ELE-B7 Power Systems Engineering Load Flow (Gauss-Seidel Method ) Dr. Ramadan El-Shatshat Classification of buses: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Different buses at the network can be classified as: 1. 2. 3. PQ bus The Load Buses (PQ bus) The Generator Bus (PV bus) The Slack or Swing Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 PQ bus Slack Bus PV bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 PV bus PV bus PV bus 1. PV bus PQ bus The Load Buses (PQ bus) A non-generator bus. The active and reactive powers are specified at this bus. The voltage magnitude and phase angle are unknown. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pig and Qig Bus i Pid and Qid Pi and Qi are known & | Vi | and δ i are unknown Generators Power: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Delivered Power: Pig = 0 and Qig = 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pid and Qid are known In practice, the load real power is known from measurement, load forecasting or historical record and the reactive power is assumed based on 0.85 pf. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi ,sch = Pig − Pid = − Pid Dr. A.M. Gaouda UAE University ELEC572, 04/05 and Gaouda Qi ,sch = Qig − Qid = −Dr.UAE QA.M.id University ELEC572, 04/05 2. The Generator Bus (PV bus) The bus is also known as “Voltage controlled bus” because the voltage magnitude can be kept constant. At this bus the net active power and the voltage magnitude are specified. The reactive power and the voltage phase angle are unknown. Pi and | Vi | are known & Qi and δ i are unknown Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 G1 Bus i Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi,sch | Vi | NOTE: There are certain buses without generators may have voltage controlled capability. At these buses the real power generation is zero. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3. Dr. A.M. Gaouda UAE University ELEC572, 04/05 The Slack or Swing Bus Because the system losses are not known precisely before completing the power flow solution, it is not possible to specify the real power injected at every bus. Hence, The real power of one of the generator buses is allowed to swing. The swing bus supplies the slack between the scheduled real power generation and the sum of all loads and system losses. The voltage angle of the slack bus serves as a reference, δ i = 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 | Vi | and δ i are known & Pi and Qi are unknown Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTE: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Real Power losses = Total generation − Total load N PL = Dr. A.M. Gaouda UAE University ELEC572, 04/05 N N ∑P −∑P = ∑P gi i =1 di i =1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 i i =1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 In the load flow problem, we select the slack bus at which the power Pg is not scheduled. After solving the load flow problem, the difference (Slack) between the total specified power (P) going into the system at all other buses and the total output (P) plus the losses (I2R) are assigned to the slack bus. For this reason a generator bus must be selected as a slack bus. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Voltage of the swing bus is selected as a reference. Generally, the bus of the largest generator is selected as swing bus and numbered as bus 1. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Solution of Non-Linear Equations Dr. A.M. Gaouda UAE University ELEC572, 04/05 The two load flow equations are: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 n PDr.i A.M. =|Gaouda Vi | UAE University ELEC572, 04/05 Qi =| Vi | ∑| Yip || V p | cos ( δ i − δ p − γ ip ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 p =1 n ∑| Yip || V p | sin ( δ i − δ p − γ ip ) p =1 These equations provide the calculated value of net real power and net reactive power entering bus ‘i’. The equations are non-linear and only a numerical solution is possible. There are different methods could be implemented to solve these equations. Among those is the Gauss-Seidel method. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Gauss-Seidel Method Consider a system of non-linear equations having “n” unknowns x1 , x 2 , .........., x n Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 f 1 ( x1 , x 2 , .........., x n ) f 2 ( x1 , x 2 , .........., x n ) .. .. Dr. A.M. Gaouda UAE University ELEC572, 04/05 .. ..Dr. A.M...Gaouda.. UAE University 04/05, x f n ( x1 , x 2 ,ELEC572, .......... n ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Rearranging, then Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x i = f i ( x1 , x 2 , .........., x n ) 1≤ i ≤ n Dr. A.M. Gaouda UAE University ELEC572, 04/05 Eq. 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Assuming initial values, x1o , x 2o , .........., x no Substituting the initial values in Eq. 1, then Dr. A.M. Gaouda first iteration UAE University ELEC572, 04/05 first variable Dr. A.M. Gaouda UAE University ELEC572, 04/05 x = f ( x , x , .........., x ) 1 1 1 1 o 1 o 2 o n x 21 = f 21 ( x11 , x 2o , .........., x no ) x = f ( x , x , x .........., x no ) 1 3 1 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 1 1 2 o 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Or in general All values are initial values o o xDr.1o A.M. , xGaouda 2 , .........., x n UAE University ELEC572, 04/05 x 1 = x 11 from previous step and all other values are initial values x 2o , .........., x no x 1 = x 11 & x 2 = x 21 and o x 3 , .........., x no x i1 = f i 1 ( x11 , x 21 , ....., x io ,....., x no ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Where Dr. A.M. Gaouda UAE University ELEC572, 04/05 x x Dr. A.M. Gaouda 1 UAE University is the first approximation using the initial assumed values. Dr. A.M.of Gaouda ELEC572, 04/05 i UAE University i ELEC572, 04/05 The kth approximation of K th iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 xi is: x = f i ( x , x , .....x i th variable k i 1 k 1 k 2 k i −1 The changes in the magnitude of each variable the previous iteration is: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ,x Dr. A.M. Gaouda UAE University ELEC572, 04/05 k −1 i x ,x k −1 i+1 ....., x k −1 n ) k k −1 Dr. A.M. Gaouda from its value at UAE University i i ELEC572, 04/05 x Δ x i = x ik − x ik − 1 If Δ x i < ε then the solution has converged. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Where, Dr. A.M. Gaouda UAE University ELEC572, 04/05 ε Dr. A.M. Gaouda UAE University ELEC572, 04/05 is a small value ( for exmple : ε = 0.001 ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 EXAMPLE: For the following equation, find an accurate value for x up to 5 decimal places. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 x − log( x ) = 7 SOLUTION: Using Gauss-Seidel Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x = 0.5 ( 7 + log x ) x =1 x 1 = 0.5( 7 + log 1 ) = 3.5 x = 3.5 x 2 = 0.5 ( 7 + log 3.5 ) = 3.772034 x 2 = 3.772034 x 3 = 0.5( 7 + log 3.772034 ) = 3.788287 o Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 x 3 = 3.788287 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 st iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 nd iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 x 4 = 0.5( 7 + log 3.788287 ) = 3.789221 x 5 = 3.789274 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x = 3.789278 6 ε = 0.000004 Dr. A.M. Gaouda UAE University ELEC572, 04/05 EXAMPLE: For the following equations, find an x and y after 4 iterations. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x = 0.7 sin x + 0.2 cos y & y = 0.7 cos x − 0.2 sin y SOLUTION: Using Gauss-Seidel, assuming initial values Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x o = y o = 0.5 ( rad ) x 1 = 0.7 sin x o + 0.2 cos y o Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x 1 = 0.7 sin 0.5 + 0.2 cos 0.5 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x 1 = 0.51111 y 1 = 0.7 cos 0.51111 − 0.2 sin 0.5 Dr. A.M. Gaouda UAE University ELEC572, 04/05 y = 0.51465 1 x 2 = 0.516497 x 3 = 0.520211 Dr. A.M. Gaouda UAE University ELEC572, 04/05 x = 0.522520 4 Dr. A.M. Gaouda UAE University ELEC572, 04/05 y 2 = 0.510241 Dr. A.M. Gaouda UAE University ELEC572, 04/05 y 3 = 0.509722 y 4 = 0.509007 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Gauss-Seidel Method for Load Flow Analysis Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Advantages 1. 2. 3. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Simplicity Small computer memory requirement Less computational time per iteration Disadvantages Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1. 2. 3. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Slow rate of convergence, and therefore large number of iterations. Increase in the number of iterations as the number of system buses increases. The speed of convergence is affected by the selected slack bus. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 I - G-S Method when PV buses are absent Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Assuming a power system in which the voltage controlled buses are absent. If the system has n buses, then; one bus will be considered as a slack bus and the other n-1 buses are load buses (PQ-buses). Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For the Slack or Swing Bus: | Vi | and δ i = 0 are known & Pi and Qi are unknown Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The swing bus voltage is taken as a reference. It is voltage magnitude is known and its phase shift angle is set equal to zero. For (n-1) Load Buses (PQ bus): Pi and Qi are known & | Vi | and δ i are unknown Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Using Gauss-Seidel Method, we assume the initial values for the magnitude and phase shift angle of (n-1) buses. These values are updated at each iteration. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For an ‘n’ bus system Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 I bus = Ybus Vbus ….. Eq. 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For the ith bus of an ‘n’ bus system, the current entering this bus is: Dr. A.M. Gaouda UAE University ELEC572, 04/05 I i = Yi1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 n I i = Yii Vi + ∑ Yip V p ….. Eq. 3 p=1 p≠ i ⎛ ⎞ n ⎟ 1 ⎜ Vi = ⎜ I i − ∑ Yip V p ⎟ Yii ⎜ p=1 ⎟ ≠ p i ⎝ ⎠ Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 4 In power systems, power is known rather than currents. The complex power injected into the ith bus is: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Si = Pi + jQi = Vi I *i ….. Eq. 5 S i* = Vi* I i ….. Eq. 6 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 OR Dr. A.M. Gaouda UAE University ELEC572, 04/05 Ii = Substituting in Eq. 4 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi − jQi ….. Eq. 7 Vi* Dr. A.M. Gaouda UAE University ELEC572, 04/05 ⎞ ⎛ n ⎟ ⎜ 1 Pi − jQi Vi = − ∑ Yip V p ⎟ ⎜ * Yii ⎜ Vi p=1 ⎟ p≠ i ⎠ ⎝ Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 8 Rearranging in GS method “Vi” is moved to the left hand side of the equation. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Since bus 1 is the slack bus “reference”, then Vi represents n-1 set of equations for i= 2, 3, …., n. These equations will be solved using G-S method for the unknowns V2, V3, …..Vn. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTES: 1. Eq. 8 can be written as: Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTE The values for P and Q are the scheduled values for PQ Bus. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 n Y 1 Pi − jQi ip Vi = * − ∑ Vp Vi Yii p = 1 Yii ….. Eq. 9 n Ki Vi = * − ∑ Lip V p Vi p = 1 ….. Eq. 10 p≠ i P − jQi Ki = i Yii Dr. A.M. Gaouda UAE University ELEC572, 04/05 p≠ i Lip = and Dr. A.M. Gaouda UAE University ELEC572, 04/05 Yip Yii Dr. A.M. Gaouda UAE University ELEC572, 04/05 The values for Ki and Lip are computed once in the beginning and used in every iteration. 2. The voltages at all the buses in a power system are close to 1.0 pu. Therefore, we can start the G-S iteration process assuming initial values for the voltages equal to 1.0 and making zero angle. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 V2o = V3o = ...... Vno = 1∠0 3. At each step in the iteration process use the most updated values for the voltages to compute the new values for the bus voltages. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 n Ki Vi = * − ∑ Lip V p Vi p=1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 11 p≠ i Ki Vi = * − Vi Dr. A.M. Gaouda UAE University ELEC572, 04/05 Therefore, for the Vi k + 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 (kth+1) i −1 ∑L p=1 ip Vp − n Dr. A.M. Gaouda ∑L UAE University ELEC572, 04/05 p= i + 1 ip Vp iteration, Ki = − k * ( Vi ) i −1 ( k +1 ) L V ∑ ip p − p=1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 12 The most updated voltage values are from the previous iteration n k L V ∑ ip p p= i + 1 ….. Eq. 13 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The most updated voltage values are from the same iteration for i = 1 ,2 ,..........n The iteration process is continuous till the convergence occurs, i.e.; Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 | ΔVi k + 1 | = | Vi k + 1 | − | Vi k | 〈 ε ….. Eq. 14 for i = 1 ,2 ,..........n Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 4. The current and complex power at ith Dr. A.M. Gaouda UAE University ELEC572, 04/05 bus are: Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 2 3 I i = Yi1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn And Or Dr. A.M. Gaouda UAE University ELEC572, 04/05 Si = Pi + jQi = Vi I *i Linear Network Ii Dr. A.M. Gaouda UAE University ELEC572, 04/05 i Vi n Reference S i* = Pi − Qi = Vi* I i Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi = Re{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )} Qi = − Im{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )} Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The two equations are known as the rectangular form of the load flow equations. They provide the calculated value of net real power and net reactive power entering bus ‘i’. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 EXAMPLE: For the three bus system. Write the expression for the bus voltages using GS method. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 SOLUTION: The system contains 3 buses, (n=3). i- Select bus 1 as a slack bus “reference”. Dr. A.M. Gaouda UAE University ELEC572, 04/05 G1 2 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 P2 , Q 2 | V1 | = 1 and δ 1 = 0 ii- Buses 2 and 3 are load buses. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 | V1 | = 1 δ1 = 0 3 P3 , Q3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 P2 , P3 , Q2 and Q3 are known V2 ,V3 ,δ 2 and δ 3 are unknown 3 K2 V2 = * − ∑ L2 p V p V2 p = 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 p≠ 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 Y 1 P2 − jQ2 2p V2 = * −∑ Vp V2 Y22 p = 1 Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 p≠ 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTE ----------The load flow problem is solve when the mismatch is equal to zero. Calculated values = Dr. A.M. Gaouda UAE University Scheduled values ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y23 ⎤ 1 P2 − jQ2 ⎡ Y21 V3 ⎥ V2 = * − ⎢ V1 + Y22 ⎦ V2 Y22 ⎣ Y22 ….. Eq. 15 Y32 ⎤ 1 P3 − jQ3 ⎡ Y31 − ⎢ V1 + V2 ⎥ V3 = * Y33 ⎦ V3 Y33 ⎣ Y33 ….. Eq. 16 Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTE The values for P and Q are the scheduled values Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Using GS method, select the initial values for the unknowns as: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 V = V = 1∠0 o 2 o 3 Start the first iteration The most updated voltage value is the initial value Y23 o ⎤ 1 P2 − jQ2 ⎡ Y21 V = V3 ⎥ − ⎢ V1 + o * ( V2 ) Y22 Y22 ⎣ Y22 ⎦ 1 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 17 The most updated voltage value is from this iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y32 1 ⎤ 1 P3 − jQ3 ⎡ Y31 − ⎢ V1 + V = V2 ⎥ o * ( V3 ) Y33 Y33 ⎦ ⎣ Y33 1 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 18 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the second iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y23 1 ⎤ 1 P2 − jQ2 ⎡ Y21 V3 ⎥ V = − ⎢ V1 + 1 * Y22 ⎦ ( V2 ) Y22 ⎣ Y22 2 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y32 2 ⎤ 1 P3 − jQ3 ⎡ Y31 − ⎢ V1 + V2 ⎥ V = 1 * Y33 ( V3 ) Y33 ⎣ Y33 ⎦ 2 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 | ΔV22 | = | V22 | − | V21 | 〈 ε Dr. A.M. Gaouda UAE University ELEC572, 04/05 | ΔV32 | = | V32 | − | V31 | 〈 ε If Eqs. 21, 22 are not satisfied then start a new iteration. Dr. A.M. Gaouda UAE University ELEC572, 04/05 The most updated voltage value is from this iteration ….. Eq. 20 for i = 1 ,2 ,..........n Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 19 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Compare the results for convergence | ΔVi k + 1 | = | Vi k + 1 | − | Vi k | 〈 ε The most updated voltage value is from previous iteration ….. Eq. 21 ….. Eq. 22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda EXAMPLE 1: UAE University Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ELEC572, 04/05 For the system shown in the figure, the line impedances are as indicated in per unit on 100MVA base. A. Using Gauss-Seidel method find the bus voltages after 7 iterations. B. Using the bus voltages find the Slack bus real and reactive power. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 2 0.02 + j 0.04 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 G 0.01 + j 0.03 pu 110.2 MVAR V1 = 1.05 ∠0 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 138.6 MW Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 0.0125 + j 0.025 pu o Dr. A.M. Gaouda UAE University ELEC572, 04/05 256.6 MW 45.2 MVAR Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 Formulation of the Bus Admittance Matrix Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 0.02 + j 0.04 pu y12 = 1 = 10 − j 20 = y 21 0.02 + j 0.04 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 y13 = = 10 − j 30 = y 31 0.01 + j 0.03 y 23 = Ybus 0.01 + j 0.03 pu V1 = 1.05 ∠0 o pu 138.6 ⎡Y11 Y12 Y13 ⎤ ⎡ y12 + y13 = ⎢⎢Y21 Y22 Y23 ⎥⎥ = ⎢ − y 21 ⎢ ⎢Y31 Y32 Y33 ⎥ ⎢⎣ − y 31 ⎢ ⎥ Ybus = 0.0125 + j 0.025 pu 110.2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 MW Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 MW MVAR 3 1 = 16 − j 32 = y 32 0.0125 + j 0.025 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 G 256.6 45.2 MVAR Dr. A.M. Gaouda UAE University ELEC572, 04/05 − y12 y 21 + y 23 − y 32 