00:00

Power System Analysis: Critical Clearing Angle Calculation

The analysis includes calculations for determining the critical clearing angle and time in a power system scenario involving fault occurrences and maximum power transfers. It covers the derivation of expressions for power transferred and reactive power as functions of voltage angle, as well as the evaluation of maximum power limits under different conditions. The critical clearing angle is determined using power-angle equations and stability criteria. The importance of reactive power resources in limiting power transfer capacity is highlighted.

fulquet
Télécharger la présentation

Power System Analysis: Critical Clearing Angle Calculation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Exercise Session 4 Power systems

  2. Question 1 • Calculate the maximum active power Pmaxthat can be transferred from busbar 1 to busbar 2 using the voltages shown in the picture.

  3. Question 1 The equivalent scheme of the line, using reactances for P.U. computation. Selecting: ??= 16 MVA ??= 110 kV U 10 0 . kV  1    . 0 996 u 1    10 5 . U          110 kV 2 N  U b 115 U 1 N U 21 0 . kV  2    . 1 000 u 2    21 U          110 kV 2 N U bU 110 1 N

  4. Question 1 Selecting: ??= 16 MVA ??= 110 kV  10   16  2 2 2 N 2 b U U 115 kV 110 kV      1 . 0 . 0 175 x Z 1 m k MVA MVA S S N b  16   16  2 2 2 N 2 b U U 110 kV 110 kV      1 . 0 . 0 100 x Z 2 m k MVA MVA S S N b X  40 j    . 0 053 x  16  j 2 b 2 U 110 kV S MVA b

  5. Power-angle equation: U U Question 1     1 X 2 sin P P 1 2 The highest power that can be transmitted: u u p Maximum value when:    sin 1 x 2       90 (sin ) 1  x x 1 2 m j m . 1  u u . 0 996  000  1 x 2    . 3 038 p max   . 0 175 . 0 100 . 0 053 x x 1 2 m j m      . 3 038 16 MVA 48 . 60 MW max P p S max b

  6. Question 2 • The power that can be transferred by a line is, usually, limited by the reactive power resources. A line has a series reactance X=100  and the voltage at the beginning of the line is U1=115 kV. The reactive power at the end of the line is Q2=0. Derive the expressions for power transferred and reactive power Q1as the function of voltage angle . How much smaller is the maximum transferred power compared to a case where the necessary reactive power could be fed to the end of the line?

  7. Question 2 Power-angle equation: U U     1 X 2 sin P P 1 2 2 1 U U U 1 X 2     cos Q 1 X 2 2 U U 1 U     2 cos Q 2 X X     0 cos Q U U 2 2 1

  8. Question 2     0 cos Q U U 2 2 1 sin(2δ) = 2 sin(δ)cos(δ)   2 2 1 2 1 X U U U U 115  kV 1 X 2             2    2 sin sin cos sin sin P P 1 2  2 2 100 X  P    2 66 1 . MW sin P 1 2 ???2δ + ???2δ = 1    1  2 2 1 2 1 2 1 U U U U U 115 kV  2 2 2 1 X 2               cos cos sin sin Q 1 100 X X X 2    132 3 . Mvar sin Q 1

  9. Question 2  2 2 2 U 2 U U        2    2 1 0 sin 1 X Q P P P sin 1 Q  1 2 2 max 1 2 X X    50 % P smaller 2 U max 2 U U U      1 0 , Q U  U   P       sin sin 1 X 2 1 P 2 2 1 max X     X ASSUMPTION

  10. Question 3: (for help, see Power System Analysis by Grainger, ch. 16 or other book) • A generator having H = 6.0 MJ/MVA is delivering power of 1.0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output to zero. The maximum power that could be delivered is 2.5 per unit. When the fault is cleared, the original network conditions again exist. Determine the critical angle and critical clearing time. Hint: For clearing time, H= Wk/P & Wk = 1 2Jω2if using equation from lecture slides

