Calculating Maximum Active Power Transfer in Power Systems

Exercise Session 4 Power systems

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

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

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

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 + = ) 1 = 90 (sin + 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

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?

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

Question 2 = = 0 cos Q U U 2 2 1 sin(2 ) = 2 sin( )cos( ) ( ) 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 ???2 + ???2 = 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

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

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 2J 2if using equation from lecture slides

P Question 3: critical clearing angle P??? 2.5 = = 0 . 1 = sin 5 . 2 sin P P max = i o = 23 58 . 4115 . 0 rad 0 A2 Pi o = = 154 42 . 1 0 m A1 A1 = A2, Stability Criterion 0 ? ?? ( ) 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 + 4115 . 0 5 . 2 cos 154 42 . 2 = = = = arccos( ) arccos( 00274 . 0 ) 89 84 . . 1 56 rad cr 5 . 2

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 0 ? ?? J 2 W Multiply by 4 and Divide by 2 = = k H P P m m 4 J H = 2 P m ( ) ( . 1 ) 4 4 6 56 4115 . 0 H = = = 0 . 0 296 cr t s cr 2 50 1 Hz s

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.

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 the fault such that A = Pmaxsin , B =k1A, and C = k2A, with k1< k2. For stability, we must have area A1= area A2. ?1=??????? ????? ??????? ????? ??????? ????? ??????? ?????(1.0 + 4.0) = ?2=??????? ????? ?????? ????? U U = = 1 X 2 sin P P 1 2 Note: picture illustrative - not from the problem

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

Question 4: Determine the critical clearing angle A = Pmaxsin , 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

Question 4: Determine the critical clearing angle ( ) 1 ( ( ) ) = sin sin k = + cos sin cos cos k k 0 0 1 0 2 cr m m 0 2 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