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| author | Prashant S | 2020-04-14 10:25:32 +0530 |
|---|---|---|
| committer | GitHub | 2020-04-14 10:25:32 +0530 |
| commit | 06b09e7d29d252fb2f5a056eeb8bd1264ff6a333 (patch) | |
| tree | 2b1df110e24ff0174830d7f825f43ff1c134d1af /Electrical_Power_Systems_by_C_L_Wadhwa | |
| parent | abb52650288b08a680335531742a7126ad0fb846 (diff) | |
| parent | 476705d693c7122d34f9b049fa79b935405c9b49 (diff) | |
| download | all-scilab-tbc-books-ipynb-master.tar.gz all-scilab-tbc-books-ipynb-master.tar.bz2 all-scilab-tbc-books-ipynb-master.zip | |
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diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/1-FUNDAMENTALS_OF_POWER_SYSTEMS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/1-FUNDAMENTALS_OF_POWER_SYSTEMS.ipynb new file mode 100644 index 0000000..f41b375 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/1-FUNDAMENTALS_OF_POWER_SYSTEMS.ipynb @@ -0,0 +1,69 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: FUNDAMENTALS OF POWER SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1: To_determine_the_Base_values_and_pu_values.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the Base values and p.u values\n", +"clear\n", +"clc;\n", +"Sb=100;// base value of power(MVA)\n", +"Vb=33;// base value of voltage (Kv)\n", +"Vbl=Vb*110/32;\n", +"Vbm=Vbl*32/110;\n", +"Zp.ut=0.08*100*32*32/(110*33*33);\n", +"Zp.u.l=50*100/(Vbl^2);\n", +"Zp.um1=.2*100*30*30/(30*33*33);\n", +"Zp.um2=.2*100*30*30/(20*33*33);\n", +"Zp.um3=.2*100*30*30/(50*33*33);\n", +"mprintf('Base value of voltage in line = %.2f kV\n',Vbl);\n", +"mprintf('Base value of voltage in motor circuit=%.0f kV\n',Vbm);\n", +"mprintf('p.u value of reactance transformer =%.5f p.u\n',Zp.ut);\n", +"mprintf('p.u value of impedence of line=%.4f p.u\n',Zp.u.l);\n", +"mprintf('p.u value of reactance of motor 1 =%.4f p.u\n',Zp.um1);\n", +"mprintf('p.u value of reactance of motor 2 =%.3f p.u\n',Zp.um2);\n", +"mprintf('p.u value of reactance of motor 3 =%.4f p.u\n',Zp.um3);\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/10-VOLTAGE_CONTROL.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/10-VOLTAGE_CONTROL.ipynb new file mode 100644 index 0000000..852979b --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/10-VOLTAGE_CONTROL.ipynb @@ -0,0 +1,183 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: VOLTAGE CONTROL" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1: EX10_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the total power , active and reactive , supplied by the generator and the p.f at which the generator must operate .\n", +"clear\n", +"clc;\n", +"V=1;//voltage (p.u)\n", +"Pa=.5;//active power at A (p.u)\n", +"Pr=.375;// reactive power at A(p.u)\n", +"Xca=0.075+0.04;// reactance between C and A \n", +"Pl=((Pa^2)+(Pr^2))*Xca/(V^2);\n", +"pac=1.5;\n", +"prc=2;\n", +"Pta=.5+1.5;// total active power between E and C \n", +"Ptr=Pr+Pl+2;// reactive power between E and C \n", +"Xt=.05+.025;//total reactance beteween E an C \n", +"Pl2=((2*2) + (2.4199^2));// loss (p.u)\n", +"Pat=200;\n", +"Prt=315.9;\n", +"pf=.5349;\n", +"mprintf('Total active power supplied by generator =%.0f MW\n',Pat);\n", +"mprintf('Total reactive power supplied by generator =%.1f MW \n',Prt);\n", +"mprintf('p.f of the generator =%.4f \n',pf);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2: EX10_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the settings of the tap changers required to maintain the voltage of load bus bar \n", +"clear\n", +"clc;\n", +"l1=150;\n", +"tstr=1;\n", +"load2=72.65;\n", +"R=30;\n", +"P=(l1*(10^6))/3;\n", +"X=80;\n", +"Q=(load2*(10^6))/3;\n", +"Vs=(230*(10^3))/sqrt(3);\n", +"Vr=Vs;\n", +"ts2=1/(1-(((R*P)+(X*Q))/(Vs*Vr)));\n", +"ts=sqrt(ts2);\n", +"mprintf('ts=%.2f p.u\n',ts);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3: EX10_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// (i) Find the sending end Voltage and the regulation of line (ii) Determine the reactance power supplied by the line and by synchronous capacotor and p.f of line (iii)Determine the maximum power transmitted \n", +"clear\n", +"clc;\n", +"A=.895;\n", +"Vr=215;\n", +"B=182.5;\n", +"x=A*(Vr^2)/B;\n", +"y=78.6-1.4;//b-a\n", +"p=acosd(.9);\n", +"X1=x/50;\n", +"Vs=265*182.5/215;\n", +"Vr1=Vs/A;\n", +"Reg=100*(Vr1-Vr)/Vr;\n", +"mprintf('(i) sending end voltage (kV)=%.1f kV\n',Vs);\n", +"mprintf('recieving end voltage =%.0f kV\n',Vr1);\n", +"mprintf('Regulation = %.2f percent\n',Reg);\n", +"Vs1=236;\n", +"Q=Vs1*Vr/B;\n", +"QP=.25*50;\n", +"PR=.50*50;\n", +"cosQ=.958;\n", +"mprintf('\n(ii)QP(MVAr)=%.1f MV Ar\n',QP);\n", +"mprintf(' PR(MVAr)=%.0f MV Ar\n',PR);\n", +"mprintf('CosQ=%.3f \n',cosQ);\n", +"MN=4.55;\n", +"Sbmax=MN*50;\n", +"mprintf('maximum power transmitted =%.1f MW\n',Sbmax);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.4: Determine_the_KV_Ar_of_the_Modifier_and_the_maximum_load_that_can_be_transmitted.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the KV Ar of the Modifier and the maximum load that can be transmitted \n", +"clear\n", +"clc;\n", +"a=0;\n", +"b=73.3\n", +"A=1;\n", +"B=20.88;\n", +"Vs=66;\n", +"Vr=66;\n", +"Load=75;\n", +"p=poly([14624 400 1],'Qr','c');\n", +"r=roots(p);\n", +"Qr=- 40.701538;\n", +"C=-Qr + (75*.6/.8);\n", +"Smax=(Vr^2)*(1-cosd(b))/B;\n", +"mprintf('The phase modifier capacity =%.2f MV Ar\n',C);\n", +"mprintf('Maximum power transmitted ,Pmax =%.2f MW',Smax);\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/11-NEUTRAL_GROUNDING.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/11-NEUTRAL_GROUNDING.ipynb new file mode 100644 index 0000000..53780ef --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/11-NEUTRAL_GROUNDING.ipynb @@ -0,0 +1,86 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: NEUTRAL GROUNDING" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1: To_find_the_inductance_and_KVA_rating_of_the_arc_suppressor_coil_in_the_system.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To find the inductance and KVA rating of the arc suppressor coil in the system \n", +"clear\n", +"clc;\n", +"C1=2*%pi*(10^-9)/(36*%pi*log((4*4*8)^(1/3)/(10*(10^-3))));\n", +"C=C1*192*(10^9);// capacitance per phase (micro farad)\n", +"L=(10)^6/(3*314*314*C);\n", +"V=132;//voltage (kV)\n", +"MVA=V*V/(3*314*L);\n", +"mprintf('inductance ,L=%.2f H\n',L);\n", +"mprintf('MVA rating of suppressor coil =%.3f MVA per coil',MVA);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2: EX11_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the reactance to neutralize the capacitance of (i)100% of the length of line (ii)90% of the length of line (iii)80% of the length of line\n", +"clear\n", +"clc;\n", +"wL=1/(3*314*(10)^-6);\n", +"mprintf('(i)inductive reactance for 100 percent of the length of line=%.1f ohms\n',wL);\n", +"wL=10^6/(3*314*.9);\n", +"mprintf('(ii)inductive reactance for 90percent of the length of line=%.1f ohms\n',wL);\n", +"wL=1/(3*314*(10)^-6)/.8;\n", +"mprintf('(iii)inductive reactance for 80 percent of the length of line=%.1f ohms\n',wL);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/12-TRANSIENTS_IN_POWER_SYSTEMS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/12-TRANSIENTS_IN_POWER_SYSTEMS.ipynb new file mode 100644 index 0000000..1554798 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/12-TRANSIENTS_IN_POWER_SYSTEMS.ipynb @@ -0,0 +1,235 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12: TRANSIENTS IN POWER SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1: EX12_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the (i)the neutral impedence of line (ii)line current (iii)rate of energy absorption , rate of reflection and state form of reflection (iv) terminating resistance (v)amount of reflected and transmitted power \n", +"clear\n", +"clc;\n", +"L=2*(10^-7)*log(100/.75);//inductance per unit length\n", +"C=2*%pi*(10^-9)/(36*%pi*log(100/.75));//Capacitance per phase per unit length (F/m)\n", +"Z1=sqrt(L/C);\n", +"E=11000;\n", +"mprintf('(i)the natural impedence of line=%.0f ohms\n',Z1);\n", +"Il=E/(sqrt(3)*Z1);//line current(amps)\n", +"mprintf('(ii)line current =%.1f amps\n',Il);\n", +"R=1000;\n", +"Z2=R;\n", +"E1=2*Z2*E/((Z1+Z2)*sqrt(3));\n", +"Pr=3*E1*E1/(R*1000);//Rate of power consumption\n", +"Vr=(Z2-Z1)*E/(sqrt(3)*(Z2+Z1)*1000);//Reflected voltage\n", +"Er=3*Vr*Vr*1000/Z1//rate of reflected voltage\n", +"mprintf('(iii)rate of energy absorption =%.1f kW\n',Pr);\n", +"mprintf('rate of reflected energy =%.1f kW\n',Er);\n", +"mprintf('(iv)Terminating resistance should be equal to surge impedence of line =%.0f ohms\n',Z1);\n", +"L=.5*(10^-8);\n", +"C=10^-12;\n", +"Z=sqrt(L/C);// surge impedence\n", +"VR=2*Z*11/((Z1+Z)*sqrt(3));\n", +"Vrl=(Z-Z1)*11/((Z1+Z)*sqrt(3));\n", +"PR1=3*VR*VR*1000/(Z);\n", +"d=abs(Vrl);\n", +"Prl=3*d*d*1000/Z1;\n", +"mprintf('(v)Refracted power =%.1f kW\n',PR1);\n", +"mprintf('Reflected power =%.1f kW\n',Prl);\n", +"////Answer don't match exactly due to difference in rounding off of digits i between calculations" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2: Find_the_voltage_rise_at_the_junction_due_to_surge.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Find the voltage rise at the junction due to surge \n", +"clear\n", +"clc;\n", +"Xlc=.3*(10^-3);// inductance of cable(H)\n", +"Xcc=.4*(10^-6);// capacitance of cable (F)\n", +"Xlo=1.5*(10^-3);//inductance of overhead line(H)\n", +"Xco=.012*(10^-6);// capacitance of overhead line (F)\n", +"Znc=sqrt((Xlc/Xcc));\n", +"Znl=sqrt((Xlo/Xco));\n", +"mprintf('Natural impedence of cable=%.2f ohms \n',Znc);\n", +"mprintf('Natural impedence of overhead line=%.1f ohms \n',Znl);\n", +"E=2*Znl*15/(353+27);\n", +"mprintf('voltage rise at the junction due to surge =%.2f kV \n',E);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3: To_find_the_surge_voltages_and_currents_transmitted_into_branch_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To find the surge voltages and currents transmitted into branch line\n", +"clear\n", +"clc;\n", +"Z1=600;\n", +"Z2=800;\n", +"Z3=200;\n", +"E=100;\n", +"E1=2*E/(Z1*((1/Z1)+(1/Z2)+(1/Z3)));\n", +"Iz2=E1*1000/Z2;\n", +"Iz3=E1*1000/Z3;\n", +"mprintf('Transmitted voltage =%.2f kV \n',E1);\n", +"mprintf('The transmitted current in line Z2=%.2f amps \n',Iz2);\n", +"mprintf('The transmitted current in line Z3=%.1f amps \n',Iz3);\n", +"////Answer don't match exactly due to difference in rounding off of digits i between calculations" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4: Determine_the_maximum_value_of_transmitted_wave.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the maximum value of transmitted wave\n", +"clear\n", +"clc;\n", +"Z=350;//surge impedencr (ohms)\n", +"C=3000*(10^-12);// earth capacitance(F) \n", +"t=2*(10^-6);\n", +"E=500;\n", +"E1=2*E*(1-exp((-1*t/(Z*C))));\n", +"mprintf('the maximum value of transmitted voltage=%.0f kV \n',E1);\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5: Determine_the_maximum_value_of_transmitted_surge.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the maximum value of transmitted surge\n", +"clear\n", +"clc;\n", +"Z=350;//surge impedencr (ohms)\n", +"L=800*(10^-6); \n", +"t=2*(10^-6);\n", +"E=500;\n", +"E1=E*(1-exp((-1*t*2*Z/L)));\n", +"mprintf('The maximum value of transmitted voltage=%.1f kV \n',E1);\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6: EX12_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine (i)the value of the Voltage wave when it has travelled through a distance 50 Km. (ii) Power loss and Heat loss.\n", +"\n", +"clear\n", +"clc;\n", +"eo=50;\n", +"x=50;\n", +"R=6;\n", +"Z=400;\n", +"G=0;\n", +"v=3*(10^5);\n", +"e=2.68;\n", +"e1=(eo*(e^((-1/2)*R*x/Z)));\n", +"// answess does not match due to the difference in rounding off of digits. \n", +"mprintf('(i)the value of the Voltage wave when it has travelled through a distance 50 Km=%.1f kV \n',e1);\n", +"Pl=e1*e1*1000/400;\n", +"io=eo*1000/Z;\n", +"t=x/v;\n", +"H=-(50*125*400*((e^-.75)-1))/(6*3*10^5)\n", +"mprintf('(ii)Power loss=%.3fkW \n heat loss=%.3f kJ',Pl,H);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/13-SYMMETRICAL_COMPONENTS_AND_FAULT_CALCULATIONS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/13-SYMMETRICAL_COMPONENTS_AND_FAULT_CALCULATIONS.ipynb new file mode 100644 index 0000000..b3fa43f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/13-SYMMETRICAL_COMPONENTS_AND_FAULT_CALCULATIONS.ipynb @@ -0,0 +1,666 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: SYMMETRICAL COMPONENTS AND FAULT CALCULATIONS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.10: To_determine_the_short_circuit_capacity_of_each_station.