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ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 1 : THREE PHASE CIRCUITS - SOLUTION
Question 1.
A balanced star connected load of (2+j3) Ω per phase and a balanced delta-connected load of (6+j6) Ω per phase are connected in parallel to a three phase, 400V, 50Hz supply. Calculate the phase current in each load, the total current in each supply line, the total power supplied and the overall power factor.
Solution. The circuit diagram for the parallel three phase load is shown below:
VR
VY
VB
IR
IY
ZstarIB
IBR
IYB
IRY
IRstar
IYstarIBstar
Zstar
Zstar
Z∆ Z∆
Z∆
IR∆
IY∆IB∆
The three line to line voltages are defined as:
°∠= 30400VVRY °−∠= 90400VVYB °−∠= 210400VVBR
So for the star load the three phase voltages are defined as:
°∠= 03
400VVR °−∠= 120
3
400VVY °−∠= 240
3
400VVB
The star impedances are :
⎟⎠
⎞⎜⎝
⎛∠=+= −
2
3tan1332 1jZstar
The star line currents are then given by:
star
phsstar
Z
VI =
( )°−∠=
°∠×°∠==
−31.5605.64
23tan133
04001
AV
Z
VI
star
RRstar
ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
2
( )°−∠=
°∠×°−∠==
−31.17605.64
23tan133
1204001
AV
Z
VI
star
YYstar b
( )°−∠=
°∠×°−∠
== − 31.29605.6423tan133
2404001
AV
Z
VI
star
BBstar
The real and reactive power delivered to the star circuit is then given by:
( ) ( )( ) ( ) kWIVP starphsstar 62.2431.56cos05.644003cos3 =°== φ
( ) ( )( ) ( ) kVarIVQ starphsstar 92.3631.56sin05.644003sin3 =°== φ
The delta load impedance is given by:
°∠=⎟⎠
⎞⎜⎝
⎛∠=+= − 45266
6tan2666 1jZdelta
For the delta load the phase to phase current is given by:
∆
∆ =Z
VI LL
°−∠=°∠°∠=∆ 1514.47
4526
30400AIRY
°−∠=°∠°−∠=∆ 13514.47
4526
90400AIYB
°−∠=°∠
°−∠=∆ 25514.474526
210400AIBR
The delta line currents are given by:
°−∠== −∆∆ 4565.813
030 AeII jRYR °−∠=∆ 16565.81 AIY °−∠=∆ 28565.81 AIB
The real and reactive power delivered to the delta circuit is given by:
( ) ( )( ) ( ) kWIVP phsphsphs 4045cos14.474003cos3 =°== −∆∆ φ
( ) ( )( ) ( ) kVarIVQ phsphsphs 4045sin14.474003sin3 =°== −∆∆ φ
The overall input line current is given by:
°−∠=
°−∠+°−∠=+= ∆
97.490.145
4565.8131.5605.64
A
AAIII RRstarR
°−∠=°−∠+°−∠=+= ∆
97.1690.145
16565.8131.17605.64
A
AAIII YYstarY
°−∠=°−∠+°−∠=+= ∆
97.2890.145
28565.8131.29605.64
A
AAIII BBstarB
Total Power Consumption:
( ) ( )( ) ( ) kWIVP LLLtot 62.6497.49cos1454003cos3 =°== φ
( ) ( )( ) ( ) kVarIVQ LLLtot 92.7697.49sin14.474003sin3 =°== φ
The overall power factor is: ( ) 6432.097.49cos.. =°=fp
ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
3
Question 2. A three phase, 4 wire, 415V 50Hz supply has balanced voltages and rotation sequence RYB. The
following loads are all connected to the supply:
(a) A single resistance of 12 Ω between the R line and Neutral.
(b) An inductive impedance of (2 + j8) Ω between the B line and Neutral.
(c) A capacitor of 120µF between the R line and the Y line.
(d) A three-phase, delta connected induction motor (balanced load) taking a total power of 10kW at 0.75 p.f. lagging.
Calculate:
(i) the magnitude and phase angle of the current in each of the four lines coming from the supply.
(ii) The total power taken from the supply.
Solution. VR
VY
VB
IR
IY
IB
IBR
IYB
IRY
IRstar
IBstar
12ΩZMot
IR∆
IY∆IB∆
ZMot
ZMot
VNIN
(2+j8)Ω120µf
Since the unbalanced load has a neutral point connected, then the star point voltage is fixed at the 0V reference point. There is hence a neutral current flowing that needs to be calculated.
The three line to line voltages are defined as:
°∠= 30415VVRY °−∠= 90415VVYB °−∠= 210415VVBR
So for the star load the three phase voltages are defined as:
°∠= 03
415VVR °−∠= 120
3
415VVY °−∠= 240
3
415VVB
First calculate the R-phase current due to the load in part (a):
00
_ 09667.19123
0415 ∠=∠== AR
VI RN
starR
The current in the B-phase inductor in part (b) is given by:
ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
4
( )0
0
_ 0362.440558.29823
240415∠=
+−∠
== AjZ
VI BN
starB
The RY current in the capacitor in part c is given by:
( )( )0
0
_ 1206451.151205021
30415 ∠=∠== AfjZ
VI
c
RYcapRY µπ
For the inductive load the real power is 10kW at 0.75pf lagging. So the power factor angle is:
( ) 01 41.4175.0cos == −φ
So the reactive power is given by:
( ) kVarPQ 192.841.41tan000,10tan 0 === φ
Since:
∗=+= LphsIVjQPS 3
Then the three line currents feeding the induction machine load are given by:
°−∠=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
°∠+=
∗
∆ 41.4155.1803415
8192000,10_ A
jIR
°−∠=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
°−∠+=
∗
∆ 41.16155.181203415
8192000,10_ A
jIY
°∠=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
°∠+=
∗
∆ 59.7855.181203415
8192000,10_ A
jIB
So the four supply line currents become: 0
___ 82.209.26 ∠=++= ∆ AIIII RcapRYstarRR
0__ 63.11677.21 −∠=+−= ∆ AIII YcapRYY
0__ 39.5756.45 ∠=+= ∆ AIII BstarBB
0__ 31.2657.45 ∠=+= AIII starBstarRN
The total power taken from the supply is given by:
kVarjkWIVIVIVS BBYYRR 080.9473.16 +=++= ∗∗∗
ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
5
Question 3. Three unequal impedances are connected in star without a neutral connection to a 415V, 50Hz,
three phase supply with a standard RYB rotation. Assuming that the supply voltages are balanced, calculate the shift of the star point voltage from the neutral current in each of the lines. Take VRN as the reference. Check that the three currents add to zero.
ZRN = 45 + j0 Ω
ZYN = 45 – j20 Ω
ZBN = 30 + j30 Ω
Solution. The figure below shows the general circuit diagram for a star connected load.
VR
VY
VB
IR
IY
IB
ZA
ZYZB
VS
To calculate the load currents it is necessary to calculate the star point voltage. This requires a
reference point, and the voltage VRN can be taken for this point. This voltage is exactly 3/1 times the line to line voltage which is precisely known. So we can write:
0=++ BRYBRY VVV
0=++ BYR III
Also:
°∠=°∠==°°
02403
30415
3 3030V
ee
VV
jjRY
RN
°−∠=°−∠==°°
1202403
90415
3 3030V
ee
VV
jjYB
YN
°−∠=°−∠==°°
2402403
210415
3 3030V
ee
VV
jjBR
BN
Then:
R
SNRNR Z
VVI
−= , Y
SNYNY Z
VVI
−= , B
SNBNB Z
VVI
−=
Summing the three currents gives:
ELEC 4100 TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
6
B
SNBN
Y
SNYN
R
SNRNBYR Z
VV
Z
VV
Z
VVIII
−+−+−==++ 0
Or:
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
⎭⎬⎫
⎩⎨⎧
++B
BN
Y
YN
R
RNSN
BYR Z
V
Z
V
Z
VV
ZZZ
111
Now note that:
Ω= 45RZ
Ω=Ω−= °− 96.2324.492045 jY ejZ
Ω=Ω+= °4543.423030 jB ejZ
So:
Se
j
ee
eeZZZ
j
jj
jjBYR
0
00
00
34.8
4596.23
4596.23
05806.0
008419.005745.0
02357.002031.002222.0
43.42
1
24.49
1
45
1111
−
−
−
=
−=++=
++=⎭⎬⎫
⎩⎨⎧
++
Also :
0
000
0
0
0
0
0
0
6.5
28504.960
45
240
96.23
120
0
0
304.6
6163.0284.6
656.5874.4333.5
43.42
240
24.49
240
45
240
j
jjj
j
j
j
j
j
j
B
BN
Y
YN
R
RN
e
j
eee
e
e
e
e
e
e
Z
V
Z
V
Z
V
=
+=++=
++=⎟⎟⎠
⎞⎜⎜⎝
⎛++
−−
−
−
−
Then:
VjVee
e
ZZZ
Z
V
Z
V
Z
V
V
jj
j
BYR
B
BN
Y
YN
R
RN
SN
17.264.10559.10805806.0
304.6
111
00
95.133.8
6.5
+===
⎭⎬⎫
⎩⎨⎧
++
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=
−
This is the star point voltage which can now be used to calculate the line currents:
004.11039.345
14.266.105240−∠=
−−= A
jIR
0
96.23
120
110590.624.49
14.266.1052400
0
jAe
jeI
j
j
Y −∠=−−
=−
−
0
45
240
8.263815.643.42
14.266.1052400
0
jAe
jeI
j
j
B −∠=−−=−
ELEC 4100 TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 2 : THREE PHASE CIRCUITS PART 2 - SOLUTION
Question 1.
[q. 2.38 Glover and Sarma]
Three identical impedances Z∆ = 20∠60° Ω are connected in a delta to a balanced three phase 480V source by three identical line conductors with an impedance of ZL = 0.8 + j0.6 Ω per line.
(a) Calculate the line to line voltage at the load terminals.
(b) Calculate the line to line voltage at the load terminals, but with a delta connected capacitor bank with a reactance of –j20 Ω per phase, connected in parallel to the original load.
Answer:
Part (a)
First convert the delta impedance into an equivalent star impedance:
Ω∠== ∆ 0603
20
3
ZZY
Then by voltage divider rule the phase voltage across the equivalent star impedance is given by:
ANlineY
YYAN V
ZZ
ZV
+=_
3
480
6.08.060320
60320
0
0
_
V
jV YAN ++∠
∠=
0_ 96.22.243 ∠= VV YAN
Therefore the line to line voltage across the delta load is given by:
030__ 96.323.4213
0
∠==∆ VeVV jYANAB
Part (b)
With a delta connected capacitor bank the new delta impedance becomes:
( ) ( )20//6020 0 jZ −Ω∠=∆
( )( ) Ω−∠=
Ω−Ω∠Ω−Ω∠=∆
00
0
1564.38206020
206020
j
jZ
Now convert this impedance to an equivalent star impedance:
Ω−∠== ∆ 01588.123
ZZY
Then by voltage divider rule the phase voltage across the equivalent star impedance is given by:
ANlineY
YYAN V
ZZ
ZV
+=_
ELEC 4100 TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
2
3
480
6.08.01588.12
1588.120
0
_
V
jV YAN ++−∠
−∠=
0_ 35.30.264 −∠= VV YAN
Therefore the line to line voltage across the delta load is given by:
030__ 65.263.4573
0
∠==∆ VeVV jYANAB
Question 2. [q. 2.39 Glover and Sarma]
Two three phase generators supply a three phase load through separate three phase lines. The load absorbs 30kW at 0.8 power factor lagging. The line impedance is 1.4 + j1.6 Ω per phase between the generator G1 and the load, and 0.8 + j1.0 Ω per phase between the generator G2 and the load. If the generator G1 supplies 15kW at 0.8 power factor lagging with a terminal voltage of 460V line to line, determine:
(a) The voltage at the load terminals.
