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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 /Power_System_Engineering_by_S_Chakraborthy | |
| 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/Power_System_Engineering_by_S_Chakraborthy/10-STEADY_STATE_CHARACTERISTICS_AND_PERFORMANCE_OF_TRANSMISSION_LINES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/10-STEADY_STATE_CHARACTERISTICS_AND_PERFORMANCE_OF_TRANSMISSION_LINES.ipynb new file mode 100644 index 0000000..f5b0fcb --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/10-STEADY_STATE_CHARACTERISTICS_AND_PERFORMANCE_OF_TRANSMISSION_LINES.ipynb @@ -0,0 +1,1700 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.10: Line_to_line_voltage_and_Power_factor_at_sending_end.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.10 :\n", +"// Page number 135\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 125.0 // Line length(km)\n", +"P_r = 40.0*10**6 // Load at receiving end(VA)\n", +"V_r = 110.0*10**3 // Line voltage at receiving end(V)\n", +"PF_r = 0.8 // Lagging load power factor\n", +"R = 11.0 // Resistance(ohm/phase)\n", +"X = 38.0 // Inductive reactance(ohm/phase)\n", +"Y = 3.0*10**-4 // Capacitive susceptance(S)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"E_r = V_r/3**0.5 // Receiving end phase voltage(V)\n", +"Z = complex(R,X) // Total impedance(ohm/phase)\n", +"I_c1 = E_r*(Y/2)*exp(%i*90.0*%pi/180) // Current through shunt admittance at receiving end(A)\n", +"I_r = P_r/(3**0.5*V_r)*exp(%i*-acos(PF_r)) // Load current(A)\n", +"I = I_r+I_c1 // Current through series impedance(A)\n", +"E_s = I*Z+E_r // Voltage across shunt admittance at sending end(V)\n", +"E_s_ll = 3**0.5*E_s/1000.0 // Line to line voltage at sending end(kV)\n", +"I_c2 = E_s*(Y/2)*exp(%i*90.0*%pi/180) // Current through shunt admittance at sending end(A)\n", +"// Case(ii)\n", +"I_s = I_c2+I_r // Sending end current(A)\n", +"angle_Er_Es = phasemag(E_s) // Angle between E_r and E_s(°)\n", +"angle_Er_Is = phasemag(I_s) // Angle between E_r and I_s(°)\n", +"angle_Es_Is = angle_Er_Es-angle_Er_Is // Angle between E_s and I_s(°)\n", +"PF_s = cosd(angle_Es_Is) // Sending end power factor\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.10 : SOLUTION :-')\n", +"printf('\nCase(i) : Line to line voltage at sending end, E_s = %.f kV', abs(E_s_ll))\n", +"printf('\nCase(ii): Sending end power factor = %.3f \n', PF_s)\n", +"printf('\nNOTE: Answers in the textbook are incomplete')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.11: EX10_11.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.11 :\n", +"// Page number 135-137\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"R = 28.0 // Resistance(ohm/phasemag)\n", +"X = 63.0 // Inductive reactance(ohm/phasemag)\n", +"Y = 4.0*10**-4 // Capacitive susceptance(mho)\n", +"P_r = 75.0*10**6 // Load at receiving end(VA)\n", +"PF_r = 0.8 // Lagging load power factor\n", +"V_r = 132.0*10**3 // Line voltage at receiving end(V)\n", +"\n", +"// Calculations\n", +"// Case(i) Nominal T method\n", +"Z = complex(R,X) // Total impedance(ohm/phasemag)\n", +"E_r = V_r/3**0.5 // Receiving end phasemag voltage(V)\n", +"I_r = P_r/(3**0.5*V_r)*exp(%i*-acos(PF_r)) // Line current at receiving end(A)\n", +"E = E_r+I_r*(Z/2)\n", +"I_c = %i*Y*E // Capacitive current(A)\n", +"I_s = I_r+I_c // Sending end current(A)\n", +"v_drop = I_s*(Z/2) // Voltage drop(V)\n", +"E_s = E+I_s*(Z/2) // Sending end voltage(V)\n", +"E_s_kV = E_s/1000.0 // Sending end voltage(kV)\n", +"E_s_ll= 3**0.5*abs(E_s) // Sending end line voltage(V)\n", +"E_s_llkV = E_s_ll/1000.0 // Sending end line voltage(kV)\n", +"angle_Er_Es = phasemag(E_s) // Angle between E_r and E_s(°)\n", +"angle_Er_Is = phasemag(I_s) // Angle between E_r and I_s(°)\n", +"angle_Es_Is = angle_Er_Es-angle_Er_Is // Angle between E_s and I_s(°)\n", +"PF_s = cosd(angle_Es_Is) // Sending end power factor\n", +"P_s = 3**0.5*E_s_ll*abs(I_s)*PF_s // Power at sending end(W)\n", +"reg = (abs(E_s_ll)-V_r)/V_r*100 // Regulation(%)\n", +"n = (P_r*PF_r)/P_s*100 // Transmission efficiency(%)\n", +"// Case(ii) Nominal π method\n", +"I_c2 = E_r*(%i*Y/2) // Current through shunt admittance at receiving end(A)\n", +"I = I_r+I_c2 // Line current(A)\n", +"E_s_p = E_r+I*Z // Sending end voltage(V)\n", +"E_s_pkV = E_s_p/1000.0 // Sending end voltage(kV)\n", +"E_s_pll = 3**0.5*abs(E_s_p) // Sending end line voltage(V)\n", +"E_s_pllkV = E_s_pll/1000.0 // Sending end line voltage(kV)\n", +"I_c1 = E_s_p*(%i*Y/2) // Current through shunt admittance at sending end(A)\n", +"I_s_p = I+I_c1 // Sending end current(A)\n", +"angle_Er_Esp = phasemag(E_s) // Angle between E_r and E_s(°)\n", +"angle_Er_Isp = phasemag(I_s) // Angle between E_r and I_s(°)\n", +"angle_Es_Isp = angle_Er_Esp-angle_Er_Isp // Angle between E_s and I_s(°)\n", +"PF_s_p = cosd(angle_Es_Isp) // Sending end power factor\n", +"P_s_p = 3**0.5*E_s_pll*abs(I_s_p)*PF_s_p // Power at sending end(W)\n", +"reg_p = (abs(E_s_pll)-V_r)/V_r*100 // Regulation(%)\n", +"n_p = (P_r*PF_r)/P_s_p*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.11 : SOLUTION :-')\n", +"printf('\n(i) Nominal T method')\n", +"printf('\nCase(a): Voltage at sending end, E_s = %.2f∠%.2f° kV = %.1f kV (line-to-line)', abs(E_s_kV),phasemag(E_s_kV),E_s_llkV)\n", +"printf('\nCase(b): Sending end current, I_s = %.1f∠%.2f° A', abs(I_s),phasemag(I_s))\n", +"printf('\nCase(c): Power factor at sending end = %.4f (lagging)', PF_s)\n", +"printf('\nCase(d): Regulation = %.2f percent', reg)\n", +"printf('\nCase(e): Efficiency of transmission = %.2f percent \n', n)\n", +"printf('\n(ii) Nominal π method')\n", +"printf('\nCase(a): Voltage at sending end, E_s = %.2f∠%.2f° kV = %.1f kV (line-to-line)', abs(E_s_pkV),phasemag(E_s_pkV),E_s_pllkV)\n", +"printf('\nCase(b): Sending end current, I_s = %.1f∠%.2f° A', abs(I_s_p),phasemag(I_s_p))\n", +"printf('\nCase(c): Power factor at sending end = %.4f (lagging)', PF_s_p)\n", +"printf('\nCase(d): Regulation = %.2f percent', reg_p)\n", +"printf('\nCase(e): Efficiency of transmission = %.2f percent \n', n_p)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here and more approximation in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.12: Receiving_end_Voltage_Load_and_Nature_of_compensation_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.12 :\n", +"// Page number 143\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"E_s = 275.0 // Sending end voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 400.0 // Line length(km)\n", +"x = 0.05 // Inductive reactance(ohm/km)\n", +"y = 3.0*10**-6 // Line charging susceptance(S/km)\n", +"r = 0.0 // Lossless line\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"R = r*l // Total resistance(ohm/phase)\n", +"X = x*l // Inductive reactance(ohm/phase)\n", +"Y = y*l // Susceptance(mho)\n", +"Z = complex(R,X) // Total impedance(ohm/phase)\n", +"A = 1+(Y*Z/2)*%i // Line constant\n", +"E_r = E_s/abs(A) // Receiving end voltage at no load(kV)\n", +"// case(b)\n", +"Z_0 = (X/Y)**0.5 // Load at receiving end(ohm)\n", +"// Case(c)\n", +"Z_0_new = 1.2*Z_0 // New load at receiving station(ohm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.12 : SOLUTION :-')\n", +"printf('\nCase(a): Receiving end voltage on open circuit = %.1f kV', E_r)\n", +"printf('\nCase(b): Load at receiving end for flat voltage profile on line, Z_0 = %.1f Ω', Z_0)\n", +"printf('\nCase(c): Distributed inductive reactance of the line is to be increased as, Loading for new voltage profile = %.2f Ω', Z_0_new)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.13: Sending_end_voltage_and_Current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.13 :\n", +"// Page number 143-144\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_r = 220.0*10**3 // Receiving end voltage(V)\n", +"Z = complex(20,100) // Impedance(ohm/phase)\n", +"Y = %i*0.0010 // Admittance(mho)\n", +"I_r = 300.0 // Receiving end current(A)\n", +"PF_r = 0.9 // Lagging power factor\n", +"\n", +"// Calculations\n", +"V_2 = V_r/3**0.5 // Receiving end phase voltage(V)\n", +"I_2 = I_r*exp(%i*-acos(PF_r)) // Receiving end current(A)\n", +"I_C2 = (Y/2)*V_2 // Capacitive current at receiving end(A)\n", +"I = I_2+I_C2\n", +"V_1 = V_2+I*Z // Voltage across shunt admittance at sending end(V)\n", +"V_1kV = V_1/1000.0 // Voltage across shunt admittance at sending end(kV)\n", +"I_C1 = (Y/2)*V_1 // Capacitive current at sending end(A)\n", +"I_1 = I_C1+I_2 // Sending end current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.13 : SOLUTION :-')\n", +"printf('\nSending end voltage, V_1 = %.2f∠%.2f° kV', abs(V_1kV),phasemag(V_1kV))\n", +"printf('\nSending end current, I_1 = %.3f∠%.4f° A', abs(I_1),phasemag(I_1))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.14: EX10_14.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.14 :\n", +"// Page number 144\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"r = 0.1 // Resistance(ohm/km)\n", +"l = 1.4*10**-3 // Inductance(H/km)\n", +"c = 8.0*10**-9 // Capacitance(F/km)\n", +"g = 4.0*10**-8 // conductance(mho/km)\n", +"V_r = 400.0 // Receiving end voltage(kV)\n", +"x = 200.0 // Length of line(km)\n", +"\n", +"// Calculations\n", +"V_2 = V_r/3**0.5 // Receiving end phase voltage(kV)\n", +"z = r+%i*2*%pi*f*l // Total impedance(ohm/km)\n", +"y = g+%i*2*%pi*f*c // Total susceptance(mho/km)\n", +"Z_c = (z/y)**0.5 // Surge impedance(ohm)\n", +"gamma = (z*y)**0.5 // γ\n", +"// Case(i)\n", +"V_0_plus = V_2/2 // Incident voltage to neutral at receiving end(kV)\n", +"// Case(ii)\n", +"V_0_minus = V_2/2 // Reflected voltage to neutral at receiving end(kV)\n", +"// Case(iii)\n", +"gamma_l = gamma*x // γl\n", +"V_1_plus = (V_2/2)*exp(gamma_l) // Incident voltage to neutral at 200 km from receiving end(kV)\n", +"V_1_minus = (V_2/2)*exp(-gamma_l) // Reflected voltage to neutral at 200 km from receiving end(kV)\n", +"// Case(iv)\n", +"V_1 = V_1_plus+V_1_minus // Resultant voltage to neutral(kV)\n", +"V_L = abs(V_1) // Resultant voltage to neutral(kV)\n", +"V_L_ll = 3**0.5*V_L // Line to line voltage at 200 km from receiving end(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.14 : SOLUTION :-')\n", +"printf('\nCase(i) : Incident voltage to neutral at receiving end, V_0_plus = %.1f∠%.f° kV', abs(V_0_plus),phasemag(V_0_plus))\n", +"printf('\nCase(ii) : Reflected voltage to neutral at receiving end, V_0_minus = %.1f∠%.f° kV', abs(V_0_minus),phasemag(V_0_minus))\n", +"printf('\nCase(iii): Incident voltage to neutral at 200 km from receiving end, V_1_plus = (%.3f+%.2fj) kV', real(V_1_plus),imag(V_1_plus))\n", +"printf('\nCase(iv) : Resultant voltage to neutral at 200 km from receiving end, V_L = %.2f kV', V_L)\n", +"printf('\n Line to line voltage at 200 km from receiving end = %.2f kV', V_L_ll)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.15: A_B_C_D_constants.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.15 :\n", +"// Page number 145\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"L = 200.0 // Line length(km)\n", +"l = 1.20*10**-3 // Inductance(H/km)\n", +"c = 8.0*10**-9 // Capacitance(F/km)\n", +"r = 0.15 // Resistance(ohm/km)\n", +"g = 0.0 // Conductance(mho/km)\n", +"\n", +"// Calculations\n", +"z = r+%i*2*%pi*f*l // Total impedance(ohm/km)\n", +"Z = z*L // Total impedance(ohm)\n", +"y = g+%i*2*%pi*f*c // Total susceptance(mho/km)\n", +"Y = y*L // Total susceptance(mho/km)\n", +"gamma_l = (Z*Y)**0.5 // γl\n", +"alpha_l = real(gamma_l) // αl\n", +"beta_l = imag(gamma_l) // βl\n", +"Z_c = (Z/Y)**0.5 // Surge impedance(ohm)\n", +"A = cosh(gamma_l) // Constant\n", +"B = Z_c*sinh(gamma_l) // Constant(ohm)\n", +"C = (1/Z_c)*sinh(gamma_l) // Constant(S)\n", +"D = A // Constant\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.15 : SOLUTION :-')\n", +"printf('\nA = D = %.3f∠%.2f° ', abs(A),phasemag(A))\n", +"printf('\nB = %.2f∠%.3f° Ω', abs(B),phasemag(B))\n", +"printf('\nC = %.2e∠%.3f° S', abs(C),phasemag(C))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.16: Sending_end_voltage_Current_Power_factor_and_Efficiency.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.16 :\n", +"// Page number 145-146\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"V_r = 132.0*10**3 // Receiving end voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"L = 200.0 // Line length(km)\n", +"l = 1.3*10**-3 // Inductance(H/km)\n", +"c = 9.0*10**-9 // Capacitance(F/km)\n", +"r = 0.2 // Resistance(ohm/km)\n", +"g = 0.0 // Conductance(mho/km)\n", +"P_r = 50.0*10**6 // Power received(VA)\n", +"PF_r = 0.8 // Lagging power factor at receiving end\n", +"\n", +"// Calculations\n", +"z = r+%i*2*%pi*f*l // Total impedance(ohm/km)\n", +"y = g+%i*2*%pi*f*c // Total susceptance(mho/km)\n", +"Z_c = (z/y)**0.5 // Surge impedance(ohm)\n", +"gamma = (z*y)**0.5 // γ\n", +"gamma_l = gamma*L // γl\n", +"cosh_gl = cosh(gamma_l) // cosh γl\n", +"sinh_gl = sinh(gamma_l) // sinh γl\n", +"V_2 = V_r/(3**0.5) // Receiving end phase voltage(V)\n", +"I_2 = P_r/(3*V_2)*exp(%i*-acos(PF_r)) // Line current(A)\n", +"V_1 = V_2*cosh_gl+I_2*Z_c*sinh_gl // Sending end voltage(V)\n", +"V_1kV = V_1/1000.0 // Sending end voltage(kV)\n", +"I_1 = (V_2/Z_c)*sinh_gl+I_2*cosh_gl // Sending end current(A)\n", +"angle_V2_V1 = phasemag(V_1) // Angle between V_2 and V_1(°)\n", +"angle_V2_I1 = phasemag(I_1) // Angle between V_2 and I_1(°)\n", +"angle_V1_I1 = angle_V2_V1-angle_V2_I1 // Angle between V_1 and I_1(°)\n", +"PF_s = cosd(angle_V1_I1) // Sending end power factor\n", +"P_1 = 3*abs(V_1*I_1)*PF_s // Sending end power(W)\n", +"P_2 = P_r*PF_r // Receiving end power(W)\n", +"n = P_2/P_1*100 // Efficiency\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.16 : SOLUTION :-')\n", +"printf('\nSending end voltage, V_1 = %.3f∠%.4f° kV per phase', abs(V_1kV),phasemag(V_1kV))\n", +"printf('\nSending end current, I_1 = %.3f∠%.2f° A', abs(I_1),phasemag(I_1))\n", +"printf('\nPower factor = %.3f ', PF_s)\n", +"printf('\nEfficiency, η = %.2f percent', n)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.17: Values_of_auxiliary_constants_A_B_C_D.