economic load dsipatch-fixed

8/2/2019 Economic Load Dsipatch-Fixed

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Economic Load Dispatch

By Engr. Khurram Hashmi


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The need for Optimization

In a given Power system, there are a number

ofgenerating units, Each of which may run on

a different source of fuel.

The individual energy units (in kWh or MWh)

produced by these Generating Sets shall have

different production costs.

Look at the chart on the next slide


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Per MWh. Prices

173.5

3063.1

15632

14759.2

3323.8

1119.1

8570

11500

9121.3

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

H del Coal HSD RFO Gas Nuclear Im ort Iran Mixed Wind

Pak.Rs

Generation cost by Source

Rs/MWh

June 2011

Rs/MWh

Bar chart gives the cost of producing a MWh of electricity from various sources by NEPRA


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The need for Optimization

There is a need to realize the Most Economic

Mix from all these generation sources.

We need to Calculate The Power Each unit in a

System Should Generate (Pg) so that overall

we have an Overall Minimum Operating Cost

This is called Economic Load Dispatch

G1 G2 G3 G4

Total Pg


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Obtaining an equation of Price/ hour

We have an equation for the performance of each Thermal generating

Unit. This Gives the Output power in Terms of Required Heat Content in

(kCal / hour) or Tehrmal.Unit / hr

To get an equation in terms of Cost. / hourthe above is multiplied with

( price /kg * kg/kCal) = i.e. (price / cal)

Dont worry about alpha, beta and gamawell always have them pre-

defined


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A Price Equation

So now we have an equation in terms of price

Notice how books change variable names to a,b and

c. dont be bothered with this.

Usually you will be given a ,b and c already. But in

case youre not and are just given a price equation.You can finda , b andc by comparing them to the

above form


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Price Equations . .

Because we Have more than one generating units.. We will

always have more than one Price equations. One for each

unit

Upto .. N. . .

For now think of as just another variable


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Our Goal

Our Goal is to produce the Maximum amount of Powerfora Minimum Amount ofMoney.

So we minimize the Sum of all cost Functions

This is a purely mathematical procedure. And is Subjectto the constraint that

i.e. Power generated - Total Load - transmission loss = 0


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We can write this thing into one as

In simpler words :

The Lamda( ) that you can see is just a variable used.. Its called

Langrangian multiplier never mind what its called. its not important

Adding all the Price

equations we weretalking about

Add all the Pg

variables P1, P2,upto Pn


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Solving an Economic Dispatch Problem

there are three main ways to solve an economic Dispatch

problem:

Fixed Loss Dispatch Classical Economic Dispatch

Optimal Power Flow

Well look at them one at a time


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Fixed Loss Dispatch

Keeping our main equation in mind:

Here we assume that the loss in the system here, is a smallFIXEDpercentage of the generated power. Say 1%

Rest is Mathematics

Well start solving Example (9.4) Melioupolis .now. And Ill

explain how it happens on the way


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Example 9.4

The system Data is:

Were given a ,b and c

As there are two generating units. We will form TWO equations of the form:

Go ahead ..substitute a, b and c make these two equations.Youll get:

It is GIVEN that Total: Pload = 382 MW

Assuming a Ploss = 1% of this Pload = 0.01* 382 MW = 0.382 MW


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Now use this calculated data to form an equation according to

Go ahead.. Put these values..

You should get..

Now take partial differentials of this L first with respect to Pg1 then Pg2 thenlamda

Youll get three things


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Now were given a1, a2 ; b1, b2 and c1,c2

Use them to calculate this

Done?..

you should get:


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Now substitute this lamda ( )into

One by one and calculate Pg1 and Pg2

We have the individual power each unit should deliver.

economic load dsipatch-fixed

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