Xcos examples from textbooks and for blocksHEADmaster

8 files changed, 229 insertions, 0 deletions

diff --git a/Working_Examples/83/CH4/EX4.1/example_4_1.sce b/Working_Examples/83/CH4/EX4.1/example_4_1.sce
new file mode 100755
index 0000000..bc919b4
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.1/example_4_1.sce
@@ -0,0 +1,43 @@
+//Chapter 4
+//Example 4.1
+//page 103
+// to draw the per unit reactance diagram
+clear;clc;
+mvab=30; kvb=33; //MVA base and KVA base are selected
+
+gen1_mva=30; gen1_kv=10.5; gen1_x=1.6;  //Generator No.1 details
+gen2_mva=15; gen2_kv=6.6; gen2_x=1.2;  //Generator No.2 details
+gen3_mva=25; gen3_kv=6.6; gen3_x=0.56;  //Generator No.3 details
+
+t1_mva=15; t1_hv=33; t1_lv=11; t1_x=15.2; //Transformer T1 details
+t2_mva=15; t2_hv=33; t2_lv=6.2; t2_x=16; //Transformer T1 details
+
+tl_x=20.5; //Transmission line recatance
+
+//Loads are neglected as said in the problem
+
+tl_pu=(tl_x*mvab)/kvb^2;
+t1_pu=(t1_x*mvab)/kvb^2;
+t2_pu=(t2_x*mvab)/kvb^2;
+gen1_kv_base=t1_lv;
+gen1_pu=(gen1_x*mvab)/gen1_kv_base^2;
+gen2_kv_base=t2_lv;
+gen2_pu=(gen2_x*mvab)/gen2_kv_base^2;
+gen3_pu=(gen3_x*mvab)/gen2_kv_base^2;
+
+//diplaying the results on console
+
+printf('Per unit impedance of the components of the given power system are as follows :\n\n');
+
+printf('Transmission line: %0.3f \n\n',tl_pu);
+
+printf('Transformer T1: %0.3f \n\n',t1_pu);
+
+printf('Transformer T2: %0.3f \n\n',t2_pu);
+
+printf('Generator 1: %0.3f \n\n',gen1_pu);
+
+printf('Generator 2: %0.3f \n\n',gen2_pu);
+
+printf('Generator 3: %0.3f \n\n',gen3_pu);
+
diff --git a/Working_Examples/83/CH4/EX4.1/result_example_4_1.txt b/Working_Examples/83/CH4/EX4.1/result_example_4_1.txt
new file mode 100755
index 0000000..975ce31
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.1/result_example_4_1.txt
@@ -0,0 +1,18 @@
+ 
+Per unit impedance of the components of the given power system are as follows :
+
+Transmission line: 0.565 
+
+Transformer T1: 0.419 
+
+Transformer T2: 0.441 
+
+Generator 1: 0.397 
+
+Generator 2: 0.937 
+
+Generator 3: 0.437 
+
+ 
+ Execution done.   
