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Power System (Formula notes)
Work done = F.d cos 
Where F= force applied ,
d = displacement,
 = angle between F & d
Energy: It is capacity to do the work.
Unit : watt second 1w  s  1Joule  1N  m  Newton  meters 
Electrical energy : It is energy that is in charged particles in an electric field.
Electrical energy generally expressed in kilo watt hours (kwh)
1 kwh  3.6 106 J
1 2
mv (Jules)
2
Potential Energy (PE): Mgh (Jules)
eu
p
Kinetic energy (KE):
Thermal Energy: Internal energy present in system by virtue of its temperature.
Unit : Calories 1 Cal  4.186 J
Power: it is time rate of change of energy
dw du

dt
dt
u = work,
w = energy
ad
P
1 Watt  1 J / s
Unit : Watt
gr
Note: Electric motor ratings are expressed in horse power (hp)
1hp = 745.7 W and also 1 metric horse power = 735 Watt.
Electric parameter:
Let
v  2V sin t
i  2I sin(t  )
where v = instantaneous value voltage
i = instantaneous value current
V = rms value of voltage
I = rms value of voltage
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In Phasor representation
v  V 0 ,
i  I  
S = P+jQ = VI cos  + jVI sin  = VI* (for this relation Q will be positive for
lagging VAR)
Where S = complex power or apparent power
P = Active power
gr
ad
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up
Q = Reactive power
For balanced 3 phase system
P  3 |VP || I P | cos  P  3 | VL || I L | cos P
Q  3 | VP || I P | sin  P  3 |VL || I L | sin P
where VL = line voltage
VP = phase voltage
Note: in  connection VP 
VL
3
& IP  IL
 connection VP  VL & I P 
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IL
3
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Hyrdo power:
P = gWh(watt)
Where  = water density (100 kg/m3)
g = 9.81 m/s2
W = discharge rate (m3/sec)
h= head of water
Tidal power
P =  gh2 A/T (watt)
Where h = tidal head
up
A = area of basin
T = period of tidal cycle
Wind power
gr
ad
e
P = 0.5 AV3 (watt)
 = air density (1201 g/m3 at NTP)
V = Wind speed in (m/s)
A = Swept area by blade (m2)
Load Curve: It is graph between the power demands of the system w.r.t. to time.
(i)
Base Load: The unvarying loads which
occur almost the whole day.
(ii)
Peak load: The various peak demands of
load over and above the base load.
Designation
capacity
Capital cost
Fuel cost
Base load
High
Low
Typical
annual load
factor
65-75
Peak load
Low
High
5-10
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Type of plant
Nuclear,
thermal
Gas based,
small hydro,
pump storage
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Designation
capacity
Capital cost
Fuel cost
Base load
High
Low
Typical
annual load
factor
65-75
Peak load
Low
High
5-10
Type of plant
Nuclear,
thermal
Gas based,
small hydro,
pump storage
Operational factors :
Maximum demand
Connected load
1.
Demand Factor =
2.
Average load 
3.
Load factor 
4.
Diversity factor 
5.
Plant Capacity factor 
6.
Reserve Capacity = Plant capacity - max. demand
7.
Plant use factor 
Average demand
Maximum load
up
energy consumed is a given period
Hours in that time period
gr
ad
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sum of individual max demands
Maximum demand on power station
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Average demand
Installed capcity
Actual energy produced
Plant capacity  hours (the plant has been in operation)
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Power Systems Generation and Distribution and Concepts of HVDC:
Thermal Power Station : Thermal efficiency, ηThermal =
Heat equivalent of mech−energy Transmitted toTurbine shat
Heat of coal combustion
 Thermal efficiency = ηboiler × ηturbine
 Overall efficiency, ηoverall =
Heat equvivalent of electrical o/p
Heat of combustion of coal
Output in k.cal
Input in k.cal
η=
up
 Overall efficiency = Thermal efficiency × Electrical efficiency.
 Energy output = coal consumption × calorific value.
= coal consumption × 6500 k.cal
gr
ad
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Water Power equation: Water Head : The difference of water level is called the water head.
 Gross Head : The total head between the water level at inlet and tail race is called as gross
head
 Rated Head : Head utilized in doing work on the turbine
 Net Head : It is the sum of the Rated Head and the loss of head in guide passage and
entrance
H = Head of water in metre
Q = Quantity of water in m3 /sec or lit/sec.
W = specific gravity of water
= 1 kg/lit when ‘Q’ represented in lit/sec.
= 100 kg/m3 when ‘Q’’ represented in m3 /sec.
η = efficiency of the system.
Effective work done
� = WQH × η kg- m/sec.
 Metric output =
WQH × η
75
(H.P)
1H.P = 75 kg-m/sec
 Metric output in watt =
 Output =
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WQH
102
× η kw
WQH × η
75
× 735.5
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 Volume of water available per annum = catchment area × Annual Rainfall
→ Electric energy generated = weight × head × overall η.