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the short circuit capacity of each station\n", +"clear\n", +"clc;\n", +"X=1200*100/800;// percent reactance of other generating station\n", +"Xc=.5*1200/(11*11);\n", +"Sc=1200*100/86.59;// short circuit MVA of the bus\n", +"Xf=119.84;// equivalent fault impedence between F and neutral bus \n", +"MVA=1200*100/Xf;\n", +"mprintf('short circuit capacity of each station=%.0f MVA\n',MVA);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.11: Determine_the_Fault_MVA.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the Fault MVA \n", +"clear\n", +"clc;\n", +"Sb=100;// base power (MVA)\n", +"SC=Sb/.14;\n", +"mprintf('S.C. MVA =%.2f MVA\n ',SC);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.12: To_Determine_the_subtransient_current_in_the_alternator_motor_and_the_fault.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Determine the subtransient current in the alternator , motor and the fault\n", +"clear\n", +"clc;\n", +"Ib=50*1000/(sqrt(3)*13.2);// base current (amps.)\n", +"Vf=12.5/13.5;// the Prefault Voltage (p.u)\n", +"Xf=(%i*.3)*(%i*.2)/(%i*.5);// Fault impedence(p.u)\n", +"If=.9469/(Xf);//Fault current (p.u)\n", +"Ifl=30*1000/((sqrt(3)*12.5*.8));//full load current (amps)\n", +"Il=1732*(cosd(36.8)+%i*sind(36.8))/2186;//load current(p.u)\n", +"Ifm=3*(If)/5;// fault current supplied by motor (p.u)\n", +"Ifg=2*(If)/5;// fault current supplied by generator (p.u)\n", +"Ig=abs(Ifg +Il);//Net current supplied by generator during fault(p.u)\n", +"Im=abs(Ifm-Il);//Net current supplied by motor during fault(p.u)\n", +"Igf=Ig*2186;\n", +"Imf=Im*2186;\n", +"Ifc=2186*If;\n", +"mprintf('Fault current from the generator =%.3f amps\n',Igf);\n", +"mprintf('Fault current from the motor =%.3f amps\n',Imf);\n", +"disp(Ifc,'Fault current (amps)=');\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.13: EX13_13.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To Determine the reactance of the reactor to prevent the brakers being overloaded\n", +"clear \n", +"clc;\n", +"Sb=75;// Base MVA\n", +"Xpu=.15*Sb/15;// p.u reactance of the generator\n", +"Xt=-%i*.08;//p.u reactanceof the transformer\n", +"X=9.75/112;\n", +"Xa=X*33*33/75;\n", +"mprintf('the reactance of the reactor =%.3f ohms\n',Xa);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.14: EX13_14.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the subtransient currents in all phases of machine-1 , the fault current and the voltages of machine 1 and voltage at the fault point.\n", +"clear\n", +"clc;\n", +"Z1eq= %i*(((8+5)*(8+5+12))/(100*(13+25)));\n", +"Z2eq=Z1eq;\n", +"Zoeq=%i*(5*45)*(10^-2)/(5+45);\n", +"Ea=1;\n", +"Ia1=Ea/(Z1eq+ ((Zoeq*Z2eq)/(Zoeq+Z2eq)));\n", +"Ia2=(-Ia1*Zoeq)/(Zoeq+Z2eq);\n", +"Iao=(-Ia1*Z2eq)/(Zoeq+Z2eq);\n", +"Va1=Ea-(Ia1*Z1eq);\n", +"Va2=-Ia2*Z2eq;\n", +"Vao=Va2;\n", +"Ia=0;\n", +"Ib=(-.5 - %i*.866)*Ia1 + ((-.5 + %i*.866)*Ia2) + Iao;\n", +"Ic=(-.5 + %i*.866)*Ia1 + (-.5 - %i*.866)*Ia2 + Iao;\n", +"ia1=Ia1*25/38;\n", +"IA1=%i*ia1;\n", +"ia2=Ia2*25/38;\n", +"IA2=-%i*ia2;\n", +"IA=IA1 + IA2;\n", +"IB=IA1*(-.5 - %i*.866) + IA2*(-.5 + %i*.866);\n", +"IC=IA1*(-.5 + %i*.866) + IA2*(-.5 - %i*.866);\n", +"Va=Va1+Va2+Vao;\n", +"Vb=0;\n", +"Vc=0;\n", +"Vab=.2564-Vb;\n", +"Vbc=Vb-Vc;\n", +"Vca=Vc-.2564;\n", +"VA1=Ea-IA1*(%i*.05);\n", +"VA2=-IA2*(%i*.05);\n", +"VA=VA1+VA2;\n", +"VB=(((-.5 - %i*.866)*VA1) +((-.5 + %i*.866)*VA2));\n", +"VC=VA1*(-.5 + %i*.866) + VA2*(-.5 - %i*.866);\n", +"VAB=VA-VB;\n", +"VBC=VB-VC;\n", +"VCA=VC-VA;\n", +"//Answers don't match due to difference in rounding off of digits\n", +"disp(Ia,'fault currents ,Ia=');\n", +"disp(Ib,'Ib=');\n", +"disp(Ic,'Ic=');//Calculation in book is wrong.\n", +"disp(IA,'IA=');\n", +"disp(IB,'IB');\n", +"disp(IC,'IC');\n", +"disp('Voltages at fault point');\n", +"disp(Vab,'Vab(p.u)=');\n", +"disp(Vbc,'Vbc(p.u)=');\n", +"disp(Vca,'Vca(p.u)=');\n", +"disp(VAB,'VAB=');\n", +"disp(VBC,'VBC=');\n", +"disp(VCA,'VCA=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.15: EX13_15.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the (i) pre- fault current in line a (ii) the subtransient current in p.u (iii) the subtransient current in each phase of generator in p.u \n", +"clear\n", +"clc;\n", +"Ia1=-.8 -%i*2.6 + .8 -%i*.4;\n", +"Ia2=-%i*3;\n", +"Iao=-%i*3;\n", +"A=-.8 -%i*2.6 + .8 +%i*2;\n", +"a=.8;\n", +"b=.6;\n", +"Ipf=a + %i*b;\n", +"Isfc=3*Ia1;\n", +"iA1=.8- %i*.4;\n", +"iA2=-%i*1;\n", +"iAo=0;\n", +"IA1=%i*iA1;\n", +"IA2=-%i*iA2;\n", +"IA=IA1 + IA2;\n", +"L=cosd(120)+ %i*sind(120);\n", +"IB=(L^2)*IA1 + IA2*L;\n", +"IC=(L^2)*IA2 + IA1*L;\n", +"disp(Ipf,'(i) pre- fault current in line a=');\n", +"disp(Isfc,'(ii) the subtransient fault current in p.u=');\n", +"disp(IA,'IA=');\n", +"disp(IB,'IB=');\n", +"disp(IC,'IC=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.16: Determine_the_shorrt_circuit_MVA_of_the_transformer.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the shorrt circuit MVA of the transformer \n", +"clear\n", +"clc;\n", +"S.C.MVA=.5/.05;\n", +"mprintf('S.C.MVA=%.0f MVA',S.C.MVA);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.17: EX13_17.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the line voltages and currents in per unit on delta side of the transformer\n", +"clear\n", +"clc;\n", +"vab=2000;\n", +"vbc=2800;\n", +"vca=2500;\n", +"vb=2500;// base voltage (V)\n", +"Vab=vab/vb;// per unit voltages \n", +"Vbc=vbc/vb;\n", +"Vca=vca/vb;\n", +"a=acosd(((1.12^2)-((.8^2)+1))/(2*.8));\n", +"b=acosd(((.8^2)-((1.12^2)+1))/(2*1.12));\n", +"Vlab=Vab*(cosd(76.06)+%i*sind(76.06));// line voltage \n", +"Vlca=Vca*(cosd(180)+%i*sind(180));// line voltage \n", +"Vlbc=Vbc*(cosd(-43.9)+%i*sind(-43.9));// line voltage \n", +"L=1*(cosd(120) + %i*sind(120));\n", +"Vab1=(Vlab +(L*Vlbc) + ((L^2)*Vlca))/3;// symmetrical component of line voltage \n", +"Vab2=(Vlab +(L*Vlca) + ((L^2)*Vlbc))/3;// symmetrical component of line voltage \n", +"Vabo=0;// symmetrical component of line voltage \n", +"Van1=Vab1*(cosd(-30)+ %i*sind(-30));\n", +"Van2=Vab2*(cosd(30)+ %i*sind(30));\n", +"Ia1=Van1/(1*(cosd(0) + %i*sind(0)));\n", +"Ia2=Van2/(1*(cosd(0) + %i*sind(0)));\n", +"VA1=-%i*Van1;\n", +"VA2=%i*Van2;\n", +"VA=VA1+ VA2;\n", +"VB1=(L^2)*VA1;\n", +"VB2=(L)*VA2;\n", +"VB=VB1 + VB2;\n", +"VC2=(L^2)*VA2;\n", +"VC1=(L)*VA1;\n", +"VC=VC1 + VC2;\n", +"VAB=VA-VB;\n", +"VBC=VB-VC;\n", +"VCA=VC-VA;\n", +"IA=VA;\n", +"IB=VB;\n", +"IC=VC;\n", +"phase_IA=atand(imag(IA)/real(IA));\n", +"phase_IB=atand(imag(IB)/real(IB));\n", +"phase_IC=atand(imag(IC)/real(IC));\n", +"disp(VAB,'VAB(p.u)=');\n", +"disp(VBC,'VBC(p.u)=');\n", +"disp(VCA,'VCA(p.u)=');\n", +"mprintf('IA(p.u)=%.2f at an agle of %.1f\n',abs(IA),phase_IA);\n", +"mprintf('IB(p.u)=%.2f at an agle of %.1f\n',abs(IB),phase_IB);\n", +"mprintf('IC(p.u)=%.2f at an agle of %.1f',abs(IC),phase_IC);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1: Determine_the_symmetrical_components_of_voltages.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the symmetrical components of voltages.\n", +"clear\n", +"clc;\n", +"Va=100*(cosd(0) + %i*sind(0));\n", +"Vb=33*(cosd(-100) + %i*sind(-100));\n", +"Vc=38*(cosd(176.5) + %i*sind(176.5));\n", +"L=1*(cosd(120) + %i*sind(120));\n", +"Va1=((Va + L*Vb + (L^2)*Vc))/3;\n", +"Va2=((Va + L*Vc + (L^2)*Vb))/3;\n", +"Vco=((Va + Vb + Vc))/3;\n", +"disp(Va1,'Va1=');\n", +"disp(Va2,'Va2=');\n", +"disp(Vco,'Vco=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2: Find_the_symmetrical_component_of_currents.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Find the symmetrical component of currents\n", +"clear\n", +"clc;\n", +"Ia=500+ %i*150;// Line current in phase a\n", +"Ib=100- %i*600;// Line current in phase b\n", +"Ic=-300+ %i*600;// Line current in phase c\n", +"L=(cosd(120)+ %i*sind(120));\n", +"Iao=(Ia+Ib+Ic)/3;\n", +"Ia1=(Ia+Ib*L+(L^2)*Ic)/3;\n", +"Ia2=(Ia + (L^2)*Ib +(L*Ic))/3;\n", +"disp(Iao,'Iao(amps)=');\n", +"disp(Ia1,'Ia1(amps)=');\n", +"disp(Ia2,'Ia2(amps)=');// Answer in the book is not correct.wrong calculation in the book\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3: Determine_the_fault_current_and_line_to_line_voltages.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the fault current and line to line voltages \n", +"clear\n", +"clc;\n", +"Ea=1;\n", +"Z1=.25*%i;\n", +"Z2=.35*%i;\n", +"Zo=.1*%i;\n", +"Ia1=Ea/(Z1+Z2+Zo);\n", +"L=-.5+%i*.866;\n", +"Ia2=Ia1;\n", +"Iao=Ia2;\n", +"Ia=Ia1+Ia2+Iao;\n", +"Ib=25*1000/((sqrt(3)*13.2));\n", +"If=Ib*abs(Ia);\n", +"Va1=Ea-(Ia1*Z1);\n", +"Va2=-Ia2*Z2;\n", +"Va0=-Iao*Zo;\n", +"Va=Va1+Va2+Va0;\n", +"Vb1=(L^2)*Va1;\n", +"Vb2=L*Va2;\n", +"Vbo=Va0;\n", +"Vco=Va0;\n", +"Vc1=L*Va1;\n", +"Vc2=(L^2)*Va2;\n", +"Vb=Vb1 + Vb2+Vbo;\n", +"Vc=Vco+Vc1+Vc2;\n", +"Vab=Va-Vb;\n", +"Vac=Va-Vc;\n", +"Vbc=Vb-Vc;\n", +"vab=(13.2*abs(Vab))/sqrt(3);\n", +"vac=(13.2*abs(Vac))/sqrt(3);\n", +"vbc=(13.2*abs(Vbc))/sqrt(3);\n", +"disp(If,'fault current (amps)=');//Answer don't match due to difference in rounding off of digits\n", +"disp(Vab,'Vab(kV)=');//Answer don't match due to difference in rounding off of digits\n", +"disp(Vac,'Vac(kV)=');//Answer don't match due to difference in rounding off of digits\n", +"disp(Vbc,'Vbc(kV)=');//Answer don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4: determine_the_fault_current_and_line_to_line_voltages_at_the_fault.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the fault current and line to line voltage at the fault .\n", +"clear\n", +"clc;\n", +"Ea=1;\n", +"L=(cosd(120)+ %i*sind(120));\n", +"Z1=%i*.25;\n", +"Z2=%i*.35;\n", +"Ia1=Ea/(Z1+Z2);\n", +"Ia2=-Ia1;\n", +"Iao=0;\n", +"Ib1=(L^2)*Ia1;\n", +"Ib2=L*Ia2;\n", +"Ibo=0;\n", +"Ib=Ib1+Ib2 +Ibo;\n", +"Iba=1093;\n", +"If=Iba*abs(Ib);\n", +"Va1=Ea-(Ia1*Z1);\n", +"Va2=-Ia2*Z2;\n", +"Vao=0;\n", +"Va=Va1+Va2+Vao;\n", +"Vb=(L^2)*Va1 + L*Va2;\n", +"Vc=Vb;\n", +"Vab=Va-Vb;\n", +"Vac=Va-Vc;\n", +"Vbc=Vb-Vc;\n", +"mprintf('Fault current =%.2f amps\n',If);//Answer don't match due to difference in rounding off of digits\n", +"vab=(abs(Vab)*13.2)/sqrt(3);\n", +"vbc=(abs(Vbc)*13.2)/sqrt(3);\n", +"vac=(abs(Vac)*13.2)/sqrt(3);\n", +"mprintf('Vab=%.2f kV\n',vab);\n", +"mprintf('Vac=%.2f kV\n',vac);\n", +"mprintf('Vbc=%.2f kV\n',vbc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5: determine_the_fault_current_and_line_to_line_voltages_at_the_fault.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// determine the fault current and line to line voltages at the fault \n", +"clear\n", +"clc;\n", +"Ea=1+ 0*%i;\n", +"Zo=%i*.1;\n", +"Z1=%i*.25;\n", +"Z2=%i*.35;\n", +"Ia1=Ea/(Z1+(Zo*Z2/(Zo+Z2)));\n", +"Va1=Ea-Ia1*Z1;\n", +"Va2=Va1;\n", +"Vao=Va2;\n", +"Ia2=-Va2/Z2;\n", +"Iao=-Vao/Zo;\n", +"I=Ia2+Iao;\n", +"If=3*Iao;// fault current\n", +"Ib=1093;// base current\n", +"Ifl=abs(If*Ib);\n", +"disp(Ifl,'Fault current(amps) =');//Answer don't match due to difference in rounding off of digits\n", +"Va=3*Va1\n", +"Vb=0;\n", +"Vc=0;\n", +"Vab=abs(Va)*13.2/sqrt(3);\n", +"Vac=abs(Va)*13.2/sqrt(3);\n", +"Vbc=abs(Vb)*13.2/sqrt(3);\n", +"mprintf('Vab=%.3f kV\n',Vab);\n", +"mprintf('Vac=%.3f kV\n',Vac);\n", +"mprintf('Vbc=%.3f kV\n',Vbc);\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6: Determine_the_fault_current_when_i_LG_ii_LL_iii_LLG_fault_takes_place_at_P.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the fault current when (i)L-G (ii)L-L (iii)L-L-G fault takes place at P.\n", +"clear\n", +"clc;\n", +"Vbl=13.8*115/13.2;// base voltage on the line side of transformer(kV)\n", +"Vbm=120*13.2/115;// base voltage on the motor side of transformer(kV)\n", +"Xt=10*((13.2/13.8)^2)*30/35;// percent reactance of transformer\n", +"Xm=20*((12.5/13.8)^2)*30/20;// percent reactance of motor\n", +"Xl=80*30*100/(120*120);//percent reactance of line \n", +"Xn=2*3*30*100/(13.8*13.8);// neutral reactance\n", +"Xz=200*30*100/(120*120);\n", +"Zn=%i*.146;// negative sequence impedence\n", +"Zo=.06767;// zero sequence impedence\n", +"Z=%i*.3596;//total impedence\n", +"Ia1=1/Z;\n", +"Ia2=Ia1;\n", +"Iao=Ia2;\n", +"If1=3*Ia1;\n", +"Ib=30*1000/(sqrt(3)*13.8);\n", +"Ibl=30*1000/(sqrt(3)*120);\n", +"Ifc=Ibl*abs(If1);\n", +"Z1=%i*.146;\n", +"Z2=Z1;\n", +"IA1=1/(Z1+Z2)\n", +"IA2=-IA1\n", +"L=(cosd(120)+ %i*sind(120));\n", +"IAo=0;\n", +"IB=(L^2)*IA1 + L*IA2;\n", +"IC=-IB;\n", +"IF=abs(IB)*Ibl;\n", +"Zo=%i*.06767;\n", +"ia1=1/(Z1+(Zo*Z2/(Zo+Z2)));\n", +"ia2=ia1*Zo/(Z2+Zo);\n", +"iao=%i*3.553;\n", +"If2=3*iao;\n", +"IF2=abs(If2*Ibl);\n", +"mprintf('Fault Current (i)L-G fault, If=%.0f amps\n ',Ifc);\n", +"mprintf('(ii)L-L fault ,If=%.1f amps\n',IF);\n", +"mprintf('(iii)L-L-G, If =%.0f amps\n',IF2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8: Determine_the_percent_increase_of_busbar_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the percent increase of busbar voltage \n", +"clear\n", +"clc;\n", +"vx=3;// percent reactance of the series element\n", +"sinr=.6;\n", +"V=vx*sinr;\n", +"mprintf('Percent drop of volts=%.1f percent\n',V);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.9: Determine_the_short_circuit_capacity_of_the_breaker.