(b) The voltage at the terminals of the generator G2.
(c) The real and reactive power supplied by the generator G2.
Assume balanced operation for the circuit.
Answer :
Part (a)
The per phase equivalent circuit is shown below:
VG1_AN VG2_AN
IG1 IG2
Zline1 Zline2
Zload
IL
The current IG1 can be calculated based on the power delivered by this generator as:
( )011 ..cos
..3fp
fpV
PI
LL
G−−∠=
( )( ) ( ) 0011 87.3653.23.8.0cos
8.04603
15 −∠=−∠= − AkW
IG
The load voltage is then simply the generator voltage less the voltage drop across the line impedance, as:
ELEC 4100 TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
3
( )( )0
00
11_1
73.29.216
6.14.187.3653.2303
460
−∠=
+−∠−∠=
−=
V
jA
IZVV GlineANGL
The line to line load voltage is then:
030 27.277.37330
∠== VeVV jLAB
Part (b)
The load current can be calculated based on the power rating and the load voltage, as:
( )01 ..cos..3
fpfpV
PI
AB
L−−∠=
( )( ) ( )( ) 0010 6.3963.57.8.0cos73.28.07.3753
30 −∠=−−∠= − AkW
IL
Hence the current IG2 can be calculated as:
( ) ( )
0
0012
49.4114.34
87.3653.256.3963.57
−∠=
−∠−−∠=−=
A
AAIII GLG
Therefore the terminal voltage at generator 2 is given by:
21_2 lineGLANG ZIVV +=
( )( )jVV ANG +−∠+−∠= 8.049.4114.3473.29.216 00_2
0_2 63.07.259 −∠= VV ANG
The line to line generator voltage is then:
030_2_2 37.298.4493
0
∠== VeVV jANGABG
Part c:
The power delivered by the second generator is given by:
( )( )
kVarjkW
IVS GANGG
4.1712.20
49.4114.3463.07.25933 002_22
+=
∠−∠== ∗
ELEC 4100 TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
4
Question 3. [q. 2.40 Glover and Sarma]
Two balanced Y-connected loads are connected in parallel, one drawing 15kW at 0.6 power factor lagging, and the other drawing 10kVA at 0.8 power factor leading. The loads are supplied by a balanced three phase 480V source.
(a) Determine the power factor for the combined load and state whether this load is lagging or leading.
(b) Determine the magnitude of the line current from the source.
(c) A delta connected capacitor bank is now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the delta to ensure that the power factor as seen by the source is unity? Give your answer in Ω.
(d) Determine the magnitude of the current in each capacitor and also the line current from the source.
Answer :
Part (a)
The total real and reactive power delivered to the combined load is:
( )( ) kWPtot 23000,158.0000,10 =+=
( ) ( )( ) ( )( )
kVar
Qtot
14
200006000
6.0costan000,158.0cossin000,10 11
=+−=
+−= −−
Hence the power factor is :
8542.023
14tancos.. 1 =⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛= −
kW
kVarfp lagging
Part (b)
The total apparent power is given by:
033.3193.261423 ∠=+= kVarjkWS
Therefore the load current is given by:
( )0
0
0*
33.3139.3204803
33.3193.26
3−∠=
∠−∠=⎟
⎟⎠
⎞⎜⎜⎝
⎛= A
V
kVA
V
SI
LL
L
Part (c)
The reactive power supplied by the capacitor bank must match the reactive power drawn by the load. Therefore the required reactive power is given by:
C
LLc X
VkVarQ
314 ==
ELEC 4100 TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
5
So :
( ) Ω=== 37.49
000,14
4803
14
3
kVar
VX LL
C
Part (d)
The current in each capacitor is given by:
AX
VI
C
LLC 72.9
37.49
480 =Ω
==
The line current in this case is then simply determined by the real power supplied.
So:
( ) AkW
V
PI
LL
totL 66.27
4803
23
3===
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 3 : PER UNIT SYSTEM - SOLUTION Question 1.
A single Phase 50kVA, 2400/240V, 60Hz distribution transformer is used as a step down transformer at the load end of a 2400V feeder whose series impedance is (1.0 + j2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the high voltage (i.e. primary) side. The transformer is delivering rated total power at 0.8 power factor lagging, and at rated secondary voltage. Neglecting the transformer excitation current, determine:
(a) The voltage at the transformer primary terminals,
(b) The voltage at the sending end of the feeder,
(c) The real and reactive power delivered to the sending end of the feeder.
Work in the Per Unit System, using the transformer ratings as base quantities.
Answer :
First Determine the base quantities.
Sbase = 50kVA
Vbase1 = 2400V
Vbase2 = 240V
Therefore:
AV
SI
base
basebase 833.20
2400
000,50
11 ===
AV
SI
base
basebase 333.208
240
000,50
22 ===
Ω=== 2.1158333.20
2400
1
11
base
basebase I
VZ
Ω=== 152.1333.208
240
2
22
base
basebase I
VZ
So the per unit impedances become:
upjj
Z
ZZ
base
eqeqpu .0217.000868.0
2.115
5.21
1
_1_1 +=+==
upjj
Z
ZZ
base
linepuline .01736.000868.0
2.115
21
1_ +=+==
Now the load power is given by:
( ) 01 9.36508.0cos50 ∠=∠= − kVAkVASload
Or:
( ) 01 9.36..0.18.0cos50 ∠=∠= − upkVASload
So the load current is given by:
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
2
00
0
__ 9.36..0.1
00.1
9.36..0.1 −∠=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∠∠=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
∗∗
upup
V
SI
puload
loadpuload
The load current is then:
09.36333.208 −∠= AIload
(a) The voltage at the transformer primary terminals is then given by:
eqpupuloadpupu ZIVV _1__2_1 +=
( )( )0217.000868.09.360.100.1 00_1 jV pu +−∠+∠=
upV pu .69.002.1 0_1 ∠=
The transformer primary voltage is then:
01 69.02448 ∠= VV
(b) The supply voltage is given by:
( )pulineeqpupuloadpupuS ZZIVV __1__2_ ++=
( )( )03906.0001736.09.360.100.1 00_ jV puS +−∠+∠=
upV puS .15.1037.1 0_ ∠=
The supply voltage is then: 015.12489 ∠= VVS
(c) The supply real and reactive power is then given by:
∗= loadSS IVS
( )( )00 9.360.115.1037.1 ∠∠=SS
upjupSS .6387.0..8169.002.38037.1 0 +=∠=
So the real and reactive power are:
kWP 845.40=
kVarQ 936.31=
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
3
Question 2. Three zones of a single phase distribution level circuit are identified in figure 1. The zones are
connected by transformers T1 and T2, whose ratings are also shown. Using base values of 3MVA and 11kV in zone 1, draw the per unit circuit and determine the per-unit impedances and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected.
Zone 1 Zone 2 Zone 3
T1 T22MVA3MVA
11 kV/6.6 k V 7.2 kV /3.3 kVXeq = 0.1 p.u. Xeq = 0.12 p.u.
Xline = j0.2 ΩXload = j2.9 Ω
Rload = 5.2 ΩVs = 13 kV
Figure 1 : Three Zone Distribution System for question 2.
Answer :
Choose Bases:
Sbase = 3MVA
Vbase1 = 11kV
Vbase2 = 6.6kV
( )( )
kVkV
kVkVVbase 025.3
2.7
3.36.63 ==
This requires:
AV
SI
base
basebase 727.272
11 ==
AV
SI
base
basebase 545.454
22 ==
AV
SI
base
basebase 736.991
33 ==
Similarly:
Ω== 333.402
11
base
basebase S
VZ
Ω== 520.142
22
base
basebase S
VZ
Ω== 050.32
33
base
basebase S
VZ
So the per unit impedance values are:
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
4
( ) ..9508.0075.13
_ upjZ
ZZ
base
loadpuload +==
..1.01_ upjZ puTeq =
..01377.02
_ upjZ
ZZ
base
linepuline ==
..2142.023
6.62.7
12.022
2_22_ up
MVA
MVA
kV
kV
S
S
V
VZZ
rate
base
base
raterateTpuTeq =⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
The per unit supply voltage is then:
..182.111
13_ up
kV
kVV pus ==
The per unit equivalent circuit is then given by:
Zone 1 Zone 2 Zone 3
XT1eq p.u. =j0.1 p.u.
XT2eq =j0.2142 p.u.
Xline p.u. = j0.01377pu
Xload p.u. =j0.9508 p.u
Rload p.u. = 1.705 p.u.
Vs = 1.182 p.u.
The load current is then given by:
puloadpuTpulinepuT
puspuload ZXXX
VI
__2__1
__ +++
=
( ) 0
00
_ 87.36131.20182.1
705.19508.001377.02142.01.00182.1
∠∠=
++++∠=
jI puload
..87.365546.0 0_ upI puload −∠=
So the load current is :
087.360.550 −∠= AIload
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
5
Question 3. A balanced Y-connected voltage source with Eab = 480∠0° V is applied to a balanced ∆ load
with Z∆ = 30∠40° ohms. The line impedance between the source and the load is ZL = 1∠85° p.u. for each phase. Calculate the per-unit and actual current in phase a of the line using Sbase3φ = 100kVA and VbaseLL = 600V.
Answer:
Define the base quantities as:
kVASbase 1003 =φ
kVASbase 333.331 =φ
VVbaseLL 600=
VVVbaseLN 412.3463
600 ==
Ω=== 6.31
2
3
2
φφ base
baseLN
base
baseLLbase S
V
S
VZ
AV
SI
baseLL
basebase 225.96
33 == φ
So :
00
_ 0..8.0600
0480 ∠=∠= upE puab and 0_ 30..8.0 −∠= upE pua
00
_ 40..333.86.3
4030 ∠=∠=∆ upZ pu
0__ 40..7778.2
3∠== ∆ up
ZZ pu
puY
00
_ 85..2778.06.3
851 ∠=∠= upZ puline
So the total impedance seen by the source is:
0___ 78.43..9807.2 ∠=+= upZZZ pulinepuYputot
Therefore the supply current is given by:
..78.732684.078.43..9807.2
30..8.0 00
0
_
__ up
up
up
Z
VI
putot
puapua −∠=
∠−∠==
So the actual load current is given by:
078.7383.25 −∠= AIa
ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
6
Question 4. A balanced Y-connected voltage source with Eag = 277∠0° V is applied to a balanced Y load in
parallel with a balanced ∆ load, where ZY = 30 + j10 ohms and Z∆ = 45 – j25 ohms. The Y load is solidly grounded. Using base values of Sbase1φ = 10kVA and VbaseLN = 277 V, calculate the source current Ia in per-unit and in amperes.
