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Sometimes the parameters for two elements in the same circuit
(network) are quoted in per-unit on a different base. The changing
base impedance is given as,
Changing base impedance (Znew]
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
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BEF 23803 - Lecture 14 - Per Unit Analysis of Three Phase System.ppt
- 1. Lecture 14 Per Unit Analysis of Three Phase System
- 2. 2 After completing this unit you will be able to: • Apply per unit models of power system components such as generators, transformers, transmission lines, etc. to analyse a balanced three-phase power system. Learning Outcomes
- 3. 3 Per-Unit System for Three Phase System For a given single-line (one-line) diagram of a power network, all component parameters are expressed in 3- quantity whether it is the rating (capacity) expressed as MVA or voltage as kV. Lets begin with 3- base quantity of base base base I V S 3 where Vbase = line voltage, Ibase = line or phase current Per phase base impedance, base base base I V Z 3 This is line-to-neutral impedance. (i) (ii)
- 4. 4 Combining (i) and (ii) yields, base base base base base base MVA kV V S V Z 2 3 3 where kVbase and MVAbase are 3- qualtities
- 5. 5 Sometimes the parameters for two elements in the same circuit (network) are quoted in per-unit on a different base. The changing base impedance is given as, Changing base impedance (Znew] 2 2 base OLD base NEW NEW OLD base OLD base NEW kV MVA Z pu Z MVA kV
- 6. 6 Determine the per-unit values of the following single-line diagram and draw the impedance diagram. From the impedance diagram obtained find the transmission line current. Example XT1 = 0.1 p.u 5 MVA Xg = 16% 100 MVA 275 kV/132 kV 50 MVA 132 kV/66 kV Transmission line j 3.48 XT2 = 0.04 p.u Load 40 MW, 0.8 p.f. lagging
- 7. 7 Chosen base: Always choose the largest rating, therefore Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV Solution Per-unit calculations Generator G1: 2 2 base OLD base NEW NEW OLD base OLD base NEW kV MVA Z pu Z MVA kV 32 . 0 50 100 275 275 16 . 0 ) ( 2 2 pu X g p.u.
- 8. 8 Transformer T1: 1 . 0 ) ( 1 pu XT p.u. Transmission line TL: base base base MVA kV Z 2 actual pu base Z Z Z where Therefore, 0195 . 0 132 100 4 . 3 2 2 .) . ( j j kV MVA Z Z base base actual u p TL p.u.
- 9. 9 Transformer T2: 2 2 base OLD base NEW NEW OLD base OLD base NEW kV MVA Z pu Z MVA kV 08 . 0 50 100 66 66 04 . 0 2 2 .) . ( 2 u p T X Now, we have all the impedance values in per-unit with a common base and we can now combine all the impedances and determine the overall impedance.
- 10. 10 Inductive load: A 39 . 437 8 . 0 10 66 3 10 40 cos 3 6 6 LL L V MVA I Magnitude of load current is Since the load is inductive, so we can write Given: 8 . 0 cos Therefore, 87 . 36 8 . 0 cos 1 A 87 . 36 39 . 437 L I
- 11. 11 87 . 36 12 . 87 87 . 36 39 . 437 0 3 10 66 3 ) ( L Phase actual L I V Z Therefore, 2 . 1 6 . 1 56 . 43 87 . 36 12 . 87 ) ( .) . ( j Z Z Z base actual L u p L and 56 . 43 100 662 2 base base base MVA kV Z Therefore,
- 12. 12 Load G j 0.32 p.u. j 0.1 p.u. j 0.0195 p.u. Transformer T1 Transformer T2 Transmission Line TL j 0.08 p.u. 1.6 p.u.. j 1.2 p.u. Generator XT1 = 0.1 p.u 5 MVA Xg = 16% 100 MVA 275 kV/132 kV 50 MVA 132 kV/66 kV Transmission line j 3.48 XT2 = 0.04 p.u Load 40 MW, 0.8 p.f. lagging
- 13. 13 Solution Transmission line current p.u. 06 . 47 426 . 0 j1.72 1.6 0 1 .) . ( .) . ( .) . ( u p total u p S u p TL Z V I A 39 . 437 10 132 3 10 100 3 3 6 ) ( base base base TL V MVA I Therefore, A 06 . 47 3 . 186 39 . 437 06 . 47 426 . 0 .) . ( ) ( base u p TL actual TL I I I
- 14. 14 END