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Lecture 27 – ECE504PS – Spring 2007
Goals of this lecture are to:
1. increase our familiarity with PSAT,
2. explore the π model of an AC line at different loadings, including SIL and Pmax,
3. look at the relationship between voltage angle and magnitude versus real and
reactive power flow,
4. investigate properties of parallel circuits, and
5. introduce TSAT.
Further exploration of SIL (Surge Impedance Loading)
In Lecture 25 we discussed a five bus-system. Two parallel lines were modeled. One
line was modeled using a single section and the other was modeled with two sections.
For this lecture, the second line will be divided into 6 sections using five buses. We will
compare these two models as we explore surge impedance loading. The data for one line
is:
 Series Impedance: 0.013+j0.1 pu
 Admittances: 0+j0.2 pu {Is this B or B/w???}
 Rating 1 of 300 MVA
When we divide the line into six sections, the data for each section becomes:
 Series Impedance: 0.00216666666666667+ 0.0166666666666667 pu
 Admittances: 0+j 0.0333333333333333 pu {Is this B or B/2???}
 Rating 1 of 300 MVA
1.0000
0.00
-3.35
0.00
-6.69
-0.0104
1.0014
-0.0166
-0.00
3.35
500.00
0.00
-0.00
6.69
0.00
-10.02
500.00
-20.02
Swing230
-0.00
0.00
1.0022
-0.0187
1.0025
0.00
-0.00
Mid5
-0.00
-3.35
Mid4
3.35
-0.0166
1.0022
0.00
Mid3
-0.00
-6.69
Mid2
Mid1
-0.0104
-10.00
0.00
0.00
-10.02 6.69
Gen230
0.0000
0.00
-0.00
-10.00
1.0014
-20.02
0.00
Unloaded lines
-0.0001
1.0000
Eliminate the two 13.8 buses and add a generator to the Gen230 Bus (Remember to
change Gen230 from a load bus to a Gen bus –What happens if you leave it as a load
bus?) Set Gen230 and Swing230 to control their own voltages to 1.0 pu. Set the MW
Output of the generator at Gen230 to zero. Add a 500 MW constant power load to
Swing230. Solve the powerflow and you should get something that looks like:
Unloaded lines: The two line models yield almost identical results. The voltage rises as
we approach the middle of an unloaded line; thus, the segmented line produces slightly
more reactive power because it models this rise. Both ends of the lines are producing
Page 1 of 8
Lecture 27 – ECE504PS – Spring 2007
10 MVARS (0.1 pu); thus, the entire line is producing 0.2 pu reactive power – This tells
us that the susceptance in the model is the total “B” for the line. Note that real power
flows from higher angles to lower angles and that Vars flow from higher to lower
voltages.
-6.47
600.00
1.0000
Heavy loaded lines
-287.85
299.51
-3.30
38.37
292.68 -290.74
-50.05 61.67
2.8663
0.9950
5.7599
0.9920
294.63 -292.68
-38.37 50.05
500.00
0.00
290.74 -288.81
-76.66
-61.67 73.19 146.24
Swing230
-294.62
8.6694
0.9910
296.58
-26.64
Mid5
26.64
11.5831
0.9920
-296.58
Mid4
298.53
-14.90
Mid3
14.90
Mid2
-298.53
14.4892
0.9950
73.06
Mid1
Gen230
300.49
-3.17
0.0000
17.3761
1.0000
Next increase the generation at Gen230 up to 600 MW and solve the powerflow to see
Heavy loaded lines: Again the two line models yield almost identical results. Both lines
consume about 12 MW of real power and 70 MW of reactive power while transferring
about 300 MW. The voltage drops as we approach the middle of an unloaded line; thus,
the segmented line consumes slightly more reactive power because it models this drop.
Note that real power flows from higher angles to lower angles and that Vars flow from
higher to lower voltages.
