Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
UNIVERSITÄT DUISBURG – ESSEN
FACHGEBIET ELEKTRISCHE ANLAGEN UND NETZE
Exercises
Accompanying the Lecture Series
Electric Power Systems I-III
Prof. Dr.-Ing. habil. Istvan Erlich
1
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
TABLE OF CONTENTS
1
The Three-Phase System .......................................................................................................... 4
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
2
Symmetrical Components ....................................................................................................... 20
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
3
Span of an overhead line ............................................................................................................. 32
Self and mutual inductances in a conductor – earth loop ........................................................ 34
Inductive coupling in a single-circuit overhead line ................................................................. 36
Inductive coupling in a double-circuit overhead line ............................................................... 38
Capacitive coupling in an overhead line.................................................................................... 39
Overhead line ............................................................................................................................... 40
Parameter of overhead lines, cables and other elements of the transmission system ........... 41
Transmission Line Performance ............................................................................................ 43
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
5
Current and power calculation in symmetrical components................................................... 21
Impedance in symmetrical components .................................................................................... 22
Unsymmetrical currents in symmetrical components .............................................................. 23
Fault conditions in natural and symmetrical components I .................................................... 24
Fault conditions in natural and symmetrical components II .................................................. 25
Sequence networks I.................................................................................................................... 26
Sequence networks II .................................................................................................................. 27
Zero Sequence equivalent circuit of transformers ................................................................... 28
Load current calculation - symmetrical voltage ....................................................................... 29
Load current calculation – unsymmetrical voltage .................................................................. 30
Transmission Line Parameters ............................................................................................... 31
3.1
3.2
3.3
3.4
3.5
3.6
3.7
4
Working with complex numbers .................................................................................................. 5
Phasor representation of electrical quantities............................................................................. 6
Circuit analysis using phasor representation.............................................................................. 7
Impedance and power in RLC circuit ......................................................................................... 8
Calculation of equivalent circuit parameters.............................................................................. 9
Alternative schemes for power transmission ............................................................................ 10
The complex operator a .............................................................................................................. 11
Symmetrical three-phase load I ................................................................................................. 12
Symmetrical three-phase load II ................................................................................................ 13
Unsymmetrical load I .................................................................................................................. 14
Unsymmetrical load II ................................................................................................................ 15
Reactive power compensation .................................................................................................... 16
Symmetrical/unsymmetrical loads ............................................................................................. 17
Voltage drop, transmission losses with/without compensation ............................................... 18
Working with per-unit (p.u.) quantities .................................................................................... 19
Equivalent circuit and phasor diagram of a transmission line ............................................... 44
Compensation of a medium voltage line.................................................................................... 45
Transmission losses I ................................................................................................................... 46
Transmission losses II ................................................................................................................. 47
Maximum power transmission and steady state stability ........................................................ 48
Voltage stability ........................................................................................................................... 49
Calculations on a 750-kV long-range overhead transmission line .......................................... 51
Wave impedance, propagation coefficient, wave velocity and travel time ............................. 53
Traveling waves ........................................................................................................................... 54
Effect of a short-circuited line on a traveling wave .................................................................. 55
Traveling waves ........................................................................................................................... 56
Generators ............................................................................................................................... 58
5.1
5.2
5.3
5.4
5.5
Parameters and characteristic values of generators I .............................................................. 59
Control variables in a synchronous generator .......................................................................... 60
Parameters and characteristic values of generators II ............................................................ 61
Limits of capacitive loading for a generator ............................................................................. 62
Operational behavior of synchronous generators .................................................................... 63
2
Exercises
Electric Power Systems I - III
5.6
5.7
6
Transformer dimensioning ......................................................................................................... 69
Determination of transformer equivalent circuit parameters ................................................. 70
Working in per unit quantities ................................................................................................... 71
Transformers in parallel operation ........................................................................................... 72
Impedance transformation ......................................................................................................... 73
Equivalent circuit of three phase MV/LV transformer ........................................................... 74
Unsymmetrical load on a transformer I .................................................................................... 75
Unsymmetrical load on a transformer II .................................................................................. 76
Zero-sequence equivalent circuit of transformers.................................................................... 77
Equivalent circuit of a three winding transformer .................................................................. 79
Behavior of a transformer under load ....................................................................................... 80
Behavior of a three winding transformer under unsymmetrical load .................................... 81
Zero-sequence network of a power system including transformers ....................................... 82
Parallel operation of transformers............................................................................................. 83
Voltage magnitude and phase angle control of transformers.................................................. 84
Autotransformer.......................................................................................................................... 85
Transformer in open delta connection ...................................................................................... 86
Effect of primary fault on secondary voltages .......................................................................... 87
Transformer in „Scott“ connection ........................................................................................... 89
Instrument transformers (IT) .................................................................................................... 90
power flow analysis ................................................................................................................. 92
7.1
7.2
7.3
7.4
7.5
8
Generator loading capability diagram ...................................................................................... 64
Steady-state and transient stability............................................................................................ 66
Transformers ........................................................................................................................... 68
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
6.18
6.19
6.20
7
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Power flow solution by Gauss – Seidel method......................................................................... 93
Power flow analysis by Newton – Raphson method ................................................................. 95
Regulating transformer and load sharing I .............................................................................. 96
Regulating transformer and load sharing II ............................................................................. 97
Representation of the regulating transformer in a load flow study ........................................ 98
Short Circuit analysis.............................................................................................................. 99
8.1
Comparison of a short circuit in R-L circuit and a generator .............................................. 101
8.2
Symmetrical short-circuit in a generator – transformer unit ............................................... 102
8.3
Short-circuit current computation: internal emf method vs. equivalent voltage source
method I ................................................................................................................................................. 103
8.4
Short-circuit current computation: internal emf method vs. equivalent voltage source
method II ................................................................................................................................................ 104
8.5
Short-circuit in a generator connected to a group of motors I .............................................. 105
8.6
Short-circuit in a generator connected to motors II............................................................... 106
8.7
Symmetrical short-circuit current computation..................................................................... 107
8.8
Mechanical and thermal stress caused by the short-circuit current..................................... 108
8.9
Calculation of short circuit currents in a meshed network ................................................... 112
8.10 Short Circuit Current Limitation ............................................................................................ 114
8.11 Analysis of unsymmetrical faults ............................................................................................. 115
8.12 Effect of transformer neutral point connection on system performance during fault ........ 117
8.13 Neutral point grounding in medium voltage networks .......................................................... 119
8.14 Problem of voltage un-symmetry in three phase transmission systems ............................... 121
8.15 Low impedance grounding in cable networks ........................................................................ 122
8.16 Transient stability analysis I .................................................................................................... 124
8.17 Transient stability analysis II ................................................................................................... 125
8.18 Transient stability analysis III ................................................................................................. 126
8.19 Transient stability analysis IV.................................................................................................. 127
3
Exercises
Electric Power Systems I - III
1
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
THE THREE-PHASE SYSTEM
4
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.1 Working with complex numbers
Given:
Z 1 = 2.598 + j1.5
Z 2 = 2.828 + j 2.828
Z 3 = 2.5 + j 4.33
)
i( t ) = I ⋅ sin( ωt + φi )
a. Evaluate the following expressions and give the results both in polar and Cartesian
coordinates:
*
i. Z 1 ⋅ Z 2 ⋅Z 3
2
*
ii. Z 1 + Z 2 ⋅Z 3
iii. Z 1 ⋅ Im⋅ {Z 3 }
−1
b. Find the complex number I such that:
{ }
2 ⋅ Re{I ⋅ e }
i. i( t ) =
2 ⋅ Im I ⋅ e jωt
ii. i( t ) =
jωt
c. What is the RMS (root mean square) or “effective” value of i(t), if:
i.
The radian frequency ω = 2π.50 s-1
ii.
The radian frequency ω = 2π.60 s-1
)
iii. A positive offset value I / 2 is added to i(t)?
Fig. 1 Sine curve with and without offset
5
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.2 Phasor representation of electrical quantities
Assuming
u(t) = 141.4 sin(ωt + 300) V and
i (t)= 11.31 cos(ωt - 300) A,
determine:
a. the amplitude and the RMS (root mean square) values for the voltage and the current
b. the current and voltage phasors
c. the complex power and the power factor
d. the values of R (resistance) and X (reactance) for
i. series
ii. parallel
connections.
6
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.3 Circuit analysis using phasor representation
R
i(t)
i(t)
u(t)
L
C
u(t)
(ii)
(i)
a.
R
For circuits (i) and (ii), give the general relationship relating the voltage u(t) to the
current i(t).
b.
Assuming a sinusoidal applied voltage, show that the differential equations obtained in
(a) can be simplified to the following algebraic equations in phasor form:
U = I ⋅ ( R + jω L )
U = I ⋅( R +
1
)
jω C
for (i)
for (ii)
where U and I are the voltage and current phasors, respectively.
7
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.4 Impedance and power in RLC circuit
R
L
a
Magnetic field energy stored in the inductance L:
e
Wmf =
a) R in series with L
1
⋅ L ⋅i2
2
C
R
Electric field energy stored in the capacitance C:
a
e
Wef =
b) R in series with C
R
1
⋅C ⋅u2
2
L
e
a
C
c) R L in parallel with C
Assume the parameters in the circuits (a) – (c) above have the following values: R = 10 Ω,
L = 15.9 mH, C = 0.6366 mF, f = 50 Hz.
a. Calculate the impedance Zae (in Ω) as well as the admittance Yae (in S) across the
terminals a – e for the circuits (a) – (c).
b. Assume voltage with the value u( t ) = 2 ⋅ 220V ⋅ cos ωt is applied across the terminals
a – e. Calculate I, S, P and Q for all three connections.
c. Now, consider the circuit (a). Find a general expression for the product i ⋅ u L (uL : the
voltage drop across the inductance L), and show that this expression corresponds to
∂Wmf
∂t
d. Re-work (c) for the circuit (b) to show that the product i ⋅ uC (uC : the voltage drop across
the capacitance C) corresponds to
∂Wef
∂t
.
e. What can you deduce from the solutions (c) and (d) with regard to the nature of Q (the
reactive power)?
8
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.5 Calculation of equivalent circuit parameters
XAE
A
UA
E
UE
RE
XE
With
UE = 220 V ∠ − 30°
RE = 63.5 Ω
jXE = j110 Ω
jX AE = j10 Ω
Calculate and indicate:
a. the direction of currents (load and output currents) and voltages according to the
generation-oriented sign convention;
b. the load admittance YE;
c. the current IE;
d. the complex power absorbed by the load;
e. the voltage across jXAE and the voltage UA;
f. the real and imaginary parts as well as the active and reactive parts of the current in
location E;
g. the complete phasor diagram; (resolve the current in location E in real and imaginary
parts as well as in active and reactive parts).
9
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.6 Alternative schemes for power transmission
A generator, situated at a remote location, is intended to supply a load via a transmission line.
The available conductor material for the transmission line has an overall cross-sectional area of A
with an allowable current density J. The conductor can be divided into multiple conductors if and
when the need arises, and is insulated for a peak voltage of U0. Determine (in terms of J, A and
U0) the maximum power that can be transmitted by the line for the following alternative
configurations:
a. DC two wire schemethe load is supplied by a single dc source
b. DC three wire schemethe load is divided into two equal parts, each supplied by a source of equal magnitude,
and a common conductor is used for the return current
c. Single-phase schemethe load is supplied by a single phase ac source
d. Single-phase three wire schemethe load is divided into two equal parts, each supplied by an identical single-phase ac
source and a common conductor is used for the return current
e. Two-phase schemeThe load is supplied by a two-phase voltage system having the following relationship:
ux = û cosωt V,
uy = û sinωt V
f. Three-phase schemeThe load is supplied by a three-phase voltage system having the following relationship:
ua = û cosωt V,
ub = û cos(ωt-1200) V,
uc = û cos(ωt+1200) V
Summarize the results obtained in (a) – (f) in a tabular form and explain why, of all the possible
alternatives, the three-phase system has established itself as the standard mode in electric power
supply system.
10
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.7 The complex operator a
Using the definition:
a = 1∠1200
simplify the following expressions:
a. 1+ a + a2
b. 1- a 2 + a
c. a 2 + a + j
d. j a + a 2
e.
ea
f.
a 10
g. (1 - a 2)3
h. a – 1
11
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.8 Symmetrical three-phase load I
Z1
~
L1
UL1-L2
L2
Z2
~
L3
Z3
~
Z 1 = Z 2 = Z 3 = 30Ω∠31.80
U L1− L 2 = 3 ⋅ 220V∠30 0
The voltage source to which the above load is attached is symmetrical.
Determine:
a. The line to line voltages
b. The phase voltages
c.
U L 2− L3 ,U L 3− L1
U L1 ,U L 2 ,U L3
The line currents I L1 , I L 2 , I L 3
d. The complex powers in each phase
S L1 , S L 2 , S L3
e. The total complex three phase power absorbed by the load
S 3φ
12
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.9 Symmetrical three-phase load II
Ia
a
Zab
Ib
b
Ic
c
Zca
Zbc
The load impedances and the source voltage have the following values.
Z ab = Z bc = Z ca = (30 + j 27 )Ω
2
U a = 220V∠0 0
U b = a ⋅U a
U c = a ⋅U a
a. Calculate the currents I a , I b , I c .
b. What is the total complex power absorbed by the load and the corresponding power factor
(cosϕ)?
c. Assume that three pieces of capacitor batteries, each with a value of C= 25 µF, are
available. Determine the power factor at the load terminals, when the capacitors are
connected:
i.
as shown in (i),
ii.
as shown in (ii).
Ia
Ia
a
a
Zab
b
c
Ib
Zab
Ib
Zca b
Zbc
Ic
Zca
C
Zbc
Ic
c
C
C
(i)
(ii)
13
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.10 Unsymmetrical load I
Consider the following circuit in which an R-C load is connected between phases a and b with
phase c remaining open. V represents a voltmeter.
R
a
Xc
b
V
c
R = Xc = 100 Ω;
Ua = 220∠00 V
Calculate the voltage across the voltmeter (V) if the source voltage represents:
a. a positive-sequence
b. a negative-sequence.
14
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.11 Unsymmetrical load II
a
R
XL
b
XC
c
Uab = 38∠0 0 kV
PR = 10 MW; Q L = Q C =
PR
3
If the source voltage is known to be a symmetrical three-phase system, calculate:
a. R, XL and XC
b. The complex power absorbed by the load
c. The currents Ia, Ib, Ic and the power factor (cos ϕ) at the generator terminals. Also show
that, as seen from the generator terminals, the load can be represented by a starconnected pure resistive load and then determine the value of the corresponding
resistance per phase.
15
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.12 Reactive power compensation
An induction motor has the following ratings:
380 V, 20 kVA, 0.7 lag
During the course of the operation of the motor, it was found out that a full load on the motor
leads to the overloading of the supply line.
a. Find the value of the capacitor per phase to be installed at the terminals of the motor
which would reduce the line current to the allowable limit of 25 A, if the capacitors are to
be connected in:
i. Star
ii. Delta.
b. Determine the power factor (after compensation) at
i. the input line
ii. the motor terminals behind the capacitor bank.
16
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.13 Symmetrical/unsymmetrical loads
Two three-phase loads, one star and the other delta connected, are supplied from a symmetrical
three-phase source via a transmission line having the reactance 2 Ω/phase and negligible
resistance. Each of the three-phase loads absorbs 20 kW at unity power factor and the rated
voltage of 0.4 kV.
a.
Determine the load impedance per phase for both the delta and the star connected
loads.
b.
Determine the currents in each of the load impedances and the supply line.
c.
Calculate the complex power and the voltage at the source end of the line.
d.
Assume now that, with the source voltage held constant at the value calculated under
(c), the delta connected load is disconnected from the supply. Additionally, one of the
impedances (phase a) in the star-connected load is parallel short-circuited. The star
points of both the load and the voltage source are grounded.
i. Determine the currents in phases a, b and c of the supply line. What is the power
absorbed by the load?
ii. What is the value of the current flowing through the neutral wire?
iii. What are the values of the phase currents, when the neutral wire is disconnected?
iv. From (iii), what can you deduce regarding the size of the neutral conductor
compared to the phase conductors?
e.
Assume now that the star load is disconnected while the delta load is reconnected to
the supply. Additionally, one of the load impedances (between phases a and b) is
parallel short-circuited.
i. Determine the phase as well as the line currents.
ii. Compare the line currents calculated in e(i) and c (i) and comment on the degree of
asymmetry in both cases.
17
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.14 Voltage drop, transmission losses with/without compensation
∼
Generator:
Overhead line:
Load:
UG= 20 kV
Zb= 2.5+j3 Ω
ZL= 110+j50 Ω
a. Draw the single-phase equivalent circuit
b. Calculate:
− The line current (I)
− The active power (P)
− The reactive power (Q)
− The apparent power (S)
− The complex power (S)
− The power factor (cos ϕ)
at the load.
c. Determine the output power and the power factor at the generator terminals
d. The power factor at the load is to be increased by connecting capacitors in parallel to the
load. Determine the value of the capacitors per phase needed to improve the power factor
to 0.98, if the capacitors are to be connected in:
− Delta
− Star.
e. Compute the line losses with and without compensation.
f.
Compare the voltages at the load terminal with and without compensation.
18
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.15 Working with per-unit (p.u.) quantities
1 Ω j8 Ω
IL
39 Ω
∼
220 V ∠00
UL
j26 Ω
In the circuit shown above, a load having an impedance of 39 + j26 Ω is fed from a voltage
source through a line having an impedance of 1 + j8Ω. The effective (RMS) value of the source
voltage is 220 V.
a. Calculate the load current IL and the voltage at the load terminal UL.
b. Calculate the active and reactive power delivered to the load.
c.
Repeat the above calculation in per-unit choosing a base of 220 V for the voltage and 1500
VA for the apparent power S.
d. Compare the values obtained in c) with those obtained in a) and b).
19
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2 SYMMETRICAL COMPONENTS
20
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.1 Current and power calculation in symmetrical components
Three identical Y-connected resistors form a load bank with a three-phase rating of 400 V,
500 kVA. The load bank has applied voltage magnitudes:
Uab= 320 V
Ubc= 480 V
Uca= 400 V
The neutral of the load is not connected to the neutral of the system.
a. Find the line currents and the complex power absorbed by the load:
i. without using symmetrical components
ii. using the symmetrical components.
b. Assuming the load is supplied from the low voltage side of a Dy-connected, 10/0.4-kVtransformer, find the line voltages and currents on the high-voltage side of the
transformer both in natural and symmetrical components.
21
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.2 Impedance in symmetrical components
Consider the following circuit, which may be described by the matrix equation given below the
figure.
A
a
E
Ia
Zs
i
Ib
b
Ic
Zg
Zs
Z
Z
Zg
Z
c
Ua,A Ub,A
Zs
Z
Uc,A
 ∆Ua   Ua,A − Ua,E   Z s
 
