Power Systems K-Notes

1

Contents

Manual for K-Notes ................................................................................. 2

Transmission Lines .................................................................................. 3

Underground Cables ............................................................................. 14

Overhead Insulators .............................................................................. 16

Distribution Systems ............................................................................. 16

Per Unit System .................................................................................... 17

Load Flow Study .................................................................................... 18

Economic Power Generation ................................................................ 20

Fault Analysis ........................................................................................ 21

Power System Stability.......................................................................... 29

Power System Protection ...................................................................... 33

2014 Kreatryx. All Rights Reserved.

2

Manual for K-Notes

Why K-Notes?

Towards the end of preparation, a student has lost the time to revise all the chapters from his /

her class notes / standard text books. This is the reason why K-Notes is specifically intended for

Quick Revision and should not be considered as comprehensive study material.

What are K-Notes?

A 40 page or less notebook for each subject which contains all concepts covered in GATE

Curriculum in a concise manner to aid a student in final stages of his/her preparation. It is highly

useful for both the students as well as working professionals who are preparing for GATE as it

comes handy while traveling long distances.

When do I start using K-Notes?

It is highly recommended to use K-Notes in the last 2 months before GATE Exam

(November end onwards).

How do I use K-Notes?

Once you finish the entire K-Notes for a particular subject, you should practice the respective

Subject Test / Mixed Question Bag containing questions from all the Chapters to make best use

of it.

2014 Kreatryx. All Rights Reserved.

3

Transmission Lines Skin Effect

It is tendency of AC current to be concentrated on the surface of conductor.

Cause: Non-uniform distribution of magnitude flux linkages Due to skin effect, the effective

area of cross section of conductor decrease and hence resistances increases.

In case of DC, There is no skin effect so

DC AC

R R

With increase in frequency, skin effect increases.

With increase in r , skin effect increases.

Inductance of a Transmission line

Single Conductor

Internal inductance 0 r

8

External inductance from distance 1d to 2d

0 r 2ex1

dL ln

2 d

Total inductance 0 r 0 r 2d

ln8 2 r

0 r 0 r1

4

d dln ln

2 2 rre

r 0.7788r = Geometric mean radius (GMR)

Single phase 2 wire line

Inductance of single wire 0 rd

ln2 r

Total inductance = 1 2L L

0sys

dL ln

r

If radius of both wire is not same, assume radius of 1st wire ar & that of second wire is br

b

0sys

a

dL ln

r r

a ar 0.7788r &

b br 0.7788r

If instead of a single conductor per phase we use multiple conductor, then GMR is

replaced by self GND (Geometric Mean Distance) and d by mutual GMD.

4

Self GMD

2

1n

11 12 1n 21 22 2n n1 n2 nnfwdself GMD D D ........D D D .........D ....... D D ........D

Where ii i i

D r 0.7788r

2

' ' ' ' ' ' ' '

1m

i'm'i 1 i 2 ml mmbwdSelf GMD D D ........D .............. D ................D

Where ii i i

D r 0.7788r

Mutual GMD

Mutual GMD ' ' ' ' ' '1

mn

11 12 1m n1 n2 nmD D ...........D ................. D D .................D

Now, with these terms all the inductance expressions change to

0d

Single wire : ln2 self GMD

0mutual GMD

1 , 2 wire: ln2 Self GMD

Three phase Transmission line

Symmetrical configuration

r0ph

DL ln

2 r

r 0.7788r

5

Asymmetrical configurations

If conductors are placed horizontally or vertically.

1

3eq ab bc ca

D D D D

eq0 rph

DL ln

2 r

In case of bundled conductor, more than one conductor per phase

We replace ab ab eqD D = mutual GMD between a phase & b phase

Similarly, bc bc eq

D D

ca ca eqD D

In place of GMR, Self GMD is used

1

3

a b cSelf GMD= Self GMD Self GMD Self GMD

Example: Calculate inductance per phase of following circuit?

