ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
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
IR
IR∆
IRstar
IBR
Zstar Z∆
Z∆ IRY
VB
Zstar
Zstar
IB IBstar
IYB IYstar IB∆
IY
VY
Z∆ IY∆
The three line to line voltages are defined as: V RY = 400V ∠30°
V YB = 400V ∠ − 90°
V BR = 400V ∠ − 210°
So for the star load the three phase voltages are defined as: V R =
400V 3
∠0°
V Y =
400V 3
∠ − 120°
The star impedances are : Z star = 2 + j3 = 13∠ tan −1
3 2
The star line currents are a re then given by: I star =
V phs Z star
I Rstar =
V R Z star
=
400V ∠0° 3 × 13∠ tan −1 (3 2)°
= 64.05 A ∠ − 56.31°
1
V B =
400V 3
∠ − 240°
ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION V Y
I Ystar =
Z star
I Bstar =
V B Z star
=
=
400V ∠ − 120° 3 × 13∠ tan −1 (3 2 )° 400V ∠ − 240° 3 × 13∠ tan −1 (3 2)°
= 64.05 A ∠ − 176.31° b
= 64.05 A ∠ − 296.31°
The real and reactive power delivered to the star circuit is then given by: Pstar = 3V phs I star cos(φ ) = 3 (400 )(64.05) cos(56.31°) = 24.62kW Qstar = 3V phs I star sin (φ ) = 3 (400 )(64.05)sin (56.31°) = 36.92kVar
The delta load impedance is given by: Z delta = 6 + j 6 = 6 2 ∠ tan −1
6 6
= 6 2∠45°
For the delta load the phase to phase current is given by: I ∆ =
V LL Z ∆
I RY ∆ =
I YB∆ =
I BR∆ =
400∠30° 6 2∠45°
= 47.14 A∠ − 15°
400∠ − 90° 6 2∠45°
= 47.14 A∠ − 135°
400∠ − 210° 6 2∠45°
= 47.14 A∠ − 255°
The delta line currents are given by: I R∆ =
3 I RY ∆ e
− j 30 0
I Y ∆ = 81.65 A∠ − 165°
= 81.65 A∠ − 45°
The real and reactive power delivered to the delta circuit is given by: P∆ = 3V phs I ∆ phs − phs cos(φ ) = 3(400 )(47.14) cos( 45°) = 40kW
Q∆ = 3V phs I ∆ phs − phs sin(φ ) = 3(400)(47.14) sin(45°) = 40kVar
The overall input line current c urrent is given by: I R = I Rstar + I R∆ = 64.05 A∠ − 56.31° + 81.65 A∠ − 45°
= 145.0 A∠ − 49.97° I Y = I Ystar + I Y ∆ = 64.05 A∠ − 176.31° + 81.65 A∠ − 165°
= 145.0 A∠ − 169.97° I B = I Bstar + I B∆ = 64.05 A∠ − 296.31° + 81.65 A∠ − 285°
= 145.0 A∠ − 289.97°
Total Power Consumption:
)(145) cos(49.97°) = 64.62kW Ptot = 3V LL I L cos(φ ) = 3 (400)( )(47.14)sin(49.97°) = 76.92kVar Qtot = 3V LL I L sin (φ ) = 3(400)( The overall power factor is: p. f . = cos(49.97°) = 0.6432 2
I B∆ = 81.65 A∠ − 285°
ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION V Y
I Ystar =
Z star
I Bstar =
V B Z star
=
=
400V ∠ − 120° 3 × 13∠ tan −1 (3 2 )° 400V ∠ − 240° 3 × 13∠ tan −1 (3 2)°
= 64.05 A ∠ − 176.31° b
= 64.05 A ∠ − 296.31°
The real and reactive power delivered to the star circuit is then given by: Pstar = 3V phs I star cos(φ ) = 3 (400 )(64.05) cos(56.31°) = 24.62kW Qstar = 3V phs I star sin (φ ) = 3 (400 )(64.05)sin (56.31°) = 36.92kVar
The delta load impedance is given by: Z delta = 6 + j 6 = 6 2 ∠ tan −1
6 6
= 6 2∠45°
For the delta load the phase to phase current is given by: I ∆ =
V LL Z ∆
I RY ∆ =
I YB∆ =
I BR∆ =
400∠30° 6 2∠45°
= 47.14 A∠ − 15°
400∠ − 90° 6 2∠45°
= 47.14 A∠ − 135°
400∠ − 210° 6 2∠45°
= 47.14 A∠ − 255°
The delta line currents are given by: I R∆ =
3 I RY ∆ e
− j 30 0
I Y ∆ = 81.65 A∠ − 165°
= 81.65 A∠ − 45°
The real and reactive power delivered to the delta circuit is given by: P∆ = 3V phs I ∆ phs − phs cos(φ ) = 3(400 )(47.14) cos( 45°) = 40kW
Q∆ = 3V phs I ∆ phs − phs sin(φ ) = 3(400)(47.14) sin(45°) = 40kVar
The overall input line current c urrent is given by: I R = I Rstar + I R∆ = 64.05 A∠ − 56.31° + 81.65 A∠ − 45°
= 145.0 A∠ − 49.97° I Y = I Ystar + I Y ∆ = 64.05 A∠ − 176.31° + 81.65 A∠ − 165°
= 145.0 A∠ − 169.97° I B = I Bstar + I B∆ = 64.05 A∠ − 296.31° + 81.65 A∠ − 285°
= 145.0 A∠ − 289.97°
Total Power Consumption:
)(145) cos(49.97°) = 64.62kW Ptot = 3V LL I L cos(φ ) = 3 (400)( )(47.14)sin(49.97°) = 76.92kVar Qtot = 3V LL I L sin (φ ) = 3(400)( The overall power factor is: p. f . = cos(49.97°) = 0.6432 2
I B∆ = 81.65 A∠ − 285°
ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
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 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 120µ 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. l agging. 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
IR
IR∆
IRstar
IBR
12Ω
ZMot
ZMot 120µf
IRY
(2+j8)Ω IYB
IB
VB
ZMot
IBstar IB∆
IY
VY VN
IY∆
IN
Since the unbalanced load has a neutral point connected, then t hen 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: V RY = 415V ∠30°
V YB = 415V ∠ − 90°
V BR = 415V ∠ − 210°
So for the star load the three phase voltages are defined as: V R =
415V 3
∠0°
V Y =
415V 3
∠ − 120°
V B =
First calculate the R-phase current due to the load in part (a): I R _ star =
V RN R
=
415∠00 312
= 19.9667 A∠00
The current in the B-phase inductor in part (b) is given by: 3
415V 3
∠ − 240°
ELEC 4100
I B _ star =
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
V BN Z
415∠ − 2400
=
3 (2 + j8)
= 29.0558 A ∠44.0362
0
The RY current in the capacitor in part c is given by: I RY _ cap =
V RY Z c
=
415∠300 1 j (2π 50 )(120 µ f )
= 15.6451 A∠1200
For the inductive load the real power is 10kW at 0.75pf lagging. So the power factor angle is: φ = cos −1 (0.75) = 41.410
So the reactive power is given by: 0
Q = P tan φ = 10,000 tan 41.41 = 8.192 kVar
Since: S = P + jQ = 3V phs I L
∗
Then the three line currents feeding the induction machine load are given by: I R _ ∆ =
I Y _ ∆ =
I B _ ∆ =
10,000 + j8192 415
∗
= 18.55 A∠ − 41.41°
3 ∠0°
∗
10,000 + j8192 415
3 ∠ − 120°
10,000 + j8192 415
3 ∠120°
= 18.55 A∠ − 161.41°
∗
= 18.55 A∠78.59°
So the four supply line currents become: I R = I R _ star + I RY _ cap + I R _ ∆ = 26.09 A ∠ 2.82
0
I Y = − I RY _ cap + I Y _ ∆ = 21.77 A ∠ − 116.630 0 I B = I B _ star + I B _ ∆ = 45.56 A ∠57.39
I N = I R _ star + I B _ star = 45.57 A ∠26.310
The total power taken from the supply is given by: ∗
∗
∗
S = V R I R + V Y I Y + V B I B = 16.473kW + j 9.080kVar
4
ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION
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 V RN as the reference. Check that the three currents add to zero. Z RN = 45 + j0 Ω Z YN = 45 – j20 Ω Z BN = 30 + j30 Ω
Solution. The figure below shows the general circuit diagram for a star connected load.
VR
IR
ZA VS
ZY
IY
VY VB
ZB
IB
To calculate the load currents it is necessary to calculate the star point voltage. This requires a reference point, and the voltage V RN can be taken for this point. This voltage is exactly 1 / 3 times the line to line voltage which is precisely known. So we can write: V RY + V YB + V BR = 0 I R + I Y + I B = 0
Also: V RN =
V YN =
V BN =
V RY
3e
V YB
3e
=
j 30°
j 30°
=
V BR
3e
j 30°
=
415∠30° 3e j 30°
= 240V ∠0°
415∠ − 90° 3e j 30°
= 240V ∠ − 120°
415∠ − 210° 3e j 30°
= 240V ∠ − 240°
Then: I R =
V RN − V SN Z R
,
I Y =
V YN − V SN Z Y
,
I B =
Summing the three currents gives:
5
V BN − V SN Z B
ELEC 4100
TUTORIAL ONE : THREE PHASE CIRCUITS - SOLUTION V RN − V SN
I R + I Y + I B = 0 =
+
Z R
V YN − V SN
+
Z Y
V BN − V SN Z B
Or: 1
+
Z R
1 Z Y
+
1
V RN
V SN =
Z B
V YN
+
Z R
Z Y
+
V BN Z B
Now note that: Z R = 45Ω Z Y = 45 − j 20Ω = 49.24e
− j 23.96°
Ω
Z B = 30 + j 30Ω = 42.43e j 45°Ω
So: 1
+
Z R
1 Z Y
+
1
=
Z B
1
1
+
45
49.24e
− j 23.96 0
+
1 42.43e j 45
0
0
= 0.02222 + 0.02031e j 23.96 + 0.02357e − j 45
0
= 0.05745 − j 0.008419 0
= 0.05806e − j 8.34 S
Also : V RN Z R
+
V YN
V BN
+
Z Y
=
Z B
240e j 0 45e j 0
0
0
+
240e − j120
0
49.24e − j 23.96
0
+
0
240e − j 240
0
42.43e j 45
0
0
= 5.333e j 0 + 4.874e − j 96.04 + 5.656e − j 285
0
= 6.284 + j 0.6163 = 6.304e j 5.6
0
Then: V RN V SN =
Z R
1 Z R
=
V YN
+
Z Y
+
1 Z Y
6.304e j
+ +
V BN Z B
1 Z B
5.6 0
0
0.05806 e − j 8.3
= 108.59e j13.95 V = 105.4 + j 26.17V
This is the star point voltage which can now be used to calculate the line currents: I R =
240 − 105.6 − j 26.14 45
= 3.039 A ∠ − 11.040
0
I Y =
240e − j120 − 105.6 − j 26.14 49.24e
− j 23.96 0
= 6.590 A ∠ − j1100
0
I B =
240e − j 240 − 105.6 − j 26.14 42.43e
j 45 0
= 6.815 A ∠ − j 263.80
6
ELEC 4100
TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
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: Z Y =
Z ∆
3
=
20
0
∠60 Ω
3
Then by voltage divider rule the phase voltage across the equivalent star impedance is given by: V AN _ Y =
Z Y Z Y + Z line
V AN
20
V AN _ Y =
20
3
0 ∠60 480V 3 0 3 ∠60 + 0.8 + j 0.6
V AN _ Y = 243.2V ∠2.96
0
Therefore the line to line voltage across the delta load is given by: V AB _ ∆
=
3V AN _ Y e j 30
0
=
421.3V ∠32.960
Part (b) With a delta connected capacitor bank the new delta impedance becomes: Z ∆
(20∠60 Ω) // (− j 20) (20∠60 Ω )(− j 20Ω ) = 38.64∠ − 15 Ω = 0
=
0
Z ∆
0
20∠60
0
Ω − j 20Ω
Now convert this impedance to an equivalent star impedance: Z Y =
Z ∆
3
0
= 12.88∠ − 15 Ω
Then by voltage divider rule the phase voltage across the equivalent star impedance is given by: V AN _ Y =
Z Y Z Y + Z line
V AN
1
ELEC 4100
TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
V AN _ Y =
12.88 12.88
∠ − 15
0
480V
0
∠ − 15 + 0.8 + j 0.6
3
V AN _ Y = 264.0V ∠ − 3.350
Therefore the line to line voltage across the delta load is given by: V AB _ ∆
=
3V AN _ Y e j 30
0
=
457.3V ∠26.650
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: IG2
IG1 IL
Zline1
Zline2
Zload VG1_AN
VG2_AN
The current I G1 can be calculated based on the power delivered by this generator as: I G1 =
I G1 =
P
3V LL p. f .
