project on economic load dispatch

PROJECT ON ECONOMIC LOAD DISPATCH

ECONOMIC LOAD DISPATCH

What is economic dispatch?

The definition of economic dispatch is:

“The operation of generation facilities to produce energy at the lowest cost to reliably serve consumers, recognizing any operational limits of generation and transmission facilities”.

Most electric power systems dispatch their own generating units and their own purchased power in a way that may be said to meet this definition.

The factors influencing power generation at minimum cost are : operating efficiencies of generators fuel cost and transmission losses


if the plant is located far from the load centre, transmission losses may be considerably higher and hence the plant may be overly uneconomical. Hence, the problem is to determine the generation of different plants whereby the total operating cost is minimum. The operating cost plays an important role in the economic scheduling .The input to the thermal plant is generally measured in Btu/h, and the output is measured in MW.

The input-output curve of a thermal unit known as heat-rate curve as shown in Fig.

HEAT-RATE CURVE

(a)

fuelinput,Btu/hr

Pi, MW

FUEL-COST CURVE

(b)

The fuel cost is commonly express as a quadratic function

Ci=αi + βiPi + γiPi2 Rs/h

costCi,Rs/hr

Pi, MW

INCREMENTAL FUEL-COST CURVE

The derivative is known as the incremental fuel cost.

λRs/MWh

Pi, MW


An economic dispatch schedule for assigning loads to each unit in a plant can be prepared by

assuming various values of total plant output calculating the corresponding incremental fuel cost λ of the plant substituting the value of λ for λi in the equation for the

incremental fuel cost of each unit to calculate its output.

For a plant with two units having no transmission losses operating under economic load distribution the λ of the plant equals λi of each unit, and so

λ= dC1/dP1 = 2γ1 P1+β1 ; λ= dC2/dP2 = 2γ2 P2+β2; and


P1 + P2 = PD

or

where PD = total load demand

Solving for P1 and P2, we obtain

P1= (λ-β1)/2γ1 ; P2 = (λ-β2)/2γ2

and for Pi

Pi= (λ-βi)/2γi (coordination equation)

Thus an analytical solution can be obtained for λ as

λ = /


For a system with k generating units

Ct= C1+ C2+ C3+…………………+ Ck=

Where Ct = total fuel cost for all generating units

The total MW power input to the network from all the units is the sum

P1+ P2+…………….+ Pk=

Where P1, P2,……Pk = the individual outputs of the units injected to the network. The total fuel cost C of the system is a function of all the power plant output

The economic dispatch problem including transmission losses is defined as

Min Ct=

Subject to PD+PL- = 0 PL is the total transmission system loss


for including the effect of transmission losses is to express the total transmission loss as a quadratic function of the generator power outputs. The simplest function is

PL=

The coefficients Bij are called loss coefficients or B-coefficients

Making use of the Langrangian multiplier λ

L= Ct + λ (PD+PL- )


For minimum cost we require the derivative of L with respect to each Pi to equal zero.

Since PD is fixed and the fuel cost of any one unit varies only of the power output of that unit is varied, equation yields

=

+ λ (PD+PL- )] = 0


For each of the generating unit outputs P1, P2,……Pk. Because Ci depends on only Pi, the partial derivative of Ci can be replaced by the full derivative and above equation then gives

λ= {1/ }

for every value of i . This equation is often written in the form

λ=Li ( )

where Li is called the penalty factor of plant i

ECONOMIC LOAD DISPATCH WITH GENERATOR LIMIT

The power output of any generator should not exceed its

rating nor be below the value for stable boiler operation Generators have a minimum and maximum real power output

limits. The problem is to find the real power generation for each

plant such that cost are minimized, subject to: Meeting load demand - equality constraints Constrained by the generator limits - inequality constraints

ECONOMIC LOAD DISPATCH WITH GENERATOR LIMIT

The Kuhn-Tucker conditions

λ , Pi(min)<Pi<Pi(max)

<= λ, Pi=Pi(max)

>= λ, Pi=Pi(min)

UNIT COMMITMENT

System Constraints• The total output of all the generating units must be equal to the forecast value of the

system demand at each time-point.• The total spinning reserve from all the generating units must be greater than or equal

to the spinning-reserve requirement of the system. This can be either a fixed requirement in MW or a specified percentage of the largest output of any generating unit.