Between successive conductors, distance = 3m , Radius of each conductor = 1m

Solution

1

4ab a1b1 a1b2 a2b1 a2b2eq

D D . D . D . D

1

43 12 12 3 6m

1

4bc eq

D 3 6 6 3 4.24m

1

4ca eq

D 6 9 9 6 7.348m

eqD = mutual GMD

1

3

ab.eq bc.eq ca.eq D D D

= 5.71m

6

1

4a1a2 a2a1a

Self GMD r D D r

1

2 2 40.7788 0.01 15

= 0.341m

1

4b1b2 b2b1b

Self GMD r D D r

= 0.2467m

1

4c1c2 c2c1c

Self GMD r D D r

= 0.1528m

1

3

a bSelf GMD Self GMD Self GMD Self GMD c

= 0.2398m

0 GMDL ln2 GMD Self

75.71

2 10 ln 0.634mH/km0.2398

Remember, Inductance calculated using these formulas is per unit length.

Transposition of Transmission line

The position of different lines are changed after regular intervals to reduce radio interference in

neighboring communication lines.

Capacitance

Single Phase 2 Wire System

0 r

ab

1 2

CD

lnr r

Line to neutral capacitance

7

0 ran

1

2C

Dln

r

, 0 rbn

2

2C

Dln

r

Three phase single conductor system

0 r

ph

2C

GMDln

r

For bundled conductors

0 0ph

2C

GMDln

Self GMD

In capacitance calculations, it must always be remembered that there is no concept of r, we

simply use radius in calculating self GMD.

Performance of Transmission line

Classification of lines based on length

1) Short Line

l < 80 km or l*f < 4000 , Where f = frequency

2) Medium Line

80 km < l < 200 km

4000 < l*f < 10000

3) Long Line

l > 200 km

l*f > 10000

Modeling of transmission lines

Transmission lines are modeled as 2 port network

s R RV AV BI

s R RI CV DI

Under no load

RI 0 , s RV AV , s

R

VV

A

8

No Load Voltage sV

A and RI 0 , s RI CV

Voltage Regulation

sR

R

VV

A100%

V

This current is called as line charging current and is responsible for as effect is called as Ferranti

Effect.

Ferranti Effect

Under no-load or light load conditions receiving end voltage becomes more than sending end

voltage due to presence of line charging current.

Short transmission line

s R RV V I R jwL

R RV I Z

s R

s R

V 1 z V

I 0 1 I

A = D (symmetrical)

AD BC = 1 (reciprocal)

Approximate Voltage Regulation

For lagging pf

R R RR

IVR= Rcos Xsin

V

Rload pf=cos

For leading pf

R cos R RR

IVR R X sin

V

9

Medium Transmission Line

Normal T Model

s R

s R

YZ YZ1 Z 1

2 4V V

I IYZY 1

2

Here all problems are in actual values & not per unit length.

Nominal--Model

s R

s R

YZ1 Z

V V2

YZ YZI IY 1 1

4 2

Long transmission Line

x R R c

V V cosh X I Z sinh x

RR Rc

VI I cosh x sinh x

Z

Where xV & xI are voltage and current at distance x from receiving end.

c

R j LZ

G j C

= surge impedance

In case of long transmission line, we use all promoters per-unit length.

For loss less line, R=G=0

c

LZ

C

For distribution less line, RC = LG

For sending end, x = l

s R R c

V V cosh l I Z sinh l

Rs R

c

VI I cosh l+ sinh l

Z

10

c

s R

s R

c

cosh l Z sinh lV V

1cosh lI I

Z

A = D (symmetric)

AD - BC = 1 (reciprocal)

Power Transfer Equation

2s r

r r

V V AP cos V cos

BB

2s r

r r

V V AQ sin V sin

BB

For Short TL

B Z B Z ;

A 1 0 A A 1, =0

2

s r r

r

V V VP cos cos

Z Z

2

s r r

r

V V VQ sin sin

Z Z

If resistance of line is neglected

Z jX X and 090

2

s r s r r

r R

V V V V VP sin ; Q = cos

X X X

Remember, the last expression can be applied between any two bases in a power system as long

as transmission line connecting them is loss less.