Answer :
Define the base quantities as:
kVASbase 303 =φ
kVASbase 101 =φ
VVbaseLN 277=
VVV baseLNbaseLL 77.4793 ==
Ω== 6729.73 3
2
φbase
baseLLbase
S
VZ
AV
SI
baseLL
basebase 101.36
33 == φ
So the per unit impedance values are given by:
0_ 055.29..709.6
6729.7
2545 −∠=−=∆ upj
Z pu
0_ 435.18..121.4
6729.71030 ∠=+= upj
Z puY
Now the delta load can be converted to an equivalent star load as:
0__ 055.29..2364.2
3−∠== ∆
∆ upZ
Z puputoY
The total per-phase impedance is then given by:
puYputoYputot ZZZ ___ //∆=
0
__
___ 74.125705.1 −∠=
+=
∆
∆
puYputoY
puYputoYputot ZZ
ZZZ
The per unit source current is then given by:
0
_
__ 74.126367.0 ∠==
putot
puagpua Z
EI
So the source current is :
074.1299.22 ∠= AIa
ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 4 : TRANSFORMERS SOLUTIONS. Question 1.
a1
a2 b2
c1b1
c2
A1 B1 C1
A2 B2 C2
a3
a4 b4
c3b3
c4
A1A2
B1
B2 C1
C2
a1a2
b1
b2 c1
c2
a3a4
b3
b4 c3
c4
A1
B1
C1
a1
b1
c1
a3b3
c3
(a)
a1
a2 b2
c1b1
c2
A1 B1 C1
A2 B2 C2
a3
a4 b4
c3b3
c4
a5
a6 b6
c5b5
c6
A1A2
B1
B2 C1
C2
a1a2
b1
b2 c1
c2
a3a4
b3
b4
a5a6
b5
b6 c5
c6
c4
c3
A1
B1
C1
a1b1
c1
c3
c4b6
b5
a3a4c6
c5b4
b3
a5a6
(b)
ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS
2
a1
a2 b2
c1b1
c2
A1 B1 C1
A2 B2 C2
a3
a4 b4
c3b3
c4
a5
a6 b6
c5b5
c6
A1A2
B1
B2 C1
C2
a1a2
b1
b2 c1
c2
a3a4
b3
b4
a5a6
b5
b6 c5
c6
c4
c3
A1
B1
C1
a1
b1
c1
a3a4
b5
b6
b3
b4
c5
c6
c4
c3
a5
a6
(c)
ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS
3
Question 2. Consider the single line diagram of the power system shown below. The equipment ratings are as follows:
• Generator 1 : 750MVA, 18kV, Xeq = 0.2 p.u.
• Generator 2 : 750 MVA, 18kV, Xeq = 0.2 p.u
• Synchronous Motor 3 : 1500MVA, 20kV, Xeq = 0.2 p.u.
• 3 Phase Transformers, T1 to T4 : 750MVA, 500kV Y/20kV ∆, Xeq = 0.1 p.u.
• 3 Phase Transformer T5 : 1500MVA, 500kV Y/20kV Y, Xeq = 0.1 p.u.
Neglecting winding resistances, transformer phase shifts, and the excitation phenomena, draw the equivalent per unit reactance diagram. Use a base of 100MVA and 500kV for the 40 Ω transmission line. Determine all per unit reactance’s.
T1
1 2
3
Bus 1 Bus 2
Bus 3
j40 ohm
j25 ohm j25 ohm
T2
T3 T4T5
Answer:
The equivalent per phase, per unit circuit diagram is shown below:
Bus 1 Bus 2
Bus 3
XT1
XT3
XT2
XT4
XT5
Xline1
Xline2 Xline3
XG2
XM3
XG1
EG2
EM3
EG1
j0.0133pu j0.0133pu
j0.0133puj0.0133pu
j0.00666pu
j0.01333pu
j0.01pu j0.01pu
j0.016pu
j0.0216pu j0.0216pu
ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS
4
The impedance values in the circuit diagram are calculated as will be detailed below:
MVASbase 100=
kVV HVbase 500_ = Transmission line zones
kVV LVbase 20_ = Generator zones
( ) ( ) Ω=== 2500
100
500 22_
_ MVA
kV
S
VZ
base
HVbaseLVbase
( ) kAkV
MVA
V
SI
HVbase
baseLVbase 887.2
203
100
3 _
_ ===
So the generator per unit impedances are:
..0216.0750
100
20
182.0
2
1 upMVA
MVA
kV
kVXG =⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
..0216.0750
100
20
182.0
2
2 upMVA
MVA
kV
kVXG =⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
..01333.01500
1002.03 up
MVA
MVAX M =⎟
⎠
⎞⎜⎝
⎛=
The transformer per unit impedances are:
..01333.0750
1001.04321 up
MVA
MVAXXXX TTTT =⎟
⎠
⎞⎜⎝
⎛====
..00666.01500
1001.05 up
MVA
MVAXT =⎟
⎠
⎞⎜⎝
⎛=
The transmission line per unit impedances are:
..016.02500
401 upXline =
ΩΩ=
..01.02500
2532 upXX lineline =
ΩΩ==
Question 3. For the power system discussed in question 2, consider the case where the motor absorbs
1200MW at 0.8p.f. leading with the Bus 3 voltage at 18kV. Determine the Bus 1 and Bus 2 voltages in kV. Assume that generators 1 and 2 deliver equal real powers and equal reactive powers. Also assume a balanced three-phase system with positive-sequence sources.
Answer :
The bus 3 voltage is given by:
..09.020
018 00
3 upkV
kVV pu ∠=∠=
ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS
5
The motor current is then:
( )( ) ( ) ( )( )01
3 87.3611.488.0183
1200..cos
..3∠==∠⎟⎟
⎠
⎞⎜⎜⎝
⎛= − kA
kV
MWfp
fpV
PI
LL
The motor current in per unit:
..87.3667.16887.2
87.3611.48 00
_3 upkA
kAI pu ∠=∠=
Due to symmetry:
( )2_3_3
_5_3_3_2_1 2 linepuTpu
puTpupupupu XXI
XIVVV +++==
( )( )
( )01333.001.02
87.3667.16
00666.087.3667.1609.00
00_2_1
jj
jVV pupu
+∠+
∠+∠==
..83.187572.0 0_2_1 upVV pupu ∠==
So the bus 1 and bus 2 voltages are:
021 83.1814.15 ∠== kVVV
ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 5 : LOAD FLOW – SOLUTION. Question 1.
Answer :
(a) For sinusoidal time varying voltage and current waveforms, define:
( ) ( ) tjexVtxv ω=,
( ) ( ) tjexItxi ω=,
Then substituting into the partial differential equations gives:
( ) ( ) ( )[ ] ( ) tjtjtj exzIexLIjxrIdx
xdVe ωωω ω −=−−=
( ) ( ) ( )[ ] ( ) tjtjtj exyVexCVjxGVdx
xdIe ωωω ω −=−−=
These expressions can be simplified as:
( ) ( )xzIdx
xdV−=
( ) ( )xyVdx
xdI−=
Differentiating with respect to x:
( ) ( ) ( )xzyVdx
xdIz
dx
xVd =−=2
2
( ) ( ) ( )xzyIdx
xdVy
dx
xId =−=2
2
These expressions are separate, second order linear differential equations involving one spatial variable only.
(b) From the π-section model it can be shown that:
( )1
2
Y
YVIVV ss
sr−−=
Rearranging:
11
21Y
I
Y
YVV s
sr −⎥⎦
⎤⎢⎣
⎡+=
Comparing this expression with the general transmission line solutions provided gives:
( )dZY
C γsinh
11 =
Using this value and again comparing with the general transmission line solutions gives:
( )[ ]( ) ⎟
⎠
⎞⎜⎝
⎛=−
=2
tanh1
sinh
1cosh2
d
ZdZ
dY
CC
γγ
γ
ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION
2
Now :
( )⎥⎦
⎤⎢⎣
⎡ −−−−=−−=1
23232 Y
VYIVYVYIVYVYII ss
sssrssr
⎥⎦
⎤⎢⎣
⎡++−⎟
⎟⎠
⎞⎜⎜⎝
⎛+=
1
223
1
3 11Y
YYYV
Y
YII ssr
Equating this expression with the general transmission line solutions gives:
( )[ ] ( )[ ]( ) 213 2
tanh1
sinh
1cosh1cosh Y
d
ZdZ
ddYY
CC
=⎟⎠
⎞⎜⎝
⎛=−=−= γγ
γγ
(b) The π-section model of a transmission line is used in load flow analysis since load flow is interested only in the steady state characteristics at the voltage buses in the network. Hence it is only necessary to consider the behaviour at the terminating ends of the transmission line, not in the middle of the line. A lumped element model is sufficient to provide this information. Furthermore the full distributed model is used to predict dynamic characteristics along the line, but since this information is irrelevant for load flow it is not necessary to utilise the full model.
Question 2.
Answer :
(a) The diagonal elements are the self admittances at each node of the network, and are the sum of all admittances connected to that node. The off-diagonal elements are the mutual admittances between two nodes of a network, and are the negative values of the admittances linking the two nodes in question. The YBUS matrix is square since the network consists of N buses, and for the ith bus there are N-1 potential mutual connections, and 1 self admittance – hence the matrix is square. The matrix is symmetric since the mutual connections between buses i and k are the same as the connections between buses k and i. The matrix is sparse since in power systems there is generally a low level of connectivity between the nodes, with couplings only between a few adjacent couplings. Hence the bulk of the mutual couplings are zero, and so the matrix is sparse.
(b) The complex conjugate of the apparent power at bus i can be written as:
iii IVS ∗∗ =
k
n
kikiiii VyVjQPS ∑
=
∗∗ =−=1
Where the yik are the elements of the admittance bus.
(c) The apparent power is given by:
k
n
kikiiii VyVjQPS ∑
=
∗∗ =−=1
iiik
n
ikk
iki
ii VyVyV
jQP +=−∑
≠=
∗1
ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION
3
Rearranging gives:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−= ∑≠=
∗ k
n
ikk
iki
ii
iii Vy
V
jQP
yV
1
1
So the Gauss implementation of a voltage calculation is:
( )⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−= ∑≠=
∗+ p
k
n
ikk
ikp
i
pi
pi
ii
pi Vy
V
jQP
yV
1
1 1
The Gauss-Seidel implementation of a voltage calculation is:
( ) ⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−−−= ∑∑
+=
+−
=∗
+ pk
n
ikik
pk
i
kik
pi
pi
pi
ii
pi VyVy
V
jQP
yV
1
11
1
1 1
Since the Gauss-Seidel method uses the most recently available iteration data it generally shows a faster convergence rate than the Gauss method. Furthermore the Gauss method must store the pth and (p+1)th bus data, whereas the Gauss-Seidel discards the previous data as soon as the new data has become available. This results in a memory allocation and storage requirement advantage for the Gauss-Seidel approach, and furthermore programming the Gauss-Seidel method is simpler.
(d) The apparent power is given by:
k
n
kikiiii VyVjQPS ∑
=
∗∗ =−=1
Define:
iii VV δ∠=
ikikik yy γ∠=
Then :
( )ikikki
n
kikiii VVyjQPS γδδ +−∠=−= ∑
=
∗
1
Or:
( ) ( )ikkiki
n
kikikikki
n
kiki VVyVVyP γδδγδδ −−=+−= ∑∑
==
coscos11
( ) ( )ikkiki
n
kikikikki
n
kiki VVyVVyQ γδδγδδ −−=+−−= ∑∑
==
sinsin11
ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION
4
(e) Define the power flow mismatches at bus i as:
( ) ( )ikkiki
n
kikiikkiki
n
kikLiGiii VVyPVVyPPPf γδδγδδ −−−=−−−−=∆= ∑∑
==
coscos11
( ) ( )ikkiki
n
kikiikkiki
n
kikLiGiii VVyQVVyQQQg γδδγδδ −−−=−−−−=∆= ∑∑
==
sinsin11
So applying the Newton-Raphson method:
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆∆+
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆∆
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
−⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
+
+
p
p
p
p
p
p
p
p
p
p
VVQ
P
V
ggV
ff
VV
δδ
δ
δδδ
1
1
1
Alternatively this can be expressed as:
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆∆
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆∆
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∂∂
∂∂
∂∂
∂∂
−=⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆∆
p
p
pp
pp
p
p
p
p
VJJ
JJ
VV
ggV
ff
Q
P δδ
δ
δ43
21
Where pP∆ are the real power mismatches at all PQ and PV buses, pQ∆ are the reactive
power mismatches at all PQ buses, pδ∆ are the voltage angle corrections for all PQ and PV
buses, and pV∆ are the voltage magnitude corrections for all PQ buses.