GAS TURBINE POWER PLANT :
→ The thermal efficiency of gas turbine plant is about 22% to 25%
→ The air fuel ratio may be of the order of 60 : 1 in this ase.
 Engine efficiency ηengines =
 Thermal efficiency ηthey =
ηoverall
ηalt
ηengine
mech.η of englnd
Terms and Definitions :-
up
 Heat produced by fuel per day = coal consumption / day × caloritic value
gr
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1. Connected load :It is the sum of ratings in kilo watts of equipment installed in the consumer’s premises
2. Demand :It is the load or power drawn from the source of supply at the receiving end averaged over a
specified period.
3. Maximum Demand :Maximum demand (M.D) of a power station is the maximum load on the power station in a given
period.
4. Average load:If the number of KWH supplied by a station in one day is divided by 24 hours, then the value so
obtained is known as daily average load.
KWH deliverd in one day
24
KWH delivered in one month
Monthly average load =
30 ×24
KWH delivered in one year
Yearly average load =
365 ×24
Daily average load =
5. Plant capacity :It is the capacity or power for which a plant or station is designed. It should be slightly more than
M.D. it is equal to sum of the ratings of all the generators in a power station.
6. Firm Power :It is the power which should be always be available even under emergency
7. Prime Power :It is the maximum power (may be thermal or hydraulic or mechanical) continuously available for
conversion into electrical power.
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8. Dump power :This is the term usually used in hydro electric plants and it represents the power in excess of the
load requirements. It is made available by surplus water.
9. Spill Power :Is that power which is produced during floods in a hydro power station.
10. Cold reserve :Is that reserve generating capacity which is not in operation but can be made available for service.
11. Hot reserve :It that reserve generating capacity which is in operation but not in service
12. Spinning reserve :Is that reserve generating capacity which is connected to bus-bars and is ready to take the load.
 Load factor =
=
up
Load factor :It is defined as the ratio of number of units actually generated in a given period to the number
of units that could have been generated with maximum demand.
Average load or Average Demand
Maximum Demand.
Energy generated in a given period
(Maximum Demand) ×(Hours of operaation in the given period)
gr
ad
e
.
→ The load factor will be always less than one (< 1)
Demand factor:It is defined as the ratio of maximum demand on the station to the total connected load to
the station.
 ∴ Demand factor =
Maximum Demand on the station
Total connected load to the station
 Its value also will be always less than one (< 1)
Diversity Factor :Diversity factor may be defined as “the sum of individual maximum demand to the station
to the maximum demand on the power station”.
 Diversity factor =
Sum of individual consumers maximum demand
Maximum demand on the station.
 Its value will be always greater than one (> 1)
Plant Factor or Plant Use Factor : Plant factor =
station output in kwh
Σ (KW1 ) H1 + (KW2 ) H2 + (KW3 )H3 + ……
Where KW1, KW2, KW3 etc. are the kilowatt ratings of each generator and H1 , H2 , H3 etc.
are the number of hours for which they have been worked.
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Capacity Factor or plant capacity factor or capability factor :→ It is defined as the ratio of average demand on the station to the maximum installed capacity.
i.e. capacity factor =
Average demand on the station
Max.installed capacity of the station
→ Coincidence factor:It is the reciprocal of diversity factor and is always less than 1
Maximum demand
Plant capacity
Service hours
Operation factor =
Total duration
Actual energy produced
Use factor =
Plant capacity ×Time (hrs)the plant has been in operation
→
→
up
→ Utilization factor =
gr
ad
e
D.C. Distribution calculations
Uniformly loaded Distributor fed at one end.
→ Fig (a) shows the single lien diagram of a 2 – wire d.c. distributor AB fed at one end A and
loaded uniformly with i amperes per metre length.
A
𝑙𝑙
i
B
i
i
i
Fig. (a)
→ Then the current at point c is.
= δ𝑙𝑙 − ix amperes
= i(𝑙𝑙 – x) amperes.
→ Total voltage drop is the distributor upto point C is
x
𝑣𝑣 = ∫0 ir (𝑙𝑙 − x)dx = ir �𝑙𝑙x −
→ Voltage drop over the distributor AB
=
1
2
1
2
ir𝑙𝑙 2
x2
�
2
= IR
Where
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C
A
i𝑙𝑙 = I, the total current entering at point A
x
𝑙𝑙
dx
B
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r𝑙𝑙 = R, the total resistance of the distributor.
Uniformly loaded distributor fed at both ends.
