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# lecture7 - there are many solution combinations for...

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Unformatted text preview: there are many solution combinations for scheduling generation in practice, power plants are not located at the same distance from the load centers power plants use different types of fuel, which vary in cost from time to time ￿ in the power flow analysis, generator buses are specified by P and |V| generation capacity is more than load demand - generators can produce more than the customers can consume For interconnected systems, the objective is to find the real and reactive power scheduling so as to minimize some operating cost or cost function ￿ ￿ ￿ ￿ In practice and in power flow analysis, there are many choices for setting the operating points of generators Power Systems I ￿ ￿ Economic Dispatch ￿ ￿ ￿ dx f (x ) = 0 dx 2 f (x ) > 0 ȹ ∂f ∂f ∂f ȹ ∂f = 0 i = 1,￿, n or ∇f = ȹ ȹ ȹ ∂x , ∂x ,￿, ∂x ȹ = 0 ∂xi 2 n Ⱥ ȹ 1 for a set of parameters, the gradient of f vanishes at a local extrema and to be a local minimum, the Hessian must be a positive definite matrix (i.e. positive eigenvalues) d2 for f to be a local minimum, the second derivative must be positive at the point of the local extrema d the first derivative of f vanishes at a local extrema General cost function: f ( x1 , x2 ,￿, xn ) = C Unconstrained parameter optimization, from calculus: Power Systems I ￿ ￿ Optimization ￿ ￿ ￿ ￿ ˆ ˆ ˆ ∂ 2 f ( x1 ￿ xi ￿ xn ) H ij = ∂xi ∂x j ˆ ˆ ˆ 0 < eigen i [H(x1 ￿ x j ￿ xn )] i = 1,￿, n this condition also requires that all the eigenvalues of the Hessian matrix evaluated at the extrema to be positive x T Hx > 0 ∀x ≠ 0 a symmetrical matrix contains the second derivatives of the function f for f to be a minimum, the Hessian matrix must be positive definite The Hessian matrix Power Systems I ￿ Optimization 2 2 Power Systems I ˆ Ⱥ2 1 0Ⱥ Ⱥ x1 Ⱥ Ⱥ 8 Ⱥ Ⱥ1 4 1Ⱥ Ⱥ x Ⱥ = Ⱥ16 Ⱥ ˆ Ⱥ Ⱥ 2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ0 1 6Ⱥ Ⱥ x3 Ⱥ Ⱥ32Ⱥ Ⱥ Ⱥ ˆ Ⱥ Ⱥ Ⱥ Ⱥ evaluating the first derivatives to zero results in ∂f = 2 x1 + x2 − 8 = 0 ∂x1 ￿ ˆ Ⱥ x1 Ⱥ Ⱥ 3Ⱥ Ⱥ x Ⱥ = Ⱥ2Ⱥ ˆ Ⱥ 2 Ⱥ Ⱥ Ⱥ Ⱥ x3 Ⱥ Ⱥ5Ⱥ Ⱥ ˆ Ⱥ Ⱥ Ⱥ f (x1 , x2 , x3 ) = x1 + 2 x2 + 3x3 + x1 x2 + x2 x3 − 8 x1 − 16 x2 − 32 x3 + 110 2 Find the minimum of ∂f = x1 + 4 x2 + x3 − 16 = 0 or ∂x2 ∂f = x2 + 6 x3 − 32 = 0 ∂x3 ￿ Example the eigenvalues are all greater than zero, so it’s a minimum point Ⱥ2 1 0Ⱥ Ⱥ1.55Ⱥ Ⱥ0Ⱥ eigen Ⱥ1 4 1Ⱥ = Ⱥ 4.0 Ⱥ > Ⱥ0Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ0 1 6Ⱥ Ⱥ6.45Ⱥ Ⱥ0Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ using the MATLAB function eig(H), the eigenvalues are found Ⱥ2 1 0Ⱥ ˆ H ( x ) = Ⱥ1 4 1Ⱥ Ⱥ Ⱥ Ⱥ0 1 6Ⱥ Ⱥ Ⱥ evaluating the second derivatives and forming the Hessian matrix Power Systems I ￿ ￿ ￿ Example ˆ ˆ ˆ gi (x1 ￿ x j ￿ xn ) = 0 i = 1,￿, k subject to the equality constraints ˆ ˆ ˆ f (x1 ￿ x j ￿ xn ) minimize the cost function Such problems may be solved by the Lagrange muliplier m e th o d ￿ ￿ This type of problem arises when there are functional dependencies