• Minimum up time : Once the unit is running, it should not be turned off immediately.

• Minimum down time: Once the unit is de-committed (off), there is a minimum time before it can be recommitted.

• The output power of the generating units must be greater or equal to the minimum power of the generating units.

• The output power of the generating units must be smaller or equal to the maximum power of the generating units.

PRIORITY LIST METHOD

• Simplest of all methods, but at the same time approximate also. Units are ranked according to their full load production cost rate and committed accordingly. Has to follow a shut down algorithm for the satisfaction of the minimum uptime and down time constraints and spinning reserve constraints.

LIMITS OF USING PRIORITY LIST METHOD

• No load cost is zero.• Unit input-output characteristics is linear between zero

output and full load.• Start up cost are fixed amount.• There are no other constraints.

ECONOMIC DISPATCH – SUMMARY

Economic dispatch determines the best way to minimize the current generator operating costs.

The lambda-iteration method is a good approach for solving the economic dispatch problem. generator limits are easily handled penalty factors are used to consider the impact of losses

Economic dispatch is not concerned with determining which units to turn on/off (this is the unit commitment problem).

Economic dispatch ignores the transmission system limitations.

PROGRAM TO SOLVE ECONOMIC DISPATCH PROBLEM

Problem 4.3.

Bi0 and B00 are neglected. Assume three units are on-line and have the following characteristics:

Unit 1:

H1 = 312.5 + 8.25P1 + 0.005P2 MBtu/h

50 ≤ P1 ≤ 250 MW

Fuel Cost = 1.05 Rs/MBtu

Unit 2:

H2 = 112.5 + 8.25P2 + 0.005P2 MBtu/h

5≤ P2 ≤150 MW

Fuel Cost = 1.217 Rs/MBtu

Unit 3:

H3 = 50 + 8.25P3 + 0.005P2 MBtu/h

5≤ P2 ≤150 MW

Fuel Cost = 1.1831Rs/MBtu


a. No Losses Used in Scheduling

i. Calculate the optimum dispatch and total cost neglecting losses for PD = 190 MW.

ii. Using dispatch and the loss formula, calculate the system losses. This can also be solved using the following Matlab script file and function file which make use of the fsolve Matlab function to solve the system of equations:

b. Losses Included in Scheduling

i. Find the optimum dispatch for a total generation of PD = 190 MW using the coordination equations and the loss formula.

ii. Calculate the cost rate.

iii. Calculate the total losses using the loss formula.

iv. Calculate the resulting load supplied. This can also be solved using the following Matlab script file and function file which make use of the fsolve Matlab function to solve the system of equations, the program is also capable of computing the incremental losses and the penalty factors which are calculated in each iteration within the function file:


% PART A OF THE PROBLEM

% set global variables

global P1 P2 P3 Lambda

global P1min P2min P3min P1max P2max P3max

% define Pi_min and Pi_max

P1min = 50;

P2min = 5;

P3min = 15;

P1max = 250;

P2max = 150;

P3max = 100;

% set initial guess of powers and system lambda

P10 = 142;

P20 = 15;

P30 = 32;

Lambda0 = 10;

% solve equations in function ednoloss_eqs_p43 using fsolve

z1 = fsolve(@ednoloss_eqs_p43,[P10 P20 P30 Lambda0],optimset(’MaxFunEvals’,10ˆ2,…

’MaxIter’,10ˆ2));%,’Display’,’iter’));

clc;


% compute cost of scheduling each unit

F1 = 308.125 + 8.6625*P1 + 0.0052500*P1ˆ2;