 

 ∆Ub  =  Ub,A − Ub,E  =  Z g
 ∆U   U − U   Z
c,E 
 c   c ,A
 g
or in compact form:
Zg
∆Uabc
Zs
Zg
Z g  Ia 
 
Z g  Ib  ,

Z s  Ic 
=
Z abc Iabc
Zg
a. Prove that by transforming the voltage and current relationships into symmetrical
components, a decoupled (i.e. zero off-diagonal elements) system can be achieved.
b. Carry out the transformation and determine the diagonal elements Z 00 , Z11 und Z 22 of
the transformed matrix in terms of the given parameters.
22
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.3 Unsymmetrical currents in symmetrical components
A
a i
E
Ia
b
c
Ib =- Ic
The line currents during a short-circuit in an overhead line are known to be as follows:
Ia = 0 A
Ib = 2700 A ∠ 0°
Ic = 2700 A ∠ 180°
Calculate the corresponding current phasors in symmetrical components.
23
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.4 Fault conditions in natural and symmetrical components I
a. For a single line-to-ground (SLG) fault in phase a (depicted below), formulate the fault
conditions in natural components, transform these relationships into symmetrical
components and then deduce from these the interconnection of the sequence networks
during fault.
Ia
a
Ib
b
Ic
c
Ua
Ub
IF
Uc
E
Single-line-to-ground (SLG) fault
b. For a double line-to-ground (DLG) fault between phases b and c (depicted below),
formulate the fault conditions in natural components, transform these relationships into
symmetrical components and then deduce from these the interconnection of the
sequence networks during fault.
Ia
a
Ib
b
Ic
c
Ua
Ub
Uc
IF
E
Double-line-to-ground (DLG) fault
24
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.5 Fault conditions in natural and symmetrical components II
A DLG fault involving a fault impedance is shown in the figure below:
Ia
a
Ib
b
Ic
c
Ua
Ub
Uc
ZF
E
a. Give the fault conditions in natural components.
b. Transform then these conditions into symmetrical components.
c. Sketch the equivalent circuit in accordance with the conditions determined in (b) and
indicate U012 and I012 in the circuits.
25
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.6 Sequence networks I
A power system feeds a three phase symmetrical load via a capacitive overhead line (see figure
below), with the star point of the load grounded through the resistance R.
Ia
Ua
Ub
Ib
Ca
Ic
Cb
Cc
a
Z
b
c
Z
c
Z
s
s
s
Uc
CaE CbE CcE
R
s
Draw the equivalent circuit of the line in terms of symmetrical components.
26
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.7 Sequence networks II
A symmetrical three phase voltage source Uabc, with internal impedance Zi and the star point
grounded through the resistance Ri, feeds a symmetrical load via a lossless line, the star point
of which is also grounded through a resistance RE, see figure below.
Uqa
Zi
s
a
ZV
s
ZV
s
Uqb
Zi
b
c
Uqc
s
Zi
s
R
s
c
Ua
Ub
ZV
s
Uc
RE
s
a. Derive the relationships, which describe the situation in natural components, and
transform these into symmetrical components.
b. Based on the results obtained in (a) draw the equivalent circuit in terms of symmetrical
components.
27
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.8 Zero Sequence equivalent circuit of transformers
Find the zero sequence equivalent circuits of two winding transformers in the following
connection:
28
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.9 Load current calculation - symmetrical voltage
A balanced three-phase Y-load (shown in the figure below) is grounded through the neutral
impedance Zn. There is a mutual coupling between the three loads making up the Y-load.
Ua
Ub
Uc
Zn
Each phase has a series reactance of Zs =j12 Ω, and the mutual coupling between phases is Zm
=j4 Ω. A balanced three phase voltage of 220-V line to neutral voltage (the neutral of the source
voltage is grounded) is applied to the load.
Determine:
a. the line currents by mesh analysis without using symmetrical components.
b. the line currents using symmetrical components.
29
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
2.10 Load current calculation – unsymmetrical voltage
A three-phase unbalanced source with the following phase to neutral voltages is applied to a
load.
0
 U a   200∠25 


0
 U b  = 100∠ − 55 
 U   80∠100 0 
 c 

The load series impedance per phase is Zs = 8+j24Ω and the mutual impedance between phases
is Zm =j4Ω . The load and source neutrals are solidly grounded.
Determine:
a. The load sequence impedance matrix Z012.
b. The symmetrical components of voltage.
c. The symmetrical components of current.
d. The load phase currents.
e. The complex power delivered to the load in terms of symmetrical components.
f.
The complex power delivered to the load in terms of the power per phase.
30
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3 TRANSMISSION LINE PARAMETERS
31
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.1 Span of an overhead line
An overhead conductor (aluminium conductor steel reinforced (ACSR), DIN 48204-240/40-Al/St)
is suspended between two back-to-back towers with both points of connection having the same
height above ground. The conductor is fastened onto the towers for a maximum sag (fmax) of 10 m
at 40°C.
Span length
A = 150 m
Conductor cross section
A = 282.5 mm2
Diameter of the stranded conductor
d = 21.8 mm
Mass of the stranded conductor per unit
m ' = 0.985
length
kg
m
ε ϑ = 1.96 ⋅ 10 −5 ⋅ K −1
Coefficient of thermal expansion
Modulus of elasticity
E = 70
kN
mm 2
Stress per unit length caused by ice layer
d N

q VZ =  5 + 0.1

mm  m
at –5° C, (in accordance with DIN VDE

0210)
a. Based on the state equation :
a2  g⋅ ρ   g⋅ ρ 
εϑ(ϑ1 − ϑ2) =  1  −  2 
24 σ1   σ2 