∠ − cos
15kW 3 (460 )(0.8)
−1
( p. f .)0
∠ − cos
−1
(0.8.)0 = 23.53 A ∠ − 36.870
The load voltage is then simply the generator voltage less the voltage drop across the line impedance, as:
2
ELEC 4100
TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION V L
= V G1 _ AN − Z line1 I G1
460
=
3
0
∠0 −
(23.53 A ∠ − 36.87 )(1.4 + j1.6) 0
0
= 216.9 V ∠ − 2.73
The line to line load voltage is then: V AB
3V Le j 30
=
0
= 373.7 V ∠27.27
0
Part (b) The load current can be calculated based on the power rating and the load voltage, as: I L
=
I L
=
P
3V AB p. f .
∠ − cos
30kW 3 (375.7 )(0.8 )
−1
( p. f .)0
(
0
∠ − 2.73 − cos
−1
(0.8.)0 ) = 57.63 A ∠ − 39.60
Hence the current I G2 can be calculated as: I G 2 = I L − I G1 = (57.63 A ∠ − 39.60 ) − (25.53 A ∠ − 36.870 ) = 34.14 A ∠ − 41.49
0
Therefore the terminal voltage at generator 2 is given by: V G 2 _ AN = V L + I G1Z line 2 V G 2 _ AN = 216.9V ∠ − 2.730
+
(34.14∠ − 41.49 )(0.8 + j ) 0
V G 2 _ AN = 259.7 V ∠ − 0.630
The line to line generator voltage is then: V G 2 _ AB
=
3V G 2 _ AN e j 30
0
=
449.8V ∠29.370
Part c: The power delivered by the second generator is given by: SG 2
∗
(259.7∠ − 0.63 )(34.14∠41.49 ) 0
= 3V G 2 _ AN I G 2 = 3 =
20.12 kW + j17.4 kVar
3
0
ELEC 4100
TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
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: Ptot = (10,000)(0.8) + 15,000 = 23kW Qtot = (10,000 )sin (− cos−1 (0.8)) + 15,000 tan (cos−1 (0.6 )) = −6000 + 20000 = 14kVar
Hence the power factor is : p. f . = cos tan −1
14kVar 23kW
= 0.8542
lagging
Part (b) The total apparent power is given by: S
=
23kW + j14kVar = 26.93∠31.330
Therefore the load current is given by: I L
=
S
*
3V LL
=
26.93kVA∠ − 31.330 3 (480V ∠0
0
)
0
= 32.39 A ∠ − 31.33
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: Qc
= 14kVar =
3V LL X C
4
ELEC 4100
TUTORIAL TWO : THREE PHASE CIRCUITS PART 2 - SOLUTION
So : X C =
3V LL
=
3(480)
14kVar 14,000
=
49.37 Ω
Part (d) The current in each capacitor is given by: I C =
V LL X C
=
480 49.37 Ω
= 9.72 A
The line current in this case is then simply determined by the real power supplied. So: I L
=
Ptot
3V LL
=
23kW 3 (480)
=
27.66 A
5
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
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 excitati on 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 V base1 = 2400V V base2 = 240V
Therefore: I base1
=
I base2
=
S base V base1
Z base1 =
Z base2
=
Sbase V base2 V base1 I base1
=
V base2 I base2
50,000
=
=
=
2400
=
50,000 240
20.833 A
208.333 A
=
2400 20.8333 240 208.333
= 115.2 Ω
= 1.152 Ω
So the per unit impedances become: Z 1 _ eqpu
Z line _ pu
=
Z 1 _ eq Z base1
=
Z line Z base1
=
1 + j 2.5
=
115.2 1 + j 2 115.2
=
=
0.00868 + j 0.0217 p.u
0.00868 + j 0.01736 p.u
Now the load power is given by: Sload = 50kVA ∠ cos
−1
(0.8) = 50kVA ∠36.90
Or: Sload = 50kVA ∠ cos −1 (0.8) = 1.0 p.u.
∠36.9
So the load current is given by:
1
0
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
I load _ pu
=
∗
Sload
=
V load _ pu
1.0 p.u.
∠36.9
1.0∠0
0
0
∗
= 1.0 p.u. ∠ − 36.9
0
The load current is then: I load = 208.333 A
∠ − 36.9
0
(a) The voltage at the transformer primary terminals is then given by: V 1 _ pu
= V 2 _ pu + I load _ pu Z 1 _ eqpu
V 1 _ pu
= 1.0∠0 +
V 1 _ pu
= 1.02∠0.69
1.0∠ − 36.9 0 (0.00868 + j 0.0217 )
0
0
p.u
The transformer primary voltage is then: V 1
=
2448V ∠0.690
(b) The supply voltage is given by: V S _ pu
= V 2 _ pu + I load _ pu Z 1 _ eqpu +
V S _ pu
= 1.0∠0 +
V S _ pu
= 1.037∠1.15
Z line _ pu
1.0∠ − 36.90 (0.001736 + j 0.03906)
0
0
p.u
The supply voltage is then: V S
=
2489 V ∠1.150
(c) The supply real and reactive power is then given by: ∗
SS
= V S I load
SS
=
SS
= 1.037∠38.02
0
1.037∠1.15 1.0∠36.9 0
=
0
0.8169 p.u. + j 0.6387 p.u
So the real and reactive power are: P = 40.845kW
Q = 31.936kVar
2
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
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 perunit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches a re neglected. Zone 2
Zone 1
Zone 3
Vs = 13 kV
Rload = 5.2 Ω Xline = j0.2 Ω T1 3MVA 11 kV/6.6 k V Xeq = 0.1 p.u.
T2 2MVA 7.2 kV /3.3 kV Xeq = 0.12 p.u.
Figure 1 : Three Zone Distribution System for question 2. Answer :
Choose Bases: Sbase = 3MVA V base1 = 11kV V base2 = 6.6kV V base3
=
I base1
=
I base 2
=
I base3
=
(6.6kV )(3.3kV ) 7.2kV
=
This requires: Sbase
272.727 A
=
V base1 Sbase V base 2 Sbase V base3
=
454.545 A
=
991.736 A
Similarly: Z base1
Z base2
Z base3
=
=
=
V base1
2
=
Sbase V base 2
2
Sbase
V base3
40.333 Ω
= 14.520 Ω
2
Sbase
=
3.050 Ω
So the per unit impedance values are: 3
3.025kV
Xload = j2.9 Ω
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION Z load
=
Z eq _ T 1 pu
= j 0.1 p.u.
Z line _ pu
=
Z base3
Z line Z base 2
=
(1.075 + j 0.9508) p.u.
Z load _ pu
= j 0.01377 p.u.
V rate
Z eq _ T 2 pu = Z T 2 _ rate
2
V base 2
Sbase S rate
=
0.12
7.2kV
2
6.6kV
3 MVA 2 MVA
=
0.2142 p.u.
The per unit supply voltage is then: V s _ pu
=
13kV 11kV
= 1.182 p.u.
The per unit equivalent circuit is then given by:
Zone 2
Zone 1
Zone 3 Rload p.u. = 1.705 p.u.
Vs = 1.182 p.u.
XT1eq p.u. =
Xline p.u. =
XT2eq =
j0.1 p.u.
j0.01377pu
j0.2142 p.u.
Xload p.u. = j0.9508 p.u
The load current is then given by: I load _ pu
=
V s _ pu X T 1 _ pu + X line _ pu + X T 2 _ pu + Z load _ pu
1.182∠00
I load _ pu
=
I load _ pu
= 0.5546∠ − 36.87
j (0.1 + 0.2142 + 0.01377 + 0.9508 ) + 1.705 0
p.u.