(i) Distributor fed at both ends with equal voltages
Current supplied from each feeding point
9𝑙𝑙
2
=
x
A
𝑙𝑙
dx
C
B
C
i
i
i
→ Voltage drop upto point C =
→ Max. voltage drop
→ Min. voltage
=
=V–
IR
8
1
8
ir
2
i
i
(𝑙𝑙x − x 2 ).
up
V
IR
volts
gr
ad
e
(ii) Distributor fed at both ends with unequal voltages :The point of minimum potential C is situated at a distance x meters from the feeding point A.
Voltage drop in section AC =
x
A
VA
i
→ x=
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i
VA − VB
ir𝑙𝑙
+
irx2
2
volts.
𝑙𝑙 - x
C
i
𝑙𝑙
2
i
B
i
VB
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Transmission Lines:
→ The empirical formula for the economical voltage line to line is V = 5.5 �
𝑙𝑙
3P
+
1.6
100
where ‘V’ = line pressure in KV, l = distance of transmission in KM,
P = estimated max.KW per phase to be delivered over one pole or tower line
Performance of Lines
→ By performance of lines is meant the determination of efficiency and regulation of
lines.
The efficiency of lines is defined as
→ % efficiency =
→ % regulation =
Power delivered at the receiving end
Power sent from sending end
× 100
Power delivered at the receiving end
Power delivered at the receiving end+losses
Vr′ − Vr
Vr
× 100
× 100
up
→ % efficiency =
gr
ad
e
Where Vr ′ is the receiving end voltage under no load condition and Vr the
Receiving end voltage under full load condition.
Effect of Earth on a 3 – 𝛟𝛟 line:S. No
Line Description
R
L
1.
Length Increases
Increases
Increases
2.
Distance of separation No change
increases
Increases
3.
Radius of
increases
4.
XL
C
XC
Increases
Increases
Decreases
Increases
Decreases
Increases
Decreases
Decreases
Increases
Decreases
Symmetrical spacing.
Does not Decreases
depend
Decreases
Increases
Decreases
5.
Unsymmetrical spacing.
Does not Increases
depend.
Increases
Decreases
Increases
6.
Effect of earth is taken No change
into account
No change
No change
Increases
Decreases
7.
Height of the conductor No change
increases
No change
No change
Decreases
Increases.
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conductor Decreases
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Short Transmission Line
→ The equivalent circuit and vector diagram for a short transmission line are shown in fig.
VS =
Vr
�1 +
2Ir R cos ϕr
Vr
+
2Ir X sin ϕr
Vr
+
Ir2
Vr2
(R2 + X 2 )
→ In practice the last term under the square root sign is generally negligible; therefore.
2Ir R
cos ϕr
Vr
VS = Vr �1 + �
R + JX
1�
2
2Ir X
sin ϕr ��
Vr
Ir
IS
vS
vr
ϕr ϕa
vr
jIrX
IrR
up
vS
+
Ir
gr
ad
e
The terms within the simple brackets is small as compared to unity. Using binomial expansion
and limiting only to second term,
Vs ≃ Vr + Ir Rcos ϕr + Ir X sin ϕr
→ The receiving end voltage under no load Vr ′ is the same as the sending end voltage under full
load condition.
VS − Vr
× 100
Vr
I R
I X
= � r cos ϕr + r sin ϕr �
Vr
Vr
Ir R
Ir X
=
cos ϕr +
sin ϕr
Vr
Vr
%regulation =
Regulation per unit
× 100
= Vr cos ϕr + Vx sin ϕr
→ Where Vr and Vx are the per unit values of resistance and reactance of the line.
Vs = AVr + BIr
Is = CVr + DIr
V
A = Vs � Ir = 0
r
This means A is the voltage impressed at the sending end per volt at the receiving end when
receiving end is open. It is dimensionless.
B=
Vs
�
Ir
Vr = 0
B is the voltage impressed at the sending end to have one ampere at the short circuited receiving
end. This is known as transfer impedence in network theory.
I
C = Vs � Ir = 0
r
C is the current in amperes into the sending end per volt on the open – circuited receiving end. It
has the dimension of admittance.
I
D = s � Vr = 0
I
r
D is the current at the sending end for one ampere of current at the short circuited receiving end
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.
 The constants A, B, C, and D are related for a passive network as follows
AD – BC = 1
→ The sending end voltage and current can be written from the equivalent network as,
Vs = Vr + Ir Z
Is = I r
→ The constants for short transmission lines are,
A=1
B=Z
C=0
D=1
% regulation =
→
%η=
VS� − Vr
A
Vr
×100
up
→
Power received at the receiving end
Power received at the receiving end+losses
gr
ad
e
Where R is the resistance per phase of the line.
× 100
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Medium Length Lines:→ Transmission lines with lengths between 80 km and 160 km are categorized as medium lines
where the parameters are assumed to be lumped. .