2
2
−


σ1 − σ2
,
E
derive a relationship for the critical span aσ max, assuming an identical tensile stress σmax
for both states 1 and 2, which are defined as follows :
State 1
ϑ1 = -5°C
δ1 = δ + ∆δ additional ice layer !
State 2
ϑ2 = -20°C
δ2 = δ
State 3
ϑ3 = 40°C
δ3 = δ
32
Exercises
Electric Power Systems I - III
δ
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
is the specific density of the stranded conductor and ∆δ is the specific density of an
assumed additional ice layer; ( δ = g.ρ).
b. Determine the span aσ max by using the given values and σ max = 72.1
N
.
mm 2
In which of the states, 1 or 2, do you expect the maximum tensile stress?
c. Derive, in an analogous way, a second equation for the second span length afmax for
which the same maximum sag occurs in the states 1 and 3 (see table above). In doing
so, replace the tensile stresses σ1 and σ2 in the state equation by the maximum sag fmax
using the equation f =
δ ⋅ a2
. Is the span afmax admissible?
8⋅σ
33
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.2 Self and mutual inductances in a conductor – earth loop
A single conductor composed of solid round wire having radius r (figure below) is used for an
overhead transmission line.
r
D
0
x
I1
a. Derive an expression for the external flux linkages (per unit length of the line) between
points r (conductor radius) and D (an arbitrary point at a remote location). Also find the
internal flux linkages, i.e. the flux linkages between x = 0 and x = r, and then an
expression for the total flux linkages arising from the sum of the internal and external
flux linkages.
b. Assume a second conductor carrying the current I2 is placed in parallel to the conductor
under consideration. The center-to-center distance between the two conductors is D12.
Derive an expression for the mutual flux linkages.
c. Assume now that, in place of the single conductor described in (a), two conductor
bundles, each with the radius r and held apart by a spacer of length d, are used. Show
that the external flux linkages can still be computed using the expression derived in (a)
by merely replacing the conductor radius (r) by the equivalent radius of the bundle,
which is rb = r ⋅ d (The total current I1 is assumed to divide equally between the two
conductors). Similarly, show that the equivalent radius for a conductor bundle
comprising of three conductors (each with radius r) and placed at the vertices of an
3
equilateral triangle of side d is: rb = r ⋅ d 2
d. Replace the second conductor in (c) with an earth conductor, which carries the return
current. The mutual distance between the earth conductor (E) and the phase conductor
(i) can be computed using the formula:
34
Exercises
Electric Power Systems I - III
D iE =
2 ⋅ e − 0 . 077
µ0 ⋅ ω ⋅ κ
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
≈
930
κ [ S ⋅ cm
−1
cm
]
ω = 100 ⋅ π Hz;
µ 0 = 4πx10 − 4 H / km;
κ = conductivity of the earth
Show that the self-reactance of the conductor – earth loop per unit length can be
computed using the formula:
X i−E = X i + ω ⋅
µ0
D
930
⋅ ln iE = X i + 0.14467 lg
rb
2⋅π
rb
κ [S ⋅ cm −1 ]
cm
⋅
Ω
,
km
where Xi represents the reactance due to the internal flux linkages (refer to (b)).
e. Similarly, show that mutual-reactance between two conductor – earth loops (i and k)
placed at a distance of Dik from one another can be computed using the formula:
X ik − E = ω ⋅
µ0
D
930
⋅ ln iE = 0.14467 ⋅ lg
D ik
2⋅π
D ik
κ [S ⋅ cm −1 ]
cm
⋅
Ω
km
f. Using the following value for the earth resistance:
RE =
μ0 ⋅ ω
≈ 0.05 Ω / km
8
show that the self- and mutual-impedances of the line can be computed using the
formulae:
Z ii−E = R + R E + j( X i−E + X i )
Z ik −E = R E + jX ik −E ,
where R represents the line resistance per unit length.
35
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.3 Inductive coupling in a single-circuit overhead line
The following figure depicts the tower and conductor arrangement for a 220-kV double circuit
overhead line using two conductor bundles for each phase. In this exercise, only one of the
systems (denoted by the small letters a b c ) is to be considered.
p
q
2.5
4.91
4.5
a
4.5
b
5.15
c
a = 0.4
C
B
A
20.14
All distances in m
Phase conductors:
Shield wires
2 x 185/32 AlSt
185/32 AlSt
r = 9.6 mm
rp = 9.6 mm
R = 0.079 Ω/km (for the bundle)
Rp = 0.158 Ω/km
Xi = 0.00678 Ω/km (for the bundle)
Xip = 0.0132 Ω/km
h = 11.068 m
hp = 15.978 m
Conductivity of the earth: κ = 10 −2 S ⋅ m −1
Earth resistance RE = 0.05 Ω/km
36
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Assuming that the tower carries only one of the two systems, calculate the following:
a. the impedance matrix Z of the un-transposed line
b. the impedance matrix Z , when the line is transposed
c. the sequence impedance matrix for the transposed line
d. the operating impedance of the line per phase (operating impedance is to be
understood as the per phase impedance per unit length for an ideally transposed line
neglecting the effect of the shield wire)
e. the impedance matrix Z for a transposed line taking the influence of the shied wire into
consideration
f. the sequence impedance matrix Z for a transposed line taking the influence of the shied
wire into consideration
g. the reduction factor, i.e. the ratio of the zero-sequence current flowing through the
ground return to the overall zero-sequence current
h. the reduction factor, if instead of the above shield wire, a 70 mm2 steel conductor with
the impedance Z’PP-E = (2.39 + j2.158)Ω/km is used
i. the voltages Uc and Ub in the figure below for a transposed line of 150 km length:
I. neglecting the shield wire
II. considering the effect of the shield wire
Ip
a
b
Ib
c
Ib
∑I
Uc
∑I
L = 150 km
L = 150km
Ib = 800A
Ib = 800A
Ub
37
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.4 Inductive coupling in a double-circuit overhead line
Conductors are now assumed to have been placed on the other arm of the tower (in Exercise
3.3) leading to a double-circuit 220-kV-three-phase overhead line. Assuming the two systems are
mirror-symmetrically arranged, determine the following:
a. the impedance matrix for the un-transposed line
b. the transformation matrix from natural components into symmetrical components for a
double-circuit line
c. the sequence impedance matrix (general) for the un-transposed line
d. the sequence impedance matrix for γ-transposition
e. the sequence impedance matrix for
β-transposition. Sketch also the transposition
scheme
f. the mutual zero-sequence impedance between the two circuits determined on the basis
of conductor geometry (for β-transposition)
g. the total impedances (Z00)g, (Z11)g und (Z22)g for each system during parallel operation of
the two systems and assuming γ- transposition
h. the conductor-to-ground voltage (ULE) for the following configuration.
a
b
c
I
II
A
B
C
ULE
∑I
∑I = 500 A
l = 200 km;
38
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.5 Capacitive coupling in an overhead line
For the transmission system described in Exercise (3.3):
a. Show that for a completely transposed line the diagonal and off-diagonal elements of
the potential coefficients matrix can be computed using the following approximate
formulae:
Pnn = 18 ⋅ 10 6 ⋅ ln
Pnm = 9 ⋅ 10 6 ⋅ ln(
2 ⋅ hn
km / F
r
4 ⋅ h n2
2
D nm
+ 1) km / F
hn = 3 h a ⋅ hb ⋅ h c
b. Determine:
i. the elements of the potential coefficient matrix
ii. the matrix of the capacitance coefficients
iii. the positive-, negative- and zero-sequence components of the capacitance
coefficients
iv. the positive-, negative- and zero-sequence capacitances of the line
v. the operating capacitance of the line neglecting the effect of the shield wire.
39
Exercises
Electric Power Systems I - III
3.6
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Overhead line
The figure below shows the conductor arrangement for a 345-kV overhead line using two bundles
for each phase. The phase conductors are assumed to be ideally transposed.
d
d
d
D
D
d = 45 cm
D=8m
Each of the conductors has the following values:
r (conductor radius)
= 1.085 cm
R (conductor resistance)
= 0.119 Ω/km
Xi (internal reactance/conductor)
= 0.0132 Ω/km
ε0 =
1
⋅ 10 −9 F / m
36 ⋅ π
µ 0 = 4π ⋅ 10 −7 H / m
a. Calculate the reactance ( X 'b ), impedance ( Z 'b ), capacitance ( C 'b ), charging reactive
power ( Q 'c ), charging current ( I 'c ) all per km and the wave impedance ( Z w ), the natural
load ( Pnat ).
b. Re-evaluate the values in (a) if, instead of two conductors per phase, a single conductor
with the same combined cross sectional area is used.
c. Re-evaluate the values in (a), if instead of two, three conductor bundles were to be
used for each phase? Assume conductors are placed at the vertices (tips) of an
equilateral triangle of side 0.45 m, and that the total conductor cross-sectional area
remains the same as in (a).
d. Determine the voltage at the sending-end, when the line delivers 250 MVA load at the
rated voltage and pf = 0.85 lagging
e. Determine the no-load voltage at the receiving-end (the load end) of the line when the
rated voltage is applied at the sending-end (the generator end) of the line.
For (d) and (e), assume that the line has a length of 300 km. Use the parameters calculated
in (a). Neglect the line resistance. Also, sketch the phasor diagram!
40
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
3.7 Parameter of overhead lines, cables and other elements of the transmission system
a. The following figure depicts the conductor arrangement for a 380-kV-Overhead line.
7.97
b
8.63
7.00
c
4.33
a
0.5
3X380/50 AlSt
All values in m
r = 1.35cm
Xi =0.0132 Ω/km
R =0.0757/3 Ω/km
Determine the following:
-
rb
-
Xb (line reactance per km),
-
Cb (line capacitance per km),
-
QC (charging reactive power per km)
-
IC (charging current per km),
-
ZW (wave impedance)
-
Pnat (natural power)
41
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
b. A 220-kV three-phase, radial field cable has the following values:
ICE (capacitive earth fault current) = 36A/km,
ZW = 36Ω
Determine the line reactance and the natural power.
c. The following values of a 20-kV three-phase cable are known:
R (cable resistance) = 0.175 Ω/km
Cb=0.33µF/km
In (rated current)= 220A
tanδ (loss factor)= 0.005
l (cable length) = 10km
Determine the dielectric losses, the conductor losses and the charging current at the
rated voltage.
d. The dimensions of a cast-resin insulated bus bar arrangement are given in the figure
below. Calculate the impedance and the capacitance of the bus bar system.
Metallic screen
250mm
Cast resin εr = 4
Conductor;
κ = 3580 S/mm
35mm
15mm
e. For the following 110-kV AlSt (210/36 AlSt) bus bar system, determine:
-
the resistance, the reactance, the line capacitance, the wave impedance and
the natural power.
1.7m
3.1m
3.1m
19.3
42
Exercises
Electric Power Systems I - III
4
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
TRANSMISSION LINE PERFORMANCE
43
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
4.1 Equivalent circuit and phasor diagram of a transmission line
A IA
i
IE
Xb
R
ICE
ICA
UA
E
Cb
Cb
2
2
UE
Rated values:
SE= 600 MVA
cos ϕE = 0.95 lag.
UE =
380 kV
3
∠0°
Parameters:
IC = 0.97
A
km
X' b = 0.24
Ω
km
R' = 0.03
Ω
km
at
85°C l = 300 km
a. Determine the parameters of the nominal π equivalent circuit
b. Calculate the line voltage drop, including its real and imaginary components, for the load
given above:
i. Considering the line capacitance
ii. Neglecting the line capacitance
iii. Considering the line capacitance and assuming a load power factor of unity.
c. Determine the transmission losses for b(i) – b(iii).
d. Calculate the no-load voltage at the receiving-end of the line when the voltage at the
sending-end is 380kV.
e. Determine the maximum permissible load at a power factor of 0.95 lagging if the real
part of the voltage drop is not to exceed 10%.
44
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
4.2 Compensation of a medium voltage line
A load of 5 MVA, cos ϕE = 0.7 lagging is fed via a 10-kV overhead line (95 mm2 Al, R = 0.309
Ω/km, length l = 8.5 km) with
UE =
10
3
Xb
= 1. 3 .
R
kV
a. What is the voltage at the sending-end of the line (UA)?
b. What is the voltage at the sending-end (UA), if the voltage at the receiving-end (UE) is
maintained at the rated value and a series capacitor with k = 2 (k =XC/Xb: degree of
compensation) is installed at the receiving-end of the line?
c. What would the voltage at the input terminals of the line (UA) be, if (instead of series
compensation described in (b)) the load power factor is improved through parallel
compensation from cos ϕE = 0.7 to cos ϕ´E = 0.95 ?
d. Calculate the required capacitance and the corresponding reactive power output for (b)
(max. line current = 350 A).
e. What is the required capacitance and the corresponding reactive power output for (c)?
(Assume star connection of capacitors).
f. If the capacitors available in stock have values given in the table below, determine the
required number of units for both (d) and (e).
QCn = 50 kVar In = 8.65 A
Un = 5.78 kV
Cn = 4.77 µF
45
Exercises
Electric Power Systems I - III
4.3
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Transmission losses I
A three-phase line has the following parameters and rated values.
L=120 km
C’b=12.32 nF/km
cos ϕE = 0.9lag
SE=100 MVA
R' = 0.079 Ω/km
X’b = 0.289 Ω/km
a. Calculate the transmission losses PR for following voltages at the receiving-end of the
line (the load SE remains constant at the value given above). (The line capacitance Cb.
i. Un = 220 kV
ii. Un = 110 kV
b. The rated line voltage (Un) is 110kV. Calculate and compare the transmission losses if
the actual voltages on the line change as follows. Neglect Cb = 0.
i. Ub = 120 kV
ii. Ub = 100 kV
iii. Ub = 90 kV
c. A 140-MVA transformer with adjustable taps is connected to the receiving-end of a 220kV line (A – E). The transformer has the following rated values:
a = 231kV/115.5 kV, xps =13.6%.
Assume the voltage at the 110-kV bus-bar is to be kept constant at the rated value using
the following two options. In one case the voltage UA is increased to compensate the
voltage drop but the transformer tap remains at nominal position (231/115.5kV). In the