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.17 :\n", +"// Page number 147-148\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"L = 160.0 // Line length(km)\n", +"r = 0.15 // Resistance(ohm/km/phasemag)\n", +"l = 1.2*10**-3 // Inductance(H/km/phasemag)\n", +"c = 0.008*10**-6 // Capacitance(F/km/phasemag)\n", +"g = 0.0 // Conductance(mho/km/phasemag)\n", +"\n", +"// Calculations\n", +"// Case(i) Using convergent series(Complex angles) method\n", +"z = r+%i*2*%pi*f*l // Impedance(ohm/km)\n", +"Z = z*L // Total series impedance(ohm)\n", +"y = g+%i*2*%pi*f*c // Shunt admittance(S/km)\n", +"Y = y*L // Total shunt admittance(S)\n", +"A = 1+(Y*Z/2)+((Y*Z)**2/24) // Constant\n", +"B = Z*(1+(Y*Z/6)+((Y*Z)**2/120)) // Constant(ohm)\n", +"C = Y*(1+(Y*Z/6)+((Y*Z)**2/120)) // Constant(mho)\n", +"D = A // Constant\n", +"// Case(ii) Using convergent series(Real angles) method\n", +"gamma_l = (Z*Y)**0.5 // γl\n", +"alpha_l = real(gamma_l) // αl\n", +"beta_l = imag(gamma_l) // βl\n", +"Z_c = (Z/Y)**0.5 // Surge impedance(ohm)\n", +"A_2 = cosh(gamma_l) // Constant\n", +"B_2 = Z_c*sinh(gamma_l) // Constant(ohm)\n", +"C_2 = (1/Z_c)*sinh(gamma_l) // Constant(mho)\n", +"D_2 = A_2 // Constant\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.17 : SOLUTION :-')\n", +"printf('\nCase(i): Using convergent series(Complex Angles) method')\n", +"printf('\nA = D = %.3f∠%.1f° ', abs(A),phasemag(A))\n", +"printf('\nB = %.f∠%.1f° ohm', abs(B),phasemag(B))\n", +"printf('\nC = %.4f∠%.1f° mho \n', abs(C),phasemag(C))\n", +"printf('\nCase(ii): Using convergent series(Real Angles) method')\n", +"printf('\nA = D = %.3f∠%.1f° ', abs(A_2),phasemag(A_2))\n", +"printf('\nB = %.1f∠%.1f° ohm', abs(B_2),phasemag(B_2))\n", +"printf('\nC = %.4f∠%.1f° S \n', abs(C_2),phasemag(C_2))\n", +"printf('\nNOTE: Slight change in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.18: Sending_end_voltage_and_Current_using_convergent_series_method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.18 :\n", +"// Page number 148\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_r = 220.0*10**3 // Line voltage at receiving end(V)\n", +"Z = complex(40,200) // Impedance per phasemag(ohm)\n", +"Y = %i*0.0015 // Admittance(mho)\n", +"I_r = 200.0 // Receiving end current(A)\n", +"PF_r = 0.95 // Lagging power factor\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"A = 1+(Y*Z/2)+((Y*Z)**2/24) // Constant\n", +"B = Z*(1+(Y*Z/6)+((Y*Z)**2/120)+((Y*Z)**3/5040)) // Constant(ohm)\n", +"C = Y*(1+(Y*Z/6)+((Y*Z)**2/120)+((Y*Z)**3/5040)) // Constant(mho)\n", +"D = A // Constant\n", +"E_r = V_r/3**0.5 // Receiving end phasemag voltage(V)\n", +"I_r1 = I_r*exp(%i*-acos(PF_r)) // Line current(A)\n", +"E_s = A*E_r+B*I_r1 // Sending end voltage(V)\n", +"E_s_ll = 3**0.5*E_s/1000.0 // Sending end line voltage(kV)\n", +"// Case(b)\n", +"I_s = C*E_r+D*I_r1 // Sending end current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.18 : SOLUTION :-')\n", +"printf('\nCase(a): Sending end voltage, E_s = %.1f∠%.2f° kV (line-to-line)', abs(E_s_ll),phasemag(E_s_ll))\n", +"printf('\nCase(b): Sending end current, I_s = %.1f∠%.2f° A\n', abs(I_s),phasemag(I_s))\n", +"printf('\nNOTE: ERROR: Z = (40+j200)Ω, not Z=(60+j200)Ω as given in problem statement')\n", +"printf('\n Changes in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.19: Sending_end_voltage_and_Current_using_nominal_pi_and_nominal_T_method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.19 :\n", +"// Page number 148-149\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_r = 220.0*10**3 // Line voltage at receiving end(V)\n", +"Z = complex(40,200) // Impedance per phasemag(ohm)\n", +"Y = %i*0.0015 // Admittance(S)\n", +"I_R = 200.0 // Receiving end current(A)\n", +"PF_r = 0.95 // Lagging power factor\n", +"\n", +"// Calculations\n", +"// Case(i) Nominal π method\n", +"// Case(a)\n", +"E_r = V_r/3**0.5 // Receiving end phasemag voltage(V)\n", +"I_r = I_R*exp(%i*-acos(PF_r)) // Line current(A)\n", +"Y_2 = Y/2.0 // Admittance(S)\n", +"I_c2 = Y_2*E_r // Current through shunt admittance at receiving end(A)\n", +"I = I_r+I_c2 // Current through impedance(A)\n", +"IZ_drop = I*Z // Voltage drop(V)\n", +"E_s = E_r+IZ_drop // Sending end voltage(V)\n", +"E_s_kV = E_s/1000.0 // Sending end voltage(kV)\n", +"// Case(b)\n", +"I_c1 = E_s*Y_2 // Current through shunt admittance at sending end(A)\n", +"I_s = I+I_c1 // Sending end current(A)\n", +"// Case(ii) Nominal T method\n", +"// Case(a)\n", +"I_r_Z2 = I_r*Z/2 // Voltage drop at receiving end(V)\n", +"E = E_r+I_r_Z2 // Voltage(V)\n", +"I_c = Y*E // Current through shunt admittance(A)\n", +"I_s_2 = I_c+I_r // Sending end current(A)\n", +"I_s_Z2 = I_s_2*(Z/2) // Voltage drop at sending end(V)\n", +"E_s_2 = I_s_Z2+E // Sending end voltage(V)\n", +"E_s_2kV = E_s_2/1000.0 // Sending end voltage(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.19 : SOLUTION :-')\n", +"printf('\nCase(i): Nominal π method')\n", +"printf('\n Case(a): Sending end voltage, E_s = %.1f∠%.2f° kV', abs(E_s_kV),phasemag(E_s_kV))\n", +"printf('\n Case(b): Sending end current, I_s = %.1f∠%.2f° A', abs(I_s),phasemag(I_s))\n", +"printf('\nCase(ii): Nominal T method')\n", +"printf('\n Case(a): Sending end voltage, E_s = %.1f∠%.2f° kV', abs(E_s_2kV),phasemag(E_s_2kV))\n", +"printf('\n Case(b): Sending end current, I_s = %.1f∠%.2f° A \n', abs(I_s_2),phasemag(I_s_2))\n", +"printf('\nThe results are tabulated below')\n", +"printf('\n________________________________________________________')\n", +"printf('\nMETHOD E_s(kV) I_s(A)')\n", +"printf('\n________________________________________________________')\n", +"printf('\nRigorous √3*132.6∠16.46° 209.8∠39.42°')\n", +"printf('\nNominal π √3*%.1f∠%.2f° %.1f∠%.2f°', abs(E_s_kV),phasemag(E_s_kV),abs(I_s),phasemag(I_s))\n", +"printf('\nNominal T √3*%.1f∠%.2f° %.1f∠%.2f°', abs(E_s_2kV),phasemag(E_s_2kV),abs(I_s_2),phasemag(I_s_2))\n", +"printf('\n________________________________________________________')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1: Voltage_regulation_Sending_end_power_factor_and_Transmission_efficiency.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.1 :\n", +"// Page number 127-128\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P = 2.0*10**6 // Power delivered(W)\n", +"V_r = 33.0*10**3 // Receiving end voltage(V)\n", +"PF_r = 0.8 // Receiving end lagging power factor\n", +"R = 10.0 // Total resistance of the line(ohm)\n", +"X = 18.0 // Total inductive resistance of the line(ohm)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"I = P/(V_r*PF_r) // Line current(A)\n", +"sin_phi_r = (1-PF_r**2)**0.5 // Sinφ_R\n", +"V_s = V_r+I*R*PF_r+I*X*sin_phi_r // Sending end voltage(V)\n", +"reg = (V_s-V_r)/V_r*100 // Voltage regulation(%)\n", +"// Case(ii)\n", +"PF_s = (V_r*PF_r+I*R)/V_s // Sending end lagging power factor\n", +"// Case(iii)\n", +"loss = I**2*R // Losses(W)\n", +"P_s = P+loss // Sending end power(W)\n", +"n = P/P_s*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.1 : SOLUTION :-')\n", +"printf('\nCase(i) : Percentage voltage regulation = %.3f percent', reg)\n", +"printf('\nCase(ii) : Sending end power factor = %.2f (lag)', PF_s)\n", +"printf('\nCase(iii): Transmission efficiency, η = %.2f percent \n', n)\n", +"printf('\nNOTE: ERROR: pf is 0.8 and not 0.9 as mentioned in the textbook problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.20: EX10_20.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.20 :\n", +"// Page number 149-153\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"L = 280.0 // Line length(km)\n", +"Z = complex(35,140) // Series impedance(ohm)\n", +"Y = %i*930.0*10**-6 // Shunt admittance(S)\n", +"P_r = 40.0*10**6 // Power delivered(W)\n", +"V_r = 220.0*10**3 // Voltage at receiving end(V)\n", +"PF_r = 0.9 // Lagging power factor\n", +"\n", +"// Calculations\n", +"R = real(Z) // Resistance of the line(ohm)\n", +"// Case(a)\n", +"I_r_a = P_r/(3**0.5*V_r*PF_r)*exp(%i*-acos(PF_r)) // Receiving end current(A)\n", +"I_s_a = I_r_a // Sending end current(A)\n", +"V_r_a = V_r/3**0.5 // phasemag voltage at receiving end(V)\n", +"V_s_a = V_r_a+I_r_a*Z // Sending end voltage(V)\n", +"V_s_a_ll = 3**0.5*V_s_a // Sending end line voltage(V)\n", +"V_s_a_llkv = V_s_a_ll/1000.0 // Sending end line voltage(kV)\n", +"reg_a = (abs(V_s_a_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"loss_a = 3*abs(I_r_a)**2*R // Line loss(W)\n", +"input_a = P_r+loss_a // Input to line(W)\n", +"n_a = P_r/input_a*100 // Efficiency of transmission(%)\n", +"A_a = 1.0 // Constant\n", +"B_a = Z // Constant(ohm)\n", +"C_a = 0 // Constant(mho)\n", +"D_a = A_a // Constant\n", +"// Case(b)\n", +"V_b = V_r_a+I_r_a*Z/2 // Voltage drop across shunt admittance(V)\n", +"I_c_b = Y*V_b // Current through shunt admittance(A)\n", +"I_s_b = I_r_a+I_c_b // Sending end current(A)\n", +"V_s_b = V_b+I_s_b*Z/2 // Sending end voltage(V)\n", +"V_s_b_ll = 3**0.5*V_s_b // Sending end line voltage(V)\n", +"V_s_b_llkv = V_s_b_ll/1000.0 // Sending end line voltage(kV)\n", +"angle_V_Is_b = phasemag(I_s_b) // Angle between V_r and I_s_b(°)\n", +"angle_V_Vs_b = phasemag(V_s_b) // Angle between V_r and V_s_b(°)\n", +"angle_Is_Vs_b = angle_V_Is_b-angle_V_Vs_b // Angle between V_s_b and I_s_b(°)\n", +"PF_s_b = cosd(angle_Is_Vs_b) // Sending end power factor\n", +"P_s_b = 3**0.5*abs(V_s_b_ll*I_s_b)*PF_s_b // Sending end power(W)\n", +"n_b = P_r/P_s_b*100 // Efficiency of transmission(%)\n", +"reg_b = (abs(V_s_b_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"A_b = 1+(1.0/2)*Y*Z // Constant\n", +"B_b = Z*(1+(1.0/4)*Y*Z) // Constant(ohm)\n", +"C_b =Y // Constant(mho)\n", +"D_b = A_b // Constant\n", +"// Alternative solution for case(b)\n", +"V_s_ba = A_b*V_r_a+B_b*I_r_a // Sending end voltage(V)\n", +"V_s_ba_ll = 3**0.5*V_s_ba // Sending end line voltage(V)\n", +"V_s_ba_llkv = V_s_ba_ll/1000.0 // Sending end line voltage(kV)\n", +"I_s_ba = C_b*V_r_a+D_b*I_r_a // Sending end current(A)\n", +"angle_V_Is_ba = phasemag(I_s_ba) // Angle between V_r and I_s_b(°)\n", +"angle_V_Vs_ba = phasemag(V_s_ba) // Angle between V_r and V_s_b(°)\n", +"angle_Is_Vs_ba = angle_V_Is_ba-angle_V_Vs_ba // Angle between V_s_b and I_s_b(°)\n", +"PF_s_ba = cosd(angle_Is_Vs_ba) // Sending end power factor\n", +"P_s_ba = 3**0.5*abs(V_s_ba_ll*I_s_ba)*PF_s_ba // Sending end power(W)\n", +"n_ba = P_r/P_s_ba*100 // Efficiency of transmission(%)\n", +"reg_ba = (abs(V_s_ba_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"// Case(c)\n", +"I_c2_c = Y/2.0*V_r_a // Current through shunt admittance at receiving end(A)\n", +"I_c = I_r_a+I_c2_c // Current through impedance(A)\n", +"V_s_c = V_r_a+I_c*Z // Sending end voltage(V)\n", +"V_s_c_ll = 3**0.5*V_s_c // Sending end line voltage(V)\n", +"V_s_c_llkv = V_s_c_ll/1000.0 // Sending end line voltage(kV)\n", +"I_c1_c = V_s_c*Y/2.0 // Current through shunt admittance at sending end(A)\n", +"I_s_c = I_c+I_c1_c // Sending end current(A)\n", +"angle_V_Is_c = phasemag(I_s_c) // Angle between V_r and I_s_c(°)\n", +"angle_V_Vs_c = phasemag(V_s_c) // Angle between V_r and V_s_c(°)\n", +"angle_Is_Vs_c = angle_V_Is_c-angle_V_Vs_c // Angle between V_s_c and I_s_c(°)\n", +"PF_s_c = cosd(angle_Is_Vs_c) // Sending end power factor\n", +"P_s_c = 3**0.5*abs(V_s_c_ll*I_s_c)*PF_s_c // Sending end power(W)\n", +"n_c = P_r/P_s_c*100 // Efficiency of transmission(%)\n", +"reg_c = (abs(V_s_c_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"A_c = 1+(1.0/2)*Y*Z // Constant\n", +"B_c = Z // Constant(ohm)\n", +"C_c =Y*(1+(1.0/4)*Y*Z) // Constant(mho)\n", +"D_c = A_c // Constant\n", +"// Alternative solution for case(c)\n", +"V_s_ca = A_c*V_r_a+B_c*I_r_a // Sending end voltage(V)\n", +"V_s_ca_ll = 3**0.5*V_s_ca // Sending end line voltage(V)\n", +"V_s_ca_llkv = V_s_ca_ll/1000.0 // Sending end line voltage(kV)\n", +"I_s_ca = C_c*V_r_a+D_c*I_r_a // Sending end current(A)\n", +"angle_V_Is_ca = phasemag(I_s_ca) // Angle between V_r and I_s_c(°)\n", +"angle_V_Vs_ca = phasemag(V_s_ca) // Angle between V_r and V_s_c(°)\n", +"angle_Is_Vs_ca = angle_V_Is_ca-angle_V_Vs_ca // Angle between V_s_b and I_s_c(°)\n", +"PF_s_ca = cosd(angle_Is_Vs_ca) // Sending end power factor\n", +"P_s_ca = 3**0.5*abs(V_s_ca_ll*I_s_ca)*PF_s_ca // Sending end power(W)\n", +"n_ca = P_r/P_s_ca*100 // Efficiency of transmission(%)\n", +"reg_ca = (abs(V_s_ca_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"// Case(d).(i)\n", +"gamma_l = (Y*Z)**0.5 // γl\n", +"Z_c = (Z/Y)**0.5 // Surge impedance(ohm)\n", +"V_s_d1 = V_r_a*cosh(gamma_l)+I_r_a*Z_c*sinh(gamma_l) // Sending end voltage(V)\n", +"V_s_d1_ll = 3**0.5*V_s_d1 // Sending end line voltage(V)\n", +"V_s_d1_llkv = V_s_d1_ll/1000.0 // Sending end line voltage(kV)\n", +"I_s_d1 = V_r_a/Z_c*sinh(gamma_l)+I_r_a*cosh(gamma_l) // Sending end current(A)\n", +"angle_V_Is_d1 = phasemag(I_s_d1) // Angle between V_r and I_s_d(°)\n", +"angle_V_Vs_d1 = phasemag(V_s_d1) // Angle between V_r and V_s_d(°)\n", +"angle_Is_Vs_d1 = angle_V_Is_d1-angle_V_Vs_d1 // Angle between V_s_d and I_s_d(°)\n", +"PF_s_d1 = cosd(angle_Is_Vs_d1) // Sending end power factor\n", +"P_s_d1 = 3**0.5*abs(V_s_d1_ll*I_s_d1)*PF_s_d1 // Sending end power(W)\n", +"n_d1 = P_r/P_s_d1*100 // Efficiency of transmission(%)\n", +"reg_d1 = (abs(V_s_d1_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"A_d1 = cosh(gamma_l) // Constant\n", +"B_d1 = Z_c*sinh(gamma_l) // Constant(ohm)\n", +"C_d1 = (1/Z_c)*sinh(gamma_l) // Constant(mho)\n", +"D_d1 = A_d1 // Constant\n", +"// Case(d).(ii)\n", +"A_d2 = (1+(Y*Z/2)+((Y*Z)**2/24.0)) // Constant\n", +"B_d2 = Z*(1+(Y*Z/6)+((Y*Z)**2/120)) // Constant(ohm)\n", +"C_d2 = Y*(1+(Y*Z/6)+((Y*Z)**2/120)) // Constant(mho)\n", +"D_d2 = A_d2 // Constant\n", +"V_s_d2 = A_d2*V_r_a+B_d2*I_r_a // Sending end voltage(V)\n", +"V_s_d2_ll = 3**0.5*V_s_d2 // Sending end line voltage(V)\n", +"V_s_d2_llkv = V_s_d2_ll/1000.0 // Sending end line voltage(kV)\n", +"I_s_d2 = C_d2*V_r_a+D_d2*I_r_a // Sending end current(A)\n", +"angle_V_Is_d2 = phasemag(I_s_d2) // Angle between V_r and I_s_d(°)\n", +"angle_V_Vs_d2 = phasemag(V_s_d2) // Angle between V_r and V_s_d(°)\n", +"angle_Is_Vs_d2 = angle_V_Is_d2-angle_V_Vs_d2 // Angle between V_s_d and I_s_d(°)\n", +"PF_s_d2 = cosd(angle_Is_Vs_d2) // Sending end power factor\n", +"P_s_d2 = 3**0.5*abs(V_s_d2_ll*I_s_d2)*PF_s_d2 // Sending end power(W)\n", +"n_d2 = P_r/P_s_d2*100 // Efficiency of transmission(%)\n", +"reg_d2 = (abs(V_s_d2_ll)-V_r)/V_r*100 // Voltage regulation(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.20 : SOLUTION :-')\n", +"printf('\nCase(a): Short line approximation')\n", +"printf('\nSending end voltage, V_s = %.1f∠%.1f° kV (line-to-line)', abs(V_s_a_llkv),phasemag(V_s_a_llkv))\n", +"printf('\nVoltage regulation = %.1f percent', reg_a)\n", +"printf('\nTransmission efficiency, η = %.1f percent', n_a)\n", +"printf('\nA = D = %.f ', A_a)\n", +"printf('\nB = %.1f∠%.1f° ohm', abs(B_a),phasemag(B_a))\n", +"printf('\nC = %.f \n', C_a)\n", +"printf('\nCase(b): Nominal T method approximation')\n", +"printf('\nSending end voltage, V_s = %.1f∠%.1f° kV (line-to-line)', abs(V_s_b_llkv),phasemag(V_s_b_llkv))\n", +"printf('\nVoltage regulation = %.2f percent', reg_b)\n", +"printf('\nTransmission efficiency, η = %.1f percent', n_b)\n", +"printf('\nA = D = %.3f∠%.2f° ', abs(A_b),phasemag(A_b))\n", +"printf('\nB = %.1f∠%.1f° ohm', abs(B_b),phasemag(B_b))\n", +"printf('\nC = %.2e∠%.f° S ', abs(C_b),phasemag(C_b))\n", +"printf('\n\tALTERNATIVE SOLUTION:')\n", +"printf('\n\tSending end voltage, V_s = %.1f∠%.1f° kV (line-to-line)', abs(V_s_ba_llkv),phasemag(V_s_ba_llkv))\n", +"printf('\n\tVoltage regulation = %.2f percent', reg_ba)\n", +"printf('\n\tTransmission efficiency, η = %.1f percent', n_ba)\n", +"printf('\n\tA = D = %.3f∠%.2f° ', abs(A_b),phasemag(A_b))\n", +"printf('\n\tB = %.1f∠%.1f° ohm', abs(B_b),phasemag(B_b))\n", +"printf('\n\tC = %.2e∠%.f° S \n', abs(C_b),phasemag(C_b))\n", +"printf('\nCase(c): Nominal π method approximation')\n", +"printf('\nSending end voltage, V_s = %.f∠%.1f° kV (line-to-line)', abs(V_s_c_llkv),phasemag(V_s_c_llkv))\n", +"printf('\nVoltage regulation = %.2f percent', reg_c)\n", +"printf('\nTransmission efficiency, η = %.1f percent', n_c)\n", +"printf('\nA = D = %.3f∠%.2f° ', abs(A_c),phasemag(A_c))\n", +"printf('\nB = %.1f∠%.1f° ohm', abs(B_c),phasemag(B_c))\n", +"printf('\nC = %.2e∠%.1f° mho', abs(C_c),phasemag(C_c))\n", +"printf('\n\tALTERNATIVE SOLUTION:')\n", +"printf('\n\tSending end voltage, V_s = %.1f∠%.1f° kV (line-to-line)', abs(V_s_ca_llkv),phasemag(V_s_ca_llkv))\n", +"printf('\n\tVoltage regulation = %.2f percent', reg_ca)\n", +"printf('\n\tTransmission efficiency, η = %.1f percent', n_ca)\n", +"printf('\n\tA = D = %.3f∠%.2f° ', abs(A_c),phasemag(A_c))\n", +"printf('\n\tB = %.1f∠%.1f° ohm', abs(B_c),phasemag(B_c))\n", +"printf('\n\tC = %.2e∠%.f° S \n', abs(C_c),phasemag(C_c))\n", +"printf('\nCase(d): Long Line Rigorous Solution')\n", +"printf('\n Case(i): Using Convergent Series (Real Angles) Method')\n", +"printf('\n Sending end voltage, V_s = %.f∠%.1f° kV (line-to-line)', abs(V_s_d1_llkv),phasemag(V_s_d1_llkv))\n", +"printf('\n Voltage regulation = %.2f percent', reg_d1)\n", +"printf('\n Transmission efficiency, η = %.1f percent', n_d1)\n", +"printf('\n A = D = %.3f∠%.2f° ', abs(A_d1),phasemag(A_d1))\n", +"printf('\n B = %.f∠%.1f° ohm', abs(B_d1),phasemag(B_d1))\n", +"printf('\n C = %.2e∠%.1f° mho \n', abs(C_d1),phasemag(C_d1))\n", +"printf('\n Case(ii): Using Convergent Series (Complex Angles) Method')\n", +"printf('\n Sending end voltage, V_s = %.f∠%.1f° kV (line-to-line)', abs(V_s_d2_llkv),phasemag(V_s_d2_llkv))\n", +"printf('\n Voltage regulation = %.2f percent', reg_d2)\n", +"printf('\n Transmission efficiency, η = %.1f percent', n_d2)\n", +"printf('\n A = D = %.3f∠%.2f° ', abs(A_d2),phasemag(A_d2))\n", +"printf('\n B = %.1f∠%.1f° ohm', abs(B_d2),phasemag(B_d2))\n", +"printf('\n C = %.2e∠%.1f° mho \n', abs(C_d2),phasemag(C_d2))\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.21: Sending_end_voltage_Current_Power_factor_and_Efficiency_of_transmission.