Chp7 3 Economic Dispatch

(1)

Chapter 7 Optimal Dispatch of Generation Part III

7.5 Economic Dispatch Neglecting Losses and

Including Generator Limits

In the real world, generators have limits on their generation, both minimum and maximum generation. Thus we have the inequality constraints

(min) i (max) 1,2, ,

i i

P ≤ ≤P P i= K ng

which are in addition to the equality constraint

1 ng i D i P P = =

∑

Now the Kuhn-Tucker conditions compliment the Lagrangian conditions to include the inequality constraints as additional terms. The necessary conditions for optimal dispatch in lossless systems but including generator limits become:

( ) ( ) ( ) ( ) min max max min for for for i i i i i i i i i i i i i dC P P P dP dC P P dP dC P P dP λ λ λ = ≤ ≤ ≤ = ≥ =

The numerical solution is the same as before. Thus for an estimated value of lambda, the powers are found for each participating generator. If the power is outside the limits of a generator, that generator is "pegged" at that limit (either Pmin or Pmax), and the

generator is no longer a participating generator in the optimization of dispatched power (since its power is pegged at either the high or low limit).

Example 7.6

Find the optimal dispatch and the total cost in $/h for the thermal plants of Example 7.4 when the total load is 975 MW with the following generator limits (in MW):

1 2 3 200 450 150 350 100 225 P P P ≤ ≤ ≤ ≤ ≤ ≤

Assume the initial value of λ( )1 =6. From the coordination equations we compute the powers as follows:

(2)

( )

(

)

( )

(

)

( )

(

)

1 1 1 2 1 3 6 5.3 87.5 2 0.004 6 5.5 41.67 2 0.006 6 5.8 11.11 2 0.009 P P P − = = − = = − = = ( )1

₍

₎

975 87.5 41.67 11.11 834.72 P ∆ = − + + =

and we compute change in lambda as ( )

(

)

(

)

(

)

1 834.72 3.1632 1 1 1 2 0.004 2 0.006 2 0.009 λ ∆ = = + +

Therefore the new value of lambda is ( )2

6 3.1632 9.1632

λ = + =

Continuing for a second iteration we have ( )

(

)

( )

(

)

( )

(

)

2 1 2 2 2 3 9.163 5.3 482.89 2 0.004 9.163 5.5 305.26 2 0.006 9.163 5.8 186.84 2 0.009 P P P − = = − = = − = = and ( )2

₍

₎

975 482.89 305.26 186.84 0 P ∆ = − + + =

Thus the equality constraint is met, but the power of generator No. 1 exceeds its upper limit. Thus we change it from 482.89 to 450, thus:

( )2

₍

₎

975 450 305.26 186.84 32.89

P

∆ = − + + =

Now we continue, but we have only two generators participating in the optimal dispatch since the first generator is "pegged" at 450:

( )

(

) (

)

2 32.89 0.2368 1 1 2 0.006 2 0.009 λ ∆ = = + and the new value of lambda is

( )3

9.1632 0.2368 9.4

λ = + =

(3)

( ) ( )

(

)

( )

(

)

3 1 3 2 3 3 450 9.4 5.5 325 2 0.006 9.4 5.8 200 2 0.009 P P P = − = = − = = and ( )3

₍

₎

975 450 325 200 0 P ∆ = − + + =

Thus the equality constraint is met and the above is the optimal generation dispatch. The total generation cost is found using the powers found above thus:

2 2 2 500 5.3 450 0.004 450 400 5.5 325 0.006 325 200 5.8 200 0.009 200 8,236.25 $/h t C = + × + × + + × + × + + × + × =

We could solve this using the following Matlab commands:

cost=[500 5.3 0.004 400 5.5 0.006 200 5.8 0.009]; mwlimits=[200 450 150 350 100 225]; Pdt=975; dispatch gencost

Incremental cost of delivered power (system lambda) = 9.400000 $/MWh Optimal Dispatch of Generation:

450.0000 325.0000 200.0000

Total generation cost = 8236.25 $/h

7.6 Economic Dispatch Including Losses

When the loads and generators are in a small geographical region, transmission losses may be neglected and the optimal dispatch of generation is achieved with all plants operating at equal incremental productions cost. However, in a large system over long distances, transmission losses become significant and need to be considered. We shall use Kron's loss formula which is

0 00 1 1 1 ng ng ng L i ij j i i i j i P P B P B P B = = = =

∑∑

+

∑

+

(4)

The coefficients B are called _ij loss coefficients or B-coefficients. It is assumed —with little error— that these coefficients are constant (as long as operation is near the value where these coefficients are computed).

