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lecture4 - the nodal voltages for a given load and...

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Unformatted text preview: the nodal voltages for a given load and generation schedule ￿ ￿ ￿ known real (P) and reactive (Q) power injections known real (P) power injection and the voltage magnitude (V) ￿ ￿ ￿ known voltage magnitude (V) and voltage angle (δ) must have one generator as the slack bus takes up the power slack due to losses in the network slack bus (swing bus) ￿ generator bus ￿ load bus Types of network buses ￿ The utility wants to know the voltage profile Power Systems I ￿ ￿ Power Flow Solution n j =1 * ii j≠i Pi − jQi → Ii = Vi* j≠i n n Pi − jQi = Vi ∑ yij − ∑ yijV j * Vi j =1 j =0 Pi + jQi = V I Power Law j =0 = Vi ∑ yij − ∑ yijV j n = ( yi 0 + yi1 + yi 2 + ￿ + yin )Vi − yi1V1 − yi 2V2 − ￿ − yinVn I i = yi 0Vi + yi1 (Vi − V1 ) + yi 2 (Vi − V2 ) + ￿ + yin (Vi − Vn ) KCL Power Systems I ￿ ￿ Power Flow Equations ￿ ￿ ￿ find an iterative improvement of x[k], that is: x[k+1] = g( x[k] ) a solution is reached when the difference between two iterations is less than a specified accuracy: Ƚx[k+1] - x[k] Ƚ ≤ ε make an an initial estimate of the variable x: x[0] = initial value take a function and rearrange it into the form x = g(x) {there are several possible arrangements} ￿ ￿ x[k+1] = x[k] + α( g( x[k] ) - x[k] ) can improve the rate of convergence: α > 1 modified step: the improvement is found as acceleration factors ￿ ￿ ￿ ￿ method of successive displacements iterative steps: A non-linear algebraic equation solver Power Systems I ￿ Gauss-Seidel Method ￿ 4 x = − 1 x3 + 6 x2 + 9 = g ( x) 9 9 9 x = − x3 + 6x2 + 4 Step 1. Cast the equation into the g(x) form. Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 Power Systems I ￿ Gauss-Seidel Example x[8] = 4.0000 x[ 7 ] = 3.9988 x[ 6 ] = 3.9568 x[5] = 3.7398 x[ 4 ] = 3.3376 6 4 x[3] = g ( x[ 2 ] = 2.5173) = − 1 (2.5173)3 + 9 (2.5173) 2 + 9 = 2.8966 9 6 4 x[ 2 ] = g ( x[1] = 2.2222) = − 1 (2.2222)3 + 9 (2.2222) 2 + 9 = 2.5173 9 6 4 x[1] = g ( x[ 0 ] = 2) = − 1 (2)3 + 9 (2) 2 + 9 = 2.2222 9 Step 2. Starting with an initial guess of x[0] = 2, several iterations are performed. Power Systems I ￿ Gauss-Seidel Example 0 Power Systems I 0 0. 5 1 1. 5 2 2. 5 3 3. 5 4 4. 5 0. 5 1 Initial Value 1. 5 Iterations 3 2 x 2. 5 x 3 3. 5 x = g(x) Solution Points 2 g(x) =-1/9x +6/9x +4/9 Matlab Results Gauss-Seidel Example 4 4. 5 ￿ x[ 2 ] = 2.2778 + 1.25 [2.5902 − 2.2778] = 2.6683 6 4 g (2.2778) = − 1 (2.2778) 3 + 9 (2.2778) 2 + 9 = 2.5902 9 x[1] = 2 + 1.25 [2.2222 − 2] = 2.2778 4 g (2) = − 1 (2) 3 + 6 (2) 2 + 9 = 2.2222 9 9 x[ 0 ] = 2 Starting with an initial guess of x[0] = 2. Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 with an acceleration factor of 1.25 Power Systems I ￿ Gauss-Seidel Example x[8] = 4.0005 x[ 7 ] = 3.9978 x[ 6 ] = 4.0084 x[5] = 3.7238 x[ 4 ] = 3.1831 x[3] = 3.0801 Additional iterations Power Systems I ￿ Gauss-Seidel Example 0 Power Systems I 0 0. 5 1 1. 5 2 2. 5 3 3. 5 4 4. 5 0. 5 1 1. 5 2 x 2. 5 x 3 3. 5 x = g(x) Solution Points 2 with acceleration factor: 1.25 Initial Value Iterations 3 g(x) = -1/9x +6/9x +4/9 Matlab Results Gauss-Seidel Example 4 4. 