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.21 :\n", +"// Page number 153\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_r = 132.0*10**3 // Line voltage at receiving end(V)\n", +"P_L = 45.0*10**6 // Load delivered(VA)\n", +"PF_r = 0.8 // Lagging power factor\n", +"A = 0.99*exp(%i*0.3*%pi/180) // Constant\n", +"B = 70.0*exp(%i*69.0*%pi/180) // Constant(ohms)\n", +"C = A // Constant\n", +"D = 4.0*10**-4*exp(%i*90.0*%pi/180) // Constant\n", +"\n", +"// Calculations\n", +"E_r = V_r/3**0.5 // Receiving end phasemag voltage(V)\n", +"I_r = P_L/(3**0.5*V_r)*exp(%i*-acos(PF_r)) // Line current(A)\n", +"E_s = A*E_r+B*I_r // Sending end voltage(V)\n", +"E_s_llkV = 3**0.5*E_s/1000.0 // Sending end line voltage(kV)\n", +"I_s = C*I_r+D*E_r // Sending end current(A)\n", +"angle_Er_Es = phasemag(E_s) // Angle between E_r and E_s(°)\n", +"angle_Er_Is = phasemag(I_s) // Angle between E_r and I_s(°)\n", +"angle_Es_Is = angle_Er_Es-angle_Er_Is // Angle between E_s and I_s(°)\n", +"PF_s = cosd(angle_Es_Is) // Sending end power factor\n", +"P_s = 3*abs(E_s*I_s)*PF_s // Sending end power(W)\n", +"P_skW = P_s/1000.0 // Sending end power(kW)\n", +"P_r = P_L*PF_r // Receiving end power(W)\n", +"n = P_r/P_s*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.21 : SOLUTION :-')\n", +"printf('\nCase(i) : Sending end voltage, E_s = %.1f∠%.f° kV (line-to-line)', abs(E_s_llkV),phasemag(E_s_llkV))\n", +"printf('\nCase(ii) : Sending end current, I_s = %.1f∠%.1f° A', abs(I_s),phasemag(I_s))\n", +"printf('\nCase(iii): Sending end power, P_s = %.f kW', P_skW)\n", +"printf('\nCase(iv) : Efficiency of transmission = %.2f percent \n', n)\n", +"printf('\nNOTE: Changes in obtained answer from that textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.23: Overall_constants_A_B_C_D.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.23 :\n", +"// Page number 156\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"A_1 = 0.98*exp(%i*2.0*%pi/180) // Constant of 1st line\n", +"B_1 = 28.0*exp(%i*69.0*%pi/180) // Constant of 1st line(ohms)\n", +"C_1 = 0.0002*exp(%i*88.0*%pi/180) // Constant of 1st line(mho)\n", +"D_1 = A_1 // Constant of 1st line\n", +"A_2 = 0.95*exp(%i*3.0*%pi/180) // Constant of 2nd line\n", +"B_2 = 40.0*exp(%i*85.0*%pi/180) // Constant of 2nd line(ohms)\n", +"C_2 = 0.0004*exp(%i*90.0*%pi/180) // Constant of 2nd line(mho)\n", +"D_2 = A_2 // Constant of 2nd line\n", +"\n", +"// Calculations\n", +"A = A_1*A_2+B_1*C_2 // Constant\n", +"B = A_1*B_2+B_1*D_2 // Constant(ohm)\n", +"C = C_1*A_2+D_1*C_2 // Constant(mho)\n", +"D = C_1*B_2+D_1*D_2 // Constant\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.23 : SOLUTION :-')\n", +"printf('\nA = %.3f∠%.1f° ', abs(A),phasemag(A))\n", +"printf('\nB = %.1f∠%.f° ohm', abs(B),phasemag(B))\n", +"printf('\nC = %.6f∠%.1f° mho', abs(C),phasemag(C))\n", +"printf('\nD = %.3f∠%.1f° ', abs(D),phasemag(D))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.24: Values_of_constants_A0_B0_C0_D0.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.24 :\n", +"// Page number 156-157\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"A = 0.94*exp(%i*1.5*%pi/180) // Constant\n", +"B = 150.0*exp(%i*67.2*%pi/180) // Constant(ohm)\n", +"D = A // Constant\n", +"Y_t = 0.00025*exp(%i*-75.0*%pi/180) // Shunt admittance(mho)\n", +"Z_t = 100.0*exp(%i*70.0*%pi/180) // Series impedance(ohm)\n", +"\n", +"// Calculations\n", +"C = (A*D-1)/B // Constant(mho)\n", +"A_0 = A*(1+Y_t*Z_t)+B*Y_t // Constant\n", +"B_0 = A*Z_t+B // Constant(ohm)\n", +"C_0 = C*(1+Y_t*Z_t)+D*Y_t // Constant(mho)\n", +"D_0 = C*Z_t+D // Constant\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.24 : SOLUTION :-')\n", +"printf('\nA_0 = %.3f∠%.f° ', abs(A_0),phasemag(A_0))\n", +"printf('\nB_0 = %.f∠%.1f° ohm', abs(B_0),phasemag(B_0))\n", +"printf('\nC_0 = %.6f∠%.1f° mho', abs(C_0),phasemag(C_0))\n", +"printf('\nD_0 = %.3f∠%.1f° \n', abs(D_0),phasemag(D_0))\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.25: Maximum_power_transmitted_Receiving_end_power_factor_and_Total_line_loss.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.25 :\n", +"// Page number 163\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"z = complex(0.2,0.6) // Per phase impedance(ohm)\n", +"V_r = 6351.0 // Receiving end voltage per phase(V)\n", +"reg = 7.5/100.0 // Voltage regulation\n", +"\n", +"// Calculations\n", +"V_s = (1+reg)*V_r // Sending end voltage per phase(V)\n", +"R = real(z) // Resistance of the line(ohm)\n", +"X = imag(z) // Reactance of the line(ohm)\n", +"Z = (R**2+X**2)**0.5 // Impedance per phase(ohm)\n", +"P_m = (V_r**2/Z)*((Z*V_s/V_r)-R) // Maximum power transmitted through line(W/phase)\n", +"P_m_MW = P_m/10**6 // Maximum power transmitted through line(MW/phase)\n", +"P_m_MWtotal = 3*P_m_MW // Total maximum power(MW)\n", +"Q = -(V_r**2*X)/Z**2 // Reactive power per phase(Var)\n", +"Q_MW = Q/10**6 // Reactive power per phase(MVAR)\n", +"phi_r = atand(abs(Q_MW/P_m_MW)) // Φ_r(°)\n", +"PF_r = cosd(phi_r) // Receiving end lagging PF\n", +"I = P_m/(V_r*PF_r) // Current delivered(A)\n", +"I_KA = I/1000.0 // Current delivered(KA)\n", +"loss = 3*I**2*R // Total line loss(W)\n", +"loss_MW = loss/10**6 // Total line loss(MW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.25 : SOLUTION :-')\n", +"printf('\nMaximum power transmitted through the line, P_m = %.1f MW', P_m_MWtotal)\n", +"printf('\nReceiving end power factor = %.2f (lagging)', PF_r)\n", +"printf('\nTotal line loss = %.2f MW', loss_MW)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.26: Maximum_power_that_can_be_transferred_to_the_load.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.26 :\n", +"// Page number 163-164\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 100.0 // Length of line(km)\n", +"PF_r = 1.0 // Receiving end Power factor\n", +"Z_c = 400.0 // Characteristic impedance(ohm)\n", +"beta = 1.2*10**-3 // Propagation constant(rad/km)\n", +"V_s = 230.0 // Sending end voltage(kV)\n", +"\n", +"// Calculations\n", +"beta_L = beta*L // (rad)\n", +"beta_L_d = beta_L*180/%pi // (°)\n", +"A = cosd(beta_L) // Constant\n", +"B = %i*Z_c*sin(beta_L) // Constant\n", +"alpha_angle = phasemag(A) // α(°)\n", +"beta_angle = phasemag(B) // β(°)\n", +"V_r = V_s // Receiving end voltage due to lossless line(kV)\n", +"P_max = (V_s*V_r/abs(B))-(abs(A)*V_r**2/abs(B))*cosd(beta_angle-alpha_angle) // Maximum power transferred(MW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.26 : SOLUTION :-')\n", +"printf('\nMaximum power that can be transferred to the load at receiving end, P_max = %.f MW \n', P_max)\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2: Line_current_Receiving_end_voltage_and_Efficiency_of_transmission.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.2 :\n", +"// Page number 128-129\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 10.0 // Length(km)\n", +"V_s = 11.0*10**3 // Sending end voltage(V)\n", +"P = 1000.0*10**3 // Load delivered at receiving end(W)\n", +"PF_r = 0.8 // Receiving end lagging power factor\n", +"r = 0.5 // Resistance of each conductor(ohm/km)\n", +"x = 0.56 // Reactance of each conductor(ohm/km)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"R = r*l // Resistance per phase(ohm)\n", +"X = x*l // Reactance per phase(ohm)\n", +"E_s = V_s/3**0.5 // Phase voltage(V)\n", +"I = P/(3**0.5*V_s*PF_r) // Line current(A)\n", +"// Case(b)\n", +"sin_phi_r = (1-PF_r**2)**0.5 // Sinφ_R\n", +"E_r = E_s-I*R*PF_r-I*X*sin_phi_r // Receiving end voltage(V)\n", +"E_r_ll = 3**0.5*E_r/1000 // Receiving end line to line voltage(kV)\n", +"// Case(c)\n", +"loss = 3*I**2*R // Loss in the transmission line(W)\n", +"P_s = P+loss // Sending end power(W)\n", +"n = P/P_s*100 // Transmission efficiency(%)\n", +"// Alternate method\n", +"Z = R**2+X**2\n", +"P_A = 1.0/3*P // Load delivered(W/phase)\n", +"Q = 1.0*P*sin_phi_r/(3*PF_r) // Reactive load delivered(VAR/phase)\n", +"A = (V_s**2/3.0)-2*(P_A*R+Q*X) // Constant\n", +"B = (1/9.0)*P**2*Z/PF_r**2 // Constant\n", +"const = (A**2-4*B)**0.5 // sqrt(A^2-4B)\n", +"E_r_A = ((A+const)/2)**0.5/1000.0 // Receiving end voltage(kV/phase)\n", +"E_r_A_ll = 3**0.5*E_r_A // Receiving end line-line voltage(kV)\n", +"I_A = P/(3**0.5*E_r_A_ll*1000*PF_r) // Line current(A)\n", +"loss_A = 3*I_A**2*R // Loss in the transmission line(W)\n", +"P_s_A = P+loss_A // Sending end power(W)\n", +"n_A = P/P_s_A*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.2 : SOLUTION :-')\n", +"printf('\nCase(a): Line current, |I| = %.1f A', I)\n", +"printf('\nCase(b): Receiving end voltage, E_r = %.f V (line-to-neutral) = %.2f kV (line-to-line)', E_r,E_r_ll)\n", +"printf('\nCase(c): Efficiency of transmission = %.2f percent \n', n)\n", +"printf('\nAlternative solution by mixed condition:')\n", +"printf('\nCase(a): Line current, |I| = %.1f A', I_A)\n", +"printf('\nCase(b): Receiving end voltage, E_r = %.3f kV/phase = %.2f kV (line-line)', E_r_A,E_r_A_ll)\n", +"printf('\nCase(c): Efficiency of transmission = %.2f percent', n_A)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3: Sending_end_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.3 :\n", +"// Page number 129\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I = 200.0 // Line current(A)\n", +"PF_r = 0.8 // Receiving end lagging power factor\n", +"R = 0.6 // Total resistance of the line(ohm)\n", +"X = 1.0 // Total inductive resistance of the line(ohm)\n", +"n = 0.93 // Efficiency(%)\n", +"\n", +"// Calculations\n", +"V_r = 3*I**2*R/((3*I*PF_r/n)-3*I*PF_r) // Receiving end phase voltage(V)\n", +"sin_phi_r = (1-PF_r**2)**0.5 // Sinφ_R\n", +"V_s = V_r+I*R*PF_r+I*X*sin_phi_r // Sending end voltage(V)\n", +"V_s_ll = 3**0.5*V_s // Sending end line voltage(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.3 : SOLUTION :-')\n", +"printf('\nSending end voltage, V_s(line-line) = %.2f V', V_s_ll)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.4: Distance_over_which_load_is_delivered.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.4 :\n", +"// Page number 129\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P = 15.0*10**6 // Load delivered at receiving end(W)\n", +"PF_r = 0.85 // Receiving end lagging power factor\n", +"r = 0.905 // Resistance of each conductor(ohm/km)\n", +"V_r = 132.0*10**3 // Receiving end voltage(V)\n", +"loss_per = 7.5/100 // Loss\n", +"\n", +"// Calculations\n", +"loss = loss_per*P // Losses in line(W)\n", +"I = P/(3**0.5*V_r*PF_r) // Line current(A)\n", +"l = loss/(3*I**2*r) // Length of line(km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.4 : SOLUTION :-')\n", +"printf('\nDistance over which load is delivered, l = %.2f km', l)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: EX10_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.5 :\n", +"// Page number 130\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 20.0 // Length(km)\n", +"P = 5.0*10**6 // Load delivered at receiving end(W)\n", +"PF_r = 0.8 // Receiving end lagging power factor\n", +"r = 0.02 // Resistance of each conductor(ohm/km)\n", +"L = 0.65*10**-3 // Inductance of each conductor(H/km)\n", +"E_r = 10.0*10**3 // Receiving end voltage(V)\n", +"\n", +"// Calculations\n", +"R = r*l // Resistance per phase(ohm)\n", +"X = 2*%pi*f*L*l // Reactance per phase(ohm)\n", +"// Case(a)\n", +"I = P/(E_r*PF_r) // Line current(A)\n", +"sin_phi_r = (1-PF_r**2)**0.5 // Sinφ_R\n", +"E_s = E_r+I*R*PF_r+I*X*sin_phi_r // Sending end voltage(V)\n", +"E_s_kV = E_s/1000.0 // Sending end voltage(kV)\n", +"reg = (E_s-E_r)/E_r*100 // Voltage regulation(%)\n", +"// Case(b)\n", +"reg_new = reg/2 // New regulation(%)\n", +"E_s_new = (reg_new/100)*E_r+E_r // New value of sending end voltage(V)\n", +"tan_phi_r1 = ((E_s_new-E_r)*(E_r/P)-R)/X // tanφ_r1\n", +"phi_r1 = atan(tan_phi_r1) // φ_r1(radians)\n", +"phi_r1d = phi_r1*180/%pi // φ_r1(degree)\n", +"PF_r1 = cos(phi_r1) // Lagging power factor of receiving end\n", +"sin_phi_r1 = (1-PF_r1**2)**0.5 // Sinφ_r1\n", +"I_R_new = P/(E_r*PF_r1) // New line current(A)\n", +"I_R = I_R_new*complex(PF_r1,-sin_phi_r1)\n", +"I_c = I_R-I*complex(PF_r,-sin_phi_r) // Capacitive current(A)\n", +"I_C = imag(I_c) // Imaginary part of Capacitive current(A)\n", +"c = I_C/(2*%pi*f*E_r)*10.0**6 // Capacitance(µF)\n", +"// Case(c)\n", +"loss_1 = I**2*R // Loss(W)\n", +"n_1 = P/(P+loss_1)*100 // Transmission efficiency(%)\n", +"loss_2 = I_R_new**2*R // Loss(W)\n", +"n_2 = P/(P+loss_2)*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.5 : SOLUTION :-')\n", +"printf('\nCase(a): Sending end voltage, E_s = %.2f kV', E_s_kV)\n", +"printf('\n Voltage regulation of the line = %.1f percent', reg)\n", +"printf('\nCase(b): Value of capacitors to be placed in parallel with load, c = %.2f