The Jacobian matrix is defined by:
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
=
n
nnn
n
n
p
fff
fff
fff
J
δδδ
δδδ
δδδ
L
MMM
L
L
32
3
3
3
2
3
2
3
2
2
2
1
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
=
++
++
++
n
n
m
n
m
n
nmm
nmm
p
V
f
V
f
V
f
V
f
V
f
V
fV
f
V
f
V
f
J
L
MMM
L
L
21
3
2
3
1
3
2
2
2
1
2
2
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
=+++
+++
n
nnn
n
mmm
n
mmm
p
ggg
ggg
ggg
J
δδδ
δδδ
δδδ
L
MMM
L
L
32
2
3
2
2
2
1
3
1
2
1
3
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
∂∂−
=
++
+
+
+
+
+
+
+
+
+
+
n
n
m
n
m
n
n
m
m
m
m
m
n
m
m
m
m
m
p
V
g
V
g
V
g
V
g
V
g
V
gV
g
V
g
V
g
J
L
MMM
L
L
21
2
2
2
1
2
1
2
1
1
1
4
The Swing bus is bus 1, while buses 2 to m are the PV buses, and buses m+1 to n are the PQ buses.
The 8 derivatives in the Jacobian matrices are given by:
( ) kiVVyf
ikkikiikk
i ≠−−=∂∂− ,sin γδδδ
( )ikkiki
n
ikk
iki
i VVyf γδδδ
−−−=∂∂− ∑
≠=
sin1
ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION
5
( ) kiVyV
fikkiiik
k
i ≠−−=∂∂− ,cos γδδ
( ) ( )ikiiiikkik
n
kik
i
i VyVyV
f γγδδ coscos1
+−−=∂∂− ∑
=
( ) ikVVyg
ikkikiikk
i ≠−−−=∂∂− ,cos γδδδ
( )ikkiki
n
ikk
iki
i VVyg γδδδ
−−=∂∂− ∑
≠=
cos1
( ) ikVyV
gikkiiik
k
i ≠−−=∂∂− ,sin γδδ
( ) ( )iiiiiikkik
n
kik
i
i VyVyV
g γγδδ sinsin1
−−−=∂∂− ∑
=
(f) Since the Newton Raphson method uses a first order Taylor series approximation of the non-linear power flow equations to iteratively find a solution, it has a much faster convergence rate than the Gauss or Gauss Seidel methods. These latter methods are limited by the sparsity of the admittance bus matrix, which limits the rate that corrective terms can propagate through the solution.
(g) The Swing Bus is needed to condition the YBUS admittance matrix so as to make solutions to the power flow problem possible. Without conditioning it may be possible to have many solutions to the load flow problem which satisfy the constraints. Hence by fixing one bus with respect to earth potential one of these many solution cases is selected. The Swing Bus also serves the purpose of carrying the slack or net power from the rest of the network.
ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 7 : SYMMETRICAL COMPONENTS - SOLUTIONS Question 1.
Determine the symmetrical components of the following line currents : (a) Ia = 5∠900, Ib = 5∠3400, Ic = 5∠2000, and (b) Ia = 50, Ia = j50, Ic = 0.
Answer
(a) The symmetrical component currents are given by:
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
200
340
90
2
2
2
2
2
1
0
5
5
5
1
1
111
3
1
1
1
111
3
1
j
j
j
c
b
a
e
e
e
aa
aa
I
I
I
aa
aa
I
I
I
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
0
0
00
00
200
340
90
120120
120120
2
1
0
5
5
5
1
1
111
3
1
j
j
j
jj
jj
e
e
e
ee
ee
I
I
I
( )( )
( )⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
++
++
++
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
− 000
000
000
4022090
8010090
20034090
2
1
0
35
353
5
jjj
jjj
jjj
eee
eee
eee
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
+
+
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
Aj
Aj
Aj
I
I
I
4760.0
9490.4
5266.0
2
1
0
(b) The symmetrical component currents are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
50
50
1
1
111
3
1
1
1
111
3
1 090
2
2
2
2
2
1
0
j
c
b
a
e
aa
aa
I
I
I
aa
aa
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
50
50
1
1
111
3
1 0
00
00 90
120120
120120
2
1
0
j
jj
jj e
ee
ee
I
I
I
( )( )( )⎥⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
+
+
+
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
− 0
0
0
30
210
90
2
1
0
1350
1350
1350
j
j
j
e
e
e
I
I
I
( )( )
( ) ⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
+
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
Aj
Aj
Aj
I
I
I
333.8100.31
333.8233.2
667.16667.16
2
1
0
ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
2
Question 2. One line of a three phase generator is open circuited, while the other two are short-circuited to
ground. The line currents are Ia = 0, Ib = 1000A∠900, and Ic = 1000A∠-300. Find the symmetrical components of these currents. Also find the current into ground.
Answer
The symmetrical component currents are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
− 0
0
30
90
2
2
2
2
2
1
0
1000
1000
0
1
1
111
3
1
1
1
111
3
1
j
j
c
b
a
e
e
aa
aa
I
I
I
aa
aa
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−
0
0
00
00
30
90
120120
120120
2
1
0
1000
1000
0
1
1
111
3
1
j
j
jj
jj
e
e
ee
ee
I
I
I
( )( )( ) ⎥⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
+
+
+
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
00
00
00
9030
150210
3090
2
1
0
31000
31000
31000
jj
jj
jj
ee
ee
ee
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
2
1
0
303.333
1507.666
303.333
A
A
A
I
I
I
The ground current is the sum of the b and c phase currents and is given by:
( ) ( ) 03090
2
1
0
301000500866100000
∠=+=+=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=+= − AAjee
I
I
I
III jjcbgnd
Question 3. Given the line to ground voltages Vag = 280V∠00, Vbg = 290V∠-1300, and Vcg = 260V∠1100,
calculate (a) the sequence components of the line to ground voltages, denoted VLg0, VLg1, and VLg2. (b) the line to line voltages Vab, Vbc, Vca. (c) The sequence components of the line to line voltages
VLL0, VLL1, and VLL2. Also verify the following general relation : VLL0 = 0, 011 303 ∠= LLgLL VV , and
021 303 −∠= LLgLL VV .
Answer
(a) The symmetrical components of the line to ground voltages are given by:
ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
3
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
0
0
0
110
130
0
2
2
2
2
2
1
0
260
290
280
1
1
111
3
1
1
1
111
3
1
j
j
j
cg
bg
ag
Lg
Lg
Lg
e
e
e
aa
aa
V
V
V
aa
aa
V
V
V
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
0
0
0
00
00
110
130
0
120120
120120
2
1
0
260
290
280
1
1
111
3
1
j
j
j
jj
jj
Lg
Lg
Lg
e
e
e
ee
ee
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−∠
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
2
1
0
43.7987.24
63.673.275
11.7855.7
V
V
V
V
V
V
Lg
Lg
Lg
(b) The line to line voltages are calculated according to:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
0
0
0110
110130
1300
49.146491.442
80.101550.476
47.25613.516
280260
260290
290280
00
00
00
V
V
V
ee
ee
ee
VV
VV
VV
V
V
V
jj
jj
jj
agcg
cgbg
bgag
ca
bc
ab
(c) The symmetrical components of the line to line voltages are given by:
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
0
0
0
49.146
80.101
47.25
2
2
2
2
2
1
0
491.442
550.476
613.516
1
1
111
3
1
1
1
111
3
1
j
j
j
cg
bg
ag
LL
LL
LL
e
e
e
aa
aa
V
V
V
aa
aa
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
∠=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
2
1
0
43.4907.43
37.2357.477
0
V
V
V
V
V
V
LL
LL
LL
So :
000
0
1
1 303307321.163.673.275
37.2357.477 ∠=∠=−∠
∠=V
V
V
V
Lg
LL
000
0
2
2 303307321.143.7987.24
43.4907.43−∠=−∠=
−∠∠
=V
V
V
V
Lg
LL
Question 4. The voltages given in question 3 are applied to a balanced Y load consisting of (12+j16) ohms
per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate I0, I1, and I2, the sequence components of the line currents. Then calculate the line currents Ia, Ib, and Ic from the sequence components, and compare with the line currents calculated directly from the network equations.
ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
4
Answer
The sequence networks are shown below:
IL1
VLg1
12+j16 Ω
Positive Sequence
Negative Sequence
Zero Sequence
IL2
IL0
VLg2
VLg0
12+j16 Ω
12+j16 Ω
The three sequence currents can be calculated as:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+−∠
+−∠
+∠
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
0
0
2
2
1
1
0
0
2
1
0
30.26243.1
76.59787.13
98.24378.0
1612
43.7987.241612
63.673.2751612
11.7855.7
A
A
A
j
Vj
Vj
V
Z
VZ
VZ
V
I
I
I
Lg
Lg
Lg
L
L
L
The line currents are then given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
0
0
120120
120120
2
1
0
2
2
30.26243.1
76.59787.13
98.24378.0
1
1
111
1
1
111
00
00
A
A
A
ee
ee
I
I
I
aa
aa
I
I
I
jj
jj
L
L
L
c
b
a
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
∠
−∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
87.560.13
87.1765.14
13.530.14
A
A
A
I
I
I
c
b
a
From the network equations directly:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
∠
−∠
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+∠
+−∠
+∠
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
0
0
87.560.13
87.1765.14
13.530.14
1612
1102601612
1302901612
0280
A
A
A
j
Vj
Vj
V
Z
VZ
VZ
V
I
I
I
cg
bg
ag
c
b
a
This matches the result calculated using the symmetrical component model.
ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
5
Question 5. As shown in figure 1, a balanced three-phase, positive sequence source with VAB = 480V∠00 is
applied to an unbalanced ∆ load. Note that one leg of the ∆ is open. Determine (a) the load currents IAB and IBC. (b) the line currents IA, IB, IC, which feed the ∆ load. (c) the zero, positive, and negative sequence components of the line currents.
Ia
Ib
Ic
Ea
Ec
Eb
(18+j10)Ω
(18+j10)Ω
Vab=480V 00
Ibc
Iab
Figure 1: Network for Question 5.
Answer
(a) The load currents are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−∠
−∠
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+−∠
+∠
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
05.149311.23
05.29311.23
01018
1204801018
0480
0
0
0
0
0
A
A
j
j
Z
VZ
V
I
I
I
bc
ab
ca
bc
ab
(b) The line currents are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
−∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
95.30311.23
05.179376.40
05.29311.23
A
A
A
I
II
I
I
I
I
bc
abbc
ab
c
b
a
(c) The sequence currents are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
−∠
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
2
2
2
2
2
1
0
95.30311.23
05.179376.40
05.29311.23
1
1
111
3
1
1
1
111
3
1
A
A
A
aa
aa
I
I
I
aa
aa
I
I
I
c
b
a
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
2
1
0
95.60459.13
055.59917.26
0
A
A
A
I
I
I
ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 8 : THREE PHASE FAULTS - SOLUTION Question 1.
Equipment ratings for the 4-bus system shown in figure 1 are as follows:
• Generator G1 : 500MVA, 13.8kV, X’ = 0.20 p.u. • Generator G2 : 750MVA, 18.0kV, X’ = 0.18 p.u. • Generator G3 : 1000MVA, 20.0kV, X’ = 0.17p.u. • Transformer T1 : 500MVA, 13.8kV delta/500kV star, X = 0.12 p.u. • Transformer T2 : 750MVA, 18kV delta/500kV star, X = 0.10 p.u. • Transformer T3 : 1000MVA, 20kV delta/500kV star, X = 0.10 p.u. • Each transmission line : X = 50 ohms.