among the parameters to be found The problem Power Systems I ￿ ￿ ￿ Equality Constraints in Optimization ￿ ￿ New cost function ∂L = gi = 0 ∂λi k ∂L ∂f gi = + ∑ λi =0 ∂xi ∂xi i =1 ∂xi The necessary conditions for finding the local minimum i =1 L = f + ∑ λi gi k introduce k-dimensional vector λ for the undetermined quantities Lagrange Multiplier method Power Systems I ￿ Equality Constraints in Optimization ￿ f ( x, y ) = x 2 + y 2 The minimum distance is obtained by minimizing the distance squared (x − 8)2 + ( y − 6)2 = 25 or 2 2 g ( x, y ) = ( x − 8) + ( y − 6) − 25 Use the Lagrange multiplier method to determine the minimum distance from the origin of the x-y plane to a circle described by Power Systems I ￿ Example [ 2 2 2 2 g ( x, y ) = ( x − 8) + ( y − 6 ) − 25 = 0 Power Systems I L = f + λ ⋅ g = x 2 + y 2 + λ ( x − 8) + ( y − 6 ) − 25 ∂L = 2 x + λ (2 x − 16 ) = 0 or 2 x (λ + 1) = 16λ ∂x ∂L = 2 y + λ (2 y − 12 ) = 0 or 2 y (λ + 1) = 12λ ∂y ∂L 2 2 = ( x − 8) + ( y − 6 ) − 25 = 0 ∂λ f ( x, y ) = x 2 + y 2 Example 2 ȹ 3 (x − 8) + ȹ x − 6 ȹ − 25 = 0 ȹ ȹ 4 Ⱥ 25 2 x − 25 x + 75 = 0 → x = 4 & x = 12 16 extrema : (4,3), λ = 1 and (12,9 ), λ = −3 x=4 min → y=3 2 substituting for y in the third equation yields 16λ 12λ 3 = →y= x 2x 2y 4 eliminating λ from the first two equations Power Systems I ￿ ￿ Example ￿ ￿ ￿ ￿ ￿ 2 λ ȹ λ ȹ f (λ ) = 100ȹ + 75 = 0 ȹ + 200 λ +1 ȹ λ + 1 Ⱥ ȹ ∂f ȹ ȹ ∂λ ȹ Ⱥ ȹ () substitute the first two equations into the third equation the third equation is non-linear and in terms of a single variable, λ − Δf λ[k ] λ[k +1] = λ[k ] + Δλ[k ] Δλ[k ] = [k ] 6λ y= λ +1 rewrite the first two equations in terms of λ 8λ x= λ +1 ￿ The equations are solved iteratively Newton-Raphson method is superior One possible way of casting the last example into an iterative process: In many problems, a direct solution using Lagrange m u l t i p l i e r m e th o d i s n o t p o s s i b l e Power Systems I ￿ Iterative Techniques Starting with an estimated value of λ, a new value is found in the direction of steepest descent The process is repeated until the error,Δf(λ) is less than a specified accuracy This algorithm is known as the gradient method Numerical results of previous example, starting with a initial value λ = 0.4 iter Δf J Δλ λ x y 1 26.02 -72.89 0.357 0.400 2.286 1.713 2 7.393 -36.87 0.201 0.757 3.447 2.585 3 1.097 -26.66 0.041 0.958 3.913 2.935 4 0.034 -25.05 0.001 0.999 3.997 2.998 5 0.000 -25.00 0.000 1.000 4.000 3.000 Power Systems I ￿ ￿ ￿ ￿ Iterative Techniques operating efficiency of prime mover and generator fuel costs transmission losses may be located in an area with high fuel costs may be located far from the load centers and transmission losses are high The problem is to determine generation at different plants to minimize the total operating costs ￿ ￿ The most efficient generator in the system does not guarantee minimum costs ￿ ￿ ￿ Factors influencing the minimum cost of power generation Power Systems I ￿ ￿ ￿ Operating Costs ￿ ￿ cost Ci, \$/hr Fuel-Cost Curve dCi dPi = β i + 2γ i Pi The derivative is known as the incremental