PowerFlow Equations: GaussSeidel.

11252016, 03:14 PM
(This post was last modified: 01052017 03:45 PM by Ángel Martin.)
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PowerFlow Equations: GaussSeidel.
PowerFlow Equations: GaussSeidel. [PFEGS ]
From the author’s Engineering Collection, included in the ETSII5 module. This program calculates the unknown 2nvariables of a “n”bus power system, when the admittance matrix and the remaining 2nvariables are known. In a powerflow system there is one main complex equation for each bus relating the active and reactive power to the voltages of all the other buses, as follows: Si = Pi + jQi = Vi . I*i ; i = 1, 2, .. n This results in the following 2 power equations per each bus  as function of the voltages on all the other system buses, where arg(Yij) are the lag angles between voltage and current and the Yij are the elements of the admittance matrix of the system: Pi = SUM Vi Vj Yij cos[ arg(Vi) – arg(Vj) – arg(Yij) ] ; j = 1, 2…n Qi = SUM Vi Vj Yij sin[ arg(Vi) – arg(Vj) – arg(Yij)] ; j = 1,2,...n The program prompts for the number of the voltagecontrolled buses (PV, those with voltage known) –excluding the slack bus (which this program assumes there’s only one!). The slack node should be the first ones entered in the bus sequence. This node’s voltage is used as reference for all other angles. Also the number of Load buses (PQ) is defined as a consequence of the number of PV voltage controlled buses, since the sum of all buses (including the slack) must be equal to the total” n”. Example 1. Calculate the voltage angles on each node, the reactive power in the two generation nodes and the active power in the slack node of a threebus power system with the following known variables: Y11 = 20j ; Y21= 10j ; Y22 = 20j ; Y31 = 10j ; Y32 = 10j ; Y33 = 20j PD1 = 1; QD1 = 0.5; V1 = 1 PD2 = QD2 = 0; PG2 = 1.5 V2 = 1 PD3 = QD3 = 1; PG3 = 0 ; V3 = 1 The results are as follows: PG1 = 0.5 pu.; QG1 = 0.523 pu. QG2 = 0.057 pu ; arg(V2) = 3.822 deg QG3 = 1.036 pu ; arg(V3) = 0.957 deg Example 2. A twobus power system has the following variables known: Y11 = 1.841631, <) = 80.48561 Y12 = 1.904443, <) = 99.197819 Y22 = Y11 V1=1.05, arg(V1) =0) SD1 = 1.15 + 0.31j SD2 = 0.45 + 0.2j a) If QG2 =0, obtain V2, arg(2), PG1 and QG1 b) If V2=1, obtain QG2, arg(2), PG1 and QG1 (one voltagecontrolled bus) The solutions are: a) PG1 = 1.6233 pu. ; QG1 = 0.5357 pu V2 = 0.8912 ; arg(V2) = 13.6293 deg b) PG1 = 1.6171 pu ; QG1 = 0.3171 pu; QG2 = 0.1645 pu ; arg(V2) = 12.9669 deg Note. The program uses 5+ n^2 + 7n data registers, therefore the maximum size is n = 14 buses ( with 299 data registers needed). CL users could use the Extended Registers functions to increase that boundary to 28 buses (using 985 data registers!) 

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