VOLTAGE CONTROL WITH SHUNT CAPACITANCE ON RADIAL
DISTRIBUTION LINE WITH HIGH R/X FACTOR
A Thesis by
Hong-Tuan Nguyen Vu
Electrical Engineer, Polytechnic University of HCMC, 1993
Submitted to the College of Engineering
and the faculty of the Graduate School of
Wichita State University
in partial fulfillment of
the requirements for the degree of
Master of Science
December 2005
VOLTAGE CONTROL WITH SHUNT CAPACITANCE ON RADIAL
DISTRIBUTION LINE WITH HIGH R/X FACTOR
I have examined the final copy of this Thesis form and content and recommend
that it be accepted in partial fulfillment of the requirement for the degree of
Master of Electrical and Computer Engineering.
_____________________________
Dr Ward T. Jewell, Committee Chair
We have read this Thesis
and recommend its acceptance:
_____________________________
Dr Andrew Acker, Committee Member
_____________________________
Dr Paul K. York, Committee Member
ii
DEDICATION
To my parents who sacrifice their whole lives to encourage me to learn and to my
family, my wife Huyen and three kids Chi, Emily, and Luke, for their lack of my
care, as I was busy to complete this Thesis.
iii
ACKNOWLEDGMENTS
I would like to thank my advisor, Dr. Ward Jewell, who led me from the
very first year in my Graduate School, for his patience as well as the professional
advices to help me to obtain this achievement.
iv
ABSTRACT
A previous thesis [1] showed that the voltage on a distribution line with
high R/X ratio will actually decrease as shunt capacitance is added. This thesis
develops a new distribution feeder model and confirms this unexpected result.
Then the IEEE radial test feeders are modeled to further study the effect and
provide additional insight into voltage control on feeders with high R/X ratio.
v
TABLE OF CONTENTS
Chapter
Page
CHAPTER I INTRODUCTION
……………………………………………01
CHAPTER II METHODOLOGY
……………………………………………02
2.1
2.2
Radial Distribution Feeder ……………………………02
Problem Formulation
……………………………02
CHAPTER III EXAMPLE SYSTEM
……………………………………08
CHAPTER IV LOAD IMPEDANCES
……………………………………11
CHAPTER V VOLTAGE CONTROL
……………………………………13
5.1
5.2
5.3
Example 1
Example 2
Example 3
LIST OF REFERENCES
APPENDIXES
……………………………………………14
……………………………………………15
……………………………………………16
……………………………………………………19
……………………………………………………………21
APPENDIX A: IEEE 37-NODE TEST FEEDERS …………………………..22
APPENDIX B: LINE IMPEDANCES CALCULATIONS …………………….24
APPENDIX C: EXAMPLES CALCULATIONS
vi
……………………………27
LIST OF TABLES
Table
TABLE 1
Page
LIST OF R/X FACTOR OF EACH LINE SEGMENTS ON
IEEE 37-NODE TEST FEEDERS SEGMENTS ………….09
TABLE 2
SPOT LOADS OF IEEE 37-NODE TEST FEEDERS ……11
vii
LIST OF FIGURES
Figure
Page
Figure 1
Feeder Model
………………………………………………….…3
Figure 2
Equivalent Circuit Model ……………………………………………..3
Figure 3
Combined All Impedances in Zt Model ……………………………...4
Figure 4
Receiving End Phasor Diagrams ………………………………….…7
Figure 5
One line Diagram of IEEE 37-Node Test Feeder ………………….10
Figure 6
Results for Example 1 ……………………………………………….14
Figure 7
Results for Example 2 ……………………………………………….15
Figure 8
Results for Example 3 ……………………………………………….16
viii
CHAPTER I
INTRODUCTION
As early as 1910, capacitor banks were applied to improve both quality
and quantity of electric power transmission and distribution. Still today they are
used to correct power factor in order to lower line losses, to raise power transfer
capabilities, and to improve a line’s voltage profile. They can help control power
flow through tie lines and reduce operating costs. In most cases, capacitor banks
play the role of reactive power generator; however they may be considered as
capacitive reactive loads too.
