Introduction to Programming in Python for Power System Analysis

Lukmanul Hakim

Why Python ?

• Procedural Programming

• Object-oriented Programming

• Complex Number

• Numerical Libraries

• Support web-programming

• And many more…..

How to write the code

• Python Shell Interactive Mode

• Script Mode Create a file with .py extension and running the interpreter

C:\>python your_file.py

• To make life easier, please use an IDE (Integrated Development Environment) like Eclipse, Solaris Studio, Aptana, etc….

Let’s start !

• Go to www.python.org Downloads

• www.numpy.org

• www.scipy.org

• www.matplotlib.org

• www.liclipse.com

• www.aptana.com

Create your first program in this text editor. People usually start by writing this line: Print “hello world!”

Save your first program as ex_1.py To run the program from this user interface, you just simply click on RUN and RUN MODULE or just by pressing F5 button

Create your first program in this text editor. People usually start by writing this line: Print “hello world!”

Save your first program as ex_1.py To run the program from this user interface, you just simply click on RUN and RUN MODULE or just by pressing F5 button

Let’s get started.

• Case 2 Make Python prints number from 1 to 10

• Create a new file New Window

• Write your code

• Save the file as ex_2.py

• Run the module

• TIPS: use your internet browser to obtain the sample code of the same case

Case 3

• Multiply two matrices of dimensions (3 x 3) and (3 x 1) with their elements are randomly generated

• TIPS :

– Use NUMPY to help you

– Browse the Internet for any example

– Find out how to use random number in Python

Case 4

2 4 1

4 2 4

5 2 6

MATRIX A =

6

2

5

MATRIX B =

SOLVE X FROM A.X = B

Case 5 – Solve current for this circuit

Tips: I = R-1.V

A bit of Object-oriented Programming

• Power system components such as Bus, Generator, Load, Transformer, and Line can be modeled mathematically

• These physical objects can also be modeled as software objects

• Let’s apply this programming style for our electrical engineering problems

• TIPS: Each bus provides connection to any components

Let’s build OO Model for Case 5

• We have two batteries

• We have four nodes/buses

1. Connecting B1 and R1

2. Connecting R1, R2, and R3

3. Connecting B2 and R3

• We have three resistors

Case 6 – Object Model

• Bus – Has id/number of type integer – Has battery connected to it [list] – Has resistors connected to it [list]

• Battery – Has voltage [type: float] – Connected to a bus

• Resistor – Has value [type: float] – Connected to two buses

Results of Case 6

Problems…..

• Although object-oriented approach was adopted, the code was not flexible… Different problem/case/circuit results in different program

• Because we are LAZY to write different code for different problem with similar calculation method, let us create an input file and leave the program for different input data….

Remember: The Node Voltage Method

• Convert voltage sources in series with a resistor to an equivalent current source with the resistor in parallel

• Change resistor values to conductances.

• Select a reference node(E0)

• Assign unknown voltages (E1)(E2) ... (EN)to remaining nodes.

• Write a KCL equation for each node 1,2, ... N. The positive coefficient of the first voltage in the first equation is the sum of conductances connected to the node. The coefficient for the second voltage in the second equation is the sum of conductances connected to that node. Repeat for coefficient of third voltage, third equation, and other equations. These coefficients fall on a diagonal.

• All other coefficients for all equations are negative, representing conductances between nodes. The first equation, second coefficient is the conductance from node 1 to node 2, the third coefficient is the conductance from node 1 to node 3. Fill in negative coefficients for other equations.

• The right hand side of the equations is the current source connected to the respective nodes.

• Solve system of equations for unknown node voltages.

1 2 3 4

0

Node 1: IB1 + I12 = 0

IB1 + (( V1 – V2 ) / R1 ) = 0

Node 2: I21 + I23 + I20 = 0

( V2 – V1 ) / R1 ) + ( V2 – V3 ) / R3 ) + ( V2 – V0 ) / R2 ) = 0

Node 4: IB2 + I43 = 0

IB2 + (( V4 – V3 ) / R5 ) = 0

Node 3: I32 + I34 + I30 = 0

( V3 – V2 ) / R3 ) + ( V3 – V4 ) / R5 ) + ( V3 – V0 ) / R4 ) = 0

(1/R1)*V1 – (1/R1)*V2 + IB1 = 0

– (1/R1)*V1 + ((1/R1)+(1/R2)+(1/R3))* V2 – (1/R3)*V3 = 0

– (1/R2)*V2 + ((1/R3)+(1/R4)+(1/R5))* V3 – (1/R5)*V4 = 0

– (1/R5)*V3 + (1/R5)*V4 + IB2 = 0

V1 = VB1

V4 = VB2

Let’s establish a system of linear equations of this class of problem

Matrix form is preferable….. A.x = b

(1/R1) – (1/R1) 0 0 1 0

– (1/R1) (1/R1)+(1/R2)+(1/R3) – (1/R3) 0 0 0

0 – (1/R2) (1/R3)+(1/R4)+(1/R5) – (1/R5) 0 0

0 0 – (1/R5) (1/R5) 0 1

1 0 0 0 0 0

0 0 0 1 0 0

V1

V2

V3

V4

IB1

IB2

0

0

0

0

VB1

VB2

x = b =

A =

Input Data…

Results of Case 7

Let’s go back to Case 6 and Check!