Xd. On neglecting the armature resistance the steady state short circuit model of an alternator will be as shown in 2(a). At the instant of short circuit the DC off set current appears in all the 3 phases of stator. This dc offset current can induce currents in rotor field winding and damper winding by transformer action. The increase in field current and damper winding current will set up flux in a direction to augument the main flux. This effect can be represented by two reactances in parallel with Xa as shown in fig. 2.(b). Here Xf represents the flux created by induced current in the field winding and Xdw represents the flux created by induced currents in the damper winding. The combined effect of all the three reactances is to reduce the total reactance of the machine and so the short circuit current is very high in this state which is

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called as sub-transient state. The total reactance under this condition is called subtransient reactance and denoted by Xd”. ∴ X d" = X l +

1 1 1 1 . . . . (1) + + X a X f X dw

Fig. 2 : Equivalent circuit of an alternator under subtransient, transient and steady state conditions The induced currents in the damper winding disappears after few cycles from the instant of fault. Because the time constant of the damper winding is smaller than the field winding. This effect is equivalent to open circuited Xdw and this state is called transient state. The transient state of the alternator is shown in fig. 2(c). The total reactance in transient state is called transient reactance and is denoted by Xd’. ∴ X d' = X l +

1 1 1 . . . . (2) + Xa X f

The transient state will exist for few cycles and then steady state conditions are achieved. Because, the effect of field winding current will also die out in a short time depending on its time constant. This effect is equivalent to open circuited Xl and this state is referred as steady state. In steady state the total reactance is given by sum of X l and Xa. ∴ X d = X l + X a . . . . (3) From equations (1), (2) and (3) we can say that the subtransient reactance of the machine is smallest and steady state reactance of the machine is highest among the reactances. Therefore Xd”<Xd’<Xd. Let |I| = RMS value of steady state current.

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|I’| = RMS value of transient current excluding dc component |I”| = RMS value of subtransient current excluding dc component ∴ X d" =

X d' =

Xd =

Eg I" Eg I

'

Eg I

=

=

=

Eg oa / 2 Eg

ob / 2 Eg oc / 2

The interrupting capacity of the circuit breaker is determined using subtransient reactance for generators and transient reactance for motors. --------------------------------------------------------------------------------------------------------INTERNAL VOLTAGES OF LOADED SYNCHRONOUS MACHINES UNDER TRANSIENT CONDITIONS Consider a generator connected to a bus with voltage Vt. Let IL be the current delivered by the generator. The circuit model of a syn. generator operator operating under steady state conditions supplying a load current IL is shown in fig. 3.5. Here Eg is the induced emf under loaded condition and Xd is the direct axis synchronous reactance of the machine. Now, E g = Vt + jI L X d

Let a short circuit occur at the terminals of the generator while delivering the load current IL. Now in order to study the subtransient state the Eg and Xd should be replaced by Eg” and Xd” a shown in fig. 3.6. In order to study the transient state the E g and Xd of Fig. 3.5 should be replaced by Eg’ and Xd’ as shown in fig. 3.6.

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Here, Eg” = Subtransient internal voltage. Eg’ = Transient internal voltage. In fig. (3.6) and (3.7) the prefault current is I L when the switch is not closed. The fault condition can be represented by closing the switch. In fig. (3.6) when switch is closed subtransient fault current will in the fault path. In fig. (3.7) when the switch is closed, transient fault current will flow in the fault path. With reference to fig. (3.6) and (3.7) if prefault load current is IL then we can write, E g" = Vt + jI L X d" E g' = Vt + jI L X d' It is important to note that the value of Eg” and Eg’ are functions of prefault load current. Therefore Eg” and Eg’ have to be estimated for each value of load current. The syn. motors have internal emfs and reactances similar to that of a generator except that the current direction is reversed. Hence for short circuit studies the circuit models similar to that of fig. (3.5) to (3.7) can be used for syn. motor with reversed direction of current. Therefore for syn. Motor, E g" = Vt − jI L X d" E g' = Vt − jI L X d' ----------------------------------------------------------------------------------------------------------Symmetrical (Balanced) 3 phase faults When a fault occurs in a power system network, the currents flowing is determined by its internal emf of the machines in the system, by their internal impedances and by impedances in the network between the machines and the fault. When the fault current is equal in all the phases the fault is called symmetrical fault. The fault current will be symmetrical only in three phase faults in which all the three phases are shunted to ground. The symmetrical fault can be analysed on per phase basis using reactance diagram or by using per unit reactance diagram. The symmetrical fault analysis has to be performed separately for subtransient, transient and steady state conditions of the fault, because the reactances and internal emfs of the synchronous machines will be different in each state.

