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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
Ⱥ2 1 0Ⱥ Ⱥ x1 Ⱥ Ⱥ 8 Ⱥ
Ⱥ1 4 1Ⱥ Ⱥ x Ⱥ = Ⱥ16 Ⱥ
ˆ
Ⱥ Ⱥ 2 Ⱥ Ⱥ Ⱥ
Ⱥ
Ⱥ0 1 6Ⱥ Ⱥ x3 Ⱥ Ⱥ32Ⱥ
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 = Ⱥ2Ⱥ
ˆ
Ⱥ 2 Ⱥ Ⱥ Ⱥ
Ⱥ x3 Ⱥ Ⱥ5Ⱥ
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 Ⱥ2 1 0Ⱥ Ⱥ1.55Ⱥ Ⱥ0Ⱥ
eigen Ⱥ1 4 1Ⱥ = Ⱥ 4.0 Ⱥ > Ⱥ0Ⱥ
Ⱥ Ⱥ Ⱥ
Ⱥ
Ⱥ Ⱥ
Ⱥ0 1 6Ⱥ Ⱥ6.45Ⱥ Ⱥ0Ⱥ
Ⱥ Ⱥ Ⱥ
Ⱥ
Ⱥ Ⱥ using the MATLAB function eig(H), the eigenvalues are found Ⱥ2 1 0Ⱥ
ˆ
H ( x ) = Ⱥ1 4 1Ⱥ
Ⱥ
Ⱥ
Ⱥ0 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 kdimensional 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 xy 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 nonlinear 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
NewtonRaphson 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,
$/hr FuelCost Curve dCi dPi = β i + 2γ i Pi The derivative is known as the incremental fuel cost Ci = α i + β i Pi + γ i Pi 2 Pi, MW
Pi, MW
The fuel cost is commonly express as a quadratic function fuel
input,
Btu/hr HeatRate Curve Generator heat rate curves lead to the fuel cost curves Power Systems I Operating Costs n gen
i =1 i Demand i =1 ∑P = P n gen the total demand is equal to the sum of the generators’ output;
the equality constrant i =1 Ctotal = ∑ Ci = ∑αi + β i Pi + γ i Pi 2 n gen minimize the objective or cost function over all plants
a quadratic cost function is used for each plant The simplest problem is when system losses and
generator limits are neglected Power Systems I Economic Dispatch → ∂Ctotal
=λ
∂Pi ∂Ctotal dCi
=
= λ ∀i = 1,, ng
∂Pi
dPi dCi
λ=
= β i + 2γ i Pi
dPi i =1 Ctotal = ∑ Ci n gen ∂L ∂Ctotal
=
+ λ (0 − 1) = 0 →
∂Pi
∂Pi L = Ctotal n gen
ȹ
ȹ + λ ȹ PDemand − ∑ Pi ȹ
ȹ
ȹ i =1
Ⱥ
ȹ A typical approach using the Lagrange multipliers Power Systems I Economic Dispatch i Demand i =1 ∑P = P n gen λ − βi
2γ i λ − βi
∑ 2γ = PDemand
i =1
i n gen Pi = gen βi
PDemand + ∑
i =1 2γ i
λ=
n
1
∑ 2γ
i =1
i n gen rearranging and combining the equations to solve for λ n gen
ȹ
dL ȹ = ȹ PDemand − ∑ Pi ȹ = 0 →
ȹ
dλ ȹ i =1
Ⱥ
ȹ the second condition for optimal dispatch Power Systems I Economic Dispatch PDemand = 800 MW C3 = 200 + 5.8 P3 + 0.009 P32 C2 = 400 + 5.5 P2 + 0.006 P22 C1 = 500 + 5.3P + 0.004 P 2 [$ / 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.
 Fall '11
 THOMASBALDWIN

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