A three phase short circuit occurs at bus 1, where the pre-fault voltage is 525kV. Pre-fault load current is negligible.
Draw the positive sequence reactance diagram in per unit on a 1000MVA base, 20kV base in the zone of generator G3.
Determine:
(a) The Thevenin reactance in per unit at the fault : [0.2670] (b) The transient fault current in per unit and kA : [-j3.933, -j4.541kA] (c) Contributions to the fault current from G1 and from line 1-2. [-j1.896, -j2.647kA]
1
2
3
Bus 1 Bus 3
Bus 4
j50 ohm j50 ohm
T3 T3
T2
Bus 2
j50 ohm
Figure 1 : Four Bus Power System.
Answer:
The positive sequence per unit network is shown below. The per unit values are determined as follows:
MVASbase 1000=
kVVbase 203 = Zone of Generator 3.
kVkVkV
kVVbase 50020
20
5004 == Zone of Transmission lines.
kVkVkV
kVVbase 18500
500
182 == Zone of Generator 2.
ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
2
kVkVkV
kVVbase 8.13500
500
8.131 == Zone of Generator 1.
( ) ( ) Ω=== 250
1000
500 224
4 MVA
kV
S
VZ
base
basebase
( ) kAkV
MVA
V
SI
base
basebase 155.1
5003
1000
3 4
4 ===
Bus 1
Bus 4
XT1 XT3
X24
X12 X23 XG3
XG2
XG1
EG3
EG2
EG1
j0.24 pu j0.1pu
j0.2pu
j0.24pu
j0.2pu j0.2puj0.4pu j0.17pu
Bus 3Bus 2
XT2j0.133pu
So applying these base values to the generators:
( ) ..4.0500
10002.01 upXG =⎟
⎠
⎞⎜⎝
⎛=
( ) ..24.0750
100018.02 upXG =⎟
⎠
⎞⎜⎝
⎛=
..17.03 upXG =
Similarly for the transformers:
( ) ..24.0500
100012.01 upXT =⎟
⎠
⎞⎜⎝
⎛=
( ) ..1333.0750
10001.02 upXT =⎟
⎠
⎞⎜⎝
⎛=
..1.03 upXT =
For the transmission lines:
..2.0250
50242312 upXXX ====
ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
3
Part (a)
The Thevenin equivalent impedance of the network when viewed from voltage bus 1 is:
( ) ( ) ( )[ ]332322241211 //// GTGTTGTh XXXXXXXXXX ++++++=
( ) ( ) ( )[ ]24.01333.02.0//17.01.02.02.0//4.024.0 jjjjjjjjjXTh ++++++=
( ) ( )4583.0//64.0 jjXTh =
..2670.0 upjXTh =
Part (b)
The pre-fault voltage, neglecting pre-fault currents is:
..005.1500
0525 00
upkV
kVVF ∠=∠=
So the fault current is:
..933.32670.0
..005.1 0
upjj
up
Z
VI
Th
FF −=∠==
kAjIF 541.4−=
Part (c)
Using the current divider rule:
..641.164.04583.0
4583.01 upj
jj
jII FG −=⎟⎟
⎠
⎞⎜⎜⎝
⎛
+=
kAjIG 896.11 −=
..292.264.04583.0
64.02 upj
jj
jII FG −=⎟⎟
⎠
⎞⎜⎜⎝
⎛
+=
kAjIG 647.21 −=
ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
4
Question 2. For the above described power system, consider the case where a balanced 3-phase short circuit
occurs at bus 2 where the pre-fault voltage is 525kV (neglect the pre-fault current).
Determine –
(a) The Thevenin equivalent impedance of the network viewed from the fault location : [0.1975 p.u.]
(b) The fault current in per unit and in kA [-j5.3155 p.u., -j6.138kA] (c) The contribution to the fault from lines 1-2, 2-3 and 2-4. [-j1.44, -j2.58, -j2.21 kA]
Answer:
Part (a) For faults on bus 2, the Thevenin equivalent impedance is given by: ( ) ( ) ( )332322241211 //// GTGTTGTh XXXXXXXXXX ++++++=
( ) ( ) ( )24.01333.02.0//17.01.02.0//2.04.024.0 jjjjjjjjjXTh ++++++=
( ) ( ) ( )5733.0//47.0//84.0 jjjXTh =
..1975.0 upjXTh =
Part (b) The pre-fault voltage, neglecting pre-fault currents is:
..005.1500
0525 00
upkV
kVVF ∠=∠=
So the fault current is:
..3155.51975.0
..005.1 0
upjj
up
Z
VI
Th
FF −=∠==
kAjIF 1379.6−=
Part (c) The contribution to the fault from line 12 is given by:
kAjupjj
I 443.1..25.184.0
005.1 0
12 −=−∠=
kAjupjj
I 580.2..234.247.0
005.1 0
23 −=−∠=
kAjupjj
I 115.2..8315.15733.0
005.1 0
24 −=−∠=
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 9 : FAULT STUDIES Question 1.
The single-line diagram and equipment ratings of a three phase electrical system are given below. The inductor connected to the neutral of generator 3 has a reactance of 0.05 p.u. using the ratings of generator 3 as a base. Draw the positive, negative and zero sequence network diagrams for the system using a 1000MVA base, and a 765kV base in the zone of line 1-2. Neglect the effects of ∆-Y transformer phase shifts.
Line 1 - 33
1
2
Bus 1 Bus 3
T1
T3
T2
Line 1 - 2 Line 2 - 3Bus 2
4
T4
Transformers:
• T1 : 1000MVA, 15 kV ∆ / 765 kV Y , X = 0.1 p.u. • T2 : 1000MVA, 15 kV ∆ / 765 kV Y , X = 0.1 p.u. • T3 : 500MVA, 15 kV ∆ / 765 kV Y , X = 0.12 p.u. • T4 : 750MVA, 15 kV ∆ / 765 kV Y , X = 0.11 p.u.
Transmission Lines :
• 1-2 : 765 kV, X1 = 50 Ω, X0 = 150 Ω. • 1-3 : 765 kV, X1 = 40 Ω, X0 = 100 Ω. • 2-3 : 765 kV, X1 = 40 Ω, X0 = 100 Ω.
Synchronous Generators :
• G1 : 1000MVA, 15 kV, X1 = X2 = 0.18 p.u., X0 = 0.07 p.u. • G2 : 1000MVA, 15 kV, X1 = X2 = 0.20 p.u., X0 = 0.10 p.u. • G3 : 500MVA, 13.8 kV, X1 = X2 = 0.15 p.u., X0 = 0.05 p.u. • G4 : 750MVA, 13.8 kV, X1 = 0.30 p.u. X2 = 0.40 p.u., X0 = 0.10 p.u.
Answer:
The three sequence networks for the system are shown below. The per unit impedance values are calculated as follows:
MVASbase 1000=
kVVbaseHV 765= Zone of Transmission Lines.
kVVbaseLV 15= Zone Generators.
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
2
( ) ( ) Ω=== 23.585
1000
765 22
MVA
kV
S
VZ
base
baseHVbaseHV
( ) kAkV
MVA
V
SI
baseHV
basebaseHV 7547.0
7653
1000
3===
The per unit sequence impedances of the generators are then given by: ..18.01_1 upXG = ..18.02_1 upXG = ..07.00_1 upXG =
..20.01_2 upXG = ..20.02_2 upXG = ..10.00_2 upXG =
( ) ..2539.0500
1000
15
8.1315.0
2
1_3 upXG =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
( ) ..2539.0500
1000
15
8.1315.0
2
2_3 upXG =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
( ) ( )
..3385.0..2539.0..08464.0
500
1000
15
8.1305.03
500
1000
15
8.1305.0
22
0_2
upupup
XG
=+=
⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
( ) ..3386.0750
1000
15
8.133.0
2
1_4 upXG =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
( ) ..4514.0750
1000
15
8.1340.0
2
2_4 upXG =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
( ) ( ) ..1129.075015
8.131.0
2
0_4 upXG =⎟⎠
⎞⎜⎝
⎛=
The per unit sequence impedances of the transformers are then given by: ..1.01 upXT =
..1.02 upXT =
..24.0500
1000
15
1512.0
2
3 upXT =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
..1467.0750
1000
15
1511.0
2
4 upXT =⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
The per unit sequence impedances of the transmission lines are then given by:
..08544.023.585
502_121_12 upXX === ..2563.0
23.585
1500_12 upX ==
..06835.023.585
402_131_13 upXX === ..1709.0
23.585100
0_13 upX ==
..06835.023.585
402_231_23 upXX === ..1709.0
23.585
1000_23 upX ==
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
3
Bus 1
XT1 XT3
X12_1 X23_1
XG3_1
XG2_1
XG1_1
EG3
EG2
EG1
j0.1 pu j0.24pu
j0.20pu
j0.06835pu
j0.18pu j0.254pu
Bus 3
Bus 2
XT2j0.1pu
X13_1
j0.08544 pu
EG4
XG4_1j0.339pu
XT4j0.147pu
j0.06835pu
Positive Sequence Network.
Bus 1
XT2 XT3
X12_2 X23_2
XG3_2
XG2_2
XG1_2
j0.1 pu j0.24pu
j0.20pu
j0.06835pu
j0.18pu j0.254pu
Bus 3
Bus 2
XT2j0.1pu
X13_2
j0.08544 pu
XG4_2j0.451pu
XT4j0.147pu
j0.06835pu
Negative Sequence Network.
Bus 1
XT1 XT3
X12_0 X23_0
XG3_0
XG2_0
XG1_0
j0.1 pu j0.24pu
j0.10pu
j0.1709pu
j0.07pu j0.339pu
Bus 3
Bus 2
XT2j0.1pu
X13_0
j0.2563 pu
XG4_0j0.113pu
XT4j0.147pu
j0.1709pu
Zero Sequence Network.
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
4
Question 2. Faults at bus 1 in question 1 are of interest. Determine the Thevenin equivalent impedance of
each sequence network as viewed from the fault bus. The pre-fault voltage is 1.0 p.u. Pre-fault load currents and ∆-Y transformer phase shifts are neglected.
Answer:
The first step towards obtaining the Thevenin equivalent networks for the sequence networks above is to simplify the networks using a Y-∆ transformation. Recall that the Y-∆ transformation is of the form:
ZB
ZA
ZC
ZCA
ZBC
ZAB
CABCAB
CAABA ZZZ
ZZZ
++=
C
ACCBBAAB Z
ZZZZZZZ
++=
CABCAB
BCABB ZZZ
ZZZ
++=
A
ACCBBABC Z
ZZZZZZZ
++=
CABCAB
BCCAC ZZZ
ZZZ
++=
B
ACCBBACA Z
ZZZZZZZ
++=
So the three sequence networks can be simplified to the form: Bus 1
EG3
EG2
EG1
j0.4939pu
j0.30pu
j0.06835pu
j0.28pu
Bus 3
Bus 2
j0.08544 pu
EG4
j0.4860pu
j0.06835pu
Bus 1
EG3
EG2
EG1
j0.4939pu
j0.7605pu
j0.06835pu
j0.28pu
Bus 3
j0.1733 pu
EG4
j0.4860pu
j0.6083pu
Positive Sequence.