So the load current is : I load = 550.0 A ∠ − 36.870
4
=
1.182∠00 2.131∠36.87
0
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
Question 3. A balanced Y-connected voltage source with E ab = 480∠0° V is applied to a balanced ∆ load with Z ∆ = 30∠40° ohms. The line impedance between the source and the l oad is Z L = 1∠85° p.u. for each phase. Calculate the per-unit and actual current in phase a of the line using Sbase3 φ = 100kVA and V baseLL = 600V. Answer:
Define the base quantities as: Sbase3φ = 100kVA Sbase1φ
=
33.333kVA
V baseLL
=
600V 600
V baseLN =
Z base
=
I base
=
3
V baseLL
V = 346.412V
2
Sbase3φ
2
V baseLN
=
S base1φ
Sbase3φ
3V baseLL
=
3.6 Ω
= 96.225 A
So : E ab _ pu
480∠00
=
Z ∆ _ pu
=
Z Y _ pu
=
600 30∠400 3.6
Z ∆ _ pu
3
=
Z line _ pu
=
1∠850 3. 6
=
0.8 p.u.∠0
0
= 8.333 p.u.∠40
and
E a _ pu
=
0.8 p.u.∠ − 30
0
2.7778 p.u.∠400
=
0.2778 p.u.∠850
So the total impedance seen by the source is: Z tot _ pu
= Z Y _ pu + Z line _ pu =
2.9807 p.u.∠43.780
Therefore the supply current is given by: I a _ pu
=
V a _ pu Z tot _ pu
=
0.8 p.u.∠ − 300 0
2.9807 p.u.∠43.78
So the actual load current is given by: I a
=
25.83 A ∠ − 73.780
5
=
0.2684∠ − 73.780 p.u.
0
ELEC 4100
TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION
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 Z Y = 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 I a in per-unit and in amperes. Answer :
Define the base quantities as: Sbase3φ = 30kVA = 10kVA
Sbase1φ
V baseLN = 277V V baseLL
3V baseLN = 479.77V
=
Z base
=
I base
=
V baseLL
2
=
3Sbase3φ Sbase3φ
3V baseLL
7.6729 Ω
= 36.101 A
So the per unit impedance values are given by: Z ∆ _ pu
=
Z Y _ pu
=
45 − j 25 7.6729 30 + j10 7.6729
=
6.709 p.u.∠ − 29.0550
=
4.121 p.u.∠18.4350
Now the delta load can be converted to an equivalent star load as: Z ∆toY _ pu
=
Z ∆ _ pu
3
=
2.2364 p.u.∠ − 29.0550
The total per-phase impedance is then given by: Z tot _ pu
= Z ∆toY _ pu // Z Y _ pu
Z tot _ pu
=
Z ∆toY _ pu Z Y _ pu Z ∆toY _ pu + Z Y _ pu
= 1.5705 ∠ − 12.74
The per unit source current is then given by: I a _ pu
=
E ag _ pu Z tot _ pu
=
0.6367 ∠12.740
So the source current is : I a
=
22.99 A ∠12.74
0
6
0
ELEC 4100
TUTORIAL FOUR : THREE PHASE TRANSFORMERS
ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 4 : TRANSFORMERS SOLUTIONS . Question 1. C1 A1
B1
C1
B2 A2
B2
A2 a1
b1
A1
A1 C2
B1
C2
B1
c1 b2 a2
b2
c2
b3
c3
a2
C1
c1
a1
c1
a1 c2
b1
b1 a3
b4 a4 b4
a4
c3 c3
a3 c4
b3
c4
b3
a3
(a)
A1
B1
C1 C1
B2 A2
A2 a1
B2 b1
C1
A1
A1 C2
B1
C2
B1
c1 b2 a2
c1
c2
b1 a2 a3
b2
c2
b3
c3
a4 a5
b4
c4
b5
c5
a3 c4
b6
a6
b6
c6
c3
c3
b3
a6
a1
b1
b4 a4
c1
a1
b6
b4
c5
a6
a5
c4
b5 c5
a5 c6
b5
(b) 1
b3
c6
a4
a3
ELEC 4100 A1
B1
TUTORIAL FOUR : THREE PHASE TRANSFORMERS C1 B2 A2
A2
B2
C2
a1
b1
c1
a2
b2
a3
b3
C2
b2
c3
b4
c4
b5
c5
b1 c3
a3
c6
b6
c3 b3
c4
c4
c5
c5
b4
a5 c6
b5
b3
(c)
2
a4
a5 b5
a6
a6 b6
a1
c2
b1
b6
a6
c1
a1
c2
a4
a5
B1
c1
b4
a4
C1
A1 B1
a2
A1
C1
c6
a3
ELEC 4100
TUTORIAL FOUR : THREE PHASE TRANSFORMERS
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, T 1 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
T2
Bus 1
Bus 2
j40 ohm
1
2 j25 ohm
j25 ohm
T5
T3
T4
Bus 3
3 Answer:
The equivalent per phase, per unit circuit diagram is shown below: Bus 1
XT1 j0.0133pu
EG1
XG1 j0.0216pu
j0.0133pu
XT3
XT2
Xline1 j0.016pu j0.01pu j0.01pu
Xline2 j0.00666pu
Xline3 XT5
XM3 EM3
3
EG2
j0.0133pu
Bus 3 j0.01333pu
Bus 2
j0.0133pu
XT4
XG2 j0.0216pu
ELEC 4100
TUTORIAL FOUR : THREE PHASE TRANSFORMERS
The impedance values in the circuit diagram are calculated as will be detailed below: Sbase
= 100 MVA
V base _ HV = 500kV
Transmission line zones
V base _ LV = 20kV
Generator zones
Z base _ LV = I base _ LV =
(V
)
2
base _ HV
=
Sbase Sbase
=
3V base _ HV
(500kV )2 100 MVA
=
100 MVA 3 (20kV )
2500Ω
=
2.887kA
So the generator per unit impedances are: X G1 = 0.2
X G 2
= 0.2
X M 3
= 0.2
2
18kV
100 MVA
20kV
= 0.0216 p.u.
750 MVA 2
18kV
100 MVA
20kV
= 0.0216 p.u.
750 MVA
100 MVA 1500 MVA
= 0.01333 p.u.
The transformer per unit impedances are: X T 1 = X T 2
X T 5
= 0.1
= X T 3 = X T 4 = 0.1
100 MVA 1500 MVA
100 MVA 750 MVA
= 0.01333 p.u.
= 0.00666 p.u.
The transmission line per unit impedances are: 40Ω
X line1
=
X line 2
= X line3 =
2500Ω
= 0.016 p.u.
25Ω 2500Ω
= 0.01 p.u.
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: V 3 pu
=
18kV ∠00 20kV
= 0.9∠0
4
0
p.u.
ELEC 4100
TUTORIAL FOUR : THREE PHASE TRANSFORMERS
The motor current is then: I 3
P
=
∠ cos
3 (V LL )( p. f .)
−1
( p. f .) =
1200 MW 3 (18kV )(0.8)
= 48.11kA∠36.87
The motor current in per unit: I 3 _ pu
=
48.11kA∠36.87
0
2.887kA
= 16.67∠36.87
0
p.u.
Due to symmetry: V 1 _ pu
= V 2 _ pu = V 3 _ pu + I 3 _ pu X T 5 _ pu +
V 1 _ pu
= V 2 _ pu = 0.9∠0 +
0
+
V 1 _ pu
2
( X
T 3 _ pu
+
X line 2 )
(16.67∠36.87 )( j 0.00666) 0
16.67∠36.870 2 0
= V 2 _ pu = 0.7572∠18.83
So the bus 1 and bus 2 voltages are: V 1 = V 2
I 3 _ pu
0
= 15.14kV ∠18.83
5
( j 0.01 + j 0.01333)
p.u.
0
ELEC 4100
TUTORIAL FIVE : LOAD FLOW - SOLUTION
ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 5 : LOAD FLOW – SOLUTION . Question 1. Answer :
(a) For sinusoidal time varying voltage and current waveforms, define: v( x, t ) = V ( x )e jω t jω t
i( x, t ) = I ( x )e
Then substituting into the partial differential equations gives: dV ( x )
e jω t
dx dI ( x )
e jω t
dx
= [− rI ( x ) − jω LI ( x )]e
jω t
jω t
= [− GV ( x ) − jω CV ( x )]e
jω t
= − zI ( x )e
jω t
= − yV ( x )e
These expressions can be simplified as: dV ( x ) dx dI ( x ) dx
= − zI ( x )
= − yV ( x )
Differentiating with respect to x: d 2V ( x ) dx
2
d 2 I ( x ) dx
2
= − z
= − y
dI ( x ) dx
= zyV ( x )
dV ( x ) dx
= zyI ( x )
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: V r = V s −
( I s − V sY 2 ) Y 1
Rearranging: V r = V s 1 +
Y 2
−
I s
Y 1
Y 1
Comparing this expression with the general transmission line solutions provided gives: Y 1 =
1 Z C sinh (γ d )
Using this value and again comparing with the general transmission line solutions gives: Y 2 =
[cosh (γ d ) − 1] 1 γ d tanh = 2 Z C sinh (γ d ) Z C 1
ELEC 4100
TUTORIAL FIVE : LOAD FLOW - SOLUTION
Now : I r = I s − Y 2V s − Y 3V r = I s − Y 2V s − Y 3 V s −
I r = I s 1 +
Y 3
( I s − Y 2V s ) Y 1
Y − V sY 3 1 + Y 2 + 2
Y 1
Y 1
Equating this expression with the general transmission line solutions gives: Y 3 = Y 1 [cosh (γ d ) − 1] =
[cosh (γ d ) − 1] 1 γ d tanh = 2 Z C sinh (γ d ) Z C
= Y 2
(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 ne twork. 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 th network consists of N buses, and for the i 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: ∗
∗
S i = V i I i ∗
∗
S i = Pi − jQi = V i
n
∑ yik V k k =1
Where the yik are the elements of the admittance bus. (c)
The apparent power is given by: ∗
∗
S i = Pi − jQi = V i
n
∑ yik V k k =1
Pi − jQi ∗
V i
n
=
∑ yik V k + yiiV i k =1 k ≠ i
2
ELEC 4100
TUTORIAL FIVE : LOAD FLOW - SOLUTION
Rearranging gives: V i =
1
Pi − jQi