This thesis extends the work of a previous thesis [1]. In this thesis, it was
shown that injecting reactive power into lines with high R/X ratio (R/X > 2),
sometimes cause the receiving voltage VR to decrease, not increase, as
expected. That thesis [1] seeks to further explain these results, and also to
extend the basic understanding of the operation of capacitors on a power
system.
1
CHAPTER II
METHODOLOGY
2.1
Radial Distribution Feeder
A radial distribution feeders supplies all loads from only one substation.
The feeder may have the following components:
2.2
•
Three-phase main feeder
•
Three-phase, two-phase, or single-phase laterals.
•
Step-type voltage regulators.
•
In-line transformers
•
Shunt capacitor banks
•
Distribution transformers
•
Loads
Problem Formulation:
A radial distribution feeder was modeled as shown in Figure 1. Variables
are source voltage Vs, line impedances R and j*X, and load impedances Rl and
j*Xl. The shunt capacitor, which was paralleled with the load to inject rated
reactive power, Qc, into the feeder, has rated capacitive reactance Xc. The
negative sign on j*Xc indicates that capacitive reactive power Qc has the
opposite sign to inductive reactive power Ql. The receiving end voltage Vl is
supplied to the load, and the shunt capacitor is connected in parallel with the
load. The source voltage is ideal; constant magnitude and infinite power.
2
~
Figure 1. Feeder Model
The equivalent circuit in Figure 2 puts the load in parallel with the
capacitor, creating the new load with the complex impedance Z.
Figure 2. Equivalent circuit model
Equation 1 gives the new load impedance Z of Figure 2.
3
Z=
j . xc . ( Rl
Rl
j Xl
jXl )
( 01)
xc
The equivalent circuit in Figure 3 combines the load, complex impedance
Z, in series with the line impedance to form the new total complex impedance Zt,
which is calculated in equation 2.
Figure 3. Combined All Impedances In Zt Model
Zt =
j. xc . ( Rl jXl)
Rl j Xl xc
4
(R
jX)
j . xc . ( Rl jXl)
Zt =
Rl j Xl xc
jRl. xc
Zt =
Xl. xc
(R
jX) .
Rl j Xl xc
Rl j Xl xc
R. Rl jR. Xl jR. xc
Rl j Xl xc
Xl. xc
Zt =
jRl. X
X . Xl X . xc
Rl j Xl xc
R. Rl X . Xl X . xc
j . Rl. xc
RXl R. xc
Rl. X
(02)
Rl j Xl xc
Applying Ohm's law for load Z and again for total line and load Zt:
Vl
=
Z .I
Vs
=
Zt . I
Assuming that Vs is the base voltage, its per-unit value will be Vs = 1 pu
with an angle of 0, so the per unit receiving end voltage Vl/Vs does not depend
on the current I, because the current I is eliminated from both numerator and
denominator, as shown in equation 3.
Vl
Vs
j.x .(Rl j. Xl)
=
c
. RRl
. XXl
.
.
Xlx
Xx
c
c
. RXl
.
. RlX
.
j. Rlx
Rx
c
c
(03)
Equation 03 gives the receiving end voltage in per unit
The capacitive reactive power, Qc, at nominal capacitor voltage is used to
rate a capacitor. Capacitive reactance, Xc, can be calculated from voltage and
reactive power. The equation for the shunt capacitor reactive power is:
5
Qc =-kV2/Xc
Where:
Xc: Shunt Capacitive reactance (Ohm)
Qc: rated capacitive reactive power (MVAr)
kV: rated capacitor voltage (kV)
The capacitive reactance Xc is given by equation 4.
2
xc
kV
Qc
(Ω )
( 4)
The value of capacitive reactance Xc will have the opposite sign from
inductive reactance Xl, and from Qc, which is positive when supplied to the
system.