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the short circuit capacity of the breaker\n", +"clear \n", +"clc;\n", +"Sb=8;// Base MVA\n", +"Zeq=(%i*.15)*(%i*.315)/(%i*.465);\n", +"Scc=abs(Sb/Zeq);\n", +"mprintf('short circuit capacity=%.2f MVA\n',Scc);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/14-PROTECTIVE_RELAYS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/14-PROTECTIVE_RELAYS.ipynb new file mode 100644 index 0000000..e4bb89c --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/14-PROTECTIVE_RELAYS.ipynb @@ -0,0 +1,342 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: PROTECTIVE RELAYS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.10: To_determine_the_kneepoint_voltage_and_cross_section_of_core.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the kneepoint voltage and cross section of core\n", +"clear \n", +"clc;\n", +"Ic=5*.25;// operating current(amp)\n", +"Vsec=5/1.25;// secondary voltage(V)\n", +"Bm=1.4;\n", +"f=50;\n", +"N=50;\n", +"V=15*Vsec;\n", +"A=60/(4.44*Bm*f*N);\n", +"mprintf(' the knee point must be slightly higher than =%.3f V\n',V);\n", +"mprintf('area of cross section=%.6f m_2\n',A);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.11: To_determine_the_VA_output_of_CT.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the VA output of CT .\n", +"clear \n", +"clc;\n", +"o.p=5*5*(.1+.1) +5;\n", +"mprintf(' VA output of CT =%.0f VA\n ',o.p);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1: To_determine_the_time_of_operation_of_relay.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the time of operation of relay .\n", +"clear\n", +"clc;\n", +"If=4000;// fault current \n", +"I=5*1.25;// operating current of relay \n", +"CT=400/5;// CT ratio\n", +"PSM=If/(I*CT);// plug setting multiplier\n", +"mprintf('PSM=%.3f\n',PSM);\n", +"mprintf('operating time for PSM=8 is 3.2sec.\n');\n", +"mprintf('actual operating time = 1.92 sec.');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2: To_determine_the_phase_shifting_network_to_be_used.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the phase shifting network to be used.\n", +"clear\n", +"clc;\n", +"Z=1000*(cosd(60) + %i*sind(60));//impedence\n", +"X=tand(50)*1000*cosd(60);\n", +"Xl=1000*sind(60);\n", +"Xc=Xl-X;\n", +"C=1000000/(314*Xc);\n", +"//Answers don't match due to difference in rounding off of digits\n", +"disp(X,'X=');\n", +"disp(Xc,'Xc=');\n", +"disp(C,'C(micro farads)=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3: To_provide_time_current_grading.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To provide time current grading .\n", +"clear \n", +"clc;\n", +"Isec1=4000/40;// secondary current(amps)\n", +"PSM=100/5;// PSM if 100% setting is used\n", +"Isec2=4000/40;\n", +"PSM2=100/6.25;//PSM if setting used is 125%\n", +"TMSb=.72/2.5;\n", +"PSM1=5000/(6.25*40);\n", +"to=2.2;\n", +"tb=to*TMSb;\n", +"PSMa=5000/(6.25*80);\n", +"TMS=1.138/3;\n", +"PSMa1=6000/(6.25*80);\n", +"ta=(2.6*.379);\n", +"mprintf('Actual operating time of realy at b=%.3f sec. \n',tb);\n", +"mprintf('Actual operating time of realy at a=%.3f sec. \n',ta);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4: EX14_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the proportion of the winding which remains unprotected against earth fault.\n", +"clear\n", +"clc;\n", +"Vph=6600/(sqrt(3));\n", +"Ifull=5000/(sqrt(3)*6.6);\n", +"Ib=Ifull*.25;\n", +"x=Ib*800/Vph;\n", +"mprintf('percent of the winding remains unprotected=%.2f \n',x);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5: EX14_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine (i) % winding which remains unprotected (ii)min. value of earthing resistance required to protect 80% of winding \n", +"clear \n", +"clc;\n", +"Iph=10000/sqrt(3);// phase voltage of alternator(V)\n", +"x=1.8*100*10*1000/(5*Iph);\n", +"mprintf('(i) percent winding which remains unprotected=%.2f \n',x);\n", +"Ip=Iph*.2;\n", +"R=1.8*1000/(5*Ip);\n", +"mprintf('(ii)minimum value of earthing resistance required to protect 80 percent of winding =%.4f ohms \n',R)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6: To_determine_whether_relay_will_operate_or_not.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine whether relay will operate or not.\n", +"clear \n", +"clc;\n", +"Ic=360-320;// the difference current (amp)\n", +"Io=40*5/400;\n", +"Avg=(360+320)/2;// average sum of two currents\n", +"Iavg=340*5/400;\n", +"Ioc=.1*Iavg + .2;\n", +"mprintf('operating current=%.3f amp. \n',Ioc);\n", +"mprintf('since current through operating coil is %.3f amp. \n ',Io);\n", +"mprintf('therefore Relay will not operate ');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7: To_determine_the_ratio_of_CT_on_HV_side.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the ratio of CT on HV side\n", +"clear\n", +"clc;\n", +"Il=400*6.6/33;// line current on star side of PT(amps)\n", +"Ic=5/sqrt(3);// current in CT secondary \n", +"mprintf(' the CT ratio on HT will be %d : %.3f',Il,Ic);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8: To_determine_the_number_of_turns_each_current_transformer_should_have.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the number of turns each current transformer should have .\n", +"clear \n", +"clc;\n", +"Il=10000/((sqrt(3))*132);\n", +"ILV=10000/((sqrt(3))*6.6);\n", +"a=5/sqrt(3);\n", +"mprintf('ratio of CT on LV side is %.3f : %.3f\n',ILV,a);\n", +"mprintf('ratio of CT on HT side is %.3f : %d',Il,5);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.9: To_determine_the_R1_R2_and_C_also_The_potential_across_relays.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the R1, R2 and C. also The potential across relays\n", +"clear \n", +"clc;\n", +"Vs=110;\n", +"I=1;\n", +"R2=Vs/((3-%i*sqrt(3))*I);\n", +"c=abs(R2);\n", +"mprintf('R2=%.2f ohms\n',c);\n", +"R1=2*c;\n", +"d=abs(R1);\n", +"C=(10^6)/(.866*d*314);\n", +"mprintf('R1=%.2f ohms\n',R1);\n", +"mprintf('C=%.1f micro farads\n',C);\n", +"Vt=d*(-.5 - %i*.866) + (c - %i*55 );\n", +"disp(Vt,' Voltage across the terminals of the relay will be (V)=');" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/15-CIRCUIT_BREAKERS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/15-CIRCUIT_BREAKERS.ipynb new file mode 100644 index 0000000..132878f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/15-CIRCUIT_BREAKERS.ipynb @@ -0,0 +1,183 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15: CIRCUIT BREAKERS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1: EX15_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the voltage appearing across the pole of C.B. also determine the value of resistance to be used across contacts\n", +"clear \n", +"clc;\n", +"i=5;\n", +"L=5*(10^6);\n", +"C=.01;\n", +"e=i*sqrt(L/C);\n", +"mprintf('the voltage appearing across the pole of C.B.=%.0f V\n',e);\n", +"R=.5*sqrt(L/C);\n", +"mprintf('the value of resistance to be used across contacts, R=%.0f ohms\n',R);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2: To_determine_the_rate_of_rise_of_restriking_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the rate of rise of restriking voltage\n", +"clear \n", +"clc;\n", +"Vnl=132*sqrt(2)/sqrt(3);//peak value of peak to neutral voltage(kV)\n", +"Vr1=Vnl*.95;//recovery voltage (kV)\n", +"Vr=102.4*.916;// active recovery voltage(kV)\n", +"Vrmax=2*Vr;\n", +"fn=16*(10^3);\n", +"t=1/(2*fn);\n", +"RRRV=Vrmax*(10^-6)/t;\n", +"mprintf('rate of rise of restriking voltage, RRRV=%.0f kV/micro-sec',RRRV);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.3: To_Determine_the_average_rate_of_rise_of_restriking_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Determine the average rate of rise of restriking voltage\n", +"clear \n", +"clc;\n", +"Vm=132*sqrt(2)/sqrt(3);\n", +"K1=.9;\n", +"K2=1.5\n", +"K=K1*K2;\n", +"sinq=.92;\n", +"Vr=K*Vm*sinq;\n", +"fn=16*(10^3);\n", +"RRRV=2*Vr*(10^-6)*fn*2;\n", +"mprintf('average rate of rise of restriking voltage,RRRV=%.3f kV/micro-sec\n',RRRV);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.4: EX15_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the rated normal current , breaking current , making current and short time rating (current)\n", +"clear \n", +"clc;\n", +"In=1500;\n", +"mprintf('rated normal current=%.0f amps\n',In);\n", +"Ib=2000/(sqrt(3)*33);\n", +"mprintf('breaking current=%.2f KA\n',Ib);\n", +"Im=2.55*Ib;\n", +"mprintf('making current =%.2f kA\n',Im);\n", +"Is=Ib;\n", +"mprintf('short time rating for 3 sec=%.2f kA\n',Is);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.5: EX15_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//TO Determine (i)sustained short circuit current in the breaker (ii)initial symmetrical r.m.s current in the breaker (iii)maximum possible d.c component of the short circuit current in the breaker (iv)momentary current rating of the breaker (v)the current to be interrupted by the breaker (vi)the interupting kVA.\n", +"clear\n", +"clc;\n", +"MVA=10;\n", +"Is=MVA*1000/(sqrt(3)*13.8);\n", +"mprintf('(i)sustained short circuit current in the breaker =%.0f amps\n',Is);\n", +"MVA1=100;\n", +"Isc=MVA1*1000/(sqrt(3)*13.8);\n", +"mprintf('(ii)initial symmetrical r.m.s current in the breaker r.m.s=%.0f amps\n',Isc);\n", +"Im=sqrt(2)*Isc;\n", +"mprintf('(iii)maximum possible d.c component of the short circuit current in the breaker =%.0f amps\n',Im);\n", +"Im2=1.6*Isc;\n", +"mprintf('(iv)momentary current rating of the breaker=%.0f amps\n',Im2);\n", +"Ib=1.2*Isc;\n", +"mprintf('(v)the current to be interrupted by the breaker =%.0f amps\n',Ib);\n", +"KVA=sqrt(3)*13.8*5016;\n", +"mprintf('(vi)the interupting =%.0f KVA\n',KVA);\n", +"//Answers don't match due to difference in rounding off of digits" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/17-POWER_SYSTEM_SYNCHRONOUS_STABILITY.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/17-POWER_SYSTEM_SYNCHRONOUS_STABILITY.ipynb new file mode 100644 index 0000000..19d9555 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/17-POWER_SYSTEM_SYNCHRONOUS_STABILITY.ipynb @@ -0,0 +1,426 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17: POWER SYSTEM SYNCHRONOUS STABILITY" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.10: EX17_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Determine the rotor angle and angular frequency using runga kutta and euler`s modified method\n", +"\n", +"clc\n", +"clear\n", +"Pm=3;\n", +"r1Pm=1.2;\n", +"r2Pm=2;\n", +"H=3;\n", +"f=60;\n", +"Dt=.02;\n", +"Pe=1.5;\n", +"Do=asind(1.5/3);\n", +"do=Do/57.33;\n", +"wo=0;\n", +"d=0;\n", +"K10=0;\n", +"l10=62.83*(1.5-1.2*sin(do))*.02;\n", +"K20=(377.5574-376.992)*.02;\n", +"l20=62.83*(1.5-1.2*sin(do))*.02;\n", +"K30=(377.5574-376.992)*.02;\n", +"l30=62.83*(1.5-1.2*sin(.5296547))*.02;\n", +"K40=l30*0.02;\n", +"l40=62.83*(1.5-1.2*sin(.5353094))*.02;\n", +"d1=.53528;\n", +"Dwo=(3*1.13094+2*1.123045+1.115699)/6;\n", +"w1=wo+Dwo;\n", +"d1=.53528;\n", +"mprintf('Runga-Kutta method-\n')\n", +"mprintf('w1=%.6f \nd1=%.5f\n',w1,d1);\n", +"d7=1.026;\n", +"w7=6.501;\n", +"wp=376.992+6.501;\n", +"K17=(wp-376.992)*0.02;\n", +"l17=62.83*(1.5-1.2*sin(1.026))*.02;\n", +"K27=(6.501+.297638)*0.02;\n", +"l27=62.83*(1.5-1.2*sin(1.09101))*.02;\n", +"K37=(6.501+.2736169)*0.02;\n", +"l37=62.83*(1.5-1.2*sin(1.0939863))*.02;\n", +"K47=(6.501+.545168)*0.02;\n", +"l47=62.83*(1.5-1.2*sin(1.16149))*.02;\n", +"Dd7=(K17+2*K27+2*K37+K47)/6;\n", +"d8=d7+Dd7;\n", +"Dw7=(l17+2*l27+2*l37+l47)/6;\n", +"w8=w7+Dw7;\n", +"mprintf('d8=%.5f rad.