E&CE 261: Power Systems
STEPS TO SOLVE POWER FLOW ANALYSIS: FOR DUMMIES
1. Represent the system by its one line diagram
The point of this is to just identify all the buses in the system and see how all the
impedances relate between them. Label all the buses and write all the data that
has been given. Normally, this diagram is already given.
Example
2. Convert all quantities to Per Unit
Often, the values that we are given are not in per unit with respect to one common
base value, so we need to find all the parameters that we are given with respect to
one common base value. This base value we compute with respect to is normally
explicitly specified, but if not we can assume one and move on.
Example Cont.
If we consider the above example, with V1 and V3 already given in per unit values and
Sbase = 100 MVA
In Bus 2: The base is: -(400 + j250)/100 = -4 – j2.5 pu
In Bus 3: The base us: +(200)/100 = 2 pu
à P2 = - 4 Q2 = - 2.5
à P3 = 2
3. Draw the Impedance Diagram
Now that all the values have been expressed in terms of one common per unit
base, we can represent the power system with inductors.
-
-
-
A Generator is represented with a source and an inductor, XL
o The value of XL will always be given, but not in terms of the
common per unit base value – we need to calculate the base value
in step 2
Transmission Lines are represented by an impedance – we need to
calculate the base value in step 2
A Motor is represented with a source and an inductor, XM
o The value of XM will always be given, but not in terms of the
common per unit base value – we need to calculate the base value
in step 2
Transformers are represented by an inductor – we need to calculate the
base value in step 2
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E&CE 261: Power Systems
Example Cont.
In the example above we are not given enough data to find this step (We don’t know the
voltage power, rated voltage, and inductance/impdedance)
4. Obtain the Ybus matrix.
We now need to find the relationships between all the bus lines. We need to
calculate the self admittance and mutual admittance. For mutual admittance we
multiply by negative 1.
Example Cont.
-1
-1
Y11 = (0.02 + j0.04) + (0.01 + j0.03) = 20 –j50
-1
Y12 = Y21 = - [0.02 + j0.04] = -10 + j20
-1
Y13 = Y31 = - [0.01 + j0.03] = -10 + 30j
-1
-1
Y22 = (0.02 + j0.04) + (0.0125 + j0.025) = 26 – j52
-1
Y23 = Y32 = - [0.0125 + j0.025] = -16 + j32
-1
-1
Y33 = (0.01 + j0.03) + (0.0125 + j0.025) = 26 - j62
Note:
Y11 = - [Y12 + Y13]
Y22 = - [Y12 + Y23]
Y33 = - [Y23 + Y13]
Putting this all together we get:
IMPORTANT STEP: It is very useful to covert these values to polar form (|Vij|, θij):
Note: Angles are in radians for this example, but for consistency use degrees.
5. Classify the buses as follows:
(Delta is the voltage angle)
Bus Type
Given Parameters
Unknown Parameters
Slack Bus
V, δ
P, Q
Generator Bus
P, |V|
Q, δ
Load Bus
P, Q
V, δ
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E&CE 261: Power Systems
6. Start answering the missing variables, by assumptions (unless it is specified
otherwise):
a. Slack, assume nothing
b. Generator, assume δ = 0
c. Load, assume V = 1 pu, δ = 0
Example Cont.
Bus Number
Type
Given
Unknown
Given
Parameters
to Use
Required to
Approximate
1
Slack
V1, δ1
P1, Q1
V1, δ1
-
2
Load
P2, Q2
|V2|, δ2
P2, Q2
|V2|, δ2
3
Voltage
P3, |V3|
Q3, δ3
P3, |V3|
δ3
Assume (for now) that:
|V2| = 1 pu
δ2 = 0
δ3 = 0
7. Find approximations for the Real and Reactive Power that we are given, using the
assumed and given values for voltage/angles/admittance. Find the difference in
this with the value that was actually given.
Example Cont.
We now need equations for P2, Q2, and P3:
We know all these parameters so we can solve for the first approximation of P2, P3, and
Q2
We find:
P2 = -1.14
P3 = 0.5616
Q2 = -2.28
Since we know P2, Q2, and P3, we can find ∆P2, ∆Q2, and ∆P3:
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E&CE 261: Power Systems
∆value = Given Value – Approximated Value
8. Write the Jacobian Matrix for the first iteration of the Newton Raphson Method.
This is in the form:
[∆
∆values] = [Jacobian Matrix] * [∆
∆ for Unknown Parameters]
Example Cont.
So in this case we know ∆P2, ∆Q2, and ∆P3 and need to find the Jacobian partial
derivatives for the unknown values: δ2, δ3, |V2|,
So this means the Jacobian matrix is a 3x3 matrix, so we need to find 9 partial
derivatives.
We can do this as follows:
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E&CE 261: Power Systems
So we can now write the matrix as follows:
9. Solve for the unknown differences, using Cramer’s Rule.
Example Cont.
We can solve for Solving for ∆δ2, ∆ δ3, ∆|V2| using Cramer’s Rule, we get:
So we must now alter our previous approximations for δ2, δ3, |V2|
Unknown Value new = Unknown Value old + Solved ∆Value
10. We now need to repeat step 7 – 9 iteratively until we obtain an accurate value for
the unknown differences as the à 0. Normally we only do 2 iterations. We then
solve for all the other unknown parameters.
Example Cont.
Repeat Steps 7 – 9 for ∆δ2, ∆ δ3, ∆|V2|
We find:
So we still need to find Q3, Q1, and P1.
We can do this as follows:
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E&CE 261: Power Systems
We have now fully solved the power system!
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