second case the transformer tap setting is changed to an = 231/127 kV to keep the
voltage at the load at 110 kV. Calculate and compare the line real and reactive losses
(PR and Qx) in both cases. (The line capacitance is to be considered, and in both cases
the line delivers 100 MVA at cosϕE = 0.9 lag).
A
E
p
220 kV
s
110 kV
46
Exercises
Electric Power Systems I - III
4.4
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Transmission losses II
A three-phase line has the following parameters and rated values.
L=120 km
C’b ≅ 0
cos ϕE = 0.7lag
SE=100 MVA
R' = 0.079 Ω/km
X’b = 0.289 Ω/km
UE=220 kV
For this line, the influence of the power factor (cos ϕE) on line transmission losses is to be
investigated.
a. Derive an expression relating the line real power losses to the power flow (real and
reactive power flows).
b. Give the losses for the given power flow both as an absolute value and as a percentage
of the real power at the receiving-end.
c. Determine the losses if cos ϕE is improved through parallel compensation to
0.8/0.9/0.95.
d. Determine the losses at the various power factors in (c) due only to the active/reactive
current.
47
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
4.5 Maximum power transmission and steady state stability
Consider the transmission system shown in the figure below in which a generator supplies power
to a grid via a transformer and an overhead line. The generator is modelled as a voltage source
in series with a reactance xd. The value of the generator internal voltage is assumed to be
u p = 1.0 p .u . The transformer and the overhead line are both represented with their reactance.
The reactances have the following values: xd = 1 p.u., xps = 0.1 p.u., xL = 2.0 p.u.
Generator
Transformer
N
∼
jxd
up
jxps
uG
jxL
Transmission
line
uN
The grid is considered to be an infinite bus, i.e. both the amplitude and the phase angle of the
voltage uN are constant and independent of the power flow. Assume u N = 1.0∠0 0 p .u .
a. Using basic power equations derive an expression for the maximum possible real power
pmax which the generator can deliver to the grid.
b. What is the power factor corresponding to the real power determined in (a) at the grid
terminals (bus N).
c. Now, the reactance of the transmission line is to be compensated by inserting a series
capacitor into the line in such a way that the transmission system maintains a 30 %
reserve (stability margin) when the generator delivers pmax (calculated in (a)) to the grid.
What is the power factor at the grid interconnection point corresponding to this load?
48
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
4.6 Voltage stability
U2
U1
Load:
S2= P2+jQ2=200 MW+j70 Mvar
Grid
Overhead line:
(infinite bus)
Zb= R+jX=10 +j100 Ω
The voltage at the sending-end of the line (U1) is 400 kV.
a. Find an implicit expression describing U 2 as a function of S2, i.e. derive the expression:
*
U 2 = U1 − Z b ⋅
S2 / 3
(1)
*
U2
b. Take the voltage U2 as the phase reference, i.e. set U2 = U2 and then solve (1) analytically, i.e.
derive the relationship:
U 21,2
2 A − U 12
= −
±
2
(2 A − U )
2 2
1
4
− A2 − B 2
(2)
where:
A=(RP2+XQ2)/3
B=(XP2-RQ2)/3
c. Plotting (2) using the given values results in the following curve. As can be seen from the plot,
there are two values for U2 corresponding to P2 = 200 MW. Which of the two values corresponds to
a stable operating point?
49
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400
350
Line to line voltage [kV]
300
250
200
150
100
50
0
0
100
200
300
Load [MW]
400
500
600
Voltage at the receiving-end of the line as a function of load power
d. Indicate how the phase angle of U1 at the given load and the value of U2 determined in (c) can be
calculated.
50
Exercises
Electric Power Systems I - III
4.7
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Calculations on a 750-kV long-range overhead transmission line
A 750-kV overhead line has the following values:
Rated voltage
Un = 750 kV
Length
l = 1100 km
Phase conductors
Four ACSR bundles, each with 400 mm2
cross-section
Resistance (of the bundle)
R = 0.0198 Ω/km
Reactance (of the bundle)
Xb = 0.289 Ω/km
Susceptance (of the bundle)
ωCb = 4.12 10-6 S/km
Conductance (of the bundle) Gb = 0.4 X 10-6 S/km
(for rainy weather)
a. Calculate the wave impedance (ZW), the propagation coefficient, the wave length, the
wave propagation velocity for the line.
b. Determine the following three equivalent circuits and compare the circuit parameters:
i. The short line approximation
ii. The nominal π-equivalent circuit
iii. The exact π-equivalent circuit.
c. Determine the sending-end voltage, the real and reactive losses when the line delivers
PE = 2000 MW at 750 kV and unity power factor at its receiving-end.
d. Calculate the voltage midway along the line when the line delivers power corresponding
to the wave impedance load at its receiving end at the rated voltage.
e. Derive a formula for the resulting wave impedance of the line (as a function of the
capacitor) when a series capacitor is inserted into the line at an equidistant point
between the sending and receiving ends of the line.
Hint:
Solution of the wave equation for sinusoidal voltages and currents:
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U A ( x ) = UE cosh( γx ) + IE Z w sinh( γx )
I A ( x ) = IE cosh( γx ) +
UE
sinh( γx )
Zw
Wave (surge) impedance
Zw =
Zb
Yb
=
R + jX b
G b + j ωC b
Propagation coefficient:
γ = α + jβ = Z b Y b =
(R + jX b )(Gb + jωC b )
52
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FG Elektrische Anlagen und Netze
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4.8 Wave impedance, propagation coefficient, wave velocity and travel time
An overhead line has the following parameters:
L =1.1 mH/km
C = 12.25 nF/km
l = 500 km
The line is connected to a cable with the following parameters:
L =0.45 mH/km
C = 0.25µF/km
l = 50 km
Determine:
a.
The wave impedance (ZW), propagation coefficient (χ), and travel time (τ) both for the
overhead line and the cable
b.
The reflection and refraction coefficients for the voltage as well as for the current at the
junction overhead line – cable.
53
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Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
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4.9 Traveling waves
A DC source voltage U0 with an internal resistance Rs is connected to a lossless line having a
wave impedance Rc. The line is terminated with a resistance RR. The travel time of the wave
along the line is τ.
A
B
Rs
Ri
U0
RR
a. Draw a lattice (Bergeron) diagram showing the voltage and the current in the line for the
period from t = 0 to t = 10τ.
b. Determine the receiving-end voltage (the voltage at B) at t= 0, 2τ, 4τ, and 6τ, and hence
at t=2nτ, where n is any positive integer.
c. Determine the steady-state voltage at the receiving-end of the line in terms of U0, Rs,
RR, and Rc.
d. Verify the result in part (c) by analyzing the system as a simple DC circuit in steadystate. (Note that the line is assumed to be lossless; note also how inductances and
capacitances behave in a DC circuit.)
e. Assuming U0=100kV, Rc=10 Ω, find the numerical values for the steady-state voltage at
B, if:
i. Ri =0 Ω, RR=1000 Ω
ii. Ri =10 Ω, RR=1000 Ω
iii. Ri =0 Ω, RR→ ∞
iv. Ri =10 Ω, RR→ ∞
54
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4.10 Effect of a short-circuited line on a traveling wave
Zc
Ri
RR
U0
150 km
400 km
The generator has an internal resistance Ri = 0, and the transmission line is assumed to be
lossless. The elements shown in the figure have the following values: U0 = 4kV, RR = 80Ω, Zc =
100Ω, wave velocity v = 3.105km/s. The generator is equipped with a relay that trips
instantaneously when the current reaches 250A.
The transmission line shown in the figure above is short-circuited at t = 0 at a location 150 km
away from the generator.
55
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FG Elektrische Anlagen und Netze
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4.11 Traveling waves
Two lines with wave impedances of 100 Ω and 500 Ω, respectively, are connected in series, as
shown in the figure below. The travel time of the wave in lines 1 and 2 are τ and 2τ, respectively.
While the far end of line 2 is open-circuited, line 1 is connected (at its sending-ending) to a
voltage source of 100 kV amplitude, which is assumed to remain constant during the period
under consideration.
A
Z1=100Ω
τ
B
Z2=500Ω
2τ
C
U=100kV
a. Sketch the lattice diagram in the space provided below for the period 0 ≤ t ≤ 12τ,
indicating all reflections and refractions.
A
0
B
C
x
1
2
3
4
5
6
7
8
9
10
11
12
t/τ
56
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b. Using the Bergeron method, calculate and then plot against time voltages and currents
at points A, B and C for 0 ≤ t ≤ 12τ.
57
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5 GENERATORS
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5.1 Parameters and characteristic values of generators I
A cylindrical rotor synchronous generator has the following ratings and parameters:
Sn
= 125 MVA
xd
= 1.8 p.u.
Cosϕn
= 0.8 lag
x’d
= 0.264 p.u.
Tm
= 6.31s
x”d
= 0.156 p.u.
Un
= 10.5 kV
a. Specify the operational conditions under which each of the reactance given in the above
table is valid! Sketch the corresponding equivalent circuits and indicate the ranges of
their validity.
b. Determine the generator excitation voltage (Polradspannung), the sub-transient and
transient voltages for no load and rated load pre-fault operating conditions.
c. Assume the generator is synchronized to a large network. After synchronization, with
the excitation remaining unchanged, the turbine input power is slowly increased to 50%
of the generator rated power. Neglecting the armature resistance, determine the power
angle (Polradwinkel) and the power factor at which the generator is operating. Is the
generator over excited or under excited?
d. For a three-phase fault occurring at the generator terminals, calculate (neglecting the
generator resistance):
i. the steady-state (Dauerkurzschlussstrom)
ii. the transient and
iii. the sub-transient short-circuit currents.
59
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5.2 Control variables in a synchronous generator
A cylindrical rotor synchronous generator is connected to an infinite bus as shown in the figure
below.
G
~
Generator
Bulk power system
The synchronous reactance and the rated values of the generator are as follows:
Sn = 1111 MVA
xd = 2.8 p.u
Ra = 0
Un = 24 kV
Cosϕn= 0.95 lag
a. Assume the generator is synchronized to the network as shown in the figure above and
delivers PG = 1050 MW at a power factor of 0.95 lagging. The voltage at the terminals of the
generator is maintained at the rated value.
Calculate:
−
The excitation (generated) voltage
−
The power angle
−
The maximum power (Pmax)
b. The mechanical power is now reduced by 20% compared to the setting in (a). The excitation
current remains unchanged.
Calculate:
−
The excitation (generated) voltage
−
The power angle
−
The maximum power (Pmax)
−
The power factor at the terminals of the generator.
c. The mechanical input power is restored to the value in (a), whereas the degree of excitation is
reduced by 10 % (with the value in (a) as a base).
Calculate:
−
The excitation (generated) voltage
−
The power angle
−
The maximum power (Pmax)
−
The power factor at the terminals of the generator.
Is the generator now overexcited or under-excited? Justify your answer.
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5.3 Parameters and characteristic values of generators II
In continuation of Exercise 5.1:
a. Derive a mathematical relationship, which relates the rotor speed during a short-circuit
at the generator terminals and determine the rate of change of speed per second. Prior
to the fault, the generator was supplying the rated load.
b. Determine the negative-sequence current, if - during operation at the rated load - one of
the phase conductors experiences an interruption. As a simplifying approximation, the
other two phase currents are assumed to maintain their pre-fault values. Would this
level of continuous negative sequence current load be permissible?
c. Neglecting the effects of any controller action, determine the theoretical maximum
steady state and transient output power. Assume the power increase takes place from
both no load and rated load initial conditions.
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5.4 Limits of capacitive loading for a generator
The cylindrical rotor synchronous generator considered in Exercises 5.1 – 5.2 is now connected
to a remote infinite bus through an overhead line and transformer (see figure below). In addition
to the generator data given under Exercise 5.1, the following transformer and line parameters are
known:
Transformer xps
Line
= 0.15 p.u.
aTr
= 10.5/380 kV
Sn
=125 MVA
C’b
= 14 nF/km
G
a. Sketch the equivalent circuit of the generator connected to a transformer and an
overhead transmission line operating on no load. Neglecting the series impedance of
the line and the transformer as well as the shunt conductance of the line, describe how
the system behaves for Xd ≤ Xc and Xd > Xc. Deduce from this the limit of the maximum
permissible capacitive loading, if generator self-excitation is to be averted! Why is selfexcitation a dangerous phenomenon?
b. Calculate the length of a 400-kV overhead line, which would cause the generator to
self-excite. How would this value change if the transmission line is a double-circuit line?
(Assume the breaker at the far end of the line to be open.)
62
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5.5 Operational behavior of synchronous generators
The terminal characteristic of the synchronous generator considered in Exercises 5.1 – 5.3 is to
be studied in this exercise. Recall that the rated values of the machine are:
Sn =125 MVA
Un = 10.5 kV
Cosϕn = 0.8 lag
xd = 1.8 p.u
a. Show that the per unit voltage at the terminals of the generator under load is given by
the following relationship:
u = up
 x ⋅i
1−  d w
 up