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The fault currents and voltages in the various parts of the system can be determined by any one of the following methods : 1. Using Kirchhoff’s voltage and current law 2. Using Thevenin’s Theorem 3. By forming Bus impedance matrix. Thevenin equivalent circuit.

If =

VT ZT + Z f

Where Zf = fault impedance VT = Thevenin voltage ZT = Thevenin impedance ----------------------------------------------------------------------------------------------------------SYMMETRICAL FAULT ANALYSIS USING BUS IMPEDANCE MATRIX The bus impedance matrix can be used to estimate the fault at any point of the system. Usually this method is useful for large system. Consider the following matrix equation matrix equation relating the bus voltages, currents injected to the buses and bus impedance matrix for a n-bus system. Zbus I = V . Z 1k . Z 1n   I 1  V1   Z 11 Z 12 Z . Z 2k . Z 2 n   I 2  V2   21 Z 22  . . . . . .  .   .     =   . Z kk . Z kn   I k  Vk   Z k1 Z k 2  . . . . . .  .   .       . Z nk . Z nn   I n  Vn   Z n1 Z n 2 Where I1, I2,. . . . In are currents injected to buses 1,2,. . . . n respectively. and V1, V2,. . . . . Vn are voltages at buses 1,2,. . . . n respectively. Let a three phase fault occur in bus k. The prefault voltage at bus k be Vpf. Let the prefault condition be represented by a source of value Vpf as shown in fig. (a). The fault condition can be represented by a source of Vpf in series with prefault voltage source as shown in fig. (b), so that the total

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bus voltage in bus k is zero. Due to this fault all other bus voltages will change. Let the change in bus voltage be ∆V1, ∆V2, . . . . . ∆Vn.

(a) Prefault condition

(b) Fault condition

Let us assume that the prefault current is zero. Therefore the prefault voltages of all the buses including bus-k will be 1 p.u. When a three phase fault occurs in bus-k the fault current If will flow away from the bus-k and so the current injected to bus-k is –I f. Now if we replace all other sources by zero value sources then the current injected to all other buses will be zero. Under this condition the only source in the network is –Vpf and current injected to bus-k is –If. Also the voltages in all other buses are only the change in voltages. condition can be represented by the following matrix equation.  Z 11 Z  21  .   Z k1  .   Z n1

Z 12 Z 22 . Zk2 . Z n2

. . . . . .

Z 1k Z 2k . Z kk . Z nk

. . . . . .

This

Z 1n   0   ∆V1  Z 2 n   0   ∆V2  .  .   .    = Z kn  − If  ∆Vk  .  .   .      Z nn   0   ∆Vn 

From the above matrix equation, we get, ∆V1 = − I f Z 1k ∆V2 = − I f Z 2 k . − V pf = − I f Z kk ∆Vn = − I f Z kn The fault current in bus–k, I f =

V pf Z kk

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In general the change in bus-q voltage due to a three phase fault in bus-k is given by ∆Vq = − I f Z qk The voltage at a bus during a fault in bus-k is given by sum of prefault bus voltage and change in bus voltage. Since the system is unloaded system the prefault voltage at all buses be Vpf = 1.0 p.u. (i.e. same as bus –k voltage). ∴V1 = V pf − I f Z 1k = 1∠0° − I f Z 1k V2 = V pf − I f Z 2 k = 1∠0° − I f Z 2 k Vk = V pf − V pf = 0 Vn = V pf − I f Z nk = 1∠0° − I f Z nk The postfault voltages at the buses can be determined as shown above. Then the fault current flowing through the lines can be estimated from the knowledge of line impedances. A sample calculation for fault current in the line-qr is shown below.

