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diff --git a/Working_Examples/83/CH7/EX7.5/example_7_5.sce b/Working_Examples/83/CH7/EX7.5/example_7_5.sce new file mode 100755 index 0000000..aed5fe2 --- /dev/null +++ b/Working_Examples/83/CH7/EX7.5/example_7_5.sce @@ -0,0 +1,69 @@ +//Chapter 7 +//Example 7.5 +//page 264 +//To find savings when losses are coordinated +clear;clc; + +function [lamdaprev,Pg]=optimum(Pd) + n=2; //number of generators + Alpha=[0.02 0.04]; + Beta=[16 20]; + lamda=20; //initial guess for lambda + lamdaprev=lamda; + eps=1; //tolerance + deltalamda=0.25; //increment in lamda + Pgmax=[200 200]; + Pgmin=[0 0]; + Pg=100*ones(n,1); + while abs(sum(Pg)-Pd)>eps + for i=1:n + Pg(i)=(lamda-Beta(i))/Alpha(i); + if Pg(i)>Pgmax(i) then + Pg(i)=Pgmax(i); + end + if Pg(i)<Pgmin(i) then + Pg(i)=Pgmin(i); + end + end + if (sum(Pg)-Pd)<0 then + lamdaprev=lamda; + lamda=lamda+deltalamda; + else + lamdaprev=lamda; + lamda=lamda-deltalamda; + end + end +endfunction + +//the above function "optimum" doesn't coordinate losses + +//case(i) when the losses are included but not coordinated +[lamda_case1,Pg_case1]=optimum(237.04); +//since Pg2 does not supply transmission losses and the losses are supplied only by Pg1 +Pg2_1=Pg_case1(2); +//to get Pg1 we will solve Pg1+Pg2=0.001*Pg1^2+237.04 +//the above equation can be wriiten as (0.001*Pg1^2) - Pg1 +(237.04-Pg2) =0 +p=poly([0.001 -1 (237.04+Pg2_1)],"Pg1"); +Pg1_1=roots(p); +Pg1_1=Pg1_1(1); + +printf('\ncase(i) when the losses are included but not coordinated'); +printf('\nPg1=%0.2f MW Pg2=%0.2f MW',Pg1_1,Pg2_1); + +//case(ii) when the losses are also coordinated +//we have the solution for case(ii) from example_7_4 +Pg1_2=128.57; Pg2_2=125; //case(ii) + +printf('\n\ncase(ii) when the losses are coordinated'); +printf('\nPg1=%0.2f MW Pg2=%0.2f MW',Pg1_2,Pg2_2); + +//saving at plant 1 is +saving1=integrate('0.02*Pg1+16','Pg1',Pg1_2,Pg1_1); +printf('\n\nSaving at plant 1 due to loss coordination is = Rs %0.2f/hr',saving1); + +//saving at plant 2 is +saving2=integrate('0.04*Pg2+20','Pg2',Pg2_2,Pg2_1); +printf('\n\nSaving at plant 2 due to loss coordination is = Rs %0.2f/hr',saving2); + +//net savings achieved +printf('\n\nThe net saving achieved by coordinating losses while scheduling the recieved load of 237.04MW is Rs %0.2f/hr',saving1+saving2);
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