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤ ⎢ − 10 + j 20 26 − j 52 − 16 + j 32 ⎥ Dr. A.M. Gaouda ⎢ ⎥ UAE University ELEC572, 04/05 ⎢⎣ − 10 + j 30 − 16 + j 32 26 − j 62 ⎥⎦ − y13 ⎤ − y 23 ⎥⎥ y 31 + y 32 ⎥⎦ Dr. A.M. Gaouda UAE University ELEC572, 04/05 Classification of buses: Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Bus 1: Slack Bus 0.02 + j 0.04 pu G V1 = 1.05∠0 pu o V1 = 1.05∠0 o pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 0.01 + j 0.03 pu 138.6 MW V2 ,V3 ,δ 2 and δ 3 are unknown P2 ,d = 256.6 MW P3 ,d = 138.6 MW 110.2 MVAR 3 P2 , P3 , Q2 and Q3 are known Dr. A.M. Gaouda UAE University ELEC572, 04/05 256.6 MW 0.0125 + j 0.025 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Buses 2 and 3: Load Buses (PQ bus) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q2 ,d = 110.2 MVAR Q3 ,d = 45.2 MVAR 45.2 MVAR Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pig and Qig G Pi ,sch = Pgi − Pdi & Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qi ,sch = Q gi − Qdi Dr. A.M. Gaouda UAE University ELEC572, 04/05 S i ,sch = Pi ,sch + jQi ,sch Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d ) Bus i Pid and Qid Dr. A.M. Gaouda UAE University ELEC572, 04/05 A.M. Gaouda ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2Dr.UAE ,d )University = pu ELEC572, 04/05 Base MVA Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 2 ,sch S 2 ,sch ( 0 − 256.6 ) + j ( 0 − 110.2 ) Dr. A.M. Gaouda = pu UAE University 100 MVA ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Reminder The bus admittance matrix is ⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤ ⎥ ⎢ 26 − j 52 − 16 + j 32 ⎥ Dr. A.M. Gaouda⎢ − 10 + j 20 UAE University ⎢ − 10 + j 30 − 16 + j 32 26 - j 62 ⎦⎥ ELEC572, 04/05⎣ S 2 ,sch = − 2.566 − j 1.102 pu S 3 ,sch = − 1.386 − j 0.452 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Using GS method, select the initial values for the unknowns as: Start the first iteration V2o = V3o = 1∠0 Y23 o ⎤ 1 P2 ,sch − jQ2 ,sch ⎡ Y21 1 V2 = V3 ⎥ − ⎢ V1 + o * ( V2 ) Y22 Y22 Y22 ⎣ ⎦ − jQ Dr. A.M. Gaouda UAE University ELEC572, 04/05 S *i ,sch = Pi ,sch 1 Dr. A.M. GaoudaV 2 UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 i ,sch ⎤ − 2.566 + j 1.102 ⎡ − 10 + j 20 − 16 + j 32 −⎢ = 1.05 + 1.0 ⎥ * 26 − j 52 26 − j 52 ( 1.0 ) ⎦ ⎣ 26 − j 52 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 OR, to simplify the calculations, we have: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Reminder The bus admittance matrix is Yip 1 Pi − jQi − ∑ Vp Vi = * Vi Yii p = 1 Yii n p≠ i Dr. A.M. Gaouda UAE University ELEC572, 04/05 n K2 V2 = * − ∑ L2 p V p V2 p = 1 [ Dr. A.M. Gaouda UAE University ELEC572, 04/05 p≠ 2 K2 o V = − L V + L V 21 1 23 3 ( V2o )* 1 2 ⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤ ⎥ ⎢− + ⎢ 10 j 20 26 − j 52 − 16 + j 32 ⎥ ⎢⎣ − 10 + j 30 − 16 + j 32 26 - j 62 ⎥⎦ ] Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The values for Ki and Lip are computed once in the beginning and used in every iteration. Y21 L21 = Y22 P − jQ2 K2 = 2 Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L21 = -0.3846 K2 = -0.0367 - j0.031 V1 = 1.05 ∠0 o pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 and Dr. A.M. Gaouda UAE University ELEC572, 04/05 V = 0.9825 − j 0.310 1 2 and Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y23 L23 = Y22 L23 = -0.6154 V3o = 1∠0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 [ K3 1 L V L V V31 = − + 31 1 32 2 ( V3o )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 P3 − jQ3 K3 = Y33 and Dr. A.M. Gaouda UAE University ELEC572, 04/05 V1 = 1.05 ∠0 o pu and Y32 Y33 V21 = 0.9825 − j 0.310 Dr. A.M. Gaouda UAE University ELEC572, 04/05 K 2 , K 3 , L21 , L23 , L31 , L32 constants will be the same. K2 1 V = − L V + L V 21 1 23 3 ( V21 )* [ ] [ ] Dr. A.M. Gaouda UAE University ELEC572, 04/05 L32 = L32 = -0.5310 - 0.0354i UAE University − j 0.0353 V31 = ELEC572, 1.0011 04/05 Start the second iteration V32 = and Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 2 Y31 Y33 L31 = - 0.4690 + 0.0354i K3 = -0.0142 - j0.0164 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L31 = ] Dr. A.M. Gaouda UAE University ELEC572, 04/05 K3 2 − L V + L V 31 1 32 2 ( V31 )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 VDr.22A.M.=Gaouda 0.9816 − j 0.0520 UAE University ELEC572, 04/05 V32 = 1.0008 − j0.0459 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the third iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 V23 = Dr. A.M. Gaouda UAE University K 2 , K 3 ,Dr.UAE LA.M. ,Gaouda L23 , L31 , L32 constants will be the same.ELEC572, 21 04/05 University [ ELEC572, 04/05 K2 2 L V L V − + 21 1 23 3 ( V22 )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] [ K3 3 V = − L V + L V 31 1 32 2 ( V32 )* 3 3 Start the fourth iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 = 1.0004 - j0.0488 K 2 , K 3 ,Dr.LA.M. L23 , L31 , L32 constants will be the same. 21 ,Gaouda Dr. A.M. Gaouda [ UAE University ELEC572, 04/05 ] V24 = K2 3 − L V + L V 21 1 23 3 ( V23 )* V34 = K3 4 − L V + L V 31 1 32 2 ( V33 )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] = 0.9808 - j0.0578 [ ] UAE University ELEC572, 04/05 = 0.9803 - j0.0594 Dr. A.M. Gaouda UAE University ELEC572, 04/05 = 1.0002 - j0.0497 After 7 iterations, V27 = 0.9800 − j0.0600 = 0.98183∠ − 3.5035 o pu Dr. A.M. Gaouda Dr. A.M. Gaouda UAE University ELEC572, 04/05 UAE University ELEC572, 04/05 V = 1.0000 − j 0.0500 = 0.00125∠ − 2.8624 pu 7 3 o Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda B. Using the bus voltages find the Slack bus real and reactive power. UAE University Dr. A.M. Gaouda ELEC572, 04/05 UAE University ELEC572, 04/05 1 2 0.02 + j 0.04 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 G 0.01 + j 0.03 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 256.6 MW 0.0125 + j 0.025 pu 110.2 MVAR V1 = 1.05∠0 o pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 138.6 MW 45.2 MVAR Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 V1 = 1.05 + j 0.0 o pu V2 = 0.9800 − j0.0600 = 0.98183∠ − 3.5035 o pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 o Dr. A.M.= Gaouda ∠ − V3 = 1.0000 − j 0.0500 0 . 00125 2 . 