µF', c)\n", +"printf('\nCase(c): Transmission efficiency in part(a), η_1 = %.2f percent', n_1)\n", +"printf('\n Transmission efficiency in part(b), η_2 = %.1f percent', n_2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.6: Voltage_regulation_Sending_end_voltage_Line_loss_and_Sending_end_power_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.6 :\n", +"// Page number 130-131\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 10.0 // Line length(km)\n", +"Z_l = 0.5*exp(%i*60.0*%pi/180) // Load impedance(ohm/km)\n", +"P = 316.8*10**3 // Load side power(W)\n", +"PF_r = 0.8 // Load side power factor\n", +"E_r = 3.3*10**3 // Load bus voltage(V)\n", +"\n", +"// Calculations\n", +"Z_line = Z_l*l // Load impedance(ohm)\n", +"I_r = P/(E_r*PF_r)*exp(%i*-acos(PF_r)) // Line current(A)\n", +"sin_phi_r = (1-PF_r**2)**0.5 // Sinφ_R\n", +"E_s = E_r+I_r*Z_line // Sending end voltage(V)\n", +"reg = (abs(E_s)-abs(E_r))/abs(E_r)*100 // Voltage regulation(%)\n", +"R = real(Z_line) // Resistance of the load line(ohm)\n", +"loss = abs(I_r)**2*R // Loss in the transmission line(W)\n", +"loss_kW = loss/1000.0 // Loss in the transmission line(kW)\n", +"P_s = P+loss // Sending end power(W)\n", +"angle_Er_Es = phasemag(E_s) // Angle between V_r and V_s(°)\n", +"angle_Er_Ir = acosd(PF_r) // Angle between V_r and I_r(°)\n", +"angle_Es_Is = angle_Er_Es+angle_Er_Ir // Angle between V_s and I_s(°)\n", +"PF_s = cosd(angle_Es_Is) // Sending end power factor\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.6 : SOLUTION :-')\n", +"printf('\nVoltage regulation = %.2f percent', reg)\n", +"printf('\nSending end voltage, E_s = %.f∠%.1f° V', abs(E_s),phasemag(E_s))\n", +"printf('\nLine loss = %.f kW', loss_kW)\n", +"printf('\nSending end power factor = %.2f ', PF_s)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.7: Nominal_pi_equivalent_circuit_parameters_and_Receiving_end_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.7 :\n", +"// Page number 132-133\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_s = 66.0 // Voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 150.0 // Line length(km)\n", +"r = 0.25 // Resistance of each conductor(ohm/km)\n", +"x = 0.5 // Inductive reactance of each conductor(ohm/km)\n", +"y = 0.04*10**-4 // Capacitive admittance(s/km)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"R = r*l // Total resistance(ohm)\n", +"X = x*l // Inductive reactance(ohm)\n", +"Y = y*l // Capacitive resistance(s)\n", +"Y_2 = Y/2 // 1/2 of Capacitive resistance(s)\n", +"// Case(b)\n", +"Z = complex(R,X) // Total impedance(ohm)\n", +"A = 1+(Y*exp(%i*90.0*%pi/180)*Z/2) // Line constant\n", +"V_R_noload = V_s/abs(A) // Receiving end voltage at no-load(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.7 : SOLUTION :-')\n", +"printf('\nCase(a): Total resistance, R = %.1f ohm', R)\n", +"printf('\n Inductive reactance, X = %.1f ohm', X)\n", +"printf('\n Capacitive resistance, Y = %.1e s', Y)\n", +"printf('\n Capacitive resistance, Y/2 = %.1e s', Y_2)\n", +"printf('\nCase(b): Receiving end voltage at no-load, V_R = %.2f kV', V_R_noload)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.8: Voltage_Current_and_Power_factor_at_sending_end.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.8 :\n", +"// Page number 133-134\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"V_r = 132.0*10**3 // Line voltage at receiving end(V)\n", +"L = 100.0 // Line length(km)\n", +"r = 0.17 // Resistance(ohm/km/phase)\n", +"l = 1.1*10**-3 // Inductance(H/km/phase)\n", +"c = 0.0082*10**-6 // Capacitance(F/km/phase)\n", +"P_L = 70.0*10**6 // Load at receiving end(W)\n", +"PF_r = 0.8 // Lagging load power factor\n", +"\n", +"// Calculations\n", +"E_r = V_r/3**0.5 // Receiving end phase voltage(V)\n", +"I_r = P_L/(3**0.5*V_r*PF_r)*exp(%i*-acos(PF_r)) // Receiving end current(A)\n", +"R = r*L // Total resistance(ohm/phase)\n", +"X = 2*%pi*f*l*L // Inductive reactance(ohm/phase)\n", +"Z = complex(R,X) // Total impedance(ohm/phase)\n", +"Y = 2*%pi*f*c*exp(%i*90.0*%pi/180)/L // Shunt admittance of line(mho/phase)\n", +"E = E_r+I_r*(Z/2) // Voltage across shunt admittance(V/phase)\n", +"I_s = I_r+E*Y // Sending end current(A)\n", +"E_s = E+I_s*(Z/2) // Sending end voltage(V/phase)\n", +"E_s_ll = 3**0.5*abs(E_s)/1000 // Sending end line to line voltage(kV)\n", +"angle_Er_Es = phasemag(E_s) // Angle between E_r and V_s(°)\n", +"angle_Er_Is = phasemag(I_s) // Angle between E_r and I_s(°)\n", +"angle_Es_Is = angle_Er_Es-angle_Er_Is // Angle between E_s and I_s(°)\n", +"PF_s = cosd(angle_Es_Is) // Sending end power factor\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.8 : SOLUTION :-')\n", +"printf('\nVoltage at sending end, E_s = %.2f∠%.2f° V/phase = %.f kV (line-to-line)', abs(E_s),phasemag(E_s),E_s_ll)\n", +"printf('\nCurrent at sending end, I_s = %.1f∠%.1f° A', abs(I_s),phasemag(I_s))\n", +"printf('\nSending end power factor = %.3f (lagging)', PF_s)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.9: Sending_end_voltage_Current_and_Transmission_efficiency.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 3.9 :\n", +"// Page number 134\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"E_r = 66.0*10**3 // Line voltage at receiving end(V)\n", +"l = 120.0 // Line length(km)\n", +"r = 0.1 // Resistance(ohm/km/phase)\n", +"x = 0.3 // Inductive reactance(ohm/km/phase)\n", +"y = 0.04*10**-4 // Capacitive susceptance(S/km/phase)\n", +"P_L = 10.0*10**6 // Load at receiving end(W)\n", +"PF_r = 0.8 // Lagging load power factor\n", +"\n", +"// Calculations\n", +"R = r*l // Total resistance(ohm/phase)\n", +"X = x*l // Inductive reactance(ohm/phase)\n", +"Y = y*l // Susceptance(mho)\n", +"Z = complex(R,X) // Total impedance(ohm/phase)\n", +"V_r = E_r/3**0.5 // Receiving end phase voltage(V)\n", +"I_r = P_L/(3**0.5*E_r*PF_r)*exp(%i*-acos(PF_r)) // Load current(A)\n", +"V_1 = V_r+I_r*(Z/2) // Voltage across capacitor(V)\n", +"I_c = %i*Y*V_1 // Charging current(A)\n", +"I_s = I_r+I_c // Sending end current(A)\n", +"V_s = V_1+I_s*(Z/2) // Sending end voltage(V/phase)\n", +"V_s_ll = 3**0.5*abs(V_s)/1000.0 // Sending end line to line voltage(kV)\n", +"angle_Vr_Vs = phasemag(V_s) // Angle between V_r and V_s(°)\n", +"angle_Vr_Is = phasemag(I_s) // Angle between V_r and I_s(°)\n", +"angle_Vs_Is = angle_Vr_Vs-angle_Vr_Is // Angle between V_s and I_s(°)\n", +"PF_s = cosd(angle_Vs_Is) // Sending end power factor\n", +"P_s = 3*abs(V_s*I_s)*PF_s // Sending end power(W)\n", +"n = P_L/P_s*100 // Transmission efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 3.9 : SOLUTION :-')\n", +"printf('\nSending end voltage, |V_s| = %.f V/phase = %.3f V (line-to-line)', abs(V_s),V_s_ll)\n", +"printf('\nSending end current, |I_s| = %.2f A', abs(I_s))\n", +"printf('\nTransmission efficiency = %.2f percent \n', n)\n", +"printf('\nNOTE: ERROR: Calculation mistake in finding sending end power factor')\n", +"printf('\n Changes in the obtained answer from that of textbook is due to more precision')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/11-OVERHEAD_LINE_INSULATORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/11-OVERHEAD_LINE_INSULATORS.ipynb new file mode 100644 index 0000000..48e4b0b --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/11-OVERHEAD_LINE_INSULATORS.ipynb @@ -0,0 +1,486 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: OVERHEAD LINE INSULATORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1: Ratio_of_capacitance_Line_voltage_and_String_efficiency.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.1 :\n", +"// Page number 183\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_1 = 9.0 // Potential across top unit(kV)\n", +"V_2 = 11.0 // Potential across middle unit(kV)\n", +"n = 3.0 // Number of disc insulators\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"K = (V_2-V_1)/V_1 // Ratio of capacitance b/w pin & earth to self capacitance\n", +"// Case(b)\n", +"V_3 = V_2+(V_1+V_2)*K // Potential across bottom unit(kV)\n", +"V = V_1+V_2+V_3 // Voltage between line and earth(kV)\n", +"V_l = 3**0.5*V // Line voltage(kV)\n", +"// Case(c)\n", +"eff = V/(n*V_3)*100 // String efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.1 : SOLUTION :-')\n", +"printf('\nCase(a): Ratio of capacitance b/w pin & earth to self-capacitance of each unit, K = %.2f ', K)\n", +"printf('\nCase(b): Line voltage = %.2f kV', V_l)\n", +"printf('\nCase(c): String efficiency = %.f percent', eff)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2: Mutual_capacitance_of_each_unit_in_terms_of_C.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.2 :\n", +"// Page number 183-184\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"m = 10.0 // Mutual capacitance of top insulator in terms of C\n", +"\n", +"// Calculations\n", +"X = 1+m // Mutual capacitance in terms of C\n", +"Y = (1.0+2)+m // Mutual capacitance in terms of C\n", +"Z = (1.0+2+3)+m // Mutual capacitance in terms of C\n", +"U = (1.0+2+3+4)+m // Mutual capacitance in terms of C\n", +"V = (1.0+2+3+4+5)+m // Mutual capacitance in terms of C\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.2 : SOLUTION :-')\n", +"printf('\nMutual capacitance of each unit:')\n", +"printf('\n X = %.f*C', X)\n", +"printf('\n Y = %.f*C', Y)\n", +"printf('\n Z = %.f*C', Z)\n", +"printf('\n U = %.f*C', U)\n", +"printf('\n V = %.f*C', V)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3: EX11_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.3 :\n", +"// Page number 184\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 3.0 // Number of insulators\n", +"\n", +"// Calculations\n", +"V_1 = 155.0/475.0 // Potential across top unit\n", +"V_2 = 154.0/155.0*V_1 // Potential across middle unit\n", +"V_3 = 166.0/155.0*V_1 // Potential across bottom unit\n", +"eff = 100/(n*V_3) // String efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.3 : SOLUTION :-')\n", +"printf('\nVoltage across top unit, V_1 = %.3f*V', V_1)\n", +"printf('\nVoltage across middle unit, V_2 = %.3f*V', V_2)\n", +"printf('\nVoltage across bottom unit, V_3 = %.2f*V', V_3)\n", +"printf('\nString efficiency = %.2f percent', eff)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.4: Line_to_neutral_voltage_and_String_efficiency.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.4 :\n", +"// Page number 184-185\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_3 = 17.5 // Voltage across line unit(kV)\n", +"c = 1.0/8 // Shunt capacitance = 1/8 of insulator capacitance\n", +"n = 3.0 // Number of insulators\n", +"\n", +"// Calculations\n", +"K = c // String constant\n", +"V_1 = V_3/(1+3*K+K**2) // Voltage across top unit(kV)\n", +"V_2 = (1+K)*V_1 // Voltage across middle unit(kV)\n", +"V = V_1+V_2+V_3 // Voltage between line & earth(kV)\n", +"eff = V*100/(n*V_3) // String efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.4 : SOLUTION :-')\n", +"printf('\nLine to neutral voltage, V = %.2f kV', V)\n", +"printf('\nString efficiency = %.2f percent', eff)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.5: Value_of_line_to_pin_capacitance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.5 :\n", +"// Page number 185\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 8.0 // Number of insulators\n", +"\n", +"// Calculations\n", +"A = 1.0/(n-1) // Line to pin capacitance\n", +"B = 2.0/(n-2) // Line to pin capacitance\n", +"C = 3.0/(n-3) // Line to pin capacitance\n", +"D = 4.0/(n-4) // Line to pin capacitance\n", +"E = 5.0/(n-5) // Line to pin capacitance\n", +"F = 6.0/(n-6) // Line to pin capacitance\n", +"G = 7.0/(n-7) // Line to pin capacitance\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.5 : SOLUTION :-')\n", +"printf('\nLine-to-pin capacitance are:')\n", +"printf('\n A = %.3f*C', A)\n", +"printf('\n B = %.3f*C', B)\n", +"printf('\n C = %.3f*C', C)\n", +"printf('\n D = %.3f*C', D)\n", +"printf('\n E = %.3f*C', E)\n", +"printf('\n F = %.3f*C', F)\n", +"printf('\n G = %.3f*C', G)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6: EX11_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.6 :\n", +"// Page number 186\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"m = 6.0 // Mutual capacitance\n", +"n = 5.0 // Number of insulators\n", +"\n", +"// Calculations\n", +"E_4 = (1+(1/m)) // Voltage across 4th insulator as percent of E_5(%)\n", +"E_3 = (1+(3/m)+(1/m**2)) // Voltage across 3rd insulator as percent of E_5(%)\n", +"E_2 = (1+(6/m)+(5/m**2)+(1/m**3)) // Voltage across 2nd insulator as percent of E_5(%)\n", +"E_1 = (1+(10/m)+(15/m**2)+(7/m**3)+(1/m**4)) // Voltage across 1st insulator as percent of E_5(%)\n", +"E_5 = 100/(E_4+E_3+E_2+E_1+1) // Voltage across 5th insulator as percent of E_5(%)\n", +"E4 = E_4*E_5 // Voltage across 4th insulator as percent of E_5(%)\n", +"E3 = E_3*E_5 // Voltage across 3rd insulator as percent of E_5(%)\n", +"E2 = E_2*E_5 // Voltage across 2nd insulator as percent of E_5(%)\n", +"E1 = E_1*E_5 // Voltage across 1st insulator as percent of E_5(%)\n", +"eff = 100/(n*E1/100) // String efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.6 : SOLUTION :-')\n", +"printf('\nVoltage distribution as a percentage of voltage of conductor to earth are:')\n", +"printf('\n E_1 = %.2f percent', E1)\n", +"printf('\n E_2 = %.2f percent', E2)\n", +"printf('\n E_3 = %.1f percent', E3)\n", +"printf('\n E_4 = %.1f percent', E4)\n", +"printf('\n E_5 = %.2f percent', E_5)\n", +"printf('\nString efficiency = %.f percent \n', eff)\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.7: EX11_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.7 :\n", +"// Page number 186-187\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 3.0 // Number of insulators\n", +"C_1 = 0.2 // Capacitance in terms of C\n", +"C_2 = 0.1 // Capacitance in terms of C\n", +"\n", +"// Calculations\n", +"// Without guard ring\n", +"e_2_a = 13.0/13.3 // Potential across middle unit as top unit\n", +"e_1_a = 8.3/6.5*e_2_a // Potential across bottom unit\n", +"E_a = 1+(1/(8.3/6.5))+(1/e_1_a) // Voltage in terms of e_1\n", +"eff_a = E_a/n*100 // String efficiency(%)\n", +"e1_a = 1/E_a // Voltage across bottom unit as a % of line voltage\n", +"e2_a = 1/(8.3/6.5)*e1_a // Voltage across middle unit as a % of line voltage\n", +"e3_a = 1/e_1_a*e1_a // Voltage across top unit as a % of line voltage\n", +"// With guard ring\n", +"e_2_b = 15.4/15.5 // Potential across middle unit as top unit\n", +"e_1_b = 8.3/7.7*e_2_b // Potential across bottom unit\n", +"E_b = 1+(1/(8.3/7.7))+(1/e_1_b) // Voltage in terms of e_1\n", +"eff_b = E_b/n*100 // String efficiency(%)\n", +"e1_b = 1/E_b // Voltage across bottom unit as a % of line voltage\n", +"e2_b = 1/(8.3/7.7)*e1_b // Voltage across middle unit as a % of line voltage\n", +"e3_b = 1/e_1_b*e1_b // Voltage across top unit as a % of line voltage\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.7 : SOLUTION :-')\n", +"printf('\nWithout guard ring:')\n", +"printf('\n Voltage across