Most loads are inductive, Figure 4a, and added compensation supplies
reactive power, Figure 4b. In the case of overcompensation, the reactive power
supplied by the added capacitive reactance is greater than that consumed by the
load. The combination becomes capacitive, as shown in Figure 4c.
6
a/ Inductive load
b/ Power factor correction 1 > pf’ > pf, adding shunt capacitor: I’ = I+Ic
(c) Overcompensation
Figure 4. Receiving End Phasor Diagrams
(a) Without adding shunt capacitor.
(b) With adding shunt capacitor.
(c) Overcompensated
7
CHAPTER III
EXAMPLE SYSTEM
A feeder from the IEEE 37-node test feeder [5] is used as the example
system in the rest of this thesis. Simple positive-sequence line impedances were
calculated from the 37 node feeder data.
IEEE 37 Node Test Feeder
The positive-sequence R and X are needed for each feeder segment of
the 37-node test feeder. From the available feeder data [5], the impedance matrix
[Zabc] is placed into the modified Carson and the Kron reduction equations [6]:
Zs
1.
Zm
1.
3
3
Zabc 0 , 0
Zabc 1 , 1
Zabc 0 , 1
Zabc 1 , 2
Zabc 2 , 2
Zabc 2 , 0
The positive sequence impedance, Z11, is
:
Z11 = Zs – Zm
The calculations for each segment are presented in Appendix B. The
results are shown in Table 1.
8
TABLE 1:
R/X FACTOR OF EACH LINE SEGMENT
701
702
702
702
703
703
704
704
705
705
706
707
707
708
708
709
709
710
710
711
711
713
714
720
720
727
730
733
734
734
737
738
744
744
799
702
705
713
703
727
730
714
720
742
712
725
724
722
733
732
731
708
735
736
741
740
704
718
707
706
744
709
734
737
710
738
711
728
729
701
0.18
0.08
0.07
0.25
0.05
0.11
0.02
0.15
0.06
0.05
0.05
0.14
0.02
0.06
0.06
0.11
0.06
0.04
0.24
0.08
0.04
0.10
0.10
0.17
0.11
0.05
0.04
0.11
0.12
0.10
0.08
0.08
0.04
0.05
0.35
Line Segment Data
Positive Seq.
(Ω/mile)
r
x
722 0.36 0.31
724 1.75 0.58
723 0.97 0.52
722 0.36 0.31
724 1.75 0.58
723 0.97 0.52
724 1.75 0.58
723 0.97 0.52
724 1.75 0.58
724 1.75 0.58
724 1.75 0.58
724 1.75 0.58
724 1.75 0.58
723 0.97 0.52
724 1.75 0.58
723 0.97 0.52
723 0.97 0.52
724 1.75 0.58
724 1.75 0.58
723 0.97 0.52
724 1.75 0.58
723 0.97 0.52
724 1.75 0.58
724 1.75 0.58
723 0.97 0.52
723 0.97 0.52
723 0.97 0.52
723 0.97 0.52
723 0.97 0.52
724 1.75 0.58
723 0.97 0.52
723 0.97 0.52
724 1.75 0.58
724 1.75 0.58
721 0.24 0.22
Config.
Length
(mile.)
Node B
Node A
OF IEEE 37-NODE TEST FEEDERS
9
Positive Seq.
