Lecture 9 – Load Flow, Economic Dispatch 1

Lecture 9

Load Flow, Economic Dispatch The load flow equations are of the form,

( )( )

,

,

P P V

Q Q V

δδ

=

=

Where 1

1

1

nb

npq

VV

V

δδ

δ −

=

=

�

�

Where nb is the number of buses, ng is the number of generator buses and nb – bg = npq is the number of PQ buses. The NR iterative formula is,

( )( )

11

1

,

,

k kk k

k k k k

P P VJ

V V Q Q V

δδ δδ

+−

+

− = + −

And the power mismatches at iteration k are,

( )( )

,

,

k k

k k

P P Vm

Q Q V

δ

δ

− = −

J is the Jacobian, P P

VJ

Q QV

δ

δ

∂ ∂ ∂ ∂ =∂ ∂ ∂ ∂

The iterations continue until the stopping condition is met,

m ε<

Lecture 9 – Load Flow, Economic Dispatch 2

Typically this algorithm converges to an accuracy of 0.001 pu in 3 or 4 iterations beginning at a flat voltage profile, 0 0δ = and 0 1V = pu. However, in heavily loaded systems the NR may diverge. There is some doubt as to whether the cause is numerical or physically unrealizable. In general, systems are designed not to operate at these loading levels. Nevertheless, cases have occurred throughout the world where there is no physically operating point. Specifically, consider the Jacobian, which is the hardest part to calculate. Since finding J and performing the operation 1J m− is time consuming, there are alternative ways to tackle this problem. Normally the inverse of the Jacobian is not calculated explicitly. The Jacobian is sparse, i.e. approximately 95% of its elements are zero, whereas the inverse of the Jacobian is full, i.e. all elements are non-zero. Solving J x m∆ = by a technique called LU decomposition and forward and backward substitution is much more efficient in terms of computational time and storage. This approach retains sparsity. Another way of solving this is by exploiting the decoupling properties of the Jacobian.

0

0

P P PV

JQ Q Q

V V

δ δ

δ

∂ ∂ ∂ ∂ ∂ ∂ =∂ ∂ ∂ ∂ ∂ ∂

�

This means that / 0P V∂ ∂ � and / 0Q δ∂ ∂ � . In other words the real power injections are very little sensitive to the voltage magnitudes and the reactive power injections are very little sensitive to the phase angles. The number of calculations is considerably reduced. E.g. if nb = 100 and npq = 80, in the couple case there are (nb-1+npq)=169 equations and unknowns, and in the uncoupled case there two sets: one with (nb-1)=99 equations and unknowns and another with npq = 80 equations and unknowns. The decoupling of ,P δ from ,Q V is possible because of the following. Recall that the real power injection at bus i is given by,

( )1

cosnb

i ij i j i j ijj

P Y VV δ δ ψ=

= − −∑

Where ijj

ij ijY Y e ψ= is the ijth element of the network admittance matrix Y . Similarly,

Lecture 9 – Load Flow, Economic Dispatch 3

( )1

sinnb

i ij i j i j ijj

Q Y VV δ δ ψ=

= − −∑

Consider the J-term /i kP V∂ ∂ and let us show that it is approximately zero near the flat voltage profile ( ) ( ), 0, 1Vδ = .

( )

( ) ( )cos

2 cos cos

ik i i k iki nb

kk k kk jk j k j kjkj k

Y V i kP

Y V Y V i kV

δ δ ψ

ψ δ δ ψ≠

− − ∀ ≠∂ = + − − ∀ =∂

∑

Therefore when 1V = and 0δ =

( )

( ) ( )cos

2 cos cos

ik iki nb

kk kk jk kjkj k

Y i kP

Y Y i kV

ψ

ψ ψ≠

− ∀ ≠∂ = + − ∀ =∂

∑

But ikψ is obtained from ikj

ik ik ikY Y e yψ= = − , that is the negative of the admittance of line connecting bus i to bus k, and is equal to