The economic dispatch problem is to minimize the overall generation cost C , which is a _i

function of the plant outputs

(

)

1 2 1 ng t i i n i i i i C C P P α β γ = = = = + +

∑

Note: If all generators participate in this minimization, then n=ng and either equation above may be used. If only ng<n generators are participating, then the first equation is used.

This minimization is to be performed subject to the constraint that the total generation must equal the demand plus the losses thus

1 ng i D L i P P P = = +

∑

satisfying the inequality constraints:

(min) i (max) 1, 2, ,

i i

P ≤ ≤P P i= K ng

where P_i_{( )}_min and P_i₍_max₎ are the minimum and maximum generating limits, respectively, for plant i.

Using the Lagrange multiplier and adding additional terms to include the inequality constraints, we have

(max)

(

(max)

)

(min)

(

( )min

)

1 1 1 ng ng ng t D L i i i i i i i i i i C λ P P P µ P P µ P P = = =   = + _ + − _+ − + − 

∑



∑

L

The constraints should be understood to mean that µ_i₍_max₎ =0 when P_i < P_i₍_max₎ and (min) 0

i

µ = when P_i > P_i_{( )}_min . Thus if the power is not beyond the limits both µ_i₍_min₎ =0 and µ_i₍_max₎ =0 thus L has only the equality constraint. If a generator's power is beyond one of the limits, then that limit is the power to which the generator is "pegged", and the remaining generators participate in the opti mization of the load dispatch. These

(5)

( ) ( ) ( ) ( ) max max min min 0 0 0 0 i i i i i i i P P P P P λ µ µ ∂ = ∂ ∂ = ∂ ∂ _{= −} ₌ ∂ ∂ _{= −} ₌ ∂ L L L L

The last two equations imply that the power should not go beyond its limit, and when it is within its limits then µ_i_{( )}_min =µ_i₍_max₎ =0 and the Kuhn-Tucker function becomes the same as the Lagrangian one.

The first equation results in

0 1 0 t L i i C P P λ P   ∂ ₊ ₊∂ _{− =}   ∂ _ ∂ _

Since C_t =C₁+C₂+L+C_ng the above equation becomes

1,2, , ** i L i i C P i ng P λ P λ ∂ ₊ ∂ ₌ ₌ ∂ ∂ K The term L i P P ∂

∂ is known as the incremental transmission loss. The second condition is

given by 1 ng i D L i P P P = = +

∑

which is precisely the equality constraint imposed by the losses and load demand. The classical form of the previous equation is

1 1,2, , 1 i L _i i C i ng P _P P λ    _{ ∂} = =  _∂  ∂  ₋   _∂    K or 1,2, , i i i C L i ng P λ ∂ = = ∂ K

where L is known as the_i penalty factor of plant i and is given by 1

1 L i P P      _∂   ₋   _∂    . Thus the

effect of transmission losses is to introduce a penalty factor with a value that depends on the location of the plant with respect to the loads. In the above equation it is clear that the

(6)

most economic cost of dispatch is obtained when the incremental cost multiplied by the corresponding penalty factor are equal for all the participating plants.

The incremental transmission loss can be found from the loss equation and is given by 0 1 2 ng L ij j i i j P B P B P ₌ ∂ ₌ ₊ ∂

∑

and the incremental production cost is given by 2 i i i i i dC P dP = γ +β

Using these two equations in the equation marked ** (reproduced below for convenience)

1,2, , ** i L i i C P i ng P λ P λ ∂ ₊ ∂ ₌ ₌ ∂ ∂ K we have 0 1 2 2 ng i i i ij j i j P B P B β γ λ λ λ = + +