(Ω)
R
X
0.07 0.06
0.13 0.04
0.07 0.04
0.09 0.08
0.08 0.03
0.11 0.06
0.03 0.01
0.15 0.08
0.11 0.04
0.08 0.03
0.09 0.03
0.25 0.08
0.04 0.01
0.06 0.03
0.11 0.04
0.11 0.06
0.06 0.03
0.07 0.02
0.42 0.14
0.07 0.04
0.07 0.02
0.10 0.05
0.17 0.06
0.31 0.10
0.11 0.06
0.05 0.03
0.04 0.02
0.10 0.05
0.12 0.06
0.17 0.06
0.07 0.04
0.07 0.04
0.07 0.02
0.09 0.03
0.08 0.08
R/X
1.16
3.02
1.88
1.16
3.02
1.88
3.02
1.88
3.02
3.02
3.02
3.02
3.02
1.88
3.02
1.88
1.88
3.02
3.02
1.88
3.02
1.88
3.02
3.02
1.88
1.88
1.88
1.88
1.88
3.02
1.88
1.88
3.02
3.02
1.09
799
724
722
707
712
701
713
742
704
720
705
702
714
706
729
744
727
703
718
725
728
730
732
708
709
731
736
733
710
775
734
740
735
737
738
711
741
Figure 5: One line diagram of IEEE 37-Node Test Feeder
10
CHAPTER IV
LOADS IMPEDANCES
Load data for a distribution feeder is commonly modeled as constant real
power, P, and reactive power, Q. This load power data is converted to a constant
impedance model by assuming a load voltage, and from the load data shown in
Table 2 [5]:
TABLE 2: SPOT LOADS OF IEEE 37-NODE TEST FEEDERS
Node Load
Model
701 D-PQ
712 D-PQ
713 D-PQ
714
D-I
718
D-Z
720 D-PQ
722
D-I
724
D-Z
725 D-PQ
727 D-PQ
728 D-PQ
729
D-I
730
D-Z
731
D-Z
732 D-PQ
733
D-I
734 D-PQ
735 D-PQ
736
D-Z
737
D-I
738 D-PQ
740 D-PQ
741
D-I
742
D-Z
744 D-PQ
Total
Ph-1 Ph-1 Ph-2 Ph-2 Ph-3
MW MVAr MW MVAr MW
140
70
140
70
350
0
0
0
0
85
0
0
0
0
85
17
8
21
10
0
85
40
0
0
0
0
0
0
0
85
0
0
140
70
21
0
0
42
21
0
0
0
42
21
0
0
0
0
0
42
42
21
42
21
42
42
21
0
0
0
0
0
0
0
85
0
0
85
40
0
0
0
0
0
42
85
40
0
0
0
0
0
0
0
42
0
0
0
0
85
0
0
42
21
0
140
70
0
0
0
126
62
0
0
0
0
0
0
0
85
0
0
0
0
42
8
4
85
40
0
42
21
0
0
0
727 357 639 314 1091
11
Ph-4
MVAr
175
40
40
0
0
40
10
0
0
21
21
0
40
0
21
0
21
40
0
0
0
40
21
0
0
530
Apply Ohm's law to calculate the load impedances:
S = P - j*Q
Then Z =
V
I
V
S = V*I = V.
Z
2
So, Z =
Hence,
Note:
V
P
j.Q
Rl
=
Re( Z ) (Ω )
(05)
Xl
=
Im( Z ) (Ω )
(06)
V is in kV
P is in MW, and Q in MVAr
12
CHAPTER V
VOLTAGE CONTROL
In this section, the effects of shunt capacitance on the receiving end
voltage are illustrated in three examples. Because of its relatively high R/X ratio,
phase a of the feeder at node 744 of the IEEE 37-node test feeder was used in
the examples. Calculations for all three examples are presented in Appendix C.
13
5.1 Example 1:
Substitute the feeder data into equations 1, 2, 3, 5, and 6. Then plot the
graph of Vr, the receiving end (load) voltage and power factor versus the rating of
capacitor Qc. These are shown in Figure 6. Note that the capacitor rating is not
the actual capacitive reactive power injected into the line, because reduced Vr
lowers the actual Qc. So in some cases, a greater value of rated Qc may supply
less capacitive reactive power than a smaller capacitor.