Wave Propagation

Due to continuous energy transfer between L & C elements of a transmission line we consider

energy propagation from sending to receiving end & hence wave propagation.

cZ = Surge impedance or characteristic impedance.

11

= Propagation constant.

R j L G j C

For loss less line

R = G = O

j LC

j

= attenuation constant

= phase constant

In case of loss less line 0 (no attenuation)

LC

Velocity of wave1

LC , Where L & C are per unit length

Wavelength 2 2

LC

Surge Impedance Loading

When load impedance = surge impedance

2

r(L L)

L

C

VP

Z

= Surge Impedance Loading

If L CZ Z

R R CV I Z

According to long line

X R RV V cosh x+V sinh x x

R2V e

So no term containing yxe & hence no reflected wave & hence whenever surge impedance is

connected at load, there is no reflection.

12

Surge Traversal

Theremin equivalent circuit

When surge voltage V is induced on the line & line can be represented as Theremin

equivalent circuit shown.

CZ = Characteristic impedance of line

LZ = load impedance or characteristics impedance of second

line connected in series to first.

Transmitted voltage L2L C

ZV 2V potential divider

Z Z

Incident Voltage 1V V

Reflected voltage V

2

V V V [Voltage continuity]

L C

L C

Z ZV V

Z Z

Reflected currentC

V

Z

, Refracted current 2

L

V

Z , Incident current

C

V

Z

Reflection coefficient: L C

L C

Z ZV

V Z Z

Refraction coefficient: 2 L

L C

V 2Z

V Z Z

Voltage Control

Usually in case of lagging loads, the voltage at receiving end falls below sending end voltage

and to boost the receiving end voltage we connect a shunt capacitor at receiving end.

Similarly, in case of leading loads, receiving end voltage is higher than sending end voltage

so we connect a shunt reactor to avoid over-voltage.

Usually in GATE, we need to calculate rating of capacitor for voltage control & it is illustrated

through a question shown below:

13

Example: A three phase overhead lines has a resistance & reactance of 5 & 20 respectively.

The load at receiving end is 30MW, 0.85 pf lagging at 33kv & we connect a compensating

equipment at receiving end to maintain voltage at each end equal to 33 Kv. Find rating of

compensating equipment?

Solution:

Assuming base (MVA) = 30 MVA

Base voltage = 33 kv

pu power = 30Mw

1pu30MVA

Base impedance = 2V

36.3S

pu impedance = 05 j20

0.568 75.9636.3

2

S R R

R

V V VP cos cos

Z Z 0Z 0.568 ; =75.96

21 1 1

1 cos 75.96 cos75.960.568 0.568

cos 75.96 0.81

040.11

S R RRV V V

Q sin sinZ Z

21 1 1

sin 75.96 40.11 sin75.960.568 0.568

RQ = - 0.645 pu

L LQ P tan 11 tan cos 0.85

L

Q = 0.6197 pu

R L C CQ Q Q Q 1.2647pu

CQ 1.2647 30 = 37.9 MVAR

So, we can observe the steps involved

Step 1 : Calculate from LP as capacitor does not consume any real power.

Step-2 : Calculate RQ using calculated above

Step-3 : R L CQ Q Q than Calculate CQ

14

Power Factor Correction

Usually, to improve the supply side power factor we connect a capacitor device like capacitor

bank or synchronous condenser (synchronous motor under over excited condition).

Suppose, initially a load of real power 1P & lagging pf 1cos is connected & we want to

improve pf to 2cos lagging 2 1 & we connect a capacitive device which consumes real

power CP & thus net real power after connection.

2 1 CP P P

2 2 2 1 1 1Q P tan ; Q P tan

C 1 1 2 2Q P tan P tan

In case of capacitor bank, CP 0 1 2P P

If we wish to calculate capacitance per phase

(in both voltage control & pf correction )

2

C phQ 3 CV C

2

ph

QC

3 V

Underground Cables Insulation resistance

R

R ln2 l r

L = length of cable

R = Outer radius (sheath radius)

R = conductor radius

Capacitance Model

CC : Core capacitance

SC : Core to sheath capacitance

15

Capacitance per phase

ph S CC C 3C

Calculating SC & CC

1) Any of two cores or conductors are connected to

sheath & capacitance is measured between remaining

core & sheath.

