GSLF_Fig 3

Applying the Nodal Equation in Power System The current source in power systems comes from generated MW and MVAR. If our convention is an inductive Mvar for a lagging current, then apparent power in MVA=(complex conjugate V)* (I). By looking at the figure on the left, we assume that E has a positive angle and I has a negative angle with respect to the horizontal reference line. The angle of S is -(a+b). So S= P -jQ. It means that an inductive power has a negative Mvar. Note that E and V are synonymous notation for voltage.

What is an Iterative Process?
Suppose you want to get the square root of 100 just by division, a number being divided by another number. A simple code below shows how to perform an iterative process. An initial estimate next=1 is used. The correct answer is 10 and 1 is far from the answer. Then that initial estimate is used to divide the number. If the initial estimate is 10, the answer will be 10. But the answer is not correct, so we get the average of the initial estimate and the quotient. The difference between the initial estimate and the answer is called an error. This error must be less than the allowable error (tolerance) and we consider the answer as true to be correct. As the iteration continues, the error becomes smaller and smaller. This computation is called an iterative process. If the computation does not reach an answer it is said to be divergent, else it is convergent.

#include "stdio.h" //replace the " with angle brackets #include "math.h" //incompatible with html coding void main(void) {double error=4, answer, next=1,number=100, toler=0.01; int iter=0; while (error>toler) {iter++; answer=(next + number/next)/2; error=fabs(next-answer); printf("iteration:%2d next=%7.3f ",iter,next); next=answer; printf("answer=%7.3f error=%7.3f\n",answer,error); } } iteration: 1 next= 1.000 answer= 50.500 error= 49.500 iteration: 2 next= 50.500 answer= 26.240 error= 24.260 iteration: 3 next= 26.240 answer= 15.026 error= 11.215 iteration: 4 next= 15.026 answer= 10.840 error= 4.185 iteration: 5 next= 10.840 answer= 10.033 error= 0.808 iteration: 6 next= 10.033 answer= 10.000 error= 0.033 iteration: 7 next= 10.000 answer= 10.000 error= 0.000

Gauss-Siedel Loadflow (GSLF)

GSLF_Eq. 4 on the left is the representation of GSLF_Eq. 3. Substituting Eq. 4 in Eq. 5 yields Eq. 6. This equation can be visualized using a sample 4-bus system as shown on Fig. (4).

Ybus for power system is in complex number. to get the real part, multiply and divide with its conjugate. This will make the denominator in real numbers only. GSLF_Fig 4

By definition, the off diagonal has negative of the line admittance and the diagonal is the sum of admittance connected to the node. With this procedure, we do not use sin() and cos(). Since bus-4 has no connection to bus-1, so there is no entry for Y(1,4). This is also true with Y(4,1). For a power system with something like 1000 nodes, each node can have an average of 4 connections thus the Ybus can have a 95% zero values. Note that the diagonal is always nonzero.

This set of simultaneous equations is considered non-linear, compared with GSLF_Eq. (3) of the previous chapter. The state variables are the voltages and these are the unknown. When the E on the denominator of the lefthand side is transferred to the right hand side, you have a product-- which makes the equation non-linear. GSLF_Fig 7

We can write a general equation for nodes 1 to 4 from Eq. (7),
GSLF_Fig 8
The summation above includes when i=j. Mark this equation as all power system algorithms are based on it. This equation is called "injected power on bus i".

A method called Gauss-Siedel can solve the unknown voltage, first by assuming a guess say E=1.0 for all. Then that value is improved by substituting them to the rest of the equation. By examining node 3, in Eq. (7),for example, E(3) is on the left side as the denominator of (P -jQ) and as the diagonal of the right side.

To apply the Gauss-Siedel iteration, we equate the YiiEi (diagonal) as the unknown voltage. Let us transpose and solving for E(3), and generalize with subscripts i,j,k as shown in GSLF_Eq. 9. Here, i is the node we are solving and j's represents the nodes connecting node-i and k represents the iteration counter. The left side of node-i are the voltages we already updated and the right side of the diagonal have voltages whose value are computed GSLF_Eq. 9.

from the previous iteration represented by superscript (k-1). The voltage at the denominator of (P -jQ), which is also for node-i has its value based on (k-1). Since we do not store the old voltage from the updated voltage, the programming aspect will just be simple. If we use rectangular coordinates for Yij as (g +jb) and E as (e +jf)

we can avoid the use of trigonometric sin(), cos() and atan() functions. GSLF_Eq. 10 represents the dissected portion for (RePQE +j ImPQE). Pi here is the injected power which is the generated MW minus the demand MW. Since the denominator is a conjugate of Ei, it is equated to (e -jf). To remove the j part from the denominator, it must be multiplied and divided by its conjugate which is (e +jf). GSLF_Eq. 10.

This portion is the summation Yij*Ej. It does not include the diagonal part. In the for loop, it exclude j=i. GSLF_Eq. 11.

Re= RePQE - ReSumYE      GSLF_Eq. 11.
Im= ImPQE - ImSumYE

Performing the last part
GSLF_Eq. 13.

Program Listing_1: Gauss-Siedel Loadflow
This is a beginner's code because the Ybus matrix is full or in 2 dimension. Node numbering is contiguous starting from 1.