→ The two configurations are known as nominal – T and nominal – π respectively.
vS
R
jX
+
2
2
IS
R
jX
+
2
2
vC
IC
Ir
R + jX
vr
Y = jwc
vS
Y
2
Nominal – T
vS
R
2
jIr X/2
up
IS
X
2
IS
gr
ad
e
IC
vr
vc
jIS
Vr ′ =
−j
�
wc
R jX
j
+ −
2 2 wc
Vr′ − Vr
|VS |�
% of regulation =
%η=
Ir
P+3
Vr
P
× 100
R
�I 2 + I52 �
2 r
A, B, C, D constant for nominal – T
A=1+
YZ
2
B = Z �1 +
C=Y
D = �1 +
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YZ
�
2
YZ
�
2
× 100
I𝑙𝑙
Y
2
Ir
IC1
vr
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Nominal – 𝛑𝛑
Z = R + jx
IS
IC2
Y
2
vS
→
Vr ′ =
=
jwc
2
Y
2
=
I𝑙𝑙
jwc
2
Ir
IC1
vS
IC & IC
vr
IS
Ir
−2j
�
ωC
j
R+jX− ωC
�2
|VS |�
% regulation =
%η=
I𝑙𝑙
vr
I𝑙𝑙R
Vr LVr
× 100
Vr
P
×100
P+3 I2𝑙𝑙 R
A=1+
B=Z
Y𝑍𝑍
2
C = Y �1 +
YZ
�
4
YZ
�
2
gr
ad
e
D = �1 +
up
A, B, C, D constants for nominal – 𝛑𝛑
Long Transmission Lines:→ In case the lines are more than 160 km long
I + ∆I
V + ∆ V1 I + ∆I Z∆X
I
vS V + ∆V
vr
V
∆X
X
→ Let Z = series impedence per unit length
Y = shunt admittance per unit length
𝑙𝑙 = length of line
Z = zl = total series impedence
Y = yl = total shunt admittance.
V = Aerx + Be−rx
I
(Aerx − Be−rx )
ZC
V +I Z
V −I Z
V = r 2 r c erx + r 2r C e−rx
1 V +I Z
V −I Z
I = � r 2 r C erx − r 2 r C e−rx �
ZC
I=
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V1 I
C∆X
∆X
Y∆X
jI𝑙𝑙X
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z
r+jωL
ZC = � = �
y
g+jωC
→ The propagation constant r = ∝ + jβ ; the real part is known as attenuation constant and the
quadrature component β the phase constant and is measured in radians per unit length.
Vr + Ir ZC
2
V=
e∝x . ejβx +
Vr −Ir ZC
2
e−∝x . e−jβx
Vs = Vr cos hrl + Ir Zc sin hrl
IS = Vr
sin hrl
+
Zc
A = cosh rl
B = Zc sinh rl
C=
Ir cos hrl
sinh rl
Zc
up
D = cosh rl
The equivalent Circuit Representation of a Long Line equivalent – 𝛑𝛑 Representation
Z1 =
Z sinhr𝑙𝑙
r𝑙𝑙
gr
ad
e
VS
y1
2
y1
2
=
y Tanhr𝑙𝑙/2
2
r𝑙𝑙/2
equivalent – T Representation of Long Line.
IS
VS
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Z′
2
=
Z
2
Tanhr𝑙𝑙/2
r𝑙𝑙/2
Y1 = Y
sinh r𝑙𝑙
r𝑙𝑙
Vr
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Constants for Two networks in Tandem
IS
vS
A1 , B1
C1 , D1
A
A B
equivalent �
�=� 1
C1
C D
IS 1
A2 , B2
V
B1 A2
��
D1 C2
C 2 , D2
B2
�
D2
Constants for networks in parallel
A1 , B1
C1 , D1
IS 2
Ir
vr
A2 , B2
C 3 , D2
A=
Equivalent
Single
Network
Parameters
A1 B2 + A2 B1
B1 + B2
B .B
B = B 1+ B2
1
A=D=
2
A1 B2 + A2 B1
B1 + B2
C = C1 + C2 +
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vr
up
IS
gr
ad
e
vS
Ir
I
=
D1 B2 + D2 B1
B1 + B2
(A1 − A2 )(D2 − D1 )
B1 + B2
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FAULTS:
→ Percentage reactance %𝑋𝑋 =
𝐼𝐼𝐼𝐼
𝑉𝑉
I=full load current
× 100
V= phase voltage
X= reactance in ohms per phase
→ Alternatively percentage reactance (%X) (an also be expressed in terms of KVA and KV
under
%X=
(𝐾𝐾𝐾𝐾𝐾𝐾)𝑋𝑋
10(𝐾𝐾𝐾𝐾)2
Where X is the reactance in ohms.