5 ￿ ￿ xn = cn + g n ( x1 , x2 , ￿ , xn ) ￿ x2 = c2 + g 2 ( x1 , x2 , ￿ , xn ) x1 = c1 + g1 ( x1 , x2 , ￿ , xn ) Rearrange each equation for one of the variables f n ( x1 , x2 , ￿ , xn ) = cn ￿ f 2 ( x1 , x2 , ￿ , xn ) = c2 f1 ( x1 , x2 , ￿ , xn ) = c1 Consider a system of n equations Power Systems I ￿ ￿ Gauss-Seidel for a System of Equations ￿ ￿ ￿ 0 2 0 n k +1] [ [ , x2k +1] , ￿ , xnk +1] ) in the Gauss-Seidel method, the updated values of the variables calculated in the preceding equations are used immediately in the solution of the subsequent equations 1 (x[ find the results in a new approximate solution 0 1 (x[ ], x[ ],￿, x[ ] ) assume an approximate solution for the independent variables s t e ps Power Systems I ￿ Gauss-Seidel for a System of Equations Power Systems I j =0 ∑y n ij Pi − jQi n + ∑ yijV j * Vi j =1 j ≠ i ⇒ Vi[ k +1] = The Gauss-Siedel form j≠i Pi − jQi → Ii = Vi* n n Pi − jQi = Vi ∑ yij − ∑ yijV j * Vi j =1 j =0 Pi + jQi = V I * ii T h e e q u a t io n Vi = ￿ ￿ The Power Flow Equation j =0 ∑y n ij Pi − jQi n + ∑ yijV j[ k ] Vi*[ k ] j =1 j≠i ￿ ￿ j≠i j≠i ￿ ￿ ￿ for generation the powers are positive for loads the powers are negative the scheduled power is the sum of the generation and load powers the real and reactive powers are scheduled for the load buses that is, they remain fixed the currents and powers are expressed as going into the bus n Ⱥ *[k ] Ⱥ [k ] n Ⱥ Ⱥ [k +1] [k ] Ⱥ Ⱥ Qi = −ℑȺVi ȺVi ∑ yij − ∑ yij V j Ⱥ Ⱥ Ⱥ j =0 j =1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ n Ⱥ *[k ] Ⱥ [k ] n Ⱥ Ⱥ [k +1] [k ] Ⱥ Ⱥ Pi = ℜȺVi ȺVi ∑ yij − ∑ yij V j Ⱥ Ⱥ Ⱥ j =0 j =1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Rewriting the power equation to find P and Q Power Systems I ￿ Power Injections j =0 ∑y n ij j≠i n Ⱥ *[k ] Ⱥ [k ] n Ⱥ Ⱥ [k +1] [k ] Ⱥ Ⱥ Qi = −ℑȺVi ȺVi ∑ yij − ∑ yij V j Ⱥ Ⱥ Ⱥ j =0 j =1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ n Ⱥ *[k ] Ⱥ [k ] n Ⱥ Ⱥ [k +1] [k ] Ⱥ Ⱥ = ℜȺVi ȺVi ∑ yij − ∑ yij V j Ⱥ Ⱥ Pi Ⱥ j =0 j =1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Vi [k +1] = Pi [sch ] − jQi[sch ] n + ∑ yij V j[k ] Vi*[k ] j =1 j≠i j≠i The complete set of equations become: Power Systems I ￿ Solution by Gauss-Seidel Ⱥ Ⱥ Ⱥ Ⱥ n Ⱥ *[k ] Ⱥ [k ] [k +1] [k ] Ⱥ Ⱥ = ℜȺVi Vi Yii + ∑ Yij V j Ⱥ Pi Ⱥ Ⱥ j =1 Ⱥ Ⱥ Ⱥ Ⱥ j ≠i Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ n Ⱥ Ⱥ Qi[k +1] = −ℑȺVi*[k ] ȺVi [k ]Yii + ∑ Yij V j[k ] Ⱥ Ⱥ Ⱥ Ⱥ j =1 Ⱥ Ⱥ Ⱥ Ⱥ j ≠i Ⱥ Ⱥ Ⱥ Ⱥ n Pi [sch ] − jQi[sch ] − ∑ Yij V j[k ] Vi*[k ] j =1, j ≠ i Vi [k +1] = Yii Rewriting the equations in terms of the Y-Bus Power Systems I ￿ Solution by Gauss-Seidel ￿ ￿ ￿ ￿ ￿ the voltage magnitude and angle must be estimated in per unit, the nominal voltage magnitude is 1 pu the angles are generally close together, so an initial value of 0 degrees is appropriate Since both components (V & δ) are specified for the slack bus, there are 2(n - 1) equations which must be solved iteratively For the load buses, the real and reactive powers are known: scheduled System characteristics Power Systems I ￿ Solution by Gauss-Seidel ￿ ￿ ￿ ￿ e [ k +1] i = Vi [ sch ] 2 ( − fi ) [ k +1] 2 Vi = ei + j f i the real power is scheduled the reactive power is computed based on the estimated voltage values the voltage is computed by Gauss-Seidel, only the imaginary part is kept the complex voltage is found from the magnitude and the iterative imaginary part For the generator buses, the real power and voltage magnitude are known Power Systems I ￿ Solution by Gauss-Seidel V1 = 1.05∠0° 0.01 + j0.03 0.0125 + j0.025 0.02 + j0.04 0.452 pu 1.386 pu 1.102 pu 2.566 pu Using the Gauss-Seidel method, determine the phasor values of the voltage at the load buses 2 and 3, accurate to 2 decimal places Power Systems I ￿ Example ...
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