Gauss-Iterative Method to Solve Power Flow Problems | Electrical Engineering

Gauss-iterative method is essentially a trial and error method and provides a sufficiently accurate numerical solution to power flow problem.

The steps in solving the problem by this method are:

1. Guess the initial solution approximately.

2. This guessed solution is used in conjunction with the original equation to compute a new and more accurate estimate for the solution.

3. The second estimate is used to find the third estimate.

The process is continued until the difference between successive estimates is extremely negligible.

The above method has two drawbacks:

1. It may require an extremely large number of iterations to reach sufficiently close to the solution, but with the use of computers it may not be thought of as a drawback.

2. The process may not converge, if the solution does not exist or if the iteration process is divergent.

Gauss-Seidel method is a modification to Gauss-iteration method. This modification reduces the number of iterations. In this method the values of unknowns immediately replace the previous values in the next step while in case of Gauss method, the calculated values replace the earlier values only at the end of the iteration.

Because of it Gauss-Seidel method converges much faster than the Gauss method, i.e., number of iterations required to obtain solution is much less in the Gauss-Seidel method compared to the Gauss method. This is why Gauss-Seidel method is very popular.

Gauss-Seidel method is of the simplest iterative methods and has been in use since early days of digital computer methods of analysis.

It has got the following advantages:

(i) It is very simple method.

(ii) Computing costs are less.

Sometimes it is used to determine the initial solution for the other iterative processes.

The Gauss and Gauss-Seidel methods of iterative solution of nonlinear algebraic equations are illustrated in simple cases by the following examples:

Example 1:

Find the value of x, accurate up to 4 decimal places in the equation x² –  5x + 6 = 0 using Gauss-iterative method.

Solution:

Given equation is X² –  5x + 6 = 0

or x = 0.2 (X² + 6)

Let the initial guess is such that:

Since the values of x⁽³¹⁾ and x⁽³²⁾ are same up to 4 decimal places, the iteration need not be continued further and 1.999792348 can be taken as the value of x.

Example 2:

Obtain the values of x and y by Gauss-iterative method for the following equations:

3x + xy – 2 = 0 and

3y – xy + 2 = 0

Solution:

The given equations can be written as –

x = 0.66666 –  0.33333 xy

and y = – 0.66666 + 0.33333 xy

Using the usual method, it is seen that the solution to the given equations is x = 1 and y = – 1

Let us start iteration by assuming values of x and y such that –

x⁽⁰⁾ = 0.5 and y⁽⁰⁾ = – 0.5

Then:

Iteration 1:

x⁽¹⁾ = 0.66666 – 0.33333 x 0.5 x (- 0.5) = 0.75

y^{1) =-0.66666 + 0.33333 x 0.5 x (-0.5) = -0.75

Iteration 2:

x⁽²⁾ = 0.66666 – 0.33333 x 0.75 x (-0.75) = 0.854166667

y⁽²⁾ =-0.66666 + 0.33333 x 0.75 x (-0.75) = -0.854166667

Iteration 3:

x⁽³⁾ = 0.66666 – 0.33333 x 0.854166667 x (-0.854166667) = 0.909866898

y⁽³⁾ =-0.66666 + 0.33333 x 0.854166667 x (-0.854166667) = -0.909866898

Iteration 4:

x⁽⁴⁾ = 0.66666 – 0.33333 x 0.909866898 x (-0.909866898) = 0.942619256

y⁽⁴⁾ =-0.66666 + 0.33333 x 0.909866898 x (-0.909866898) = -0.942619256

Substitute the values obtained of x and y and continue steps x⁽⁵⁾, y⁽⁵⁾; x⁽⁶⁾, y⁽⁶⁾ x⁽⁷⁾, y⁽⁷⁾ etc. till final results obtained are quite steady.

It will be observed that the speed of convergence in this method is increased or the numbers of iterations required are lesser.

Acceleration Factor:

The convergence process can be significantly accelerated in the above methods by the application of acceleration factor.

If ∆ x^(r) and ∆ y^(r) indicate the variable change in the (r –  1)th and r^th iteration, the difference variables can be written as:

If the difference is multiplied by the acceleration factor a which is larger than unity (usually between 1.5 and 1.7 for the solution of power flow equations), the accelerated iterates are obtained as follows:

The accelerated values of x and y are used in the iteration process instead of x, y.