1 S CC C 2C

2) All three cores are connected together & capacitance is

measured between any core & sheath.

2 SC 3C

2

S

CC

3

2

C 1

C2C C

3

1 2

C

C CC

2 6

1 2ph S C3C C

C C 3C2 6

3) Any one of core is connected to sheath & capacitance is

measured between remaining 2 cores.

C S

3

3C CC

2 2

ph 3C 2C

16

Dielectric loss in a UG cable

2ph ph

P 3 C V tan

where tan loss tangent

ph

1tan

c R

R = Insulation resistance

Overhead Insulators For suspension type string insulator, the model for 3-discs looks like as shown.

Let m

S

Cm

C

2 1V V 1 m

23 1V V 1 m 3m The voltage of disc nearest to the conductor is highest.

String efficiency string voltage

No. of discs voltage across bottom disc

1 2 3

3

V V V100%

3 V

Distribution Systems Sources fed from both ends

1) Assume AI from AV

2) Calculate AI from

A B A 1 1 A 1 2 2 A 1 2 3 3V V I I r I I I r I I I I r

3) Substitute AI in A 1 A 1 2 A 1 2 3I I , I I I & I I I I & check for sign change.

4) Node for minimum potential = Node for sign change

5) Calculate minimum potential by KVL

Example: Refer Kuestion power systems for that.

17

Per Unit System In pu system, energy quantity is expressed as a ratio of some based value.

Absolute value or Actual value

pu value=Base value

Percentage value = pu value x 100%

Base value

1 - System

base base base baseS , V , I , Z

Out of these, 2 value must be known, to convert entire system into pu system.

base base baseS V I

2

base base base

base base

base base base

S V VI , Z

V I S

Usually, we assume baseS & baseV as known.

3 - System

baseS & baseV are assumed

basebase

base

SI

3 V

baseV = line to line voltage

baseS = 3 phase power

For start connection

2base base base

base

base base base

V ph V / 3 VZ

I ph I S

For delta connection

2base base base

base

base basebase

V ph V 3VZ

I ph SI / 3

In per unit system, equivalent impedance of transformer referred to primary or

secondary in same.

18

Change of base

If base of system is changed from base base base baseV old , S old to V new ,S new

2

base base

pu pu

base base

V old S newZ new Z old

V new S old

Load Flow Study Power System Matrices

BUS

Y matrix

10 12 12

BUS 12 20 12 23 23

23 23 30

y y y 0

Y y y y y y

0 y y y

OBSERVATIONS

1) The diagonal elements are sum of all admittance connected to that particular bus.

2) The off-diagonal elements are negative of admittance connected between two buses.

3) If two buses are not connected to each other than that elements is zero.

4) BUSY Matrix is a symmetrical matrix.

5) Most of the elements are zero & hence it is a sparse matrix.

Total number of zero elements% sparsity =

Total number of elements

BUS

Z matrix

1

BUS BUSZ Y

BUSZ matrix used in fault analysis.

Suppose a 3 phase SC fault occurs on bus k then fault current

pre fault

f

kk f

V ,kI

Z Z

pre faultk Pr e fault voltage at bus 'k'V ,

kkZ = elements of BUSZ matrix.

fZ = fault impedance

Due to fault voltage at other buses are also affected.

19

1 1

k k

BUS BUSBUSf

n n

V I0

. .0

. ..

V I.

. Z . V ZI

. ..

V I.

0

k f kkV I Z

ff fkk

VI Z 0

Z

"j jk fV Z I

jk

j fkk

ZV V

Z

Post fault voltage at bus j

jk

jjf fkk

ZV V V

Z

If there is generator connected to bus j then current supplied by generator.

g jf

"d

E VI

jX

Classification of buses

At each bus, there are 4 parameter: V , ,P,Q .