2

 − x d ⋅ ib ,


up= generator internal emf;
(where u = terminal voltage;
components of the load current, respectively;
iw, ib = real and reactive
xd = synchronous reactance. All
values in per unit.)
b. Assuming that the excitation as well as the real part of the load current are held at their
respective rated values, show that the terminal voltage as a function only of the reactive
current is given by the relationship:
u = 2.08 - xd . ib
c. With the excitation remaining unchanged, derive a relationship (similar to the one
derived in part (b)), which relates the real current (instead of the reactive current) to the
terminal voltage. The reactive current is now assumed to remain constant at the rated
value.
d. Based on the relationship given in part (a), small voltage changes around the operating
point arising from real and reactive current variations can be determined using the
relationship:
 ∂u
∆u = 
 ∂i w
0
0
 ∂u 

 ⋅ ∆ib = c w ⋅ ∆i w + c b ⋅ ∆ib
 ⋅ ∆i w + 

 ∂ib 
Calculate the coefficients cw and cb, and confirm that the influence of the reactive
component of the load current (ib) on the terminal voltage (u) is significantly higher than
that of the real current (iw), i.e. c b > c w .
e. Determine the necessary change in the generator excitation (generator internal emf up)
as a function of the change in the reactive current, if the terminal voltage is to be held
constant at the rated value.
63
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FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
5.6 Generator loading capability diagram
For this exercise the parameters given in Exercise 5.1 are to be used.
a. The generator is assumed to be operating on an infinite bus ( U = U∠0° = const. ). Derive
an expression, which relates the armature current to the rotor angle (δp) and the degree
of excitation (ε = Up/U).
b. Show how the active and reactive powers can be derived from the current loci and
determine the power scale of the diagram. Also, derive from the current relationship
characteristic curves of the active power vs. rotor angle (P=f(δp)) und reactive power
versus rotor angle (Q = f(δp)).
c. The loading capability diagram (also known as the operation chart) for this generator is
to be drawn in the following steps:
Step 1: Draw the phasor diagram of the machine.
Up =
Un
3
∠0 0 − jX d ⋅ Ia = 6.06 kV∠0 0 − j1.59 Ω ⋅ 6.87 kA∠143.13 0 Ω
Up
jXdIa
Ia
Diagram 1: Phasor Diagram
Ua
Rotating the mirror image (with respect to the imaginary axis) of the above diagram by 900 gives the following diagram.
64
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jXdIa
P
Ia
Up
Ua
Diagram 2: Operating Chart
i.
Point P represents the rated operating point. Starting from P, draw the constant
excitation circle, the circle for constant armature current, the locus of operating points
for constant real power (P), and the locus of operating points for constant reactive
power (Q). Additionally, locate the locus of operating points for unity power factor and
constant power factor.
ii.
Modify Diagram 2 by multiplying (re-scaling) all lengths by U/Xd. The locus of P (real
power) and Q (reactive power) is a circle having the form: (x-a)2+(y-b)2=r2. Determine
the center of the circle (a, b) and its radius (r).
iii.
Finally, draw the load chart of the generator in the following steps:
•
Along the P axis mark the point for maximum power of the turbine, which is
PTmax≈1.1 Pn
•
Mark the circular MVA arc corresponding to the maximum permissible armature
current (Ia,max=1.2 Ia,r)
•
Construct the arc corresponding to maximum permissible excitation, which in this
case corresponds to ε = 3.2.
•
Make a reasonable assumption with regard to the under-excitation limit and mark it
on the diagram to complete the loading chart.
65
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FG Elektrische Anlagen und Netze
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5.7 Steady-state and transient stability
The following figure represents a section of a power system
380kV
G
220kV
G
Sk“ =13.5GVA
G
G
Generators:
Transformers:
Lines:
Sn = 235 MVA
Un = 15.75 kV
Cos ϕ = 0.85lag
xd = 1.88
x’d = 0.285
Tm=11.54 s
Sn = 235 MVA
Xb = 0.25Ω/km
410/15.75 kV
(per system)
xps = 1.88
l = 300km
Transformers:
Sn = 630MVA
400/231kV
xps = 13%
The initial condition is as follows: rated real power, 0.95 lagging power factor and rated voltage at
the generator terminals.
a. Determine the voltage phase angle difference between network voltage and the
generator internal emf (power angle δ). Identify the available stability reserve.
b. Derive a mathematical expression for the generator internal emf as a function of the
power angle (δ), if the real power output of the generators is slowly increased from the
rated value to the stability limit:
i. without generator terminal voltage control
ii. with the generator terminal voltage held constant
c. How does the steady-state stability margin change, if, instead of the double line, one
system is used to transmit the power given in part (a):
i. without generator terminal voltage control
ii. with the generator terminal voltage held constant ?
d. Determine the transient power versus power angle characteristic of the generator!
66
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e. Assume now that from a situation where both lines were operating in parallel, one of the
lines is switched off. Using the equal area criterion, determine the maximum swing
angle.
f.
Assuming rated initial condition, determine the critical clearing time for a three-phase
fault occurring at the terminals of the generator.
67
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6
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TRANSFORMERS
68
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FG Elektrische Anlagen und Netze
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6.1 Transformer dimensioning
A single phase transformer 50 Hz with circular core (see figure) has the following data :
Primary voltage
Up =
110 kV
Secondary voltage
Us =
10 kV
Rated power
Sn =
4.4 MVA
Maximal core flux density
B=
1.8 T
Maximum current density in conductors
Jmax = 4 A/mm2
Space factor of winding
f=
66 %
Specific resistance of copper
σ=
2.14 10-8 Ω.m
hw
400
30
40
50
40
The mechanical construction principle is shown in the above figure (distances in mm).
a. Calculate the number of turns required on both the primary and the secondary sides.
b. Calculate the cross-section of the copper wire for both the primary and the secondary
windings.
c. Should the inner coil (close to the core) be used for primary or secondary voltage?
Why?
d. Calculate the copper losses of the transformer at nominal power.
e. Calculate the height hW of the windings (see figure).
f. What is the per unit value of the short circuit voltage uk of this transformer ?
Hint: µ0 = 4π 10-7 H/m
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6.2 Determination of transformer equivalent circuit parameters
The no-load and short-circuit tests carried out on a 15/0.4-kV, 400-kVA transformer provided the
following results:
Test
Current
Power
Applied voltage
Open-circuit
Open-circuited/
short-circuited side
low voltage side
0.2 A
19.2 W
rated voltage
Short-circuit
high voltage side
rated current
7 kW
28 V
(line to line voltage)
Determine:
a. the parameters of the equivalent circuit referred to the high voltage side
b. the parameters of the equivalent circuit referred to the low voltage side
c. the efficiency of the transformer, when the transformer supplies half the rated power at a
power factor of unity (cos phi =1).
70
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6.3 Working in per unit quantities
Generator
T1
Transmission
line
∼
UG: 13.2 kV
5 MVA
13.2 kV ∆/132 kV Y
10%
2+j10Ω
T2
Load
10 MVA
300+j0 Ω
138 kV Y/69kV ∆
8%
Use a common base of 138 kV and 10 MVA.
a. Draw the equivalent circuit and determine all impedance values in per unit
b. Calculate the currents through T1, the transmission line, T2 and the load both in per unit
and Amperes
c. Determine the power consumed by the load both in per unit and MVA.
71
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6.4 Transformers in parallel operation
Two transformers (T1 and T2) with the following rated values are to be operated in parallel.
Transformer
Rated power
voltage
uK
Copper loss
T1
400 kVA
24/0.4 kV
6%
6720 W
T2
160 kVA
24/0.4 kV
4%
1792 W
a. Determine the load on each transformer (in percent of the rating of each transformer)
when transformers T1 and T2 in a parallel operation jointly supply the load current IL = 600
A.
b. Determine the maximum load current which does not lead to the overloading of any of the
transformers
c. Assume now the attached load draws power corresponding to the rated power of both
transformers at a power factor of 0.85 lagging (i.e., 400 kVA + 160 kVA = 560 kVA, cos
phi = 0.85 lag).
i.
Determine IT1 (the current through T1) and IT2 (the current through T2).
ii. The solution in (i) will show that T2 is overloaded by over 30% while T1 is not yet
fully loaded. Now determine the tap change (in percent) on transformer T1 which
would reduce the load on T2 to its rated value (with the total load current attached
to both transformers remaining unchanged).
iii. What is the new load current on T1?
iv. Show that the current through the external load remains at the value before the tap
change.
72
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6.5 Impedance transformation
Three identical single-phase transformers have each the following rated values:
1.2kV/220 V, Sn=7.2 kVA, xps = 0.05 p.u., R = 0
The transformers are to be connected for three-phase operation. A symmetrical three phase load
with an impedance 5 + j 0 Ω/phase is connected to the secondary terminals. Determine the total
impedance referred to the primary as well as to the secondary sides both in Ohms and p. u. for
the following transformer connections:
a. Y-Y
b. Y-∆
c. ∆-Y
d. ∆-∆.
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6.6 Equivalent circuit of three phase MV/LV transformer
A 1-MVA MV/LV transformer has the following characteristic data:
Primary voltage
Unp =
10 kV
Secondary voltage
Uns =
0.4 kV
Rated power
SnT =
1000 kVA
Short circuit voltage
uk =
6%
Copper losses
Pk =
10 kW
No load current
IL=
0.7 A /10kV
No load power
PL=
2.3 kW
Vector group
Dy5
a. Sketch the connection of the windings indicating the direction of flux linkages in each
winding.
b. Calculate the leakage impedance and the shunt admittance referred to the primary side
c. Give the per phase equivalent circuit neglecting the shunt branch
d. By using the π-equivalent circuit (instead of the one obtained in (c)), the need for
referring impedances to one side or the other can be obviated. Derive such a πequivalent circuit of this transformer in the following steps:
i. Determine the nodal bus admittance matrix of the circuit obtained in (c) and substitute
any referred quantity with its original value
ii. Sketch an arbitrary π circuit with unknown parameters and formulate the general
relationships for the elements of its nodal bus admittance matrix in terms of the
parameters of the π circuit.
iii. Solve for the parameters of the π circuit in (ii) by equating the corresponding elements
of the two bus admittance matrices (obtained in (i) and (ii)).
74
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6.7 Unsymmetrical load on a transformer I
The following single-phase load occurs in a three-phase, 25MVA, 231/115.5, Yy0 transformer:
2.5 MVA, 110 / 3 kV, cos ϕ = 1
Star points
Primary
Secondary
Open
Solidly grounded
a. Calculate the currents in the primary windings without using symmetrical components.
b. Discuss the effect on the transformer of the unsymmetrical loading.
c. Assume that the transformer contains an additional delta-connected tertiary winding.
What are now the currents in all three sets of transformer windings?
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6.8 Unsymmetrical load on a transformer II
The following load occurs on the transformer described in exercise 6.7:
Secondary load:
cos ϕ = 1
Symmetrical load:
5 MVA, 110 kV,
Single-phase load (in Phase a):
1 MVA, 110 kV/ 3 , cos ϕ = 1
Tertiary load
Symmetrical load:
1 MVA, 10 kV,
cos ϕ = 1
For the purposes of this exercise, the transformer is to be considered ideal.
a. Determine the primary currents IA, IB, IC.
b. Verify the results obtained in (a) using the relationship:
S(PRIMARY)= S(SECONDARY)+S(TERTIARY).
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6.9 Zero-sequence equivalent circuit of transformers
The value of the zero-sequence impedance of a star-connected transformer winding is essentially
determined by the connection of the star point. Derive the zero-sequence equivalent circuit of
transformer windings in the following connection:
a. Star, Star point open;
IA0
ZP
A
B
C
b. Star, Star point solidly grounded;
IA0
ZP
A
B
C
c. Star, Star point grounded through impedance Zn;
IA0
ZP
A
B
C
Zn
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d. Primary as in (c), secondary: delta connected.
IA0
ZP
A
B
C
Zn
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6.10 Equivalent circuit of a three winding transformer
A three winding transformer used for power station supply has the following characteristic data:
Primary voltage
Unp =
20 kV
Secondary voltage
Uns =
6.3 kV
Tertiary voltage
Unt =
6.3 kV
Primary rated power
Snp =
40 MVA
Secondary rated power
Sns =
20 MVA
Tertiary rated power
Snt =
20 MVA
Short circuit voltage prim.-sec.
ukps=
8.7 %
Short circuit voltage prim.-tertiary
ukpt=
8.7 %
Short circuit voltage sec.-tertiary
ukst=
15 %
Resistance prim.-sec.
rps=
0.525 %
Resistance prim.-tertiary
rpt=
0.525 %
Resistance sec.-tertiary
rst=
0.88 %
No load current
IL=
4.62 A / 20kV
No load losses
PL =
29 kW
a. Sketch the positive-sequence equivalent circuit of the transformer.
b. Calculate the values of the elements of the equivalent circuit referred to the primary,
secondary and tertiary voltage levels.
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6.11 Behavior of a transformer under load
A three winding transformer for power house load supply has the following characteristic data:
Rated voltages
Un =
231/115.5/10.5 kV
Rated apparent power
Sn =
140/140/42 MVA
Relative losses
Pv/Sn
1.1 10-2
No load current
iL
1%
Impedance prim.-sec.
zps
13.6 %
Copper losses prim.-sec.
rps
0.4 %
a. Sketch the equivalent circuit valid for nominal operation of the transformer with no load
on the 10-kV side.
b. Sketch the phasor diagram of primary voltage Up = f (cosϕ), cosϕ = 0ind ..1 .. 0cap for
nominal values of voltage and current on the secondary side.
c.
Derive then from this curve a diagram for the relative voltage drop up – us = f (cosϕ) for
cosϕ = 0ind ..1 .. 0cap
d. What is the equation giving the total losses of n identical transformers operated in
parallel as a function of the total current Ig ? Determine the current Iu at which switching
from n to n-1 transformers should take place so that the total loss is minimized?
e. The transformer under consideration is to be operated in parallel with a 250-MVA