bottom unit, e_1 = %.2f*E', e1_a)\n", +"printf('\n Voltage across bottom unit, e_2 = %.2f*E', e2_a)\n", +"printf('\n Voltage across bottom unit, e_3 = %.2f*E', e3_a)\n", +"printf('\n String efficiency = %.1f percent \n', eff_a)\n", +"printf('\nWith guard ring:')\n", +"printf('\n Voltage across bottom unit, e_1 = %.2f*E', e1_b)\n", +"printf('\n Voltage across bottom unit, e_2 = %.2f*E', e2_b)\n", +"printf('\n Voltage across bottom unit, e_3 = %.3f*E', e3_b)\n", +"printf('\n String efficiency = %.2f percent', eff_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.8: EX11_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.8 :\n", +"// Page number 187-188\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 3.0 // Number of insulators\n", +"\n", +"// Calculations\n", +"V_1 = 0.988 // Voltage across top unit as middle unit\n", +"V_3 = 1.362 // Voltage across bottom unit as middle unit\n", +"V_2 = 1/(V_1+1+V_3) // Voltage across middle unit as % of line voltage to earth\n", +"V1 = V_1*V_2*100 // Voltage across top unit as % of line voltage to earth\n", +"V2 = V_2*100 // Voltage across middle unit as % of line voltage to earth\n", +"V3 = V_3*V_2*100 // Voltage across bottom unit as % of line voltage to earth\n", +"eff = 100/(n*V3/100) // String efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.8 : SOLUTION :-')\n", +"printf('\nCase(a): Voltage across top unit as a percentage of line voltage to earth, V_1 = %.2f percent', V1)\n", +"printf('\n Voltage across middle unit as a percentage of line voltage to earth, V_2 = %.2f percent', V2)\n", +"printf('\n Voltage across bottom unit as a percentage of line voltage to earth, V_3 = %.2f percent', V3)\n", +"printf('\nCase(b): String efficiency = %.2f percent', eff)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.9: Voltage_on_the_line_end_unit_and_Value_of_capacitance_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 4: OVERHEAD LINE INSULATORS\n", +"\n", +"// EXAMPLE : 4.9 :\n", +"// Page number 188\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 3.0 // Number of insulators\n", +"V = 20.0 // Voltage across each conductor(kV)\n", +"c = 1.0/5 // Capacitance ratio\n", +"\n", +"// Calculations\n", +"V_2 = 6.0/5.0 // Voltage across middle unit as top unit\n", +"V_1 = V/(1+2*V_2) // Voltage across top unit(kV)\n", +"V_3 = V_2*V_1 // Voltage across bottom unit(kV)\n", +"C_x = c*(1+(1/V_2)) // Capacitance required\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 4.9 : SOLUTION :-')\n", +"printf('\nCase(a): Voltage on the line-end unit, V_3 = %.2f kV', V_3)\n", +"printf('\nCase(b): Value of capacitance required, Cx = %.3f*C', C_x)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/12-MECHANICAL_DESIGN_OF_OVERHEAD_LINES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/12-MECHANICAL_DESIGN_OF_OVERHEAD_LINES.ipynb new file mode 100644 index 0000000..5033193 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/12-MECHANICAL_DESIGN_OF_OVERHEAD_LINES.ipynb @@ -0,0 +1,620 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12: MECHANICAL DESIGN OF OVERHEAD LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.10: Sag_in_inclined_direction_and_Vertical_direction.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.10 :\n", +"// Page number 201-202\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 250.0 // Span(m)\n", +"d = 1.42 // Diameter(cm)\n", +"w = 1.09 // Dead weight(kg/m)\n", +"wind = 37.8 // Wind pressure(kg/m^2)\n", +"r = 1.25 // Ice thickness(cm)\n", +"f_m = 1050.0 // Maximum working stress(kg/sq.cm)\n", +"\n", +"// Calculations\n", +"w_i = 913.5*%pi*r*(d+r)*10**-4 // Weight of ice on conductor(kg/m)\n", +"w_w = wind*(d+2*r)*10**-2 // Wind load of conductor(kg/m)\n", +"w_r = ((w+w_i)**2+w_w**2)**0.5 // Resultant pressure(kg/m)\n", +"a = %pi*d**2/4.0 // Area(cm^2)\n", +"T_0 = f_m*a // Tension(kg)\n", +"S = w_r*L**2/(8*T_0) // Total sag(m)\n", +"vertical_sag = S*(w+w_i)/w_r // Vertical component of sag(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.10 : SOLUTION :-')\n", +"printf('\nCase(i) : Sag in inclined direction = %.f m', S)\n", +"printf('\nCase(ii): Sag in vertical direction = %.2f m', vertical_sag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.11: Sag_in_still_air_Wind_pressure_Ice_coating_and_Vertical_sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.11 :\n", +"// Page number 202-203\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"a = 120.0 // Area(mm^2)\n", +"ds = 2.11 // Diameter of each strand(mm)\n", +"W = 1118.0/1000 // Weight of conductor(kg/m)\n", +"L = 200.0 // Span(m)\n", +"stress = 42.2 // Ultimate tensile stress(kg/mm^2)\n", +"wind = 60.0 // Wind pressure(kg/m^2)\n", +"t = 10.0 // Ice thickness(mm)\n", +"\n", +"// Calculations\n", +"n = 3.0 // Number of layers\n", +"d = (2*n+1)*ds // Overall diameter of conductor(mm)\n", +"u = stress*a // Ultimate strength(kg)\n", +"T = u/4.0 // Working stregth(kg)\n", +"// Case(a)\n", +"S_a = W*L**2/(8*T) // Sag in still air(m)\n", +"// Case(b)\n", +"area = d*100*10.0*10**-6 // Projected area to wind pressure(m^2)\n", +"w_w = wind*area // Wind load/m(kg)\n", +"w_r = (W**2+w_w**2)**0.5 // Resultant weight/m(kg)\n", +"S_b = w_r*L**2/(8*T) // Total sag with wind pressure(m)\n", +"w_i = 0.915*%pi/4*((d+2*t)**2-(d**2))/1000.0 // Weight of ice on conductor(kg/m)\n", +"area_i = (d+2*t)*1000.0*10**-6 // Projected area to wind pressure(m^2)\n", +"w_n = wind*area_i // Wind load/m(kg)\n", +"w_r_c = ((W+w_i)**2+w_n**2)**0.5 // Resultant weight/m(kg)\n", +"S_c = w_r_c*L**2/(8*T) // Total sag with wind pressure and ice coating(m)\n", +"S_v = S_c*(W+w_i)/w_r_c // Vertical component of sag(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.11 : SOLUTION :-')\n", +"printf('\nCase(a) : Sag in still air, S = %.2f m', S_a)\n", +"printf('\nCase(b) : Sag with wind pressure, S = %.2f m', S_b)\n", +"printf('\n Sag with wind pressure and ice coating, S = %.2f m', S_c)\n", +"printf('\n Vertical sag, S_v = %.2f m \n', S_v)\n", +"printf('\nNOTE: ERROR: calculation mistake in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1: Weight_of_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.1 :\n", +"// Page number 198\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"u = 5758.0 // Ultimate strength(kg)\n", +"S = 2.0 // Sag(m)\n", +"s = 2.0 // Factor of safety\n", +"L = 250.0 // Span length(m)\n", +"\n", +"// Calculations\n", +"T = u/s // Allowable max tension(kg)\n", +"w = S*8.0*T/L**2 // weight(kg/m)\n", +"l = L/2 // Half span length(m)\n", +"half_span = l+(w**2*l**3/(6*T**2)) // Half span length(m)\n", +"total_length = 2*half_span // Total length(m)\n", +"weight = w*total_length // Weight of conductor(kg)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.1 : SOLUTION :-')\n", +"printf('\nWeight of conductor = %.2f kg', weight)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2: Point_of_maximum_sag_at_the_lower_support.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.2 :\n", +"// Page number 198\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 250.0 // Span length(m)\n", +"h = 10.0 // Difference in height(m)\n", +"r = 1.0 // Radius of conductor(cm)\n", +"w = 2.5 // Weight of conductor(kg/m)\n", +"wind = 1.2 // Wind load(kg/m)\n", +"s = 3.0 // Factor of safety\n", +"tensile = 4300.0 // Maximum tensile strength(kg/sq.cm)\n", +"\n", +"// Calculations\n", +"W = (w**2+wind**2)**0.5 // Total pressure on conductor(kg/m)\n", +"f = tensile/s // Permissible stress in conductor(kg/sq.cm) \n", +"a = %pi*r**2 // Area of the conductor(sq.cm)\n", +"T = f*a // Allowable max tension(kg)\n", +"x = (L/2)-(T*h/(L*W)) // Point of maximum sag at the lower support(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.2 : SOLUTION :-')\n", +"printf('\nPoint of maximum sag at the lower support, x = %.2f metres', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3: Vertical_sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.3 :\n", +"// Page number 198-199\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"a = 2.5 // Cross-sectional area(sq.cm)\n", +"L = 250.0 // Span(m)\n", +"w_c = 1.8 // Weight of conductor(kg/m)\n", +"u = 8000.0 // Ultimate strength(kg/cm^2)\n", +"wind = 40.0 // Wind load(kg/cm^2)\n", +"s = 3.0 // Factor of safety\n", +"\n", +"// Calculations\n", +"d = (4.0*a/%pi)**0.5 // Diameter(cm)\n", +"T = u*a/s // Allowable max tension(kg)\n", +"w_w = wind*d/100.0 // Horizontal wind force(kg)\n", +"w_r = (w_c**2+w_w**2)**0.5 // Resultant force(kg/m)\n", +"S = w_r*L**2/(8*T) // Slant sag(m)\n", +"vertical_sag = S*(w_c/w_r) // Vertical sag(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.3 : SOLUTION :-')\n", +"printf('\nVertical sag = %.3f metres', vertical_sag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4: Height_above_ground_at_which_the_conductors_should_be_supported.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.4 :\n", +"// Page number 199\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"a = 110.0 // Cross-sectional area(sq.mm)\n", +"w_c = 844.0/1000 // Weight of conductor(kg/m)\n", +"U = 7950.0 // Ultimate strength(kg)\n", +"L = 300.0 // Span(m)\n", +"s = 2.0 // Factor of safety\n", +"wind = 75.0 // Wind pressure(kg/m^2)\n", +"h = 7.0 // Ground clearance(m)\n", +"d = 2.79 // Diameter of copper(mm)\n", +"n = 7.0 // Number of strands\n", +"\n", +"// Calculations\n", +"dia = n*d // Diameter of conductor(mm)\n", +"w_w = wind*dia/1000.0 // Horizontal wind force(kg)\n", +"w = (w_c**2+w_w**2)**0.5 // Resultant force(kg)\n", +"T = U/2.0 // Allowable tension(m)\n", +"l = L/2.0 // Half-span(m)\n", +"D = w*l**2/(2*T) // Distance(m)\n", +"height = h+D // Height above ground at which the conductors should be supported(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.4 : SOLUTION :-')\n", +"printf('\nHeight above ground at which the conductors should be supported = %.2f metres', height)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5: Permissible_span_between_two_supports.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.5 :\n", +"// Page number 199\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"w_w = 1.781 // Wind pressure on conductor(kg/m)\n", +"w_i = 1.08 // Weight of ice on conductor(kg/m)\n", +"D = 6.0 // Maximum permissible sag(m)\n", +"s = 2.0 // Factor of safety\n", +"w_c = 0.844 // Weight of conductor(kg/m)\n", +"u = 7950.0 // Ultimate strength(kg)\n", +"\n", +"// Calculations\n", +"w = ((w_c+w_i)**2+w_w**2)**0.5 // Total force on conductor(kg/m)\n", +"T = u/s // Allowable maximum tension(kg)\n", +"l = ((D*2*T)/w)**0.5 // Half span(m)\n", +"L = 2.0*l // Permissible span between two supports(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.5 : SOLUTION :-')\n", +"printf('\nPermissible span between two supports = %.f metres \n', L)\n", +"printf('\nNOTE: ERROR: Horizontal wind load, w_w = 1.781 kg/m, not 1.78 kg/m as mentioned in problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6: EX12_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.6 :\n", +"// Page number 199-200\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"a = 0.484 // Area of conductor(sq.cm)\n", +"d = 0.889 // Overall diameter(cm)\n", +"w_c = 428/1000.0 // Weight(kg/m)\n", +"u = 1973.0 // Breaking strength(kg)\n", +"s = 2.0 // Factor of safety\n", +"L = 200.0 // Span(m)\n", +"t = 1.0 // Ice thickness(cm)\n", +"wind = 39.0 // Wind pressure(kg/m^2)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"l = L/2.0 // Half span(m)\n", +"T = u/s // Allowable maximum tension(kg)\n", +"D_1 = w_c*l**2/(2*T) // Maximum sag due to weight of conductor(m)\n", +"// Case(ii)\n", +"w_i = 913.5*%pi*t*(d+t)*10**-4 // Weight of ice on conductor(kg/m)\n", +"w = w_c+w_i // Total weight of conductor & ice(kg/m)\n", +"D_2 = w*l**2/(2*T) // Maximum sag due to additional weight of ice(m)\n", +"// Case(iii)\n", +"D = d+2.0*t // Diameter due to ice(cm)\n", +"w_w = wind*D*10**-2 // Wind pressure on conductor(kg/m)\n", +"w_3 = ((w_c+w_i)**2+w_w**2)**0.5 // Total force on conductor(kg/m)\n", +"D_3 = w_3*l**2/(2*T) // Maximum sag due to (i), (ii) & wind(m)\n", +"theta = atand(w_w/(w_c+w_i)) // θ(°)\n", +"vertical_sag = D_3*cosd(theta) // Vertical sag(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.6 : SOLUTION :-')\n", +"printf('\nCase(i) : Maximum sag of line due to weight of conductor, D = %.2f metres', D_1)\n", +"printf('\nCase(ii) : Maximum sag of line due to additional weight of ice, D = %.2f metres', D_2)\n", +"printf('\nCase(iii): Maximum sag of line due to (i),(ii) plus wind, D = %.2f metres', D_3)\n", +"printf('\n Vertical sag = %.2f metres', vertical_sag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.7: Point_of_minimum_sag.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.7 :\n", +"// Page number 200\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 428/1000.0 // Weight(kg/m)\n", +"u = 1973.0 // Breaking strength(kg)\n", +"s = 2.0 // Factor of safety\n", +"l = 200.0 // Span(m)\n", +"h = 3.0 // Difference in tower height(m)\n", +"\n", +"// Calculations\n", +"T = u/s // Allowable maximum tension(kg)\n", +"x_2 = (l/2.0)+(T*h/(W*l)) // Point of minimum sag from tower at higher level(m)\n", +"x_1 = l-x_2 // Point of minimum sag from tower at lower level(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.7 : SOLUTION :-')\n", +"printf('\nPoint of minimum sag, x_1 = %.1f metres', x_1)\n", +"printf('\nPoint of minimum sag, x_2 = %.1f metres', x_2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.8: Clearance_between_conductor_and_water_at_a_point_midway_between_towers.