+
diff --git a/Working_Examples/83/CH4/EX4.2/example_4_2.sce b/Working_Examples/83/CH4/EX4.2/example_4_2.sce
new file mode 100755
index 0000000..31701df
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.2/example_4_2.sce
@@ -0,0 +1,43 @@
+//Chapter 4
+//Example 4.2
+//page 104
+// To draw the per unit reactance diagram when pu values are specified based on euipment rating
+clear;clc;
+mvab=30; kvb=11; //MVA base and KVA base are selected in the circuit of generator 1
+
+gen1_mva=30; gen1_kv=10.5; gen1_x=0.435;  //Generator No.1 details
+gen2_mva=15; gen2_kv=6.6; gen2_x=0.413;  //Generator No.2 details
+gen3_mva=25; gen3_kv=6.6; gen3_x=0.3214;  //Generator No.3 details
+
+t1_mva=15; t1_hv=33; t1_lv=11; t1_x=0.209; //Transformer T1 details
+t2_mva=15; t2_hv=33; t2_lv=6.2; t2_x=0.220; //Transformer T1 details
+
+tl_x=20.5; //Transmission line recatance
+
+//Loads are neglected as said in the problem
+
+tl_pu=(tl_x*mvab)/t1_hv^2;
+t1_pu=t1_x*(mvab/t1_mva);
+t2_pu=t2_x*(mvab/t2_mva);
+gen1_pu=gen1_x*(mvab/gen1_mva)*(gen1_kv/kvb)^2;
+gen2_kv_base=t2_lv;
+gen2_pu=gen2_x*(mvab/gen2_mva)*(gen2_kv/gen2_kv_base)^2;
+gen3_kv_base=t2_lv;
+gen3_pu=gen3_x*(mvab/gen3_mva)*(gen3_kv/gen3_kv_base)^2;
+
+//diplaying the results on console
+
+printf('Per unit impedance of the components of the given power system are as follows :\n\n');
+
+printf('Transmission line: %0.3f \n\n',tl_pu);
+
+printf('Transformer T1: %0.3f \n\n',t1_pu);
+
+printf('Transformer T2: %0.3f \n\n',t2_pu);
+
+printf('Generator 1: %0.3f \n\n',gen1_pu);
+
+printf('Generator 2: %0.3f \n\n',gen2_pu);
+
+printf('Generator 3: %0.3f \n\n',gen3_pu);
+
diff --git a/Working_Examples/83/CH4/EX4.2/result_example_4_2.txt b/Working_Examples/83/CH4/EX4.2/result_example_4_2.txt
new file mode 100755
index 0000000..30c5370
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.2/result_example_4_2.txt
@@ -0,0 +1,32 @@
+ 
+Per unit impedance of the components of the given power system are as follows :
+
+Transmission line: 0.565 
+
+Transformer T1: 0.418 
+
+Transformer T2: 0.440 
+
+Generator 1: 0.396 
+
+Generator 2: 0.936 
+
+Generator 3: 0.437 
+
+Per unit impedance of the components of the given power system are as follows :
+
+Transmission line: 0.565 
+
+Transformer T1: 0.419 
+
+Transformer T2: 0.441 
+
+Generator 1: 0.397 
+
+Generator 2: 0.937 
+
+Generator 3: 0.437 
+
+ 
+ Execution done.   
+
diff --git a/Working_Examples/83/CH4/EX4.3/example_4_3.sce b/Working_Examples/83/CH4/EX4.3/example_4_3.sce
new file mode 100755
index 0000000..9c72377
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.3/example_4_3.sce
@@ -0,0 +1,24 @@
+//Taking Base value MVA and KVA
+clear;clc;
+mvab=645; //Base MVA in 3-phase
+kvb=24; //Base KV,line-to-line
+
+vl=24/kvb; //Load voltage
+xs=1.2;
+xs=(xs*mvab)/kvb^2; // xs converted to its pu
+
+//since the generator is operating at full load & 0.9pf
+pf_angle=acos(0.9);
+Ia=1*(cos(pf_angle)-%i*sin(pf_angle)); //load current
+//to find excitation emf
+ef=vl+%i*xs*Ia;
+delta=atand(imag(ef)/real(ef));//positive for leading
+ef=abs(ef)*kvb; //pu to actual unit conversion
+  if(delta>0) then lead_lag='leading';
+   else lead_lag='lagging';
+  end