yii
V i
n
∑ yik V k
−
∗
k =1 k ≠ i
So the Gauss implementation of a voltage calculation is: p +1
V i
p
n
p
1
Pi − jQi
yii
( )
=
−
p ∗ V i
∑ yik V k
p
k =1 k ≠ i
The Gauss-Seidel implementation of a voltage calculation is: p +1
V i
p
Pi − jQi
yii
(V )
=
i −1
p
1
−
p ∗
∑
p +1 yik V k
n
k =1
i
∑ yik V k
−
p
k = i +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 th th p and (p+1) 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: ∗
∗
S i = Pi − jQi = V i
n
∑ yik V k k =1
Define: V i = V i ∠δ i yik = yik ∠γ ik
Then : n
∗
S i = Pi − jQi =
∑ yik V i V k ∠(δ k − δ i + γ ik ) k =1
Or: n
Pi =
∑
n
yik V i V k cos(δ k − δ i + γ ik ) =
k =1
∑ yik V i V k cos(δ i − δ k − γ ik ) k =1
n
Qi = −
∑ k =1
n
yik V i V k sin (δ k − δ i + γ ik ) =
∑ yik V i V k sin(δ i − δ k − γ ik ) k =1
3
ELEC 4100 (e)
TUTORIAL FIVE : LOAD FLOW - SOLUTION
Define the power flow mismatches at bus i as: f i = ∆Pi = PGi − P Li −
n
n
k =1
k =1
∑ yik V i V k cos(δ i − δ k − γ ik ) = Pi − ∑ yik V i V k cos(δ i − δ k − γ ik ) n
g i = ∆Qi = QGi − Q Li −
∑
n
yik V i V k sin (δ i − δ k − γ ik ) = Qi −
k =1
∑ yik V i V k sin(δ i − δ k − γ ik ) k =1
So applying the Newton-Raphson method: p +1
p
δ
=
p +1
V
δ
−
p
V
∂ f
∂ f
∂δ ∂g
∂V ∂g
∂δ
∂V
−1
p
p
∆P
=
p
∆Q
p
δ
∆δ
+
p
p
∆V
V
Alternatively this can be expressed as: p
∆P
p
∆Q
Where
p
∆P
∂ f
∂ f
∂δ =− ∂g
∂V ∂g
∂δ
∂V
p
∆δ
=
p
p
J 2
p
J 4
J 1
∆V
J 3
p
∆δ
p
∆V
p p
are the real power mismatches at all PQ and PV buses,
power mismatches at all PQ buses,
p
∆δ
p
∆Q
are the reactive
are the voltage angle corrections for all PQ and PV
p
buses, and ∆V are the voltage magnitude corrections for all PQ buses. The Jacobian matrix is defined by: ∂ f 2
−
p
J 1 =
p J 3
=
−
∂δ 2 ∂ f 3
−
∂ f n
∂δ 2
∂δ 2
−
∂g m +1
−
∂g m + 2
∂δ 2 ∂δ 2
−
∂g n ∂δ 2
−
∂ f 2
−
∂δ 3 ∂ f 3
−
∂ f n
−
∂δ 3
∂δ 3 −
∂g m +1
−
∂g m + 2
−
∂δ 3 ∂δ 3
∂g n ∂δ 3
∂ f 2
−
−
∂δ n ∂ f 3
−
∂ f n
p
J 2 =
∂δ n
−
−
∂δ n −
∂g m +1
−
∂g m + 2
−
∂δ n
−
∂δ n ∂g n ∂δ n
p J 4
=
−
−
∂ f 2 ∂ V m +1 ∂ f 3 ∂ V m +1 ∂ f n ∂ V m +1 ∂g m +1 ∂ V m +1 ∂g m + 2 ∂ V m +1 ∂g n ∂ V m +1
− −
−
− −
−
∂ f 2 ∂ V m + 2 ∂ f 3 ∂ V m + 2 ∂ f n ∂ V m + 2 ∂g m +1 ∂ V m + 2 ∂g m + 2 ∂ V m + 2 ∂g n ∂ V m + 2
− −
−
∂ f 2 ∂ V n ∂ f 3 ∂ V n ∂ f n ∂ V n
−
∂g m +1
−
∂g m + 2
∂ V n
−
∂ V n ∂g n ∂ V n
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: −
−
∂ f i ∂δ k ∂ f i ∂δ i
= yik V i V k sin (δ i − δ k − γ ik ),
i ≠ k
n
=−
∑ yik V i V k sin(δ i − δ k − γ ik ) k =1 k ≠ i
4
ELEC 4100
−
−
−
−
−
−
∂ f i ∂ V k
∂ f i ∂ V i ∂g i ∂δ k ∂g i ∂δ i ∂g i ∂ V k ∂g i ∂ V i
TUTORIAL FIVE : LOAD FLOW - SOLUTION
= yik V i cos(δ i − δ k − γ ik ),
i ≠ k
n
=
∑ yik V k cos(δ i − δ k − γ ik ) + yii V i cos(γ ik ) k =1
= − yik V i V k cos(δ i − δ k − γ ik ),
k ≠ i
n
=
∑ yik V i V k cos(δ i − δ k − γ ik ) k =1 k ≠ i
= yik V i sin (δ i − δ k − γ ik ),
k ≠ i
n
=
∑ yik V k sin(δ i − δ k − γ ik ) − yii V i sin(γ ii ) k =1
(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.
5
ELEC 4100
TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 7 : SYMMETRICAL COMPONENTS - SOLUTIONS Question 1. 0
Determine the symmetrical components of the following line currents : (a) I a = 5∠90 , I b = 0 0 5∠340 , I c = 5∠200 , and (b) I a = 50, I a = j50, I c = 0. Answer
(a) The symmetrical component currents are given by: I 0 I 1
=
1 3
1
1
1
1
a
a 2 I b
2
I 2
1 a
I 0
1
1
1
e j120
I 1
=
1 3
a
( ( (
5
1
=
3
0
0
j 90 0
1
1
5e j 90
1
a
a2
5e j 340
2
a
5e j
5e j 90
e j −120
0
j120 0
e
j 340 0
0
5e j 200
0
j 200 0
I 0
+ j 0.5266 A
I 1
= + j 4.9490 A
I 2
− j 0.4760 A
=
200 0
) ) )
I 2
I 1
0
0
5e j 340
e +e +e 3 0 0 0 5 e j 90 + e j100 + e j 80 3 0 0 0 5 e j 90 + e j 220 + e − j 40 3
I 0
0
1
1 a
I c
1
1 e j −120
I 2
I a
(b) The symmetrical component currents are given by: I 0 I 1
=
1 3
1
1
1
1
a
a 2 I b
I 2
1 a
I 0
1
I 1
=
1 3
1
2
a
=
j120 0
e
e
0
j −120 0
e j120
(16.667 + j16.667 ) A
I 1 I 2
j 210
0
− j 30
0
+e
+e
=
j 90 0
1
1
50
1
a
a2
50e j 90
1 a2
a
0
50e j 90
0
I 0
=
+e
1
50
I 2
I 1
3
1
(1 3 50 (1 3 50 (1 3 50
I 0
1
I c
1
1 e j −120
I 2
I a
0
) ) )
(2.233 − j8.333) A (31.100 − j8.333) A 1
0
0
ELEC 4100
TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
Question 2. One line of a three phase generator is open circuited, while the other two are short-circuited to 0 0 ground. The line currents are I a = 0, I b = 1000A∠90 , and I c = 1000A∠-30 . Find the symmetrical components of these currents. Also find the current into ground. Answer
The symmetrical component currents are given by: I 0 I 1
=
1 3
=
1
I a
1
a
a
2
I b
1
3
2
a
1000
I 0 =
1000
3
0
e j −120
0
e j120
(e
j 90 0
(
+e
1
a
a
1 a2
a
1000 e j 90
0
0
2
1000 e j 90
0
1000 e − j 30
0
− j 30
0
1000 e − j 30 0
0
333.3 A ∠30 =
1
0
e j 210 + e − j150 3 0 0 1000 e − j 30 + e j 90 3
I 0
1
0
(
I 2
1
1
e j120
3
1
=
I c
1
1 e j −120
I 2
I 1
1
1
I 0
I 1
1
1 a
I 2
I 1
1
0
) ) )
0
0
666.7 A ∠ − 150 0
I 2
333.3 A ∠300
The ground current is the sum of the b and c phase currents and is given by: I 0 I gnd = I b + I c
=
I 1
(
= 1000 e
j 90
0
+e
−
j 30
0
)
=
(866 + j500) A = 1000 A ∠300
I 2
Question 3. 0
0
0
Given the line to ground voltages Vag = 280V∠0 , Vbg = 290V∠-130 , and Vcg = 260V∠110 , calculate (a) the sequence components of the line to ground voltages, denoted V Lg0, V Lg1, and V Lg2. (b) the line to line voltages V ab, V bc, V ca. (c) The sequence components of the line to line voltages V LL0, V LL1, and V LL2. Also verify the following general relation : VLL0 = 0, V LL1 = 3V LLg1∠300 , and V LL1 = 3V LLg 2∠ − 300 . Answer
(a) The symmetrical components of the line to ground voltages are given by:
2
ELEC 4100
TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
1
1
1
V ag
1
a
a
2
V bg
V Lg 2
1 a2
a
V cg
V Lg 0
1
1
1
e j120
V Lg 0 =
V Lg1
=
V Lg1
1 3
1 3
0
3
0
0
7.55V ∠78.110
1
1
a
a
1 a2
a
260e j110
e j
280e j 0
2
0
290e − j130 260e j110
0
0
0
290e − j130
120 0
V Lg 0
1
280e j 0
e j −120
1 e j −120
=
1
1
V Lg 2
V Lg1
=
1
0
0
275.73V ∠ − 6.630 24.87 V ∠ − 79.430
V Lg 2
(b) The line to line voltages are calculated according to: V ab =
V bc V ca
V ag
− V bg
V bg
− V cg
V cg
− V ag
280e j 0
0
− 290e
290e − j130
=
0
260e j110
0
− j130
− 260e
0
516.613V ∠25.47 0
0 j110
− 280e
=
j 0 0
476.550V ∠ − 101.800 442.491V ∠146.490
(c) The symmetrical components of the line to line voltages are given by: V LL 0 =
V LL1
1 3
1
1
1
V ag
1
a
a
2
V bg
1 a
V LL 2
2
a
=
1 3
V cg
1
1
1
1
a
a
1 a2
a
516.613 e j 25.47
2
0
476.550 e − j101.80 442.491e j146.49
0
0
0 V
V LL 0 =
V LL1
477.57 V ∠23.37 0 43.07 V ∠ 49.430
V LL 2
So : V LL1 V Lg1 V LL 2 V Lg 2
=
=
477.57 V ∠23.37 0 0
275.73V ∠ − 6.63
43.07 V ∠ 49.430 24.87 V ∠ − 79.430
= 1.7321∠30
0
=
0
3∠30 0
= 1.7321∠ − 30 =
3∠ − 300
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 I 0, I 1, and I 2, the sequence components of the line currents. Then calculate the line currents I a, I b, and I c from the sequence components, and compare with the line currents calculated directly from the network equations.