transformer having the values: Un=231/115.5kV, zps = 15.5 %, rps = 0.3%. Calculate:
i. the load carried by each transformer for a total load of 390 MVA (S1n + S2n)
ii. the ratio of copper loss to the rated power for each of the transformers the
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6.12 Behavior of a three winding transformer under unsymmetrical load
231 kV
10.5 kV
115.5 kV
IA
a
IB
b
IC
c
Ib
Ic
The following short circuit currents were measured during a double-phase-to-ground fault
occurring on a three-phase, three winding, 231/115.5 kV /10.5 kV transformer (see figure above).
For the purposes of this exercise, the transformer can be considered ideal.
Sort-circuit currents
Ia
0
Ib
5400 A ∠ 0°
Ic
5400 A ∠ 240°
Give a general expression for the short circuit currents on the 220 kV side in general as well as
their concrete values both in the natural and the symmetrical components.
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6.13 Zero-sequence network of a power system including transformers
The zero-sequence network of a section of a power system is to be determined.
a. Which vector groups of transformers permit the transmission of zero-sequence systems
between primary and secondary sides?
b. Comment whether or not this transmission of zero-sequence systems is desirable.
c. Sketch the zero-sequence the equivalent circuit of the power system given in the figure
below and calculate the values of the impedance elements.
4 units
Sn=235 MVA
Un=231/15.75 kV
xps=14.5%
S“k=2500 MVA
Z0→∞
B
Sn=235 MVA
Un=15.75 kV
x‘‘d=19%
15 kV
100 km
t
110 kV
s
~
p
Z11+Z’11=j0.32 Ω/km
Z00+Z’00‘=j1.12 Ω/km
S“k=4500 MVA
C
220kV
50 km
220kV
50 km
t
4 transformers
Sn=250/250/83 MVA
Un=231/115.5/10.5 kV
xps=15.5%
xpt=14.8%
xst=9%
s
30 km
2 transformers
t
110kV
p
s
380kV
p
D
Z0=j50 Ω
S“k=10 000 MVA
Sn=630/630/210 MVA
Un=400/231/31.5 kV
xps=10.5%
xpt=16.6%
xst=11.3%
A
S“k=3000 MVA
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6.14 Parallel operation of transformers
Two transformers, Transformer 1 and Transformer 2, having the power ratings Sn1 and Sn2,
respectively, are to be operated in parallel.
a. If the two transformers jointly carry the load S, determine the complex power through
Transformer 1 (S1) and Transformer 2 (S2) in relation to their short-circuit voltage (uk).
b. Show that the two transformers share the load in an ideal manner, i.e. in the
ratio
S1 S n1
=
, only if they have equal short circuit voltage.
S 2 S n2
c. Determine the load in each of the parallel connected transformers and the circulating
current, if any, occurring under each of the following conditions and decide, on the basis
of the results obtained, if parallel operation is permissible or not. (In each case, assume
equal R/X ratio.)
i. different rated powers :
Transformer 1
Dy5
S = 45 MVA
uk = 10%
a= 110/10
Transformer 2
Dy5
S = 35 MVA
uk = 10%
a= 110/10
ii. different short circuit voltages :
Transformer 1
Dy5
S = 40 MVA
uk = 10%
a= 110/10
Transformer 2
Dy5
S = 40 MVA
uk = 12%
a= 110/10
iii. different transformation ratios :
Transformer 1
Dy5
S = 40 MVA
uk = 10%
a= 110/10
Transformer 2
Dy5
S = 40 MVA
uk = 10%
a= 120/10
iv. different vector groups :
Transformer 1
Dy5
S = 40 MVA
uk = 10%
a= 110/10
Transformer 2
Yy6
S = 40 MVA
uk = 10%
a= 110/10
83
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6.15 Voltage magnitude and phase angle control of transformers
Two buses of a system (Bus A and B) are connected through two parallel lines (Line 1 and 2).
The lines have the following impedances:
Line 1 = j 0,1 p. u.; Line 2 = j0,1 p. u.
Bus B is a load bus with a load current of I = 1,0 p. u. ∠ − 30° . The voltage in bus A is so adjusted
that the voltage at Bus B remains always constant at U = 1.0 p. u. ∠0° . Calculate the real and
reactive power absorbed by the load and the share of Line 1 and 2:
a. For the situation described.
b. When a regulating transformer, installed at Bus B end of the Line 2, injects an emf with
a value 0.03 p.u.
c. When the transformer introduced in (c), instead of the voltage magnitude, causes the
voltage phase angle to increase by 20.
84
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6.16 Autotransformer
A single-phase transformer has the following ratings:
30 kVA, 10,6 kV/0,4 kV.
This transformer is reconnected as an autotransformer for 11 kV/10,6 kV operation.
a. Sketch the connection of the windings.
b. Find the rating of the transformer as an autotransformer and the currents in both
windings.
c. The efficiency of the transformer before the reconnection and under rated conditions at
a load power factor of 1 was 97%. What is the efficiency of the transformer as an
autotransofrmer?
d. What would be the rating of the transformer for 11/0.4 kV-connection?
e. Compare the results in (b) und (d) and deduce from this comparison the most suitable
field of application for an autotransformer.
f.
Assume now that the polarity of the windings in the interconnection for 11 kV/10.6 kV
operation was wrong (the primary and secondary windings were not connected additive
cumulatively). What is the effect on the transformer of the wrong connection?
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6.17 Transformer in open delta connection
Three identical single-phase transformers, each rated 5 kVA, are connected in ∆-∆ to supply a
rated load of 15 kVA at 380 V. It turned out later that the actual load on the transformer never
exceeds 10 kVA and that only two of the three single-phase transformers would be enough to
supply the load. Accordingly, the transformer between the phases a and c is to be removed and
the transformer to be operated in open ∆. The operation of the transformer under this mode and
at a load of 10 kVA and cosϕ=1 is to be investigated.
a. Show that the three line-to-line voltages at the load terminal remain symmetrical in the
open ∆ mode!
b. Calculate the currents in the primary and secondary windings and in the input and
output lines.
c. What is the power transferred by each of the transformers?
d. Is there any additional constraint that needs to be introduced, when a ∆-∆ connected is
modified for an operation in open ∆ ?
86
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6.18 Effect of primary fault on secondary voltages
A symmetrical voltage source with its star point solidly grounded is to be connected to the primary
side of a 10/0.4-kV, core type transformer, as shown in the figure below. The arrangement of the
primary and secondary windings is also given in the figure.
A
a
B
b
C
c
Primary
Secondary
a. Assuming that the primary and secondary windings are connected as follows,
Connect only the middle phase (phase B) of the transformer primary winding to the
corresponding phase of the voltage source.
i.
Determine the secondary voltages Ua, Ub and Uc and the primary voltages UA and UC
on open-circuit
ii. Calculate the secondary currents, if a star-connected load impedance with its star
point grounded and having a value of 10+j0 Ω /phase is connected to the secondary
terminals
iii. Comment on the possibility of power transfer between any one phase (in this case
phase B) of the primary winding that is connected to the source and the three
secondary windings.
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b. Assuming that the secondary winding remains connected as in (a), disconnect phase B
of the primary from the source and connect phases A and C to the source. What
voltages can be measured on open-circuit at the secondary terminals, if:
i. The primary star point remains solidly grounded
ii. The primary star point is not grounded?
c. With the secondary winding connection remaining unchanged, now the connection of
the primary windings is converted to delta. What secondary voltages (on open-circuit)
can one measure, if on the primary side the connection linking the beginning of the coil
phase A to the end of phase B is removed, as shown in the figure below? Would this
transformer be able to supply a three-phase load connected to the secondary terminals?
Explain.
A
B
C
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6.19 Transformer in „Scott“ connection
Two single-phase transformers in a so-called „Scott“ connection supply a two phase load from a
three-phase supply (figure below).
A
IA
B
IB
N1
U2
3 N1
N2
N1
C
IC
Transformer 2
N2
U1
Transformer 1
Make the following assumptions:
-
The transformer is to be considered ideal
-
The supply line has a line-to-line voltage of 6 kV
-
N1/N2=13
a. What are the primary voltages in transformers 1 and 2?
b. What are the secondary voltages U1 and U2 on no load?
c. Assume now that transformer 1 supplies a 50 kVA load at a power factor of 0.8 lagging
at the rated voltage. Transformer 2 is unloaded. Determine the currents IA, IB, and IC in
the input line?
d. With the load on transformer 1 remaining unchanged, load transformer 2 with 50 kVA at
0.8 lagging and rated voltage. Re-evaluate the currents IA, IB, and IC in the input line?
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6.20 Instrument transformers (IT)
The following figure shows the schematic diagram of connection for a current transformer (CT)
and a potential transformer (PT), which collectively are called instrument transformers.
Primary circuit
A
CT- connection
V
PT- connection
a. If a CT is accidentally open-circuited, it would momentarily be destroyed, and similarly if
a PT were inadvertently to be short-circuited, it would be damaged. Explain why.
b. The total impedance on the CT due to lead resistance and metering load (known as
“burden”) strongly affects the measurement accuracy.
i. Give the equivalent circuit of the CT, neglecting the primary resistance and leakage
reactance; Comment on the justification of neglecting the primary impedance.
ii. Using the equivalent circuit, relate the measurement error (both magnitude and angle)
to ZB (burden).
iii. A CT with the ratio 250:5 A is known to have the following parameters (referred to the
secondary side):
Zs = 0.03 + j0.05 Ω
Z’m = 7 + j10 Ω
For a primary current of 150 A, determine the magnitude and angle error when the
burden on the CT is:
iii-1) 0.06 + j0 Ω;
iii-2) 0.12 + j 0 Ω;
iii-3) 3+j0 Ω
c. In practical cases Zm does not remain constant as a result of iron core saturation. In this
exercise the effect of the core saturation is to be considered. The following pair of
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values are points taken from the magnetization characteristic of the 250:5 CT
considered above.
I’e/A
0
0,28
0,55
1
4
6.5
8
Es/V
0
5.7
11.3
11.4
11.45
11.5
12
For a primary current of 200 A, compute the magnitude and phase angle errors first
neglecting core saturation and then considering the core saturation and compare the
results obtained. Make the following assumptions: ZB = 3 + j 0 Ω; the impedance angle of
Z’m is assumed to remain constant at 550.
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7 POWER FLOW ANALYSIS
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7.1 Power flow solution by Gauss – Seidel method
a. Assume the following simultaneous equations are given:
2x + xy = 1
2y - xy = -1
Solve for x and y using:
i. The Jacobi method
ii. The Gauss – Seidel method
iii. The Gauss – Seidel method using an acceleration factor of 1.6.
b. A power flow analysis using the Gauss – Seidel method is to be carried out for the
network, the one-line diagram of which is given below. All other necessary data are also
given in the tables following the diagram.
1
2
3
4
Line data (Referred to the 230-kV side)
From
To
R
(Ω)
X
(Ω)
Y/2
(mS)
1
2
5.34
26.66
0.097
1
3
3.91
19.68
0.073
2
4
3.91
19.68
0.073
3
4
6.72
33.64
0.121
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Bus data
Generation
Bus
Bus type
P
(MW)
1
Slack
-
2
Load
3
Load
4
Voltage controlled
Load
Q
(Mvar)
-
P
(MW)
50
Q
(Mvar)
30.99
Voltage
(kV)
230
0
0
170
105.35
230
0
0
200
123.94
230
318
-
80
49.58
234.6
Var limit at bus 4: Qmin = -200 Mvar; Qmax =50 Mvar
i. Assemble the nodal bus admittance matrix of the network
ii. Formulate the power flow equations
iii. Solve the power flow equations using the Gauss – Seidel method for the first iteration.
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7.2 Power flow analysis by Newton – Raphson method
a. Using the Newton – Raphson method, solve for x and y of the following nonlinear
equations:
F1(x, y) = 4y sin x + 0.6 = 0
F2(x, y) = 4y2 – 4y cos x + 0.3 = 0
Initial values: x(0) = 0 , y(0) = 1
b. Consider again the power system given in exercise 7.1 (b). With the objective of
carrying out a power flow study using the polar form of the power flow equations,
determine:
i. the number of rows and columns of the Jacobi matrix
ii. algorithms for the determination of the elements of the Jacobi matrix and the initial
elements of the matrix
iii. the initial power mismatch and give a brief outline of the solution procedure
iv. the first iteration solution to the power flow problem.
c. Derive the algorithm for the load flow study using the decoupled form of the Newton –
Raphson method and determine the first iteration solution.
d. Solve the power flow equations using the DC power flow model.
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7.3 Regulating transformer and load sharing I
Two transformers (Ta and Tb) are connected in parallel to supply a load of 0.8 + j 0.6 p.u. at
a voltage of U2 = 1.0∠00 p.u. Transformer Ta has a voltage ratio equal to 1 p.u. on both
primary and secondary sides on no load. The tap on the second transformer (Tb) is
adjusted to yield a secondary voltage of 1.05 p.u. on open-circuit. Both transformers have
an impedance of j0.1 p.u.
a. Find the complex power transmitted to the load through each transformer using:
i. the circulating current method (approximate solution)
ii. the nodal bus admittance matrix (Ybus) method (exact solution)
b. Repeat (a) except that Tb now includes a transformer having the same turns ratio as Ta
but a regulating transformer with a phase shift of 30.
c. Show how the presence of a phase-shifting transformer causes the [Ybus] of the entire
network to become unsymmetrical.
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7.4 Regulating transformer and load sharing II
Two Transformers, Transformer a and Transformer b, having per unit reactances 0.2 and 0.4,
respectively, are operated in parallel. Transformer a operates at nominal tap-setting, while the tap
on transformer b can be changed under load.
Assume a load connected to the secondary terminals of the parallel connected transformers
draws 1.05 p.u. at a power factor of 0.707 lagging and rated voltage. Determine the primary
voltage and the complex power delivered by each of the transformers, when the transformation
ratio of transformer b is:
a. 1 p.u. at a phase angle of zero, i.e. nominal setting
b. 1.05 p.u. at a phase angle of zero
c. 1.0 at a phase angle of 30
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7.5 Representation of the regulating transformer in a load flow study
Consider the power system considered in exercise (7.1). Assume now that a regulating