F2 = 136.9125 + 10.04025*P2 + 0.0060850*P2ˆ2;

F3 = 59.155 + 9.760575*P3 + 0.0059155*P3ˆ2;

FT = F1+F2+F3;

% output

disp(’Problem 4.3.’)

disp(’------------’)

disp(’Part a - i’)

disp(’P1 P2 P3 Lambda’)

disp(num2str(z1))

disp([’Cost of Dispatching Unit 1: ’ num2str(F1)])



disp([’Total Cost of ED: ’ num2str(F1+F2+F3)])


% PART ii OF THE PROBLEM

P = [P1; P2; P3];

B = [ 1.36255*10ˆ(-4) 1.75300*10ˆ(-5) 1.83940*10ˆ(-4);

1.75400*10ˆ(-5) 1.54480*10ˆ(-4) 2.82765*10ˆ(-5);

1.83940*10ˆ(-4) 2.82765*10ˆ(-4) 1.61470*10ˆ(-3)];

PLOSS = P’*B*P

Function File:

function F = ednoloss_eqs_p43(z)

global P1 P2 P3 Lambda


% Unknown variables

P1 = z(1);

P2 = z(2);

P3 = z(3);

Lambda = z(4);

if P3 < P3min

P3 = P3min;

else

if P3 > P3max

P3 = P3max;

end

end


if P2 < P2min

P2 = P2min;

else

if P2 > P2max

P2 = P2max;

end

end

if P1 < P1min

P1 = P1min;

else

if P1 > P1max

P1 = P1max;

end

end

% dL/dPi and dL/dLambda

F(1,1) = 8.6625 + 0.0105*P1 - Lambda;

F(2,1) = 10.04025 + 0.01217*P2 - Lambda;

F(3,1) = 9.760575 + 0.011831*P3 - Lambda;

F(4,1) = P1 + P2 + P3 - 190;


Execution of the Matlab Script file yields the solution for the problem as described below:

Problem 4.3.

------------

Part a - i

P1 P2 P3 Lambda

143.9936 11.02567 34.98076 10.17443

Cost of Dispatching Unit 1: 1664.3235



Total Cost of ED: 2320.5021

PLOSS =

6.8484


% PART B OF THE PROBLEM

% set global variables

global P1 P2 P3 Lambda IL1 IL2 IL3 PF1 PF2 PF3 PLOSS B PLOAD


% define Pi_min and Pi_max

P1min = 50;

P2min = 5;

P3min = 15;

P1max = 250;

P2max = 150;

P3max = 100;

% define system load

PLOAD = 190;

% define b-matrix

B = [ 1.36255*10ˆ(-4) 1.75300*10ˆ(-5) 1.83940*10ˆ(-4);

1.75400*10ˆ(-5) 1.54480*10ˆ(-4) 2.82765*10ˆ(-5);

1.83940*10ˆ(-4) 2.82765*10ˆ(-4) 1.61470*10ˆ(-3)];


% set initial guess of powers and system lambda

% this values are the results from part one of the problem

P10 = 143.9936;

P20 = 11.02567;

P30 = 34.98076;

Lambda0 = 10.17443;

% solve equations in function ednoloss_eqs_p43 using fsolve

z1 = fsolve(@dispatchloss_eqs_p43,...

[P10 P20 P30 Lambda0 1 1 1 1 1 1 1 1 1 1 1 1],optimset(’MaxFunEvals’,10ˆ2,...

’MaxIter’,10ˆ2));%,’Display’,’iter’));

clc;

% compute cost of scheduling each unit

F1 = 308.125 + 8.6625*P1 + 0.0052500*P1ˆ2;

F2 = 136.9125 + 10.04025*P2 + 0.0060850*P2ˆ2;

F3 = 59.155 + 9.760575*P3 + 0.0059155*P3ˆ2;

FT = F1+F2+F3;

% output


disp(’Problem 4.3.’)

disp(’...................................................................’)

disp(’Part b - i’)

disp(’----------’)

disp(’P1 P2 P3 Lambda’)

disp(num2str([z1(1,1) z1(1,2) z1(1,3) z1(1,4)]))




disp([’Total Cost of ED: ’ num2str(F1+F2+F3)])

disp(’...................................................................’)

