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E&CE 261: Power Systems

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STEPS TO SOLVE POWER FLOW ANALYSIS: FOR DUMMIES

1. Represent the system by its one line diagramThe point of this is to just identify all the buses in the system and see how all theimpedances relate between them. Label all the buses and write all the data thathas been given. Normally, this diagram is already given.

Example

2. Convert all quantities to Per UnitOften, the values that we are given are not in per unit with respect to one commonbase value, so we need to find all the parameters that we are given with respect toone common base value. This base value we compute with respect to is normallyexplicitly 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 andSbase = 100 MVA

In Bus 2: The base is: -(400 + j250)/100 = -4 j2.5 pu P2 = - 4 Q2 = - 2.5In Bus 3: The base us: +(200)/100 = 2 pu P3 = 2

3. Draw the Impedance DiagramNow that all the values have been expressed in terms of one common per unitbase, we can represent the power system with inductors.

- A Generator is represented with a source and an inductor, XLo 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 valuein step 2

- Transmission Lines are represented by an impedance we need tocalculate 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 thecommon per unit base value we need to calculate the base valuein step 2

- Transformers are represented by an inductor we need to calculate thebase value in step 2


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Example Cont.

In the example above we are not given enough data to find this step (We dont know thevoltage 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 tocalculate the self admittance and mutual admittance. For mutual admittance wemultiply by negative 1.

Example Cont.

Y11 = (0.02 + j0.04)-1

+ (0.01 + j0.03)-1

= 20 j50Y12 = Y21 = - [0.02 + j0.04]

-1= -10 + j20

Y13 = Y31 = - [0.01 + j0.03]-1

= -10 + 30jY22 = (0.02 + j0.04)

-1+ (0.0125 + j0.025)

-1= 26 j52

Y23 = Y32 = - [0.0125 + j0.025]-1

= -16 + j32Y33 = (0.01 + j0.03)

-1+ (0.0125 + j0.025)

-1= 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|, ij):

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, P, Q

Generator Bus P, |V| Q,

Load Bus P, Q V,


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6. Start answering the missing variables, by assumptions (unless it is specifiedotherwise):

a. Slack, assume nothing

b. Generator, assume = 0

c. Load, assume V = 1 pu, = 0

Example Cont.

Bus Number Type Given UnknownGiven

Parametersto Use

Required toApproximate

1 Slack V1, 1 P1, Q1 V1, 1 -

2 Load P2, Q2 |V2|, 2 P2, Q2 |V2|, 2

3 Voltage P3, |V3| Q3,3 P3, |V3| 3

Assume (for now) that:|V2| = 1 pu

2 = 0

3 = 0

7. Find approximations for the Real and Reactive Power that we are given, using theassumed and given values for voltage/angles/admittance. Find the difference inthis 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 = -1.14P3 = 0.5616Q2 = -2.28

Since we know P2, Q2, and P3, we can find P2, Q2, and P3:


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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: 2, 3, |V2|,

So this means the Jacobian matrix is a 3x3 matrix, so we need to find 9 partialderivatives.We can do this as follows:


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So we can now write the matrix as follows:

9. Solve for the unknown differences, using Cramers Rule.

Example Cont.

We can solve for Solving for 2, 3, |V2| using Cramers Rule, we get:

So we must now alter our previous approximations for 2, 3, |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 forthe unknown differences as the 0. Normally we only do 2 iterations. We thensolve for all the other unknown parameters.

Example Cont.

Repeat Steps 7 9 for 2, 3, |V2|

We find:

So we still need to find Q3, Q1, and P1.We can do this as follows:


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We have now fully solved the power system!

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