For explaining the application of Gauss-Seidel method for power flow studies, let it be assumed that all buses other than the swing or slack bus are P-Q or load buses. At slack bus both V and δ are specified and they remain fixed throughout. There are (n – 1) buses where P and Q are given. Initially we assume the magnitudes and angles at these (n – 1) buses and update these voltages at every step of iteration.
Bus 1 is a slack bus with specified V, hence in above equation V1 is taken as specified.
Thus Eq. (6.69) represents a set of (n – 1) equations for i = 2, 3, …, n which are to be solved simultaneously for V2, V3, V4 … Vn.
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In the Gauss method, we assume the voltage for all the buses except the slack bus where the voltage magnitude and phase angle are specified and remain fixed. Normally, we set i.e., assume the voltage magnitude and phase angle of these buses equal to that of the slack bus and work in per unit system.
The assumed bus voltage and the slack bus voltage along with P and Q are substituted in RHS of the Eq. (6.69) to obtain new set of bus voltages. After the entire iteration is complete, the new set of bus voltages is again substituted along with the specified slack bus voltage in the RHS of Eq. (6.69) to obtain a new set of bus voltages.
The process is continued till:
where r is an iteration count and ԑ is a very small number which depends upon the system accuracy and is normally equal to 0.0001 etc.
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Thus the process is continued till the mod of the bus voltage obtained at the current iteration less the value of bus voltage at the previous iteration is smaller than a chosen very small number and in this way we obtain the solution, i.e., magnitude and phase angle of voltage.
Gauss iterative method, explained above, is much slower to converge and may sometimes fail to do so.
In case of Gauss-Seidel method, the value of bus voltages calculated for any bus immediately replace the previous values in the next step while in case of Gauss method, as stated earlier, the calculated bus voltages replace the earlier value only at the end of the iteration. Due to this Gauss- Seidel method converges much faster than that of Gauss method compared to Gauss method.
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Gauss-Seidel method was one of the most common methods employed for solving power flow equations.
It has the following advantages and disadvantages:
Advantages:
1. Simplicity of technique.
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2. Small computer memory requirement.
3. Less computational time per iteration.
Disadvantages:
1. Slow rate of convergence resulting in larger number of iterations.
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2. Increase in number of iterations directly with the increase in the number of buses.
3. Effect on convergence due to choice of slack bus.
Because of the above drawbacks, use of Gauss-Seidel method is limited only to systems with smaller number of buses.
Gauss-Seidel Method of Solution of Power Flow Problems on Digital Computers:
Let us continue to consider the case where all buses except slack bus are load or P-Q buses.
Block 1. Reading of Primitive Y Matrix:
With the load profile known at each bus (PDi and QDi being known) PGi and QGi are allocated to all generating stations.
While active and reactive generations are allocated to the slack bus, these are permitted to vary during iterative computation. This is necessary because the magnitude and angle of voltage are specified at this bus and only two variables can be specified at any bus.
With this step, bus injections (Pi + j Qi) are known at all buses other than the slack bus.
Block 2. Formation of Bus Admittance Matrix Ybus:
With the line and shunt admittance data stored in computer, bus admittance matrix Ybus is formed by using the rules for self and mutual admittances. Ybus can also be formed by using Eq. (6.27) where input data is in the form of primitive matrix Y and singular connection matrix A.
Block 3. Iterative Computation of Bus Voltages:
To start the iterations a set of initial voltage values is to be assumed. Since, in a power system the voltage spread is not too wide, it is normal practice to assume a flat voltage profile for all nodal voltages except the slack bus 1. Initially all voltages are set equal to (1 + j 0) except the voltage of the slack bus which is fixed.
It is to be noted that voltages and admittances in Eq. (6.69) are complex quantities. In most of the computers the complex number operations are available. If complex number operations are not available in a computer, Eq. (6.69) can be converted into 2 (n – 1) equations in real unknowns. (ei fi or Vi δi) by writing,
It is possible to obtain significant reduction in computer time if all arithmetic operations are performed in advance when they do not change with iterations. Since P, Q and Y at a bus do not change with iterations, the term [Pi – (j Qi /Yii)] can be evaluated before-hand.
In order to save computer time it is more appropriate to write Eq. (6.69) in the following form:
The values of Ai and Bik are computed once in the beginning and then used in every iteration.
Now for the (r + 1)th iteration, the voltage Eq. (6.71) becomes –
The second term on the RHS of the above Eq. (6.74) is obvious because the voltage prior to bus i should correspond to the value as computed during the current iteration.
The iterative process is continued till the change in magnitude of the bus voltages in successive iterations become less than a certain pre-specified tolerance level for all buses, i.e.,
To reduce the number of iterations, an acceleration factor α = 1.6 is considered to be a good value for power flow studies. A wrong choice of a may indeed slowdown convergence or even cause the method to diverge.
