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lecture8 - i = 1, 2,￿, k u j ( x1 , x2 , ￿ xn ) ≤ 0 j...

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Unformatted text preview: i = 1, 2,￿, k u j ( x1 , x2 , ￿ xn ) ≤ 0 j = 1,2, ￿ , m and the inequality constraints gi ( x1 , x2 ,￿ xn ) = 0 subject to the equality constraints ) The Lagrange multiplier is extended to include the inequality constraints by introducing the m-dimensional vector µ of undetermined quantities ￿ ￿ ( Practical problems contain inequality as well as equality constraints Minimize the cost function f x1 , x2 , ￿ xn Power Systems I ￿ ￿ ￿ Inequality Constraints in Optimization j = 0 l = 1,￿, n ∂L = gi = 0 i = 1,￿, k ∂xl ∂λi ∂L = u j ≤ 0 j = 1,￿, m µ j u j = 0 & µ j > 0 j = 1,￿, m ∂µ j ∂L The resulting necessary conditions for contrained local minima of L are the following i L = f + ∑ λi gi + ∑ µ j u j The unconstrained cost function becomes Power Systems I ￿ ￿ Kuhn-Tucker Method ￿ f ( x, y ) = x 2 + y 2 The minimum distance is obtained by minimizing the distance squared u ( x, y ) = 2 x + y ≥ 12 constrained by (x − 8)2 + ( y − 6)2 = 25 or 2 2 g ( x, y ) = ( x − 8) + ( y − 6 ) − 25 Use the Kuhn-Tucker 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 Power Systems I ∂L ∂L = 2 x + 2λ ( x − 8) + 2 µ = 0 = 2 y + 2λ ( y − 6 ) + µ = 0 ∂x ∂y ∂L ∂L 2 2 = ( x − 8) + ( y − 6) − 25 = 0 = 2 x + y − 12 = 0 ∂λ ∂µ The resulting necessary conditions for constrained local minima of L = x 2 + y 2 + λ ( x − 8) + ( y − 6 ) − 25 + µ [2 x + y − 12] [ L = f + λ ⋅ g + µ ⋅u The cost function f ( x, y ) = x 2 + y 2 g ( x, y ) = ( x − 8) + ( y − 6 ) − 25 = 0 u ( x, y ) = 2 x + y ≥ 12 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 Generators have a minimum and maximum real power output limits Meeting load demand - equality constraints Constrained by the generator limits - inequality constraints ← Pi (min) < Pi < Pi (max) ← Pi = Pi (max) ← Pi = Pi (min) dCi dPi = λ dCi dPi ≤ λ dCi dPi ≥ λ The Kuhn-Tucker conditions ￿ ￿ The problem is to find the real power generation for each plant such that cost are minimized, subject to: ￿ The power output of any generator should not exceed its rating nor be below the value for stable boiler operation Power Systems I ￿ ￿ ￿ Economic Dispatch with Generator Limits PDemand = 975 MW 100 ≤ P3 ≤ 225 150 ≤ P2 ≤ 350 200 ≤ P ≤ 450 1 C3 = 200 + 5.8P3 + 0.009 P32 C2 = 400 + 5.5P2 + 0.006 P22 C1 = 500 + 5.3P + 0.004 P 2 [$ / MWhr ] 1 1 Neglecting system losses, find the optimal dispatch and the total cost in $/hr for the three generators and the given load demand and generation limits Power Systems I ￿ Example 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 2(0.006 ) 2γ i 8.5 − 5.8 P3 = = 150 MW 2(0.009 ) PDemand = 800 MW = 400 + 250 + 150 MW n gen Example ￿ [1] P2 = 6 .0 − 5 .5 = 41.7 2(0.006) (834.7) = 3.1632 1 1 1 2 (0.004 ) + 2 (0.006 ) + 2 (0.009 ) λ[2 ] = 6.0 + 3.1632 = 9.1632 ΔP [1] Δλ[1] = [1] = (∂P ∂λ ) ΔP [1] = 975 − (87.5 + 41.7 + 11.1) = 834.7 initial guess λ[1] = 6.0 [1] 6.0 − 5.3 = 87.5 P= 1 2(0.004 ) [1] 6.0 − 5.8 = 11.1 P3 = 2(0.009 ) Solve using iterative methods Power Systems I ￿ Example 9.16 − 5.5 = = 305 2(0.006) 9.16 − 5.3 = 483 2(0.004 ) [2 ] 9.16 − 5.8 = 187 P3 = 2(0.009 ) [2 ] P= 1 [ FIXED ] →P 1 = 450max (33) = 0.2368 1 1 2 (0.006 ) + 2 (0.009 ) Power Systems I λ[3] = 9.1632 + 0.2368 = 9.4 ΔP [2 ] Δλ[2 ] = [2 ] = (∂P ∂λ ) ∴ ΔP [2 ] = 975 − ( 450 + 305 + 187) = 33 P = 483 > 450max 1 [2 ] ΔP [2 ] = 975 − ( 483 + 305 + 187) = 0 P2 [2 ] λ[2 ] = 9.1632 Example [3] P3 = P ( [3] ) ( [3] ) 2 100 ≤ P3 = 200 ≤ 225 Power Systems I (325) + 0.006(325)2 + 400 + 5.5 2 + 200 + 5.8(200) + 0.009(200) = 8,236.25 $ / hr Ctotal = 500 + 5.3(450) + 0.004(450) 150 ≤ P2 = 325 ≤ 350 = 450 9.4 − 5.8 = 200 2(0.009 ) [ FIXED ] 1 ΔP [3] = 975 − ( 450 + 325 + 200) = 0 λ = 9.4 [3] 9.4 − 5.5 P2 = = 325 2(0.006) [3] Example transmission line losses are a major factor losses affect the optimum dispatch of generation simplest form: Kron’s loss formula: ￿ ￿ n gen i =1 j =1 j =1 PL = ∑ ∑ Pi Bij Pj + ∑ B0 j Pj + B00 n gen n gen i =1 j =1 PL = ∑ ∑ Pi Bij Pj One common practice for including the effect of transmission losses is to express the total transmission loss as a quadratic function of the generator power n gen n gen o u tp u t s ￿ ￿ For large interconnected system where power is transmitted over long distances with low load density areas Power Systems I ￿ ￿ Economic Dispatch including Losses they are assumed to be constant reasonable accuracy is expected when actual operating conditions are close to the base case conditions used to compute the coefficients ￿ ￿ the generation equals the total load demand plus transmission losses each plant output is within the upper and lower generation limits inequality constraints The economic dispatch problem is to minimize the overall generation cost, C, which is a function of plant output Constraints: ￿ ￿ Bij are called the loss coefficients Power Systems I ￿ ￿ ￿ Economic Dispatch including Losses n gen i demand i =1 ∑P = P + Plosses i =1 2 Power Systems I Pi < Pi (max ) : µi (max) = 0 i =1 + ∑ µi (min ) (Pi − Pi (min ) ) n gen L = Ctotal Pi > Pi (min ) : µi (min) = 0 n gen ȹ ȹ ngen + λ ȹ Pdemand + Plosses − ∑ Pi ȹ + ∑ µi ( max ) (Pi ( max ) − Pi ) ȹ ȹ i =1 i =1 ȹ Ⱥ The resulting optimization equation u : Pi (min ) ≤ Pi ≤ Pi ( max ) i = 1,￿, ngen g: n gen i =1 f : Ctotal = ∑ Ci = ∑αi +β i Pi + γ i Pi n gen Economic Dispatch including Losses ∂L = Pi − Pi (min) = 0 ∂µi (min) ∂L =0 ∂λ ∂L = Pi − Pi (max) = 0 ∂µi (max) The minimum of the unconstrained function is found w he n : ∂L =0 ∂Pi Power Systems I ￿ Economic Dispatch including Losses ) ∂L = 0 = PD + PL − ∑ Pi ∂λ i =1 n gen i =1 ∴ ∑ Pi = PD + PL n gen ȹ dCi dCi dCi 1 ∂PL ȹ ∴ λ= +λ = ȹ ȹ 1 − ∂P ∂P ȹ dP = Li dP ȹ dPi ∂Pi ȹ L i Ⱥ i i ( ȹ ȹ ∂L ∂Ctotal ∂PL =0= + λ ȹ 0 + − 1ȹ ȹ ȹ ∂Pi ∂Pi ∂Pi ȹ Ⱥ dCi ∂Ctotal ∂ C1 + C2 + ￿ + Cngen = = dPi ∂Pi ∂Pi When generator limits are not violated: Power Systems I ￿ Economic Dispatch including Losses ￿ ￿ 1 ∂PL 1− ∂Pi The effect of transmission losses introduces a penalty factor that depends on the location of the plant The minimum cost is obtained when the incremental cost of each plant multiplied by its penalty factor is the same for all plants Li = The incremental transmission loss equation becomes the p e n a l t y fa c t o r Power Systems I ￿ The Penalty Factor n gen j =1 n gen dCi = β i + 2γ i P dPi Power Systems I j ≠i β i ȹ ȹ γ i ȹ 1 ȹ ȹ + Bii ȹ Pi + ∑ Bij Pj = 2 ȹ 1 − B0i − ȹ λ Ⱥ ȹ Ⱥ ȹ λ j =1 n gen Rearrange the equation dCi ∂PL λ= +λ = β i + 2γ i Pi + 2λ ∑ Bij Pj + λB0i dPi ∂Pi j =1 gen ∂PL = 2∑ Bij Pj + B0i ∂Pi j =1 n i =1 j =1 PL = ∑ ∑ Pi Bij Pj + ∑ B0 j Pj + B00 n gen n gen The Penalty Factor EP=D Ⱥ γ 1 Ⱥ λ + B11 Ⱥ Ⱥ B21 Ⱥ Ⱥ ￿ Ⱥ B n1 Ⱥ Ⱥ or ￿ γ2 + B22 ￿ λ ￿ ￿ ￿ Bn 2 B12 β1 Ⱥ Ⱥ Ⱥ B1n Ⱥ Ⱥ 1 − B01 − λ Ⱥ Ⱥ P Ⱥ 1 Ⱥ Ⱥ Ⱥ Ⱥ β 2 Ⱥ B2 n Ⱥ Ⱥ P2 Ⱥ = 1 Ⱥ1 − B02 − Ⱥ λ Ⱥ Ⱥ Ⱥ ￿ Ⱥ 2 Ⱥ ￿ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ γn P β n Ⱥ Ⱥ + Bnn Ⱥ Ⱥ n Ⱥ Ⱥ1 − B0 n − λ Ⱥ Ⱥ λ Ⱥ Ⱥ Ⱥ Extending the equation to all plants results in the following linear equations (in matrix form) Power Systems I ￿ The Penalty Factor ￿ ￿ ￿ Pi [k ] λ − βi = 2 [k ] 2(γ i + λ Bii ) [k ] n gen Ⱥ γ i + Bii β i Ⱥ ȹ ∂Pi ȹ ∑ ȹ ∂λ ȹ =∑ Ⱥ 2(γ + λ[k ]B )2 Ⱥ Ⱥ i =1 Ⱥ i i =1 ȹ Ⱥ ii Ⱥ Ⱥ n gen the solution of the simultaneous equations reduces to i =1 PL = ∑ Bii Pi 2 n gen The simultaneous linear equation is solved, E P = D Then the iterative process is continued using the gradient method If an approximate loss formula is used To find the optimal dispatch for an estimated value of λ[1] Power Systems I ￿ Economic Dispatch including Losses ￿ ￿ 10 ≤ P2 ≤ 80 10 ≤ P3 ≤ 70 PDemand = 150 MW Ploss = 0.000218 P 2 + 0.000228 P22 + 0.000179 P32 1 real power loss and total load demand C2 = 180 + 6.3P2 + 0.009 P22 C3 = 140 + 6.8P3 + 0.007 P32 C1 = 200 + 7.0 P + 0.008P 2 [$ / hr ] 10 ≤ P ≤ 85 MW 1 1 1 fuel costs and plant output limits F i n d th e o p ti m a l d i s p a t c h a n d t h e t o t a l c o s t i n $ / h r Power Systems I ￿ Example [k ] 2 2 2 Power Systems I Ploss = 2.886 Ploss = 0.000218(51.3) + 0.000228(78.5) + 0.000179(71.2 ) [0] λ[ k ] − βi λ = 8.0 Pi = 2 2(γ i + λ[ k ] Bii ) 8.0 − 7.0 [1] P= = 51.3 1 2 2(0.008 + 8.0 ⋅ 0.000218) 8.0 − 6.3 P2[1] = = 78.5 2 2(0.009 + 8.0 ⋅ 0.000228) 8.0 − 6.8 [1] P3 = = 71.2 2 2(0.007 + 8.0 ⋅ 0.000179 ) First iteration with initial guess at 8.0 Example [k ] 3 Newton-Raphson γ i + Bii β i =∑ J the Jacobian 2 [k ] γ i =1 2 ( i + λ Bii ) 0.008 + 0.000218 × 7.0 0.009 + 0.000228 × 6.3 J [1] = + + 2 2 2(0.008 + 8.0 × 0.000218) 2(0.009 + 8.0 × 0.000228) 0.007 + 0.000179 × 6.8 = 152.5 2 2(0.007 + 8.0 × 0.000179 ) − 48.1 Δλ[1] = = −0.316 the improvement 152.5 λ[ 2 ] = λ[1] + Δλ[1] = 8.0 − 0.316 = 7.684 Power Systems I [k ] ȹ ∂Pi ȹ = ∑ ȹ ȹ i =1 ȹ ∂λ Ⱥ 3 ΔP[1] = 150 + 2.9 − (51.3 + 78.5 + 71.2 ) = −48.1 ΔP[ k ] = PD + PL[ k ] − (P[ k ] + P2[ k ] + P3[ k ] ) 1 Example Power Systems I Ploss = 1.717 (35.4 )2 + 0.000228(64.4 )2 + 0.000179(52.8)2 Ploss = 0.000218 λ[ k ] − β i Pi [ k ] = λ[ 2 ] = 7.684 2 [k ] 2(γ i + λ Bii ) 7.684 − 7.0 [2] P= = 35.4 1 2 2(0.008 + 7.684 ⋅ 0.000218) 7.684 − 6.3 [2] P2 = = 64.4 2 2(0.009 + 7.684 ⋅ 0.000228) 7.684 − 6.8 [2] P3 = = 52.8 2 2(0.007 + 7.684 ⋅ 0.000179 ) Second iteration Example [k ] 3 Newton-Raphson γ i + Bii βi =∑ J the Jacobian 2 [k ] γ i =1 2 ( i + λ Bii ) 0.008 + 0.000218 × 7.0 0.009 + 0.000228 × 6.3 J [2] = + + 2 2 2(0.008 + 7.68 × 0.000218) 2(0.009 + 7.68 × 0.000228) 0.007 + 0.000179 × 6.8 = 154.6 2 2(0.007 + 7.68 × 0.000179 ) − 0.84 Δλ[ 2 ] = = −0.00543 the improvement 154.6 λ[ 3] = λ[ 2 ] + Δλ[ 2 ] = 7.684 − 0.0.00543 = 7.679 Power Systems I [k ] ȹ ∂Pi ȹ = ∑ ȹ ȹ i =1 ȹ ∂λ Ⱥ 3 ΔP[ 2 ] = 150 + 1.7 − (35.4 + 64.4 + 52.8) = −0.84 ΔP[ k ] = PD + PL[ k ] − (P[ k ] + P2[ k ] + P3[ k ] ) 1 Example 2 2 Power Systems I Ploss = 1.699 Ploss = 0.000218(35.09 ) + 0.000228(64.14 ) + 0.000179(52.48) 2 λ[ k ] − β i Pi [ k ] = λ[ 3] = 7.679 2 [k ] 2(γ i + λ Bii ) 7.679 − 7.0 P [ 3] = = 35.09 1 2 2(0.008 + 7.679 ⋅ 0.000218) 7.679 − 6.3 [ 3] P2 = = 64.14 2 2(0.009 + 7.679 ⋅ 0.000228) 7.679 − 6.8 [ 3] P3 = = 52.48 2 2(0.007 + 7.679 ⋅ 0.000179 ) Third iteration Example Newton-Raphson [k ] 3 γ i + Bii βi =∑ J the Jacobian 2 [k ] γ i =1 2 ( i + λ Bii ) 0.008 + 0.000218 × 7.0 0.009 + 0.000228 × 6.3 J [ 3] = + + 2 2 2(0.008 + 7.679 × 0.000218) 2(0.009 + 7.679 × 0.000228) 0.007 + 0.000179 × 6.8 = 154.6 2 2(0.007 + 7.679 × 0.000179 ) − 0.0174 Δλ[ 3] = = −0.000113 the improvement 154.6 λ[ 4 ] = λ[ 3] + Δλ[ 3] = 7.684 − 0.0.000113 = 7.6789 Power Systems I [k ] ȹ ∂Pi ȹ = ∑ ȹ ȹ i =1 ȹ ∂λ Ⱥ 3 ΔP[ 3] = 150 + 1.697 − (35.09 + 64.14 + 52.48) = −0.0174 ΔP[ k ] = PD + PL[ k ] − (P[ k ] + P2[ k ] + P3[ k ] ) 1 Example 2 2 Power Systems I Ploss = 1.699 Ploss = 0.000218(35.09 ) + 0.000228(64.13) + 0.000179(52.47 ) 2 λ[ k ] − β i Pi [ k ] = λ[ 4 ] = 7.6789 2 [k ] 2(γ i + λ Bii ) 7.6789 − 7.0 P [ 3] = = 35.09 1 2 2(0.008 + 7.6789 ⋅ 0.000218) 7.6789 − 6.3 [ 3] P2 = = 64.13 2 2(0.009 + 7.6789 ⋅ 0.000228) 7.6789 − 6.8 [ 3] P3 = = 52.47 2 2(0.007 + 7.6789 ⋅ 0.000179 ) Forth iteration Example Power Systems I + 140 + 6.8(52.47 ) + 0.007(52.47 ) = 1592.65 $ / hr 2 2 + 180 + 6.3(64.13) + 0.009(64.13) Ctotal = 200 + 7.0(35.09 ) + 0.008(35.09 ) 2 ΔP[ 2 ] = 150 + 1.699 − (35.09 + 64.13 + 52.47 ) = 0.0 ΔP[ k ] = PD + PL[ k ] − (P[ k ] + P2[ k ] + P3[ k ] ) 1 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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