Bus 1
j0.4939pu
j0.30pu
j0.06835pu
j0.28pu
Bus 3
Bus 2
j0.08544 pu
j0.5981pu
j0.06835pu
Bus 1
j0.4939pu
j0.7605pu
j0.06835pu
j0.28pu
Bus 3
j0.1733 pu
j0.6083puj0.5981pu
Negative Sequence.
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
5
Bus 1
j0.1 pu
j0.1709pu
j0.07pu j0.339pu
Bus 3
Bus 2
j0.1pu
j0.2563 pu j0.09116puj0.1709pu
Bus 1
j0.1 pu
j0.1709pu
j0.07pu j0.339pu
Bus 3
j0.5063pu
j0.09116puj0.8652pu
j0.3376pu
Zero Sequence.
So from these simplified networks, the Thevenin equivalent impedances can be derived looking in at bus 1, as:
( ) ( )_ 1 0.28 // 0.7605 // 0.06835 // 0.1733 0.4939 // 0.4860 // 0.6083THZ j j j j j j j⎡ ⎤= +⎣ ⎦
_1 0.1069THZ j=
And:
( ) ( )_ 2 0.28 // 0.7605 // 0.06835 // 0.1733 0.4939 // 0.5981// 0.6083THZ j j j j j j j⎡ ⎤= +⎣ ⎦
_ 2 0.1097THZ j=
And:
( ) ( )_ 0 0.1 // 0.5063// 0.1709 // 0.8652 0.3376 // 0.09116THZ j j j j j j⎡ ⎤= +⎣ ⎦
_ 0 0.0601THZ j=
Question 3. For a bolted three phase fault, the fault current is given by:
0 2 0I I= = ,
00
1_1
1 09.355 . . 90
0.1069F
TH
VI p u
Z j
∠= = = ∠ −
Similarly :
1 9.355 . .a b cI I I I p u= = = =
So in ampere:
1 7.06a b cI I I I kA= = = =
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
6
Question 4. For a single line to ground fault the sequence networks are connected in series as:
I1 ZTh1
VF
I2
I0
V1
V2
V0
ZTh2
ZTh0
Positive Sequence
Negative Sequence
Zero Sequence
Hence the sequence currents are:
00
0 1 2_1 _ 2 _ 0
1 03.614 . . 90
0.2767F
TH TH TH
VI I I p u
Z Z Z j
∠= = = = = ∠ −+ +
And:
0b cI I= =
013 10.84 . . 90aI I p u= = ∠ −
In amperes. 08.183 . 90aI kA= ∠ −
The sequence voltages are:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
0
0
0
0
90
90
90
0
2
1
0
2
1
0
2_
1_
0_
2
1
0
614.3
614.3
614.3
1097.000
01069.00
000601.0
0
0.1
0
00
00
00
0
0
j
j
j
j
TH
TH
TH
F
e
e
e
j
j
j
e
V
V
V
I
I
I
Z
Z
Z
V
V
V
V
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
7
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
0
0
2
1
0
3965.0
6137.0
2172.0
j
j
j
e
e
e
V
V
V
Hence the phase to ground voltages are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
0
0
00
00
0
0
0
120120
120120
2
1
0
2
2
3965.0
6137.0
2172.0
1
1
111
1
1
111
j
j
j
jj
jj
c
b
a
e
e
e
ee
ee
V
V
V
aa
aa
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+−
−+−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
00
00
120120
120120
3965.06137.02172.0
3965.06137.02172.0
0
jj
jj
c
b
a
ee
ee
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠−
−∠=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−
−−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
4.110..9336.0
4.110..9336.0
0
8749.03258.0
8749.03258.0
0
up
up
j
j
V
V
V
c
b
a
Question 5. The fault impedance in per unit is:
30
0.0513585.23FZ = =
The connection of the sequence networks is then as shown below:
I1 ZTh1
VF
I2
I0
V1
V2
V0
ZTh2
ZTh0
Positive Sequence
Negative Sequence
Zero Sequence
3ZF
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
8
So for a single line to ground fault through this impedance:
00
0 1 2_1 _ 2 _ 0
1 02.322 . . 90
3 0.4306F
TH TH TH F
VI I I p u
Z Z Z Z j
∠= = = = = ∠ −+ + +
And:
0b cI I= =
013 6.967 . . 90aI I p u= = ∠ −
In amperes. 05.258 . 90aI kA= ∠ −
The sequence voltages are:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
0
0
0
0
90
90
90
0
2
1
0
2
1
0
2_
1_
0_
2
1
0
322.2
322.2
322.2
1097.000
01069.00
000601.0
0
0.1
0
00
00
00
0
0
j
j
j
j
TH
TH
TH
F
e
e
e
j
j
j
e
V
V
V
I
I
I
Z
Z
Z
V
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
0
0
2
1
0
2547.0
7518.0
1396.0
j
j
j
e
e
e
V
V
V
Hence the phase to ground voltages are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
0
0
00
00
0
0
0
120120
120120
2
1
0
2
2
2547.0
7518.0
1396.0
1
1
111
1
1
111
j
j
j
jj
jj
c
b
a
e
e
e
ee
ee
V
V
V
aa
aa
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+−
−+−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
00
00
0
120120
120120
0
2547.07518.01396.0
2547.07518.01396.0
3575.0
jj
jj
j
c
b
a
ee
ee
e
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠−
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−
−−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
00
114..9542.0
114..9542.0
0..3575.0
8717.03882.0
8717.03882.0
3575.00
up
up
up
j
j
e
V
V
V j
c
b
a
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
9
Question 6. For a bolted line to line fault the sequence networks are connected as shown below:
I1 ZTh1
VF
I2
I0
V1
V2
V0
ZTh2
ZTh0
Positive Sequence
Negative Sequence
Zero Sequence
The sequence currents are therefore:
0 0I =
00
1 2_1 _ 2
1 04.617 . . 90
0.2166F
TH TH
VI I p u
Z Z j
∠= − = = = ∠ −+
And:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
−
−
0
0
00
00
90
90
120120
120120
2
1
0
2
2
617.4
617.4
0
1
1
111
1
1
111
j
j
jj
jj
c
b
a
e
e
ee
ee
I
I
I
aa
aa
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
..997.7
..997.7
0
up
up
I
I
I
c
b
a
In amperes:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
kA
kA
I
I
I
c
b
a
035.6
035.6
0
The sequence voltages are:
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
10
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
0
00
90
900
2
1
0
2
1
0
2_
1_
0_
2
1
0
617.4
617.4
0
1097.000
01069.00
000601.0
0
0.1
0
00
00
00
0
0
j
jj
TH
TH
TH
F
e
e
j
j
j
e
V
V
V
I
I
I
Z
Z
Z
V
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
5064.0
5064.0
0
5064.0
4936.01
0
2
1
0
V
V
V
Hence the phase to ground voltages are given by:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
5064.0
5064.0
0
1
1
111
1
1
111
00
00
120120
120120
2
1
0
2
2
jj
jj
c
b
a
ee
ee
V
V
V
aa
aa
V
V
V
( )( )⎥⎥
⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
00
00
120120
120120
0
5064.0
5064.0
0..0128.1
jj
jj
c
b
a
ee
ee
up
V
V
V
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
180..5064.0
180..5064.0
0..0128.1
up
up
up
V
V
V
c
b
a
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
11
Question 7. Recall that for a bolted double line to ground fault, the sequence networks are connected together
as shown below:
I1 ZTh1
VF
I2
I0
V1
V2
V0
ZTh2
ZTh0
Positive Sequence
Negative Sequence
Zero Sequence
Therefore the positive sequence fault current is given by:
0601.0//1097.01069.0
0.1
// 0_2_1_1 jjjZZZ
VI
THTHTH
F
+=
+=
00
1 90..862.61457.0
00.1 −∠=∠= upj
I
The negative and zero sequence currents can be determined by current divider rule, as:
1097.00601.0
0601.090862.6 0
2_0_
0_12 jj
j
ZZ
ZII
THTH
TH
+−∠−=
+−=
02 90..429.2 ∠= upI
Similarly:
1097.00601.0
1097.090862.6 0
2_0_
2_10 jj
j
ZZ
ZII
THTH
TH
+−∠−=
+−=
00 90..433.4 ∠= upI
The phase currents are therefore:
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
12
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
0
0
0
90
90
90
2
2
429.2
862.6
433.4
1
1
111
j
j
j
c
b
a
e
e
e
aa
aa
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+−
−+−
−+−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−−
00
000
120120
12012090
429.2862.6433.4
429.2862.6433.4
429.2862.6433.4
jj
jjj
c
b
a
ee
eee
I
I
I
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−
−−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
0
0
0
000
6.39
4.140
6.129
4.2309090
..44.10
..44.10
0
44.10
44.10
0
046.865.6
046.865.6
0
j
j
j
jjj
c
b
a
eup
eup
e
ee
j
je
I
I
I
So in amperes:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
6.39
4.140
879.7
879.7
0
j
j
c
b
a
ekA
ekA
I
I
I
The sequence voltages are:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−∠−
−∠
−∠−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
2
1
0
2
1
0
2_
1_
0_
2
1
0
90429.2
90862.6
90433.4
1097.000
01069.00
000601.0
0
0.1
0
00
00
00
0
0
0
j
j
j
e
V
V
V
I
I
I
Z
Z
Z
V
V
V
V
j
TH
TH
TH
F
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
0
0
2
1
0
0..2664.0
0..2664.0
0..2664.0
02664.0
7335.01
02664.0
up
up
up
V
V
V
The phase voltages are then:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
2664.0
2664.0
2664.0
1
1
111
1
1
111
00
00
120120
120120
2
1
0
2
2
jj
jj
c
b
a
ee
ee
V
V
V
aa
aa
V
V
V
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
13
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ ∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0..7993.0 0up
V
V
V
c
b
a
Question 8. The phase voltages and phase fault currents in the above cases are:
Three Phase Fault.
1 7.06a b cI I I I kA= = = =
Single Line to Ground Fault.
08.183 . 90aI kA= ∠ −
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠−
−∠=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
4.110..9336.0
4.110..9336.0
0
up
up
V
V
V
c
b
a
Single Line to Ground Fault Through an Impedance.
013 6.967 . . 90aI I p u= = ∠ −
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠−
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
114..9542.0
114..9542.0
0..3575.0
up
up
up
V
V
V
c
b
a
Line to Line Fault.
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
..997.7
..997.7
0
up
up
I
I
I
c
b
a
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
−∠
∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0
180..5064.0
180..5064.0
0..0128.1
up
up
up
V
V
V
c
b
a
Double Line to Ground Fault.
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∠
∠=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
6.39..44.10
4.140..44.10
0
up
up
I
I
I
c
b
a
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ ∠
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
0
0
0..7993.0 0up
V
V
V
c
b
a
The following observations apply:
• The Double Line to Ground Fault leads to the worst case fault current.
• Line to Line voltages lead to an increase in the un-faulted phase voltage.
ELEC 4100 TUTORIAL NINE : FAULT STUDIES
14
Question 9. Now since:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
2
1
0
2
2
1
1
111
I
I
I
aa
aa
I
I
I
c
b
a
Then:
[ ] 0210 =++= IIIIa : Then via KCL, the three sequence currents must form a node.
And:
[ ] 0'''' 210 =++= IIIIa : Then via KCL, the three sequence currents must form a node.
Similarly:
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
3
3
3
0
0
1
1
111
3
1
1
1
111
3
1
'
'
''
2
2
'
'
'
2
2
'22
'11
'00
aa
aa
aaaa
cc
bb
aa
V
V
VV
aa
aa
V
V
V
aa
aa
V
V
V
The three sequence voltages are therefore equal in magnitude:
I2
V00'
I0
I1V11'
V22'
I2'
I0'
I1'
ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 10 : TRANSIENT STABILITY - SOLUTIONS
Question 1.
