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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 nonlinear algebraic equation solver Power Systems I GaussSeidel 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 GaussSeidel 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 GaussSeidel 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 GaussSeidel 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 GaussSeidel 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 GaussSeidel 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 GaussSeidel 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 GaussSeidel for a System of Equations 0
2 0
n k +1] [
[
, x2k +1] , , xnk +1] ) in the GaussSeidel 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 GaussSeidel 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 GaussSiedel 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 GaussSeidel 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 YBus Power Systems I Solution by GaussSeidel 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 GaussSeidel 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 GaussSeidel, 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 GaussSeidel 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 GaussSeidel 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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 Fall '11
 THOMASBALDWIN
 Volt

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