Power system analysis notes PPT ,pu methods, reactance diagram

Power system analysis

1. References: Saadat,H ., Power System Analysis , McGraw-Hill book company, NewYork , 1999 Stevenson,W.D ., and Grainger,J.J ., Power S ystem Analysis , McGraw-Hill book company , NewYork , 1995 Glover,J.D.Sarma,M.S , and Overbye,T.J ., Power System Analysis and Design , Cengage Learning,USA,2012 Weedy,B.M.,Cory,B.J.,Jenkins,N.,Ekanayake,J.B.,and Strbac,G ., Electric Power Systems , John Wiley & Sons, Ltd., UK,2012 Overbye,T.J , Lecture notes, University of Illinois,USA Lectures on Power System Analysis 01 Dr. Kassim Al- Anbarri B.Sc., M.Sc ., Ph.D (Electrical Power Engineering) Affiliations : Electrical Engineering Department,College of Engineering , Mustansiriyah University Bab Al- Muadham Campus, Al Mustansiriya 10052, Baghdad, Iraq Email: [email protected] [email protected] [email protected]

2. Modelling of Power System Components Transformer(single phase, three phase) Transmission Lines(Over head , Under ground Cables ) Load ( Con. Power, Current, Impedance) Generators

3. Zone 1 Zone 2 High voltage(tension) side HV Low voltage(tension) side LV 33kV 132 kV n 1 n 2 Power source Load X 2 R 2 R 1 V 2 V 1 X 1 Single phase transformer Referring to one side yield   Power source X 1 R 1 R m X m R ׳ 2 X ׳ 2 Load Neglecting the magnetization branch yield   Power source X 1 R 1 R ׳ 2 X ׳ 2 Load

4. Power source X=X 1 + X ’ 2   R=R 1 + R ’ 2   Load Power source X=X 1 + X ’ 2   Load Since R<<X, the simplified equivalent circuit of a single phase transformer Where X is the equivalent transformer reactance referred to one side

5. V S Sending end voltage R= Resistance of the line = Length of line * resistance per unit length R L V R Receiving end voltage L= Inductance of the line = Length of line * inductance per unit length C= Capacitance of the line = Length of line * capacitance per unit length C/2 C/2 For short lines <50 km , Capacitance and resistance are neglected Transmission Line π model  

6. L V S =ǀV s ǀ ∠δ s Sending end voltage V R =ǀV R ǀ ∠δ R Receiving end voltage L= Inductance of the line = Length of line * inductance per unit length  

7. + E 60 V     - I R 12 Ω There are four quantities E,I,Z ( R,X ), S ( P,Q ) In per unit system, we have to select the bases for only two quantities, the remaining two should be calculated by the available laws( Ohm,s law) * for the above circuit using actual units The power dissipated in the resistance  

8. * for solving the previous problem by per unit system Suppose that we select the EMF source and the resistance as a base quantities The base values for the remaining quantities should be calculated as: For the previous circuit To calculate the current I Converting the results obtained from per unit to actual units     + E 0.5 pu     - I R 0.4 pu

9. Problem 2.10 , pp.83 of Ref[2] A single-phase system shown below has two transformers A-B And B-C connected by a line B feeding a load at the receiving end C. The ratings and parameter values of the components are: Transformer A-B : 500 V/1.5 kV, 9.6 kVA , leakage reactance= 5% Transformer B-C : 1 .2 kV/120 V, 7.2 kVA , leakage reactance= 4% Line B : series impedance =(0.5+j 3.0)Ω Load C : 120 V,6 kVA at 0.8 power factor lagging   Determine the value of the load impedance in ohms and the actual ohmic impedances of the two transformers referred to both their primary and secondary sides. Choosing 1.2 kV as the voltage base for circuit B and 10 kVA as the system wide kVA base, express all system impedances in per unit What value of sending –end voltage corresponds to the given loading conditions?