8624 pu UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Using the rectangular form of the load flow equations, then the net active and reactive powers at 1th bus are: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi = Re{ V1* ( Y11 V1 + Y12 V2 + Y13 V3 )} Reminder The bus admittance matrix is ⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤ ⎥ ⎢ 26 − j 52 − 16 + j 32 ⎥ Dr. A.M. Gaouda⎢ − 10 + j 20 UAE University ⎢ − 10 + j 30 − 16 + j 32 26 - j 62 ⎦⎥ ELEC572, 04/05⎣ Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qi = − Im{ V1* ( Y11 V1 + Y12 V2 + Y13 V3 )} P1 − jQ1 = V1* ( Y11 V1 + Y12 V2 + Y13 V3 ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi − jQi = Dr. A.M. Gaouda UAE University ELEC572, 04/05 4.0938 - j1.8894 1 P1 = 4.0938 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 G Dr. A.M. Gaouda UAE University 0.01 + j 004/05 .03 pu ELEC572, P1 = 409.38 MVA 256.6 MW 0.0125 + j 0.025 pu 110.2 V1 = 1.05∠0 o pu Base MVA=100 Q1 = 188.94 MVA 2 0.02 + j 0.04 pu Q1 = 1.8894 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 MVAR 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 138.6 MW 45.2 MVAR Dr. A.M. Gaouda UAE University ELEC572, 04/05 Problem 1: The line impedances are as indicated in per unit on 100MVA base. Using GaussSeidel method: 1. Classify each bus 2. find bus admittance matrix 2. find bus 2 voltage after the first iteration. 3. find bus 1 real and reactive power. NOTE: select the initial value for bus 2 voltage as: V2o = 1∠ − 22.0169 o SG 2 = 0.25 + j 0.6459 pu G1 1 S D1 G2 V1 = 1.0∠0 o pu V2 Z L = j 0.5 pu 2 S D2 = 1.0 + j 0.5 pu Ybus = -j2.0 j2.0 j2.0 -j2.0 S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d ) S 2 ,sch = - 0.7500 + j0.1459 pu Using GS method, select the initial values for the unknowns as: V1 = 1.0∠0 o pu V2o = 1∠ − 22.0169 o Start the first iteration V21 = 1 P2 ,sch − jQ2 ,sch ( V2o )* Y22 ⎡ Y21 −⎢ V1 ⎢⎣ Y22 ⎤ ⎥ ⎥⎦ V21 = 1∠ − 22.0238 o = 0.9271 - j0.3749 Power at bus 1 P1 − jQ1 = V1* ( Y11 V1 + Y12 V2 ) P1 + jQ1 = 0.7500 +j0.1459 II. Modifying G-S Method when PV buses are present Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Assuming a power system has n buses, then; one bus will be considered as a slack bus and the other buses are load buses (PQ-buses) and voltage controlled buses (PV-buses). Let the system buses be numbered as: Dr. A.M. Gaouda UAE University ELEC572, 04/05 i=1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Salck bus i = 2 , 3 , ..... , m PV − buses i = m + 1 , m + 2 , ...... , n Dr. A.M. Gaouda UAE University ELEC572, 04/05 PQ − buses Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For the voltage controlled buses, Pi and | Vi | are known & Qi and δ i are unknown Dr. A.M. Gaouda UAE University ELEC572, 04/05 | Vi | = | Vi | Specified Qi ,min 〈 Qi 〈 Qi ,max ….. Eq. 23 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ….. Eq. 24 The second requirement for the voltage controlled bus may be violated if the bus voltage becomes too high or too small. It is to be noted that we can control the bus voltage by controlling the bus reactive power. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Therefore, during any iteration, if the PV-bus reactive power violates its limits then set it according to the following rule. Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qi 〉 Qi ,max set Qi = Qi ,max Qi 〈 Qi ,min set Qi = Qi ,min Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qi ,min 〈 Qi 〈 Qi ,max Dr. A.M. Gaouda UAE University ELEC572, 04/05 NOTE For PQ − bus And treat this bus as PQ-bus. Pi and Qi are known & | Vi | and δi are unknown Load flow solution when PV buses are present Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 a. Calculate Qi In the polar form, n Qi =| Vi | For the (kth+1) iteration, Dr. A.M. Gaouda UAE University ELEC572, 04/05 ( k +1 ) i Q Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 = | Vi |speci + | Vi |speci ∑| Yip || V p | sin ( δ i − δ p − γ ip ) p =1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 i −1 ( k +1 ) (k ) ( k +1 ) | Y | | V | sin ( δ − δ − γ ip ) ∑ ip p i p p=1 n (k ) (k ) (k ) | Y | | V | sin ( δ − δ − γ ip ) ∑ ip p i i p= i Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For p = 1 to ( i − 1 ), use | V | & δ p of ( k + 1 ) iteration Dr. A.M. Gaouda UAE University p ELEC572, 04/05 th Dr. A.M. Gaouda UAE University ELEC572, 04/05 For p = i to n , use | V p | & δ p of ( k th ) iteration Dr. A.M. Gaouda UAE University ELEC572, 04/05 Set | Vi | = | Vi | speci Dr. A.M. Gaouda UAE University ELEC572, 04/05 In the rectangular form, Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qi = − Im{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )} b. Check Qik+1 to see if it is within the limits Qi ,min 〈 Qi 〈 Qi ,max Dr. A.M. Gaouda UAE University ELEC572, 04/05 Case 1: If the reactive power limits are not violated, calculate Vi k + 1 Vi k +1 Ki = − k * ( Vi ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 i −1 ∑ Lip V p=1 ( k +1 ) p − n Dr. A.M. Gaouda UAE University ELEC572, 04/05 • Use the most updated value of Qi to calculate Ki . k L V ∑ ip p = |Vik + 1 |∠δ i Dr. A.M. Gaouda UAE University ELEC572, 04/05 p= i + 1 k +1 • New Voltage A.M. Gaouda magnitudeDr. and UAE University ELEC572, 04/05 angle are obtained Use Dr. A.M. Gaouda UAE University ELEC572, 04/05 | Vi | speci and δ ik + 1 For the PV-bus voltage. Reset the magnitude |Vi Dr. A.M. Gaouda UAE University ELEC572, 04/05 k +1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 | = |Vi |Speci Dr. A.M. Gaouda UAE University ELEC572, 04/05 Voltage magnitude is known for PV bus, therefore the new calculated magnitude will not be used. ∠δ ik + 1 Vi k + 1 = |Vi |Speci Dr. A.M. Gaouda UAE University ELEC572, 04/05 Only the calculated angle will be updated and used. Case 2: If the reactive power limits are violated, Qik + 1 〉 Qi ,max Or set Qik + 1 = Qi ,max Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qik + 1 〈 Qi ,min Dr. A.M. Gaouda UAE University ELEC572, 04/05 set Qik + 1 = Qi ,min Consider this bus as a PQ-Bus, calculate bus voltage Vi k + 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Vi k + 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Ki = − k * ( Vi ) Vi i −1 ( k +1 ) L V ∑ ip p − k +1 p=1 = |Vi n k L V ∑ ip p p= i + 1 k +A.M. 1 Gaouda k + 1 ∠δ Dr. iUAE University | ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The PV-bus becomes PQ-bus and both Voltage magnitude and angle are calculated and Dr. A.M. Gaouda UAE University used ELEC572, 04/05 EXAMPLE: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Each line has an impedance of 0.05+j0.15 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 Line Data for the 5 buses Network 4 From Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 5 G1 Dr. A.M. Gaouda UAE University ELEC572, 2 04/05 G Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 To Dr. A.M. Gaouda Bus UAE University ELEC572, 04/05 R X 1 2 0.0500 0.1500 2 3 0.0500 0.1500 2 4 0.0500 0.1500 3 4 0.0500 0.1500 1 5 Dr. A.M. Gaouda 0.0500 0.1500 UAE University 4 5 0.0500 ELEC572, 04/05 0.1500 The shunt admittance is neglected Bus Data for the the 5 buses Network Before load flow solution Load MW Bus No. Bus code VoltDr. A.M. GaoudaVolt UAE University Mag.ELEC572, Angle 04/05 1 Slack 1.0200 0 100 50 ? ? 0 2 PV 1.0200 ? 0 0 200 ? 3 PQ ? ? 50 20 PQ ? ? PQ ? ? Dr. A.M. Gaouda 4 UAE University ELEC572, 04/05 5 Load MVAR Gen.Dr. A.M.Gen. Gaouda UAE University MW ELEC572, MVAR 04/05 Q Min. Q Max. Inject MVAR 0 0 20 60 0 0 0 0 0 0 50 Dr. A.M. Gaouda UAE University 20 ELEC572, 04/05 0 0 0 0 50 20 0 0 0 0 Dr. 0 A.M. Gaouda UAE University ELEC572, 04/05 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For the ‘5’ bus system Construct the bus admittance matrix Ybus Find Q2 ,δ 2 ,V3 ,V4 and V5 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qmax = 0.6 pu Qmin = 0.2 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 4 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 G1 G2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 5 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 SOLUTION: 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 A.M. Gaouda 4 Dr. UAE University ELEC572, 04/05 Ybus Construction y = 1 1 Dr. A.M. Gaouda = 2 − j 6 = UAE University z 0.05 +ELEC572, j 0.15 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 Y11 = y12 + y15 = 4 − j 12 G2 G1 5 1 Y22 = y 21 + y 23 + y 25 = 6 − j 18 Y = y + y 34 = 4 − j 12 Dr. A.M. Gaouda UAE University 33ELEC572, 04/05 32 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y44 = y43 + y 45 = 4 − j 12 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Ybus = Dr. A.M. Gaouda UAE University ELEC572, 04/05 -2.0 + J6.0 6.0 -J18.0 -2.0 + J6.0 0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y15 = − y15 = −2 + j 6 Y13 = Y14 = 0 Y55 = y51 + y52 + y54 = 6 − j 18 4.0 - J12.0 -2.0 + J6.0 0 0 -2.0 + J6.0 Y12 = − y12 = −2 + j 6 Dr. A.M. Gaouda UAE University ELEC572, 04/05 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University 0 ELEC572, 04/05 0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 -2.0 + J6.0 -2.0 + J6.0 0 -2.0 + J6.0 6.0 -J18.0Dr. A.M. Gaouda UAE University ELEC572, 04/05 The net scheduled power injected at each bus is: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 4 Bus No. Load MW Bus code Volt Mag. Volt Angle Load MVAR 1 Dr. A.M. Gaouda UAE University 2 ELEC572, 04/05 Slack 1.0200 0 100 PV 1.0200 ? 0 3 PQ ? ? 50 20 4 PQ ? ? 50 5 PQ ? ? 50 Gen. MW Gen. MVAR Q Min. Q Max. Inject MVAR 2 G1 G2 5 1 Dr.50A.M. Gaouda ? UAE University 200 0 ELEC572, 04/05 ? 0 0 0 ? 20 60 0 0 0 0 0 0 20 0 0 0 0 0 20 0 0 0 0 0 S 1 ,sch = ( P1 , g − P1 ,d ) + j ( Q1 , g − Q1 ,d ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 1 ,sch = ( P1 , g − 1.0 ) + j ( Q1 , g − 0.5 ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d ) S 2 ,sch = ( 2.0 − 0 ) + j ( Q2 , g − 0 ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 ,sch S = ( 0 − 0.5 ) + j ( 0 − 0.2 ) Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 3 ,sch = − 0.5 − j 0.2 S 4 ,sch = − 0.5 − j0.2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 5 ,sch = − 0.5 − j 0.2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda The known values are: UAE University Dr. A.M. Gaouda UAE University ELEC572, 04/05 4.0 - J12.0 ELEC572, 04/05 V1 = 1.02∠0 o -2.0 + J6.0 0 0 -2.0 + J6.0 |V2 |spec = 1.02 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The bus admittance matrix is -2.0 + J6.0 6.0 -J18.0 -2.0 + J6.0 0 -2.0 + J6.0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 0 0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 -2.0 + J6.0 -2.0 + J6.0 0 -2.0 + J6.0 6.0 -J18.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q2 ,min = 0.2 Q2 ,max = 0.6 and Using GS method, select the initial values for the unknowns as: V3o =V4o = V5o = 1∠0o and Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 δ 2o = 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the first iteration Bus 2 is PV Bus Q2 ,min 〈 Q2 〈 Q2 ,max Check Q2 is within the limits Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn ) Q21 = − Im{ V2* ( Y21 V1 + Y22 V2o + Y23 V3o + Y24 V4o + Y25 V5o )} Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q = 0.2448 1 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q2 ,min 〈 Q2 〈 Q2 ,max i .e .; Dr. A.M. Gaouda UAE University ELEC572, 04/05 0.20 〈 0.2448 〈 0.6 The reactive power limits are not violated, Calculate: [ K2 1 o o o − L V + L V + L V + L V V2 = 21 1 23 3 24 4 25 5 ( V2o )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] The values for Ki and Lip are computed once in the beginning and used in every iteration. Y L21 = 21 Y22 P2 − jQ2 K2 = Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y L23 = 23 Y22 Y L24 = 24 Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 S 2 ,sch = 2.0 + j 0.2448 L21 = -0.3333 K2 = 0.0456 + j0.0959 L23 = -0.3333 V = 1.0555∠5.1113 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Reset the magnitude Therefore, Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 2 o |V | = |V2 |Speci = 1.02 1 2 δ 21 = 5.1113o Dr. A.M. Gaouda UAE University ELEC572, 04/05 V21 = 1.02∠5.1113o L24 = 0.0 L25 = Y25 Y22 L25 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Voltage magnitude is known and fixed for a PV bus, therefore the new calculated magnitude willDr.not be A.M. Gaouda UAE University used. ELEC572, 04/05 Bus 3 is PQ Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 V31 = P3 − jQ3 K3 = Y33 [ K3 1 o o − L V + L V + L V + L V 31 1 32 2 34 4 35 5 ( V3o )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 L31 = K3 = -0.0275 - j0.0325 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y31 Y33 L31 = 0.0 L32 = Y32 Y33 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L34 = L32 = -0.5000 L35 = L34 = -0.5000 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Bus 4 is PQ Bus [ K4 = P4 − jQ4 Y44 K4 = -0.0275 - j0.0325 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y L41 = 41 Y44 L41 = 0.0 Y42 L42 = Y44 V = 0.9631∠ - 1.5489 1 4 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L43 = Y43 Y44 L43 = -0.5000 L42 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y35 Y33 L35 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 K4 1 1 o L V L V L V L V V = − + + + 41 1 42 2 43 3 45 5 ( V4o )* 1 4 ] Y34 Y33 V31 = 0.9806∠0.7559o Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 o ] L45 = Y45 Y44 L45 = -0.5000 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Bus 5 is PQ Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 [ Dr. A.M. Gaouda UAE University ELEC572, 04/05 K5 1 3 1 L V L V L V L V V = − + + + 51 1 52 2 53 3 54 4 ( V5o )* 1 5 K5 = -0.0183 - 0.0217i L53 = 0.0 L52 = -0.3333 L51 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 V51 = 0.9812∠ - 0.0031o Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] L54 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the second iteration Bus 2 is PV Bus Check Q2 is within the limits Dr. A.M. Gaouda UAE University ELEC572, 04/05 0.2 〈 Q2 〈 0.6 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q22 = − Im{ V21* ( Y21 V1 + Y22 V21 + Y23 V31 + Y24 V41 + Y25 V51 )} Q22 = 0.0290 The reactive power limits are violated Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q2 〈 Qi ,min And treat this bus as PQ-bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 set Q2 = Qi ,min = 0.2 S 2 ,sch = 2.0 + j 0.2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Use the most updated value of Q2 to calculate the constant K2 Dr. A.M. Gaouda UAE University All Buses 2, 3, 4 and 5 are PQ Buses. Find the bus voltages using GS method ELEC572, 04/05 Problem 2: A. The line impedances are as indicated in per unit on 100MVA base. The line charging susceptances are neglected. Using Gauss-Seidel method find the power flow solution of the system. Ignoring the limits of Q3. 1 2 0.02 + j 0.04 pu G 400 MW 0.0125 + j0.025 pu 0.01 + j 0.03 pu 250 MVAR V1 = 1.05 ∠0 o pu 3 | V3 | = 1.04 200 MW G 1 1 y12 = = 10 − j 20 = y 21 0.02 + j 0.04 1 y13 = = 10 − j 30 = y 31 0.01 + j 0.03 y 23 = 2 0.02 + j 0.04 pu G 250 V1 = 1.05 ∠0 o pu MVAR 3 | V3 | = 1.04 200 MW Ybus ⎡Y11 Y12 Y13 ⎤ ⎡ y12 + y13 ⎥ ⎢ = ⎢Y21 Y22 Y23 ⎥ = ⎢ − y 21 ⎢ ⎢Y31 Y32 Y33 ⎥ ⎢⎣ − y 31 ⎥ ⎢ Ybus = MW 0.0125 + j 0.025 pu 0.01 + j 0.03 pu 1 = 16 − j 32 = y 32 0.0125 + j 0.025 400 − y12 y 21 + y 23 − y 32 ⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤ ⎢ − 10 + j 20 26 − j 52 − 16 + j 32 ⎥ ⎢ ⎥ ⎢⎣ − 10 + j 30 − 16 + j 32 26 − j 62 ⎥⎦ G − y13 ⎤ − y 23 ⎥⎥ y 31 + y 32 ⎥⎦ Classification of buses: 1 2 0.02 + j 0.04 pu Bus 1: Slack Bus G 250 V1 = 1.05 ∠0 pu MVAR 3 Bus 2: Load Bus (PQ bus) | V3 | = 1.04 200 MW V2 and δ 2 are unknown P2 and Q2 are known S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d ) S 2 ,sch = Base MVA pu ( 0 − 400 ) + j ( 0 − 250 ) pu 100 S 2 ,sch = − 4 − j 2.5 pu Bus 3: Voltage Controlled Bus (PV bus) |V3 | and Pg ,3 are known 0.0125 + j 0.025 pu 0.01 + j 0.03 pu o V1 = 1.05∠0 o pu Q3 ,sch and δ 3 are unknown P3 ,sch = 2.0 pu 400 MW G Using GS method, select the initial values for the unknowns as: V1 = 1.05∠0 o pu V2o = 1∠0 | V3 | = 1.04 Start the first iteration Bus 2 is PQ Bus n K2 V2 = * − ∑ L2 p V p V2 p = 1 p≠ 2 [ K2 o L V L V V = − + 21 1 23 3 ( V2o )* 1 2 K2 = - 0.0692 - j0.0423 L21 = - 0.3846 ] L23 = -0.6154 V21 = 0.9746 - j0.0423 δ 3o = 0 o Bus 3 is PV Bus Calculate and Check Q3 is within the limits Q3 ,min 〈 Q3 〈 Q3 ,max Q31 = − Im{ V3* ( Y31 V1 + Y32 V21 + Y33 V3o )} Q31 = j1.1600 K3 1 1 V3 = − L V + L V 31 1 32 2 ( V3o )* [ K3 = 0.0274 + j0.0208 L31 = -0.4690 + j0.0354 ] L32 = -0.5310 - j0.0354 V31 = 1.0378 - j0.0052 = 1.0378∠ - 0.2854o Reset the magnitude |V31 | = |Vi |Speci = 1.04 V31 = 1.04∠ - 0.2854o V31 = 1.0400 - j0.0052 Voltage magnitude is fixed for a PV bus, therefore the new calculated magnitude will not be used. Start the second iteration Bus 2 is PQ Bus K 2 , L21 , L23 are constants and will be the same. [ K2 1 L V L V V = − + 21 1 23 3 ( V21 )* V22 = 0.9711 - j0.0434 2 2 Bus 3 is PV Bus Calculate and check Q3 is within the limits ] Q3 ,min 〈 Q3 〈 Q3 ,max Q32 = − Im{ V3* ( Y31 V1 + Y32 V22 + Y33 V31 )} Q32 = j1.3881 [ K3 2 V = − L V + L V 31 1 32 2 ( V31 )* 2 3 ] L31 and L32 are constants and will be the same. K 3 is changed as Q3 change P3 − jQ32 K3 = 0.0305 + j0.0194 K3 = Y33 V32 = 1.0391 - j0.0073 = 1.0391∠ - 0.4028o Reset the magnitude V32 = 1.04∠ - 0.4028o = 1.0400 - j0.0073 EXAMPLE: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Each line has an impedance of 0.05+j0.15 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 Line Data for the 5 buses Network 4 From Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 5 G1 Dr. A.M. Gaouda UAE University ELEC572, 2 04/05 G Dr. A.M. Gaouda UAE University ELEC572, 04/05 1 To Dr. A.M. Gaouda Bus UAE University ELEC572, 04/05 R X 1 2 0.0500 0.1500 2 3 0.0500 0.1500 2 4 0.0500 0.1500 3 4 0.0500 0.1500 1 5 Dr. A.M. Gaouda 0.0500 0.1500 UAE University 4 5 0.0500 ELEC572, 04/05 0.1500 The shunt admittance is neglected Bus Data for the the 5 buses Network Before load flow solution Load MW Bus No. Bus code VoltDr. A.M. GaoudaVolt UAE University Mag.ELEC572, Angle 04/05 1 Slack 1.0200 0 100 50 ? ? 0 2 PV 1.0200 ? 0 0 200 ? 3 PQ ? ? 