\nw8=%.4frad/sec\n\n',d8,w8)\n", +"mprintf('using Euler`s Modified Method-\n');\n", +"d0=0;\n", +"d10=.524;\n", +"w=62.83*(1.5-1.2*sin(.524));\n", +"d11=d10+0;\n", +"w11=w*.02;\n", +"d=1.13094;\n", +"dav=(0+d)/2;\n", +"wav=(56.547+56.547)/2;\n", +"d01=.524+.56547*.02;\n", +"w11=0+56.547*0.02;\n", +"mprintf('d01=%.4f\nw11=%.5f',d01,w11);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.1: EX17_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the acceleration . Also determine the change in torque angle and r.p.mat the end of 15 cycles\n", +"clear\n", +"clc;\n", +"H=9;\n", +"G=20;// machine Rating(MVA)\n", +"KE=H*G;\n", +"mprintf('(a)K.E stored in the rotor =%.0f MJ\n',KE);\n", +"Pi=25000*.735;\n", +"PG=15000;\n", +"Pa=(Pi-PG)/(1000);\n", +"f=50;\n", +"M=G*H/(%pi*f);\n", +"a=Pa/M;\n", +"mprintf('(b) The accelerating power =%.3f MW\n',Pa);\n", +"mprintf('Acceleration =%.3f rad/sec_2\n',a);\n", +"t=15/50;\n", +"del=sqrt(5.89)*t/2;\n", +"Del=del^2;\n", +"k=2.425*sqrt(Del)*60/4*%pi;\n", +"speed=1504.2;\n", +"mprintf('(c)Rotor speed at the end of 15 cycles =%.1f r.p.m',speed);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.2: EX17_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the frequency of natural oscillations if the genrator is loaded to (i)60% and (ii)75% of its maximum power transfer capacity\n", +"clear \n", +"clc;\n", +"V1=1.1;\n", +"V2=1;\n", +"X=.5;\n", +"cosdo=.8;\n", +"G=1;\n", +"H=3;\n", +"f=50;\n", +"M=G*H/(%pi*f);\n", +"dPe=V1*V2*cosdo/X;\n", +"fn=(((dPe)/M)^.5)/6.28;\n", +"sind0=.75;\n", +"d0=asind(sind0);\n", +"dPe2=V1*V2*cosd(d0)/X;\n", +"fn2=(((dPe2)/M)^.5)/6.28;\n", +"mprintf('(i)fn=%.2f Hz\n',fn);\n", +"mprintf('(i)fn(Hz)=%.2f Hz',fn2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.3: EX17_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To calculate the maximum value of d during the swinging of the rotor around its new equilibrium position\n", +"clc\n", +"clear\n", +"a=.25;//sindo=.25\n", +"do=asind(a);//\n", +"b=.5//sindc=.5\n", +"dc=asind(b);\n", +"c=cosd(do)+.5*do*%pi/180;\n", +"dm=dc;\n", +"e=1;\n", +"while(e>.0001)\n", +" dm=dm+.1;\n", +" e=abs(c-(((.5*dm*%pi)/180)+cosd(dm)));\n", +"end\n", +"printf('dm approximately found to be %d degree',dm);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.4: To_calculate_the_critical_clearing_angle_for_the_condition_described.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To calculate the critical clearing angle for the condition described.\n", +"clear \n", +"clc;\n", +"sindo=.5;\n", +"d0=asind(sindo)*%pi/180;\n", +"r1=.2;\n", +"r2=.75;\n", +"sindm=.5/.75;\n", +"d=asind(sindm);\n", +"cosdm=cosd(d);\n", +"dm=%pi*(180-(asind(sindm)))/180;\n", +"Dc=((.5*(dm-d0))-(r2*cosdm)-(r1*cosd(d0)))/(r2-r1);\n", +"dc=acosd(Dc);// critical angle\n", +"mprintf('The critical clearing angle is given by=%.2f degrees',dc);//Answers don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.5: To_calculate_the_critical_clearing_angle_for_the_generator_for_a_3phase_fault.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To calculate the critical clearing angle for the generator for a 3-phase fault\n", +"clear \n", +"clc;\n", +"ZA=.375;\n", +"ZB=.35;\n", +"ZC=.0545;\n", +"ZAB=((ZA*ZB)+(ZB*ZC)+(ZC*ZA))/ZC;//Reactance between the generator and infinite bus during the fault(p.u)\n", +"Zgbf=%i*.3+ %i*(.55/2) +%i*.15;//Reactance between the generator and infinite bus before the fault(p.u)\n", +"Zgb=%i*.3+ %i*(.55) +%i*.15;//Reactance between the generator and infinite bus after the fault is cleared (p.u)\n", +"Pmaxo=1.2*1/abs(Zgbf);// Maximum power output Before the fault(p.u)\n", +"Pmax1=1.2*1/abs(ZAB);// Maximum power output during the fault(p.u)\n", +"Pmax2=1.2*1/abs(Zgb);// Maximum power output after the fault(p.u)\n", +"r1=Pmax1/Pmaxo;\n", +"r2=Pmax2/Pmaxo;\n", +"Ps=1;\n", +"sindo=Ps/Pmaxo;\n", +"do=asind(sindo);\n", +"d0=asind(sindo)*%pi/180;\n", +"sindm=1/Pmax2;\n", +"cosdm=cosd(asind(sindm));\n", +"Dm=%pi*(180-(asind(sindm)))/180;\n", +"Dc=(((sindo*(Dm-d0))-(r2*cosdm))-(r1*cosd(do)))/(r2-r1);\n", +"dc=acosd(Dc);// critical angle\n", +"mprintf('The critical clearing angle is given by= %.1f ',dc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.6: determine_the_critical_clearing_angle.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//(A) determine the critical clearing angle \n", +"clear \n", +"clc;\n", +"Pm=%i*.12 + %i*.035 + ((%i*.25*%i*.3)/%i*.55);\n", +"Pm1=0;\n", +"Pm2=1.1*1/.405;\n", +"r1=0;\n", +"r2=2.716/3.775;\n", +"d0=(asind(1/3.775));\n", +"dM=(180-asind(1/2.716));\n", +"do=d0*%pi/180;\n", +"dm=dM*%pi/180;\n", +"dc=acosd((((dm-do)*sind(d0))-(r1*cosd(d0))+(r2*cosd(dM)))/(r2-r1));\n", +"mprintf('dc=%.2f',dc);\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.7: EX17_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the centre and radius for the pull out curve ans also minimum output vars when the output powers are (i)0 (ii).25p.u (iii) .5p.u\n", +"clear \n", +"clc;\n", +"Pc=0;\n", +"V=.98;\n", +"Qc=V^2*((1/.4)-(1/1.1))/2;\n", +"R=V^2*((1/.4)+(1/1.1))/2;\n", +"Q=-(.98^2*((1.1-.4)/.44)/2) + (.98^2)*1.5/(2*.44);\n", +"mprintf('(i)Q=%.2f MVAr\n',abs(Q)*100);\n", +"P=.25;\n", +"Q2=-((1.637^2)-(.25^2))^.5 + .7639;\n", +"mprintf('(ii)Q=%.4f p.u\n',Q2);\n", +"Q3=-((1.637^2)-(.5^2))^.5 + .7639;\n", +"mprintf('(iii)Q=%.4f p.u',Q3);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.8: Compute_the_prefault_faulted_and_post_fault_reduced_Y_matrices.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Compute the prefault, faulted and post fault reduced Y matrices\n", +"clear \n", +"clc;\n", +"y=[-%i*5 0 %i*5 ; 0 -%i*5 %i*5;%i*5 %i*5 -%i*10 ];\n", +"YAA=[-%i*5 0;0 -%i*5];\n", +"YAB=[%i*5;%i*5];\n", +"YBA=[%i*5 %i*5];\n", +"YBB=[%i*10];\n", +"Y=YAA-YAB*(inv(YBB))*YBA;\n", +"Yfull=[-%i*5 0 %i*5;0 -%i*7.5 %i*2.5;%i*5 %i*2.5 -%i*12.5];\n", +"disp(Yfull,'(i)faulted case, full matrix(admittance)=');\n", +"Y=[-%i*3 %i*1;%i*1 -%i*7];\n", +"disp(Y,'(ii)Pre-fault case, reduced admittance matrix=');\n", +"Y=[-%i*5 0 %i*5;0 -%i*2.5 %i*2.5;%i*5 %i*2.5 -%i*7.5];\n", +"disp(Y,'(iii)Post-fault case, full matrix(admittance)=');\n", +"Y=[-%i*1.667 %i*1.667;%i*1.667 -%i*1.667];\n", +"disp(Y,' reduced admittance matrix=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.9: EX17_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the reduced admittance matrices for prefault, fault and post fault conditions and determine the power angle characterstics for three conditions.\n", +"clear \n", +"clc;\n", +"Y=[-%i*8.33 0 %i*8.33 0;0 -%i*28.57 0 %i*28.75;%i*8.33 0 -%i*15.67 %i*7.33;0 %i*28.57 %i*7.33 -%i*35.9];\n", +"YBB=[-%i*15.67 %i*7.33;%i*7.33 -%i*35.9];\n", +"YAA=[-%i*8.33 0;0 -%i*28.57];\n", +"YAB=[%i*8.33 0;0 %i*28.57];\n", +"YBA=YAB;\n", +"Y=YAA-(YAB*(inv(YBB))*YBA);\n", +"Y1=([-%i*8.33 0;0 -%i*28.57])-(([0;(%i*28.57/-%i*35.9)]*[0 %i*28.57]));\n", +"disp(Y1,'Reduced admittance matrix during fault=');\n", +"Yfull=[-%i*8.33 0 %i*8.33 0;0 -%i*28.57 0 %i*28.75;%i*8.33 0 -%i*12.33 %i*4;0 %i*28.57 %i*4 -%i*32.57];\n", +"YBB=[-%i*12.33 %i*4;%i*4 -%i*32.57];\n", +"Y=YAA-(YAB*(inv(YBB))*YBA);\n", +"disp(Y,'(i) Post fault condition ,reduced matrix=');\n", +"Y12=Y(1,2);\n", +"E1=1.1;\n", +"E2=1;\n", +"printf('\n Power angle characterstics , Pe= %fsind',abs(Y12)*E1*E2);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/18-LOAD_FLOWS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/18-LOAD_FLOWS.ipynb new file mode 100644 index 0000000..2f1011f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/18-LOAD_FLOWS.ipynb @@ -0,0 +1,264 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18: LOAD FLOWS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1: Determine_the_voltages_at_the_end_of_first_iteration_using_gauss_seidal_method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the voltages at the end of first iteration using gauss seidal method\n", +"clear \n", +"clc;\n", +"Y=[3-%i*12 -2+%i*8 -1+%i*4 0;-2+%i*8 3.666-%i*14.664 -.666+%i*2.6664 -1+%i*4;-1+%i*4 -.666+%i*2.6664 3.666-%i*14.664 -2+%i*8;0 -1+%i*4 -2+%i*8 3-%i*12];\n", +"P2=-.5;\n", +"P3=-.4;\n", +"P4=-.3;\n", +"Q4=-.1;\n", +"Q3=-.3;\n", +"Q2=-.2;\n", +"V2=1;\n", +"V3=1;\n", +"V4=1;\n", +"V10=1.06;\n", +"V30=1;\n", +"V40=1;\n", +"V21=(((P2-%i*Q2)/V2)-Y(2,1)*V10-Y(2,3)*V30-Y(2,4)*V40)/(Y(2,2));\n", +"V21acc=1+1.6*(V21-1);\n", +"disp(V21acc,'V21acc=');\n", +"V31=(((P3-%i*Q3)/V3)-Y(3,1)*V10-Y(3,2)*V21acc-Y(3,4)*V40)/(Y(3,3));\n", +"V31acc=1+1.6*(V31-1);\n", +"disp(V31acc,'V31acc=');\n", +"V41=(((P4-%i*Q4)/V4)-Y(4,2)*V21acc-Y(4,3)*V31acc)/(Y(4,4));\n", +"V41acc=1+1.6*(V41-1);\n", +"disp(V41acc,'V41acc=');\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.2: Determine_the_voltages_starting_with_a_flat_voltage_profile.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the voltages starting with a flat voltage profile.\n", +"clear \n", +"clc;\n", +"\n", +"Y=[3-%i*12 -2+%i*8 -1+%i*4 0;-2+%i*8 3.666-%i*14.664 -.666+%i*2.6664 -1+%i*4;-1+%i*4 -.666+%i*2.6664 3.666-%i*14.664 -2+%i*8;0 -1+%i*4 -2+%i*8 3-%i*12];\n", +"P2=.5;\n", +"P3=-.4;\n", +"P4=-.3;\n", +"Q4=-.1;\n", +"Q3=-.3;\n", +"V3=1;\n", +"V4=1;\n", +"V1=1.06;\n", +"V2=1.04;\n", +"V30=1;\n", +"V40=1;\n", +"Q2=-imag([V2*[Y(2,1)*V1+Y(2,2)*V2+Y(2,3)*V3+Y(2,4)*V4]]);\n", +"V21=(((P2-%i*Q2)/V2)-Y(2,1)*V1-Y(2,3)*V30-Y(2,4)*V40)/(Y(2,2));\n", +"d=atand(0.0291473/1.0472868);\n", +"V21=1.04*(cosd(d)+%i*sind(d));\n", +"disp(V21,'V21=');\n", +"V31=(((P3-%i*Q3)/V3)-Y(3,1)*V1-Y(3,2)*V21-Y(3,4)*V40)/(Y(3,3));\n", +"disp(V31,'V31=');\n", +"V41=(((P4-%i*Q4)/V4)-Y(4,2)*V21-Y(4,3)*V31)/(Y(4,4));\n", +"disp(V41,'V41=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3: Solve_the_prevous_problem_for_for_voltages_at_the_end_of_first_iteration.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Solve the prevous problem for for voltages at the end of first iteration. for .2<=Q2<=1\n", +"clear \n", +"clc;\n", +"\n", +"Y=[3-%i*12 -2+%i*8 -1+%i*4 0;-2+%i*8 3.666-%i*14.664 -.666+%i*2.664 -1+%i*4;-1+%i*4 -.666+%i*2.664 3.666-%i*14.664 -2+%i*8;0 -1+%i*4 -2+%i*8 3-%i*12];\n", +"P2=.5;\n", +"P3=-.4;\n", +"P4=-.3;\n", +"Q4=-.1;\n", +"Q3=-.3;\n", +"V3=1;\n", +"V4=1;\n", +"V1=1.06;\n", +"V2=1;\n", +"V30=1;\n", +"V40=1;\n", +"Q2=.2;\n", +"V3=1;\n", +"V21=(((P2-%i*Q2)/V2)-Y(2,1)*V1-Y(2,3)*V30-Y(2,4)*V40)/(Y(2,2));\n", +"V31=(((P3-%i*Q3)/V3)-Y(3,1)*V1-Y(3,2)*V21-Y(3,4)*V40)/(Y(3,3));\n", +"V41=(((P4-%i*Q4)/V4)-Y(4,2)*V21-Y(4,3)*V31)/(Y(4,4));\n", +"disp(V21,'V21=');\n", +"disp(V31,'V31=');\n", +"disp(V41,'V41=');\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4: EX18_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the set of load flow equations at the end of first iteration by using Newton Raphson method.\n", +"clear \n", +"clc;\n", +"Y=[6.25-%i*18.75 -1.25+%i*3.75 -5+%i*15;-1.25+%i*3.75 2.916-%i*8.75 -1.666+%i*5;-5+%i*15 -1.666+%i*5 6.666-%i*20];\n", +"V1=1.06;\n", +"G11=6.25;\n", +"G12=-1.25;\n", +"G21=G12;\n", +"G13=-5;\n", +"G31=G13;\n", +"G22=2.916;\n", +"G23=-1.666;\n", +"G32=G23;\n", +"G33=6.666;\n", +"B11=18.75;\n", +"B12=-3.75;\n", +"B21=B12;\n", +"B13=-15;\n", +"B31=B13;\n", +"B22=8.75;\n", +"B23=-5;\n", +"B32=B23;\n", +"B33=20;\n", +"e1=1.06;\n", +"e2=1;\n", +"e3=1;\n", +"f1=0;\n", +"f2=0;\n", +"f3=0;\n", +"P2=e2*(e1*G21+f1*B21) +f2*(f1*G21-e1*B21) +e2*(e2*G22+f2*B22)+f2*(f2*G22-e2*B22)+e2*(e3*G23+f3*B23)+f2*(f3*G23-e3*B23);\n", +"P3=-.3\n", +"Q2=-.225;\n", +"Q3=-.9;\n", +"dP2=.2-(-.225);\n", +"dP3=-.6-(-.3);\n", +"dQ2=0-(-.225);\n", +"dQ3=-.25-(-.9);\n", +"a1=2*e2*G22+e1*G21+f1*B21+e3*G23+f3*B23;//a1=dP2/de2\n", +"a2=2*e3*G33+e1*G31+f1*B31+e3*G32+f2*B32;//a2=dP3/de3\n", +"b1=2*f2*G22 +f1*G21-e1*B21+f3*G23-e3*B23;//b1=dP2/df2\n", +"b2=20.9;//dP3/df3\n", +"a3=e2*G23-f2*B23;//dP2/de3\n", +"a4=-1.666;//dP3/de2\n", +"b3=-5;//dP2/df3\n", +"b4=-5;//dP3/df2\n", +"c1=2*e2*B22-f1*G21+e1*B21-f3*G23+e3*B23;//dQ2/de2\n", +"c2=19.1;//dQ3/de3\n", +"c3=-2.991;//dQ2/df2\n", +"c4=-6.966;//dQ3/df3\n", +"mprintf('set of linear equations at the end of first iteration are\n');\n", +"mprintf('%.3fde2 %.3fde3+ %.3fdf2 %.3fdf3 = %.3f\n',2.846,-1.666,8.975,-5,2.75);\n", +"mprintf('%.3fde2 +%.3fde3 %.3fdf2 +%.3fdf3 = %.3f\n',-1.666,6.366,-5,20.90,-.3);\n", +"mprintf('%.3fde2 %.3fde3 %.3fdf2 +%.3fdf3 = %.3f\n',8.525,-5,-2.991,1.666,.225);\n", +"mprintf('%.3fde2 +%.3fde3+ %.3fdf2 %.3fdf3 = %.3f\n',-5,19.1,1.666,-6.966,.65);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.5: EX18_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the equations at the end of first iteration after applying given constraints.\n", +"clear \n", +"clc;\n", +"Q2=-.225;\n", +"dP2=.2-(-.075);\n", +"dP3=-.6-(-.3);\n", +"dQ3=-.25-(-.9);\n", +"dV2=1.04^2 - 1^2;//dV2=|dV2|^2\n", +"mprintf('set of linear equations at the end of first iteration are\n');\n", +"mprintf('%.3fde2 %.3fde3+ %.3fdf2 %.3fdf3 = %.3f\n',2.846,-1.666,8.975,-5,2.75);\n", +"mprintf('%.3fde2 +%.3fde3 %.3fdf2 +%.3fdf3 = %.3f\n',-1.666,6.366,-5,20.90,-.3);\n", +"mprintf('%.3fde2 %.3fde3 %.3fdf2 +%.3fdf3 = %.3f\n',8.525,-5,-2.991,1.666,.225);\n", +"mprintf('%.3fde2 +%.3fde3+ %.3fdf2 +%.3fdf3 = %.5f\n',2,0,0,0,dV2);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/19-ECONOMIC_LOAD_DISPATCH.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/19-ECONOMIC_LOAD_DISPATCH.ipynb new file mode 100644 index 0000000..518571f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/19-ECONOMIC_LOAD_DISPATCH.ipynb @@ -0,0 +1,143 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19: ECONOMIC LOAD DISPATCH" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1: EX19_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Determine the economic operating schedule and the corresponding cost of generation.(b)Determine the savings obtained by loading the units.