5 files changed, 149 insertions, 0 deletions
diff --git a/1904/CH8/EX8.1/8_1.sce b/1904/CH8/EX8.1/8_1.sce
new file mode 100755
index 000000000..fabe90c7a
--- /dev/null
+++ b/1904/CH8/EX8.1/8_1.sce
@@ -0,0 +1,24 @@
+//To Determine the Capacitor Size to improve the power factor to a 700kVA Load
+//Page 390
+clc;
+clear;
+
+SL=700; //Load in kVA
+pf1=65/100; //Power Factor
+PL=SL*pf1; //Real Power
+//From the Table of Power Factor Correction
+CR=0.74; //Co-relation factor
+CS=PL*CR; //Capacitor Size
+
+CSr=360; //Next Higher Standard Capacitor Size
+
+CRn=CSr/PL; //New Co-Relation Factor
+
+//From the table by linear interpolation
+pfr=93; //In Percentage
+pfn=pfr+(172/320);
+
+printf('\a) The Correction Factor is %g\n',CR)
+printf('b) The Capacitor Size Required is %g kVAr\n',CS)
+printf('c) Resulting power factor if the next higher standard capacitor size is used is %g percent\n',pfn)
+ 
diff --git a/1904/CH8/EX8.2/8_2.sce b/1904/CH8/EX8.2/8_2.sce
new file mode 100755
index 000000000..3badcf151
--- /dev/null
+++ b/1904/CH8/EX8.2/8_2.sce
@@ -0,0 +1,31 @@
+//To determine the Capacitor bank required to correct power factor of induction motor
+//Page 393
+clc;
+clear;
+
+Vll=4.16; //Line to Line Voltage in kV
+Pr=(500*0.7457); //Rating of motor in kW
+pf1=0.75; //Initial Power Factor
+pfn=0.9; //Improved Power Factor
+eff=0.88; //Efficiency
+P=Pr/eff; //Input Power of Induction Motor
+Q1=P*tand(acosd(pf1)); //Reactive Power
+Q2=P*tand(acosd(pfn)); //REactive power of motor after power factor improvement
+f=60; //Frequency of supply
+w=2*%pi*f; //Angular Frequency
+Qc=Q1-Q2; //Reactive Power of Capacitor
+Il=Qc/(sqrt(3)*Vll);
+
+//Capacitor Connectd in Delta
+Ic1=Il/(sqrt(3));
+Xc1=Vll*1000/Ic1; //Reactance of each capacitor
+C1=(10^6)/(w*Xc1); //Capacitance in Micro Farad
+
+//Capacitor Connected in Wye
+Ic2=Il;
+Xc2=Vll*1000/(sqrt(3)*Ic2); //Reactance of each capacitor
+C2=(10^6)/(w*Xc2); //Capacitance in Micro Farad
+
+printf('\na) Rating of Capacitor Bank is %g kVAr\n',Qc)
+printf('b) The Value of Capacitance if there are connected in delta is %g micro F\n',C1)
+printf('c) The Value of Capacitance if there are connected in wye is %g micro F\n',C2)
diff --git a/1904/CH8/EX8.3/8_3.sce b/1904/CH8/EX8.3/8_3.sce
new file mode 100755
index 000000000..351fdcab7
--- /dev/null
+++ b/1904/CH8/EX8.3/8_3.sce
@@ -0,0 +1,19 @@
+//To determine the power factors of a 2.4 kV phase circuit feeder
+//Page 396
+clc;
+clear;
+
+V=2.4; //Voltage in kV
+I=200; //Load Current
+P=360; //Real Load in kW
+S1=V*I; //Total Load in kVA
+pf1=P/S1; //Power Factor
+Q1=S1*sind(acosd(pf1)); //Reactive Load
+
+Qc=300; //Capacitor Size
+
+Q2=Q1-Qc; //The New Reactive Load
+pf2=P/sqrt((P^2)+((Q1-Qc)^2)); //Improved Power Factor
+
+printf('\na) The Uncorrected power factor and reactive load is %g and %g kVAr\n',pf1,Q1)
+printf('b) The New Corrected factor after the introduction of capacitor unit is %g\n',pf2)
diff --git a/1904/CH8/EX8.4/8_4.sce b/1904/CH8/EX8.4/8_4.sce
new file mode 100755
index 000000000..09b843f85
--- /dev/null
+++ b/1904/CH8/EX8.4/8_4.sce
@@ -0,0 +1,28 @@
+//To determine the necessity of additional capacitors
+//Page 398
+clc;