∑

+ =

or (divide by 2λ and pull out β from the summation) _ii 0 1, 1 1 * 2 ng i i ii i ij j i j j i B P B P B γ β λ ₌ _≠ λ  ₊  ₊ ₌  ₋ ₋       

∑

 

Expanding the above equation for all plants and detailing it into full matrix form we have

1 1 11 12 1 01 1 2 2 2 21 22 2 02 1 2 0 1 1 1 2 1 ng ng ng ng ng ng ng ngng ng B B B B P P B B B B P B B B B γ β λ λ γ β λ λ γ β λ λ  ₊   ₋ ₋               ₊   ₋ ₋     _ _=          _ _         + − −         L L M M M O M M L or in short matrix form

EP=D

To find the optimal dispatch first an estimated guess of λ, say λ( )1 is made. The simultaneous linear equations above are solved using Matlab (P = E\D). Then the

iterative process is continued using the gradient method. Thus, from equation * above we have P at the kth iteration expressed as _i

( ) ( )

(

)

( ) ( ) ( )

(

)

0 1 2 *** 2 k k k i i ij j k j i i _k i ii B B P P B λ β λ γ λ ≠ − − − = +

∑

(7)

( ) ( )

(

)

( ) ( ) ( )

(

)

( ) 0 1 1 1 2 2 k k k ng ng i i ij j k j i k D i _k L i i _i _ii B B P P P P B λ β λ γ λ ≠ = = − − − = = + +

∑

or

( )

( )k ( )k D L f λ =P +P

Expanding the left side above in a Taylor series and keeping only first order terms results in

( )

( )k

( )

( )k ( )k D L df f P P d λ λ λ λ   +_ _∆ = +   or ( ) ( ) ( )

( )

( ) ( )

( )

( ) ( ) ( ) 1 # ng k k D L i k i k k k k k i P P P df d P df d P dP d λ λ λ λ λ λ = + − ∆ =       ∆ =       ∆ =      

∑

where ( ) ( ) ( ) 1 ng k k k D L i i P P P P = ∆ = + −

∑

+ + and ( )

₍

₎

( ) ( )

(

)

0 2 1 1 1 2 2 k k ng ng i i ii i i j i ij j i k i i i ii B B B P P B γ β γ λ _γ _λ ≠ = = − − − ∂   ₌ _{+ + +} _∂    ₊

∑

This last equation is derived by differentiation with respect to λ of *** above and summing the result

2 recall that u v u u v v v ∂ − ∂  _∂ ₌     _{ }   . Now we have ( )k 1 ( )k ( )k λ + =λ + ∆λ

The process is repeated till ∆P( )k is less than a specified accuracy. If we simplify the loss equations and use the simpler

(8)

2 1 ng L ii i i P B P = =

∑

(where we assume B_ij =0 and B₀₀ =0), the solution of the simultaneous equations simplifies to ( ) ( ) ( )

(

)

## 2 k k _i i _k i ii P B λ β γ λ − = +

which are known as the coordination equations and equation +++ simplifies to ( ) ( )

(

)

2 1 1 ### 2 k ng ng i i ii i k i i i ii P B B γ β λ _γ _λ = = ∂ −   ₌ _∂    ₊

∑

The iteration process remains the same using these simpler functions. Example 7.7

The fuel cost in $/h of three thermal plants of a power system are 2 1 1 1 2 2 2 2 2 3 3 3 200 7.0 0.008 180 6.3 0.009 140 6.8 0.007 C P P C P P C P P = + + = + + = + +

where the powers are in MW. The plant outputs are subject to the following limits 1 2 3 10 MW 85 MW 10 MW 80 MW 10 MW 70 MW P P P ≤ ≤ ≤ ≤ ≤ ≤

For this example, assume the power loss is given by the simplified expression ( ) 0.0218 12( ) 0.0228 2( )2 0.0179 32( )

L pu pu pu pu

P = P + P + P

where the loss coefficients are specified per unit on a 100-MVA base. Determine the optimal dispatch of generation when the total system load is 150 MW.

First we change the power loss equation above to actual units instead of pu:

2 2 2

1 2 3

0.000218 0.000228 0.000179 L

P = P + P + P

where the powers are now in MW.