2 2

1 ik ikik

ik ik ik ik

r jxyr jx r x

−= − = − + +

And since rik << xik in transmission lines we can further simplify to,

902 2 2 2

jik ikik

ik ik ik ik

x xy j er x r x

=+ +

�

Consequently ( )cos 0ikψ � . A similar procedure is followed to show that / 0Q δ∂ ∂ � . Fast Decoupled Method (FDLF) The assumptions, in order to calculate the Jacobian, are,

( )

1

1ik ik

i j

Vr x

δ δ

=<<

− <<

Lecture 9 – Load Flow, Economic Dispatch 4

Shunt elements are ignored The mismatches are calculated without these simplifications. Hence the Jacobian matrix is constant for all iterations,

0kJ J k= ∀ DC Load Flow This method is not iterative, therefore is only approximate and the solution is not the exact solution of the load flow. The assumptions are the same as the fast decoupled load flow,

( )( )( )

190

1 ,

cos 1 ,

sin ,

ik ik ik

i j

i j

i j i j

Vr x

rad i j

i j

i j

ψ

δ δ

δ δ

δ δ δ δ

=<< → =

− << ∀

− = ∀

− = − ∀

Then the load flow equations are linear,

( )1

1

'

nb

i i jj ij

P ix

P B

δ δ

δ=

= − ∀

=

∑

There is a relation between the fast decoupled load flow and the DC load flow, and that is that the matrix 'B is the one used in the FDLF which can be expressed as,

( )( )

1

1

,' 00 '' ,

k kk

k k k

P P VBB V Q Q V

δδδ

+

+

− ∆ = ∆ −

Where ''B is the sub-block of 'B corresponding to the PQ buses. Applications of LF

1. Transmission planning. When adding new lines, transformers, and control devices you would have to run load flows.

2. Optimal power flow (OPF). We want to minimize cost of generation or minimize losses in the network subject to the load flow equations and to inequality constraints on the voltages, generation levels, and flows.

Lecture 9 – Load Flow, Economic Dispatch 5

3. Generation planning. Addition of new generating plants must satisfy LF and inequalities.

4. Reliability. How robust or fragile is the network when losing an element? Therefore the operation and planning of the power system must not be vulnerable to the loss of any single major component.

Economic Dispatch A network has generators and loads,

Network

Pg1

Pgng

Pd1+jQd1

Pdnp+jQdnp

Economic dispatch (ED), or more generally Optimal Power Flow (OPF), is a way of formulating the operation of a power system as an optimization problem with constraints.

( )1

minimizeg

ng

i giP iC P

=∑

Where Ci(Pgi) is the cost in $/hour of generating Pgi MW at generator i.

( ) $ /i giC P h

giPmingiP max

giP

The curve depends on the efficiency of the plant and the cost of fuel. The goal of a central operator is to find the required Pgi’s so that the operating cost is minimized. The simplest ED is formulated as follows,

Lecture 9 – Load Flow, Economic Dispatch 6

( )1

ming

ng

i giP iC P

=∑

subject to,

1min max

ng

gi di

gi gi gi

P P

P P P=

=

≤ ≤

∑

This formulation ignores the power flows in the network. And Pd is the total demand of all loads. The solution of this problem is as follows. For a general equation f(x) the minimum is obtain as,

( )minx

f x

The necessary condition for minimum is that 0dfdx

= . And the sufficient condition for a

minimum is that 2

2 0d fdx

> .

For a two variable function f(x1,x2), the necessary conditions are,

1 2

0; 0df dfdx dx

= =

And the sufficient condition is that the Hessian must be positive definite,

2 2

21 1 2

2 2

21 2 2

0

f fx x x

Hf f

x x x

∂ ∂ ∂ ∂ ∂ = > ∂ ∂ ∂ ∂ ∂

Consider for example ( ) ( ) ( )2 2

1 21 2f x x x= − + − , then,

( ) ( )1 2

2 21 2,

min 1 2x x

x x− + −

The necessary conditions are,

Lecture 9 – Load Flow, Economic Dispatch 7

( )

( )

1 11

2 22

2 1 0 1

2 2 0 2

f x xxf x xx

= − = ⇒ =∂

= − = ⇒ =∂

x1

x2

1

2

If we include now a constraint of the form x1+x2 = 5, then the minimum is obtain by the Lagrange formulation,

