Power System Analysis Lecture on Power Flow and Modeling Voltage Dependent Load

ECE 476 Power System Analysis Lecture 14: Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu

Announcements Read Chapter 6, Chapter 12.4 and 12.5 Quiz today on HW 6 HW 7 is 6.50, 6.52, 6.59, 12.20, 12.26; due October 22 in class (no quiz) Power and Energy scholarships will be decided on Monday; application on website; apply to Prof. Sauer; Grainger Awards due on Nov 1; application on website; apply to Prof. Sauer energy.ece.illinois.edu 1

Solving Large Power Systems The most difficult computational task is inverting the Jacobian matrix inverting a full matrix is an order n3 operation, meaning the amount of computation increases with the cube of the size size this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix using sparse matrix methods results in a computational order of about n1.5. this is a substantial savings when solving systems with tens of thousands of buses 2

Newton-Raphson Power Flow Advantages fast convergence as long as initial guess is close to solution large region of convergence Disadvantages each iteration takes much longer than a Gauss-Seidel iteration more complicated to code, particularly when implementing sparse matrix algorithms Newton-Raphson algorithm is very common in power flow analysis 3

Modeling Voltage Dependent Load So far we've assumed that the load is independent of the bus voltage (i.e., constant power). However, the power flow can be easily extended to include voltage depedence with both the real and reactive l is done by making P and Q a function of oad. This V : Di Di i n + + = ( cos sin ) ( ) 0 V V G B P P V i k ik ik ik ik Gi Di i = 1 k n + = ( sin cos ) ( ) 0 V V G B Q Q V i k ik ik ik ik Gi Di i = 1 k 4

Voltage Dependent Load Example In previous two bus example now assume the load is constant impedance, so 2 = ) 2.0 + = x P ( ) (10sin 0 V V 2 2 2 2 2 2 = ( 10cos + (10) 1.0 + = ( ) x ) 0 Q V V V 2 2 2 2 2 Now calculate the power flow Jacobian 10 ( ) 10 2 V + + cos sin 10sin 4.0 V V V + 2 2 2 20 2 2.0 = x J 10cos V 2 2 2 2 5

Voltage Dependent Load, cont'd 0 1 (0) = = x Again set 0, guess v Calculate 2 ) 2.0 + (10sin V V 2.0 1.0 2 2 V 2 + (0) = = x f( ) 2 2 ( 10cos + ) (10) 1.0 V V 2 2 2 2 10 0 4 (0) = J x ( ) 12 1 0 1 10 0 4 2.0 1.0 0.1667 0.9167 (1) = = x Solve 12 6

Voltage Dependent Load, cont'd With constant impedance load the MW/Mvar load at bus 2 varies with the square of the bus 2 voltage magnitude. This if the voltage level is less than 1.0, the load is lower than 200/100 MW/Mvar 160.0 MW 120.0 MVR -160.0 MW -80.0 MVR Line Z = 0.1j 0.894 pu -10.304 Deg One 1.000 pu Two 160.0 MW 120.0 MVR 160 MW 80 MVR 7

Dishonest Newton-Raphson Since most of the time in the Newton-Raphson iteration is spend calculating the inverse of the Jacobian, one way to speed up the iterations is to only calculate/inverse the Jacobian occasionally known as the Dishonest Newton-Raphson an extreme example is to only calculate the Jacobian for the first iteration ( 1) ( ) Honest: + = x x + ( ) -1 v ( ) v v v = x x - ( J x f x ) ( ) ( 1) ( ) v (0) -1 ( ) v v - ( J x f x Dishonest: ) ( ) ( ) v for a solution f x Both require ( ) 8

Dishonest Newton-Raphson Example Use the Dishonest Newton-Raphson to solve 2 = = ( ) f x -2 0 x 1 (0) ( dx ) df x ( ) v ( ) v = ( ) x f x 1 x ( ) v ( ) 2 v = (( ) -2) x x (0) 2 1 x + ( 1) ( ) v ( ) 2 v v = (( ) -2) x x x (0) 2 9