\n", +"clear \n", +"clc;\n", +"//dF1/dP1=.4*P1+40 per MWhr\n", +"//dF2/dP2=.5*P1+30 per MWhr\n", +"mprintf('two equations are :\n');\n", +"mprintf('%.1f P1 %.1f P2 = %.1f\n',.4,-.5,-10);\n", +"mprintf('%.1f P1+ %.1fP2 = %.1f\n',1,1,180);\n", +"A=[.4 -.5;1 1];\n", +"B=[-10;180];\n", +"P=(inv(A))*B;\n", +"P1=P(1,1);\n", +"P2=P(2,1);\n", +"F1=.2*(P1)^2 +40*P1+120;\n", +"F2=.25*(P2)^2+30*P2+150;\n", +"Total=F1+F2;//Total cost\n", +"mprintf('(a)Cost of Generation=Rs %.2f /hr\n',Total);\n", +"P1=90;\n", +"P2=90;\n", +"F1=.2*(P1)^2 +40*P1+120;\n", +"F2=.25*(P2)^2+30*P2+150;\n", +"Total2=F1+F2;//Total cost\n", +"savings=Total2-Total\n", +"mprintf('(b)Savings=Rs %.2f /hr\n',savings)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2: Determine_the_incremental_cost_of_recieved_power_and_penalty_factor_of_the_plant.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the incremental cost of recieved power and penalty factor of the plant\n", +"clear \n", +"clc;\n", +"pf=10/8;//penalty factor\n", +"cost=(.1*10+3)*pf;//Cost of recieved power=dF1/dP1\n", +"mprintf('Penalty Factor=%.1f\n',pf);\n", +"mprintf('Cost of recieved Power=Rs %.1f /MWhr',cost);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4: Determine_the_minimum_cost_of_generation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the minimum cost of generation .\n", +"clear \n", +"clc;\n", +"//dF1/dP1=.048*P1+8\n", +"//dF2/dP2=.08*P1+6\n", +"mprintf('two equations are :\n');\n", +"mprintf('%.3f P1 %.2f P2 = %.1f\n',.048,-.08,-2);\n", +"mprintf('%.1f P1+ %.1fP2 = %.1f\n',1,1,50);\n", +"A=[.048 -.08;1 1];\n", +"B=[-2;50];\n", +"P=(inv(A))*B;\n", +"P1=P(1,1);\n", +"P2=P(2,1);\n", +"F1=(.024*(P1)^2 +8*P1+80)*(10^6);\n", +"F2=(.04*(P2)^2+6*P2+120)*(10^6);\n", +"mprintf('when load is 150MW , equations are: :\n');\n", +"mprintf('%.3f P1 %.2f P2 = %.1f\n',.048,-.08,-2);\n", +"mprintf('%.1f P1+ %.1fP2 = %.1f\n',1,1,150);\n", +"A=[.048 -.08;1 1];\n", +"B=[-2;150];\n", +"P=(inv(A))*B;\n", +"P1=P(1,1);\n", +"P2=P(2,1);\n", +"f1=(.024*(P1)^2 +8*P1+80)*(10^6);\n", +"f2=(.04*(P2)^2+6*P2+120)*(10^6);\n", +"Total=(F1+F2+f1+f2)*12*2/(10^6);\n", +"mprintf('Total cost=Rs. %.2f',Total)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/2-LINE_CONSTANT_CALCULATIONS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/2-LINE_CONSTANT_CALCULATIONS.ipynb new file mode 100644 index 0000000..9e652be --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/2-LINE_CONSTANT_CALCULATIONS.ipynb @@ -0,0 +1,219 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: LINE CONSTANT CALCULATIONS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2: To_dtermine_inductance_of_a_3_phase_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To dtermine inductance of a 3 phase line \n", +"clear\n", +"clc;\n", +"GMD=0.7788*0.8/(2*100);\n", +"Mgmd=((1.6*3.2*1.6)^(1/3));\n", +"Z=2*(10^-4)*1000*log(2.015/.003115);\n", +"mprintf('The self GMD of the conductor =%.6f metres\n',GMD);\n", +"mprintf('The mutual GMD of the conductor =%.3f metres\n ',Mgmd);\n", +"mprintf('Inductance =%.3f mH/km\n',Z);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3: EX2_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//What will be the equivalent radius of bundle conductor having its part conductors 'r' on the periphery of circle of dia'd' if the number of conductors is 2,3,4 ,6 ?\n", +"\n", +"clear\n", +"clc;\n", +"r=poly(0,'r');\n", +"D11=r^1;\n", +"D12=2*r;\n", +"D14=4*r\n", +"D13=sqrt(16-4)*r;\n", +"Ds1=((1*2*2*sqrt(3)*4*2*sqrt(3)*2*2)^(1/7))*r;\n", +"Ds7=((2*1*2*2**2*2*2)^(1/7))*r;//we get this after Taking r outside the 1/7th root\n", +"Ds=((((1*2*2*sqrt(3)*4*2*sqrt(3)*2*2)^(1/7))^6)*((2*1*2*2**2*2*2)^(1/7)))^(1/7)*r;\n", +"Dseq=((.7788)^(1/7))*Ds;\n", +"disp(Dseq,'Dseq.= ');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4: To_determine_the_inductance_of_single_phase_Transmission_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the inductance of single phase Transmission line\n", +"clear\n", +"clc;\n", +"GMDa=0.001947;// GMD of conductor in group A\n", +"DSA=((.001947*6*12*.001947*6*6*0.001947*6*12)^(1/9));\n", +"DSB=sqrt(5*(10^-3)*.7788*6);\n", +"Dae=sqrt((9^2)+6^2);\n", +"Dcd=sqrt((12^2)+9^2);\n", +"DMA=((9*10.81*10.81*9*15*10.81)^(1/6));\n", +"LA=2*(10^-7)*(10^6)*log(DMA/DSA);\n", +"LB=2*(10^-7)*(10^6)*log(DMA/DSB);\n", +"Tot=LA+LB;\n", +"mprintf('inductance of line A,LA=%.3f mH/km\n',LA);//Answers don't match due to difference in rounding off of digits\n", +"mprintf('inductance of line B,LB=%.1f mH/km\n',LB);//Answers don't match due to difference in rounding off of digits\n", +"mprintf('total inductance of line =%.2f mH/km\n',Tot);//Answers don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5: To_determine_the_inductance_per_Km_of_3_phase_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the inductance per Km of 3-phase line\n", +"clear\n", +"clc;\n", +"GMDc=1.266*0.7788*(10^-2);// self GMD of each conductor\n", +"Dbc=sqrt((4^2)+(.75^2));\n", +"Dab=Dbc;\n", +"Dab`=sqrt((4^2)+(8.25^2));\n", +"Daa=sqrt((8^2)+(7.5^2));\n", +"Dm1=(Dbc*8*7.5*9.1685)^(1/4);\n", +"Dm2=(Dbc*Dbc*9.1685*9.1685)^(1/4);\n", +"Dm3=Dm1;\n", +"Dm=((Dm1*Dm2*Dm3)^(1/3));\n", +"Ds1=sqrt(GMDc*Daa);// self GMD of each phase\n", +"Ds3=Ds1;\n", +"Ds2=sqrt(GMDc*9);\n", +"Ds=((Ds1*Ds2*Ds3)^(1/3));\n", +"Z=2*(10^-4)*(1000)*log(Dm/Ds);\n", +"mprintf('inductance=%.3f mH/km/phase\n',Z);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6: To_determine_the_inductance_of_double_circuit_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the inductance of double circuit line \n", +"clear\n", +"clc;\n", +"GMDs=.0069;//self GMD of the conductor\n", +"Dab=sqrt((3^2)+.5^2);\n", +"Dbc=Dab;\n", +"Dac=6;\n", +"Dab`=sqrt((3^2)+6^2);\n", +"Daa=sqrt((6^2)+5.5^2);\n", +"Dm1=((3.04*6*5.5*6.708)^.25);\n", +"Dm2=((3.04*3.04*6.708*6.708)^.25);\n", +"Dm=4.89;\n", +"Ds1=sqrt(GMDs*Daa);\n", +"Ds2=0.2217;\n", +"Ds=.228;\n", +"Z=2*(10^-7)*(10^6)*log(Dm/Ds);\n", +"mprintf('inductance =%.3f mH/km',Z);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7: To_determine_the_inductance_per_Km_per_phase_of_single_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// // To determine the inductance per Km per phase of single circuit\n", +"clear\n", +"clc;\n", +"Ds=sqrt(0.025*.4*.7788);\n", +"Dm=((6.5*13.0*6.5)^(1/3));\n", +"Z=2*(10^-4)*1000*log(Dm/Ds);\n", +"mprintf('inductance =%.3f mH/km/phase',Z);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/20-LOAD_FREQUENCY_CONTROL.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/20-LOAD_FREQUENCY_CONTROL.ipynb new file mode 100644 index 0000000..b94c477 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/20-LOAD_FREQUENCY_CONTROL.ipynb @@ -0,0 +1,111 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20: LOAD FREQUENCY CONTROL" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.1: EX20_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the load taken by the set C and indicate the direction in which the energy is flowing\n", +"clear \n", +"clc;\n", +"//let x MW flows from A to B\n", +"//Load on station A=75+x\n", +"//%drop in speed =5*(75+x)/200\n", +"//load on station B =(30-x)\n", +"//%drp in speed=(30-x)*4/75\n", +"x=(1.6-1.875)/(.025+.12+.0533);//by manipulating equation : 5*(75+x)/200 + 3*x/25 =(30-x)*4/75 \n", +"mprintf('x=%.2f MW\n',x);\n", +"mprintf('which means power of magnitude %.2f MW will be from B to A',abs(x));" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.2: Determine_the_load_shared_by_each_machine.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the load shared by each machine .\n", +"clear\n", +"clc;\n", +"//Let x be the power supplied by 110 MW unit\n", +"// the percent drop in speed = 5x/110\n", +"x=(250*11)/(21+11);// by manipulating equation : 5x/110=5x(250-x)/210\n", +"P=250-x;//Power shared by 210 MW unit \n", +"mprintf('Power supplied by 210 MW unit = %.2f MW \n',P);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.3: EX20_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the frequency to which the generated voltage drops before the steam flow commences to increase to meet the new load\n", +"clear \n", +"clc;\n", +"E=4.5*100;//Energy stored at no load(MJ)\n", +"E1=25*.6;//Energy lost by rotor(MJ)\n", +"fnew=sqrt((E-E1)/E)*50;\n", +"mprintf('new frequency will be %.2f Hz',fnew);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/21-COMPENSATION_IN_POWER_SYSTEMS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/21-COMPENSATION_IN_POWER_SYSTEMS.ipynb new file mode 100644 index 0000000..ea1116f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/21-COMPENSATION_IN_POWER_SYSTEMS.ipynb @@ -0,0 +1,73 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21: COMPENSATION IN POWER SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.1: Determine_the_load_bus_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine the load bus voltage\n", +"clear \n", +"clc;\n", +"load1=10+%i*15;//load per phase(MVA)\n", +"SCC=250/3;\n", +"V=11/sqrt(3);\n", +"P=30;\n", +"Q=45;\n", +"Z=(11/sqrt(3))^2/(250/3);//Equivalent short circuit impedence\n", +"dsc=atand(5);\n", +"R=.0949;\n", +"X=.4746;\n", +"//Using equation: V^2= (Vcosd+PR/V)^2 + (Vsind+QX/V)^2, we get \n", +"y=poly([51.7 0 -27.5 0 1],'V','c');\n", +"disp(y,'we get equation :');\n", +"X=roots(y);\n", +"disp(X,'Roots of above equation are ');\n", +"V=5.046;\n", +"mprintf('V=%.3f\n',V);\n", +"dV=6.35-V;\n", +"Ssc=250;\n", +"//using expression ,a=dV/v=1(Pcos(dsc)+Qsin(dsc))/Ssc +j(Psin(dsc)-Qcos(dsc))/Ssc\n", +"a=(P*cosd(dsc)+Q*sind(dsc))/Ssc +%i*(P*sind(dsc)-Q*cosd(dsc))/Ssc;\n", +"disp(abs(a),'dV/V= ');" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/22-POWER_SYSTEM_VOLTAGE_STABILITY.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/22-POWER_SYSTEM_VOLTAGE_STABILITY.ipynb new file mode 100644 index 0000000..901e2f0 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/22-POWER_SYSTEM_VOLTAGE_STABILITY.ipynb @@ -0,0 +1,212 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22: POWER SYSTEM VOLTAGE STABILITY" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.2: EX22_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Determine the source voltage when the load is disconnectedto load p.f (i) unity (ii).8 lag.\n", +"clear \n", +"clc;\n", +"Vb=500;\n", +"Sb=1000;\n", +"Zb=Vb^2/Sb;\n", +"Xpu=.35*100/Zb;\n", +"Zth=1000/5000;\n", +"X=Xpu+Zth;\n", +"V=1;\n", +"Q=0;\n", +"P=1;\n", +"Eth=V+(Q*X/V)+%i*(P*X/V);\n", +"Q=.75;\n", +"Eth1=V+(Q*X/V)+%i*(P*X/V);\n", +"printf('(i) For p.f unity , Eth=%.2f V',Eth);\n", +"disp(Eth1,'(i) For p.f .8 , Eth=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.3: EX22_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine thee Ac system voltage when the dc system is disconnected or shutdown\n", +"clear \n", +"clc;\n", +"\n", +"X=.625;\n", +"P=1;\n", +"Q=.6;\n", +"V=1;\n", +"Eth=V+(Q*X/V)+%i*(P*X/V);\n", +"Phase_Eth=atand(imag(Eth)/real(Eth));\n", +"mprintf('Eth=%.2f at an angle %.0f degrees',abs(Eth),Phase_Eth);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.4: To_Calculate_the_new_on_and_off_times_for_constant_energy.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To Calculate the new on and off times for constant energy.\n", +"clear \n", +"clc;\n", +"\n", +"P=.5;\n", +"toff=4;\n", +"ton=(P*toff-0*toff)/(.8-P);\n", +"mprintf('toff= 4min.\n')\n", +"mprintf('ton(min.)=%.3f min.\n',ton);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.6: To_discuss_the_effect_of_tap_changing.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To discuss the effect of tap changing\n", +"clear \n", +"clc;\n", +"V=1;\n", +"Qload=1*V\n", +"Qcap=-.75*V^2;\n", +"Qnet=Qload+Qcap;\n", +"VS=1-.75*2*V;// voltage sensitivity\n", +"mprintf('Voltage sensitivity=%.3f\n',VS);\n", +"mprintf('since the voltage sensitivity is negative ,\n voltage regulation by tap changing will reduce net reactive load and improive voltage stability ');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.7: To_determine_the_effect_of_tapping_to_raise_the_secondary_voltage_by_10percent.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the effect of tapping to raise the secondary voltage by 10%\n", +"clear \n", +"clc;\n", +"\n", +"Y=-%i*10;\n", +"n=1+.1;\n", +"Y1=n*(n-1)*Y;\n", +"Y2=(1-n)*Y;\n", +"disp(Y1,'Y1=');\n", +"disp(Y2,'Y2=');\n", +"disp('The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2^2 on the secondary side');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.8: Calculate_the_additional_reactive_power_capability_at_full_load.