3
ELEC 4100
TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
Answer
The sequence networks are shown below: IL1
12+j16 Ω
VLg1 Positive Sequence IL2
12+j16 Ω
VLg2 Negative Sequence IL0
12+j16 Ω
VLg0 Zero Sequence
The three sequence currents can be calculated as: I L0 I L1
=
V Lg 0
7.55V ∠78.110
Z 0 V Lg1
12 + j16 275.73V ∠ − 6.630
Z 1 V Lg 2
I L 2
=
0.378 A ∠24.980 =
12 + j16 24.87 V ∠ − 79.430
Z 2
13.787 A ∠ − 59.760 1.243 A ∠26.300
12 + j16
The line currents are then given by: 1
1
1
I L0
1 a2
a
I L1
I c
1
a 2 I L 2
I a
14.0 A ∠ − 53.13
I a I b
I b
=
a
1 =
1
1 e − j120 1
e j120
0
0
e
1
0.378 A ∠24.980
j120 0
13.787 A ∠ − 59.760
e − j120
0
1.243 A ∠26.30 0
0
=
14.5 A ∠176.87 0 13.0 A ∠56.87 0
I c
From the network equations directly: I a I b I c
=
V ag
280V ∠0 0
Z V bg
12 + j16 290V ∠ − 1300
Z V cg Z
=
12 + j16 260V ∠1100
14.0 A ∠ − 53.130 =
14.5 A ∠176.87 0 13.0 A ∠56.87 0
12 + j16
This matches the result calculated using the symmetrical component model.
4
ELEC 4100
TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS
Question 5. 0
As shown in figure 1, a balanced three-phase, positive sequence source with V AB = 480V∠0 is applied to an unbalanced ∆ load. Note that one leg of the ∆ is open. Determine (a) the load currents I AB and I BC . (b) the line currents I A, I B, I C, which feed the ∆ load. (c) the zero, positive, and negative sequence components of the line currents.
Ia (18+j10)Ω
Vab= 480V 00
Ea Ec
Iab Eb
Ibc
Ib
(18+j10) Ω
Ic Figure 1: Network for Question 5. Answer
(a) The load currents are given by: I ab I bc
480∠00
V ab Z V bc
=
Z 0
I ca
=
18 + j10 480∠ − 1200 18 + j10 0
23.311 A ∠ − 29.050 =
23.311 A ∠ − 149.050 0
(b) The line currents are given by: I a I b
23.311 A ∠ − 29.050
I ab =
I bc − I ab
=
23.311 A ∠30.950
− I bc
I c
40.376 A ∠ − 179.050
(c) The sequence currents are given by: I 0 I 1
=
I 2
I 2
3
1
1
I a
1
a
a
2
I b
1 a2
a
=
1 3
I c
1
1
1
23.311 A ∠ − 29.050
1
a
a
2
40.376 A ∠ − 179.05
1 a2
a
0 A
I 0 I 1
1
1
=
26.917 A ∠ − 59.055
0
13.459 A ∠60.950
5
0
23.311 A ∠30.950
ELEC 4100
TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
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) (b) (c)
The Thevenin reactance in per unit at the fault : [0.2670] The transient fault current in per unit and kA : [-j3.933, -j4.541kA] Contributions to the fault current from G1 and from line 1-2. [-j1.896, -j2.647kA] Bus 1
1
Bus 3
Bus 2
T3
3
j50 ohm
j50 ohm
T3
j50 ohm Bus 4
T2 2 Figure 1 : Four Bus Power System. Answer:
The positive sequence per unit network is shown below. The per unit values are determined as follows: Sbase
= 1000 MVA
V base3
=
V base 4
=
V base 2
=
20kV 500kV 20kV 18kV 500kV
Zone of Generator 3. 20kV = 500kV
Zone of Transmission lines.
500kV = 18kV
Zone of Generator 2.
1
ELEC 4100
TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
V base1 =
13.8kV
Z base 4
=
I base 4
=
500kV
500kV = 13.8kV
(V base4 )2 Sbase
Sbase
3V base4
=
(500kV )2 1000 MVA
3 (500kV )
Bus 1
EG1
= 250Ω
1000 MVA
=
Zone of Generator 1.
= 1.155kA
Bus 2
Bus 3
j0.4pu
j0.24 pu
j0.2pu
j0.2pu
j0.1pu
XG1
XT1
X12
X23
XT3
j0.2pu
X24 Bus 4
j0.133pu
XT2
j0.24pu
XG2 EG2
So applying these base values to the generators: X G1 = (0.2 )
1000 500 1000
(0.18)
X G 2
=
X G 3
= 0.17 p.u.
750
= 0.4 p.u.
= 0.24 p.u.
Similarly for the transformers: X T 1 = (0.12 )
(0.1)
1000 500
1000
X T 2
=
X T 3
= 0.1 p.u.
750
= 0.24 p.u.
= 0.1333 p.u.
For the transmission lines: X 12
= X 23 = X 24 =
50 250 2
= 0.2 p.u.
j0.17pu
XG3
EG3
ELEC 4100
TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
Part (a) The Thevenin equivalent impedance of the network when viewed from voltage bus 1 i s: X Th
=
( X G1 + X T 1 ) // [ X 12 + ( X 24 + X T 2 + X G 2 ) // ( X 23 + X T 3 + X G3 )]
X Th
=
( j 0.24 + j 0.4) // [ j 0.2 + ( j 0.2 + j 0.1 + j 0.17 ) // ( j 0.2 + j 0.1333 +
X Th
=
( j 0.64) // ( j 0.4583 )
X Th
= j 0.2670 p.u.
Part (b) The pre-fault voltage, neglecting pre-fault currents is: V F =
525kV ∠00 500kV
= 1.05∠0
0
p.u.
So the fault current is: I F =
V F Z Th
=
0 1.05∠0 p.u.
j 0.2670
= − j 3.933 p.u.
I F = − j 4.541 kA
Part (c) Using the current divider rule: I G1 = I F
j 0.4583 j 0.4583 + j 0.64
= − j1.641 p.u.
I G1 = − j1.896 kA I G 2 = I F
j 0.64 j 0.4583 + j 0.64
= − j 2.292 p.u.
I G1 = − j 2.647 kA
3
j 0.24 )]
ELEC 4100
TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION
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) (b) (c)
The Thevenin equivalent impedance of the network viewed from the fault location : [0.1975 p.u.] The fault current in per unit and in kA [-j5.3155 p.u., -j6.138kA] 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: X Th = ( X G1 + X T 1 + X 12 ) // ( X 24 + X T 2 + X G 2 ) // ( X 23 + X T 3 + X G 3 ) X Th
=
( j 0.24 + j 0.4 + j 0.2) // ( j 0.2 + j0.1 + j0.17 ) // ( j 0.2 + j 0.1333 +
X Th
=
( j 0.84) // ( j 0.47 ) // ( j 0.5733)
X Th
= j 0.1975 p.u.
Part (b) The pre-fault voltage, neglecting pre-fault c urrents is: 525kV ∠00
V F =
500kV
= 1.05∠0
0
p.u.
So the fault current is: I F =
V F Z Th
0
=
1.05∠0 p.u. j 0.1975
= − j 5.3155 p.u.
I F = − j 6.1379 kA
Part (c) The contribution to the fault from line 12 is given by: I 12 I 23 I 24
=
=
=
1.05∠0
0
j 0.84
1.05∠00 j 0.47
1.05∠00 j 0.5733
− j1.25 p.u. = − j1.443 kA
− j 2.234 p.u. = − j 2.580 kA
− j1.8315 p.u. = − j 2.115 kA
4
j 0.24 )
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
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. Bus 1
Bus 3
Line 1 - 3
1
T3 3
Bus 2 Line 1 - 2
Line 2 - 3
4
T1 T4
T2 2 Transformers: •
T1 : 1000MVA, • T2 : 1000MVA, • T3 : 500MVA, • T4 : 750MVA, Transmission Lines :
15 kV ∆ / 765 kV Y , 15 kV ∆ / 765 kV Y , 15 kV ∆ / 765 kV Y , 15 kV ∆ / 765 kV Y ,
1-2 : 765 kV, X1 = 50 Ω, X1 = 40 Ω, • 1-3 : 765 kV, • 2-3 : 765 kV, X1 = 40 Ω, Synchronous Generators : •
• • • •
G1 : 1000MVA, G2 : 1000MVA, G3 : 500MVA, G4 : 750MVA,
15 kV, 15 kV, 13.8 kV, 13.8 kV,
X = 0.1 p.u. X = 0.1 p.u. X = 0.12 p.u. X = 0.11 p.u.
X0 = 150 Ω. X0 = 100 Ω. X0 = 100 Ω. X1 = X2 = 0.18 p.u., X0 = 0.07 p.u. X1 = X2 = 0.20 p.u., X0 = 0.10 p.u. X1 = X2 = 0.15 p.u., X0 = 0.05 p.u. X1 = 0.30 p.u. X2 = 0.40 p.u., X 0 = 0.10 p.u.
Answer:
The three sequence networks for the system are shown below. The per unit impedance values are calculated as follows: Sbase = 1000 MVA V baseHV = 765kV
Zone of Transmission Lines.
V baseLV = 15kV
Zone Generators.
1
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
Z baseHV
=
I baseHV =
(V baseHV )2
=
Sbase Sbase
(765kV )2 1000 MVA
=
3V baseHV
= 585.23Ω
1000 MVA
3 (765kV )
= 0.7547kA
The per unit sequence impedances of the generators are then given by: X G1 _ 1 = 0.18 p.u. X G1 _ 2 = 0.18 p.u. X G1 _ 0 = 0.07 p.u. X G 2 _ 1 = 0.20 p.u.
X G 2 _ 2 2
X G 3 _ 1 = (0.15)
13.8
(0.15)
13.8
(0.05)
13.8
X G 3 _ 2
X G 2 _ 0
=
=
1000
15
500 2
1000
15
500 2
1000
15
500
=
0.20 p.u.
X G 2 _ 0
=
0.10 p.u.
= 0.2539 p.u.
= 0.2539 p.u.
(0.05)
+3
13.8
2
15
1000 500
= 0.08464 p.u. + 0.2539 p.u. = 0.3385 p.u.
X G 4 _ 1 = (0.3)
X G 4 _ 2
X G 4 _ 0
=
=
1000
15
(0.40) (0.1)
2
13.8
750
13.8
2
15
13.8
1000 750
= 0.3386 p.u.
= 0.4514 p.u.
2
(750 ) = 0.1129 p.u.
15
The per unit sequence impedances of the transformers are then given by: X T 1 = 0.1 p.u. X T 2 X T 3
X T 4
= 0.1 p.u.
= 0.12
= 0.11
15
2
15 15 15
1000 500
2
1000 750
= 0.24 p.u.
= 0.1467 p.u.