+printf('Excitation emf= %0.2f kV at an angle %0.3f (%s) \n\n',ef,delta,lead_lag);
+//to find reactive power drawn by load
+Q=vl*abs(imag(Ia));
+Q=Q*mvab; //pu to actual unit conversion
+printf('Reactive power drawn by laod= %d MVAR',Q);
diff --git a/Working_Examples/83/CH4/EX4.3/result_example_4_3.txt b/Working_Examples/83/CH4/EX4.3/result_example_4_3.txt
new file mode 100755
index 0000000..17267b8
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.3/result_example_4_3.txt
@@ -0,0 +1,5 @@
+ 
+Excitation emf= 47.86 kV at an angle 37.331 (leading) 
+
+Reactive power drawn by laod= 281 MVAR 
+
diff --git a/Working_Examples/83/CH4/EX4.4/example_4_4.sce b/Working_Examples/83/CH4/EX4.4/example_4_4.sce
new file mode 100755
index 0000000..064e191
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.4/example_4_4.sce
@@ -0,0 +1,50 @@
+//Taking Base value MVA and KVA
+clear;clc;
+global mvab
+mvab=645; //Base MVA in 3-phase
+kvb=24; //Base KV,line-to-line
+vt=24/kvb; //Terminal voltage
+xs=1.2;
+xs=(xs*mvab)/kvb^2; // xs converted to its pu
+
+//since the generator is operating at full load & 0.9pf
+pf_angle=acos(0.9);
+Ia=1*(cos(pf_angle)-%i*sin(pf_angle)); //load current
+//to find excitation emf
+ef=vt+%i*xs*Ia;
+ef=abs(ef);
+P=1*0.9; //at Full load
+
+/////// writing an inline function /////////////////
+function [pf,lead_lag,Q]=excitation_change(P,ef,vt,xs)
+sin_delta=(P*xs)/(ef*vt);
+delta=asind(sin_delta);
+ef0=ef*(cosd(delta)+(%i*sind(delta)));
+Ia=(ef0-vt)/(%i*xs);
+Ia_mag=abs(Ia);Ia_ang=atand(imag(Ia)/real(Ia)); // Magnitude and angle of Ia
+pf=cosd(abs(Ia_ang));
+if(Ia_ang>0) then lead_lag='leading';
+  elseif (Ia_ang==0) then lead_lag='unity pf' 
+    else lead_lag='lagging';
+  end
+Q=vt*Ia_mag*sind(abs(Ia_ang));
+Q=abs(Q)*mvab;
+endfunction
+////////////////////////////////////////////////////////
+
+
+// First Case when Ef is increased by 20% at same real load now
+ ef1=ef*1.2;
+[pf1,lead_lag1,Q1]=excitation_change(P,ef1,vt,xs);
+disp("Case (i): When Ef is increased by 20% ");
+printf('\n\tPower factor pf= %0.2f %s \n',pf1,lead_lag1);
+printf('\tReactive power drawn by the load = %0.1f MVAR \n',Q1);
+
+//Second Case when Ef is decreased by 20% at same real load now
+ ef2=ef*0.8;
+[pf2,lead_lag2,Q2]=excitation_change(P,ef2,vt,xs);
+disp("Case (ii): When Ef is decreased by 20% ");
+printf('\n\tPower factor pf= %0.2f %s \n',pf2,lead_lag2);
+printf('\tReactive power drawn by the load = %0.1f MVAR \n',Q2);
+
+disp('The answers given here are exact values.Textbook answers has an approximation of upto 2 decimal places on Xs,Ia,pf. ');
diff --git a/Working_Examples/83/CH4/EX4.4/result_example_4_4.txt b/Working_Examples/83/CH4/EX4.4/result_example_4_4.txt
new file mode 100755
index 0000000..7e86950
--- /dev/null
+++ b/Working_Examples/83/CH4/EX4.4/result_example_4_4.txt
@@ -0,0 +1,14 @@
+
+ 
+ Case (i): When Ef is increased by 20%    
+
+	Power factor pf= 0.75 lagging 
+	Reactive power drawn by the load = 511.2 MVAR 
+ 
+ Case (ii): When Ef is decreased by 20%    
+
+	Power factor pf= 1.00 lagging 
+	Reactive power drawn by the load = 19.5 MVAR 
+ 
+ The answers given here are exact values.Textbook answers has an approximat 
+      ion of upto 2 decimal places on Xs,Ia,pf.