𝑉𝑉
𝑋𝑋
100
)
%𝑋𝑋
Isc= = I × (
up
→ If X is the only reactance element in the circuit then short circuit currenr is given by
gr
ad
e
i.e short circuit current is obtained by multiplying the full load current by 100/%X
Short- circuit KVA=Base KVA ×
100
%𝑋𝑋
Symmetrical components in terms of phase currents:-
→ The unbalanced phase current in a 3-phase system can be expressed in terms
of symmetrical components as under.
�����⃗ �����⃗ �����⃗
𝐼𝐼���⃗
𝑅𝑅 = 𝐼𝐼𝑅𝑅1 +𝐼𝐼𝑅𝑅2 +𝐼𝐼𝑅𝑅0
�����⃗ �����⃗
𝐼𝐼���⃗𝑌𝑌 = 𝐼𝐼�����⃗
𝑌𝑌1 +𝐼𝐼𝑌𝑌2 +𝐼𝐼𝑌𝑌0
���⃗
�����⃗ �����⃗
𝐼𝐼𝐵𝐵 = 𝐼𝐼�����⃗
𝐵𝐵1 +𝐼𝐼𝐵𝐵2 +𝐼𝐼𝐵𝐵0
���� ����
Where The positive phase current (𝐼𝐼����
𝑅𝑅1 , 𝐼𝐼𝑌𝑌1 , &𝐼𝐼𝐵𝐵1 )
���� ����
Negative phase sequence currents (𝐼𝐼����
𝑅𝑅2 , 𝐼𝐼𝑌𝑌2 , &𝐼𝐼𝐵𝐵2 ) and
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���� ����
Zero phase sequence currents ( 𝐼𝐼����
𝑅𝑅0 , 𝐼𝐼𝑌𝑌0 , &𝐼𝐼𝐵𝐵0 )
→ The operator ‘a’is one, which when multiplied to a vector rotates the vector through 1200
in the anticlockwise direction.
→ A=-0.5+j 0.866
;
a2=-0.5- j 0.866
a3=1
→ Properties of operator ‘a’:
1+a+a2=0
a-a2=j √3
up
0
�����⃗
�����⃗
������⃗
→ Positive sequence current 𝐼𝐼�����⃗
𝐵𝐵1 in phase B leads 𝐼𝐼𝑅𝑅1 by 120 and therefore 𝐼𝐼𝐵𝐵1 = 𝑎𝑎 𝐼𝐼𝑅𝑅1
2�����⃗
similarly, positive sequence current in phase Y is 2400 ahead of 𝐼𝐼�����⃗
𝑌𝑌1 = a 𝐼𝐼𝑅𝑅1
�����⃗ �����⃗ �����⃗
𝐼𝐼���⃗
𝑅𝑅 = 𝐼𝐼𝑅𝑅1 + 𝐼𝐼𝑅𝑅2 + 𝐼𝐼𝑅𝑅0
gr
ad
e
2 �����⃗
�����⃗ �����⃗
�����⃗ �����⃗
𝐼𝐼���⃗𝑌𝑌 = 𝐼𝐼�����⃗
𝑌𝑌1 + 𝐼𝐼𝑌𝑌2 + 𝐼𝐼𝑌𝑌0 = a 𝐼𝐼𝑅𝑅1 + 𝑎𝑎 𝐼𝐼𝑅𝑅2 + 𝐼𝐼𝑅𝑅0
2 �����⃗
�������⃗
�����⃗ �����⃗ �����⃗ �����⃗
𝐼𝐼���⃗
𝐵𝐵 = a 𝐼𝐼𝑅𝑅1 + a 𝐼𝐼𝑅𝑅1 + 𝐼𝐼𝑅𝑅0 = 𝐼𝐼𝐵𝐵0 +𝐼𝐼𝐵𝐵1 +𝐼𝐼𝐵𝐵2
→ Zero sequence current:
2
2
���⃗ ���⃗ �����⃗
�����⃗
�����⃗
𝐼𝐼���⃗
𝑅𝑅 + 𝐼𝐼𝑌𝑌 + 𝐼𝐼𝐵𝐵 = 𝐼𝐼𝑅𝑅1 (1+a+a ) + 𝐼𝐼𝑅𝑅2 (1+a+a ) + 3𝐼𝐼𝑅𝑅0
= 3 𝐼𝐼�����⃗
𝑅𝑅0
1
���⃗ ���⃗
∴ �����⃗
𝐼𝐼𝑅𝑅0 = [ 𝐼𝐼���⃗
𝑅𝑅 + 𝐼𝐼𝑅𝑅 + 𝐼𝐼𝑅𝑅 ]
3
→ Positive sequence current :
3
3
2
4
2
���⃗ 2 ���⃗ �����⃗
�����⃗
�����⃗
𝐼𝐼���⃗
𝑅𝑅 +a 𝐼𝐼𝑌𝑌 +a 𝐼𝐼𝐵𝐵 = 𝐼𝐼𝑅𝑅1 (1+a +a ) + 𝐼𝐼𝑅𝑅2 (1+a +a ) + 𝐼𝐼𝑅𝑅0 (1+a+a )
∴ 𝐼𝐼⃗R1 = =
1
[
3
������⃗1
= 3𝐼𝐼𝑅𝑅
���⃗ 2 ���⃗
𝐼𝐼���⃗
𝑅𝑅 +a 𝐼𝐼𝑌𝑌 +a 𝐼𝐼𝐵𝐵 ]
→ Negative sequence current:2 ���⃗
4
2
3
3
2
���⃗ �����⃗
�����⃗
�����⃗
𝐼𝐼���⃗
𝑅𝑅 +a 𝐼𝐼𝑌𝑌 +a 𝐼𝐼𝐵𝐵 = 𝐼𝐼𝑅𝑅1 (1+a +a ) + 𝐼𝐼𝑅𝑅2 (1+a +a ) + 𝐼𝐼𝑅𝑅0 (1+a +a)
������⃗2
= 3 𝐼𝐼𝑅𝑅
1
2 ���⃗
���⃗
∴ 𝐼𝐼⃗R2 = [ 𝐼𝐼���⃗