The per unit sequence impedances of the transmission lines are then given by: 50 150 X 12 _ 1 = X 12 _ 2 = X 12 _ 0 = = 0.08544 p.u. = 0.2563 p.u. 585.23 585.23 40 100 = 0.06835 p.u. = 0.1709 p.u. X 13 _ 1 = X 13 _ 2 = X 13 _ 0 = 585.23 585.23 40 100 X 23 _ 1 = X 23 _ 2 = X 23 _ 0 = = 0.06835 p.u. = 0.1709 p.u. 585.23 585.23
2
ELEC 4100
TUTORIAL NINE : FAULT STUDIES Bus 1
j0.06835pu
j0.18pu j0.1 pu
XG1_1
EG1
XT1
Bus 3
X13_1 j0.08544 pu
XT3
j0.06835pu
X12_1
X23_1
XT4 j0.147pu
Bus 2
XG3_1
XG4_1 j0.339pu
XT2
j0.1pu
EG3
j0.24pu j0.254pu
EG4 XG2_1
j0.20pu
EG2 Positive Sequence Network. Bus 1 j0.18pu j0.1 pu
XG1_2
XT2
Bus 3
j0.06835pu
j0.24pu j0.254pu
X13_2 j0.08544 pu
XT3
j0.06835pu
X12_2
X23_2 Bus 2
XT4 j0.147pu
XG4_2 j0.451pu
XT2
j0.1pu
XG3_2
XG2_2
j0.20pu
Negative Sequence Network. Bus 1 j0.07pu
j0.1 pu
XG1_0
XT1
j0.1709pu
Bus 3 j0.24pu
X13_0 j0.2563 pu
X12_0
j0.1709pu
X23_0 Bus 2
j0.1pu
XT3 XT4 j0.147pu
XT2
XG2_0
XG4_0
j0.10pu
j0.113pu
Zero Sequence Network.
3
j0.339pu
XG3_0
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
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:
ZA ZCA
ZAB
ZB ZC
Z A
=
Z B
=
Z C =
ZBC
Z AB Z CA
Z AB
Z AB + Z BC + Z CA Z AB Z BC
=
Z BC =
Z AB + Z BC + Z CA Z CA Z BC
Z CA
Z AB + Z BC + Z CA
=
Z A Z B + Z B Z C + Z C Z A Z C Z A Z B + Z B Z C + Z C Z A Z A Z A Z B + Z B Z C + Z C Z A Z B
So the three sequence networks can be simplified to the form: Bus 1
j0.06835pu
j0.28pu
Bus 1
Bus 3 j0.4939pu
EG3
j0.06835pu
j0.28pu
Bus 3 j0.4939pu
EG3
j0.1733 pu j0.08544 pu
EG1
j0.06835pu
EG1 j0.7605pu j0.6083pu
Bus 2 j0.4860pu
j0.30pu
j0.4860pu
EG2
EG4
EG2
EG4
Positive Sequence. Bus 1
j0.06835pu
j0.28pu
Bus 1
Bus 3 j0.4939pu
j0.06835pu
j0.28pu
Bus 3 j0.4939pu
j0.1733 pu j0.08544 pu
j0.06835pu j0.7605pu j0.6083pu
Bus 2 j0.30pu
j0.5981pu
Negative Sequence. 4
j0.5981pu
ELEC 4100
TUTORIAL NINE : FAULT STUDIES Bus 1
j0.07pu
Bus 1
Bus 3
j0.1709pu
j0.1 pu
j0.339pu j0.2563 pu
j0.1709pu
j0.07pu
j0.1709pu
Bus 3
j0.1 pu
j0.09116pu
j0.339pu j0.8652pu
j0.09116pu
Bus 2 j0.1pu
j0.5063pu
j0.3376pu
Zero Sequence.
So from these simplified networks, the Thevenin equivalent impedances can be derived looking in at bus 1, as:
( j0.28) // j0.7605 // ({ j0.06835 // j0.1733} + { j0.4939 // j0.4860 // j0.6083} )
ZTH _ 1
=
ZTH _1
=
j 0.1069
ZTH _ 2
=
( j0.28) // j0.7605 // ({ j0.06835 // j0.1733} + { j0.4939 // j0.5981// j0.6083} )
ZTH _ 2
=
And:
j 0.1097
And: ZTH _ 0
=
ZTH _ 0
=
( j0.1) // j0.5063 // ({ j0.1709 // j0.8652} + { j0.3376 // j0.09116} ) j 0.0601
Question 3. For a bolted three phase fault, the fault current is given by: I 0 1
=
=
I 2
=
V F
0, 1∠00
I=
ZTH _1
j 0.1069
= 9.355
. . ∠ − 90p0u
Similarly : Ia
=
Ib
=
Ic
=
I1
= 9.355
=
Ib
=
Ic
=
I1
= 7.06
p. u.
So in ampere: Ia
kA
5
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
Question 4. For a single line to ground fault the sequence networks are connected in series as: I1
ZTh1 V1
VF Positive Sequence I2
ZTh2 V2
Negative Sequence
I0
ZTh0 V0
Zero Sequence Hence the sequence currents are: 0
=I
1
=I
2
Ib
=
I c
=
0
V F
=I
ZTH _1 + ZTH _ 2 + ZTH _ 0
=
1∠00 j0.2767
= 3.614
. .p∠u− 900
And:
I = 3 1I= 10.84 .p .u∠ − 900
a
In amperes. a
0 .kA ∠ − 90
I 8.183 =
The sequence voltages are: 0
V 0 V 1
=
V 2
V 2
0
0
I 0
0
Z TH _ 1
0
I 1
0
0
−
0 0
V 0 V 1
V F
Z TH _ 0
=
j 0 0
1. 0 e 0
−
Z TH _ 2 I 2
j 0.0601
0
0
3.614e − j 90
0
j 0.1069
0
3.614e −
0
0
6
0
j 90 0
j 90 0
j 0.1097 3.614e −
ELEC 4100
TUTORIAL NINE : FAULT STUDIES j 0 0
V 0 V 1
− 0.2172e =
j 0 0
0.6137e
j 0 0
− 0.3965e
V 2
Hence the phase to ground voltages are given by: 1
V a V b V c
=
1
1
V 0
2
a
V 1
1 a 1
a 2 V 2
a
V a
1 =
1 − j120
1 e
0
j120 0
e
1
j 0 0
1
− 0.2172e
j120 0
j 0 0
e
e−
0.6137e
j120 0
j 0 0
− 0.3965e
0 − j120
0
j120 0
V b
= − 0.2172 + 0.6137e
V c
− 0.2172 + 0.6137e
V a
0
0
V b
= − 0.3258 − j 0.8749 =
0.9336 p.u. ∠ − 110.40
V c
− 0.3258 + j 0.8749
− 0.9336 p.u.∠110.4
j120 0
− 0.3965e
− 0.3965e
− j120
0
0
Question 5. The fault impedance in per unit is: Z F =
30 585.23
=
0.0513
The connection of the sequence networks is then as shown below: I1
ZTh1 V1
VF Positive Sequence I2
ZTh2 V2
Negative Sequence
I0
ZTh0 V0
Zero Sequence 7
3ZF
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
So for a single line to ground fault through this impedance: 0
=I
1
=I
2
=I
V F
1∠00
=
ZTH _1 + ZTH _ 2 + ZTH _ 0 + 3 ZF
j0.4306
=
0 2.322 . .p ∠u− 90
And: Ib
=
I c
=
0
I = 3 1I= 6.967 .p .u∠ − 900
a
In amperes. a
0 .kA ∠ − 90
= I 5.258
The sequence voltages are: V 0 V 1
0 =
V 2
V F
0
I 0
0
Z TH _ 1
0
I 1
0
0
−
0 =
1.0 e j 0
0
Z TH _ 2 I 2 j 90 0
j 0.0601
0
0
2.322e −
0
j 0.1069
0
2.322e − j 90
0
0
0
j 90 j 0.1097 2.322e −
0
−
0
V 2
j 0 0
− 0.1396e
V 0 V 1
0
0
V 0 V 1
Z TH _ 0
=
j 0 0
0.7518e
j 0 0
V 2
− 0.2547e
Hence the phase to ground voltages are given by: V a V b V c V a
1
1
V 0
1 a2
a
V 1
1 =
1
a
a 2 V 2
1 =
1 j120 0
1 e−
e j120
1
= − 0.1396 + 0.7518e
V c
− 0.1396 + 0.7518e
V a
0.3575e j 0
− j120
j120
0
0
0
j120 0
j 0 0
e
e− j120
0.7518e
0
j 0
− 0.2547 e
0
0.3575e j 0
V b
j 0 0
− 0.1396e
1
0
j120 0
− 0.2547e
− 0.2547e
− j120
0
0.3575 p.u. ∠00
V b
= − 0.3882 − j 0.8717 =
0.9542 p.u. ∠ − 1140
V c
− 0.3882 + j 0.8717
− 0.9542 p.u.∠114
8
0
0
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
Question 6. For a bolted line to line fault the sequence networks are connected as shown below: I1
ZTh1 V1
VF Positive Sequence I2
ZTh2 V2
Negative Sequence
I0
ZTh0 V0
Zero Sequence The sequence currents are therefore: I 0 1
=
0
= −I
2
=
V F
I ZTH _1 + ZTH _ 2
1∠00
=
j 0.2166
=
4.617 . . ∠p−u900
And: 1
I a I b I c I a I b
=
1
1 a 1
1 I 0
2
a
I 1
a 2 I 2
a
1 =
1 e 1
0 = − 7.997 p.u.
I c
7.997 p.u.