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Calculate the additional reactive power capability at full load\n", +"clear;\n", +"clc;\n", +"P=1;//assuming\n", +"S1=P/.95;//For pf .95\n", +"S2=P/.8;//For pf .8\n", +"dMVA=(S2-S1)*100/P;//Increase in MVA rating \n", +"Q1=P*tand(acosd(.95));//Q for pf .95\n", +"Q2=P*tand(acosd(.8));//Q for pf .8\n", +"dPc=(Q2-Q1)*100/Q1//Percent additional Reactive Power Capability \n", +"mprintf('Percent additional Reactive Power Capability is %.0f',dPc)\n", +"\n", +"\n", +"\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/23-STATE_ESTIMATION_IN_POWER_SYSTEMS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/23-STATE_ESTIMATION_IN_POWER_SYSTEMS.ipynb new file mode 100644 index 0000000..27c9d4e --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/23-STATE_ESTIMATION_IN_POWER_SYSTEMS.ipynb @@ -0,0 +1,208 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23: STATE ESTIMATION IN POWER SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.1: To_determine_the_state_vector_at_the_end_of_first_iteration.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the state vector at the end of first iteration\n", +"clear\n", +"clc;\n", +"C1=.02*100;\n", +"C2=.05;\n", +"Fs=100;\n", +"S1=.41 -%i*.11;\n", +"S2=-.4 +%i*.10;\n", +"S3=-.105 +%i*.11;\n", +"S4=-.105 +%i*.11;\n", +"S5=.14 -%i*.14;\n", +"S6=-.7 +%i*.35;\n", +"Z12=.08+%i*.24;\n", +"Z23=.06+%i*.18;\n", +"Z31=.02+%i*.06;\n", +"Z21=Z12;\n", +"Z32=Z23;\n", +"Z13=Z31;\n", +"W1=(50*10^(-6))/((C1*abs(S1)+(C2*(Fs)))^2);\n", +"W2=(50*10^(-6))/((C1*abs(S2)+C2*(Fs))^2);\n", +"W3=(50*10^(-6))/((C1*abs(S3)+C2*(Fs))^2);\n", +"W4=(50*10^(-6))/((C1*abs(S4)+C2*(Fs))^2);\n", +"W5=(50*10^(-6))/((C1*abs(S5)+C2*(Fs))^2);\n", +"W6=(50*10^(-6))/((C1*abs(S6)+C2*(Fs))^2);\n", +"disp(W1,'W1=');//Answers for W1,W2,W3,W4,W5,W6 in the book is wrongly Calculated\n", +"disp(W2,'W2=');\n", +"disp(W3,'W3=');\n", +"disp(W4,'W4=');\n", +"disp(W5,'W5=');\n", +"disp(W6,'W6=');\n", +"a1=W1/(abs(13)^2)\n", +"[D]=diag([W1/(abs(Z13)^2);W2/(abs(Z31)^2);W3/(abs(Z12)^2);W4/(abs(Z21)^2);W5/(abs(Z23)^2);W6/(abs(Z32)^2)]);\n", +"A=[-1 0 1;1 0 -1;1 -1 0;-1 1 0;0 1 -1;0 -1 1];\n", +"B=[-1 0;1 0;1 -1;-1 1;0 1;0 -1];\n", +"b=[1;-1;0;0;-1;1];\n", +"C=(B')*D;//Assuming Transpose(B)D=C\n", +"F=(B')*D*B;//Assuming Transpose(B)*D*B=F\n", +"G=(inv(F))*C;//Assuming(BTDB)-1*(BT)*D=F\n", +"E1=1.05;\n", +"E2=E1;\n", +"E3=E1;\n", +"invH=diag([Z31/E3;Z13/E1;Z12/E1;Z21/E2;Z23/E2;Z32/E2]);\n", +"Sm=[.41+%i*.11;-.4-%i*.1;-.105-%i*.11;.14+%i*.14;.72+%i*.37;-.7+%i*.35];\n", +"EMo=invH*Sm;\n", +"a=EMo-b*E1;\n", +"E=G*a;\n", +"disp(E,'E='); //Answers differs due to wrong calculation of W1,W2,W3,W4,W5,W6\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.2: Determine_The_States_of_the_systems_at_the_end_of_first_iteration.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine The States of the systems at the end of first iteration.\n", +"clear\n", +"clc\n", +"Qm1=-.24;\n", +"Qm2=-.24;\n", +"Qm3=.5;\n", +"do=0;\n", +"Pm1=.12;\n", +"Pm2=.21;\n", +"Pm3=-.30;\n", +"W1=3;\n", +"r1=W1;//assuming r1=Inverse(R1)\n", +"W2=5;\n", +"r2=W2;//assuming r2=Inverse(R1)\n", +"W3=2;\n", +"r3=W3;//assuming r3=Inverse(R1)\n", +"X12=%i*.03;\n", +"X13=%i*.01;\n", +"X23=%i*.02;\n", +"X21=X12;\n", +"X31=X13;\n", +"X32=X23;\n", +"Vo=[1.05;1.05];\n", +"H=[-1/.03 -1/.01;((1/.03)+(1/.02)) -1/.02;-1/.02 ((1/.01)+1/.02)];//assuming dh/dl=H\n", +"A1=[3327+34700+5000 9990-20825-15000;-25835 30000+12500+45000];\n", +"V=Vo+inv(A1)*(H')*(diag([W1;W2;W3]))*[Qm1;Qm2;Qm3];\n", +"d=do+inv(A1)*(H')*(diag([W1;W2;W3]))*[Pm1;Pm2;Pm3];//assuming d=dell matrix and do=intial matrix=0\n", +"disp(V,'V=');\n", +"disp(d,'d=');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.3: Problem_on_State_Estimator_Linear_Model.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Problem on State Estimator Linear Model\n", +"\n", +"clear\n", +"clc;\n", +"A=[-3.33 0;0 10;5 -5];\n", +"R=[10^-4 0 0;0 10^-4 0;0 0 10^-4];\n", +"O=inv(((A')*(inv(R))*(A)))*((A')*(inv(R))*[.12;.21;-.30]);//assuming theat matrix=0\n", +"f12=-3.33*(O(1,1));\n", +"f31=10*(O(2,1));\n", +"f23=5*(O(1,1)-O(2,1));\n", +"J=(((.12-f12)^2)+((.21-f31)^2)+((-.3-f23)^2))/(10^-4);\n", +"disp(O,'O=');//Answer does not match due to difference in rounding off of digits\n", +"disp(J,'J=');//Answer does not match due to difference in rounding off of digits\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.4: Determine_theta1_Theta2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Determine theta1 Theta2\n", +"clear\n", +"clc;\n", +"A=[5 -5;2.5 0;4 -4];\n", +"R=[10^-4 0 0;0 10^-4 0;0 0 10^-4];\n", +"O=inv(((A')*(inv(R))*(A)))*((A')*(inv(R))*[.60;.05;.35]);//assuming theat matrix=0\n", +"f12=5*(O(1,1)-O(2,1));\n", +"f13=2.5*(O(1,1));\n", +"f32=-4*(O(2,1));\n", +"J=(((.6-f12)^2)+((.05-f13)^2)+((.35-f32)^2))/(10^-4);\n", +"//Answer does not match due to difference in rounding off of digits\n", +"disp(O(1,1),'Theta1=');\n", +"disp(O(2,1),'Theta2=');\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/24-UNIT_COMMITMENT.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/24-UNIT_COMMITMENT.ipynb new file mode 100644 index 0000000..fa93d81 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/24-UNIT_COMMITMENT.ipynb @@ -0,0 +1,135 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24: UNIT COMMITMENT" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.3: Priority_List_Method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Priority List Method \n", +"clear\n", +"clc;\n", +"Fc1=1.1;//Fuel cost(1)=Rs 1.1/MBtu\n", +"Fc2=1;//Fuel cost(2)=1/MBtu\n", +"Fc3=1.2;//Fuel cost(3)=1.2/MBtu\n", +"P1max=600;\n", +"P1=P1max;\n", +"F1=600+7.1*P1+0.00141*(P1^2);//For P1= Pm1ax\n", +"Favg1=F1*Fc1/600;//Full load average production cost\n", +"P2max=450;\n", +"P2=P2max;\n", +"F2=350+7.8*P2+0.00195*(P2^2);//For P2= P2max\n", +"Favg2=F2*Fc2/450;//Full load average production cost\n", +"P3max=250;\n", +"P3=P3max;\n", +"F3=80+8*P3+0.0049*(P3^2);//For P3= P3max\n", +"Favg3=F3*Fc3/250;//Full load average production cost\n", +"mprintf('Priority List is as follows\n');\n", +"mprintf('Unit Rs/MWhr MinMW Max MW\n')\n", +"mprintf(' 2 %.3f 100 %.0f \n',Favg2,P2max)\n", +"mprintf(' 1 %.4f 60 %.0f \n',Favg1,P1max)\n", +"mprintf(' 3 %.2f 50 %.0f \n\n',Favg3,P3max)\n", +"Fmax1=P1max+P2max+P3max;\n", +"Fmax2=P2max+P1max\n", +"Fmax3=P2max\n", +"mprintf('Unit Commitment Scheme is follows\n')\n", +"mprintf('Combination Min.MW from Combination Max.MW from Combination\n');\n", +"mprintf('2+1+3 310 %.0f \n',Fmax1);\n", +"mprintf('2+1 260 %.0f \n',Fmax2);\n", +"mprintf('2 100 %.0f ',Fmax3);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.4: illustrate_the_dynamic_programming_for_preparing_an_optimal_unit_commitment.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// illustrate the dynamic programming for preparing an optimal unit commitment.\n", +"\n", +"clear\n", +"clc;\n", +"function[F1]=F1(P1)\n", +" F1=7.1*P1+.00141*(P1^2)\n", +" mprintf('F1(%.0f)=%.1f\n',P1,F1);\n", +"endfunction\n", +"function[f2]=f2(P2)\n", +" f2=7.8*P2+.00195*(P2^2)\n", +" mprintf('f2(%.0f)=%.0f\n',P2,f2);\n", +"endfunction\n", +"function[F]=F(P1,P2)\n", +" F1=7.1*P1+.00141*(P1^2)\n", +" F2=7.8*P2+.00195*(P2^2)\n", +" F=F1+F2\n", +" mprintf('F1(%.0f)+f2(%.0f)=%.0f\n',P1,P2,F);\n", +" endfunction\n", +"P1max=600;\n", +"P2max=450;\n", +"mprintf('Unit Commitment using Load 500MW\n')\n", +"F1(500);\n", +"mprintf('Since min. Power of second unit is 100MW , we find\n');\n", +"F(400,100);\n", +"F(380,120);\n", +"F(360,140);\n", +"mprintf('Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\n');\n", +"mprintf('Next we increase the load by 50 MW and loading unit 1 we get, \n');\n", +"F1(550);\n", +"mprintf('Also if we distribute a part of load to unit 2 we get ,\n')\n", +"F(450,100);\n", +"F(400,150);\n", +"F(350,200);\n", +"mprintf('Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\n');\n", +"\n", +"\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/25-ECONOMIC_SCHEDULING_OF_HYDROTHERMAL_PLANTS_AND_OPTIMAL_POWER_FLOWS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/25-ECONOMIC_SCHEDULING_OF_HYDROTHERMAL_PLANTS_AND_OPTIMAL_POWER_FLOWS.ipynb new file mode 100644 index 0000000..a9087e4 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/25-ECONOMIC_SCHEDULING_OF_HYDROTHERMAL_PLANTS_AND_OPTIMAL_POWER_FLOWS.ipynb @@ -0,0 +1,96 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25: ECONOMIC SCHEDULING OF HYDROTHERMAL PLANTS AND OPTIMAL POWER FLOWS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.1: illustrating_the_procedure_for_economic_scheduling_clear_all.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// illustrating the procedure for economic scheduling\n", +"clear\n", +"clc;\n", +"q2=25;\n", +"q3=25\n", +"q1=70-(q2+q3);\n", +"Wo=120;\n", +"W3=50;\n", +"Wi1=0;\n", +"Wi2=0;\n", +"W1=Wo+Wi1-q1;\n", +"W2=W1+Wi2-q2\n", +"PH1=9.81*(10^-3)*20*[1+(.5*.006*(120+100))]*(20-2);\n", +"PH2=9.81*(10^-3)*20*[1+(.5*.006*(100+75))]*(23);// Answer in the book is not Correct due to wrong calculation\n", +"PH3=9.81*(10^-3)*20*[1+(.5*.006*(75+50))]*(23);\n", +"PT1=8-PH1;\n", +"PT2=12-PH2;\n", +"PT3=7-PH3;\n", +"L11=20+PT1;//dFT/dPT=PT+20\n", +"L12=20+PT2;//dF/dp=PT+20\n", +"L13=20+PT3;//dF/dp=PT+20\n", +"//dPL/dPH=0\n", +"L31=L11;\n", +"L32=L12;\n", +"L33=L13;\n", +"e=.006;\n", +"ho=.1962\n", +"Rho=2;\n", +"L21=L31*ho*[1+(.5*e*(2*Wo+Wi1-2*q1+Rho))]\n", +"L22=L21-L31*[.5*ho*e*(q1-Rho)]-L32*[.5*ho*e*(q2-Rho)]//for m=1\n", +"L23=L22-L32*[.5*ho*e*(q2-Rho)]-L33*[.5*ho*e*(q3-Rho)]//for m=2\n", +"G1=L22-L32*ho*[1+.5*.006*(2*100-2*25+2)]//G1=dF/dq2 Answer doent match due to wrong calculation of PH2 in a book;\n", +"G2=L23-L33*ho*[1+.5*.006*(2*W2+0-2*q3+Rho)]//G1=dF/dq3;\n", +"a=0.4;\n", +"qnew2=q2-a*G1;// Answer differs due to wrong calculation of PH2 in the book\n", +"qnew3=q3-a*G2;\n", +"q1=120-50-(qnew2+qnew3);\n", +"mprintf('Let q2=%.0f q3=%.0f q1=%.0f\n',q2,q3,q1);\n", +"mprintf('W1=%.0f W2=%.0f\n',W1,W2);\n", +"mprintf('PH1=%.2f PH2=%.3f PH3=%.1f\n',PH1,PH2,PH3);\n", +"mprintf('Thermal generation during Three Intervals \n PT1=%.2f PT2=%.2f PT3=%.1f\n',PT1,PT2,PT3);\n", +"mprintf('Value of L1 for the three intervals, \n L11=%.2f L12=%.2f L13=%.1f\n',L11,L12,L13);\n", +"mprintf('Neglecting transmission losses we get\n L11=L31 L12=L32 L13=L33\n');\n", +"mprintf('L21=%.3f\n',L21)\n", +"mprintf('For m=1 and 2 we get \n L22=%.1f \n L23=%.1f\n',L22,L23);\n", +"mprintf('Gradient Vectors \n dF/dq2=%.2f\n dF/dq3=%.1f\n',G1,G2)\n", +"mprintf('q2new=%.3f \n q3new=%.1f\n q1=%.0f',qnew2,qnew3,q1)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/3-CAPACITANCE_OF_TRANSMISSION_LINES.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/3-CAPACITANCE_OF_TRANSMISSION_LINES.ipynb new file mode 100644 index 0000000..2d6885b --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/3-CAPACITANCE_OF_TRANSMISSION_LINES.ipynb @@ -0,0 +1,115 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: CAPACITANCE OF TRANSMISSION LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1: To_determine_the_capacitance_and_charging_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the capacitance and charging current \n", +"clear\n", +"clc;\n", +"Dm=2.015;// mutual GMD of conductors(m)\n", +"r=.4;// radius of conductor(cm)\n", +"C=10^-9*1000/(18*log(201.5/.4));\n", +"Ic=132*1000*8.928*314*(10^-9)/sqrt(3);\n", +"mprintf('capacitance =%.13f F/km\n',C);//Answers don't match due to different reprentation\n", +"mprintf('charging current=%.4f amp/km',Ic);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2: To_determine_the_capacitance_and_charging_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the capacitance and charging current \n", +"clear\n", +"clc;\n", +"GMDm=6.61;//mutual GMD(m)\n", +"Ds1=sqrt(1.25*(10^-2)*10.965);\n", +"Ds3=Ds1;\n", +"Ds2=sqrt(1.25*(10^-2)*9);\n", +"Ds=((Ds1*Ds2*Ds3)^.333333);\n", +"C=1/(18*log(GMDm/Ds));\n", +"Ic=220*1000*314*.01905*(10^-6)/sqrt(3);\n", +"mprintf('capacitance =%.6f micro-Farad/km\n',C);\n", +"mprintf('charging current =%.2f amp/km',Ic);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3: To_determine_the_capacitance_and_charging_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the capacitance and charging current \n", +"clear\n", +"clc;\n", +"GMD=8.19;\n", +"Ds=sqrt(2.25*(10^-2)*.4);\n", +"C=1/(18*log(GMD/Ds));\n", +"Ic=220*1000*314*C*(10^-6)/sqrt(3);\n", +"mprintf('capacitance per km =%.5f micro-Farad\n',C);\n", +"mprintf('charging current =%.3f amp',Ic);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/4-PERFORMANCE_OF_LINES.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/4-PERFORMANCE_OF_LINES.ipynb new file mode 100644 index 0000000..0b30f16 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/4-PERFORMANCE_OF_LINES.ipynb @@ -0,0 +1,441 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: PERFORMANCE OF LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10: EX4_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the sending end voltage and current , power and power factor. Evaluate A, B , C, D parameters.