After computing bus voltages (magnitudes and phase angles) for all the buses, injected powers Si and line flows are computed using the nodal voltages. Equation (6.57) is used to determine Si and conjugate of Si gives injected power Si.
The last step in power flow study is the calculation of line flows and line losses.
Computations of Line Flows and Line Losses:
Fig. 6.18 shows a line connecting ith and kth buses. Here we assume the normal π representation of transmission line.
Current flowing from bus i towards bus k,
where Vi and Vk are the bus voltages at the buses i and k respectively which are already calculated from the power flow studies.
The power flow in the line i-k at the bus i is given as –
Similarly, the power flow in the line i-k at the bus k is given as –
Thus power flows over all the lines can be computed.
The power losses in the (I – k)th line are given by sum of the power flows determined from above Eqs. (6.77) and (6.78) i.e. power losses in the (I – k)th line = Sik + Ski. Total transmission losses can be computed by summing all the line flows (i.e., Sik + Ski. for all i, k).
It may be noted that the slack bus power can also be determined by summing the power flows on the lines terminating at the slack bus. This concludes the power flow study for the case of P-Q buses only.
Treatment to Voltage Controlled Buses in Gauss-Seidel Method:
In a power system, some of the buses are voltage controlled buses where P and V are specified while Q and 8 are unknowns and are to be determined. However, usually limit of reactive power, i.e., Qmax and Qmin to hold the generation voltage within limits are also given.
So far power flow study in a system with generation or voltage controlled buses with G-S method the values of Q and δ are to be updated in every GS iteration through appropriate bus equations and therefore computational procedure needs some modifications.
Let the buses be numbered as –
i = 1 slack or swing bus
i = 2, 3 … m voltage controlled or P-V buses
i = m + 1, m + 2, m + 3, …, n load or P-Q buses
For the voltage controlled buses, the bus voltage Vi must be equal to the specified voltage Vi specified in magnitude and the value of reactive power Qi (for i = 2, 3, …, m) must lie between the limits.
Thus, the conditions to be satisfied are –
The second requirement may be violated, i.e., the reactive power generation at the bus may be lesser than the minimum reactive power generation permissible (because of stability problem) or it may be more than the maximum reactive power generation permissible (because of rotor heating).
If during any iteration Qi violates the specified limits, the voltage controlled bus is made to act as a load bus for that iteration only and the reactive power to be substituted in the expression for that iteration will correspond to the limit it has violated, i.e., if it is less than Qmin, the value to be substituted will be Qmin. If calculated value of Q lies within the limits, this value of Q will be substituted in the expression.
The violation of reactive power limit could be due to the specified voltage either being too low or too high. Since Vi specified can be obtained only by controlling reactive power, therefore, it is possible that we have specified voltage beyond the capability of the reactive power generation of the generator.
If the value of Q for a generator bus is calculated after assuming magnitude equal to Vspecified for that bus and phase angle corresponding to the value in that iteration and if Q so obtained violates the limits, the value of voltage should be taken corresponding to the voltage in that iteration and not Vspecified. Otherwise the voltage should correspond to specified voltage and phase angle as in that iteration.
In view of above observations the procedure of computation for voltage controlled or generation buses is an under –
(i) Calculate reactive power generation using Eq. (6.60) which is reproduced below:
At the beginning of (r + 1)th iteration, Vi(r) , i.e., magnitude of Vi obtained during rth iteration may not necessarily be equal to Vi specified. We have this value of Vi(r)
The values of Vk and δk to be used in Eq. (6.80) are those obtained during (r + 1)th iteration for (k = 1 to i – 1) and those obtained during rth iteration (for k = i to n). Moreover, for every iteration Vi must be equal to Vi specified. In fact, for the (r + 1)th iteration one can write from Eq. (6.79) –
(ii) If Qi(r + 1) is found to lie within limits, i.e., Qi min < Qi(r + 1) < Qi max calculate new value of Ai and then use Eq. (6.74) for computation of Vi(r + 1) using Vi specified and δi(r) for the magnitude and phase angle of Vi(r + 1). Reset vi(r + 1) to Vi specified but retain the phase angle δi(r + 1) and continue to the next bus.
If Qi(r + 1) < Qi min, set Qi(r + 1) = Qi min and treat bus i as a P-Q bus. Compute Ai(r + 1) and Vi(r + 1) from Eqs. (6.72) and (6.74) respectively.
If Qi(r + 1) > Qi max set Qi(r + 1) = Qi max and treat bus i as P-Q bus. Compute Ai(r+1) and Vi(r + 1) from Eqs. (6.72) and (6.74) respectively.
Now all the computational steps are summarized in the detailed flow chart shown in Fig. 6.19 which serves as basis for the reader to write his own computer programme.