% PART ii of Part B of the Problem

CR1 = Lambda*(1/PF1);



disp(’Part b - ii’)


disp(’-----------’)

disp([’Penalty Factor of unit 1:’ num2str(PF1)])



disp([’Cost Rate of unit 1 :’ num2str(CR1)])



disp(’...................................................................’)

disp(’Part b - iii’)

disp(’------------’)

P = [z1(1,1); z1(1,2); z1(1,3)];

PLOSS1 = P’*B*P

disp(’...................................................................’)

disp(’Part b - iv’)

disp(’------------’)

disp(’Resulting load supplied: Psupplied = Ploss_calculated + Pload’)

Psupplied = PLOSS1 + PLOAD


Function File:

function F = dispatchloss_eqs_p43(z)

global P1 P2 P3 Lambda IL1 IL2 IL3 PF1 PF2 PF3 PLOSS B PLOAD


% Unknown variables

P1 = z(1);

P2 = z(2);

P3 = z(3);

Lambda = z(4);

IL1 = z(5);

IL2 = z(6);

IL3 = z(7);

PF1 = z(8);

PF2 = z(9);

PF3 = z(10);

PLOSS = z(11);

% bound powers to their limits

if P3 < P3min

P3 = P3min;

else


if P3 > P3max

P3 = P3max;

end

end

if P2 < P2min

P2 = P2min;

else

if P2 > P2max

P2 = P2max;

end

end

if P1 < P1min

P1 = P1min;

else

if P1 > P1max

P1 = P1max;

end

end


IL1 = 2*(B(1,1)*P1 + B(1,2)*P2 + B(1,3)*P3);

IL2 = 2*(B(2,1)*P1 + B(2,2)*P2 + B(2,3)*P3);

IL3 = 2*(B(3,1)*P1 + B(3,2)*P2 + B(3,3)*P3);

PF1 = 1/(1 - IL1);

PF2 = 1/(1 - IL2);

PF3 = 1/(1 - IL3);

PLOSS = (P1ˆ2)*B(1,1) + P1*B(1,2)*P2 + P1*B(1,3)*P3+ ...

P2*B(2,1)*P1 + (P2ˆ2)*B(2,2) + P2*B(2,3)*P3+ ...

P3*B(3,1)*P1 + P3*B(3,2)*P2 + (P3ˆ2)*B(3,3);

F(1,1) = PF1*(8.6625 + 0.0105*P1) - Lambda;

F(2,1) = PF2*(10.04025 + 0.004*P2) - Lambda;

F(3,1) = PF3*(9.760575 + 0.011831*P3) - Lambda;

F(4,1) = PLOAD + PLOSS - (P1+P2+P3) ;


The results of the execution of the script file, shown below, yield the solution to the problem:

Problem 4.3.

...................................................................

Part b - i

----------

P1 P2 P3 Lambda

143.8125 35.70524 14.94276 10.778




Total Cost of ED: 2372.5356

...................................................................

Part b - ii

-----------

Penalty Factor of unit 1:1.0482



Cost Rate of unit 1 :10.2826




...................................................................

Part b - iii

------------

PLOSS1 =

4.5121

...................................................................

Part b - iv

------------

Resulting load supplied: Psupplied = Ploss_calculated + Pload

Psupplied =

194.51


Does including loss formula matter?

The inclusion of losses does matter; it can be seen from the results above that the consideration of losses in the scheduling changes the optimum economic dispatch, increases the total generation cost (since the total demand will now include the power dissipated in the resistive elements) and also the system lambda; the inclusion of losses it’s also a more realistic representation of what is happening in the real power system, which has losses.

THANK YOU

project on economic load dispatch

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