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 mdimensional
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 KuhnTucker 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 KuhnTucker method to determine the minimum
distance from the origin of the xy 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 KuhnTucker 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 ȹ1 − 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 NewtonRaphson 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 NewtonRaphson 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 NewtonRaphson [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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 Fall '11
 THOMASBALDWIN
 Elementary algebra, Electric power transmission

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