18.28
139.91 -139.48
-18.27 18.25
1.3526
1.0000
2.7095
1.0000
140.34 -139.91
-18.28 18.27
500.00
0.00
139.48 -139.05
-18.25 18.20
222.36
36.39
Swing230
-140.34
4.0705
1.0000
140.77
-18.26
Mid5
18.26
5.4356
1.0000
-140.77
Mid4
141.21
-18.23
Mid3
18.23
Mid2
-141.21
6.8049
18.19
Mid1
Gen230
1.0000
-138.59
141.19
-18.18
141.65
-18.17
0.0000
Lines at SIL
1.0000
-36.35
282.84
8.1783
1.0000
In lecture 26 we noted that SIL occurs when I = √(B/X). In this instance I would equal
1.4142 pu. At 1.0 pu voltage this would correspond to 1.4142 pu power in each line, so
the generator will need to produce 282.84 MW, to load the lines to SIL as shown below:
Lines at SIL: Again the two line models yield almost identical results. The lines
produce no net reactive power (a second verification that the susceptance in the model is
the total “B” for the line). The voltage profile of the segmented line is flat. This very
slight difference comes from the 2.6 MW of losses in each line due to the line resistance.
Note that real power flows from higher angles to lower angles.
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Lecture 27 – ECE504PS – Spring 2007
500.00
0.00
11.3280
1003.24 -1003.24 1003.24
331.10 -331.10 661.45 -661.46
-1003.24 -1503.24
990.35 1977.10
Swing230
0.8512
26.5574
-1003.24
Mid5
1003.24
-0.00 -0.00
0.7478
44.9175
-1003.24
Mid4
1003.24
-331.10 331.10
0.7099
63.2777
0.7478
-1003.24
Mid3
1003.24
-661.46 661.45
Mid2
-1003.24
78.5071
987.12
Mid1
Gen230
1.0000
-1000.00
1000.00
987.12
1003.24
990.34
0.0000
Pure X lines (R=0)
0.8512
1977.10
2003.24
89.8351
1.0000
Pmax: In earlier lectures we looked at the swing equation. We talked about the maximum
power transfer in the swing equation as Pmax = | V1|∙|V2|/X. For our lines Pmax = 10 pu as
X = 1 u and V1 = V2 = 1. What will the voltage in the middle of the line become? At
Pmax, V1 & V2 are 90º apart. The middle of the line should is a voltage divider, (or
average of V1 & V2). The voltage at the middle of the line will approach 1/sqrt(2) or
0.707 pu. {Add figure below:}
Further exploration of π model: Consider the pi model of a transmission line with the
sending side voltage, V1, and load side voltage, V2. Assume the angle on V2 is zero and
the angle for V1 is . That is |V2| = |V2|0º and V1 = |V1|∙cos + j∙|V1|∙sin. Let the
current into the sending side be I1 and the current to the load be I2. {Draw figure below:}
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Lecture 27 – ECE504PS – Spring 2007
The sending side voltage equals the load side voltage plus the current times the series
reactance:
V1 = V2 + (R+jX)∙I2
Solving for I2 yields:
I2 = {V1 - V2}/(R+jX)
Multiply top and bottom by R-jX and doing a little algebra yields:
I2 = {(R-jX)∙V1 - [R–jX]∙V2}/(R2+X2)
Substitute [|V1|∙cos + j∙|V1|∙sin] for V1 and separate real and imaginary terms
I2 = {(R-jX)∙ [|V1|∙cos + j∙|V1|∙sin]
-[R –jX]∙|V2|}/(R2+X2)
{Remember that |V2| = |V2|0º}
I2 = {(R∙|V1|∙cos+X∙|V1|∙sin-jX∙|V1|∙cos+jR∙|V1|∙sin]
-R∙|V2|+jX∙|V2|}/(R2+X2)
Remember that the complex power, S, equals the voltage time the complex conjugate of
the current. The complex power to the load is therefore:
S2 = V2∙I2*
S2 = |V2|∙I2* {Remember that |V2| = |V2|0º}
We can substitute the above equation for I2 into our complex power equation:
S2 = |V2|∙{R∙|V1|∙cos+X∙|V1|∙sin-jX∙|V1|∙cos+jR∙|V1|∙sin]
-R∙|V2|+jX∙|V2|}/(R2+X2)*
S2 = |V2|∙{R∙|V1|∙cos+X∙|V1|∙sin+jX∙|V1|∙cos-jR∙|V1|∙sin]
-R∙|V2|-jX∙|V2|}/(R2+X2)
S2 = {R∙|V1|∙|V2|∙cos+X∙|V1|∙|V2|∙sin+jX∙|V1|∙|V2|∙cos-jR∙|V1|∙|V2|∙sin]
-R∙|V2|∙|V2|-jX∙|V2|∙|V2|}/(R2+X2)
S2 = {R∙|V1|∙|V2|∙cos+X∙|V1|∙|V2|∙sin-R∙|V2|∙|V2|}/(R2+X2)+
+j{X∙|V1|∙|V2|∙cos-R∙|V1|∙|V2|∙sin-X∙|V2|∙|V2|}/(R2+X2)
S2 = {R∙(|V1|∙cos- |V2|)∙|V2|+X∙|V1|∙|V2|∙sin}/(R2+X2)+
+j{X∙(|V1|∙cos-|V2|)∙|V2|-R∙|V1|∙|V2|∙sin}/(R2+X2)
The real part of the complex power is:
P = R∙(|V1|∙cos-|V2|)∙|V2|+X∙|V1|∙|V2|∙sin}/(R2+X2)
Which simplifies to a familiar form when R<<X:
P = |V1|∙|V2|∙sin/X
Provided X>>R, power flows from higher angles to lower angles as we saw in the earlier
examples.
The imaginary part of the complex power is:
Q = j{X∙(|V1|∙cos-|V2|)∙|V2|-R∙|V1|∙|V2|∙sin}/(R2+X2)
When R<<X this simplifies to:
Q = j{(|V1|∙cos-|V2|)∙|V2|}/X
If  is relatively small cos ≈ 1 and the imaginary part becomes:
Q = j{(|V1|-|V2|)∙|V2|}/X
Provided X>>R and cos ≈ 1, reactive power flows from higher voltages to lower
voltages as we saw in the earlier examples.
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Lecture 27 – ECE504PS – Spring 2007
Let us explore the situations where these are not true:
1. If R is significant the term, R∙(|V1|∙cos-|V2|)∙|V2| becomes significant. In this
instance real power tend to flow from higher to lower voltages in addition to from
higher to lower angles.
2. If R is significant the term, -R∙|V1|∙|V2|∙sin term becomes significant. In this
instance, in addition to flowing from higher to lower voltages, Vars tend to flow
from lower to higher angles!
-27.89
-136.11
X and R values reversed for this line (Z = 0.1+j0.01)
Power flows from higher to lower voltages
Vars flow from lower to higher angle
0.0000
1.0500
7.2948
1.0000
To evaluate the above thoughts, imagine that we enter data for two parallel lines with a
series impedance of Z = 0.01 + j0.1 pu, but accidentally reverse R and X for the first line:
The solved powerflow will look like:
47.19
138.04
-190.40
100.00
100.00
0.00
X and R entered properly for this line (Z = 0.01+J0.01)
-125.96
73.60
21.23
211.64
LoadBus
Power flows from higher to lower angles
Vars flow from higher to lower voltage
GenBus
127.89
-54.29
This figure clearly shows that real power can flow from a lower angle to a higher angle in
the theoretical case of R>>X. To make clear that the Vars can flow from a lower to a
higher voltage, the same case is rerun with the load voltage reduced to 0.95 pu. The
results of this powerflow are shown at the top of the next page. Now you can see that the
Vars can flow from a lower to a higher voltage in the theoretical case of R>>X. I have
observed Vars flow from lower to higher voltages on heavy load 7#8 copper 115 kV
transmission lines. To get an idea of how big this conductor is imagine that #8 wires in a
house are used for 40 Amp circuits, (#14 for 15 Amp circuits, #12 for 20 Amp circuits …
you get the idea).