fuel cost Ci = α i + β i Pi + γ i Pi 2 Pi, MW Pi, MW The fuel cost is commonly express as a quadratic function fuel input, Btu/hr Heat-Rate Curve Generator heat rate curves lead to the fuel cost curves Power Systems I ￿ Operating Costs ￿ ￿ ￿ n gen i =1 i Demand i =1 ∑P = P n gen the total demand is equal to the sum of the generators’ output; the equality constrant i =1 Ctotal = ∑ Ci = ∑αi + β i Pi + γ i Pi 2 n gen minimize the objective or cost function over all plants a quadratic cost function is used for each plant The simplest problem is when system losses and generator limits are neglected Power Systems I ￿ Economic Dispatch → ∂Ctotal =λ ∂Pi ∂Ctotal dCi = = λ ∀i = 1,￿, ng ∂Pi dPi dCi λ= = β i + 2γ i Pi dPi i =1 Ctotal = ∑ Ci n gen ∂L ∂Ctotal = + λ (0 − 1) = 0 → ∂Pi ∂Pi L = Ctotal n gen ȹ ȹ + λ ȹ PDemand − ∑ Pi ȹ ȹ ȹ i =1 Ⱥ ȹ A typical approach using the Lagrange multipliers Power Systems I ￿ Economic Dispatch i Demand i =1 ∑P = P n gen λ − βi 2γ i λ − βi ∑ 2γ = PDemand i =1 i n gen Pi = gen βi PDemand + ∑ i =1 2γ i λ= n 1 ∑ 2γ i =1 i n gen rearranging and combining the equations to solve for λ n gen ȹ dL ȹ = ȹ PDemand − ∑ Pi ȹ = 0 → ȹ dλ ȹ i =1 Ⱥ ȹ the second condition for optimal dispatch Power Systems I ￿ ￿ Economic Dispatch PDemand = 800 MW C3 = 200 + 5.8 P3 + 0.009 P32 C2 = 400 + 5.5 P2 + 0.006 P22 C1 = 500 + 5.3P + 0.004 P 2 [\$ / MWhr ] 1 1 Neglecting system losses and generator limits, find the optimal dispatch and the total cost in \$/hr for the three generators and the given load demand Power Systems I ￿ Example n gen Power Systems I gen βi PDemand + ∑ .3 .5 .8 800 + 0.5008 + 0.5012 + 05018 i =1 2γ i . λ= = = \$8.5 / MWhr n 1 1 1 1 0.008 + 0.012 + 0.018 ∑ 2γ i =1 i 8.5 − 5.3 P= = 400 MW 1 2(0.004 ) λ − βi 8.5 − 5.5i Pi = P2 = = 250 MW 2γ i 2(0.006 ) 8.5 − 5.8 P3 = = 150 MW 2(0.009 ) PDemand = 800 = 400 + 250 + 150 Example 0 Power Systems I 7 7. 5 8 8. 5 9 9. 5 10 10. 5 Example \$ /MWh 200 400 600 P , MW Fuel Cost Curves 800 1000 1200 Δλ = [k ] ȹ ∂f ȹ ∂λ ȹ ȹ ȹ Ⱥ Δf (λ ) [k ] [k ] PD − f (λ ) = 1 1 1 + + 2γ 1 2γ 2 2γ 3 x = λ → λ[k +1] = λ[k ] + Δλ[k ] 1 ȹ ∂f ȹ = dPi = ȹ ∂λ ȹ ∑ dλ ∑ 2γ ȹ Ⱥ i Δy = Δf (λ ) = PD − f (λ ) λ − βi f (λ ) = ∑ 2γ i Solve again using iterative methods Power Systems I ￿ Example [0 ] [0 ] (659.7) =1 = 2.5 1 1 2 (0.004 ) + 2 (0.006 ) + 2 (0.009 ) Power Systems I ȹ ∂f ȹ ∂λ ȹ ȹ ȹ Ⱥ λ[1] = 6.0 + 2.5 = 8.5 Δλ = [0 ] Δf (λ ) [0 ] λ[0 ] − β i λ[0 ] = 6.0 f (λ ) = ∑ 2γ i [0 ] 6.0 − 5.3 [0 ] 6.0 − 5.5 P= = 87.5 P2 = = 41.7 1 2(0.004 ) 2(0.006 ) [0 ] 6.0 − 5.8 P3 = = 11.1 Δf = 800 − (87.5 + 41.7 + 11.1) 2(0.009) Example [1] [1] [1] [1] Δf (λ ) = Power Systems I ȹ ∂f ȹ ∂λ ȹ ȹ ȹ Ⱥ λ[2 ] = 8.5 + 0 = 8.5 Δλ = [1] 1 2 (0.004 ) + 1 2 (0.006 ) ( 0) + 1 2 (0.009 ) =0 λ[1] − β i λ = 8.5 f (λ ) = ∑ 2γ i [1] 8.5 − 5.3 [1] 8.5 − 5.5 P= = 400 P2 = = 250 1 2(0.004 ) 2(0.006 ) [1] 8.5 − 5.8 P3 = = 150 Δf = 800 − (400 + 250 + 150) 2(0.009 ) Example ...
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## This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.

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