I qr =

Vq − Vr Z qr

= (Vq − Vr ) y qr

-------------------------------------------------------------------------------------------------SELECTION OF CIRCUIT BREAKERS The circuit breakers are protective devices which are used in power system to automatically open the faulty part of the system in the event of a fault. Since the power system is predominantly inductive in nature, the interruption of heavy current when the CB open its contacts is associated with large voltage induced across its contacts which inturn results in sparking at the contacts. Hence in CB the amount of current it has to interrupt is an important criteria. The CB for a particular application (or load) is selected based on the following ratings : 1. Normal working power level specified as rated interrupting current or rated interrupting kVA. 2. The fault level specified as either the rated short circuit interrupting current or rated short circuit interrupting MVA. 3. Momentary current rating. 4. Normal working voltage. 10

5. Speed of circuit breaker. The speed of CB is the time between the occurrence of the fault to the extinction of the arc. It is normally specified in cycles of power frequency. The standard speed of CBs are 8, 5, 3 or 1½ cycles. Short circuit interrupting MVA= 3 V pfL I fL where V pfL = Magnitude of prefault voltage at fault point in kV. I fL = Magnitude of line value of short circuit interrupting current at the fault in kA. or Short circuit interrupting MVA= V pfL , pu x I fL , pu xMVAb where V pfL , pu = Magnitude of prefault voltage at fault point in pu. I fL , pu = Magnitude of line value of short circuit interrupting current at the fault in pu. Note : Here, the short circuit interrupting MVA is a 3 phase power rating. ---------------------------------------------------------------------------------------------------------SHORT CIRCUIT CAPACITY (SCC) The short circuit capacity (SCC) of a bus of a network is defined as the product of the magnitudes of the prefault voltage and the post fault current. The SCC is also known as the fault level. |SCC| = |V0| x |IF| VA where V0 = prefault voltage in volts and IF = post fault current in amps.

SCC 3Φ =

S b , 3Φ Z f , pu

MVA

The fault current (short circuit current) can be found as follows:

If =

SCC 3Φ x10 6 3 xVL ,b x10 3

where Sb = base volt ampere in VA Vb = base voltage in volts SCC3Φ = 3 phase Short circuit capacity in VA --------------------------------------------------------------------------------------------------------Algorithm for short circuit studies : Short circuit studies for larger systems need systematic algorithm so that a digital computer can be used. Consider an n-bus system shown in fig. 3.6 which is having a steady load. The following steps are to be followed.

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Fig. 3.6 Sample n-bus system Step 1 : Obtain prefault voltages at all buses and currents in all transmission lines by conducting load flow study. Let the prefault bus voltage vector be

0 VBus

V10   0 V2  =  .  . . .. . (1)    .  V 0   n

Let us assume that a fault occurs at rth bus through a fault impedance Zf. The postfault bus voltage vector is given by, f 0 VBus = V Bus + ∆V . . . . . .(2)

where ∆V is the vector of changes in bus voltages caused by the fault. Step 2 : The next step is to draw the passive Thevenin’s network for the system with generators replaced by transient or subtransient reactances with their emfs shorted as shown in fig. 3.7.

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Step 3 : Next we excite the Thevenin’s network with –Vr0 in series with Zf as shown in fig. The vector ∆V comprises the bus voltages of this network. ∆V = Z J . . . . (3) bus

f

where ZBUS is the bus impedance matrix of Thevenin’s network and is given by,

Z Bus

 Z 11  . =  .   Z n1

. . . .