transformer is inserted between bus 3 and the load, as shown in the figure below. The
variable tap is on the network side of the transformer and the reactance of the transformer
(referred to the 230-kV) side is 105.8 Ω. The resistance is to be neglected. A value of 1 is to
be used as an initial estimate for t (t is the ratio of the actual tap-setting to the nominal tapsetting).
The variable tap (t) of the transformer is to be regarded as a state variable in a load flow
study using the Newton – Raphson method. The voltage magnitude at bus 5 is specified to
be 225 kV
a. Write the mismatch equations in a symbolic form
b. Write equations for the jocabian elements of the column corresponding to variable t (the
partial derivative with respect to t) and evaluate them using the initial estimates.
c. Write equations for P and Q mismatches at buses 4 and 5 and evaluate them for the
first iteration.
1
2
3
4
5
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8 SHORT CIRCUIT ANALYSIS
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Summary of important short-circuit terminology:
Initial short circuit current (Ik“)
Short circuit capacity (Sk“)
Equivalent voltage source at the fault location ( U " =
c ⋅ Un
3
)
The first amplitude of the short-circuit current (ip) and the surge factor (κ)
i p = κ ⋅ 2 ⋅ I "k
κ = 1.02 + 0.98 ⋅ e −3 R / X
DC component of the fault current (id.c.)
i d.c. = 2 ⋅ I "k ⋅ e −2 π⋅f ⋅R ⋅t / X
Voltage factor c;
High- and medium voltage: cmax=1.1;
cmin=1.0
Low voltage:
cmin=0.95
cmax= 1.05;
Short-circuit interrupting current (Ib) and the decay factor (µ)
I b = μ ⋅ I"k
Distance short circuit: μ
=1
Short-circuit in the vicinity of a generator:
µ = 0.84 + 0.26 ⋅ e −0.26⋅I
µ = 0.71 + 0.51 ⋅ e −0.30⋅I
"
kG
"
kG
/ I rG
for t min = 0.02s
/ I rG
for t min = 0.05s
µ = 0.62 + 0.72 ⋅ e −0.32⋅I
"
kG
/ I rG
for t min = 0.10s
µ = 0.56 + 0.94 ⋅ e −0.38⋅I
"
kG
/ I rG
for t min ≥ 0.25s
"
t min = Minimum delay; I kG
= Generator short circuit current;
I rG = Genertator rated current
Short-circuit interrupting power(Sb)
Steady-state short-circuit current (Ik)
Short-circuit duration (Tk)
Thermal equivalent short-circuit current (Ith) and the parameters m and n
m: thermal component of the dc current; f(κ, Tk)
n: thermal component of the ac current; f(Ik“/Ik, Tk)
I th = I "k ⋅ m + n
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8.1 Comparison of a short circuit in R-L circuit and a generator
A 50-Hz alternating voltage having an RMS value of 215 V is supplied to a series R – L circuit by
closing the switch (S) in the figure below. The resistance and the inductance values are15 Ω and
0.27 H, respectively.
a. Find the value of the dc component of current upon closing the switch if the
instantaneous value of the voltage at switching is 100 V.
b. What is the instantaneous value of the voltage, which will produce the maximum dc
component of current upon closing the switch?
c. What is the instantaneous value of the voltage, which will result in the absence of any
dc component of current upon closing the switch?
d. If the switch is closed when the instantaneous voltage is zero, find the instantaneous
current 0.5, 1.5, and 5.5 cycles later.
e. Neglecting any possible DC component in the short-circuit currents, how does the shortcircuit current versus time curve in an R – L circuit differ from a typical time behaviour of
a current flowing when a synchronous generator is short-circuited at its terminals? How
can this difference be explained?
S
R =15 Ω
U =215V
L =0.27 H
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8.2 Symmetrical short-circuit in a generator – transformer unit
A generator connected to a transformer is rated 100 MVA, 18 kV, with reactances of xd” = 19%,
xd’ = 26%, and xd= 130%. A three-phase transformer rated 100 MVA, 220 Y/10.5∆ kV, xps= 10%
is connected to the generator. The generator was operating at no load and rated voltage when a
three-phase short circuit occurs between the transformer and its generator side breaker. Find (in
kA):
a. the sustained short-circuit current in the breaker,
b. the transient, the sub-transient and the initial symmetrical RMS currents
c. the maximum possible dc component of the short-circuit current in the breaker.
If the location of the short-circuit is assumed to be the high-voltage side of the transformer, find
d. the initial symmetrical RMS current in the transformer windings on the high voltage side
e. the initial symmetrical RMS current in the transformer on the low voltage side
f.
the breaker interrupting current for a minimum time delay of 0.1 s.
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8.3 Short-circuit current computation: internal emf method vs. equivalent voltage source
method I
A 50-Hz generator is rated 500 MVA, 20 kV, with xd”= 0.2 per unit. Prior to the fault the generator
was supplying a purely resistive load of 400 MW at 20 kV connected directly across the terminals
of the generato. If all three phases of the load are short-circuited simultaneously, find the initial
symmetrical RMS current in the generator, using:
a. the internal emf method
b. the equivalent voltage source method (Thévenin theorem).
.
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8.4 Short-circuit current computation: internal emf method vs. equivalent voltage source
method II
A synchronous generator is rated 30 MVA, 13.2 kV, and has the sub-transient reactance of 20%.
This generator is supplying a group of motors connected via a line having a reactance of 10% on
the base of the machine ratings. The motors are drawing 20 MW at 0.8 power factor leading and
a terminal voltage of 12.8 kV when a symmetrical three-phase fault occurs at the motor terminals.
The equivalent sub-transient reactance of the motors is assumed to be 20 % (on the base of the
generator rating).
a. Find the sub-transient currents in the generator, the motor, and the fault by using the
internal voltages of the machines.
b. Solve this problem using the Thévenin theorem. Take the pre-fault load into
consideration. How does the pre-fault load current affect the current flowing into the fault
location?
104
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8.5 Short-circuit in a generator connected to a group of motors I
A 625-kVA generator with xd” = 0.20 per unit (base: 625 kVA, 2.4 kV) is connected to a bus
through a circuit breaker, as shown in the figure below. Connected through circuit breakers to the
same bus are three synchronous motors rated 187.5 kW, 2.4 kV, 1.0 power factor, 90%
efficiency, with xd” = 0.2 per unit (base: 187.5 kVA, 2.4 kV). The motors are operating at full load,
unity power factor and rated voltage, with the load equally divided among the machines.
a. Draw the impedance diagram with the impedances in per unit on a base of 625 kVA, 2.4
kV.
b. Find the symmetrical short-circuit current in amperes, which must be interrupted by
breakers A and B of a three-phase fault at P. Simplify the calculations by neglecting the
pre-fault current.
c. Repeat part (b) for a three-phase fault at point Q.
d. Repeat part (b) for a three-phase fault at point R.
R
A
Q
P
B
105
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8.6 Short-circuit in a generator connected to motors II
A 25-MVA, 13.8-kV generator with xd” = 15% is connected through a transformer to a bus which
supplies four identical motors, as shown in the figure below. The sub-transient reactance xd” of
each motor is 20% on a base of 5 MVA, 6.9-kV. The three-phase rating of the transformer is 25
MVA, 13.8/6.9 kV, with a leakage reactance of 10%. The bus voltage of the motor is 6.9 kV when
a three-phase fault occurs at point P.
Determine
a. the sub-transient current in the fault,
b. the sub-transient current in breaker A.
P
A
106
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8.7 Symmetrical short-circuit current computation
110kV
G1
T3
220kV
220kV
T6 110kV
T1
T4
T5
T2
G2
T7
Q
2X240/40-Al/St
T9
G9
T8
50km,
double line
SS2
SS3
240/40-Al/St
50km,
double line
SS1
SS4
Data of the network components:
Xb’=0.298 Ω/km (220-kV overhead line; for the bundle in each system)
Xb’=0.393 Ω/km (110-kV overhead line; for each system)
Overhead line 220 kV: R/X = 0.26
Overhead line 110 kV: R/X = 0.3
G1, G2:
10.5 kV; 100 MVA; xd“ =0.16; RsG/Xd“=0.05;
G9:
21 kV; 225 MVA; xd“ =0.19; RsG/Xd“=0.05;
T1, T2:
120 MVA; uk =10%; ür=115 kV/10.5 kV
T3 …T8:
200 MVA; uk =12%; ür=240 kV/110 kV
T9:
250 MVA; uk =10%; ür=112 kV/21 kV; R/X = 0.03
Q:
SkQ“=20 GVA
a. Calculate the initial short-circuit current (Ik“) fort the fault location shown (SS4).
b. Compute the first amplitude of the fault current.
c. Calculate the fault interrupting current for the maximum interrupting time delay tv = 0.25 s
d. Calculate the maximum dc component of the short-circuit current
e. Estimate the significance of considering the resistive components of all impedances.
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8.8 Mechanical and thermal stress caused by the short-circuit current
21 kV
G
110 kV
T
G:
UrG = 21 kV; SrG = 225 MVA; xd“ = 0.18; xd = 2; R/X= 0.05
T:
SrT = 250 MVA; uk = 10%
The generator is connected to the transformer through rectangular-shaped aluminium (Al) bars.
The expected load current necessitates the use of three Al bars, each with the dimension 200
mm X 15 mm. All the bars are arranged in a horizontal plane, with the centre-to-centre distance
between the phases (the main distance) am = 350 mm and the distance between the conductors
within each phase (sub distance) as =15 mm. The following schematic shows the bars for one of
the three phases.
1m
Side view
a. Which type of fault causes the maximum mechanical stress on the Al-bars?
b. Determine the combined force exerted on the bars of the middle phase by the fault
currents flowing in the other two phases for the fault location on the generator side
terminals of the transformer (shown above). Identify the appropriate curve from the family
of curves given in Figure 1 and then obtain the correction factors for each of the mutual
distances appearing in your formula. (Correction factors are needed to determine the
effective distance when conductors having a large dimension are used.)
c. Additionally, what is the force exerted on the outer bar of the middle phase due to the
currents flowing in the other bars of the same phase?
d. Neglecting the phase difference between the forces obtained under (b) and (c), calculate
the total force exerted on the mounting during the short-circuit. Is the force vertically or
horizontally directed?
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e. Determine whether or not the Al bars are capable of withstanding the thermal stress due
to the short-circuit current, if the generator protection clears the fault 0.2 s after the onset
of the fault. (In accordance with DIN VDE 0102, the sustained short-circuit current (Ik) for
this system is assumed to be 176% of the rated generator current. (The bars are designed
for a short-circuit current density (Sthr) of 87 A/mm2, which leads to a temperature rise
from the operating temperature of 650C to the maximum permissible short-circuit
temperature of 2000C within 1 s). Use Figures 1 and 2 to determine the factors m and n.
109
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k1j
kij
Figure 1: Correction factor for main distance (am) and sub distance (as)
110
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Figure 2
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Tk.f
Parameter: I“k/Ik
I“k/Ik = 1
n
Tk
Figure 3
f.Tk
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8.9 Calculation of short circuit currents in a meshed network
A short-circuit analysis is to be carried out for the network given below
F1
G1
G2
F2
~
L1
~
110 kV
T1
T3
T4
T2
F3
A
B
L3
30 kV
kV
L4
T5
T6
F4
10 kV
kV
L2
F5
M
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Data:
S”k = 4000 MVA
G1, G2: 31.5 MVA, 6.3 kV, x”d = 0.12
T1, T2 : 31.5 MVA, 31.5/6.3 kV, xps = 0.12
T3, T4: 31.5 MVA, 110/31.5 kV, xps = 0.12
T5, T6: 16 MVA, 31.5/10.5 kV, xps = 0.11
M : 12 MW, 10kV, cos ϕ =0.8, x’M = 0.3
L1 : X’b = 0.4 Ω/km, 15 km
L2 : X’b = 0.4 Ω/km, 2.5 km
Calculate the initial symmetrical short circuit current I”k for the fault locations F1 ... F5, and
identify the contributions by the generator (I”kG), the grid (I”kN) and the motors (I”kM). All reactances
should be referred to the 110-kV level. The currents should also be given in their original voltage
level. For fault locations F1 and F2, the motors can be neglected. Additionally neglect the
impedances of cables L3 and L4 in all cases.
113
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8.10 Short Circuit Current Limitation
a. List and discuss the available options for short circuit current limitation.
b. Discuss the effectiveness of the two alternatives measures for reducing short-circuit
currents given below. For fault location F1 and F2 the contribution of the motors towards
the fault current can be neglected:
◊ Measure 1: Coupling between SS A and B open.
◊ Measure 2: Use of inductances for short circuit current limitation with parallel Is (surge
current)-limiter between transformer T3 and bus bar A, as well as T4 and SS B
respectively, with bus bar coupling between SS A and B closed. Rated data of short
circuit current limitation inductances : xD=10%, Un=30kV In=630 A
114
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8.11 Analysis of unsymmetrical faults
A synchronous machine is connected through a three-phase transformer and transmission line to
a large system, as shown in the Figure 1 below:
G1
T1
1
3
2
Sk“= 10 GVA
X0→ ∞
Xn
Fig. 1. Network
Ratings and parameters of the synchronous machine, the transformer and the line are as follows:
Generator G1: 100 MVA, 13.8 kV;
Xd” = X1 = X2 = 0.19 Ω, X0 = 0.067 Ω, Xn = 0.095 Ω
Transformer T1: 100 MVA, 13.8 kV ∆/380 kV Y; Xps = 115.5 Ω (referred to the 380-kV side)
Transmission line: 100 MVA, 380 kV
X1 = X2 = 21.66 Ω ; X0 = 72.2 Ω
F
F
A
B
A
B
C
C
IA
IB
IC
IA
IB
IC
E
E
a) Symmetrical fault
b) Single line to ground (SLG) fault
F
F
A
B
A
B
C
C
IA
IB
IC
IA
IB
IC
E
c) Double line to ground (DLG) fault
E
d) Double line (DL) fault
Fig. 2. Fault types
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For various fault locations, the fault current and the voltages at the lines not directly affected by
the fault are to be computed. The fault types to be considered are shown in the Fig. 2 (fault types
(a) – (d).) and the pre-fault current is to be neglected in all cases.
a. Formulate the fault conditions for all four fault types.
b. Transform the fault conditions (formulated in (a)) into symmetrical components, and
then deduce from the result the interconnection of the sequence networks for each fault
type.
c. For a bolted (Zf = 0) single line-to-ground fault occurring on phase A at bus 3, determine
the following:
i.
the sub-transient current to ground at the fault location
ii. the line-to-ground voltages at the fault location as well as at the terminals of the
generator
iii. the sub-transient current out of phase c of the generator.