10. 500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Fig. Problem 2.10 Line

11. * The load impedance in Ohm can be calculated as follow : Where S is the complex power in VA Substitute (3) in (1) yields: The Load impedance in Ohms can be calculated from the following equation: The complex power at the load is 6 kVA with 0.8 PF lagging S=6000 ∠ (cos -1 0.8) VA =6000 ∠ 36.8 VA    

12. * The Ohmic impedance of transformer T A-B :           The data of transformer A-B: 500 V/1.5 kV, 9.6 kVA , leakage reactance= 5% We have to calculate the base impedance Z B for each zone. The power rating and the voltage rating of the unit (transformer) have to be taken as the Base power( S B ) and Base voltage( V B ), respectively. For the low voltage side of the transformer T A-B (i.e. Zone A) S B = 9.6 kVA, V B =500 V The Ohmic impedance of T A-B viewed from low voltage side Z LV Z LV = Z T in per uint * Z B = j0.05*26.042= j1.30208 Ω For the high voltage side of the transformer T A-B (i.e. Zone B) S B = 9.6 kVA,V B =1.5 kV The Ohmic impedance of T A-B viewed from high voltage side Z HV Z HV = Z T in per uint * Z B = j0.05*234.375= j11.7187 Ω Verification   500 V 1.5 kV n 1 n 2 1 3 Zone B Zone A

13. * The Ohmic impedance of transformer T B-C :           The data of transformer B-C: 1 .2 kV/120 V, 7.2 kVA , leakage reactance= 4% We have to calculate the base impedance Z B for each zone. The power rating and the voltage rating of the unit (transformer) have to be taken as the Base power( S B ) and Base voltage( V B ), respectively. For the high voltage side of the transformer T B-C (i.e. Zone B) S B = 7.2 kVA,V B =1.2 kV The Ohmic impedance of T B-C viewed from high voltage side Z HV Z HV = Z T in per uint * Z B = 0.04*200= 8 Ω   For the low voltage side of the transformer T B-C (i.e. Zone C) S B = 7.2 kVA,V B =120 V The Ohmic impedance of T B-C viewed from low voltage side Z LV Z LV = Z T in per uint * Z B = 0.04*2= 0.08 Ω Verification   1.2 kV 120 V n 1 n 2 10 1 Zone C Zone B

14. Referring Z Load to Zone B by multiplying it with square of turns ratio of T B-C Referring Z T B-C to Zone B by multiplying it with square of turns ratio     Z ref Load     X ref TB-C     Z Line 0.5+j 3.0 Ω   500 V 1.5 kV n 1 n 2 T A-B Zone B Zone A

15. Referring Z ref Load to Zone A by multiplying it with square of turns ratio of T A-B Referring Z T B-C to Zone A by multiplying it with square of turns ratio of T A-B Referring Z Line to Zone A by multiplying it with square of turns ratio of T A-B Referring Z T A-B to Zone A by multiplying it with square of turns ratio of T A-B  

16. By taking the voltage at the load (at zone C)as a reference phasor Referring V R to Zone B then referring the result to Zone A yields: Referring I Load to Zone B then referring the result to Zone A yields:  

17. By taking the voltage at the load (at zone C)as a reference phasor Referring V R to Zone B then referring the result to Zone A yields: Referring I Load to Zone B then referring the result to Zone A yields:  

18. By Choosing 1.2 kV as the voltage base for circuit B ( V B B )and 10 kVA as S B , the voltage base for other zones can be calculated as follow: * For Zone C * For Zone A   500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Line

19. The per unit impedances of power system components should be corrected to the new bases by using the following formula * for Load     500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Line

20. The per unit impedances of power system components should be corrected to the new bases by using the following formula * for Load (Another approach)   500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Line

21. * for transformer T B-C or     500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Line

22. * for Line Since the impedance of the line is given in actual units, The impedance have to be converted to per unit values by calculating the Base Impedance as follow :       500 V 1.5 kV n 1 n 2 Load Zone A T A-B Zone B Zone C T B-C 1.2 kV 120 V n 1 n 2 Line

23. Per unit system procedure for scaling 3 phase power system