Plotting of Received End Voltages
1
Received Voltage Per Unit
0.98
0.96
0.94
V
pf
c
c
1
1
0.92
0.9
0.88
0.86
0.84
0.82
0.8
0
20
40
60
80
100
120
140
Q
c 1
Rating Capacitor (MVA)
Figure 6. Results for Example 1.
14
160
180
200
5.2 Example 2:
Increase the R/X ratio from its original value of 3 to 31. The results, shown
in Figure 7 show that the higher value of k results in the receiving end voltage to
starts decreasing earlier.
Plotting of Received End Voltages
1
Received Voltage Per Unit
0.97
0.94
0.91
V
c
pf
c
1
1
0.88
0.85
0.82
0.79
0.76
0.73
0.7
0
20
40
60
80
100
120
140
Q
c 1
Rating Capacitor (MVA)
Figure 7. Results for example 2
15
160
180
200
5.3 Example 3:
Example 3 repeats the case simulated in a previous thesis [1]. This
example verifies agreement between the procedure developed in this thesis with
the previous work. Results are shown in Figure 8.
1
Plotting of Received End Voltages
1
Received Voltage Per Unit
0.97
0.94
0.91
V
c
pf
c
1
1
0.88
0.85
0.82
0.79
0.76
0.73
0.70
0.7
0
0
20
40
60
80
100
120
140
Q
c 1
Added Capacitive Power (KVAr)
Figure 8. Results for example 3.
16
160
180
200
200
In all three examples, the line’s R/X ratio is greater than 2. In each,
receiving end voltage Vr is observed increasing to its maximum value, and then
decreasing, as the capacitor rating increases. Also in each, the plot of power
factor has an M shape.
In example 1, the concept of power factor correction capacitance, or
reactive power generator can explain the behavior of the shunt capacitor within
the first interval, 0 < Qc < 26.6. The reactive power produced by the shunt
capacitor, compensated for the reactive power consumed by the load, which
corrects the power factor, and increases pf to unity.
In the second interval, 26.6 < Qc < 83.7, the capacitor reactive power flow
is greater than that consumed by the load, so the power factor becomes leading
thus and starts to decrease from unity. The voltage continues to increase with
increasing capacitive reactive power.
In the last interval, 83.7 < Qc < 200, the system becomes
overcompensated; the capacitive reactance is greater than that consumed by the
load and the line. The surplus capacitive reactance can be considered as a
capacitive load. Then the capacitive component, Xc, of the load impedance Z
becomes smaller and forces Z to do so as well. The voltage drop on the line
inscreases, and the receiving end voltage starts dropping with increasing
capacitor rating. The actual capacitive reactive power injected is decreasing due
to the decrease in the receiving end voltage, returning the power factor to unity.
In example 2, the R/X ratio is increased by lowering the line reactance X.
That change also reduces the demand of reactive power that combine the
17
reactive load and the reactive power loss on the line. So the case of
overcompensation comes earlier results in the receiving end voltage start
dropping as soon as the rating capacitor is 30 MVAr.
Example 3 repeats an example from an earlier thesis [1]. The results were
the same as in [1]. This was used to verify the model developed in previous
chapters.
Future work:
o
As shown in example 2, the receiving end voltage drops earlier with
increasing R/X. Further work should be done to quantify this effect.
o
Study the effects of placing more than one capacitor on the line.
18
LIST OF REFERENCES
19
REFERENCES
[1]
D. K. Mehdi and W. Jewell, “Reactive power on Radial Distribution
Systems: Voltage Control and Line Losses,” Wichita State University,
Electrical Computer Engineering Department 2004.
[2]
M.H. Shwedhi and M.R. Sultan, “Power Factor Correction Capacitors;
Essentials and Cautions,” IEEE Power Engineering Society Summer
Meeting, vol. 3, pp 1317-1322, 2000.
[3]
B. Milosevic and M. Begovic, “Capacitor Placement for Conservative
Voltage Reduction on Distribution Feeders,” IEEE Transactions on Power
Delivery, vol. 19, No. 3, July 2004.
