1. Project on Economic Load dispatch

4. Heat-Rate Curve (a) fuel input, Btu/hr Pi, MW

5. Fuel-Cost Curve (b) The fuel cost is commonly express as a quadratic function Ci=αi+ βiPi+ γiPi2 Rs/h cost Ci, Rs/hr Pi, MW

6. incremental fuel-cost curve The derivative is known as the incremental fuel cost. λ Rs/MWh Pi, MW

8. calculating the corresponding incremental fuel cost λ of the plant

9. 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

10. Economic Load dispatch 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 = /

11. Economic Load dispatch 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

12. Economic Load dispatch 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 Bijare called loss coefficients or B-coefficients Making use of the Langrangian multiplier λ L= Ct + λ (PD+PL- )

13. Economic Load dispatch 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

14. Economic Load dispatch 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

16. The problem is to find the real power generation for each plant such that cost are minimized, subject to:

17. Meeting load demand - equality constraints

20. The total spinning reservefrom 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.

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

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

23. The output powerof the generating units must be greater or equalto the minimum powerof the generating units.

25. Unit input-output characteristics is linear between zero output and full load.

26. Start up cost are fixed amount.

27. There are no other constraints.

29. The lambda-iteration method is a good approach for solving the economic dispatch problem.

30. generator limits are easily handled

31. penalty factors are used to consider the impact of losses

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

34. program to solve economic dispatch problem 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 fsolveMatlab 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 fsolveMatlab 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:

35. program to solve economic dispatch problem % 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;

36. program to solve economic dispatch problem % 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([’Cost of Dispatching Unit 2: ’ num2str(F2)]) disp([’Cost of Dispatching Unit 3: ’ num2str(F3)]) disp([’Total Cost of ED: ’ num2str(F1+F2+F3)])

37. program to solve economic dispatch problem % 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 global P1min P2min P3min P1max P2max P3max % 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

38. program to solve economic dispatch problem 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;

39. program to solve economic dispatch problem 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 Cost of Dispatching Unit 2: 248.3527 Cost of Dispatching Unit 3: 407.8259 Total Cost of ED: 2320.5021 PLOSS = 6.8484

40. program to solve economic dispatch problem % PART B OF THE PROBLEM % set global variables global P1 P2 P3 Lambda IL1 IL2 IL3 PF1 PF2 PF3 PLOSS B PLOAD 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; % 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)];

41. program to solve economic dispatch problem % 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

42. program to solve economic dispatch problem 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([’Cost of Dispatching Unit 1: ’ num2str(F1)]) disp([’Cost of Dispatching Unit 2: ’ num2str(F2)]) disp([’Cost of Dispatching Unit 3: ’ num2str(F3)]) disp([’Total Cost of ED: ’ num2str(F1+F2+F3)]) disp(’...................................................................’) % PART ii of Part B of the Problem CR1 = Lambda*(1/PF1); CR2 = Lambda*(1/PF2); CR3 = Lambda*(1/PF3); disp(’Part b - ii’)

43. program to solve economic dispatch problem disp(’-----------’) disp([’Penalty Factor of unit 1:’ num2str(PF1)]) disp([’Penalty Factor of unit 2:’ num2str(PF2)]) disp([’Penalty Factor of unit 3:’ num2str(PF3)]) disp([’Cost Rate of unit 1 :’ num2str(CR1)]) disp([’Cost Rate of unit 2 :’ num2str(CR2)]) disp([’Cost Rate of unit 3 :’ num2str(CR3)]) 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

44. program to solve economic dispatch problem Function File: function F = dispatchloss_eqs_p43(z) global P1 P2 P3 Lambda IL1 IL2 IL3 PF1 PF2 PF3 PLOSS B PLOAD global P1min P2min P3min P1max P2max P3max % 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

45. program to solve economic dispatch problem 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

46. program to solve economic dispatch problem 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) ;

47. program to solve economic dispatch problem 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 Cost of Dispatching Unit 1: 1662.4798 Cost of Dispatching Unit 2: 503.1612 Cost of Dispatching Unit 3: 206.8946 Total Cost of ED: 2372.5356 ................................................................... Part b - ii ----------- Penalty Factor of unit 1:1.0482 Penalty Factor of unit 2:1.0172 Penalty Factor of unit 3:1.1384 Cost Rate of unit 1 :10.2826 Cost Rate of unit 2 :10.5956 Cost Rate of unit 3 :9.468

48. program to solve economic dispatch problem ................................................................... Part b - iii ------------ PLOSS1 = 4.5121 ................................................................... Part b - iv ------------ Resulting load supplied: Psupplied = Ploss_calculated + Pload Psupplied = 194.51

49. Economic Load dispatch 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.

50. Thank you