d. Instead of the fault described in part (c), assume now a bolted line-to-line fault at bus 3
between phases B and C. Determine the fault current at the fault location and the
currents in the lines affected by the fault at the fault location and the line-to-line voltages
at the fault bus.
e. Find the sub-transient fault currents and the line-to-line voltages at the fault when a
double line-to-ground fault with Zf = 0 occurs at the terminals of the generator (bus 1).
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8.12 Effect of transformer neutral point connection on system performance during fault
The following 220/380 kV system is given:
4 units
Sn=235 MVA
Un=231/15.75 kV
xps=14.5%
Sk‘‘=2500 MVA
Z0→∞
B
Sn=235 MVA
Un=15.75 kV
x‘‘d=19%
15 kV
100 km
t
110 kV
s
~
p
Z11’+Z11’=j0.32 Ω/km
Z00’+Z00’=j1.12 Ω/km
S‘‘k=4500 MVA
220kV
50 km
C
50 km
220kV
t
4 transformers
Sn=250/250/83 MVA
Un=231/115.5/10.5 kV
xps=15.5%
xpt=14.8%
xst=9%
s
30 km
Z11 = j0.16 Ω/km
Z00 = j0.56 Ω/km
t
p
2 transformers
Data same as station B
110kV
s
380kV
p
D
Z0=j50 Ω
S‘‘k=10 000 MVA
full transformer or auto-transformer
Sn=630/630/210 MVA
Un=400/231/31.5 kV
xps=10.5%
xpt=16.6%
xst=11.3%
A
S‘‘k=3000 MVA
Initially, the transformer in substation D is assumed to be an auto-transformer, with the star point
being solidly grounded. For a fault at the 220-kV bus bars of substation A (as indicated in the
diagram), calculate:
a. the initial single-phase (I‘‘k(1)) and three-phase (I‘‘k(3)) short-circuit currents
b. the earth-fault factor cf
c. the conductor-to-ground voltages of the unaffected phases during a single phase-toground fault using the earth-fault factor (cf) for each of the following three alternatives:
i. for the given case (i.e. star points of all transformers on the 220-kV side including the
star point of the auto-transformer in substation D solidly grounded)
ii. with the star point of the auto-transformer in substation D remaining grounded, only
one of the 220-kV star points in each of the substations A, B, and C grounded
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iii. as in (ii), but the transformer in substation D is now a full transformer and only the star
point on the 220-kV side is solidly grounded (the 380-kV winding is also star
connected and the star point is isolated)
d. the neutral point to ground voltage (UEM) of the isolated transformers in substations A, B,
and C for the scenario described under alternative c (iii) above.
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8.13 Neutral point grounding in medium voltage networks
110 kV
S“k=2500 MVA
t
p
20 kV
s
20-kV Network
100-km Overhead line
30-km Radial field cable
→XN=0.176 Ω
ZME
xps=1.26Ω
xpt=1.8Ω
xst=0.51Ω
(Referred to the 20-kV side)
IC=0.02 A/km (Overhead line)
IC=1.47 A/km (Cable)
(IC=charging current)
A 20-kV distribution network is being supplied by a 110-kV network via a 40-MVA transformer.
The transformer reactance and the charging currents in the distribution network (referred to the
20-kV side) are given in the figure above. The positive and negative sequence reactances of the
20-kV are can be neglected against the much larger transformer reactance.
a. For a single line-to-ground fault at the 20-kV bus, draw the sequence networks! Place
the total capacitance (C00→CE) of the distribution network at the 20-kV bus.
b. Assuming the neutral point of the transformer on the 20-kV side is isolated (ZME→∞),
determine the single line-to-ground fault current (I‘‘k(1)). (Use as a voltage source: Upa1 =
UNn/ 3 )
c. Assuming now resonant grounding (instead of isolated neutral point) for the
transformer:
i. Determine the reactance of the grounding coil (Petersen coil) for resonance tuning
ii. Draw the zero-sequence network assuming a damping factor of 4% and a detuning
(off-resonance factor) of -3%. Neglect the line reactance.
iii. Neglecting the positive, negative and zero-sequence reactance of the transformer,
calculate the single line-to-ground fault current (I“k(1)) for the damping and offresonance factors given under (ii) above.
d. Assume now that the resonant grounding is replaced by an impedance grounding.
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i. Determine the value of the grounding reactance, if the single line-to-ground fault
current (I“k(1)) is not to exceed 1000 A. Use a source voltage of (Upa1 = 1.05 UNn/ 3 )
and neglect the effects of the zero-sequence capacitance and damping resistance.
ii. Estimate the influence of the zero-sequence capacitance on the fault current.
e. The 20-kV winding of the transformer is now assumed to be solidly grounded (ZME=0).
i. Determine the fault current neglecting the zero-sequence line capacitance and
damping.
ii. How would the zero-sequence capacitance affect the result obtained in (i)?.
(The source voltage (Upa1) is now assumed to be = 1.1 UNn/ 3 .)
f.
Discuss briefly the results obtained in (a) – (e) in relation to the magnitude of the fault
current and identify the grounding option, which is best suited for this network.
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8.14 Problem of voltage un-symmetry in three phase transmission systems
For a 110-kV overhead transmission line, the effect of varying shunt admittance in each phase of
the line on the voltage symmetry is to be investigated.
a. The shunt elements of the line (line capacitance and conductance) are assumed to have
the following relationship:
G aE = G bE = G cE
C aE + ∆C = C bE = C cE
Derive an expression, which relates the capacitive un-symmetry to the neutral point to
ground voltage (UEM) of the line.
For which value of v (off-resonance factor) is the voltage un-symmetry maximum?
b. Re-work part (a), when the shunt elements have the following relationship:
G aE + ∆G = G bE = G cE
C aE = C bE = C cE
c. Repeat part(a) for the case, where all the three capacitances are different from one
another, i.e.
G aE = G bE = G cE
C aE ≠ C bE ≠ C cE →
C aE = C aE
C bE = C aE + ∆C bE
C cE = C aE + ∆C cE
d. The following values are obtained from a control measurement on the transmission line:
UEM (at v=0) = 11.6 kV∠2450
ICE = 395 A
d (damping coefficient)= 2.5%
Determine the capacitances CaE, CbE and CcE.
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8.15 Low impedance grounding in cable networks
The following cable loop represents part of a metropolitan distribution network.
110 kV
t
p
A20-1
s
B1
3 km
ZME
C1
4 km
2 km
D1
20 kV
S“k
3km
3 km
p
s
3 km
CB
B2
A20-2
t
C2
D2
CB: Circuit Breaker
Parameters of the transformers:
Cable parameters (for all sections):
Un = 110/20 kV
Sn = 40 MVA
Xps = 1.26 Ω
Xpt = 1.8 Ω
Xst = 0.51 Ω
(All reactance referred to the 20-kV side)
Zb=0.15+j0.1 Ω/km
Z0=0.1+j0.64 Ω/km
(Return current flows through cable screen and earth.)
Further, the following data/assumptions are given:
◊ The total line-to-earth capacitance of the 20-kV network in the zero-sequence network is
to be placed at the 20-kV bus, and to be calculated from the capacitive earth leakage
current (ICE), which has the following values:
ICE = 51A in bus A20-1 or A20-2
ICE = 85A in A20-1 with the circuit breaker to the left of bus B2 open
(needed for exercise part (b))
◊ Neglect the line capacitance in the positive and negative sequence networks
◊ Use a source voltage of Upa1 = 1.05 ⋅
UNn
3
during fault.
◊ The equivalent impedance of the 110-kV network is 0.176 Ω (referred to 20-kV side).
a. For a single line-to-ground fault in bus A, calculate the fault current I“k(1) for both
resistance grounding (RME =12 Ω) as well as reactance grounding (XME =12 Ω) for the
following alternative scenarios:
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UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
i. The coupling switch in bus bar A open, switch at bus D open, the switch connecting
the groundings of the two transformers open
ii. The coupling switch in bus bar A open, switch at bus D open, the switch connecting
the groundings of the two transformers closed
iii. The coupling switch in bus A closed, switch at bus D open, grounding not connected
to one another
b. A single line-to-ground fault is now assumed to occur in B2. The following additional
information is given: bus bar coupling in A open, transformer groundings not connected
to one another, switch at bus D closed, circuit breaker (to the left of B2) open. Calculate
the single line-to-ground fault current for both resistance grounding (RME = 12 Ω) and
reactance grounding (XME = 12 Ω).
c. Discuss the results obtained in (a) and (b) with regard to:
i. The effectiveness of the grounding
ii. The effect of the cable impedance on the fault current
iii. The effect of the line capacitance on the fault current
iv. Fault current in relation to grounding impedances, RME and XME.
123
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.16 Transient stability analysis I
An elementary principle of dynamics states that:
J⋅
d 2 δm
dt 2
= Ta
(1)
where:
J = the total moment of inertia of the rotating masses in kg.m2
δm = the angular displacement of the rotor with respect to a stationary axis in
mechanical radians
Ta = the net accelerating torque in N.m.
On the basis of (1) and neglecting the damping torque, derive the swing equation (2),
which describes the behaviour a synchronous generator subjected to a disturbance.
d 2δ
dt
2
=
π⋅ f
⋅ ( p m − pe )
H
(2)
where:
δ
= the angular displacement of the rotor (in electrical radians) with respect to a
reference frame rotating at synchronous speed
H = the inertia constant in seconds
pm = shaft mechanical power in per unit
pe = electrical output power in per unit
124
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.17 Transient stability analysis II
a. A generator in a thermal power plant has the following values:
Pn = 150 MW, cos ϕn= 0.85, Un = 15 kV, n = 1500 rpm, J =75 ·103 kgm2 (the total
moment of inertia of the rotating masses).
Calculate:
i. the stored kinetic energy of the rotating masses at nominal speed
ii. the inertia constant H (in seconds)
iii. the inertia constant M (in joule-seconds/mechanical radian).
b. A generator with the rating of 100 MVA has an inertia constant H = 8.0 s.
i. What is the stored kinetic energy of the rotating masses when the generator runs at
synchronous speed?
ii. At what speed would a 40-tonne lorry need to travel in order for the lorry to have this
amount of kinetic energy?
iii. What should the total mass of a train travelling at 100 km/h be to have the same
kinetic energy?
c. A 4-pole, 100-MVA, 11-kV, 50-Hz generator has an inertia constant of 8.0 s.
i. What angular acceleration does the rotor experience, if the mechanical power input
increases in a step from the initial value of 50 MW to 80 MW?
ii. What values would the rotor angle and rotor speed assume if this acceleration were
to be maintained constant at this value for 10 cycles?
d. A 50-Hz generator in a power system supplies 1 p.u. power into an infinite bus via
transformers and transmission lines. A fault in one of the transmission lines reduces the
maximum transferable power from the pre-fault value of 2.0 p.u. to 0.5 p.u. If the
maximum transferable power upon clearance of the fault is 1.5 p.u, what is the critical
fault clearing angle?
125
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.18 Transient stability analysis III
a. Assuming a pre-fault generator load of 1 p.u, calculate the critical fault clearing angle for
a solid three-phase fault occurring at point P in the system given below. The fault is
cleared by disconnecting the faulted line in both ends.
j0.15
j0.15
j0.1
j0.15
j0.28
UN=1.0
j0.15
∼
E=1.2
j0.15
j0.14
j0.14
Infinite bus
j0.15
b. A solid three- phase fault is assumed to occur at point P in the system given below. All
the necessary data are given in the figure and the system frequency is 50 Hz. The fault
is cleared by disconnecting the faulted line in both ends. What are the critical fault
clearing angle and critical fault clearing time for a pre-fault generator load of 1 p.u.
j0.15
j0.1
UN=1.0
j0.5
j0.05
∼
E =1.2 p.u
H=4s
j0.4
P
Infinite bus
c. The generator in the figure below has an inertia constant H = 4 s and feeds 1 p.u. power
into the infinite bus. Further data for the system are: UN = Ut = 1.0 p.u., xd’ = 0.25 p.u., xt =
0.10 p.u., xl = 0.5 p.u., f = 50 Hz.
xd’
xt
xl
UN
∼
Up’
Ut
xl
Infinite bus
i. What is the value of the emf (Up’) behind the transient reactance?
ii. Calculate the maximum power that can be transmitted if:
− the topology of the system is as shown in the figure.
− When one of the lines experiences a three-phase fault midway along the line.
i.
What is the accelerating power immediately after the fault described in (ii)?
ii. What is the value of the rotor angle if the acceleration is maintained constant at the
value calculated in (iii) for 0.05 s?
iii. What is the accelerating power at the rotor angle calculated in (iv)?
iv. Discuss the difference between the values obtained in (iii) and (v)!
126
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.19 Transient stability analysis IV
The following single-line diagram shows a generator connected through two parallel transmission
lines to a large metropolitan system considered as an infinite bus.
All relevant constants and parameters of the system are given in the single-line diagram and in
the table following it, and resistances are to be neglected in all cases.
G
1
T1
3
2
T2
Infinite bus
H=5s
H→∞
Equipment
Generator
Transformers
(T1 and T2)
Lines
Positive and negative
sequence reactance
0.12
0.10 (each)
Zero sequence
reactance
0.05
0.10 (each)
0.2 (each)
0.5 (each)
Prior to the fault, the machine was delivering 1.4 per unit power, with both the generator terminal
voltage and the infinite bus voltage maintained at 1.0 per unit.
a. Symmetrical fault
A symmetrical fault is assumed to occur at the sending-end terminal of one of the lines, as shown
in the single-line diagram. The fault is cleared 0.25 s after its inception by simultaneously opening
the circuit breakers at both ends of the faulted line. Subsequently, a successful re-closure takes
place after a further 0.5 s, which restores the system to its original configuration.
Using the equal area criterion, determine whether or not the generator remains in synchronism.
b. Double line-to-ground fault
Starting from the same pre-fault condition as (a) a double line-to-ground fault occurs at the same
location described in (a). Find the equation relating the electrical output power to the power angle
(P vs. δ curve) and show that, for this particular fault, the generator would not lose synchronism,
even if the fault were not to be cleared. Determine the maximum angle, which the generator will
reach, in the process of swinging subsequent to this fault.
Hint: for numerical integration of the swing equation, use the following algorithm:
127
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
δn = δn −1 + ∆δn (in electrical radians )
∆δ n = ∆δ n −1 + k ⋅ Pa,n −1
k=
π⋅f
(∆t )2
H
Pa, n = Pm, n − Pel, n
(where pa = accelerating power; pm = input mechanical power; pel = output electrical power)
128