[4]
R.T. Saleh and A.E. Emanuel, “Optimum Shunt Capacitor for Power
Factor Correction at Busses with Lightly Distorted Voltage,” IEEE
Transactions on Power Delivery, vol PWRD-2, No. 1, pp 165 – 173,
January 1987.
[5]
IEEE 34, 37, 123 – node test feeders please visit the website
http://www.ewh.ieee.org/soc/pes/dsacom/testfeeders.html
[6]
William H. Kersting, “Distribution System Modeling and Analysis”, CRC
Press, 2002.
[7]
J. D. Glover and M.S. Sarma, “Power System Analysis and Design”, 3rd
Edition, Brooks/Cole, 2002.
20
APPENDICES
21
APPENDIX A
IEEE 37 NODE TEST FEEDER
PHASE IMPEDANCE AND ADMITTANCE MATRICES
Configuration 721
Z (R +jX) in ohms per mile
0.2926
0.1973
0.0673 -0.0368
0.0337 -0.0417
0.2646
0.0673 -0.0368
0.1900
0.2926
0.1973
B in micro Siemens per mile
159.7919
0.0000
0.0000
159.7919
0.0000
159.7919
Configuration 722
Z (R +jX) in ohms per mile
0.4751
0.2973
0.1629 -0.0326
0.1234 -0.0607
0.4488
0.1629 -0.0326
0.2678
0.4751
B in micro Siemens per mile
127.8306
0.0000
0.0000
127.8306
0.0000
127.8306
22
0.2973
Configuration 723
Z (R +jX) in ohms per mile
1.2936
0.6713
0.4871
0.2111
0.4585
0.1521
1.3022
0.6326
0.4871
0.2111
1.2936
0.6713
B in micro Siemens per mile
74.8405
0.0000
0.0000
74.8405
0.0000
74.8405
Configuration 724
Z (R +jX) in ohms per mile
2.0952
0.7758
0.5204
0.2738
0.4926
0.2123
2.1068
0.7398
0.5204
0.2738
2.0952
0.7758
B in micro Siemens per mile
60.2483
0.0000
0.0000
60.2483
0.0000
60.2483
23
APPENDIX B
FIND POSITIVE-SEQUENCE R AND X
FOR EACH FEEDER SEGMENT OF 37-NODE TEST FEEDER SEGMENTS
AND CALCULATE THEIR RATIO R/X:
Configuration 721:
Zabc
0.2926 i. 0.1973 0.0673 i. 0.0368 0.0337 i. 0.0417
0
0.2646 i. 0.1900 0.0673 i. 0.0368
0
0.2926 i. 0.1973
0
Using the modified Carson and the Kron reduction [6]:
Equation 4.68 =>
Zs
1.
3
Zabc 0 , 0
Zabc 1 , 1
Zabc 2 , 2
Zs = 0.283 + 0.195i
Zabc 1 , 2
Zabc 2 , 0
Zm = 0.045 0.025i
Equation 4.69 =>
Zm
1.
3
Zabc 0 , 1
Positive sequence impedance, Z11:
Z11
Zs
Z11 = 0.238 + 0.219i
Zm
Ω /mile
Configuration 722:
Zabc
0.4751 i. 0.2973 0.1629 i. 0.0326 0.1234 i. 0.0607
0
0.4488 i. 0.2678 0.1629 i. 0.0326
0
0.4751 i. 0.2973
0
Using the modified Carson and the Kron reduction [6]:
Equation 4.68 =>
Zs
1.
3
Zabc 0 , 0
Zabc 1 , 1
Zabc 2 , 2
Equation 4.69 =>
24
Zs = 0.466 + 0.287i
Zm
1.