I a
0
In amperes:
I b I c
1
= − 6.035 kA
6.035 kA
The sequence voltages are:
9
1
− j120
e j120
0
0
e j
0
120 0
e − j120
0
4.617e
− j 90
− 4.617e
0
− j 90
0
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
V 0 V 1
0 =
V F
V 0
0
I 0
0
Z TH _ 1
0
I 1
0
0
−
0 =
1.0 e j 0
0
0
0
0
0
j 0.1069
0
4.617e− j 90
0
0
j 0.1097
−
0
V 0 =
− 4.617e
0
1 − 0.4936
V 2
Z TH _ 2 I 2
j 0.0601
0
V 2
V 1
0
0
V 2
V 1
Z TH _ 0
0.5064
=
0.5064
0.5064
Hence the phase to ground voltages are given by: 1
V a V b
=
V c
=
1 a2
a
V 1
a
1
1
( 0.5064 (e 0.5064 e
− j120
0
e j120
1
j120 0
−e
−e
− j120
0
) )
1.0128 p.u. ∠00 =
1
1 e − j120
=
a 2 V 2
j120 0
V a
V c
V 0
1.0128 p.u. ∠00
V c
V b
1
1
V a V b
1
0.5064 p.u. ∠ − 180
0
0.5064 p.u. ∠1800
10
0
0
0
j120 0
e
e− j120
0
0.5064 0.5064
0
− j 90
0
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
Question 7. Recall that for a bolted double line to ground fault, the sequence networks are connected together as shown below: I1
ZTh1 V1
VF Positive Sequence I2
ZTh2 V2
Negative Sequence
I0
ZTh0 V0
Zero Sequence
Therefore the positive sequence fault current is given by: I 1 =
V F Z TH _ 1 + Z TH _ 2 // Z TH _ 0
I 1 =
1.0∠0
=
1.0 j 0.1069 + j 0.1097 // j 0.0601
0
j 0.1457
= 6.862 p.u. ∠ − 90
0
The negative and zero sequence currents can be determined by current divider rule, a s: I 2
= − I 1
I 2
=
I 0
= − I 1
I 0
=
Z TH _ 0 Z TH _ 0 + Z TH _ 2
= −6.862∠ − 90
0
= −6.862∠ − 90
0
j 0.0601 j 0.0601 + j 0.1097
2.429 p.u. ∠900
Similarly: Z TH _ 2 Z TH _ 0 + Z TH _ 2
4.433 p.u. ∠900
The phase currents are therefore:
11
j 0.1097 j 0.0601 + j 0.1097
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
I a I b
=
I c
1
4.433e j 90
1 a2
a
6.862e− j 90
1
a2
2.429e j 90
a
I a I b
0
0
− 4.433 + 6.862 − 2.429 =e
− j 90
0
− 4.433 + 6.862e
− j120
0
j120 0
I c
− 4.433 + 6.862e
I a I b
0
1
1
j120 0
− 2.429e
− 2.429e
− j120
0 =e
− j 90
0
0
− 6.65 − j8.046 = e
I c
0
− j 90
0
0
j 230.4 0
10.44e
10.44e j129.6
− 6.65 + j8.046
=
0
j140.4 0
10.44 p.u. e
10.44 p.u. e j 39.6
So in amperes: 0
I a I b
=
7.879kA e j
140.4 0
7.879kA e j 39.6
I c
0
The sequence voltages are: V 0 V 1
0 =
V 2
V F
0
I 0
0
Z TH _ 1
0
I 1
0
0
−
0 =
1.0 e j 0
V 2
0
0
− 4.433∠ − 90
0
j 0.1069
0
6.862∠ − 900
0
0
j 0.1097
− 2.429∠ − 90
−
1 − 0.7335
0.2664 p.u. ∠00 =
0.2664 ∠00
V 2
0.2664 p.u. ∠00 0.2664 p.u. ∠00
The phase voltages are then: 1
V a V b V c
=
1
1
V 0
2
a
V 1
1 a 1
a
0
0
0.2664 ∠00 =
Z TH _ 2 I 2
j 0.0601
0
V 0 V 1
0
0
V 0 V 1
Z TH _ 0
a 2 V 2
1 =
1
1 e 1
12
1
− j120
e j120
0
0
e j
0.2664
120 0
e − j120
0
0.2664 0.2664
0
0
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
V a
0.7993 p.u. ∠0
V b
0
0
=
0
V c
Question 8. The phase voltages and phase fault currents in the above cases are: Three Phase Fault. =
Ia
Ib
=
Ic
=
I1
= 7.06
kA
Single Line to Ground Fault. V a a
I 8.183 =
0 .kA ∠ − 90
V b
0 =
V c
0.9336 p.u. ∠ − 110.40 − 0.9336 p.u.∠110.4
0
Single Line to Ground Fault Through an Impedance. 0.3575 p.u. ∠00
V a I = 3 1I= 6.967 .p .u∠ − 900
a
V b
=
0.9542 p.u. ∠ − 1140
V c
− 0.9542 p.u.∠114
V a
1.0128 p.u. ∠00
0
Line to Line Fault. 0
I a I b
= − 7.997 p.u.
I c
V b
7.997 p.u.
=
0.5064 p.u. ∠ − 1800 0.5064 p.u. ∠1800
V c
Double Line to Ground Fault. I a I b I c
0 =
10.44 p.u. ∠140.4
0.7993 p.u. ∠00
V a 0
V b
10.44 p.u. ∠39.60
V c
=
0 0
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.
13
ELEC 4100
TUTORIAL NINE : FAULT STUDIES
Question 9. Now since: 1
I a I b
=
I c
1
1 a
2
a
1
1 I 0 a
I 1
a 2 I 2
Then: I a
=
[ I 0 + I 1 + I 2 ] = 0
: Then via KCL, the three sequence currents must form a node.
And: I a ' = [ I 0 '+ I 1 '+ I 2 '] = 0 : Then via KCL, the three sequence currents must form a node.
Similarly: 1
V 00' V 11'
=
1 3
V 22 '
1
1
a
a
2
1 a2
a
1
1
V aa ' V bb '
=
V cc '
1 3
1
1
a
a
2
1 a2
a
1
V aa '
V aa '
0
=
0
V aa ' V aa '
3 3 3
The three sequence voltages are therefore equal in magnitude:
I0
I0' V00'
I1 V11' I2
I1'
I2' V22'
14
ELEC 4100
T
TEN : TRANSIENT STABILITY SOLUTIONS
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: EV
P
=
P
=
P
=
X
sin δ
+ X + X
(1.28)(1.0 ) sin δ 0.3 + 0.1 + (0.2 ) // (0.1 + 0.2 ) 2.462 sin δ
During the fault, a star-delta transform can be applied to simplify the circuit as shown below: X X
X
X'
X
The X
is given by: X
=
X
=
X
=
(X
+
X
)X
+
(X
+
X
)X
+
X
X
X
( 0.3 + 0.1) 0.2 + (0.3 + 0.1 )0.1 + 0.1 (0.2 ) 0.1 1.4
So the faulted electrical power delivered to the infinite bus is given by: P
EV
sin δ
=
X
P
=
P
=
1.28) (1.0 1.0 ) (1.28
sin δ
1.4 0.9143 sin δ
The post-fault electrical power delivered to the infinite bus is given by: P
EV =
X
+
X
+
sin δ X
ELEC 4100
T P
TEN : TRANSIENT STABILITY SOLUTIONS
2.133sin 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: P
δ
δ
2.462 sin δ
=
=
P
=
1.0
⎛ 1.0 ⎞ = sin ⎜ ⎟ ⎝ 2.462 ⎠ 23.96
=
Similarly: P
δ
δ
=
(
2.133 sin 180
−
δ
)
=
P
=
1.0
⎛ 1.0 ⎞ = 180 − sin ⎜ ⎟ ⎝ 2.133 ⎠ =
152
Now apply the equal area criterion.
∫ (1.0 (δ
−
−δ
)
0.9143 sin δ dδ
cos δ ) + 2.133 co
0.48 0.4835 35
−
δ
δ
=
∫ (2.133 sin δ
=
cos δ − 0.9143 co
=
)
1.0 d δ
−
0.9143 )cos δ (2.133 − 0.
1.21 1.219 9 cos cos δ
⎛ −0.4835 ⎞ = cos ⎜ ⎟ ⎝ 1.219 ⎠ =
113.4
This is the critical clearing angle.
Question 2. Again apply the equal angle criterion, but with: δ
=
51.2
So the equal area criterion requires that:
∫ (1.0 (δ δ
−
−δ
+
)
0.9143 sin δ d δ
cos δ ) + 0.9143 co
2.133 co cos δ
=
δ
=
∫ (2.133 sin δ
cos δ − 0.9143 co +
)
cos = −2.133 co
δ
1.0 d δ
−
( 2.133 − 0.9143 )cos δ
cos + 2.133 co
cos + 0.9143 co
δ
δ
ELEC 4100
T
TEN : TRANSIENT STABILITY SOLUTIONS
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:
⎡ y − f ( x )⎤ ⎦ =x +⎣
x
f
'
(x )
So:
+ (1 − 2.133sin δ
=δ
δ
⎡⎣2.017 − δ
)
− 2.133cos δ
⎤⎦
Solving Iteratively: δ δ
=
=
1.0 ,
δ
=
1.1545 ,
δ
=
1.1704 ,
δ
=
,
δ
=
1.1545
1.1546 ,
δ
=
1.1545
δ
=
1.1545 , 1.1545
So after 8 iterations the solution is: δ
=
66.15
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. 1.0
=
0.03183
d δ dt
+ 0.01
d δ
+ 2.462 sin
δ
dt
Or: 31.42
d δ =
dt
+ 0.3142
d δ dt
+ 77.35 sin
δ
Now consider small deviations in the rotor angle. If the rotor angle changes from
δ
then:
(
sin δ + ∆δ
) = sin (δ ) cos ( ∆δ ) + cos (δ ) sin ( ∆δ ) = sin (δ ) + cos (δ ) δ
So: 31.42
d δ =
dt
+ 0.3142
d δ dt
+ 77.35 sin
δ
Changes to: 31.42
=
d
(δ + δ ) dt
+ 0.3142
( +δ )
d δ
dt
+ 77.35 ⎡⎣sin (δ ) + cos (δ ) δ ⎤⎦
to
δ + ∆δ ,
ELEC 4100
T
0=
d δ dt
+ 0.3142
d δ dt
TEN : TRANSIENT STABILITY SOLUTIONS
+ 77.35 cos
(δ ) δ
Taking the Laplace transform: 0
= δ ( s ) ⎡⎣ s + 0.3142 s + 77.35 cos (δ ) ⎤⎦
Solving for the roots of the quadratic equation: s
,s
=
0.0987 − 309.4cos (δ
−0.3142 ±
)
2
For the system to be stable, it is necessary that: −0.3142 +
0.0987
− 309.4 cos
(δ ) < 0
( ) < 0.3142
0.0987 − 309.4 cos δ
( ) < 0.0987
0.0987 − 309.4 cos δ
( ) >0
309.4 cos δ
For
( )>0,
cos δ
it is necessary to have
constraint on the system.
−90
<
δ
< +90
. This is the small signal stability
ELEC 4100
TUTORIAL ELEVEN : PROTECTION SOLUTIONS
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 P 1, 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.
B1
B12
Bus 3
Bus 2
Bus 1
P3
B23
B21
L1
P2
B32
B34
L3
L2
Bus 4
P1
B43
B4
L4
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).
