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lecture4

# lecture4 - Power Flow Solution l The utility wants to know...

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Power Systems I Power Flow Solution l The utility wants to know the voltage profile u the nodal voltages for a given load and generation schedule l Types of network buses u load bus n known real (P) and reactive (Q) power injections u generator bus n known real (P) power injection and the voltage magnitude (V) u slack bus (swing bus) n known voltage magnitude (V) and voltage angle ( δ ) n must have one generator as the slack bus n takes up the power slack due to losses in the network

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Power Systems I l KCL l Power Law ( ) ( ) ( ) ( ) i j V y y V V y V y V y V y y y y V V y V V y V V y V y I n j j ij n j ij i n in i i i in i i i n i in i i i i i i i = + + + + = + + + + = = = 1 0 2 2 1 1 2 1 0 2 2 1 1 0 L L L Power Flow Equations i j V y y V V jQ P V jQ P I I V jQ P n j j ij n j ij i i i i i i i i i i i i = = = + = = 1 0 * * *
Power Systems I Gauss-Seidel Method l A non-linear algebraic equation solver u method of successive displacements u iterative steps: n take a function and rearrange it into the form x = g( x ) {there are several possible arrangements} n make an an initial estimate of the variable x : x [0] = initial value n find an iterative improvement of x [ k ] , that is: x [ k +1] = g( x [ k ] ) n a solution is reached when the difference between two iterations is less than a specified accuracy: x [ k +1] - x [ k ] ε u acceleration factors n can improve the rate of convergence: α > 1 n modified step: the improvement is found as x [ k +1] = x [ k ] + α ( g( x [ k ] ) - x [ k ] )

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Power Systems I Gauss-Seidel Example l Find the root of the equation: f ( x ) = x 3 - 6 x 2 + 9 x - 4 = 0 u Step 1. Cast the equation into the g ( x ) form.
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lecture4 - Power Flow Solution l The utility wants to know...

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