(9)

( )

(

)

( )

(

)

( )

(

)

1 1 1 2 1 3 8.0 7.0 51.3136 MW 2 0.008 8.0 0.000218 8.0 6.3 78.5292 MW 2 0.009 8.0 0.000228 8.0 6.8 71.1575 MW 2 0.007 8.0 0.000179 P P P − = = + × − = = + × − = = + ×

The real power loss is

( )1

₍

₎

2

₍

₎

2

₍

₎

2

0.000218 51.3136 0.000228 78.5292 0.000179 71.1575 2.886 L

P = + + =

Since P_D =150 MW, the error is (from equation ##)

( )1

₍

₎

150 2.8864 51.3136 78.5292 71.1517 48.1139

P

∆ = + − + + = −

and from equation ### : ( )

(

)

(

)

(

)

1 3 2 2 1 2 0.008 0.000218 7.0 0.009 0.000228 6.3 2 0.008 8.0 0.000218 2 0.009 8.0 0.000228 0.007 0.000179 6.8 152.4924 2 0.007 8.0 0.000179 i i P λ = ∂ + × + ×  _{ =} ₊ ₊ _∂    + × + × + × ₌ + ×

∑

From equation # we have:

( )1 48.1139

0.31552 152.4924

λ −

∆ = = −

Hence the new estimate for λ is ( )2

8.0 0.31552 7.6845

λ = − =

Repeating the above steps it is found at the fourth step that the change in lambda becomes very small and convergence is declared at

( )4

7.6789

λ λ= = And the optimal dispatch for λ=7.6789 is given by

( )

(

)

( )

(

)

( )

(

)

4 1 4 2 4 3 7.6789 7.0 51.3136 MW 2 0.008 7.6789 0.000218 7.6789 6.3 64.1317 MW 2 0.009 7.6789 0.000228 7.6789 6.8 52.4767 MW 2 0.007 7.6789 0.000179 P P P − = = + × − = = + × − = = + ×

The real power loss is

( )4

₍

₎

2

₍

₎

2

₍

₎

2

0.000218 35.0907 0.000228 64.1317 0.000179 52.4767 1.699 L

P = + + =

and the total fuel cost is

(

)

(

)

(

)

(

)

(

)

(

)

2 2 2 200 7.0 35.0907 0.008 35.0907 180 6.3 64.1317 0.009 64.1317 140 6.8 52.4767 0.007 52.4767 1592.65 $/h t C = + + + + + + + + =

(10)

The "dispatch" program can be used to solve example 7.7. The program is designed so the loss coefficients are specified in pu. The loss coefficients are arranged in a matrix form with the reserved name "B". The base MVA must be specified in the reserved variable name "basemva". If the base MVA is not specified, it is set to 100 MVA. The following Matlab commands are then used to solve example 7.7:

cost=[200 7.0 0.008 180 6.3 0.009 140 6.8 0.007]; mwlimits=[10 85 10 80 10 70]; Pdt=150; B=[0.0218 0 0 0 0.0228 0 0 0 0.0179]; basemva=100; dispatch gencost

35.0907 64.1318 52.4767

Total generation cost = 1592.65 $/h

It is very impressive how much less time it takes to solve example 7.7 using

the Matlab code…

Example 7.8

Figure 7.7 page 295 of the text shows the one-line diagram of a 5-bus 3-generator power system. Given the loss coefficients (pu on a 100 MVA base) are as follows:

[

]

[

]

0 00 0.0218 0.0093 0.0028 0.0093 0.0228 0.0017 0.0028 0.0017 0.0179 0.0003 0.0031 0.0015 0.00030523 B B B     =       = =

Cost functions, generator limits, and total loads are as given in example 7.7. Use "dispatch" program to find the optimal dispatch of generation.

(11)

The following Matlab commands are used: cost=[200 7.0 0.008 180 6.3 0.009 140 6.8 0.007]; mwlimits=[10 85 10 80 10 70]; Pdt=150; B=[0.0218 0.0093 0.0028 0.0093 0.0228 0.0017 0.0028 0.0017 0.0179]; B0=[0.0003 0.0031 0.0015]; B00=0.00030523; basemva=100; dispatch gencost

33.4701 64.0974 55.1012