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.8 :\n", +"// Page number 200-201\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"h_1 = 50.0 // Height of tower P1(m)\n", +"h_2 = 80.0 // Height of tower P2(m)\n", +"L = 300.0 // Horizontal distance b/w towers(m)\n", +"T = 2000.0 // Tension in conductor(kg)\n", +"w = 0.844 // Weight of conductor(kg/m)\n", +"\n", +"// Calculations\n", +"h = h_2-h_1 // Difference in height of tower(m)\n", +"x_2 = (L/2.0)+(T*h/(w*L)) // Point of minimum sag from tower P2(m)\n", +"x_1 = (L/2.0)-(T*h/(w*L)) // Point of minimum sag from tower at lower level(m)\n", +"P = (L/2.0)-x_1 // Distance of point P(m)\n", +"D = w*P**2/(2*T) // Height of P above O(m)\n", +"D_2 = w*x_2**2/(2*T) // Height of P2 above O(m)\n", +"mid_point_P2 = D_2-D // Mid-point below P2(m)\n", +"clearance = h_2-mid_point_P2 // Clearance b/w conductor & water(m)\n", +"D_1 = w*x_1**2/(2*T) // Height of P1 above O(m)\n", +"mid_point_P1 = D-D_1 // Mid-point above P1(m)\n", +"clearance_alt = h_1+mid_point_P1 // Clearance b/w conductor & water(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.8 : SOLUTION :-')\n", +"printf('\nClearance between conductor & water at a point midway b/w towers = %.2f m above water\n', clearance)\n", +"printf('\nALTERNATIVE METHOD:')\n", +"printf('\nClearance between conductor & water at a point midway b/w towers = %.2f m above water', clearance_alt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.9: Sag_at_erection_and_Tension_of_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES\n", +"\n", +"// EXAMPLE : 5.9 :\n", +"// Page number 201\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 300.0 // Span(m)\n", +"T_still = 45.0 // Temperature in still air(°C)\n", +"a = 226.0 // Area(mm^2)\n", +"d = 19.53/10 // Overall diameter(cm)\n", +"w_2 = 0.844 // Weight of conductor(kg/m)\n", +"u = 7950.0 // Ultimate strength(kg)\n", +"alpha = 18.44*10**-6 // Co-efficient of linear expression(/°C)\n", +"E = 9.32*10**3 // Modulus of elasticity(kg/mm^2)\n", +"t = 0.95 // Ice thickness(cm)\n", +"wind = 39.0 // Wind pressure(kg/m^2)\n", +"T_worst = -5.0 // Temperature in worst condition(°C)\n", +"\n", +"// Calculations\n", +"w_i = 915.0*%pi*t*(d+t)*10**-4 // Weight of ice on conductor(kg/m)\n", +"w_w = wind*(d+2*t)*10**-2 // Wind load of conductor(kg/m)\n", +"w_1 = ((w_2+w_i)**2+w_w**2)**0.5 // Total force on conductor(kg/m)\n", +"t = T_still-T_worst // Temperature(°C)\n", +"l = L/2.0 // Half span(m)\n", +"T = u/2.0 // Allowable tension(kg)\n", +"A = 1.0 // Co-efficient of x^3\n", +"B = a*E*(alpha*t+((w_1*l/T)**2/6))-T // Co-efficient of x^2\n", +"C = 0 // Co-efficient of x\n", +"D = -(w_2**2*l**2*a*E/6) // Co-efficient of constant\n", +"T_2_sol = roots([A,B,C,D]) // Roots of tension of a line\n", +"T_2_s = T_2_sol(3) // Feasible solution of tension of \n", +"T_2 = 1710.0 // Tension in conductor(kg). Obtianed directly from textbook\n", +"sag = w_2*l**2/(2*T_2) // Sag at erection(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 5.9 : SOLUTION :-')\n", +"printf('\nSag at erection = %.2f metres', sag)\n", +"printf('\nTension of the line, T_2 = %.f kg (An app. solution as per calculation) = %.f kg (More correctly as standard value)', T_2_s,T_2)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/13-INTERFERENCE_OF_POWER_LINES_WITH_NEIGHBOURING_COMMUNICATION_CIRCUITS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/13-INTERFERENCE_OF_POWER_LINES_WITH_NEIGHBOURING_COMMUNICATION_CIRCUITS.ipynb new file mode 100644 index 0000000..620f104 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/13-INTERFERENCE_OF_POWER_LINES_WITH_NEIGHBOURING_COMMUNICATION_CIRCUITS.ipynb @@ -0,0 +1,146 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: INTERFERENCE OF POWER LINES WITH NEIGHBOURING COMMUNICATION CIRCUITS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1: Mutual_inductance_between_the_circuits_and_Voltage_induced_in_the_telephone_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 6: INTERFERENCE OF POWER LINES WITH NEIGHBOURING COMMUNICATION CIRCUITS\n", +"\n", +"// EXAMPLE : 6.1 :\n", +"// Page number 206\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"d = 4.0 // Spacing b/w conductors(m)\n", +"D = 2.0 // Distance of telephone line below conductor(m)\n", +"s = 60.0/100 // Spacing b/w telephone line(m)\n", +"r = 2.0 // Radius of power line(mm)\n", +"I = 150.0 // Current in power line(A)\n", +"\n", +"// Calculations\n", +"D_ac = (D**2+((d-s)/2)**2)**0.5 // Distance b/w a & c(m)\n", +"D_ad = (D**2+(((d-s)/2)+s)**2)**0.5 // Distance b/w a & d(m)\n", +"M = 4.0*10**-7*log(D_ad/D_ac)*1000 // Mutual inductance b/w circuits(H/km)\n", +"V_CD = 2.0*%pi*f*M*I // Voltage induced in the telephone line(V/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 6.1 : SOLUTION :-')\n", +"printf('\nMutual inductance between the circuits, M = %.e H/km', M)\n", +"printf('\nVoltage induced in the telephone line, V_CD = %.2f V/km', V_CD)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2: Induced_voltage_at_fundamental_frequency_and_Potential_of_telephone_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 6: INTERFERENCE OF POWER LINES WITH NEIGHBOURING COMMUNICATION CIRCUITS\n", +"\n", +"// EXAMPLE : 6.2 :\n", +"// Page number 206-207\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 160.0 // Length of line(km)\n", +"V = 132.0*10**3 // Line voltage(V)\n", +"P = 25.0*10**6 // Load delivered(W)\n", +"PF = 0.8 // Lagging power factor\n", +"r = 5.0/1000 // Radius of power line conductor(m)\n", +"d = 4.0 // Spacing b/w conductors(m)\n", +"OS = 6.0 // Distance(m)\n", +"OT = 6.5 // Distance(m)\n", +"CT = 18.0 // Distance(m)\n", +"\n", +"// Calculations\n", +"AO = 3**0.5*d/2.0 // Distance A to O(m). From figure E6.2\n", +"AS = OS+AO // Distance A to S(m)\n", +"AT = AO+OT // Distance A to T(m)\n", +"OB = d/2.0 // Distance O to B(m)\n", +"BS = (OB**2+OS**2)**0.5 // Distance B to S(m)\n", +"BT = (OB**2+OT**2)**0.5 // Distance B to T(m)\n", +"M_A = 0.2*log(AT/AS) // Mutual inductance at A(mH/km)\n", +"M_B = 0.2*log(BT/BS) // Mutual inductance at B(mH/km)\n", +"M = M_B-M_A // Mutual inductance at C(mH/km)\n", +"I = P/(3**0.5*V*PF) // Current(A)\n", +"E_m = 2.0*%pi*f*M*I*10**-3*l // Induced voltage(V)\n", +"V_A = V/3**0.5 // Phase voltage(V)\n", +"h = AO+CT // Height(m)\n", +"V_SA = V_A*log10(((2*h)-AS)/AS)/log10(((2*h)-r)/r) // Potential(V)\n", +"H = CT // Height(m)\n", +"V_B = V_A // Phase voltage(V)\n", +"V_SB = V_B*log10(((2*H)-BS)/BS)/log10(((2*H)-r)/r) // Potential(V)\n", +"V_S = V_SB-V_SA // Total potential of S w.r.t earth(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 6.2 : SOLUTION :-')\n", +"printf('\nInduced voltage at fundamental frequency, E_m = %.1f V', E_m)\n", +"printf('\nPotential of telephone conductor S above earth, V_S = %.f V \n', V_S)\n", +"printf('\nNOTE: ERROR: Changes in obtained answer is due to precision and calculation mistakes in textbook')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/14-UNDERGROUND_CABLES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/14-UNDERGROUND_CABLES.ipynb new file mode 100644 index 0000000..68da1da --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/14-UNDERGROUND_CABLES.ipynb @@ -0,0 +1,740 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: UNDERGROUND CABLES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.10: Maximum_stress_in_the_two_dielectrics.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.10 :\n", +"// Page number 215-216\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"e_1 = 3.6 // Inner relative permittivity\n", +"e_2 = 2.5 // Outer relative permittivity\n", +"d = 1.0 // Conductor diameter(cm)\n", +"d_1 = 3.0 // Sheath diameter(cm)\n", +"D = 5.0 // Overall diameter(cm)\n", +"V_l = 66.0 // Line Voltage(kV)\n", +"\n", +"// Calculations\n", +"V = V_l/3**0.5*2**0.5 // Peak voltage on core(kV)\n", +"g1_max = 2*V/(d*(log(d_1/d)+e_1/e_2*log(D/d_1))) // Maximum stress in first dielectric(kV/km)\n", +"g_max = 2*V/(d_1*(e_2/e_1*log(d_1/d)+log(D/d_1))) // Maximum stress in second dielectric(kV/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.10 : SOLUTION :-')\n", +"printf('\nMaximum stress in first dielectric, g_1_max = %.2f kV/cm', g1_max)\n", +"printf('\nMaximum stress in second dielectric, g_max = %.2f kV/cm', g_max)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.11: EX14_11.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.11 :\n", +"// Page number 216-217\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 85.0 // Line Voltage(kV)\n", +"g_max = 55.0 // Maximum stress(kV/cm)\n", +"\n", +"// Calculations\n", +"V_1 = 0.632*V // Intersheath potential(kV)\n", +"d = 0.736*V/g_max // Core diameter(cm)\n", +"d_1 = 2*V/g_max // Intersheath diameter(cm)\n", +"D = 3.76*V/g_max // Overall diameter(cm)\n", +"d_un = 2*V/g_max // Core diameter of ungraded cable(cm)\n", +"D_un = 2.718*d_1 // Overall diameter of ungraded cable(cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.11 : SOLUTION :-')\n", +"printf('\nDiameter of intersheath, d_1 = %.2f cm', d_1)\n", +"printf('\nVoltage of intersheath, V_1 = %.2f kV, to neutral', V_1)\n", +"printf('\nConductor diameter of graded cable, d = %.2f cm', d)\n", +"printf('\nOutside diameter of graded cable, D = %.2f cm', D)\n", +"printf('\nConductor diameter of ungraded cable, d = %.2f cm', d_un)\n", +"printf('\nOutside diameter of ungraded cable, D = %.2f cm', D_un)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.12: Equivalent_star_connected_capacity_and_kVA_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.12 :\n", +"// Page number 219\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"c = 0.3 // Capacitance b/w any 2 conductor & sheath earthed(µF/km)\n", +"l = 10.0 // Length(km)\n", +"V = 33.0 // Line Voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"C_eq = l*c // Capacitance b/w any 2 conductor & sheath earthed(µF)\n", +"C_p = 2.0*C_eq // Capacitance per phase(µF)\n", +"kVA = V**2*2*%pi*f*C_p/1000.0 // Three-phase kVA required(kVA)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.12 : SOLUTION :-')\n", +"printf('\nEquivalent star connected capacity, C_eq = %.f µF', C_eq)\n", +"printf('\nkVA required = %.1f kVA', kVA)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.13: Charging_current_drawn_by_a_cable_with_three_cores.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.13 :\n", +"// Page number 219\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 11.0*10**3 // Line Voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"C_c = 3.7 // Measured capacitance(µF)\n", +"\n", +"// Calculations\n", +"C_0 = 2*C_c // Capacitance(µF)\n", +"I_ch = 2*%pi*f*C_0*V/3**0.5*10**-6 // Charging current per phase(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.13 : SOLUTION :-')\n", +"printf('\nCharging current drawn by a cable = %.2f A', I_ch)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.14: EX14_14.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.14 :\n", +"// Page number 219-220\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"c_s = 0.90 // Capacitance b/w all conductors(µF)\n", +"C_0 = 0.4 // Capacitance b/w two conductor(µF)\n", +"V = 11.0*10**3 // Line Voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"C_s = c_s/3.0 // Capacitance measured(µF)\n", +"C_c = (C_0-C_s)/2.0 // Capacitance(µF)\n", +"C_a = 3.0/2*(C_c+(1/3.0)*C_s) // Capacitance b/w any two conductors(µF)\n", +"C_b = 2.0*C_c+(2.0/3)*C_s // Capacitance b/w any two bounded conductors and the third conductor(µF)\n", +"C_o = 3.0*C_c+C_s // Capacitance to neutral(µF)\n", +"I_c = 2.0*%pi*f*C_o*V/3**0.5*10**-6 // Charging current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.14 : SOLUTION :-')\n", +"printf('\nCase(a): Capacitance between any two conductors = %.3f µF', C_a)\n", +"printf('\nCase(b): Capacitance between any two bounded conductors and the third conductor = %.1f µF', C_b)\n", +"printf('\nCase(c): Capacitance to neutral, C_0 = %.2f µF', C_o)\n", +"printf('\n Charging current taken by cable, I_c = %.3f A \n', I_c)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in textbook answer')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.15: Charging_current_drawn_by_cable.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.15 :\n", +"// Page number 220-221\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 13.2*10**3 // Line Voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"C_BC = 4.2 // Capacitance b/w two cores(µF)\n", +"\n", +"// Calculations\n", +"C_n = 2.0*C_BC // Capacitance to neutral(µF)\n", +"V_ph = V/3**0.5 // Operating phase voltage(V)\n", +"I_c = 2.0*%pi*f*C_n*V/3**0.5*10**-6 // Charging current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.15 : SOLUTION :-')\n", +"printf('\nCharging current drawn by cable, I_c = %.2f A', I_c)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.16: EX14_16.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.16 :\n", +"// Page number 222-223\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 33.0*10**3 // Line Voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 4.0 // Length(km)\n", +"d = 2.5 // Diameter of conductor(cm)\n", +"t = 0.5 // Radial thickness of insulation(cm)\n", +"e_r = 3.0 // Relative permittivity of the dielectric\n", +"PF = 0.02 // Power factor of unloaded cable\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"R = r+t // External radius(cm)\n", +"e_0 = 8.85*10**-12 // Permittivity\n", +"C = 2.0*%pi*e_0*e_r/log(R/r)*l*1000 // Capacitance of cable/phase(F)\n", +"// Case(b)\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"I_c = V_ph*2.0*%pi*f*C // Charging current/phase(A)\n", +"// Case(c)\n", +"kVAR = 3.0*V_ph*I_c // Total charging kVAR\n", +"// Case(d)\n", +"phi = acosd(PF) // Φ(°)\n", +"delta = 90.0-phi // δ(°)\n", +"P_c = V_ph*I_c*sind(delta)/1000 // Dielectric loss/phase(kW)\n", +"// Case(e)\n", +"E_max = V_ph/(r*log(R/r)*1000) // RMS value of Maximum stress in cable(kV/cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.16 : SOLUTION :-')\n", +"printf('\nCase(a): Capacitance of the cable, C = %.3e F/phase', C)\n", +"printf('\nCase(b): Charging current = %.2f A/phase', I_c)\n", +"printf('\nCase(c): Total charging kVAR = %.4e kVAR', kVAR)\n", +"printf('\nCase(d): Dielectric loss/phase, P_c = %.2f kW', P_c)\n", +"printf('\nCase(e): Maximum stress in the cable, E_max = %.1f kV/cm (rms)', E_max)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1: Insulation_resistance_per_km.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.1 :\n", +"// Page number 211\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.5 // Core diameter(cm)\n", +"t = 1.25 // Insulation thickness(cm)\n", +"rho = 4.5*10**14 // Resistivity of insulation(ohm-cm)\n", +"l = 10.0**5 // Length(cm)\n", +"\n", +"// Calculations\n", +"D = d+2*t // Overall diameter(cm)\n", +"R_i = rho/(2*%pi*l)*log(D/d) // Insulation resistance(ohm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.1 : SOLUTION :-')\n", +"printf('\nInsulation resistance per km, R_i = %.2e ohm\n', R_i)\n", +"printf('\nNOTE: ERROR: Mistake in final answer in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2: Insulation_thickness.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.2 :\n", +"// Page number 211\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R = 495.0*10**6 // Insulation resistance(ohm/km)\n", +"d = 3.0 // Core diameter(cm)\n", +"rho = 4.5*10**14 // Resistivity of insulation(ohm-cm)\n", +"\n", +"// Calculations\n", +"l = 1000.0 // Length of cable(m)\n", +"r_2 = d/2.0 // Core radius(cm)\n", +"Rho = rho/100.0 // Resistivity of insulation(ohm-m)\n", +"r1_r2 = exp((2*%pi*l*R)/Rho) // r1/r2\n", +"r_1 = 2*r_2 // Cable radius(cm)\n", +"thick = r_1-r_2 // Insulation thickness(cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.2 : SOLUTION :-')\n", +"printf('\nInsulation thickness = %.1f cm', thick)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3: Capacitance_and_Charging_current_of_single_core_cable.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.3 :\n", +"// Page number 212\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 66.0*10**3 // Line Voltage(V)\n", +"l = 1.0 // Length of cable(km)\n", +"d = 15.0 // Core diameter(cm)\n", +"D = 60.0 // Sheath diameter(cm)\n", +"e_r = 3.6 // Relative permittivity\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"C = e_r/(18.0*log(D/d))*l // Capacitance(µF)\n", +"I_ch = V/3**0.5*2*%pi*f*C*10**-6 // Charging current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.3 : SOLUTION :-')\n", +"printf('\nCapacitance of single-core cable, C = %.3f µF', C)\n", +"printf('\nCharging current of single-core cable = %.2f A', I_ch)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4: EX14_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.4 :\n", +"// Page number 212\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_l = 132.0 // Line Voltage(kV)\n", +"g_max = 60.0 // Maximum Line Voltage(kV)\n", +"\n", +"// Calculations\n", +"V = V_l/3**0.5*2**0.5 // Phase Voltage(kV)\n", +"d = 2*V/g_max // Core diameter(cm)\n", +"D = 2.718*d // Overall diameter(cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.4 : SOLUTION :-')\n", +"printf('\nMost economical diameter of a single-core cable, d = %.1f cm', d)\n", +"printf('\nOverall diameter of the insulation, D = %.3f cm\n', D)\n", +"printf('\nNOTE: Slight change in obtained answer due to precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6: Conductor_radius_and_Electric_field_strength_that_must_be_withstood.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.6 :\n", +"// Page number 212-213\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 11.0*10**3 // Line Voltage(V)\n", +"dia_out = 8.0 // Outside diameter(cm)\n", +"\n", +"// Calculations\n", +"D = dia_out/2.0 // Overall diameter(cm)\n", +"d = (D)/2.718 // Conductor diameter(cm)\n", +"r = d/2 // Conductor radius(cm)\n", +"g_m = 2*V/(d*log(D/d)*10) // Maximum value of electric field strength(kV/m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.6 : SOLUTION :-')\n", +"printf('\nConductor radius, r = %.3f cm', r)\n", +"printf('\nElectric field strength that must be withstood, g_m = %.f kV/m', g_m)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7: EX14_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.7 :\n", +"// Page number 214\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R_3 = 1.00 // Cable radius(cm)\n", +"R_1 = 2.5 // Cable radius(cm)\n", +"\n", +"// Calculations\n", +"R_2 = (R_1*R_3)**0.5 // Location of intersheath(cm)\n", +"alpha = R_1/R_2 // α\n", +"ratio = 2.0/(1+alpha) // Ratio of maximum electric field strength with & without intersheath\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.7 : SOLUTION :-')\n", +"printf('\nLocation of intersheath, R_2 = %.2f cm', R_2)\n", +"printf('\nRatio of maximum electric field strength with & without intersheath = %.3f ', ratio)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8: Maximum_and_Minimum_stress_in_the_insulation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.8 :\n", +"// Page number 215\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 33.0 // Line Voltage(kV)\n", +"D_2 = 2.0 // Conductor diameter(cm)\n", +"D_1 = 3.0 // Sheath diameter(cm)\n", +"\n", +"// Calculations\n", +"R_2 = D_2/2 // Conductor radius(cm)\n", +"R_1 = D_1/2 // Sheath radius(cm)\n", +"g_max = V/(R_2*log(R_1/R_2)) // RMS value of maximum stress in the insulation(kV/cm)\n", +"g_min = V/(R_1*log(R_1/R_2)) // RMS value of minimum stress in the insulation(kV/cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.8 : SOLUTION :-')\n", +"printf('\nMaximum stress in the insulation, g_max = %.2f kV/cm (rms)', g_max)\n", +"printf('\nMinimum stress in the insulation, g_min = %.2f kV/cm (rms)', g_min)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.9: EX14_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 7: UNDERGROUND CABLES\n", +"\n", +"// EXAMPLE : 7.9 :\n", +"// Page number 215\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.5 // Conductor diameter(cm)\n", +"D = 6.0 // Sheath diameter(cm)\n", +"V_l = 66.0 // Line Voltage(kV)\n", +"\n", +"// Calculations\n", +"alpha = (D/d)**(1.0/3) // α\n", +"d_1 = d*alpha // Best position of first intersheath(cm)\n", +"d_2 = d_1*alpha // Best position of second intersheath(cm)\n", +"V = V_l/3**0.5*2**0.5 // Peak voltage on core(kV)\n", +"V_2 = V/(1+(1/alpha)+(1/alpha**2)) // Peak voltage on second intersheath(kV)\n", +"V_1 = (1+(1/alpha))*V_2 // Voltage on first intersheath(kV)\n", +"stress_max = 2*V/(d*log(D/d)) // Maximum stress without intersheath(kV/cm)\n", +"stress_min = stress_max*d/D // Minimum stress without intersheath(kV/cm)\n", +"g_max = V*3/(1+alpha+alpha**2) // Maximum stress with intersheath(kV/cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 7.9 : SOLUTION :-')\n", +"printf('\nMaximum stress without intersheath = %.2f kV/cm', stress_max)\n", +"printf('\nBest position of first intersheath, d_1 = %.2f cm', d_1)\n", +"printf('\nBest position of second intersheath, d_2 = %.3f cm', d_2)\n", +"printf('\nMaximum stress with intersheath = %.2f kV/cm', g_max)\n", +"printf('\nVoltage on the first intersheath, V_1 = %.2f kV', V_1)\n", +"printf('\nVoltage on the second intersheath, V_2 = %.2f kV \n', V_2)\n", +"printf('\nNOTE: Changes in the obtained answer is due to more precision here') " + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/15-CORONA.ipynb b/Power_System_Engineering_by_S_Chakraborthy/15-CORONA.ipynb new file mode 100644 index 0000000..9a3fc8f --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/15-CORONA.ipynb @@ -0,0 +1,445 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15: CORONA" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1: Minimum_spacing_between_conductors.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.1 :\n", +"// Page number 227\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 30.0/10 // Diameter of conductor(cm)\n", +"delta = 0.95 // Air density factor\n", +"m = 0.95 // Irregularity factor\n", +"E = 230.0 // Line voltage(kV)\n", +"g_0 = 30.0/2**0.5 // Breakdown strength of air(kV/cm)\n", +"\n", +"// Calculations\n", +"E_0 = E/3**0.5 // Disruptive critical voltage(kV)\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"D = exp(E_0/(m*delta*g_0*r))*r/100 // Minimum spacing between conductors(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.1 : SOLUTION :-')\n", +"printf('\nMinimum spacing between conductors, D = %.3f m \n', abs(D))\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook due to precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2: Critical_disruptive_voltage_and_Corona_loss.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.2 :\n", +"// Page number 227-228\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 220.0 // Operating line voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"d = 1.5 // Diameter of conductor(cm)\n", +"D = 300.0 // Distance b/w conductor(cm)\n", +"delta = 1.05 // Air density factor\n", +"g_0 = 21.1 // Breakdown strength of air(kV/cm)\n", +"m = 1.0 // Irregularity factor\n", +"\n", +"// Calculations\n", +"E = V/3**0.5 // Phase voltage(kV)\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"E_0 = m*g_0*delta*r*log(D/r) // Disruptive critical voltage to neutral(kV/phase)\n", +"E_0_ll = 3**0.5*E_0 // Line-to-line Disruptive critical voltage(kV)\n", +"P = 244.0*10**-5*(f+25)/delta*(r/D)**0.5*(E-E_0)**2 // Corona loss(kW/km/phase)\n", +"P_total = P*3.0 // Corona loss(kW/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.2 : SOLUTION :-')\n", +"printf('\nCritical disruptive voltage, E_0 = %.2f kV/phase = %.2f kV (line-to-line)', E_0,E_0_ll)\n", +"printf('\nCorona loss, P = %.2f kW/km \n', P_total)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the final answer in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.3: Corona_loss_in_fair_weather_and_Foul_weather.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.3 :\n", +"// Page number 228\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 132.0 // Operating line voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"d = 1.17 // Diameter of conductor(cm)\n", +"D = 300.0 // Distance b/w conductor(cm)\n", +"m = 0.96 // Irregularity factor\n", +"b = 72.0 // Barometric pressure(cm)\n", +"t = 20.0 // Temperature(°C)\n", +"\n", +"// Calculations\n", +"delta = 3.92*b/(273.0+t) // Air density factor\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"E_0 = 21.1*m*delta*r*log(D/r) // Critical disruptive voltage for fair weather condition(kV/phase)\n", +"E_0_foul = 0.8*E_0 // Critical disruptive voltage for foul weather(kV/phase)\n", +"E = V/3**0.5 // Phase voltage(kV)\n", +"P_fair = 244.0*10**-5*(f+25)/delta*(r/D)**0.5*(E-E_0)**2 // Corona loss for fair weather condition(kW/km/phase)\n", +"P_foul = 244.0*10**-5*(f+25)/delta*(r/D)**0.5*(E-E_0_foul)**2 // Corona loss for foul weather condition(kW/km/phase)\n", +" \n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.3 : SOLUTION :-')\n", +"printf('\nCorona loss in fair weather, P = %.3f kW/km/phase', P_fair)\n", +"printf('\nCorona loss in foul weather, P = %.3f kW/km/phase', P_foul)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.4: Corona_characteristics.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.4 :\n", +"// Page number 228-229\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 110.0 // Operating line voltage(kV)\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 175.0 // Line length(km)\n", +"d = 1.0 // Diameter of conductor(cm)\n", +"D = 300.0 // Distance b/w conductor(cm)\n", +"t = 26.0 // Temperature(°C)\n", +"b = 74.0 // Barometric pressure(cm)\n", +"m = 0.85 // Irregularity factor\n", +"m_v_local = 0.72 // Roughness factor for local corona\n", +"m_v_gen = 0.82 // Roughness factor for general corona\n", +"\n", +"// Calculations\n", +"delta = 3.92*b/(273.0+t) // Air density factor\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"E_0 = 21.1*m*delta*r*log(D/r) // Critical disruptive voltage(kV) rms\n", +"E_v_local = 21.1*m_v_local*delta*r*(1+(0.3/(delta*r)**0.5))*log(D/r) // Critical disruptive voltage for local corona(kV) rms\n", +"E_v_gen = 21.1*m_v_gen*delta*r*(1+(0.3/(delta*r)**0.5))*log(D/r) // Critical disruptive voltage for general corona(kV) rms\n", +"E = V/3**0.5 // Phase voltage(kV)\n", +"// Case(i)\n", +"P_c_i = 244.0*10**-5*(f+25)/delta*(r/D)**0.5*(E-E_0)**2 // Peek's formula for fair weather condition(kW/km/phase)\n", +"P_c_i_total = P_c_i*l*3 // Total power loss(kW)\n", +"// Case(ii)\n", +"P_c_ii = 244.0*10**-5*(f+25)/delta*(r/D)**0.5*(E-0.8*E_0)**2 // Peek's formula for stormy condition(kW/km/phase)\n", +"P_c_ii_total = P_c_ii*l*3 // Total power loss(kW)\n", +"// Case(iii)\n", +"F_iii = 0.0713 // From text depending on E/E_0\n", +"P_c_iii = 21.0*10**-6*f*E**2*F_iii/(log10(D/r))**2 // Peterson's formula for fair condition(kW/km/phase)\n", +"P_c_iii_total = P_c_iii*l*3 // Total power loss(kW)\n", +"// Case(iv)\n", +"F_iv = 0.3945 // From text depending on E/E_0\n", +"P_c_iv = 21.0*10**-6*f*E**2*F_iv/(log10(D/r))**2 // Peterson's formula for stormy condition(kW/km/phase)\n", +"P_c_iv_total = P_c_iv*l*3 // Total power loss(kW)\n", +" \n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.4 : SOLUTION :-')\n", +"printf('\nCase(i) : Power loss due to corona using Peek formula for fair weather condition, P_c = %.3f kW/km/phase', P_c_i)\n", +"printf('\n Total corona loss in fair weather condition using Peek formula = %.1f kW', P_c_i_total)\n", +"printf('\nCase(ii) : Power loss due to corona using Peek formula for stormy weather condition, P_c = %.2f kW/km/phase', P_c_ii)\n", +"printf('\n Total corona loss in stormy condition using Peek formula = %.f kW', P_c_ii_total)\n", +"printf('\nCase(iii): Power loss due to corona using Peterson formula for fair weather condition, P_c = %.4f kW/km/phase', P_c_iii)\n", +"printf('\n Total corona loss in fair condition using Peterson formula = %.2f kW',P_c_iii_total)\n", +"printf('\nCase(iii): Power loss due to corona using Peterson formula for fair weather condition, P_c = %.4f kW/km/phase', P_c_iv)\n", +"printf('\n Total corona loss in stormy condition using Peterson formula = %.1f kW \n',P_c_iv_total)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the final answer in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.5: Spacing_between_the_conductors.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.5 :\n", +"// Page number 229\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 132.0 // Operating line voltage(kV)\n", +"dia = 1.956 // Diameter of conductor(cm)\n", +"v_c = 210.0 // Disrputive voltage(kV) \n", +"g_0 = 30.0/2**0.5 // Breakdown strength of air(kV/cm)\n", +"\n", +"// Calculations\n", +"r = dia/2.0 // Radius of conductor(cm)\n", +"V_c = v_c/3**0.5 // Disrputive voltage/phase(kV)\n", +"m_0 = 1.0 // Irregularity factor\n", +"delta = 1.0 // Air density factor\n", +"d = exp(V_c/(m_0*delta*g_0*r))*r // Spacing between conductors(cm)\n", +" \n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.5 : SOLUTION :-')\n", +"printf('\nSpacing between the conductors, d = %.f cm \n', abs(d))\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.6: Disruptive_critical_voltage_and_Corona_loss.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.6 :\n", +"// Page number 229\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P_c1 = 53.0 // Total corona loss(kW)\n", +"V_1 = 106.0 // Operating line voltage(kV)\n", +"P_c2 = 98.0 // Total corona loss(kW)\n", +"V_2 = 110.9 // Operating line voltage(kV)\n", +"V_3 = 113.0 // Operating line voltage(kV)\n", +"\n", +"// Calculations\n", +"E_1 = V_1/3**0.5 // Phase voltage(kV)\n", +"E_2 = V_2/3**0.5 // Phase voltage(kV)\n", +"P_ratio = (P_c2/P_c1)**0.5\n", +"E_0 = (P_ratio*E_1-E_2)/(P_ratio-1) // Disruptive critical voltage(kV)\n", +"E_3 = V_3/3**0.5 // Phase voltage(kV)\n", +"W = ((E_3-E_0)/(E_1-E_0))**2*P_c1 // Corona loss at 113 kV(kW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.6 : SOLUTION :-')\n", +"printf('\nDisruptive critical voltage, E_0 = %.f kV', E_0)\n", +"printf('\nCorona loss at 113 kV, W = %.f kW\n', W)\n", +"printf('\nNOTE: Changes in obtained answer from textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.7: Corona_will_be_present_in_the_air_space_or_not.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.7 :\n", +"// Page number 229-230\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 3.0 // Diameter of conductor(cm)\n", +"e_r = 4.0 // Relative permittivity\n", +"d_1 = 3.5 // Internal diameter of porcelain bushing(cm)\n", +"d_2 = 9.0 // External diameter of porcelain bushing(cm)\n", +"V = 25.0 // Voltage b/w conductor and clamp(kV)\n", +"\n", +"// Calculations\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"r_1 = d_1/2.0 // Internal radius of porcelain bushing(cm)\n", +"r_2 = d_2/2.0 // External radius of porcelain bushing(cm)\n", +"g_2max = r/(e_r*r_1) // Maximum gradient of inner side of porcelain\n", +"g_1max = V/(r*log(r_1/r)+g_2max*r_1*log(r_2/r_1)) // Maximum gradient on surface of conductor(kV/cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.7 : SOLUTION :-')\n", +"printf('\nMaximum gradient on surface of conductor, g_1max = %.2f kV/cm', g_1max)\n", +"printf('\nSince, gradient exceeds 21.1 kV/cm, corona will be present')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.8: Line_voltage_for_commencing_of_corona.