elec4100tutsol

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


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


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:


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 +=++= ∗∗∗


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:


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


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

+=_


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:


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

+=

∠−∠== ∗


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 ==


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


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:


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=


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:


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


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


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


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)


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)


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


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 ∠=∠=


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


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


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


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


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


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


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


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:


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.


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.


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


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.


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 ====


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 −=


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


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.


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 ==


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.


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.


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= = = =


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


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


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= ∠ −


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

−

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


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

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


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



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


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

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


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:


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


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−∠−

−∠

−∠−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

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


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.


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


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=

+ +


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+ =


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:


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


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).


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.



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


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:


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.


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


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− − − = −


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

−=


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:


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


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:


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

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