Dishonest N-R Example, contd 1 x + ( 1) ( ) v ( ) 2 v v = (( ) -2) x x x (0) 2 We pay a price in increased iterations, but with decreased computation per iteration (0) = Guess x 1. Iteratively solving we get ( ) v ( ) v v 0 1 2 3 4 (honest) 1 1.5 1.41667 1.41422 1.41422 (dishonest) 1 1.5 1.375 1.429 1.408 x x 10

Two Bus Dishonest ROC Slide shows the region of convergence for different initial guesses for the 2 bus case using the Dishonest N-R Red region converges to the high voltage solution, while the yellow region converges to the low voltage solution 11

Honest N-R Region of Convergence Maximum of 15 iterations 12

Decoupled Power Flow The completely Dishonest Newton-Raphson is not used for power flow analysis. However several approximations of the Jacobian matrix are used. One common method is the decoupled power flow. In this approach approximations are used to decouple the real and reactive power equations. 13

Decoupled Power Flow Formulation General form of the power flow problem V ( ) v ( ) v P P V ( ) v ( ) v P x ( ) ( ) v = = f x ( ) ( ) v ( ) v ( ) v ( ) v Q x V ( ) Q Q where ( ) v + x ( ) P P P 2 2 2 D G ( ) v = P x ( ) ( ) v + x ( ) P P Gn P n Dn 14

Decoupling Approximation ( ) v ( ) v P V Q Usually the off-diagonal matrices, and are small. Therefore we approximate them as zero: V can be decoupled = ( ) v P 0 ( ) v ( ) v P x ( ) ( ) v = = f x ( ) ( ) v ( ) v ( ) v Q Q x V ( ) 0 Then the problem 1 1 ( ) v ( ) v P Q V ( ) v ( ) v ( ) v ( ) v = ( ) ( ) P x V Q x 15

Off-diagonal Jacobian Terms Justification for Jacobian approximations: 1. Usually r x, therefore G B ij ij 2. Usually is small so sin 0 ij ij Therefore P V ( ) i = + cos sin 0 V G B i ij ij ij ij j Q ( ) i = + cos sin 0 V V G B i j ij ij ij ij j 16

Decoupled N-R Region of Convergence 17

Fast Decoupled Power Flow By continuing with our Jacobian approximations we can actually obtain a reasonable approximation that is independent of the voltage magnitudes/angles. This means the Jacobian need only be built/inverted once. This approach is known as the fast decoupled power flow (FDPF) FDPF uses the same mismatch equations as standard power flow so it should have same solution The FDPF is widely used, particularly when we only need an approximate solution 18

FDPF Approximations The FDPF makes the following approximations: = 1. G 0 ij = = 2. 3. 1 V i = sin 0 cos 1 ij ij Then ( ) v ( ) v P x V Q x V Y ( ) ( ) ( ) v ( ) v 1 1 = = B V B ( ) v ( ) v = + B G B Where is just the imaginary part of the except the slack bus row/co , j bus lumn are omitted 19

FDPF Three Bus Example Use the FDPF to solve the following three bus system Line Z = j0.07 One Two 200 MW 100 MVR Line Z = j0.05 Line Z = j0.1 34.3 14.3 20 14.3 24.3 10 20 10 30 Three 1.000 pu = Y j bus 200 MW 100 MVR 20

FDPF Three Bus Example, contd 34.3 14.3 20 0.0477 0.0159 14.3 24.3 10 20 10 30 24.3 10 10 30 = = Y B j bus 0.0159 0.0389 1 = B Iteratively solve, starting with an initial voltage guess (0) (0) V V 0 0 1 1 2 2 = = 3 3 (1) 0 0 0.0477 0.0159 0.0159 0.0389 2 2 0.1272 0.1091 2 = + = 3 21

FDPF Three Bus Example, contd (1) x V V 1 1 0.0477 0.0159 0.0159 0.0389 1 1 0.9364 0.9455 2 = + = 3 n P ( ) P P i V Di Gi = + + ( cos sin ) V G B k ik ik ik ik V = i i 1 k (2) 0.1272 0.1091 0.0477 0.0159 0.0159 0.0389 0.151 0.107 0.1361 0.1156 2 = + = 3 (2) V V 0.924 0.936 2 = 3 0.1384 0.1171 0.9224 0.9338 = = Actual solution: V 22