+clear;
+
+S1=7800; //Peak Load in kVA
+T=3*2000; //Total Rating of the Transformer
+pf1=0.89; //Load Power Factor
+TC=120/100; //Thermal Capability
+Qc=1000; //Size of capacitor
+
+P=S1*pf1; //Real Load
+Q1=S1*sind(acosd(pf1)); //Reactive Load
+
+Q2=Q1-Qc; //The New Reactive Load
+pf2=P/sqrt((P^2)+((Q1-Qc)^2)); //Improved Power Factor
+
+S2=P/pf2; //Corrected Apprarent power
+
+ST=T*TC; //Transformer Capabilty
+
+pf3=P/ST; //New Corrected Power factor required
+
+Q2new=P*tand(acosd(pf3)); //Required Reactive Power
+Qcadd=Q2-Q2new; //Additional Rating of the Capacitor
+
+printf('\na) Since the Apparent Power(%g kVAr) is more than Transformer Capability (%g kVAr), \nHence Additional Capacitors are required\n',S2,ST)
+printf('b) The Rating of the Addtional capacitor is %g kVAr\n',Qcadd)
diff --git a/1904/CH8/EX8.5/8_5.sce b/1904/CH8/EX8.5/8_5.sce
new file mode 100755
index 000000000..5bc406e70
--- /dev/null
+++ b/1904/CH8/EX8.5/8_5.sce
@@ -0,0 +1,47 @@
+//To Determine the savings in kilowatt losses
+//Page 411
+clc;
+clear;
+
+// 1 is Total Loss Reduction due to Capacitors
+// 2 is Additional Loss Reduction due to Capacitor
+// 3 is Total Demand Reduction due to capacitor
+// 4 is Total required capacitor additions
+
+C90=[495165,85771,22506007,9810141]; //Characteristics at 90% Power Factor
+C98=[491738,75343,21172616,4213297]; //Characteristics at 98% Power Factor
+
+//Responsibility Factors
+RF90=1;
+RF98=0.9;
+
+SLF=0.17; //System Loss Factor
+FCR=0.2; //Fixed Charge Rate
+DC=250; //Demand Cost
+ACC=4.75; //Average Capacitor Cost
+EC=0.045; //Energy Cost
+Cd=C90-C98; //Difference in Characteristics
+
+TAS=Cd(1)+Cd(2); //Total Additional Kilowatt Savings
+
+ASDR1=Cd(1)*RF90*DC*FCR;
+ASDR2=Cd(2)*RF98*DC*FCR;
+TASDR=ASDR1+ASDR2; //Total Annual Savings due to demand
+x=27; // Cost for Per kVA of the capacity
+TASTC=Cd(3)*FCR*x; //Annual Savings due to Transmission Capacity
+TASEL=TAS*SLF*EC*8760; //Savings due to energy loss reduction
+TACAC=Cd(4)*FCR*ACC; //Annual Cost of Additional Capacitors
+Savings=TASEL+TASDR+TASTC; //Total Savings
+
+printf('\na) The Resulting additional savings in kilowatt losses due to power factor improvement at the substation buses is %g kW\n',Cd(1))
+printf('b) The Resulting assitional savings in kilowatt  losses due to the power factor improvement for feeders is %g kW\n',Cd(2))
+printf('c) The Additional Kilowatt Savings is %g kW\n',TAS)
+printf('d) The Additional savings in the system kilovoltampere capacity is %g kVA\n',Cd(3))
+printf('e) The Additional Capacitors required are %g kVAr\n',Cd(4))
+printf('f) The Annual Savings in demand reduction due to capacitors applied to distribution substation buses is approximately is %g dollars/year\n',TASDR)
+printf('g) The Annual Savings due to the additional released transmission capacity is %g dollars/year\n',TASTC)
+printf('h) The Total Anuual Savings due to the energy loss reduction is %g dollars/year\n',TASEL)
+printf('i) The Total Annual Cost of the additional capacitors is %g dollars/year\n',TACAC)
+printf('j) The Total Annual Savings is %g dollars/year\n',Savings)
+printf('k) No, Since the total net annual savings is not zero\n')
+