The pre-fault electrical power delivered to the infinite bus is given by:
δsin
_
_
EQLtd
prefaulteXXX
EVP
++
=∞
( )( )( ) ( )
δsin2.01.0//2.01.03.0
0.128.1_
+++
=prefaulteP
δsin462.2_
=prefaulteP
During the fault, a star-delta transform can be applied to simplify the circuit as shown below:
XL12
XL13
X'd
XT1
XEQ
The XEQ is given by:
( ) ( )12 13 13 12
13
d t L d t L L L
EQ
X X X X X X X XX
X
+ + + +
=
( ) ( ) ( )0.3 0.1 0.2 0.3 0.1 0.1 0.1 0.2
0.1EQX
+ + + +
=
1.4EQX =
So the faulted electrical power delivered to the infinite bus is given by:
_
sine fault
EQ
EVP
Xδ
∞
=
( ) ( )_
1.28 1.0sin
1.4e faultP δ=
_
0.9143 sine faultP δ=
The post-fault electrical power delivered to the infinite bus is given by:
_
_
sine postfault
d t L EQ
EVP
X X Xδ
∞
=
+ +
( ) ( )_
1.28 1.0sin
0.3 0.1 0.2e postfaultP δ=
+ +
ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS
2
_
2.133sine prefaultP δ=
To apply the equal area criterion, it is necessary to determine the maximum possible swing
angle, and the initial operating angle. This is done as follows:
_ 02.462 sin 1.0e prefault mP Pδ= = =
1
0
1.0sin
2.462δ
−
⎛ ⎞= ⎜ ⎟
⎝ ⎠
0
0 23.96δ =
Similarly:
( )0
_ 2.133sin 180 1.0e prefault m mP Pδ= − = =
0 1 1.0180 sin
2.133m
δ−
⎛ ⎞= − ⎜ ⎟
⎝ ⎠
0152
mδ =
Now apply the equal area criterion.
( ) ( )1
0 1
1.0 0.9143 sin 2.133sin 1.0
m
d d
δδ
δ δ
δ δ δ δ− = −∫ ∫
( ) ( )0 0 12.133cos 0.9143cos 2.133 0.9143 cos
m mδ δ δ δ δ− + − = −
10.4835 1.219cosδ− =
1
1
0.4835cos
1.219δ
−
−⎛ ⎞= ⎜ ⎟
⎝ ⎠
0
1113.4δ =
This is the critical clearing angle.
Question 2.
Again apply the equal angle criterion, but with:
0
151.2δ =
So the equal area criterion requires that:
( ) ( )max1
0 1
1.0 0.9143sin 2.133sin 1.0d d
δδ
δ δ
δ δ δ δ− = −∫ ∫
( )max 0 1 0 max 10.9143cos 0.9143cos 2.133cos 2.133cosδ δ δ δ δ δ− + − = − +
( )max max 0 1 02.133cos 2.133 0.9143 cos 0.9143cosδ δ δ δ δ+ = + − +
max max2.133cos 2.017δ δ+ =
ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS
3
This is a non-linear equation, which can be solved iteratively using the Newton-Raphson
method. Recall that the NR method approximates a solution using the local gradient of the function
as:
( )( )
1
'
p
p p
p
y f xx x
f x
+
⎡ ⎤−⎣ ⎦= +
So:
( )1
1
max max max max max1 2.133sin 2.017 2.133cos
p p p p pδ δ δ δ δ
−
+ ⎡ ⎤= + − − −⎣ ⎦
Solving Iteratively:
1
max1.0δ = , 2
max1.1704δ = , 3
max1.1546δ = , 4
max1.1545δ = ,
5
max1.1545δ = , 6
max1.1545δ = , 7
max1.1545δ = 8
max1.1545δ =
So after 8 iterations the solution is:
0
max66.15δ =
Since this rotor angle is well below the maximum possible stable rotor angle, the generator can
remain synchronised to the infinite bus.
Question 3.
Substituting in the system parameters.
2
21.0 0.03183 0.01 2.462sin
d d
dt dt
δ δδ= + +
Or:
2
231.42 0.3142 77.35sin
d d
dt dt
δ δδ= + +
Now consider small deviations in the rotor angle. If the rotor angle changes from 0
δ to 0
δ δ+ ∆ ,
then:
( ) ( ) ( ) ( ) ( )
( ) ( )
0 0 0
0 0
sin sin cos cos sin
sin cos
δ δ δ δ δ δ
δ δ δ∆
+ ∆ = ∆ + ∆
= +
So:
2
0 0
0231.42 0.3142 77.35sin
d d
dt dt
δ δδ= + +
Changes to:
( ) ( )( ) ( )
2
0 0
0 0231.42 0.3142 77.35 sin cos
d d
dt dt
δ δ δ δδ δ δ
∆ ∆
∆
+ += + + +⎡ ⎤⎣ ⎦
The difference becomes:
ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS
4
( )2
020 0.3142 77.35cos
d d
dt dt
δ δδ δ
∆ ∆
∆= + +
Taking the Laplace transform:
( ) ( )2
00 0.3142 77.35coss s sδ δ
∆⎡ ⎤= + +⎣ ⎦
Solving for the roots of the quadratic equation:
( )01 2
0.3142 0.0987 309.4cos,
2s s
δ− ± −
=
For the system to be stable, it is necessary that:
( )00.3142 0.0987 309.4cos 0δ− + − <
( )00.0987 309.4cos 0.3142δ− <
( )00.0987 309.4cos 0.0987δ− <
( )0309.4cos 0δ >
For ( )0cos 0δ > , it is necessary to have 0 0
090 90δ− < < + . This is the small signal stability
constraint on the system.
ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 11 : PROTECTION - SOLUTIONS Question 1.
The input current to a Westinghouse CO-8 relay is 10A. Determine the relay operating time for the following current tap settings (CTS) and time dial settings (TDS).
(a) CTS = 1.0, TDS = 0.5.
(b) CTS = 2.0, TDS = 1.5.
(c) CTS = 2.0, TDS = 7.
(d) CTS = 3.0, TDS = 7.
(e) CTS = 12.0, TDS = 1.
Answer: From the inverse time curves :
(a) Time to operate : 0.1s.
(b) Time to operate : 0.55s.
(c) Time to operate : 3.0s.
(d) Time to operate : 5.2s.
(e) The breaker can not operate – the current is less than the pick-up current.
Question 2. For the system shown in figure 1, directional over-current relays are used at breakers B12, B21,
B23, B32, B34 and B43. Over-current relays alone are used at B1 and B4. (a) For a fault at P1, which breakers do not operate? Which breakers should be coordinated? Repeat (a) for a fault at (b) P2, (c) P3. (d) Explain how the system is protected against bus faults.
Bus 1
B1B12
Bus 2
L1 L2
Bus 3 Bus 4
L3 L4
B21B23 B32 B34 B43 B4
P1P2P3
Figure 1
Answer : (a) For a fault at P1, only B34 and B43 should operate. If B34 fails to operate, then B23, B12
and B1 would operate as a backup. So B23, B12 and B1 must coordinate with B34 in the sequence (B34 – B23 – B12 – B1). If B43 fails to operate, B4 would operate as a backup, so B4 must coordinate with B43 in the sequence (B43 – B4).
(b) For a fault at P2, only B23 and B32 should operate. As backup protection, B12 and B1 should coordinate with B23 in the sequence (B23 – B12 – B1), and B43 and B4 should coordinate with B32 in the sequence (B32 – B43 – B4).
ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS
2
(c) For a fault at P3, only B12 and B21 should operate. As backup protection, B1 should coordinate with B12 in the sequence (B12 – B1), and B32, B43 and B4 should coordinate with B21 in the sequence (B21 – B32 – B43 – B4).
(d) Fault at Bus 1 : Breakers B1 and B21 should open.
Fault at Bus 2 : Breakers B12 and B32 should open.
Fault at Bus 3 : Breakers B23 and B43 should open.
Fault at Bus 4 : Breakers B34 and B4 should open.
Question 3. (a) Draw the protective zones for the power system shown in figure 2. (b) Which circuit breakers
should open for a fault at (i) P1, (ii) P2, (iii) P3? (c) For case (i), if circuit breaker B21a failed to operate, which circuit breakers would open as back-up?
Bus 1 Bus 2
B1
Bus 3
P1Bus 4
B12a
B12b
B13
B31
B3
B46
B32
B32
B21b
B21a
B24a B42a
B24b B42b
P2
P3
Figure 2
Answer : (a) The figure below shows the protective zones of the system in figure 2.
Bus 1 Bus 2
B1
Bus 3
P1
Bus 4B12a
B12b
B13
B31
B3
B46
B32
B32
B21b
B21a
B24a B42a
B24b B42b
P2
P3Zone 1
Zone 2 Zone 3
Zone 4
Zone 5
Zone 6
Zone 7
Zone 8
Zone 9
Zone 10
Zone 11 Zone 12
Zone 13
ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS
3
(b i) For a fault at P1, breakers in Zone 3 should operate – i.e. B12a and B21a.
(b ii) For a fault at P2, breakers in Zone 9 should operate – i.e. B21a, B21b, B23, B24a and B24b.
(b iii) For a fault at P3, breakers in both Zone 6 and Zone 9 should operate – i.e. B21a, B21b, B23, B32, B24a and B24b.
(c) If in case b(i), B21a did not operate, then back-up protection would be achieved by opening the breakers in Zone 9 – i.e. B21b, B23, B24a and B24b.
Question 4. Three-zone mho relays are used for transmission line protection of the power system shown in
figure 3. Positive sequence line impedances are given as follows:
• Line 1-2 : Z12_1 = (6+j60) Ω
• Line 2-3 : Z23_1 = (5+j50) Ω
• Line 2-4 : Z24_1 = (4+j40) Ω
Rated voltage for the high voltage buses is 500kV. Assume a 1500:5 CT ratio and a 4500:1 VT ratio at B12. (a) Determine the zone 1, zone 2 and zone 3 settings Zr1, Zr2 and Zr3 for the mho relay at B12 if zone 1 is set for 80% reach of line 1-2, zone 2 is set for 120% reach of line 1-2, and zone 3 is set to cover 120% of adjacent lines. (b) Maximum current for line 1-2 under emergency loading conditions is 1400A at 0.9 p.f. lagging. Verify that B12 does not trip during emergency loading conditions.
1
2
3
Bus 1 Bus 3
Bus 4
line 1-2
Bus 2
line 2-3
line 2-4
B1
B4
B3B12B21 B23 B32
B24
B42
Figure 3.
Answer: Part (a):
The impedance seen by the mho relay at B12 is:
( )( )
4500 /1''
' 1500 / 5
1'
15 15
LNLN
L L
LN
L
VVZ
I I
V ZZ
I
= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
Set the B12 relay zone 1 Zr1 setting for 80% reach of line 1-2, as:
ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS
4
( )1
6 600.8 0.32 3.2
15r
jZ j
+⎡ ⎤= = + Ω⎢ ⎥⎣ ⎦ secondary
Set the B12 relay zone 2 Zr2 setting for 120% reach of line 1-2, as:
( )1
6 601.2 0.48 4.8
15r
jZ j
+⎡ ⎤= = + Ω⎢ ⎥⎣ ⎦ secondary
Set the B12 relay zone 3 Zr3 setting for 100% reach of line 1-2, and 120% reach of line 2-3, as:
( )1
6 60 5 501.2 0.8 8.0
15 15r
j jZ j
+ +⎡ ⎤ ⎡ ⎤= + = + Ω⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ secondary
Part (b)
The secondary impedance viewed by B12 during emergency loading is:
0
01
5000ˆ ' 13ˆ ' 13.7 25.8
ˆ 1.4 cos 0.9 15'LN
L
VZ
I −
∠= = = ∠ Ω
∠ −
This value is well in excess of the zone 3 impedance setting, so the impedance seen by B12 during emergency loading will not trip the three zone mho relay.
