Star-Delta Transformation

1. Star-Delta Transformation

2. Examples • Star-to-Delta Page 19 Advance Electrical Eng by Morton

3. Mesh Transformation Examples From The application of Matrix Theory to EE By Lewis Pruce Chapter four page 95

4. Mesh Analysis Mesh 1 25I1 + 60I1 – 60I2 - 25I3 – 10 = 0 85 I1 – 60 I2 - 25 I3 = 10 …..(1)

5. Mesh Analysis Mesh 2 60 I2 + 40 I2 - 60 I1 - 40 I3 + 20 = 0 - 60 I1 + 100 I2 - 40 I3 = - 20 …….(2)

6. Mesh Analysis Mesh 3 40 I3 + 25 I3 + 10 I3 – 25 I1 – 40 I2 = 0 -25 I1 - 40 I2 + 75 I3 = 0 ……. (3)

7. Mesh Analysis So we have our three mesh equations from the circulating currents in the three meshes. Note we write them down in a certain order : A I1 + B I2 + C I3 = D We can represent these simultaneous equations as a matrix equation as follows: [Z]*[I] = [E]

8. Mesh Analysis [Z]*[I] = [E] Can be written with values :

9. Mesh Analysis [Z] can written like : Where Z11 represents impedance of loop 1 and does not share with any other loop. Z12 is an impedance in loop 1 that is shared with loop 2. Z13 is an impedance of the loop 1 and shared with loop 3.

10. Mesh Analysis • Similarly each row has impedances present in the respective loop but shared with mesh of suffix no. of the column.

11. Mesh Analysis • The numbers on the diagonal of the matrix are positive. These are the mesh self-impedance and are just the sum of the impedances in each mesh. • The numbers off the diagonal represent the total impedances from one mesh with respect to another i.e. Z12, Z13 etc. Note there are always two i.e. the impedance from mesh one to mesh two is the same as from mesh 2 to mesh 1. Hence for example Z31 is the same as Z13. • The numbers from one mesh with respect to another are always negative. • Note that the emf are positive when aiding a circulating current i.e. on the LHS of the mesh and negative when opposing a circulating current i.e. on the RHS of a mesh.

12. Solution using Cramers Rule Find the determinant of Z matrix: That could be done by expanding either one row or one column: For 1st element i.e 85 minors are : Is evaluated (100 x 75) – (-40 x -40) = 5900 and is called co-factor of 85

13. Solution using Cramers Rule • The procedure is repeated for each element of the row (column) chosen: • Another thing to note is that the sign of the number in the determinant has a pattern as follows:

14. Solution using Cramers Rule R = (+) (85) [(100)(75) – (-40)(-40)] (-) (-60) [(-60)(75) – (-25)(-40)] (+) (-25) [(-60)(-40) – (-25)(100)] Note the signs (+ - +) !! R =49000

15. Solution using Cramers Rule • The next step is to replace the coefficients of I1 with the numbers on the right hand side of the equation, that is the column vector of applied e.m.f's, as follows and work out its determinant: 1 = -51000 

16. Solution using Cramers Rule Then for I2 and I3 and workout their determinants. i.e: 2 = -60000 & 3= -49000 

17. Solution using Cramers Rule Then to workout different branch currents : I1 = 1/ R = -1.04 A I2 = 2/ R = -1.22 A I3 = 3/ R = -1 A

18. Solution by Matrix Inversion [Z]*[I] = [E] [I] = [Z]-1 [E] Which involves finding the inverse matrix [Z] -1 There are several methods of doing this. One method is as follows: Where CT is the adjoint of the matrix Z and ¦Z¦ is the determinant of the matrix Z

19. Solution by Matrix Inversion ¦Z¦ =49000 Finally [I] = [Z]-1 [E]

20. Solution by Matrix Inversion We can now label the currents

21. Assignment exercise • Determine Loop & branch currents using both methods;i.e. Cramers & Matrix

22. Hyperlinks • en.wikipedia.org/wiki/Mesh_analysis