( ) ( ) ( )1 2 1 2 1 2, , , 5x x f x x x xλ λ− + −�L

And the necessary condition for a minimum are,

0

0

i

ix

λ

∂ = ∀∂∂ =∂

L

L

Then, for the example,

( )

( )

( )

11

22

1 2

2 1 0

2 2 0

5 0

xx

xx

x x

λ

λ

λ

∂ = − − =∂∂ = − − =∂∂ = − + − =∂

L

L

L

which gives,

Lecture 9 – Load Flow, Economic Dispatch 8

1

2

1 2

12

22

5 3

x

x

x x

λ

λ

λ

= +

= +

+ = = +

and the final result is,

1 22; 2; 3x xλ = = =

x1

x2

2

3

Applying this method to the ED problem, the Lagrangian takes the form,

( ) ( )1 1

,ng ng

g i gi gi di i

P C P P Pλ λ= =

= − −

∑ ∑L

For a minimum, the necessary conditions are,

1

0 0

0 0

i

gi i

ng

gi di

C iP P

P P

λ

λ =

∂∂ = → − = ∀∂ ∂

∂ = → − − = ∂ ∑

L

L

Example. Consider a system with 2 generators and one load. The costs of the generators are,

21 1 1

22 2 2

20 0.01 $ /

25 0.02 $ /g g

g g

C P P h

C P P h

= +

= +

Find the generation levels minimizing the cost and satisfying the demand Pd. The necessary conditions are,

Lecture 9 – Load Flow, Economic Dispatch 9

11

22

1 2

20 0.02

25 0.04

g

g

g g d

PP

PP

P P P

λ

λ

λ

∂⇒ + =

∂∂

⇒ + =∂∂

⇒ + =∂

L

L

L

Then,

1

2

20 50 10000.02

25 25 6250.04

g

g

P

P

λ λ

λ λ

−= = −

−= = −

Consequently,

( )

( )1

2

162575

50 162550 100075 75

25 162525 62575 75

d

g d

g d

P

P P

P P

λ +=

= + −

= + −

If Pd changes by ∆Pd, then the generation is adjusted as,

1

2

2313

g d

g d

P P

P P

∆ = ∆

∆ = ∆

This is because generator 1 is a little bit cheaper than generator 2 and therefore takes a larger share of the change. So if generator 1 is cheaper, why not load it at 100% of Pd? Compare as an exercise what the cost would be by doing this. The cost will be higher than the one obtained by the Lagrange formulation. The sharing amount determines the participation factors of each generator.

Lecture 9 – Load Flow, Economic Dispatch 10

11

0 1

ng

ii

i

gi i d

iP P

α

αα

==

≤ ≤ ∀∆ = ∆

∑

Where αi is the participation factor of generator i.

∑

∑

∑

1α

ngα

10gP

0gngP

0dP

dP

dP∆

1gP∆

gngP∆

If a generator reaches the maximum capacity then its α = 0. Depending on the type of plant, the participation factors vary. For example, nuclear plants have a constant power generation and the participation factor is close to zero. Hydroelectric plants can vary their participation factor more easily. In general, if Ci is a quadratic function,

20 2

ii i i gi gi

bC C a P P= + +

Then the necessary conditions of the Lagrangian are,

1,...,i i gigi

a b P i ngP

λ∂ = + = ∀ =∂

L

Solving for Pgi,

igi

i

aPb

λ −=

Lecture 9 – Load Flow, Economic Dispatch 11

Therefore the summation of the generation yields,

1 1

ng ngi

gi di i i

aP Pb

λ= =

−= =∑ ∑

And we can solve for λ,

1

11/

ngi

di i

ng

ii

aPb

bλ =

=

−=

∑

∑

Then generator k produces,

1 1/ 1/

ng ngi

d i ki ii

gkk

aP b ab

Pb

= =

− −

=∑ ∑

And the incremental dispatch is,

( )

1

1/

1/

d kgk ng

ii

P bP

b=

∆∆ =

∑

Finally, the participation factor of generator k is,

( )

1

1/

1/

kk ng

ii

b

bα

=

=∑