Question 5. Line impedances for the power system shown in figure 4 are Z12 = Z23 = (3+j40) Ω, and
Z24 = (6+j80)Ω. Reach for the zone 3 B12 impedance relay is set for 100% of line 1-2 plus 120% of line 2-4. (a) for a bolted three phase fault at bus 4, show that the apparent primary impedance “seen” by the B12 relay is:
3212 24 24
12apparent
IZ Z Z ZI⎛ ⎞= + + ⎜ ⎟⎝ ⎠
Where ( )32 12I I is the line 2-3 to line 1-2 fault current ratio. (b) if 32 12 0.2I I > , does the B12
relay see the fault at bus 4?
NOTE: This problem illustrates the “infeed effect”. Fault currents from line 2-3 can cause the zone 3 B12 relay to under-reach. As such, remote backup of line 2-4 at B12 is ineffective.
1 3
Bus 1 Bus 3
line 1-2
Bus 2
line 2-3
B1 B3B12B21 B23 B32
B24
I12 I32
Bus 4I24B24 B42
Figure 4.
ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS
5
Answer : Part (a)
For the bolted three phase fault at bus 4 the system of figure 4 can be reduced to:
Z12 Z23
Z24V2
I24
I12 I32
V1 V3
The primary impedance seen by the B12 relay is then:
1 1 2 2 212
12 12 12 12
' ' ' ' '
' ' ' 'apparent
V V V V VZ Z
I I I I
−= = + = +
( )24 12 3212
12
' '
'apparent
Z I IZ Z
I
+= +
3212 24 24
12apparent
IZ Z Z Z
I
⎛ ⎞= + + ⎜ ⎟
⎝ ⎠
Part (b):
The apparent secondary impedance seen by B12 for the bolted three phase fault at bus 4 is:
( )' apparent
apparentV I
ZZ
N N=
Where NV and NI are the turns ratios of the potential and current transformers for B12.
( )( )
( ) ( )( )( )
12 24 32 12 32 121 3 40 6 80 1'apparent
V I V I
Z Z I I j j I IZ
N N N N
+ + + + + += =
( )( ) ( )( )( )
32 12 32 129 6 120 80'apparent
V I
I I j I IZ
N N
+ + +=
Also the zone 3 relay for B12 is set for 100% reach of line 1-2, and 120% of line 2-4. So:
( ) ( )( ) ( )3
3 40 1.2 6 80 10.2 136r
V I V I
j j jZ
N N N N
+ + + += = Ω
Comparing this expression with the case for the balanced three phase fault when 32 12 0.2I I >
shows that:
( )10.2 136
'apparentV I
jZ
N N
+>
So the apparent impedance for the three phase fault is greater than the zone 3 set point, and B12 will not trip in this case. Remote backup in this case is not effective.
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
1
ELEC 4100 ELECTRICAL ENERGY SYSTEMS
TUTORIAL 12 : TRANSMISSION LINES – SOLUTIONS
Question 1.
A single phase transmission line of 1p.u. length with distributed parameters R, L, C and G has a
step voltage E applied to the sending end of the line. The general solutions to the transmission line
partial differential equations is given by:
( ) 1 2, x xV x s k e k eγ γ−= + and ( ) 1 2
0
1, x xI x s k e k e
Z
γ γ− = −
Where: ( )( )R sL G sCγ = + + ( )( )0
R sLZ
G sC
+=
+
a) For the case where the receiving end of the transmission line is short circuited
determine the constants k1 and k2, and derive simplified expressions for the voltage
and current.
b) For the case where the receiving end of the transmission line is open circuited
determine the constants k1 and k2, and derive simplified expressions for the voltage
and current.
Answer :
Part (a)
The boundary conditions are :
( )0,V s E= , ( )1, 0V s =
Substituting these constraints into the general solutions gives:
( ) 1 20,V s k k E= + =
( ) 1 21, 0V s k e k eγ γ−= + =
From the second equation:
2
1 2k k e γ= −
Substituting this expression into the first equation gives:
( )2
21k e Eγ− = or
( ) ( ) ( )2 2 2sinh1
E e E e Ek
e e e
γ γ
γ γ γ γ
− −
−= = = −
− −
So:
( )2
1 22sinh
e Ek k e
γγ
γ= − =
The voltage expression for the transmission line is then given by:
( )( )
( ) ( )1 1,
2sinh
x xEV x s e e
γ γ
γ− − − = −
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
2
( )( )( )( )
sinh 1,
sinh
xV x s E
γ
γ
−=
The current expression for the transmission line is then given by:
( )( )
( ) ( )1 1
0
1,
2sinh
x xEI x s e e
Z
γ γ
γ− − − = +
( )( )( )( )0
cosh 1,
sinh
xEI x s
Z
γ
γ
−=
Part (b)
The boundary conditions are :
( )0,V s E= , ( )1, 0I s =
Substituting these constraints into the above expressions for the voltage and current reveals:
( ) 1 20,V s k k E= + =
( ) 1 2
0
11, 0I s k e k e
Z
γ γ− = − =
From the second expression involving the current at the receiving end of the line:
2
1 2k k e γ=
Substituting into the expression for the voltage at the sending end of the line:
( )2
21k e Eγ+ = or
( )2 21 2cosh
E e E e Ek
e e e
γ γ
γ γ γ γ
− −
−= = =
+ +
And so:
( )
2
1 21 2cosh
e E e E e Ek
e e e
γ γ γ
γ γ γ γ−= = =
+ +
Hence:
( )( ) ( )
,2cosh 2cosh
x xe E e EV x s e e
γ γγ γ
γ γ
−−= +
( )( )( )( )
cosh 1,
cosh
xV x s E
γ
γ
−=
Similarly
( )( ) ( )0
1,
2cosh 2cosh
x xe E e EI x s e e
Z
γ γγ γ
γ γ
−−
= −
( )( )( )( )0
sinh 1,
cosh
xEI x s
Z
γ
γ
−=
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
3
Question 2.
A single phase lossless transmission line of 150m length has an inductance and shunt
capacitance per unit length of L = 1µH/m and C = 11.111 pF/m. The line is terminated by a 600Ω resistance. The transmission line is struck by lightning through an effective 100Ω impedance at the sending end of the line, creating a surge voltage of 30kV peak and 50µs duration.
a) Determine the characteristic impedance, travelling wave propagation velocity and the
one-way transit time for the transmission line.
b) Draw the equivalent circuit of the transmission line under the surge voltage conditions,
and calculate the reflection coefficients at each end of the transmission line.
c) Plot the voltages at the sending and receiving ends of the line for the first 5µs.
Answer:
Part (a)
The characteristic impedance is given by:
6
0 12
1 10300
11.111 10
LZ
C
−
−
×= = = Ω
×
The travelling wave propagation velocity is :
( )( )8 1
0 12 6
1 13 10
11.111 10 1 10v ms
LC
−
− −= = = ×
× ×
This is the speed of light in free space. The one-way transit time for the transmission line is then:
( )( )
7
8 10
1505 10 0.5
3 10
mds s
v msτ µ−
−= = = × =
×
Part (b)
v(0,t) v(d,t)
30kV
Z0 = 300 Ω
τ = 0.5µs
ZL= 600 Ω
100Ω
The reflection coefficients at the sending and receiving ends of the line are then:
0
0
600 300 1
600 300 3
LR
L
Z Z
Z Z
− −Γ = = =
+ +
0
0
100 300 1
100 300 2
SL
S
Z Z
Z Z
− −Γ = = = −
+ +
The initial surge voltage on the line is then: ( )(0,0) 30 300 100 300 22.5v kV kV= Ω Ω+ Ω =
Part (c):
The Bewley lattice diagram can be developed as shown below:
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
4
τ = 0.5µs
2τ = 1.0µs
3τ = 1.5µs
4τ = 2.0µs
5τ = 2.5µs
6τ = 3.0µs
7τ = 3.5µs
22.5 kV
x = 0 x = d
8τ = 4.0µs
9τ = 4.5µs
10τ = 5.0µs
7.5 kV
-3.750 kV
-1.25 kV
0.625 kV
0.2083 kV
-0.1042 kV
-0.0347 kV
0.0174 kV
0.0058 kV
-0.0029 kV
From the lattice diagram the sending and receiving end voltages can be developed as shown
below:
v(d,t)
v(0,t)
t
t
30kV
25kV
0.5µs 1.0µs 1.5µs 2.0µs 2.5µs 3.0µs 3.5µs 4.0µs 4.5µs 5.0µs
0.5µs 1.0µs 1.5µs 2.0µs 2.5µs 3.0µs 3.5µs 4.0µs 4.5µs 5.0µs
25.83kV 25.69kV 25.72kV
25.63kV 25.73kV 25.71kV26.25kV22.5kV
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
5
Question 3.
Figure 1 below shows a single phase lossless transmission line composed of two different
sections of underground cable. The first section has a characteristic impedance 100Ω and a one-way propagation time of 0.1ms, while the second section has a characteristic impedance 400Ω and a one-way propagation time of 0.1ms. A surge voltage of 20kV is applied to the line through a 100Ω impedance, and the line is terminated with a 800Ω load impedance. Plot the voltages at the transmission line junction, and the sending and receiving ends of the total line for the first 0.6ms.
Z2=400Ω
100Ωv(0,t) v(d
1+d
2,t)20kV
Z1=100Ωτ = 0.1msτ = 0.1ms
ZL=800Ωv(d1,t)
ZS
Figure 1 : Single Phase lossless transmission line.
Answer:
The reflection coefficient at the sending end of the transmission line is:
1
1
100 1000
100 100
SS
S
Z Z
Z Z
− −Γ = = =
+ +
The reflection and refraction coefficients on the sending side of the junction are:
2 112
2 1
400 100 3
400 100 5
Z Z
Z Z
− −Γ = = =
+ +
( )2
12
1 2
2 4002 8
400 100 5
Z
Z Zβ = = =
+ +
The reflection and refraction coefficients on the receiving side of the junction are:
1 221
2 1
100 400 3
400 100 5
Z Z
Z Z
− −Γ = = = −
+ +
( )1
21
1 2
2 1002 2
400 100 5
Z
Z Zβ = = =
+ +
The reflection coefficient at the receiving end of the transmission line is:
2
2
800 400 1
800 400 3
LR
L
Z Z
Z Z
− −Γ = = =
+ +
The surge voltage entering the line is given by:
( ) 1
1
1000,0 20 10
100 100S
Zv E kV kV
Z Z= = =
+ +
The Bewley lattice diagram can now be developed as shown below:
ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
6
0.1ms
10kV
16kV6kV
5.33kV
2.133kV -3.2kV
-1.067kV
-0.427kV 0.640kV
0.213kV
x = 0 x = d1+d2x = d1
0.2ms
0.3ms
0.4ms
0.5ms
0.6ms
0.7ms
From this lattice diagram it is possible to plot the three relevant voltage profiles as shown below:
10kV
v(d1+d2,t)
v(d1,t)
v(0,t)
t
t
t
τ 2τ 3τ 4τ 5τ 6τ 7τ
τ 2τ 3τ 4τ 5τ 6τ 7τ
τ 2τ 3τ 4τ 5τ 6τ 7τ
16kV
21.33kV
16kV
18.13kV
17.06kV
18.13kV
17.7kV
17.71kV
17.92kV