. . . .

Z 1n  .  . . . . . (4) .   Z nn 

and Jf is the bus current injection vector. Since the network is injected with current –Jf only at the rth bus,

    Jf = I r   

     . . . . . (5) = −I f   .  0  0 0 .

Substituting in equation (5) in equation (3), we get the changes in voltage in rth bus ∆Vr = − Z rr I f . . . . (6) Step 4 : The voltage at the rth bus under fault is, ∆Vr = Vr0 + ∆Vr0 = Vr0 − Z rr I f . . . . (7) 0 Also, Vr = Z f I f We have from equation (6) and (7)

Solved Problems :

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1. A 3 phase transmission line operating at 33 kV and having a resistance and reactance of 5 ohm and 20 ohm respectively is connected to a generating station bus bar through a 15 MVA step up transformer which has a reactance of 0.06 pu. Two generators one 10 MVA having 0.1 pu reactance and another 5 MVA having 0.075 pu reactance are connected to the bus bars. Calculate the short circuit MVA and the fault current when a 3 phase short circuit occurs at (i) the high voltage terminals of the transformer (ii) at the load end of the transmission line. Solution : The single line diagram is drawn as

Selecting 10 MVA as base MVA and 33kV as base kV in transmission line X g1 = 0.1 10 = 0.15 5 10 X T 1 = 0.06 x = 0.04 15 2 33 Z base = = 108.9Ω 10 5 + j 20 Z T .L = = 0.046 + j 0.18366 108.9 X g 2 = 0.075 x

From the single line diagram, the reactance diagram is drawn as

When a 3 phase fault occurs at point A, We get the value of p.u. impedance as

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j 0.1xj 0.15 = j 0.04 + j 0.06 = j 0.1 p.u. j 0.25 Short circuit MVA fed into the fault is, S 10 SCC = b = = 100 MVA Z Tp .u . 0.1 Z Tp .u . = j 0.04 +

∴ Faultcurrent =

100 x10 6 3 x33x10 3

= 1749.5 Amps.

When a 3 phase fault occurs at the point B We get the value of p.u. impedance as ZTp.u. = 0.046+j0.28366 = 0.2874<80.79° Sb 10 = = 34.79 MVA Z Tp.u . 0.2874 = 34.79 x10 6 Faultcurrent = = 608.8 Amps. 3 x33 x10 3 -------------------------------------------------------------------------------------------------------Problem 2 : For the radial network shown, a 3 phase fault occurs at F. Determine the fault current and the line voltage at 11 kV bus under fault conditions.

Solution : Base MVA = 100 Base kV on Generator side = 11 Base kV on Overhead Transmission Line = 33 kV. Base kV on Cable = 6.6 kV 2 100  11  Reactance of Generator 1 = j 0.15 x x  = j1.5 p.u. 10  11 

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2

Reactance of Generator 2 = j 0.125 x

100  11  x  = j1.25 p.u. 10  11  2

Reactance of Transformer 1 = j 0.1x

100  11  x  = j1.0 p.u. 10  11  2

100  33  Reactance of Transformer 2 = j 0.0.08 x x  = j1.6 p.u. 5  33  33 2 Zbus in overhead transmission line = = 10.89Ω 100 6.6 2 Zbus in cable = = 0.4356Ω 100 30 x(0.135 + j 0.08) = 0.744 + j 0.99 p.u. 10.89 3 x(0.135 + j 0.08) = 0.93 + j 0.55 p.u. impedance of cable = 0.4356 The p.u. reactance diagram is shown below : p.u. impedance of overhead transmission line =

The fault occurs when the generators are at no load. Therefore they are considered at Vt = 1<0° j1.5 xj1.25 X eq = = j 0.682 j 2.75 Thevenin’s impedance = j0.682 + j1.0 + 0.744 + j0.99 + j1.6 + 0.93 + j0.55 = 1.674 + j4.822 ZTH = 5.1<70.8° p.u. Using Thevenin’s theorem, V 1∠0° I SC = T = = 0.196∠ − 70.8° Amp. Z TH 5.1∠70.8° I base =

100 x10 6

= 8750 Amp. 3 x6.6 x10 3 ∴ I SCactual = 0.196 x8750 = 1.715kAmp. Total impedance between F and 11 kV bus = 0.93 + j0.55 + j1.6 + 0.766 + j0.99 + j1.0 = 1.674 + j4.14 = 4.43<67.8° ohms Voltage at 11 kV bus = 4.43<67.8° x 0.196<-70.8° = 0.88<-3° = 9.68 kV. 16

--------------------------------------------------------------------------------------------------------Problem 3 : A syn. Generator and a syn motor each rated 25 MVA, 11 kV having 15% subtransient reactance are connected through transformers and a line as shown in fig. The transformers are rated 25 MVA, 11/66 kV and 66/11 kV with leakage reactance of 10 % each. The line has a reactance of 10% on a base of 25 MVA, 66 kV. The motor is drawing 15 MW at 0.8 pf lead and a terminal voltage of 10.6 when a symmetrical 3 phase fault occurs at the motor terminals. Find the subtransient current in the generator motor and fault.

Solution : Choose a base of 25 MVA, 11 kV. All reactances are given in proper bases. Therefore, the p.u. reactance diagram is

Base voltage is 11 kV at fault point. 10.6 = 0.9636 p.u. Load voltage in p.u., V p.u . = 11 15 = 18.75MVA Load MVA = 0.8 18.75 = 0.75 p.u. Load MVA in p.u.= 25 LoadMVAp.u. 0.75 = = 0.7783 Load current in p.u., I Lp.u . = Loadvoltagep.u. 0.9636 Since the pf is 0.8 lead, the pf angle is Φ = cos-10.8=36.9° ILp.u. = 0.7783<36.9° The emf behind subtransient reactance of the generator

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E g" = Vt + I L X d" = 0.9636∠0° + (0.7783∠36.9 xj 0.45) = 0.7536 + j 0.28 p.u. The emf behind the subtransient reactance of the motor E m" = Vt − I L X d" = 0.9636∠0° − (0.7783∠36.9 xj 0.45) = 1.0336 − j 0.0933 p.u. Under faulted condition, I = " g

E g" X

" totalfromgeneratortofault

=

0.7536 + j 0.28 j 0.45

= 0.6226 − j1.6746 I m" =

E m" X

" totalfrommotortofault

=

1.0336 − j 0.0933 j 0.15

= −0.6226 − j 6.8906 p.u. ∴ I f = I G" + I m" = − j8.5653 p.u. 25 x10 6

= 1312.2 Amp. 3 x11x10 3 Now Ig” = 1312.2 x (0.6226 - j1.6746) = 816.4 – j2197.4Amps. Im” = 1312.2 x (-0.6226 - j6.8906) = -816.4 – j9041.8 Amps. ∴ I f = I g" + I m" = −11239 Amps. Base current =

----------------------------------------------------------------------------------------------------------Problem 4 : Consider the 3-bus system shown in figure below. The generators are 100 MVA, with transient reactance 10% each. Both the transformers are 100 MVA with a leakage reactance of 5 %. The reactance of each of the lines to a base o 100 MVA, 110 kV is 10 %. Obtain the short circuit solution for a three phase solid short circuit on bus 3. Assume prefault voltages to be 1 p.u. and prefault currents to be zero. Solution :

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Step 1: Addition of element connected between nodes (0) and (1). This is adding an element of value 0.15 ohms. from the new bus to the reference bus. (1) ∴ Z bus = (1)[ j 0.15] Step 2 : Addition of element connected between nodes (1) and (2). This is adding an element of value 0.1 ohms from the existing bus to the new bus.  j1.5 Z bus = (1)   j1.5

j1.5  j 2.5

Step 3 : Addition of element connected between nodes (0) and node (2). This is adding an element of value 0.15 ohms from the existing bus to reference bus. Create a factious bus (l) and generate Zbus. 19

Z bus

(1)  j 0.15 = (1)  j 0.15  j 0.15

(2) j 0.15 j 0.25 j 0.25

(l) j 0.15 j 0.25 j 0.4 

Eliminating the factious node (l)  j 0.09375 j 0.05625 Z bus =    j 0.05625 j 0.09375 2 ( 0.15) Z 11 = j 0.15 − = j 0.09375 j 0 .4 ( 0.25) x(0.15) = j 0.05625 Z 21 = Z 12 = j 0.15 − j 0 .4 j ( 0.25) Z 22 = j 0.25 − = j 0.09375 j 0 .4 Step 4 : Addition of element connected between nodes (1) and (3). This is adding an element of 0.1 ohms between the existing bus (1) and the new bus (3).

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Review Questions (2 marks) 1. How do short circuits occur on a power system ? 2. Distinguish between symmetrical and unsymmetrical short circuits. 3. What are the applications of short-circuit analysis? 4. Define short circuit capacity of a power system. 5. What are the different types of fault which occur in a power system? 6. Define symmetrical fault. 7. Name any two methods of reducing short circuit current. Reviw Questions (16 marks) 1. (a) Explain the need for short circuit studies. (b) Draw the oscillogram when an unloaded alternator is subjected to symmetrical short circuit and hence explain how to obtain xd, xd’ and xd” from it. 2. With the help of a detailed flow chart, explain how a symmetrical fault can be analyzed using ZBUS. 3. What are the various types of faults ? Discuss the frequency of occurrence and severity.

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4. Explain the difference between direct axis transient reactance and direct axis sub-transient reactance of alternators. 5. Explain the method of calculation of fault currents in the transmission system using Thevenin’s theorem. 6. A Generator-transformer unit is connected to a line through a circuit breaker. The unit ratings are : Generator : 10 MVA, 6.6 kV, Xd” = 0.1 p.u., Xd’= 0.20 p.u. and Xd = 0.80 p.u. Transformer : 10 MVA, 6.9/33 kV, reactance = 0.08 p.u. The system is operating at no load at a line voltage at 30 kV, when a three phse fault occurs on the line just beyond the circuit breaker. Find (a) the initial symmetrical rms current in the breaker. (b) the maximum possible DC off-set current in the breaker (c) the momentary current rating of the breaker (d) the current to be interrupted by the breaker and the interrupting kV and (e) the sustained short circuit current in the breaker. 8. The system shown in figure is delivering 50 MVA at 11 kV, 0.8 lagging power factor into a bus which may be regarded as infinite. Particulars of various system components are : Generator : 60 MVA, 12 kV, Xd’ = 0.35 p.u. Transformers (each) : 10 MVA, 12/66 kV, reactance 0.08 p.u. Line : Reactance : 12 ohms, resistance negligible. Calculate the symmetrical current that the circuit breakers A and B will be called upon to interrupt in the event of a three phase fault occurring at F near the circuit breaker B.

9. In the system configuration of figure, the system impedance data is given below : Transient reactance of each generator = 0.15 p.u. Leakage reactance of each transformer = 0.05 p.u., Z12 = j0.1, Z13 = j0.12, Z23 = j0.08 p.u. For a solid 3 phase fault on bus 3, find all bus voltages and S C currents in each component. Assume prefault voltages to be 1 p.u. and prefault currents to be zero.

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10. For a fault (solid) location shown in figure, find the short circuit currents in line 1-2 and 1-3. Prefault system is on no-load with 1 p.u. voltage and prefault currents are zero. Use ZBUS method and compute its elements by the current injection technique.

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