50 20 PQ ? ? PQ ? ? Dr. A.M. Gaouda 4 UAE University ELEC572, 04/05 5 Load MVAR Gen.Dr. A.M.Gen. Gaouda UAE University MW ELEC572, MVAR 04/05 Q Min. Q Max. Inject MVAR 0 0 20 60 0 0 0 0 0 0 50 Dr. A.M. Gaouda UAE University 20 ELEC572, 04/05 0 0 0 0 50 20 0 0 0 0 Dr. 0 A.M. Gaouda UAE University ELEC572, 04/05 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 For the ‘5’ bus system Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Construct the bus admittance matrix Ybus Find Q2 ,δ 2 ,V3 ,V4 and V5 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Qmax = 0.6 pu Qmin = 0.2 pu Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 SOLUTION: 3 Dr. A.M. Gaouda UAE University ELEC572, 04/05 A.M. Gaouda 4 Dr. UAE University ELEC572, 04/05 Ybus Construction y = 1 1 Dr. A.M. Gaouda = 2 − j 6 = UAE University z 0.05 +ELEC572, j 0.15 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 2 Y11 = y12 + y15 = 4 − j 12 G2 G1 5 1 Y22 = y 21 + y 23 + y 25 = 6 − j 18 Y = y + y 34 = 4 − j 12 Dr. A.M. Gaouda UAE University 33ELEC572, 04/05 32 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y44 = y43 + y 45 = 4 − j 12 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Ybus = Dr. A.M. Gaouda UAE University ELEC572, 04/05 -2.0 + J6.0 6.0 -J18.0 -2.0 + J6.0 0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y15 = − y15 = −2 + j 6 Y13 = Y14 = 0 Y55 = y51 + y52 + y54 = 6 − j 18 4.0 - J12.0 -2.0 + J6.0 0 0 -2.0 + J6.0 Y12 = − y12 = −2 + j 6 Dr. A.M. Gaouda UAE University ELEC572, 04/05 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University 0 ELEC572, 04/05 0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 -2.0 + J6.0 -2.0 + J6.0 0 -2.0 + J6.0 6.0 -J18.0Dr. A.M. Gaouda UAE University ELEC572, 04/05 The net scheduled power injected at each bus is: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 3 4 Bus No. Load MW Bus code Volt Mag. Volt Angle Load MVAR 1 Dr. A.M. Gaouda UAE University 2 ELEC572, 04/05 Slack 1.0200 0 100 PV 1.0200 ? 0 3 PQ ? ? 50 20 4 PQ ? ? 50 5 PQ ? ? 50 Gen. MW Gen. MVAR Q Min. Q Max. Inject MVAR 2 G1 G2 1 5 ? 0 ? 0 20 20 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr.50A.M. Gaouda ? UAE University 200 0 ELEC572, 04/05 0 0 20 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda The known values are: UAE University Dr. A.M. Gaouda UAE University ELEC572, 04/05 4.0 - J12.0 ELEC572, 04/05 -2.0 + J6.0 0 0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 The bus admittance matrix is -2.0 + J6.0 6.0 -J18.0 -2.0 + J6.0 0 -2.0 + J6.0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 0 0 0 -2.0 + J6.0 4.0 -J12.0 -2.0 + J6.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 -2.0 + J6.0 -2.0 + J6.0 0 -2.0 + J6.0 6.0 -J18.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Using GS method, select the initial values for the unknowns as: Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the first iteration Bus 2 is PV Bus Q2 ,min 〈 Q2 〈 Q2 ,max Check Q2 is within the limits Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn ) Q21 = − Im{ V2* ( Y21 V1 + Y22 V2o + Y23 V3o + Y24 V4o + Y25 V5o )} Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q = 1 2 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Q2 ,min 〈 Q2 〈 Q2 ,max i .e .; Dr. A.M. Gaouda UAE University ELEC572, 04/05 0.20 〈 0.448 〈 0.6 The reactive power limits are not violated, Calculate: [ K2 1 o o o L V L V L V L V V2 = − + + + 21 1 23 3 24 4 25 5 ( V2o )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] The values for Ki and Lip are computed once in the beginning and used in every iteration. P2 − jQ2 K2 = Y22 Y21 L21 = Y22 K2 = 0.0456 + j0.0959 L21 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L23 = Dr. A.M. Gaouda UAE University ELEC572, 04/05 L23 = -0.3333 V21 = 1 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Reset the magnitude Therefore, Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y L24 = 24 Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L24 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 |V21 | = δ 21 = 5 V21 = 1 Y23 Y22 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y25 L25 = Y22 L25 = -0.3333 Voltage magnitude is known and fixed for a PV bus, therefore the new calculated magnitude will not be A.M. Gaouda used. Dr. UAE University ELEC572, 04/05 Bus 3 is PQ Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 V31 = P3 − jQ3 K3 = Y33 [ K3 1 o o L V L V L V L V − + + + 31 1 32 2 34 4 35 5 ( V3o )* Dr. A.M. Gaouda UAE University ELEC572, 04/05 L31 = K3 = -0.0275 - j0.0325 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y31 Y33 L31 = 0.0 L32 = Y32 Y33 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L34 = L32 = -0.5000 L35 = L34 = -0.5000 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Bus 4 is PQ Bus [ K4 = P4 − jQ4 Y44 K4 = -0.0275 - j0.0325 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y L41 = 41 Y44 L41 = 0.0 Y42 L42 = Y44 V = 0.9631∠ - 1.5489 1 4 Dr. A.M. Gaouda UAE University ELEC572, 04/05 L43 = Y43 Y44 L43 = -0.5000 L42 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Y35 Y33 L35 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 K4 1 1 o L V L V L V L V V = − + + + 41 1 42 2 43 3 45 5 ( V4o )* 1 4 ] Y34 Y33 V31 = 0.9806∠0.7559o Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 o ] L45 = Y45 Y44 L45 = -0.5000 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Bus 5 is PQ Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 [ Dr. A.M. Gaouda UAE University ELEC572, 04/05 K5 1 3 1 L V L V L V L V V = − + + + 51 1 52 2 53 3 54 4 ( V5o )* 1 5 K5 = -0.0183 - 0.0217i L51 = -0.3333 L52 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 V51 = 0.9812∠ - 0.0031o L53 = 0.0 Dr. A.M. Gaouda UAE University ELEC572, 04/05 ] L54 = -0.3333 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Start the second iteration Bus 2 is PV Bus Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 Dr. A.M. Gaouda UAE University ELEC572, 04/05 EXAMPLE: Each line has an impedance of 0.05+j0.15 Line Data for the 5 buses Network 3 4 2 5 G1 G2 1 Bus nl Bus nr R 1 2 0.0500 0.1500 2 3 0.0500 0.1500 2 4 0.0500 0.1500 3 4 0.0500 0.1500 1 5 0.0500 0.1500 4 5 0.0500 0.1500 X The shunt admittance is neglected Bus Data for the the 5 buses Network Before load flow solution Load MW Bus No. Bus code Volt Mag. Volt Angle Load MVAR Gen. MW Gen. MVAR Q Min. 1 Slack 1.0200 0 100 50 ? ? 0 2 PV 1.0200 ? 0 0 200 ? 3 PQ ? ? 50 20 0 4 PQ ? ? 50 20 5 PQ ? ? 50 20 Q Max. Inject MVAR 0 0 20 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 For the ‘5’ bus system Construct the bus admittance matrix Ybus Find Q2 ,δ 2 ,V3 ,V4 and V5 Qmax = 0.6 pu Qmin = 0.2 pu SOLUTION: Solution Ybus Construction 3 4 2 G1 G2 1 5 Bus nl Bus nr R 1 2 0.0500 0.1500 2 3 0.0500 0.1500 2 4 0.0500 0.1500 3 4 0.0500 0.1500 1 5 0.0500 0.1500 4 5 0.0500 0.1500 X Solution 3 4 2 G1 G2 1 5