3
Zabc 0 , 1
Zabc 1 , 2
Zm = 0.109 0.022i
Zabc 2 , 0
Positive sequence impedance, Z11:
Z11
Zs
Z11 = 0.358 + 0.309i
Zm
Ω / mile
Configuration 723:
Zabc
1.2936 i .0.6713 0.4871 i .0.2111 0.4585 i .0.1521
0
1.3022 i .0.6326 0.4871 i .0.2111
0
1.2936 i .0.6713
0
Using the modified Carson and the Kron reduction [6]:
Equation 4.68 =>
Zs
1.
3
Zabc 0 , 0
Zabc 1 , 1
Zabc 2 , 2
Zs = 1.296 + 0.658i
Zabc 1 , 2
Zabc 2 , 0
Zm = 0.325 + 0.141i
Equation 4.69 =>
Zm
1.
Zabc 0 , 1
3
Positive sequence impedance, Z11:
Z11
Zs
Z11 = 0.972 + 0.518i
Zm
Ω / mile
Configuration 724:
Zabc
2.0952 i .0.7758 0.5204 i .0.2738 0.4926 i .0.2123
0
2.1068 i .0.7398 0.5204 i .0.2738
0
2.0952 i .0.7758
0
Using the modified Carson and the Kron reduction [6]:
Equation 4.68 =>
Zs
1.
3
Zabc 0 , 0
Zabc 1 , 1
Zabc 2 , 2
25
Zs = 2.099 + 0.764i
Equation 4.69 =>
Zm
1.
3
Zabc 0 , 1
Zabc 1 , 2
Zm = 0.347 + 0.183i
Zabc 2 , 0
Positive sequence impedance, Z11:
Z11
Zs
Zm
Z11 = 1.752 + 0.581i
26
Ω / mile
APPENDIX C
EXAMPLES CALCULATION IN MATHCAD FORMAT
Example 1 : Node 744, IEEE 37-Node Test Feeder, using Shunt capacitor.
From Table 1 , We have:
R
0.0929
kV
4.8
X
Line impedances( Ω )
0.0308
R
( kV)
X
= 3.016
Calculate the load impedances (apply equation 04 and equation 05):
p
42
( MW )
q
21
( MVAr)
2
Z
kV
p
Z = 0.439 + 0.219i
j.q
Rl
Re( Z )
Rl = 0.439
Load impedance ( Ω )
Xl
Im( Z )
Xl = 0.219
Load impedance ( Ω )
Without adding shunt capacitor:
capacitive reactive power = zero
Q0
0
Z0
Rl j.Xl
Zt0
(R
Rl)
j .( X
Xl)
(R
Rl j .Xl
Rl) j .( X
Xl)
V0
Load impedance (Ω )
Total load & line impedance (Ω )
Received end voltage
V0 = 0.835
Add shunt capacitor Qc (MVAr); Qc delivers capacitive reactive power up
to 200 MVAr. A capacitor absorbs zero real power, Pc = 0 W, and
negative reactive power, Qc = -V^2/Xc var. Alternatively, a capacitor
delivers positive reactive power, +V^2/Xc [6]. So, we have:
27
Qc =
2
kV
xc
2
Capacitance reactance:
kV
xc
(Equation 04)
(Ω )
Qc
Load impedance (Ω ) (Applying equation 01)
j.xc .( Rl j .Xl)
Zc
(Ω )
Rl j . Xl xc
Load & line impedance (Ω ) (Applying equation 02)
Zt c
Xl.xc
R.Rl X .Xl
X .xc
j. Rl.xc
R.Xl R.xc
Rl.X
j. Xl xc
Rl
(Ω )
The received end voltage per unit: (Applying equation 03)
j .xc .( Rl j.Xl)
Vc
Xl.xc
R.Rl X .Xl X .xc
The power factor pf
Re
pfc
j . Rl.xc
R.Xl
R.xc
Rl.X
(pu)
2
Vc
Zc
Vc
2
Zc
Example 2 : Node 744, IEEE 37-Node Test Feeder, using Shunt capacitor.