𝑅𝑅 +a 𝐼𝐼𝑌𝑌 +a𝐼𝐼𝐵𝐵 ]
3
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Single Line to –Ground Fault:�⃗ R=0 and 𝐼𝐼⃗B =𝐼𝐼⃗Y=0
→ 𝑉𝑉
The sequence currents in the red phase in terms of line currents shall be :-
�����⃗
Phase voltage at fault
up
3 𝐸𝐸𝑅𝑅
���⃗
→ Fault current:- Fault current, 𝐼𝐼���⃗
𝑅𝑅 =3 𝐼𝐼0 =����⃗+𝑍𝑍
�������������⃗
𝑧𝑧0
�
1 +�
𝑍𝑍2
gr
ad
e
Since the generated emf system is of positive sequence only ,the sequence
components of emf in R-phase are:
→
The sequence voltage at the fault for R-phase are:
shorted to ground.
∴
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The phase voltages at fault are:
This is ecpected because R-phase is
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Line –To-Line fault:The condition created by this fault lead to:
→
Again taking R-phase as the reference, we have
→
Expressing in terms of sequence components of red line, we have
Also,
gr
ad
e
Fault current:
up
→
→ Phase voltages: - since the generated emf system is of positive phase sequence only,the
sequence components of emf in R-phase are:
→ The sequence voltages at the fault for R-phase are:
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gradeup
→ The phase voltages at the fault are:
→ Double Line- To – Ground Fault:-
gr
ad
e
up
The conditions created by this fault lead to :
Also,
→ Fault current:→
Phase voltages: - the sequence voltages for phase R are:
→
Now
→
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gradeup
TRANSIENTS IN SIMPLE CIRCUITS:
1. D.C sources
a) Resistance only: - As soon as switch is closed, the current in the circuit will be
determined according to ohms law.
𝐼𝐼 =
𝑉𝑉
𝑅𝑅
𝑁𝑁𝑁𝑁𝑁𝑁 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 .
b) Inductance only:- when switch s is closed, the current in the circuit will be given by
𝑖𝑖(𝑡𝑡) =
𝑉𝑉(𝑠𝑠)
𝑍𝑍(𝑠𝑠)
𝑉𝑉
t
𝐿𝐿
𝑉𝑉 1
𝑆𝑆 𝐿𝐿𝐿𝐿
= .
𝑉𝑉 1
𝐿𝐿 𝑆𝑆 2
= .
up
𝐼𝐼(𝑠𝑠) =
c) Capacitance only:- when switch s is closed, the current in the circuit is given
I(s) =
𝑉𝑉(𝑠𝑠)
𝑍𝑍(𝑠𝑠)
𝑉𝑉
𝑆𝑆
= .CS =VC
gr
ad
e
Which is an impulse of strength (magnitude)VC
d) R-L circuit: when switch s is closed, the current in the circuit is given by
𝑉𝑉(𝑠𝑠)
𝑉𝑉
1
𝑉𝑉 1/𝐿𝐿
= .
= .
𝑍𝑍(𝑠𝑠)
𝑆𝑆 𝑅𝑅+𝐿𝐿𝐿𝐿
𝑆𝑆 𝑆𝑆+𝑅𝑅/𝐿𝐿
𝑉𝑉 1
1
𝐿𝐿
= � −
�.