1
ELEC 4100
TUTORIAL ELEVEN : PROTECTION SOLUTIONS
(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) P 1, (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
B21a B21b
B12b
B24a
B42a B46
B32
B13
B1
Bus 4
Bus 2
P1
B12a
P3 B31
B24b
B42b
P2
B32
Bus 3 B3
Figure 2 Answer :
(a) The figure below shows the protective zones of the system in figure 2. B12a B12b B1
Zone 10
Zone 3
Zone 2 Bus 1
Bus 2
Bus 4
B21a
P1
B21b
Zone 4
B24a
B42a B46
B32
B13
Zone 13
Zone 1
P3
B24b
Zone 5 B31
B32
Zone 6
Bus 3 B3
Zone 7
Zone 8
2
P2
B42b
Zone 11 Zone 9
Zone 12
ELEC 4100
TUTORIAL ELEVEN : PROTECTION SOLUTIONS
(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. Bus 1 B1
B21
B12
1
Bus 3
Bus 2
line 1-2
B23
B32 line 2-3
B24 line 2-4 B42
Bus 4 B4
2
Figure 3. Answer:
Part (a): The impedance seen by the mho relay at B12 is: Z ' = Z ' =
V LN ' I L '
=
V LN ( 4500/1)
V LN
1
I L
15
I L (1500 / 5 ) =
Z
15
Set the B12 relay zone 1 Zr1 setting for 80% reach of line 1-2, as:
3
B3
3
ELEC 4100
TUTORIAL ELEVEN : PROTECTION SOLUTIONS
Z r 1
=
0.8
6 + j 60 15
=
( 0.32 + j3.2 ) Ω
secondary
Set the B12 relay zone 2 Zr2 setting for 120% reach of line 1-2, as: Z r 1
= 1.2
6 + j 60 15
=
( 0.48 + j 4.8 ) 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: Z r 1
=
6 + j 60 15
+ 1.2
5 + j 50
=
15
( 0.8 + j8.0 ) Ω
secondary
Part (b) The secondary impedance viewed by B12 during emergency loading is: 500 ˆ ˆ ' = V LN ' Z ˆ ' I L
=
3
∠0
0
1
−1
0
1.4∠ − cos 0.9 15
= 13.7 ∠25.8 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 Z 12 = 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: Zapparent = Z12 + Z24 +
I 32
Z24
I 12
Where ( I 32 I 12 ) is the line 2-3 to line 1-2 fault current ratio. (b) if I 32 I 12
>
0.2 , 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. Bus 1 B1
1
B12
Bus 3
Bus 2 I12
B21
line 1-2
B23
B32
B3
3
line 2-3
B24 B24
Figure 4.
4
I32
I24
Bus 4 B42
ELEC 4100
TUTORIAL ELEVEN : PROTECTION SOLUTIONS
Answer :
Part (a) For the bolted three phase fault at bus 4 the system of figure 4 can be reduced to: I12
I32 I24
Z12 V1
V2
Z23
Z24
V3
The primary impedance seen by the B12 relay is then: Z apparent =
V1 ' I '
=
V1 '− V2 ' I '
12
+
12
Z apparent = Z 12 +
V2 ' I '
=
Z 12
12
+
V2 ' I '
12
Z24 ( I12 '+ I32 ' ) I 12 '
Zapparent = Z12 + Z24
+
I Z24 32 I 12
Part (b): The apparent secondary impedance seen by B12 for the bolted three phase fault at bus 4 is: Z apparent ' =
Z apparent
( NV
N I )
Where NV and NI are the turns ratios of the potential and current transformers for B12. Z apparent ' =
Z apparent ' =
Z12
+
Z24 (1 + I32 I12 )
(
NV
( 9 + 6 ( I32
NI )
=
( 3 + j40 ) + ( 6 + j80 )(1 + I32 ( NV NI )
I12 )
I12 ) ) + j(120 + 80 ( I32 I12 ) )
( NV
N I )
Also the zone 3 relay for B12 is set for 100% reach of line 1-2, and 120% of line 2-4. So: Z r 3
=
( 3 + j 40 ) + 1.2 ( 6 + j80 ) 10.2 + j136 = Ω ( NV NI ) ( NV NI )
Comparing this expression with the case for the balanced three phase fault when I 32 I 12
>
0.2
shows that: Z apparent ' >
10.2 + j136
( NV
N I )
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. 5
ELEC 4100
TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
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:
V ( x, s ) = k1e−γ x
Where:
γ =
+ k2 eγ x
( R+
a) For the case determine the and current. b) For the case determine the and current.
I ( x, s ) =
and
sL)( G + sC)
=
Z 0
1
k1e−γ x − k2 eγ x Z 0
( R+ sL) ( G + sC )
where the receiving end of the transmission line is short circuited constants k1 and k2, and derive simplified expressions for the voltage where the receiving end of the transmission line is open circuited constants k1 and k2, and derive simplified expressions for the voltage
Answer :
Part (a) The boundary conditions are :
V ( 0, s ) = E ,
V (1, s ) = 0
Substituting these constraints into the general solutions gives:
V ( 0, s ) = k1 + k2 V (1, s ) = k1e − γ
=E
+ k 2e γ = 0
From the second equation: k1
= −k2e 2γ
Substituting this expression into the first equation gives: k2 (1 − e
2 γ
) = E
or k 2
=
E
=
e−γ E
(1 − e ) ( e− 2γ
γ
− eγ )
=−
e−γ E 2sinh ( γ )
So: k1
= − k 2e =
eγ E
2γ
2sinh ( γ )
The voltage expression for the transmission line is then given by:
V ( x, s ) =
E
eγ (1− x) − e−γ (1− x) 2sinh ( γ )
1
ELEC 4100
TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
sinh ( γ (1 − x ) ) V ( x, s ) = E sinh ( γ ) The current expression for the transmission line is then given by:
I( x, s) =
I ( x, s ) =
1
E
eγ (1− x) + e−γ (1− x) Z 0 2sinh ( γ ) E cosh ( γ (1 − x ) ) sinh ( γ )
Z 0
Part (b) The boundary conditions are :
V ( 0, s ) = E ,
I (1, s ) = 0
Substituting these constraints into the above expressions for the voltage and current reveals:
V ( 0, s ) = k1 + k2 I (1, s ) =
=E
1
k1 e− γ − k2 eγ = 0 Z 0
From the second expression involving the current at the receiving end of the line:
k1
= k2e2γ
Substituting into the expression for the voltage at the sending end of the line: k2 (1 + e 2γ ) = E
k 2
or
=
E 1 + e 2γ
=
e− γ E e−γ
+ eγ
And so: k 1
=
e 2γ E 1 + e 2γ
=
eγ E e −γ
+ eγ
=
e γ E 2 cosh ( γ )
Hence:
V ( x, s ) =
eγ E 2 cosh ( γ )
e
− γ x
+
e−γ E 2 cosh (γ )
x
eγ
cosh ( γ (1 − x ) ) V ( x, s ) = E cosh ( γ ) Similarly
eγ E e −γ E γ x − γ x I( x, s) = e − e Z 0 2 cosh ( γ ) 2 cosh ( γ ) 1
I ( x, s ) =
E sinh ( γ (1 − x ) ) Z 0
cosh ( γ ) 2
=
e−γ E 2 cosh ( γ )
ELEC 4100
TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
Question 2. A single phase lossless transmission line of 150m length has an inductance and shunt capacitance per unit length of L = 1 µH/m and C = 11.111 pF/m. The line is terminated by a 600 Ω resistance. The transmission line is struck by lightning through an effective 100 Ω impedance at the sending end of the line, creating a surge voltage of 30kV peak and 50 µs 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 5 µs. Answer:
Part (a) The characteristic impedance is given by:
Z 0
L
=
C
1×10−6
=
11.111 ×10 −12
= 300Ω
The travelling wave propagation velocity is :
v0
1
=
LC
1
=
(11.111×10 )(1 ×10 ) −12
−6
= 3 ×108 ms −1
This is the speed of light in free space. The one-way transit time for the transmission line is then: τ
=
d v0
(150m )
=
( 3 ×10 ms ) 8
−1
= 5 ×10−7 s = 0.5 µ s
Part (b)
Z 0 = 300 Ω
100Ω
s τ = 0.5µ v(d,t)
v(0,t)
Z L= 600 Ω
30kV The reflection coefficients at the sending and receiving ends of the line are then:
Γ R = Γ L =
− Z 0 600 − 300 1 = = Z L + Z 0 600 + 300 3 Z L
Z S − Z 0 Z S + Z 0
=
100 − 300 100 + 300
=−
1 2
The initial surge voltage on the line is then: v(0, 0) = 30 kV 300 Ω (100Ω + 300 Ω ) = 22.5 kV Part (c): The Bewley lattice diagram can be developed as shown below: 3
ELEC 4100
TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS x = d
x = 0 22.5 kV
τ = 0.5µs 7.5 kV
2τ = 1.0µs
-3.750 kV
3τ = 1.5µs -1.25 kV
4τ = 2.0µs
0.625 kV
5τ = 2.5µs 0.2083 kV
6τ = 3.0µs -0.1042 kV
7τ = 3.5µs -0.0347 kV
8τ = 4.0µs
0.0174 kV
9τ = 4.5µs 0.0058 kV
10τ = 5.0µs -0.0029 kV
From the lattice diagram the sending and receiving end voltages can be developed as shown below: v(d,t) 30kV 25.83kV
25kV
25.69kV
25.72kV
t
0.5µs
1.0µs
1.5µs
2.0µs
2.5µs
3.0µs
3.5µs
4.0µ s
4.5µs
5.0µs
v(0,t) 26.25kV
25.73kV
25.63kV
22.5kV
25.71kV
t
0.5µs
1.0µs
1.5µs
2.0µs
4
2.5µs
3.0µs
3.5µs
4.0µs
4.5µs
5.0µs
ELEC 4100
TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS
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 100 Ω and a one-way propagation time of 0.1ms, while the second section has a characteristic impedance 400 Ω and a one-way propagation time of 0.1ms. A surge voltage of 20kV is applied to the line through a 100 Ω impedance, and the line is terminated with a 800 Ω 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.
τ = 0.1ms Z 2=400Ω
τ = 0.1ms Z 1=100Ω
Z S 100Ω 20kV v(0,t)
v(d 1+d 2 ,t)
v(d 1 ,t)
Z L=800Ω
Figure 1 : Single Phase lossless transmission line. Answer:
The reflection coefficient at the sending end of the transmission line is:
Γ S =
Z S − Z 1 Z S + Z 1
=
100 − 100 100 + 100
=0
The reflection and refraction coefficients on the s ending side of the junction are:
Γ12 = β 12
=
Z 2 − Z 1 Z 2 + Z 1 2 Z 2
Z1 + Z 2
= =
400 − 100 400 + 100 2 ( 400 ) 400 + 100
=
3
=
8
5
5
The reflection and refraction coefficients on the recei ving side of the junction are:
Γ 21 = β 21
=
Z1 − Z 2 Z 2 + Z 1 2 Z 1
Z1 + Z 2
= =
100 − 400 400 + 100 2 (100 ) 400 + 100
=− =
3 5
2 5
The reflection coefficient at the receiving end of the transmission line is:
Γ R =
− Z 2 800 − 400 1 = = Z L + Z 2 800 + 400 3 Z L
The surge voltage entering the line is given by:
v ( 0, 0 ) =
Z 1 Z1 + Z S
E
=
100 100 + 100
20 kV
= 10 kV
The Bewley lattice diagram can now be developed as shown below:
5