FDPF Region of Convergence 23

DC Power Flow The DC power flow makes the most severe approximations: completely ignore reactive power, assume all the voltages are always 1.0 per unit, ignore line conductance This makes the power flow a linear set of equations, which can be solved directly 1 = B P 24

Power System Control A major problem with power system operation is the limited capacity of the transmission system lines/transformers have limits (usually thermal) no direct way of controlling flow down a transmission line (e.g., there are no valves to close to limit flow) open transmission system access associated with industry restructuring is stressing the system in new ways We need to indirectly control transmission line flow by changing the generator outputs 25

DC Power Flow Example 26

DC Power Flow 5 Bus Example One Five Four Three A A 360 MW 0 Mvar 520 MW 0 Mvar MVA MVA A MVA slack 1.000 pu 0.000 Deg 1.000 pu -4.125 Deg 1.000 pu -1.997 Deg 80 MW 0 Mvar A A MVA MVA 1.000 pu 0.524 Deg 1.000 pu -18.695 Deg Two 800 MW 0 Mvar Notice with the dc power flow all of the voltage magnitudes are 1 per unit. 27

Indirect Transmission Line Control What we would like to determine is how a change in generation at bus k affects the power flow on a line from bus i to bus j. The assumption is that the change in generation is absorbed by the slack bus 28

Power Flow Simulation - Before One way to determine the impact of a generator change is to compare a before/after power flow. For example below is a three bus case with an overload 131.9 MW 124% One Two 200 MW 100 MVR 200.0 MW 71.0 MVR 68.1 MW 68.1 MW Z for all lines = j0.1 Three 1.000 pu 0 MW 64 MVR 29

Power Flow Simulation - After Increasing the generation at bus 3 by 95 MW (and hence decreasing it at bus 1 by a corresponding amount), results in a 31.3 drop in the MW flow on the line from bus 1 to 2. 101.6 MW 100% One Two 200 MW 100 MVR 105.0 MW 64.3 MVR 3.4 MW 98.4 MW 92% Z for all lines = j0.1 Limit for all lines = 150 MVA 1.000 pu Three 95 MW 64 MVR 30

Analytic Calculation of Sensitivities Calculating control sensitivities by repeat power flow solutions is tedious and would require many power flow solutions. An alternative approach is to analytically calculate these values The power flow from bus i to bus j is V V i X j i j sin( ) P ij i j X ij ij i j ij So We just need to get P ij X P ij Gk 31

Analytic Sensitivities From the fast decoupled power flow we know = B P x 1 ( ) So to get the change in generation at bus k, just set all zeros except a minus one at position k. 0 due to a change of ( ) equal to P x = P Bus k 1 0 32

Three Bus Sensitivity Example = For the previous three bus case with Z 20 10 10 20 10 10 Hence for a change of generation at bus 3 0.1 j line 10 10 20 20 10 10 20 = = Y B j bus 1 20 10 0.0667 10 20 0 0.0333 0.0667 2 = = 1 3 0 3 to 1 P = = Then 0.667 pu 0.1 3 to 2 P = 2 to 1 P = 0.333 pu 0.333 pu 33

Balancing Authority Areas An balancing authority area (use to be called operating areas) has traditionally represented the portion of the interconnected electric grid operated by a single utility Transmission lines that join two areas are known as tie-lines. The net power out of an area is the sum of the flow on its tie-lines. The flow out of an area is equal to total gen - total load - total losses = tie-flow 34

Area Control Error (ACE) The area control error (ace) is the difference between the actual flow out of an area and the scheduled flow, plus a frequency component = + ace 10 P sched P f int Ideally the ACE should always be zero. Because the load is constantly changing, each utility must constantly change its generation to chase the ACE. 35

Automatic Generation Control Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero. Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds. 36

Power Transactions Power transactions are contracts between generators and loads to do power transactions. Contracts can be for any amount of time at any price for any amount of power. Scheduled power transactions are implemented by modifying the value of Psched used in the ACE calculation 37