𝐿𝐿 𝑆𝑆
𝑆𝑆+𝑅𝑅/𝐿𝐿 𝑅𝑅
𝑉𝑉 1
1
= � −
�
𝑅𝑅 𝑆𝑆
𝑆𝑆+𝑅𝑅/𝐿𝐿
𝑉𝑉
−𝑅𝑅
𝑖𝑖(𝑡𝑡) = �1 − 𝑒𝑒𝑒𝑒𝑒𝑒 � 𝑡𝑡��
𝑅𝑅
𝐿𝐿
I(s) =
e) R-L circuit: After the switch s is closed, current in the circuit is given by
I(s) =
1
=
𝑉𝑉
1
.
𝑆𝑆 𝑅𝑅+1/𝐶𝐶𝐶𝐶
𝑉𝑉 �𝑅𝑅𝑅𝑅�𝐶𝐶𝐶𝐶
𝑉𝑉
=
𝑆𝑆 𝑆𝑆+1/𝑅𝑅𝑅𝑅 𝑅𝑅
𝑉𝑉
(t)= .𝑒𝑒 −𝑡𝑡/𝐶𝐶𝐶𝐶
𝑅𝑅
=
i
𝑉𝑉(𝑠𝑠)
𝑍𝑍(𝑠𝑠)
.
1
𝑆𝑆+1/𝑅𝑅𝐶𝐶
→ R-L-C circuit: - After the switch S is closed, the current in the circuit is given by
𝑉𝑉
𝑉𝑉
𝐿𝐿
I(s) = 𝑆𝑆
1
𝑅𝑅+𝐿𝐿𝐿𝐿+1𝐶𝐶𝐶𝐶
1
I(s) = .(𝑆𝑆+𝑎𝑎−𝑏𝑏)(𝑆𝑆+𝑎𝑎+𝑏𝑏)
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i(t) =
where
𝑅𝑅
2𝐿𝐿
𝑉𝑉
2𝑏𝑏𝑏𝑏
𝑅𝑅2
4𝐿𝐿2
= a and �
�𝑒𝑒 −(𝑎𝑎−𝑏𝑏)+ − 𝑒𝑒 −(𝑎𝑎+𝑏𝑏)𝑡𝑡 �
−
1
𝐿𝐿𝐿𝐿
= b; then
→ There are three conditions based on the value of to
∗
∗
If
If
Case I: when b is real
→ i(t) =
𝑅𝑅2
4𝐿𝐿2
𝑅𝑅2
4𝐿𝐿2
=
<
1
𝐿𝐿𝐿𝐿
1
𝐿𝐿𝐿𝐿
1
𝐿𝐿𝐿𝐿
𝑉𝑉
𝑅𝑅2
1
2� 2 − .𝐿𝐿
𝐿𝐿𝐿𝐿
4𝐿𝐿
,b is zero
,b is imaginary
𝑅𝑅2
𝑅𝑅
1
The expression for current becomes
→ i(t)=
𝑉𝑉
2𝑏𝑏𝑏𝑏
{𝑒𝑒 −𝑎𝑎𝑎𝑎 − 𝑒𝑒 −𝑎𝑎𝑎𝑎 } which is indeterminate.
→ Now at b=0
i (t) =
𝑉𝑉
𝐿𝐿
𝑡𝑡 𝑒𝑒 −𝑎𝑎𝑎𝑎 =
𝑉𝑉𝑉𝑉𝑒𝑒
𝐿𝐿
Case III. When b is imaginary
→ i (t) =
=
𝑉𝑉
𝑉𝑉
2𝑏𝑏𝑏𝑏
𝑅𝑅2
1
2𝐿𝐿� 2 −
𝐿𝐿𝐿𝐿
4𝐿𝐿
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𝑅𝑅
𝑅𝑅2
1
�𝑒𝑒𝑒𝑒𝑒𝑒 �− �2𝐿𝐿 + �4𝐿𝐿2 − 𝐿𝐿𝐿𝐿 � +� − 𝑒𝑒𝑒𝑒𝑒𝑒 �− �2𝐿𝐿 − �4𝐿𝐿2 − 𝐿𝐿𝐿𝐿 � 𝑡𝑡� �
gr
ad
e
Case II: when b= 0
,b is real
up
𝑅𝑅2
If 4𝐿𝐿2 >
∗
− (𝑅𝑅/2𝐿𝐿)𝑡𝑡
𝑉𝑉
�𝑒𝑒 −𝑎𝑎𝑎𝑎 . 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑒𝑒 −𝑎𝑎𝑎𝑎 . 𝑒𝑒 −𝑗𝑗𝑗𝑗𝑗𝑗 � = 2𝑏𝑏𝑏𝑏 𝑒𝑒 −𝑎𝑎𝑎𝑎 . 2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
−𝑅𝑅2
𝑒𝑒 −𝑎𝑎𝑎𝑎 .2 sin�� 4𝐿𝐿2 +
1
�
𝐿𝐿𝐿𝐿
t
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A.C source:
→ R-L circuit: when switch is is closed, the current in the circuit is given by
I (s) =
=
𝑉𝑉(𝑆𝑆)
𝑍𝑍(𝑆𝑆)
𝑉𝑉𝑚𝑚 𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
�
𝐿𝐿 𝑆𝑆 2+𝜔𝜔2
𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= 𝑉𝑉𝑚𝑚 �
+
2
𝑆𝑆 2+𝜔𝜔
𝑆𝑆 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
2
𝑆𝑆 2+𝜔𝜔
+
1
� 𝑆𝑆+𝑅𝑅/𝐿𝐿
𝑆𝑆 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
2
𝑆𝑆 2+𝜔𝜔
1
� 𝑅𝑅+𝐿𝐿𝑆𝑆
→ R-L circuit connected to an ac source
I(S)
𝑅𝑅
𝐿𝐿
i(t) =
= 𝑎𝑎; 𝑡𝑡ℎ𝑒𝑒𝑒𝑒
=
𝑉𝑉𝑚𝑚
𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
�
𝐿𝐿 (𝑠𝑠+𝑎𝑎)(𝑆𝑆 2+𝜔𝜔2 )
𝑉𝑉𝑚𝑚
�(𝑅𝑅2 +𝜔𝜔2 𝐿𝐿2 )1�2
Where θ= 𝑡𝑡𝑡𝑡𝑡𝑡−1
+
𝑆𝑆 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
2
(𝑠𝑠+𝑎𝑎)(𝑆𝑆 2+𝜔𝜔 )
�
{sin(𝜔𝜔𝜔𝜔 + 𝜑𝜑 − 𝜃𝜃) − sin(𝜑𝜑 − 𝜃𝜃)𝑒𝑒 −𝑎𝑎𝑎𝑎 }
up
Let
𝜔𝜔𝜔𝜔
𝑅𝑅
gr
ad
e
Circuit Breaker ratings :
→ The value of resistor required to be connected across the breaker contacts which will
give no transient oscillations, is R= 0.5�
𝐿𝐿
𝐶𝐶