Page 5 of 8
54.43
-35.50
X and R values reversed for this line (Z = 0.1+j0.01)
Power flows from higher to lower voltages
Vars flow from lower to higher angle
0.0000
0.9500
2.4699
1.0000
Lecture 27 – ECE504PS – Spring 2007
-50.21
35.92
10.83
100.00
100.00
0.00
X and R entered properly for this line (Z = 0.01+J0.01)
-45.15
Power flows from higher to lower angles
Vars flow from higher to lower voltage
-42.10
4.64
-6.18
GenBus
LoadBus
45.57
46.33
50.00
103.75
0.0000
"From" and "To" accidentally refersed for tap ratio of 1.05
Wrong ratio causes Vars to flow to increase low side voltage
1.0000
2.8660
1.0500
Another interesting situation occurs when parallel transformers are set at different tap
ratios. Imagine that the transmission is set to run at 1.05 and the load buses at 1.0. While
entering data for parallel transformers, imagine you reverse the “from” and “to” buses
and keep the same tap ratio. See the example below.
-50.00
-91.72
1.05000
105.00
100.00
100.00
0.00
50.00
1.25
-50.00
1.25
0.00
-90.47
Highside
LoadBus
1.05000
50.00
50.80
0.0000
"From" and "To" accidentally refersed for tap ratio of 1.05
Wrong ratio causes Vars to flow to increase low side voltage
1.0200
2.8922
1.0200
It is likely in such a situation that the high-side voltage is reduced and the low side is
increased yielding a situation more like:
-50.00
-45.92
1.05000
4.88
100.00
100.00
0.00
Note that VArs are "circulating"
50.00
-45.92
-50.00
50.80
0.01
4.88
Highside
LoadBus
1.05000
Page 6 of 8
Lecture 27 – ECE504PS – Spring 2007
When two control areas are parallel to one another, the operators can get in this situation
with tap changes transformers if the voltage schedules of the two systems are not
coordinated.
Housekeeping:
Project files have the extension .psp.
Single Line Diagram files have the extension .pfd
Powerflow data files have the extension .pfb
(Powerflow data can be exported to a number of formats including GE, PTI, and BPA)
Enough powerflow stuff, on to transient stability
TSAT
Start TSAT. Click “Help” and select “Online Manual”. A one page .pdf file will open.
In the middle of the page will be blue block with two grey buttons in it, “User Manual”
and “Model Manual”. Open bookmarks. If you browse through the model manual, you
should see some things that are familiar; for example:
 At the top of page 17 is the quadratic saturation model
 Many of the parameters in the model DG0S4 on page 24 should look familiar
 Governor Type 20 on page 171
 Tables 2-1 to 2-9 on pages 76-78 are a good place to start looking at the models
Now open the user manual. Open bookmarks. We will briefly look at chapters 3, 6, 7,
and 8. Look at Figure 1-1 adobe page 10 (manual page 4). We are now at the center of
the stability world! The TSAT tool is set up to allow us analyze a system in the
following order:
1. Open 2area.tsa
2. Start with a base case and screen the contingencies of possible interest.
The intent is to find the most severe contingencies. In TSAT, click
“Options” and select “computations” There are five choices
a. Transient stability – Estimate rotor angle stability (higher is more
likely to be stable; default threshold is a 0%)
b. Damping – Estimates the damping of the oscillations (higher index
is better; default threshold is 3%)
c. Transient Voltage – Look at Figure 3-2 adobe page 18 (manual
page 12). This is the voltage dip criterion we looked at in L26.
(Anything over zero is bad; default is 0 sec duration below 0.8 pu).
Note that the default is to NOT use threshold as a percentage of
pre-fault voltage.
d. Transient Frequency – Like voltage dip this is the time a frequency
dip is below a threshold (higher is less stable; default is 1 second
duration below 59 Hz).
e. Relay Margin – The relay margin indices measure the distance
between the apparent impedance trajectories of a transmission line
Page 7 of 8
Lecture 27 – ECE504PS – Spring 2007
and the impedance circles of user-defined distance relay
characteristics (higher is better; default thresholds are zero).
3. Run base case analysis by clicking “Run” and selecting “Basecase
analysis”
a. Look at rotor angles and bus voltages with “spinner”
b. Look at Results summary, double one contingency
c. View results
4. Run Transaction analysis
Page 8 of 8