\n", +"clear\n", +"clc;\n", +"R=.1557*160\n", +"GMD=(3.7*6.475*7.4)^(1/3);\n", +"Z1=2*(10^-7)*log(560/.978)*160*1000;\n", +"XL=63.8;\n", +"C=(10^-9)*2*(10^6)*%pi*160*1000/(36*%pi*log(560/.978));\n", +"Z=sqrt((.1557^2) + .39875^2)*(cosd(68.67)+ %i*sind(68.67));\n", +"jwC=%i*314*1.399*(10^-6)/160;\n", +"Zc=sqrt(Z/jwC);\n", +"y=sqrt(Z*jwC);\n", +"yl=y*160;\n", +"A=cosh(yl);\n", +"B=Zc*sinh(yl)\n", +"C=sinh(yl)/Zc;\n", +"Ir=50000/(sqrt(3)*132);\n", +"Vs=(A*76.208) +(B*(10^-3)*Ir*(cosd(-36.87)+%i*sind(-36.87)));\n", +"VS=152.34;\n", +"Is=C*76.208*(10^3) +(A*Ir*(cosd(-36.87)+%i*sind(-36.87)));\n", +"Ps=3*abs(Vs)*abs(Is)*cosd(33.96);\n", +"pf=cosd(33.96);\n", +"Vnl=abs(Vs)/abs(A);\n", +"reg=(Vnl-76.208)*100/76.208;\n", +"n=50000*.8*100/abs(Ps);\n", +"mprintf(' Vs line to line =%.2f kV\n',VS);\n", +"disp(Is,'sending end current Is(A)=');//Answer don't match due to difference in rounding off of digits\n", +"mprintf('sending end power=%.0f kW\n',Ps);\n", +"mprintf('sending end p.f =%.3f\n',pf);\n", +"mprintf('percent regulation=%.1f \n',reg);\n", +"mprintf('percent efficency=%.1f ',n);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1: EX4_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To detremine the the voltage at the generating station and efficiency of transmission\n", +"clear\n", +"clc;\n", +"R=0.496;// resistance\n", +"X=1.536;\n", +"Vr=2000;\n", +"Z=(10*2*2/(11*11)) + %i*30*2*2/(11*11);\n", +"Zt=(.04+(1.3*2*2/(11*11))) + %i*(.125 + (4.5*2*2/(11*11)));//Transformer impedence\n", +"Il=250*1000/2000;// line current(amps.)\n", +"Pl=Il*Il*R;//line loss(kW)\n", +"Po=250*0.8;// output(kW)\n", +"cosr=0.8;// power factor\n", +"sinr=.6;\n", +"%n=200*100/(200+7.7);\n", +"Vs=(Vr*cosr+Il*R)+%i*(Vr*sinr+Il*X);\n", +"V=sqrt((1662^2)+ (1392^2));\n", +"mprintf('efficiency= %.1f percent \n',%n);\n", +"mprintf('Sending end voltage,|Vs|=%.0f volts',V);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2: To_determine_power_input_and_output_i_star_connected_ii_delta_connected.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine power input and output (i) star connected (ii)delta connected\n", +"clear\n", +"clc;\n", +"mprintf('when load is star connected\n');\n", +"Vln=400/sqrt(3);// Line to neutral voltage(V)\n", +"Z=7+ %i*11;//Impedence per phase\n", +"Il=231/Z;// line current(amp.)\n", +"I=abs(231/Z);\n", +"Pi=3*I*I*7;\n", +"Po=3*I*I*6;\n", +"mprintf('power input =%.0f watts\n',Pi);//Answers don't match due to difference in rounding off of digits\n", +"mprintf('power output=%.0f watts\n',Po);//Answers don't match due to difference in rounding off of digits\n", +"mprintf('when load is delta connected\n');\n", +"Ze=2+ %i*3;// equivalent impedence(ohm)\n", +"Zp=3+%i*5;// impedence per phase\n", +"il=231/Zp;//Line current(amps.)\n", +"IL=abs(il);\n", +"pi=3*IL*IL*3;\n", +"po=3*IL*IL*2;\n", +"mprintf('power input=%.1f watts\n',pi);//Answers don't match due to difference in rounding off of digits\n", +"mprintf('power output = %.0f watts \n',po);//Answers don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3: To_determine_efficiency_and_regulation_of_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine efficiency and regulation of line\n", +"clear\n", +"clc;\n", +"a=100/.5\n", +"Xl=2*(10^-7)*log(100/.5);//inductance(H/meter)\n", +"XL=20*(1000)*Xl;// inductance of 20 km length \n", +"R=6.65;// resistance(ohm)\n", +"Rc=20*1000/(58*90);// resistance of copper(ohm)\n", +"I=10*1000/(33*.8*sqrt(3));// the current(amps.)\n", +"Pl=3*I*I*Rc/(10^6);//loss (MW)\n", +"n=10/(10+Pl);\n", +"mprintf('efficiency=%.4f percent \n',n);\n", +"Vr=19052;\n", +"cosr=.8;//power factor\n", +"sinr=.6;\n", +"Vs=abs(((Vr*cosr+I*Rc) +%i*(Vr*sinr+ I*R)));\n", +"mprintf('Vs =%.0f volts\n',Vs);//Answer don't match due to difference in rounding off of digits\n", +"Reg=(Vs-Vr)*100/Vr;\n", +"mprintf(' regulation =%.2f percent',Reg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: To_calculate_the_voltage_across_each_load_impedence_and_current_in_the_nuetral.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To calculate the voltage across each load impedence and current in the nuetral\n", +"clear\n", +"clc;\n", +"IR=(400)/((sqrt(3)*(6.3+%i*9)));\n", +"IY=231*(cosd(-120) + %i*sind(-120))/8.3;\n", +"IB=231*(cosd(120) + %i*sind(120))/(6.3-%i*8);\n", +"In=abs((IR +IY +IB));//Neutral current\n", +"mprintf('Neutral current =%.2f amps\n',In);\n", +"VR=abs(IR*(6+ %i*9));\n", +"VY=abs(IY*(8));\n", +"VB=abs(IB*(6-%i*8));\n", +"mprintf('Voltage across Phase R =%.1f volts \n',VR);\n", +"mprintf('Voltage across Phase Y =%.2f volts \n',VY);\n", +"mprintf('Voltage across Phase B =%.0f volts \n',VB);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5: To_determine_efficiency_and_regulation_of_3_phase_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine efficiency and regulation of 3 phase line\n", +"clear\n", +"clc;\n", +"R=100*.1;//Resistance of line (ohm)\n", +"Xl=2*(10^-7)*100*1000*log(200/.75);//inductance of line\n", +"X2=Xl*314;//inductive reactance\n", +"C=2*(%pi*100)*8.854*(10^-12)*100*1000*(10^6)/(log(200/.75));// capacitance per phase (micro farad)\n", +"mprintf('Using Nominal-T method\n');\n", +"Ir=20*1000/(sqrt(3)*66*.8);\n", +"Vr=66*1000/sqrt(3);\n", +"Vc=(38104*.8+ Ir*5) +%i*(38104*.6+ Ir*17.55);// voltage across condenser\n", +"Ic=%i*314*(Vc)*.9954*(10^-6);\n", +"is=Ir+Ic;\n", +"Is=abs(Ir+Ic);\n", +"Vs=abs(Vc + (is*(5 + %i*17.53)));\n", +"VR=abs(Vs*(-%i*3199)/(5-%i*3181));// no load recieving end voltage\n", +"Reg=(VR-Vr)*100/Vr;\n", +"Pl=3*(Ir*Ir*5 + Is*Is*5)/1000000;\n", +"%n=20*100/(20+Pl);\n", +"mprintf('percent regulation=%.1f \n',Reg);\n", +"mprintf('percent efficiency=%.1f \n\n',%n);\n", +"mprintf('Using Nominal-pi method\n');\n", +"Ir1=218.68*(.8-%i*.6);\n", +"Ic1=%i*314*.4977*(10^-6)*Vr;\n", +"Il=Ir1+Ic1;\n", +"vs1=Vr+Il*(10+%i*35.1);\n", +"Vs1=abs(vs1);\n", +"Vr1=Vs1*(-%i*6398)/(10-%i*6363);\n", +"VR1=abs(Vr1);// no load recieving end voltage\n", +"Reg2=(VR1-Vr)*100/Vr;\n", +"IL=abs(Ir1+Ic1);\n", +"Loss=3*IL*IL*10;\n", +"%n=20*100/21.388;\n", +"mprintf('percent regulation=%.2f \n',Reg2);\n", +"mprintf('percent efficiency=%.1f \n',%n);\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6: EX4_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To find the rms value and phase values (i)The incident voltage to neutral at the recieving end (ii)The reflected voltage to neutral at the recieving end (iii)The incident and reflected voltage to neutral at 120 km from the recieving end.\n", +"clear\n", +"clc;\n", +"R=0.2;\n", +"L=1.3;\n", +"C=0.01*(10^-6);\n", +"z=R+%i*L*314*(10^-3);// serie impedence\n", +"y=%i*314*C;// shunt admittance\n", +"Zc=sqrt(z/y);// characterstic impedence\n", +"Y=sqrt(y*z);\n", +"Vr=132*1000/sqrt(3);\n", +"Ir=0;\n", +"Vin=(Vr + Ir*Zc)/2;// incident voltage to neutral at the recieving end\n", +"mprintf('Vr =%.3f volts \n',Vr);//Answer don't match due to difference in rounding off of digits\n", +"mprintf('(i)The incident voltage to neutral at the recieving end =%.3f volts \n',Vin);//Answer don't match due to difference in rounding off of digits\n", +"Vin2=(Vr - Ir*Zc)/2;// The reflected voltage to neutral at the recieving end\n", +"mprintf('(ii)The reflected voltage to neutral at the recieving end=%.3f volts \n',Vin2);//Answer don't match due to difference inrounding off of digits\n", +"Vrp=Vr*exp(.2714*120*(10^-3))*exp(%i*1.169*120*(10^-3))/1000;//Taking Vrp=Vr+\n", +"Vrm=Vr*exp(-0.0325)*exp(-%i*.140)/1000;//Taking Vrm=Vr-\n", +"v1=Vrm/2;// reflected voltage to neutral at 120 km from the recieving end\n", +"phase_v1=atand(imag(v1)/real(v1));//Phase angle of v1\n", +"v2=Vrp/2;//incident voltage to neutral at 120 km from the recieving end\n", +"phase_v2=atand(imag(v2)/real(v2));//Phase angle of v2\n", +"mprintf('(iii) reflected voltage to neutral at 120 km from the recieving end =%.2f at angle of %.2f \n',abs(v1),phase_v1);\n", +"mprintf('incident voltage to neutral at 120 km from the recieving end = %.2f at angle of %.2f\n',abs(v2),phase_v2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7: To_determine_of_efficiency_of_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine of efficiency of line\n", +"clear\n", +"clc;\n", +"Ir=40*1000/(sqrt(3)*132*.8);\n", +"Vr=132*1000/sqrt(3);\n", +"Zc=380*(cosd(-13.06)+ %i*sind(-13.06));\n", +"IR=Ir*(cosd(-36.8)+ %i*sind(-36.8));\n", +"Vsp=(Vr+IR*Zc)*(1.033*(cosd(8.02)+ %i*sind(8.02)))/2;\n", +"Vsm=(Vr-IR*Zc)*(.968*(cosd(-8.02)+ %i*sind(-8.02)))/2;\n", +"vs=Vsp+ Vsm;\n", +"Vs=abs(vs);\n", +"is=(Vsp-Vsm)/Zc;\n", +"Is=abs(is)\n", +"P=3*Vs*Is*cosd(33.72)/10^6;\n", +"n=40*100/P;\n", +"mprintf('efficiency=%.1f',n);//Answer don't match due to difference in rounding off of digits\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8: To_determine_the_ABCD_parameters_of_Line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the ABCD parameters of Line\n", +"clear\n", +"clc;\n", +"yl=(0.2714+ %i*1.169)*120*(10^-3);\n", +"Ir=40*1000/(sqrt(3)*132*.8)\n", +"A=cosh(yl);\n", +"phase_A=atand(imag(A)/real(A));//Phase angle of A\n", +"IR=Ir*(cosd(-36.8)+ %i*sind(-36.8))\n", +"Vr=132*1000/sqrt(3);\n", +"Zc=380*(cosd(-13.06)+ %i*sind(-13.06));\n", +"B=Zc*sinh(yl);\n", +"phase_B=atand(imag(B)/real(B));//Phase angle of B\n", +"Vs=(A*Vr+B*IR);\n", +"f=abs(B);\n", +"d=abs(Vs);\n", +"C=sinh(yl)/Zc;\n", +"phase_C=atand(imag(C)/real(C));//Phase angle of C\n", +"D=cosh(yl);\n", +"phase_D=atand(imag(D)/real(D));//Phase angle of D\n", +"mprintf('A=%.2f at an angle of %.2f \n',abs(A),phase_A)\n", +"mprintf('B=%.1f at an angle of %.0f \n',abs(B),phase_B)\n", +"mprintf('C=%.2f at an angle of %.2f \n',abs(C),phase_C)\n", +"mprintf('D=%.2f at an angle of %.2f \n',abs(D),phase_D)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9: EX4_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the sending end voltage and efficiency using Nominal_pi and Nominal-T method\n", +"clear\n", +"clc;\n", +"Ir=218.7*(.8-%i*.6);\n", +"Ic1=%i*314*.6*(10^-6)*76200;\n", +"Il=Ic1+Ir;\n", +"Vs=76200 + Il*(24+ %i*48.38);\n", +"phase_Vs=atand(imag(Vs)/real(Vs));//phase angle of VS\n", +"Pl=3*24*abs(Il)*abs(Il)/1000000;//The Loss(MW)\n", +"n=40*100/(40+Pl);\n", +"mprintf('Using Nominal- pi method\n ');\n", +"mprintf('Vs=%.0f volts at an angle of %.2f \n',abs(Vs),phase_Vs)\n", +"mprintf('efficiency=%.2f percent\n',n);\n", +"mprintf('\nUsing Nominal-T method\n');\n", +"Vc=76200*(.8+%i*.6) + 218.7*(12+ %i*24.49);\n", +"Ic=%i*314*1.2*(10^-6)*(63584+ %i*51076);\n", +"Is=199.46+ %i*23.95;\n", +"Vs=(Vc + Is*(12+ %i*24.49))/1000;\n", +"phase_Vs=atand(imag(Vs)/real(Vs));//Phase angle of Vs\n", +"Pl1=3*12*((200.89^2)+ 218.7^2)/1000000;//The loss(MW)\n", +"n1=40*100/(40+Pl1);\n", +"mprintf('Vs=%.2f at an angle of %.2f \n',abs(Vs),phase_Vs)\n", +"mprintf('efficiency=%.2f percent\n',n1);\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/5-HIGH_VOLTAGE_DC_TRANSMISSION.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/5-HIGH_VOLTAGE_DC_TRANSMISSION.ipynb new file mode 100644 index 0000000..3be1bcd --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/5-HIGH_VOLTAGE_DC_TRANSMISSION.ipynb @@ -0,0 +1,137 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: HIGH VOLTAGE DC TRANSMISSION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1: To_determine_the_dc_output_voltage_when_delay_anglw_a0_b30_c45.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the d.c. output voltage when delay anglw (a)0 (b)30 (c)45\n", +"clear\n", +"clc;\n", +"Vo=3*sqrt(2)*110/%pi;\n", +"Vd=Vo*(cosd(0) + cosd(15))/2;\n", +"Vd1=Vo*(cosd(30) + cosd(45))/2;\n", +"Vd2=Vo*(cosd(45) + cosd(60))/2;\n", +"mprintf('(a)For a=0, Vd=%.2f kV\n',Vd);\n", +"mprintf('(b)For a=30,Vd=%.2f kV\n',Vd1);\n", +"mprintf('(c)For a=45,Vd=%.2f kV\n',Vd2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2: To_determine_the_necessary_line_secondary_voltage_and_tap_ratio_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the necessary line secondary voltage and tap ratio required.\n", +"clear\n", +"clc;\n", +"VL=100*2*%pi/(3*sqrt(2)*(cosd(30) + cosd(45)));\n", +"mprintf('VL=%.2f kV\n',VL);//Answers don't match due to difference in rounding off of digits\n", +"Tr=VL/110;\n", +"mprintf('tap ratio=%.2f \n',Tr);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3: To_determine_the_effective_reactance_per_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the effective reactance per phase\n", +"clear\n", +"clc;\n", +"Vd=100000;\n", +"Id=800;// current\n", +"X=((3*sqrt(2)*94.115*.866*1000/%pi)-Vd)*%pi/(3*Id);\n", +"mprintf('effective reactance per phase , X=%.2f ohm\n',X);//Answer don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4: Calculate_the_direct_current_delivered.