Where L,C are the inductance and capacitance upto the circuit breaker
→ The average RRRV =
2Vr
π√Lc
→ Maximum value of RRRV = wn Epeak
→ Where wn = 2πfn ,
→ Natural frequency of oscillations, fn =
1
1
�
2π LC
Where L , C are the reactance and capacitance up to the location of circuit breaker
→ Frequency of demand oscillations, f =
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1
1
1
� − 2 2
2π LC
4R C
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Breaking capacity:
→ Symmetrical breaking current = r.m.s value of a.c component
x
= 2
√
→ Asymmetrical breaking current = r.m.s value of total current.
Where
𝑋𝑋 2
= �� 2� + 𝑌𝑌 2
√
X = maximum value of a.c component
Y = d.c component
→ Is the rated service line voltage in volts, then for 3-phae circuit? Breaking capacity = √3 ×
V × I × 10−6 MVA
Voltage across the string
n × voltage acrosss the unit near power conductor
Where, n = no. of insulators
gr
ad
e
Making capacity :-
up
→ String efficiency =
→ Making capacity = 2.55 × symmetrical breaking capacity.
The Universal Relay Torque Equation:-
→ The universal relay torque equation is given as follows
T = K1 I 2 + K 2 V 2 + K 3 VI(θ − τ) + K
Distance Relays:
impedance relays :
From the universal torque equation putting 𝐾𝐾3 = 0 and giving negative sign to voltage term,
it becomes
→ T = K1 I 2 − K 2 V 2 (Neglecting spring torque)
For the operation of the relay the operating toque should be greater than the
restraining torque i.e
K1 I 2 > K 2 V 2
→ Here V and I are the voltage and current quantities fed to the relay.
V2
K
→ I2 < 1�K
2
K1
�K
2
→ Z<�
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gradeup
→ Z < constant (design impedance)
This means that the impedance relay will operate only if the impedance seen by the relay
is less than a pre-specified value (design impedance). At threshold condition,
Z=�
Reactance Relay:
K1
�K
2
VI
I2
gr
ad
e
sinθ < K1 /K 3
K
Z sinθ < 1�K
3
K1
X < �K
3
up
The directional element is so designed that its maximum torque angle is 900
i.e. in the universal torque equation.
T = K1 I 2 − K 3 VI cos(θ − τ)
= K1 I 2 − K 3 VI cos(θ − 90)
= K1 I 2 − K 3 VI sin θ
For the operation of the relay
KI 2 > K 3 VI sinθ
The mho relay:-
→ In the relay the operating torque is obtained by the V – I element and restraining torque due
tot the voltage elemen
T = K 3 VI cos(θ − τ) − K 2 V 2
→ For relay to operate
K 3 VI cos (θ − τ) > K 2 V 2
V2
K
< 3�K cos(θ − τ)
VI
2
K3
Z < �K cos(θ − τ)
2
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