04-ECONOMIC DISPATCH 2

04-Economic Dispatch 2 EE570 Energy Utilization & Conservation

Professor Henry Louie

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Topics  Example 1  Example 2

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Dr. Henry Louie

Example 1  Consider two generators with the following cost curves and constraints: C1(P1) = 600 + 41P1 + 0.002P12 C2(P2) = 500 + 40P2 + 0.007P22 20  P1  400 20  P2  350

The total demand is 700MW. Find the optimal dispatch of P1 and P2

Example 1  We will solve this problem using MATLAB in an iterative process where we start by guessing the incremental cost, l and revise it while enforcing generator constraints

Example 1  Begin by defining system variables

Because this is an iterative method, We do not expect to arrive at the exact solution. However, we can set an error bound. We will make Sure that the solution is such that the total demand is met within 0.1 MW (699.9 to 700.1)

Example 1  Start with an initial guess of l. A reasonable guess is near the b coefficient of one of the cost functions. C1(P1) = 600 + 41P1 + 0.002P12 C2(P2) = 500 + 40P2 + 0.007P22

Example 1  Find an equation that relates a value of l to generation values dC1 IC1  l   0  41  0.004P1 dP1 IC2  l 

dC2  0  40  0.014P2 dP2

rearranging

P1 

l  41

0.004 l  40 P2  0.014

Recall that:

Example 1  Define variables that are used in administering the algorithm Placing bounds on what l can be Will speed up our algorithm. l will never Be negative, and we are reasonably sure It will not exceed 100. However, the upper bound can be set to a much greater value if unsure.

Example 1  We will use a “while” loop, since we do not know when the solution will be reached

Example 1  Now update l based on whether or not the previous guess of l resulted in too much or too little generation relative to demand

Note: this is within the “while” loop

Example 1  Update P1 to correspond to the new l and enforce generation limits, if needed

Example 1  Update P2 to correspond to the new l and enforce generation limits, if needed

Example 1  Display the updated generation values and end the “while” loop

“pause” lets you see the solution for this iteration. Press any key to advance to the next iteration. It is not needed to include the “pause”

Example 1  We converge on a solution at the 14th iteration:

Generator 1 outputs 400 MW (its limit) Generator 2 outputs 300.098 MW Note: P1 + P2 = 700.098MW, which is within 0.1MW of the demand

Example 1  The incremental costs are: • Generator 1: $42.6/hr • Generator 2: $44.2/hr

Incremental costs are not equal because generator 1 is at its limit

Example 1  Visualizing the results

Example 2  Consider the system without generator limits IC1  0.007PG1  4.1 $/MWh IC2  0.007PG2  4.1 $/MWh PL  0.001 PG2  50 

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Note: the incremental costs are the same

MW

PD1  300 MW PD2  50 MW

 Find the optimal generation for each generator and the power loss in the transmission line 1

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Dr. Henry Louie

Example 2  First find

PL  PG2

now find the penalty factor

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L1  L2 

Dr. Henry Louie

Example 2  First find

PL  0.002 PG2  50   0.002PG2  0.1 PG2

now find the penalty factor

L1  1 1 L2  1.1  0.002PG2

 At the solution, L1xIC1 = L2xIC2 = l 19

Dr. Henry Louie

L1  1 Li 

1 P 1 L PGi

i  2,...,m

Example 2

 We then have: IC1  0.007PG1  4.1 $/MWh IC2  0.007PG2  4.1 $/MWh L1  1 L2 

1 1.1  0.002PG2

 According to the optimal dispatch rule: 1  0.007PG1  4.1  l 1 0.007PG2  4.1  l  1.1  0.002PG2

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Dr. Henry Louie

Example 2

 Start with a guess for l, and update if needed (depending on the corresponding generation values) 1  0.007PG1  4.1  l 1 0.007PG1  4.1  l  1.1  0.002PG2

• it is easier to rewrite the equations as: PG1  PG2

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l  4.1

0.007 1.1l  4.1  0.007  0.002l Dr. Henry Louie

Example 2  Start with a guess of l = 5.0 PG1  PG2  PL 

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Dr. Henry Louie

Example 2  Start with a guess of l = 5.0 PG1  PG2 

5.0  4.1  128.6 0.007 1.1 5  4.1

0.007  0.002 5 

 82.4

PL  0.001  82.4  50   1.0 2

 Power to the load is 128.6 + 82.4 – 1 = 210 MW (the load is 350 MW)  We need more generation to meet the load, so increase l to 6.0 23

Dr. Henry Louie

Example 2  l = 6.0, what are the corresponding power and loss? PG1  PG2  PL 

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Dr. Henry Louie

Example 2  l = 6.0, what are the corresponding power and loss? PG1  PG2 

6  4.1  271.4 0.007 1.1  6   4.1

0.007  0.002 6 

 131.6

PL  0.001 131.6  50   6.7 2

 Power to the load is 271.4 + 131.6 – 6.7 = 396.3 MW (the load is 350 MW)  Now there is too much generation; we need to decrease l 25

Dr. Henry Louie

Example 2  After some more iterations, we find that l = 5.744 • PG1 = 234.90 MW • PG2 = 120.00 MW • PL = 4.9 MW

IC1 = $5.74/MWh IC2 = $4.94/MWh L2 = 1.16

Note: PG1 + PG2 = PD + PL

 This is close enough to the exact answer  Even though the generators have the same IC, due to losses, it is cheaper to have generator 1 supply more of the load 1

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300 MW 26

50 MW

Dr. Henry Louie

Economics Summary  The economic dispatch (ED) problem comes in several flavors depending on • loss considerations and generator limits

 The goal is to minimize the fuel costs needed to serve the load  We use a solution approach based on the incremental costs of generators  The solution approaches presented have used approximate system models (voltage, reactive power, line constraints, etc have been ignored) 27

Dr. Henry Louie