However somehow we can change the line impedance to force the factor
R/X up ( by lowering X)
R
0.0929
kV
4.8
X
( kV)
0.003
Line impedances( Ω )
R
X
= 30.967
28
Calculate the load impedances (applying equation 04 and equation 05):
p
42
( MW )
q
21
( MVAr)
2
kV
Z
Z = 0.439 + 0.219i
j.q
p
Rl
Re( Z )
Rl = 0.439
Load impedance ( Ω )
Xl
Im( Z )
Xl = 0.219
Load impedance ( Ω )
Without adding shunt capacitor:
capacitive reactive power = zero
Q0
0
Z0
Rl j.Xl
Zt0
(R
Rl)
j.( X
Xl)
(R
Rl j .Xl
Rl) j .( X
Xl)
V0
Load impedance (Ω )
Total load & line impedance (Ω )
Received end voltage
V0 = 0.851
Add shunt capacitor Qc:
Qc =
2
kV
xc
2
Capacitance reactance:
xc
kV
(Equation 04)
(Ω )
Qc
Load impedance (Ω ) (Applying equation 01)
Zc
j .xc .( Rl j.Xl)
Rl j . Xl xc
(Ω )
Load & line impedance (Ω ) (Applying equation 02)
29
Xl.xc
Zt c
R.Rl X .Xl
X .xc
j. Rl.xc
R.Xl R.xc
Rl.X
Rl j . Xl xc
(Ω )
The received end voltage per unit: (Applying equation 03)
j .xc.( Rl j.Xl)
Vc
Xl.xc
R.Rl X .Xl X .xc
j . Rl.xc
The power factor pf
Re
pfc
Vc
R.Xl R .xc
Rl.X
(pu)
2
Zc
Vc
2
Zc
Example 3 : Reproduce one of the cases from previous thesis
Line impedance Zabs = 0.05 pu, Zbase = 10Ω, R/X = 2.5
Sending Voltage Vs = 1.0 pu with an angle of zero, Vln = 7.62 kV
Load S = 1.0 pu with Sbase = 100 MVA
Line impedances
0.05.10
Zabs
Zabs
R
2
1
1
R = 0.464
(Ω )
X = 0.186
(Ω )
2
2.5
X
R
X
R
2.5
= 2.5
Calculate the load impedances (applying equation 04 and equation 05):
p
100.0.7
3
p = 23.333
30
( MW )
2
100
q
( 100.0.7)
2
q = 23.805
3
kV
7.62
( kV)
2
kV
Z
p
( MVAr)
Z = 1.219 + 1.244i
j .q
Rl
Re( Z )
Rl = 1.219
Load impedance ( Ω )
Xl
Im( Z )
Xl = 1.244
Load impedance ( Ω )
Without adding shunt capacitor:
capacitive reactive power = zero
Q0
0
Z0
Rl j .Xl
Zt 0
(R
Rl)
j.( X
Xl)
(R
Rl j .Xl
Rl) j .( X
Xl)
V0
Load impedance (Ω )
Total load & line impedance (Ω )
Received end voltage
V0 = 0.789
Add shunt capacitor Qc(MVAr)
Qc =
2
kV
xc
2
Capacitance reactance:
xc
kV
(Equation 04)
(Ω )
Qc
Load impedance (Ω ) (Applying equation 01)
Zc
j.xc .( Rl j .Xl)
Rl j . Xl xc
(Ω )
Load & line impedance (Ω ) (Applying equation 02)
31
Zt c
Xl.xc
R.Rl X .Xl X .xc
j. Rl.xc
R.Xl
R.xc
Rl.X
Rl j . Xl xc
(Ω )
The received end voltage per unit: (Applying equation 03)
Vc
j .xc.( Rl j .Xl)
Xl.xc
R .Rl X .Xl X .xc
j . Rl.xc
The power factor pf
Re
pfc
Vc
2
Zc
Vc
2
Zc
32
R .Xl R.xc
Rl.X
(pu)