At any bus, out of these 4 quantities any 2 are specified.

1) Slack Bus / Swing Bus/Reference Bus

V , are know quantities.

P, Q are unknown quantities.

Any extra power needed by the system is supplied by slack bus.

2) Generator Bus / PV Bus

P, V is specified

Q, are unspecified

20

3) Load Bus / PQ Bus

P & Q are constant as specified

V & are unknown

Generally, newton Raphson method is used for load flow solution and we form Jacobian

matrix, & the order of Jacobian matrix is

2n m 2 2n m 2 N = no. of total buses

M = no. of pv buses

Economic Power Generation Incremental cost

If is cost required to generate an additional unit of energy.

i

IC = Incremental cost of thi generator

ii

Gi

CIC

P

GiP = Power generated by thi generator

iC = cost of thi generator

Transmission Loss

m m

L i j ij

i 1 j 1

P PPB

i jP , P : Real power injection at th thi & j buses

ijB = loss coefficient

m : no. of generator units

Penalty Factor

i

L

Gi

L1

P1

P

For economic power sharing

iiIC L cons tant

Constant is called as incremental cost of system.

i

IC = incremental cost thi unit

iL = penalty factor of thi unit

21

From this expression, for m generator we get ( m 1) equation and thm equation is

m

Gi D LOSS

i 1

P P P

DP = total power demand

For example, refer kuestions on Power systems.

Fault Analysis Symmetrical Components

For an unbalanced 3 phase system, the analysis is done better by means of symmetrical

components.

a0V Zero sequence components

a1V Positive sequence components

a2V Negative sequence components

a0 a

2a1 b

2a2 c

V 1 1 1 V1

V 1 V3

V 1 V

; Where 0j120e

1

s pV A V

pV = phase voltage

sV = Symmetrical component

2

2

1 1 1

A 1

1

Power in terms of symmetrical components

a1 a1 a2 a2 a3 a0P 3 V I V I V I

Remember, same transformation exist for current also.

Sequence Network

Alternators

Positive Sequence Network

1Z positive sequence impedance "

1 dZ jX

22

Negative Sequence Network

a2 a2 2V I Z

2Z : Negative sequence impedance

d q

2

X " X "Z j

2

Zero Sequence Network

a0 a0 0 nV I Z 3Z

0Z = Zero sequence impedance

nZ =Neutral impedance ( in case of delta)

0 lZ jX lX = leakage reactance

Transformers


1Z = Positive sequence impedance

1 lZ jX


2Z = Negative sequence impedance

2 lZ jX

Zero sequence network

Depending upon scheme of connection, we close series or shunt connection & method

of grounding.

23

Shunt connection are closed for delta connection & series connection are closed for star

connection with grounded neutral. If primary & secondary are inter changed then circuit

becomes mirror image.

Case 1

If neutral is grounded through an impedance nZ

0 T0 nZ Z 3Z

Case 2

Case 3

Case 4

24

Case 5

Transmission Lines



Zero Sequence Network

1 2 s mZ Z Z Z

0 s mZ Z 2Z

sZ = Self impedance

mZ = mutual impedance

Remembers, all sequence networks are always drawn in per unit & never in actual values.

Fault Analysis

The following short circuit faults are considered

1. LG (Single Line to ground fault)

2. LL (Line to line fault)

3. LLG (Line to Line to Ground fault)

4. 3-phase short circuit fault.

25

3-phase short circuit fault comes under the category of symmetrical SC fault whereas other 3

faults are called as unsymmetrical SC fault.

Order of severity

LG < LL < LLG < 3 - SC

But if faults occur at terminal of alternator then LG fault is most severe.

Occurrence of SC fault

LG > LL > LLG > 3 - SC

Transient on a Transmission line

Equivalent Circuit

t ssi t i i

Rt

m mLV V

sin e sin wtZ Z

2

1 2Ltan ; Z R LR

Maximum momentary current

mmm2V

i sinZ

If resistance is neglected, 090

mmmi2V

cosZ

Short circuit model of a synchronous machine

During initial SC period for 1-2 cycles, current are induced in field & damper winding of machine

so reactance at least & called as sub transient reactance dX "

After initial sub-transient period, current in damper winding in reduced to zero, and this period

is called as transient period & reactance of machine is called as Transient Reactance dX "

26

Finally, when current in field winding is also reduced to zero, we enter steady state period &

reactance is called as steady state reactance dX .

d d dX " X ' X

Symmetrical Fault Analysis

We replace alternators by an emf source in series with sub transient reactance and emf

source under no-load is usually 01 0 pu or terminal voltage in pu.

0

t

f

base

V 0E pu

V

Transformer & Transmission lines are replaced by reactance.

The equivalent circuit can be solved either by finding thevenin equivalent across fault or

by simple network analysis & fault in SC is calculated.

ffeq f

EI

Z Z

fZ = fault impedance.

In symmetrical fault analysis, we only consider positive sequence impedance.

eq f

1

SC MVA puZ Z

base

eq f

MVA

SC MVA MVAZ Z

For example, refer to kuestion on power systems.

Selection of Circuit Breakers

Usually, circuit breakers are selected on the basis of most severe fault which is 3 phase

SC fault.

Three ratings of circuit Breakers are important.

1) Rated momentary current

Momentary current ( rms ) = sc1.6 I

scI is symmetrical SC current which we calculated in previous section.

2) Making current

Making current = sc2.54 I

27

3) Symmetrical Interrupting Current

We need to recalculate scI by using sub-transient reactance for alternators & transient

reactance for synchronous motors. Induction motors & other loads are neglected.

Then, we multiply it by a factor to calculate symmetrical interrupting current. This factor

depends on speed of circuit breakers which is measured in terms of numbers of cycles it takes to

operate.

Speed Factor

8 Cycles or slower

5 Cycles

3 Cycles

2 Cycles

1.0

1.1

1.2

1.3

Unsymmetrical Faults Analysis

Line to Ground Fault

We first draw equivalent positive, negative & zero sequence networks & calculate thevenin

impedance across fault terminals from each network.

Assuming equivalent positive, negative & zero sequence reactance are 1 2 0Z ,Z & Z

respectively.

Here all sequence currents are equal.

a1 a2 a0I I I

aa11 2 0 f

EI

Z Z Z 3Z

Fault current aI

a a1I 3I

aa

1 2 0 f

3 EI

Z Z Z 3Z

Short Circuit MVA

a1 a1SC MVA 3E , I *

28

2

a1

1 2 0 f

3 E I

Z Z Z 3Z

In pu 1 2 0 f

3SC MVA pu

Z Z Z 3Z

base

1 2 0 f

3 MVAMVA

Z Z Z 3Z

Line to Line Fault

Here, we calculate equivalent positive & negative sequence impedance 1 2Z & Z respectively.

aa1 f1 2

EI

Z Z Z

Fault current

ab f

1 2

j 3 EI

Z Z Z

Short Circuit MVA

1 2 f

3SC MVA pu

Z Z Z

base

1 2 f

3 MVAMVA

Z Z Z

Line to Line to Ground Fault

aa1

1 2 0 f

EI

Z Z Z 3Z

a1 2

a0

2 0

I ZI

Z Z

Fault Current = ao

3 I

29

Short Circuit MVA

1 2 0f

3SC MVA pu

Z Z 1 3Z Z

base

1 2 0 f

3 MVAMVA

Z Z Z 3Z

Remember, all fault analysis will be done in pu system.

Power System Stability Two types of stability are studied:

1) Steady State Stability

2) Transient Stability

Steady State Stability

For Steady State Stability

dP

0d

And for this condition to be true.

e maxP P

If power demand is greater than maximum demand than machine goes out of synchronous.

For a loss less machine, maxS

E VP

X

Transient Stability

Swing Equation

2

m e2

MdP P

dt

M = inertia constant ( MJ-S / elect - rad)

Pm= mechanical input (MW)

e

P = electrical output (MW)

= rotor angle

Another Form

2

m e2

H dP P

f dt

H = inertia constant ( MJ / MVA)

mP & eP both are in pu

30

GH

M180f

(MJ S / elect - deg)

GH

Mf

(MJ S / elect - rad)

G = machine rating (MVA)

If two alternators are swinging coherently. Then they can be replaced by a single

alternator having

eq 1 2M M M

But H cannot be added directly, they must first be on same base.

If machines are not swinging coherently, then

1 2eq

1 2

M MM

M M

Accelerating Power,

a m eP P P

In steady state m eP P

In transient, m eP P so rotor accelerate or decelerate.

Equal area criterion

For system to possess transient stability

aP d 0

There are basically 3 stages in stability analysis

Before Fault

We say maximum power transferrable is max,1P

& e max,1P P sin

During fault


e max,2P P sin

After Fault


e max,3P P sin

31

Critical clearing angle

It is the maximum value of beyond which if the fault is cleared system will be unstable. The

time instant corresponding to this angle is called as critical clearing time assuming fault occurs

at t = 0.

Case-1 : Fault occurs on TL near to bus

max,2P 0

max,3 max,1P P

Crclearing angle

By equal area criteria

2

0

m max,1P P sin d 0

1 m0

max,1

Psin

P

For critical clearing 2 max

0max

Cr 0

Crm

2Ht

f P

= Critical Clearing Time

Case-2 : Fault occurs on one of parallel lines close to bus

Before Fault

max,1 g 1 2

E VP

X X X

During Fault

max,2

eq

E VP 0

X

After Fault

max,3

g 1

E VP

X X

32

1 m0

max,1

Psin

P

1 mmax

max,3

Psin

P

For transient stability

2

0

aP d 0

c 2

c0

m m max,3P 0 d P P sin d 0

For critical Clearing

2 max

Cr 0

Crm

2Ht

f P

Case-3 : Fault occurs in middle of one of parallel lines

The equivalent reactance during the fault is highest and thus max,2P is lowest

max,1 max,3 max,2P P P

2

0

aP d 0

c 2

c0

m mmax,2 max,3P P sin d P sin P d

For critical clearing,

2 max 1 m

max,3

Psin

P

33

m max 0 max,3 max max2 0

crmax3 max2

P P cos P coscos 1

P P

This is a generic formula and can be applied to other two cases as well after substituting

value of max,1 max,2 max,3

P , P & P .

But crt can only be calculated from cr in previous two cases using expression written

before.

Power System Protection In our current protection, normally a current transformer is connected between

protected elements and the relay.

Plug setting multiplier (PSM)

Fault current

T ratio Pick up current

Usually pick up current = Relay setting x Rated secondary current of CT

Pick-up current is minimum current above which a relay operates.

Differential Relays

The current through operating coil 1 2k I I

If this current is greater than pick-up current then

it operates, else it does not operates.

1 2 pick upK I I I Trip

1 2 pick upK I I I Block

We usually provide a restraining coil to avoid relay mal-operation.

Relay operates if

r 1 21 2 pu

0

N I IK I I K I

N 2

puI = pick up current

rN = Number of turns in restraining coil

0N = Number of turns in operating coil

The ratio operating coil current & restraining coil current is called as bias of differential

relay.

34

Protection of Transmission line

1) Mho relay is at least affected by power surges& thus it is used for protection of long

transmission lines. It is inherently directional.

2) Impedance relay is used for protection of medium transmission lines.

3) Reactance relay is unaffected by ground resistance & hence used for earth fault

protection & also for short transmission Lines.

These relays are collectively called as distance relays.

Protection of Transformers

Differential relays are used for protection of large transformers and CT are always

connected in configuration opposite to power transformer.

Example: 1) If power transformer is Y then CT is Y

: 2) If power transformer is then CT is Y Y

Buccholz relay used to prevent any incipient fault below oil level in a transformer of small

KVA.

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