  11. P Question 3: critical clearing angle P??? 2.5     0 . 1  sin 5 . 2 sin P P max  i o    23 . 58 . 0 4115 rad 0 A2 Pi o       154 . 42 1 0 m A1 A1 = A2, Stability Criterion δ0 δ? δ?? δ     m          d      sin ( ) A P A max P P 1 0 2 i cr i m cr  cr     m         d      1 sin 1 ( ) max P 0 cr m cr  cr           cos cos P P 0 max P max P m cr m               cos cos ( ) 0 max  max  0 m cr     2 cos  m P cos max P P 0 max  m cr    cos 2 max P 0   cos cr max      5 . 2 cos 154 . 42 2 . 0 4115       arccos( ) arccos( . 0 00274 ) 89 84 . . 1 56 rad cr 5 . 2

  12. Question 3: critical clearing time P o    23 58 . 4115 . 0 rad 0 P??? 2.5     89 84 . . 1 56 rad cr A2 Pi 1  J     ' 2 ( ´ ) t From lecture slides 0 A1 P m 1 δ0 δ? δ?? δ  J 2 W Multiply by 4 and Divide by ω 2   k H P P m m  4  J H  2 P m    . 1         4 4 6 56  . 0 4115  H    0 . 0 296 cr t s cr  2 50 1 Hz s

  13. Question 4 • A 60-Hz generator is supplying 60% of Pmaxto an infinite bus through a reactive network. A fault occurs which increases the reactance of the network between the generator internal voltage and the infinite bus by 400%. When the fault is cleared, the maximum power that can be delivered is 80% of the original maximum value. Determine the critical clearing angle for the condition described.

  14. Question 4: (for help, see Power System Analysis by Grainger, ch. 16 or other book) Power-angle curve A before a fault, B during the fault, and C after that A = Pmaxsin δ, B =k1A, and C = k2A, with k1< k2. For stability, we must have area A1= area A2. the fault such ?1=??????? ????? ??????? ????? ??????? ????? ??????? ?????(1.0 + 4.0) = ?2=??????? ????? ?????? ????? U U     1 X 2 sin P P 1 2 Note: picture illustrative - not from the problem

  15. Question 4: Determine the critical clearing angle           sin sin sin P P k P k P max 0 2 max 2 max i m m         sin sin k 0 2 m In our case: Before the fault:     0 . 1  sin . 1 667 sin P max P 0 0 i o     36 86 . . 0 643 rad 0 During the fault: After the fault: . 1 667  ' ' 8 . 0 P P     ' 5 ' . 0 333 X X P max max  max f f 5    sin . 1 33 sin , 8 . 0 k max P k     sin . 0 333 sin , 2 . 0 k max P k 2 2 1 1

  16. Question 4: Determine the critical clearing angle A = Pmaxsin δ, B =k1A, and C = k2A. For stability, we must have area A1= area A2.         cr cr        (           ) sin sin A P B d P k P d 1 0 max 0 0 1 max i cr cr   0 0          cr |                        sin cos sin cos cos P k P k k max 0 0 1 max 0 0 1 1 0 cr cr cr  0      m m   (              (   ) sin sin ) A C d P k P d P 2 2 max max 0 i m cr m cr   cr cr      m |            (           (   cos sin ) cos cos sin ) P k P k k max 2 0 max 2 2 0 m cr cr m m cr  cr  cos A A 1 2               ( ) sin cos cos k k k k 2 1 0 0 1 0 2 cr m m  1                cos sin cos cos k k 0 0 1 0 2 cr m m ( ) k k 2 1

  17. Question 4: Determine the critical clearing angle   1           sin sin k           cos sin cos cos k k 0 0 1 0 2 0 2 cr m m m ( ) k k 2 1 Using the previously derived equations: P i          sin sin( ), arcsin( ) k 0 2 m m k max P 2   0 . 1 o o         180 arcsin 131 2 . . 2 290 rad m . 1  33 1  0 . 1                cos sin cos cos k k 0 0 1 0 2 cr m m (  ) k k 2 1     . 2  o o     2 . 0  8 . 0      sin 36 84 . 290 . 0 643 cos 36 86 . cos 131 2 . 8 . 0 2 . 0  . 0  500  o   arccos( ) 5 . 0 60 cr

More Related