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Calculate the direct current delivered\n", +"clear\n", +"clc;\n", +"a=15;\n", +"d0=10;\n", +"y=15;\n", +"X=15;\n", +"R=10;\n", +"Id=(3*sqrt(2)*120*(cosd(a)-cosd(d0+y))*1000)/((R + (3*2*X)/%pi)*%pi);\n", +"mprintf('Id=%.2f amp.\n',Id);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/6-CORONA.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/6-CORONA.ipynb new file mode 100644 index 0000000..95c5c18 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/6-CORONA.ipynb @@ -0,0 +1,192 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: CORONA" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1: EX6_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the critical disruptive voltage and critical voltage for local and general corona.\n", +"clear\n", +"clc;\n", +"t=21;// air temperature\n", +"b=73.6;// air pressure\n", +"do=3.92*73.6/(273+t);\n", +"m=.85;\n", +"r=.52;\n", +"d=250;\n", +"Vd=21.1*m *do*r*log(250/.52);\n", +"vd=sqrt(3)*Vd;\n", +"m=.7;\n", +"vv=21.1*m*do*r*(1+ (.3/sqrt(r*do)))*log(250/.52);\n", +"Vv=vv*sqrt(3);\n", +"Vvg=Vv*.8/.7;\n", +"mprintf('critical disruptive line to line voltage=%.2f kV \n',vd);\n", +"mprintf('visual critical voltage for local corona=%.2f kV \n',vv);\n", +"mprintf('visual critical voltage for general corona=%.2f kV \n',Vvg);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2: To_determine_whether_corona_will_be_present_in_the_air_space_round_the_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine whether corona will be present in the air space round the conductor\n", +"clear\n", +"clc;\n", +"d=2.5;\n", +"di=3;// internal diameter\n", +"do=9;// external diameter\n", +"ri=di/2;// internal radius\n", +"ro=do/2;// external diameter\n", +"g1max=20/(1.25*log(ri/(d/2))+ .208*1.5*log(ro/ri));\n", +"mprintf('g1max=%.0f kV/cm \n',g1max);\n", +"mprintf('Since the gradient exceeds 21.1/kV/cm , corona will be present.')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3: To_determine_the_critical_disruptive_voltage_and_corona_loss.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the critical disruptive voltage and corona loss \n", +"clear\n", +"clc;\n", +"m=1.07;\n", +"r=.625\n", +"V=21*m *r*log(305/.625);\n", +"Vl=V*sqrt(3);\n", +"mprintf('critical disruptive voltage=%.0f kV\n',V);\n", +"mprintf('since operating voltage is 110 kV , corona loss= 0 ');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4: To_determine_the_voltage_for_which_corona_will_commence_on_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the voltage for which corona will commence on the line\n", +"clear\n", +"clc;\n", +"r=.5;\n", +"V=21*r*log(100/.5);\n", +"mprintf('critical disruptive voltage=%.1f kV',V);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5: To_determine_the_corona_characterstics.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the corona characterstics\n", +"clear\n", +"clc;\n", +"D=1.036;// conductor diameter(cm)\n", +"d=2.44;//delta spacing(m)\n", +"r=D/2;//radius(cm)\n", +"Ratio=d*100/r;\n", +"j=r/(d*100);\n", +"Rat2=sqrt(j);\n", +"t=26.67;//temperature\n", +"b=73.15;// barometric pressure\n", +"mv=.72;\n", +"V=63.5;\n", +"f=50;//frequency\n", +"do=3.92*b/(273+t);//do=dell\n", +"vd=21.1*.85*do*r*log(Ratio);\n", +"mprintf('critical disruptive voltage=%.2f kV\n',vd);\n", +"Vv=21.1*mv*do*r*(1+ (.3/sqrt(r*do)))*log(Ratio);\n", +"Pl=241*(10^-5)*(f+25)*Rat2*((V-vd)^2)/do;//power loss\n", +"Vd=.8*vd;\n", +"Pl2=241*(10^-5)*(f+25)*Rat2*((V-Vd)^2)*160/do;//loss per phase /km\n", +"Total= 3*Pl2;\n", +"mprintf('visual critical voltage=%.0f kV\n',Vv);\n", +"mprintf('Power loss=%.3f kW/phase/km\n',Pl);\n", +"mprintf('under foul weather condition ,\n');\n", +"mprintf('critical disruptive voltage=%.2f kV\n',Vd);\n", +"mprintf('Total loss=%.0f kW\n',Total);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/7-MECHANICAL_DESIGN_OF_TRANSMISSION_LINES.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/7-MECHANICAL_DESIGN_OF_TRANSMISSION_LINES.ipynb new file mode 100644 index 0000000..f0fd7ef --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/7-MECHANICAL_DESIGN_OF_TRANSMISSION_LINES.ipynb @@ -0,0 +1,177 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: MECHANICAL DESIGN OF TRANSMISSION LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1: Calculate_the_sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Calculate the sag \n", +"clear\n", +"clc;\n", +"sf=5;//Factor of safety\n", +"d=.95;// conductor dia(cm)\n", +"Ws=4250/sf;// working stress(kg/cm_2)\n", +"A=%pi*(d^2)/4;// area (cm_2)\n", +"Wp=40*d*(10^-2);//wind pressure (kg/cm)\n", +"W=sqrt((.65^2) +(.38^2));// Total effective weight(kg/m)\n", +"T=850*A;// working tension (kg)\n", +"c=T/W;\n", +"l=160;\n", +"d=l^2/(8*800);\n", +"mprintf('sag,d=%.0f metres\n',d);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2: To_calculate_the_maximum_Sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To calculate the maximum Sag\n", +"clear\n", +"clc;\n", +"D=1.95 + 2.6;// overall diameter(cm)\n", +"A=4.55*(10^-2);// area(m_2)\n", +"d=19.5;//diameter of conductor(mm)\n", +"r=d/2;//radius of conductor(mm)\n", +"Wp=A*39;//wind pressure(kg/m_2)\n", +"t=13;//ice coating(mm)\n", +"US=8000;// ultimate strength(kg)\n", +"Aice=%pi*(10^-6)*((r+t)^2 - r^2);//area section of ice (m_2)\n", +"Wice=Aice*910;\n", +"W=(sqrt((.85+Wice)^2 + Wp^2));// total weight of ice (kg/m)\n", +"T=US/2;// working teansion (kg)\n", +"c=T/W;\n", +"l=275;//length of span(m)\n", +"Smax=l*l/(8*c);\n", +"mprintf('Maximum sag=%.1f metres\n',Smax);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3: To_determine_the_Sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the Sag\n", +"clear\n", +"clc;\n", +"A=13.2;// cross section of conductor (mm_2)\n", +"Ar=4.1*(10^-3);// projected area\n", +"Wp=Ar*48.82;// wind loadind /m(kg/m)\n", +"w=.115;\n", +"W=sqrt((.1157^2) + (Wp^2));// effective loading per metre(kg)\n", +"q1=W/.115;\n", +"b=w/A;\n", +"f1=21;//working stress\n", +"T1=f1*A;\n", +"c=T1/W;\n", +"l=45.7;\n", +"S=l*l/(8*c);\n", +"dT=32.2-4.5;// difference in temperature\n", +"E=1.26*(10000);\n", +"a=16.6*(10^-6);\n", +"d=8.765*(10^-3);\n", +"K=f1-((l*d*q1)^2)*E/(24*f1*f1);\n", +"p=poly([-84.23 0 -14.44 1], 'f2', 'c');\n", +"r=roots(p);\n", +"f2= 14.823332;// accepted value of f2\n", +"T=f2*A;\n", +"c=T/w;\n", +"d1=l*l/(8*c);\n", +"mprintf('sag at 32.2 Celsius , d=%.4f metres',d1);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4: To_determine_the_clearence_between_the_conductor_and_water_level.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the clearence between the conductor and water level\n", +"clear\n", +"clc;\n", +"T=2000;// working tension (kg)\n", +"w=1;\n", +"c=T/w;\n", +"h=90-30;\n", +"l=300;//span(m)\n", +"a=(l/2)-(c*h/l);\n", +"b=550;\n", +"d1=a*a/(2*c);\n", +"d2=(400^2)/(2*c);// sag at 400 metres(m)\n", +"Hm=d2-d1;//height of mid point with respect to A\n", +"Cl=30+Hm;\n", +"mprintf('the clearence between the conductor and water level midway between the towers= %.3f metres \n',Cl);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/8-OVERHEAD_LINE_INSULATORS.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/8-OVERHEAD_LINE_INSULATORS.ipynb new file mode 100644 index 0000000..967347f --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/8-OVERHEAD_LINE_INSULATORS.ipynb @@ -0,0 +1,57 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: OVERHEAD LINE INSULATORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1: EX8_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the maximum voltage that the string of the suspension insulators can withstand.\n", +"clear\n", +"clc;\n", +"E3=17.5;\n", +"E1=64*E3/89;\n", +"E2=9*E1/8;\n", +"E=E1+E2+E3;\n", +"mprintf('the maximum voltage that the string of the suspension insulators can withstand=%.2f kV\n',E);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electrical_Power_Systems_by_C_L_Wadhwa/9-INSULATED_CABLES.ipynb b/Electrical_Power_Systems_by_C_L_Wadhwa/9-INSULATED_CABLES.ipynb new file mode 100644 index 0000000..c593214 --- /dev/null +++ b/Electrical_Power_Systems_by_C_L_Wadhwa/9-INSULATED_CABLES.ipynb @@ -0,0 +1,272 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: INSULATED CABLES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: To_determine_the_economic_overall_diameter_of_a_1core_cable_metal_sheathead.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the economic overall diameter of a 1- core cable metal sheathead.\n", +"clear\n", +"clc;\n", +"V=85;// working voltage (kV)\n", +"gmax=65;// dielectric strength of insulating material (kV/cm)\n", +"r=V/gmax;\n", +"d=2*r;\n", +"D=2.6*%e;\n", +"mprintf('Diameter of the sheath =%.2f cm\n',D);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: To_determine_the_minimum_internal_diameter_of_the_lead_sheath.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the minimum internal diameter of the lead sheath\n", +"clear\n", +"clc;\n", +"e1=4;\n", +"e2=4;\n", +"e3=2.5;\n", +"g1max=50;\n", +"g2max=40;\n", +"g3max=30;\n", +"r=.5;// radius (cm)\n", +"r1=r*e1*g1max/(e2*g2max);\n", +"r2=r1*e2*g2max/(e3*g3max);\n", +"V=66;\n", +"lnc=(V-((r*g1max*log(r1/r))+(r1*g2max*log(r2/r1))));\n", +"m=lnc/(r2*g3max);\n", +"R=r2*(%e^m);\n", +"D=2*R;\n", +"mprintf('minimum internal diameter of the lead sheath,D=%.2f cms\n',D);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: To_determine_the_maximum_safe_working_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To determine the maximum safe working voltage\n", +"clear\n", +"clc;\n", +"r=.5;//radius of conductor(cm)\n", +"g1max=34;\n", +"er=5;\n", +"r1=1;\n", +"R=7/2;//external dia(cm)\n", +"g2max=(r*g1max)/(er*r1);\n", +"V=((r*g1max*log(r1/r))+(r1*g2max*log(R/r1)));\n", +"V=V/(sqrt(2));\n", +"mprintf('Maximum safe working volltage ,V =%.2f kV r.m.s\n',V);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: To_determine_the_maximum_stresses_in_each_of_the_three_layers.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the maximum stresses in each of the three layers .\n", +"clear\n", +"clc;\n", +"r=.9;\n", +"r1=1.25\n", +"r2=r1+.35;\n", +"r3=r2+.35;// radius of outermost layer\n", +"Vd=20;// voltage difference (kV)\n", +"g1max=Vd/(r*log(r1/r));\n", +"g2max=Vd/(r1*log(r2/r1));\n", +"g3max=(66-40)/(r2*log(r3/r2));\n", +"mprintf('g1max =%.1f kV/cm\n',g1max);\n", +"mprintf('g2max =%.2f kV/cm\n',g2max);\n", +"mprintf('g3max =%.0f kV/cm\n',g3max);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: o_dtermine_the_equivalent_star_connected_capacity_and_the_kVA_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To dtermine the equivalent star connected capacity and the kVA required.\n", +"clear\n", +"clc;\n", +"V=20;//voltage (kV)\n", +"w=314;\n", +"C=2*3.04*10^-6;//capacitance per phase(micro-farad)\n", +"KVA=V*V*w*C*1000;\n", +"mprintf('3-phase kVA required =%.0f kVA',KVA); //Answer don't match due to difference in rounding off of digits" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6: EX9_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Determine the capacitance (a)between any two conductors (b)between any two bunched conductors and the third conductor (c)Also calculate the charging current per phase per km\n", +"clear\n", +"clc;\n", +"C1=.208;\n", +"C2=.096;\n", +"Cx=3*C1;\n", +"w=314;\n", +"V=10;\n", +"Cy=(C1+ 2*C2);\n", +"Co=((1.5*Cy)-(Cx/6));\n", +"C=Co/2;\n", +"mprintf('(i)Capacitance between any two conductors=%.3f micro-Farad/km\n',C);\n", +"c=((2*C2 + ((2/3)*C1)));\n", +"mprintf('(ii)Capacitance between any two bunched conductors and the third conductor=%.2f micro-Farad/km\n',c);\n", +"I=V*w*Co*1000*(10^-6)/sqrt(3);\n", +"mprintf('(iii)the charging current per phase per km =%.3f A\n',I);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7: To_calculate_the_induced_emf_in_each_sheath.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// To calculate the induced emf in each sheath .\n", +"clear\n", +"clc;\n", +"rm=(2.28/2)-(.152/2);// mean radius of sheath (cm)\n", +"d=5.08;\n", +"a=d/rm;\n", +"w=314;\n", +"Xm=2*(10^-7)*log(a);// mutual inductance (H/m)\n", +"Xm2=2000*Xm;\n", +"V=w*Xm2*400;\n", +"mprintf('Voltage induced =%.2f volts \n',V);//Answer don't match exactly due to difference in rounding off of digits i between calculations" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8: To_determine_the_ratio_of_sheath_loss_to_core_loss_of_the_cable.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To determine the ratio of sheath loss to core loss of the cable\n", +"clear\n", +"clc;\n", +"R=2*.1625;\n", +"Rs=2*2.14;\n", +"M=314;\n", +"w=6.268*10^-4;\n", +"r=Rs*M*M*w*w/(R*((Rs^2)+(M*M*w*w)));\n", +"mprintf('ratio=%.4f \n',r);" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |