SHORT CIRCUIT CALCULATIONS

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SHORT-CIRCUIT ANALYSIS -OBJECTIVE TYPE QUESTIONS

1. If a positive sequence current passes through a transformer and its phase shift is 30 degrees, the negative sequence current flowing through the transformer will have a phase shift of

1. 30 deg.
2. -30deg
3. 120 deg.

Ans: (b)

2. The zero sequence impedances of an ideal star-delta connected transformer (star-grounded)

1. looking from star side is zero and looking from delta side is infinite
2. looking from star side is zero and looking from delta side is also zero
3. looking from star side is infinite and looking from delta side is zero

Ans: (a)

2. The positive, negative and zero sequence impedances of a transmission line are 0.5,0.5 and 1.1 pu respectively. The self (Zs) and mutual (Zm) impedances of the line will be given by

(a) Zs = 0.7 pu, Zm =0.2 pu

(b) Zs = 0.5 pu, Zm =0.6 pu

(c)Zs = 1 pu, Zm =0.6 pu

Ans: (a)

3. Symmetrical component method of analysis is more useful when

1. system has unsymmetrical fault and the network is otherwise balanced
2. system has symmetrical fault and the network is otherwise unbalanced
3. system has unsymmetrical fault and the network is unbalanced.

Ans: (a)

2. For a line to line fault analysis using symmetrical components,

(a) the positive and negative sequence networks at the fault point are

connected in series

(b) the positive and negative sequence networks at the fault point are

connected in parallel

(c) the positive, negative, and zero sequence networks at the fault point are

connected in parallel

Ans: (b)

3. The machine reactances used for computation of short circuit current ratings of a circuit breaker are

1. synchronous reactance
2. transient reactance
3. sub-transient reactance

Ans: (c)

2. The load currents in short-circuit calculations are neglected because

1. short-circuit currents are much lager than load currents
2. short -circuit currents are greatly out of phase with load currents

The correct alternative is

1. both (i) and (ii) are wrong
2. (i) is wrong and (ii) is correct
3. both (i) and (ii) are correct.

Ans: (c)

2. The positive sequence network of a sample power system is shown in Fig.2 and the primitive reactance of each element is marked in ohms. The element Zcc of the bus impedance matrix Zbus will be

1. 1.0
2. 0.1
3. 0.21

Ans: (c)

2. Which one of the following statements is not true?

1. Fault levels in an all a. c system are less than in an a. c system with a D.C. link operating
2. Large systems may be interconnected with d. c link of small capacity
3. Limitation on the critical length of underground cables for use in A. C no longer exists if D.C transmission by cables is used.
4. Corona loss and radio and TV interference with D.C. transmission is less

Ans. a

2. Advantage of power system interconnection is

1. Large size circuit breakers are required because of large short-circuit currents
2. Machines of one system remain in step with machines of another system
3. Fewer machines are required as reserve for operation at peak loads

Ans. c

2. The interenal voltages of a 3-phase synchronous generator correspond to

1. Positive sequence
2. Negative sequence
3. Zero sequence

These were developed to ease the calculations for unbalanced 3-phase systems and as a help to numerical solution using network analyzers. Even with present day digital computation, the symmetrical components help in solution of unbalanced systems, besides explaining many phenomena such as rotor heating in machines, neutral current etc.

The basic concept is to convert a set of three phasors into another set of three phasors with certain desirable properties. The symmetrical components (introduced by Fortescue) is only one such set, the other set is the Kimbark/Clarke components.

The unique property of symmetrical components is that they retain the concept of 3-phase system associated with each component phasor. Thus the positive sequence retains the concept of 3 balanced phasors having the same phase sequence as the original phasors whereas the negative sequence component retains the concept of 3 balanced phasors but rotating in opposite direction of rotation. The zero sequence component is balanced set of 3- coincident phasors but rotating in the same direction as the original unbalanced phasors.

The basic definition for a set of unbalanced three-phase system in terms of the sequence components is

I_a = I_0a + I_1a +I_2a

I_b = I_0b + I_1b + I_2b + I_{0 a} + a² I_{1 a} +a I_{2 a}

I_c = I_{0 c} +I_{1 c} +I_{2 c} = I_{0 a} + a I_{1 a} + a ² I_{2 a}

These equations may be written in matrix form as

I ^a,b,c = [Ts] I ^0,1,2

Where Ts is called the symmetrical component transformation matrix.

I ^0,1,2 = [Ts]^-1 I ^a,b,c

Similar transformation may be applied for unbalanced voltages also.

Illustrate these transformations by phasor additions.

If the above transformation Ts is used simultaneously on the voltage and current values of three phase network elements, then

Spq ^abc = P pq + j Qpq =[( I pq ^abc )*]^t e pq ^abc

And

Spq⁰¹² = [( I pq⁰¹²)*]^t e pq⁰¹²

Note that Spq^abc is not equal to Spq⁰¹².

Spq^abc = 3Spq⁰¹² since Spq⁰¹² refers to only phase "a" power , and similar amount of power is in phases "b" and "c" also. Thus all three phases of symmetrical components must be used. Often for computer work, the symmetrical transformation given by

T_si = (1/Ö 3) T_s

is use where T_si^*t . T_si = unity matrix. This is a property of orthogonal matrices. Further since T_si^* = T_si^-1 we can show that Spq^abc = Spq^012. However, in earlier works, and even now, the originally defined non-power invariant transformation is being used

To solve the network in terms of the sequence components, sequence components of impedance are required. These are obtained from their corresponding three phase values.

E^abc = Z^abc I ^abc

Or,

Ts E⁰¹² = Z ^abc Ts I ⁰¹²

Sequence networks have the advantage since for balanced network there is no mutual coupling between sequence component elements unlike what happens in balanced 3-phase components. Thus balanced 3-phase network can be assembled component for component in three separate sequence networks. Sequence networks of generator, transmission line, transformer, and loads should be known. Zero sequence networks for different transformer connections should also be known. The ideal earth and the neutral point should be distinguished. Relative magnitudes of sequence impedances of generator and transmission lines should be known.

• Find Z⁰¹² for rotating and non-rotating elements in terms of corresponding phase quantities
• Study the measurements of zero sequence impedances of transformer for different connections
• Find phase shifts in sequence components in D U transformation
• If the positive, negative and zero sequence impedances of a transmission line are 0.3,0.3, and 0.5 respectively, find the self and mutual impedances between the phases.

An essential part of the design of a power system is the calculation of currents which flow in the components and the resulting voltages , when the faults of various type occur. Common faults on a transmission system are;

1. LG
2. LL
3. LLG
4. LLL or LLLG
5. Open conductor
6. Simultaneous faults that may be any combination of the above five.

Faults may also occur in switchgear, transformers and machines but their nature may be different than those in transmission lines.

Faults are mostly due to lightning and switching. Most of them are temporary. LG fault is most common and LLL fault is least common. Relays provided on the system detect these faults and produce a trip signal for the circuit breaker to isolate the faulty portion form the system. Thus to determine the circuit breaker and relay operating times, fault currents & voltages have to be calculated and for many other applications.

Except for the three phase faults ,all other types of faults cause unbalanced operation, an fault currents & voltages under such conditions are required to be obtained using symmetrical components, or phase components ( the latter analysis is much more difficult even with digital computers.

The calculation of 3-phase balanced faults is relatively simple but forms the basis of determining the circuit breaker ratings.

When a sudden short circuit occurs on the electric supply system, the currents & voltages are of transient nature before they settle down to steady state values.

The fault current at any time consists of

1. Dc transient , also called the DC offset, arising out of the terms of the type

Ae^-Rt/L

2. Ac transients consisting of the terms of the type

Ae^-Rt/L Sin(w t + f ) where w = supply frequency

3. Steady state value of the type A Sin(w t + f ).

X_d" = Ö 2 .E. /Oa

X_d'= Ö 2 .E. /Ob

X_d= Ö 2 .E. /Oc

What are the typical values of these reactances for turbo and hydro generators?

For the interruption current rating of a circuit breaker (time involved being 2-5 cycles), the subtransient current is important. The transient current is important for transient stability studies lasting from 1-2 seconds to 10 seconds.

Short circuit volt-amperes =Ö 3. (short-circuit current). [Nominal Voltage (line value)]

Simplifying assumptions

1. Each machine is represented by a constant voltage behind the machine reactance, subtransient or transient depending on the situation.
2. Shunt connections i.e.; loads , line charging etc. are neglected.
3. All transformers are assumed to be at nominal tap settings.
4. The above three assumptions do not imply that the system is unloaded before the fault.. However , it is assumed that the network is unloaded and that all the generator voltages are at 1 pu. This assumption is justified because fault currents in the network are much larger than the load currents.

If assumptions 2-4 are not taken into account, pre-fault currents are added to the fault currents obtained by using the Thevenin's equivalent of the network.

The Thevenin's equivalent of the network at the point of the fault location (either in phase or sequence components) is obtained for calculating the fault currents. If the network is balanced before the fault, then at the fault point , Thevenin's sequence voltages become

E_a1= E_a

E_a2= 0

E_a0= 0.

If the network is not balanced , the above voltages may have finite values and the sequence component approach may not be useful. The performance equation for the balanced network in terms of sequence components is :

[Vs] = [Eph]- [Zs][Is],

where

[Vs]=[V0a,V1a,V2a]

[Eph] = [0,E1a,0]

[Zs] = diagonal matrix of [Z0,Z1,Z2]

[Is] = [I0a,I1a,I2a]

Depending upon the type of fault, the sequence components of currents and voltages are constrained leading to particular connections of sequence network. After the fault currents coming out of the network at the fault point are determined, the current & voltage distributions inside the network are found out . If originally, there are load currents , these are restricted to the positive sequence network , and they are superposed on the fault currents for accurate results, but this is often not done.

Here ,

I_a1 = E_a/(Z₁ +Z_F)

I_a2 = 0

I_a0= 0

Fault is assumed in phase "a". If it is in any other phase, symmetrical shifting will be required.

Study the network connections for L-L and DLG faults.

• How do you distinguish switching transient currents from sub-transient currents?
• Sketch the subtransient/transient current wave-shapes in phase b and c respectively for a synchronous generator when the current in phase a is symmetrical.
• Under what conditions can you expect a negative or zero sequence Thevenin's voltage at the fault point?

• Determine and draw the sequence network connection diagrams for

1. Single conductor opening at a point
2. Two conductors opening at a point.