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 8: CORONA\n", +"\n", +"// EXAMPLE : 8.8 :\n", +"// Page number 230\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.0 // Diameter of conductor(cm)\n", +"D = 150.0 // Spacing b/w conductor(cm)\n", +"delta = 1.0 // Air density factor\n", +"\n", +"// Calculations\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"V_d = 21.1*delta*r*log(D/r) // Disruptive critical voltage(kV/phase)\n", +"V_d_ll = 3**0.5*V_d // Line voltage for commencing of corona(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 8.8 : SOLUTION :-')\n", +"printf('\nLine voltage for commencing of corona = %.2f kV \n', V_d_ll)\n", +"printf('\nNOTE: Solution is incomplete in textbook')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/16-LOAD_FLOW_STUDY_USING_COMPUTER_TECHNIQUES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/16-LOAD_FLOW_STUDY_USING_COMPUTER_TECHNIQUES.ipynb new file mode 100644 index 0000000..537142e --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/16-LOAD_FLOW_STUDY_USING_COMPUTER_TECHNIQUES.ipynb @@ -0,0 +1,488 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16: LOAD FLOW STUDY USING COMPUTER TECHNIQUES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.1: Bus_admittance_matrix_Ybus.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.1 :\n", +"// Page number 235-236\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Z_L1 = complex(14.3,97) // Series impedance of line L1(ohm)\n", +"Z_PL1 = complex(0,-3274) // Shunt impedance of line L1(ohm)\n", +"Z_L2 = complex(7.13,48.6) // Series impedance of line L2(ohm)\n", +"Z_PL2 = complex(0,-6547) // Shunt impedance of line L2(ohm)\n", +"Z_L3 = complex(9.38,64) // Series impedance of line L3(ohm)\n", +"Z_PL3 = complex(0,-4976) // Shunt impedance of line L3(ohm)\n", +"\n", +"// Calculations\n", +"Y_S12 = 1.0/Z_L1 // Series admittance(mho)\n", +"Y_P12 = 1.0/Z_PL1 // Shunt admittance(mho)\n", +"Y_S23 = 1.0/Z_L3 // Series admittance(mho)\n", +"Y_P23 = 1.0/Z_PL3 // Shunt admittance(mho)\n", +"Y_S13 = 1.0/Z_L2 // Series admittance(mho)\n", +"Y_P13 = 1.0/Z_PL2 // Shunt admittance(mho)\n", +"Y_11 = Y_P12+Y_P13+Y_S12+Y_S13 // Admittance(mho)\n", +"Y_12 = -Y_S12 // Admittance(mho)\n", +"Y_13 = -Y_S13 // Admittance(mho)\n", +"Y_21 = Y_12 // Admittance(mho)\n", +"Y_22 = Y_P12+Y_P23+Y_S12+Y_S23 // Admittance(mho)\n", +"Y_23 = -Y_S23 // Admittance(mho)\n", +"Y_31 = Y_13 // Admittance(mho)\n", +"Y_32 = Y_23 // Admittance(mho)\n", +"Y_33 = Y_P13+Y_P23+Y_S23+Y_S13 // Admittance(mho)\n", +"Y_bus = [[Y_11, Y_12, Y_13],\n", +" [Y_21, Y_22, Y_23],\n", +" [Y_31, Y_32, Y_33]]\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.1 : SOLUTION :-')\n", +"printf('\n[Y_bus] = \n'); disp(Y_bus)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.3: Voltage_values_at_different_buses.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.3 :\n", +"// Page number 236-237\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_a = 1.0 // Voltage(p.u)\n", +"V_b = 1.0*exp(%i*-36.87*%pi/180) // Voltage(p.u)\n", +"V_c = 1.0 // Voltage(p.u)\n", +"Z_1 = complex(0,1) // Reactance(p.u)\n", +"Z_2 = complex(0,1) // Reactance(p.u)\n", +"Z_3 = complex(0,1) // Reactance(p.u)\n", +"Z_13 = complex(0,0.4) // Reactance(p.u)\n", +"Z_23 = complex(0,0.4) // Reactance(p.u)\n", +"Z_14 = complex(0,0.2) // Reactance(p.u)\n", +"Z_24 = complex(0,0.2) // Reactance(p.u)\n", +"Z_34 = complex(0,0.2) // Reactance(p.u)\n", +"Z_12 = complex(0,0) // Reactance(p.u)\n", +"\n", +"// Calculations\n", +"I_1 = V_a/Z_1 // Current injection vector(p.u)\n", +"I_2 = V_b/Z_2 // Current injection vector(p.u)\n", +"I_3 = V_c/Z_3 // Current injection vector(p.u)\n", +"I_4 = 0.0 // Current injection vector(p.u)\n", +"y1 = 1.0/Z_1 // Admittance(p.u)\n", +"y2 = 1.0/Z_2 // Admittance(p.u)\n", +"y3 = 1.0/Z_3 // Admittance(p.u)\n", +"y13 = 1.0/Z_13 // Admittance(p.u)\n", +"y23 = 1.0/Z_23 // Admittance(p.u)\n", +"y14 = 1.0/Z_14 // Admittance(p.u)\n", +"y24 = 1.0/Z_24 // Admittance(p.u)\n", +"y34 = 1.0/Z_34 // Admittance(p.u)\n", +"y12 = 0.0 // Admittance(p.u)\n", +"Y_11 = y1+y13+y14 // Equivalent admittance(p.u)\n", +"Y_12 = y12 // Equivalent admittance(p.u)\n", +"Y_13 = -y13 // Equivalent admittance(p.u)\n", +"Y_14 = -y14 // Equivalent admittance(p.u)\n", +"Y_21 = Y_12 // Equivalent admittance(p.u)\n", +"Y_22 = y2+y23+y24 // Equivalent admittance(p.u)\n", +"Y_23 = -y23 // Equivalent admittance(p.u)\n", +"Y_24 = -y24 // Equivalent admittance(p.u)\n", +"Y_31 = Y_13 // Equivalent admittance(p.u)\n", +"Y_32 = Y_23 // Equivalent admittance(p.u)\n", +"Y_33 = y3+y13+y23+y34 // Equivalent admittance(p.u)\n", +"Y_34 = -y34 // Equivalent admittance(p.u)\n", +"Y_41 = Y_14 // Equivalent admittance(p.u)\n", +"Y_42 = Y_24 // Equivalent admittance(p.u)\n", +"Y_43 = Y_34 // Equivalent admittance(p.u)\n", +"Y_44 = y14+y24+y34 // Equivalent admittance(p.u)\n", +"Y_bus = [[Y_11, Y_12, Y_13, Y_14],\n", +" [Y_21, Y_22, Y_23, Y_24],\n", +" [Y_31, Y_32, Y_33, Y_34],\n", +" [Y_41, Y_42, Y_43, Y_44]] // Bus admittance matrix\n", +"I_bus = [I_1,\n", +" I_2,\n", +" I_3,\n", +" I_4]\n", +"V = inv(Y_bus)*I_bus // Bus voltage(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.3 : SOLUTION :-')\n", +"printf('\nVoltage at bus 1, V_1 = %.4f%.4fj p.u', real(V(1,1:1)),imag(V(1,1:1)))\n", +"printf('\nVoltage at bus 2, V_2 = %.4f%.4fj p.u', real(V(2,1:1)),imag(V(2,1:1)))\n", +"printf('\nVoltage at bus 3, V_3 = %.4f%.4fj p.u', real(V(3,1:1)),imag(V(3,1:1)))\n", +"printf('\nVoltage at bus 4, V_4 = %.4f%.4fj p.u\n', real(V(4,1:1)),imag(V(4,1:1)))\n", +"printf('\nNOTE: Node equation matrix could not be represented in a single equation. Hence, it is not displayed')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.4: New_bus_admittance_matrix_Ybus.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.4 :\n", +"// Page number 237-238\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_a = 1.0 // Voltage(p.u)\n", +"V_b = 1.0*exp(%i*-36.87*%pi/180) // Voltage(p.u)\n", +"V_c = 1.0 // Voltage(p.u)\n", +"Z_1 = complex(0,1) // Reactance(p.u)\n", +"Z_2 = complex(0,1) // Reactance(p.u)\n", +"Z_3 = complex(0,1) // Reactance(p.u)\n", +"Z_13 = complex(0,0.4) // Reactance(p.u)\n", +"Z_23 = complex(0,0.4) // Reactance(p.u)\n", +"Z_14 = complex(0,0.2) // Reactance(p.u)\n", +"Z_24 = complex(0,0.2) // Reactance(p.u)\n", +"Z_34 = complex(0,0.2) // Reactance(p.u)\n", +"Z_12 = complex(0,0) // Reactance(p.u)\n", +"\n", +"// Calculations\n", +"I_1 = V_a/Z_1 // Current injection vector(p.u)\n", +"I_2 = V_b/Z_2 // Current injection vector(p.u)\n", +"I_3 = V_c/Z_3 // Current injection vector(p.u)\n", +"I_4 = 0.0 // Current injection vector(p.u)\n", +"y1 = 1.0/Z_1 // Admittance(p.u)\n", +"y2 = 1.0/Z_2 // Admittance(p.u)\n", +"y3 = 1.0/Z_3 // Admittance(p.u)\n", +"y13 = 1.0/Z_13 // Admittance(p.u)\n", +"y23 = 1.0/Z_23 // Admittance(p.u)\n", +"y14 = 1.0/Z_14 // Admittance(p.u)\n", +"y24 = 1.0/Z_24 // Admittance(p.u)\n", +"y34 = 1.0/Z_34 // Admittance(p.u)\n", +"y12 = 0.0 // Admittance(p.u)\n", +"Y_11 = y1+y13+y14 // Equivalent admittance(p.u)\n", +"Y_12 = y12 // Equivalent admittance(p.u)\n", +"Y_13 = -y13 // Equivalent admittance(p.u)\n", +"Y_14 = -y14 // Equivalent admittance(p.u)\n", +"Y_21 = Y_12 // Equivalent admittance(p.u)\n", +"Y_22 = y2+y23+y24 // Equivalent admittance(p.u)\n", +"Y_23 = -y23 // Equivalent admittance(p.u)\n", +"Y_24 = -y24 // Equivalent admittance(p.u)\n", +"Y_31 = Y_13 // Equivalent admittance(p.u)\n", +"Y_32 = Y_23 // Equivalent admittance(p.u)\n", +"Y_33 = y3+y13+y23+y34 // Equivalent admittance(p.u)\n", +"Y_34 = -y34 // Equivalent admittance(p.u)\n", +"Y_41 = Y_14 // Equivalent admittance(p.u)\n", +"Y_42 = Y_24 // Equivalent admittance(p.u)\n", +"Y_43 = Y_34 // Equivalent admittance(p.u)\n", +"Y_44 = y14+y24+y34 // Equivalent admittance(p.u)\n", +"Y_bus = [[Y_11, Y_12, Y_13, Y_14],\n", +" [Y_21, Y_22, Y_23, Y_24],\n", +" [Y_31, Y_32, Y_33, Y_34],\n", +" [Y_41, Y_42, Y_43, Y_44]] // Bus admittance matrix\n", +"K = Y_bus([1,2],1:2)\n", +"L = Y_bus([1,2],3:4)\n", +"M = Y_bus([3,4],3:4)\n", +"N = Y_bus([3,4],1:2)\n", +"inv_M = inv([M(1,1:2);M(2,1:2)]) // Multiplication of marix [L][M^-1][N]\n", +"Y_bus_new = K-L*inv_M*N // New bus admittance matrix\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.4 : SOLUTION :-')\n", +"printf('\n[Y_bus]_new = \n'); disp(Y_bus_new)\n", +"printf('\nNOTE: ERROR: Mistake in representing the sign in final answer in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.5: Bus_admittance_matrix_V1_and_V2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.5 :\n", +"// Page number 238\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_1 = 2.0 // Voltage(p.u)\n", +"I_2 = 2.0*exp(%i*45.0*%pi/180) // Voltage(p.u)\n", +"y1 = complex(0,-1.0) // Admittance(p.u)\n", +"y2 = complex(0,-2.0) // Admittance(p.u)\n", +"y12 = complex(0,-2.0) // Admittance(p.u)\n", +"\n", +"// Calculations\n", +"E_1 = I_1*y1 // Voltage element(p.u)\n", +"E_2 = I_2*y2 // Voltage element(p.u)\n", +"Y_11 = y1+y12 // Self Admittance(p.u)\n", +"Y_12 = -y12 // Mutual Admittance(p.u)\n", +"Y_21 = Y_12 // Mutual Admittance(p.u)\n", +"Y_22 = y2+y12 // Self Admittance(p.u)\n", +"Y_bus = [[Y_11, Y_12],\n", +" [Y_21, Y_22]] // Bus admittance matrix\n", +"I_bus = [I_1,\n", +" I_2]\n", +"V = inv(Y_bus)*I_bus\n", +"V_1 = V(1,1:1) // Voltage(p.u)\n", +"V_2 = V(2,1:1) // Voltage(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.5 : SOLUTION :-')\n", +"printf('\n[Y_bus] = \n'); disp(Y_bus)\n", +"printf('\nV_1 = %.3f∠%.1f° p.u', abs(V_1),phasemag(V_1))\n", +"printf('\nV_2 = %.3f∠%.1f° p.u\n', abs(V_2),phasemag(V_2))\n", +"printf('\nNOTE: ERROR: Calculation mistake in V_1 in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.6: Bus_impedance_matrix_Zbus.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.6 :\n", +"// Page number 238\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Y_bus = [[-%i*10.5, 0, %i*5.0, %i*5.0],\n", +" [0, -%i*8.0, %i*2.5, %i*5.0],\n", +" [%i*5.0, %i*2.5, -%i*18.0, %i*10.0],\n", +" [%i*5.0, %i*5.0, %i*10.0, -%i*20.0]] // Bus admittance matrix\n", +"\n", +"// Calculations\n", +"Z_bus = inv(Y_bus) // Bus impedance matrix\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.6 : SOLUTION :-')\n", +"printf('\n[Z_bus] = \n'); disp(Z_bus)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.7: Power_flow_expressions.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.7 :\n", +"// Page number 239\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Y_C = complex(0,0.1) // Shunt admittance(mho)\n", +"Z_L = complex(0,0.2) // Series impedance(mho)\n", +"\n", +"// Calculations\n", +"Y_L = 1.0/Z_L // Series admittance(mho)\n", +"Y_11 = Y_C+Y_C+Y_L+Y_L // Admittance(mho)\n", +"Y_12 = -Y_L // Admittance(mho)\n", +"Y_13 = -Y_L // Admittance(mho)\n", +"Y_21 = Y_12 // Admittance(mho)\n", +"Y_22 = Y_L+Y_L+Y_C+Y_C // Admittance(mho)\n", +"Y_23 = -Y_L // Admittance(mho)\n", +"Y_31 = Y_13 // Admittance(mho)\n", +"Y_32 = Y_23 // Admittance(mho)\n", +"Y_33 = Y_L+Y_L+Y_C+Y_C // Admittance(mho)\n", +"Y_bus = [[Y_11, Y_12, Y_13],\n", +" [Y_21, Y_22, Y_23],\n", +" [Y_31, Y_32, Y_33]] // Bus admittance matrix\n", +"S_11 = conj(Y_bus(1,1:1))\n", +"S_12 = conj(Y_bus(1,2:2))\n", +"S_13 = conj(Y_bus(1,3:3))\n", +"S_21 = S_12\n", +"S_22 = conj(Y_bus(2,2:2))\n", +"S_23 = conj(Y_bus(2,3:3))\n", +"S_31 = S_13\n", +"S_32 = S_23\n", +"S_33 = conj(Y_bus(3,3:3))\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.7 : SOLUTION :-')\n", +"printf('\nPower flow expressions are:')\n", +"printf('\nS_1 = %.1fj|V_1|^2 %.1fjV_1V_2* %.1fjV_3*', imag(S_11),imag(S_12),imag(S_13))\n", +"printf('\nS_2 = %.1fjV_2V_1* + %.1fj|V_2|^2 %.1fjV_2V_3*', imag(S_21),imag(S_22),imag(S_23))\n", +"printf('\nS_3 = %.1fjV_3V_1* %.1fjV_3V_2* + %.1fj|V_3|^2', imag(S_31),imag(S_32),imag(S_33))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.8: Voltage_V2_by_GS_method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 9: LOAD FLOW STUDY USING COMPUTER TECHNIQUES\n", +"\n", +"// EXAMPLE : 9.8 :\n", +"// Page number 242\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_1 = 1.0 // Voltage(p.u)\n", +"S_g2 = complex(0,1.0) // Complex power generated(p.u)\n", +"S_D2 = complex(0.5,1.0) // Complex power demand(p.u)\n", +"Z_L = complex(0,0.5) // Impedance(p.u)\n", +"\n", +"// Calculations\n", +"Y_L = 1.0/Z_L // Admittance(p.u)\n", +"Y_22 = Y_L // Admittance(mho)\n", +"Y_21 = -Y_L // Admittance(mho)\n", +"S_2 = S_g2-S_D2\n", +"V_2_0 = 1.0 // Initial guess\n", +"V_2_1 = 1.0/Y_22*((conj(S_2/V_2_0))-Y_21*V_1) // V_2(p.u). In 1st iteration\n", +"V_2_2 = 1.0/Y_22*((conj(S_2/V_2_1))-Y_21*V_1) // V_2(p.u). In 2nd iteration\n", +"V_2_3 = 1.0/Y_22*((conj(S_2/V_2_2))-Y_21*V_1) // V_2(p.u). In 3rd iteration\n", +"V_2_4 = 1.0/Y_22*((conj(S_2/V_2_3))-Y_21*V_1) // V_2(p.u). In 4th iteration\n", +"V_2_5 = 1.0/Y_22*((conj(S_2/V_2_4))-Y_21*V_1) // V_2(p.u). In 5th iteration\n", +"V_2_6 = 1.0/Y_22*((conj(S_2/V_2_5))-Y_21*V_1) // V_2(p.u). In 6th iteration\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 9.8 : SOLUTION :-')\n", +"printf('\nBy G-S method, V_2 = %.6f∠%.5f° p.u\n', abs(V_2_6),phasemag(V_2_6))" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/17-POWER_SYSTEM_STABILITY.ipynb b/Power_System_Engineering_by_S_Chakraborthy/17-POWER_SYSTEM_STABILITY.ipynb new file mode 100644 index 0000000..5c3a3a0 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/17-POWER_SYSTEM_STABILITY.ipynb @@ -0,0 +1,1426 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17: POWER SYSTEM STABILITY" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.10: EX17_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.10 :\n", +"// Page number 275-276\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"E_1 = 1.1 // Sending end voltage(p.u)\n", +"x_d1 = 1.0 // Reactance(p.u)\n", +"x_T1 = 0.1 // Reactance(p.u)\n", +"x_l1 = 0.4 // Reactance(p.u)\n", +"x_l2 = 0.4 // Reactance(p.u)\n", +"x_T2 = 0.1 // Reactance(p.u)\n", +"E_2 = 1.0 // Receiving end voltage(p.u)\n", +"x_d2 = 1.0 // Reactance(p.u)\n", +"x_L = 1.0 // Shunt inductor reactance(p.u)\n", +"x_C = 1.0 // Static capacitor reactance(p.u)\n", +"delta = 30.0 // δ(°)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"Z_1_a = x_d1+x_T1+(x_l1/2.0) // Reactance(p.u)\n", +"X_1_a = %i*Z_1_a\n", +"Z_2_a = x_T2+x_d2 // Reactance(p.u)\n", +"X_2_a = %i*Z_2_a\n", +"Z_3_a = -x_C // Reactance(p.u)\n", +"X_3_a = %i*Z_3_a\n", +"X_a = X_1_a+X_2_a+(X_1_a*X_2_a/X_3_a) // Transfer reactance(p.u)\n", +"P_max_a = E_1*E_2/abs(X_a) // Maximum steady state power if static capacitor is connected(p.u)\n", +"P_a = P_max_a*sind(delta) // Value of P(p.u)\n", +"Q_a = (E_1*E_2/abs(X_a))*cosd(delta)-(E_2**2/abs(X_a)) // Value of Q(p.u)\n", +"// Case(b)\n", +"Z_1_b = x_d1+x_T1+(x_l1/2.0) // Reactance(p.u)\n", +"X_1_b = %i*Z_1_b\n", +"Z_2_b = x_T2+x_d2 // Reactance(p.u)\n", +"X_2_b = %i*Z_2_b\n", +"Z_3_b = x_L // Reactance(p.u)\n", +"X_3_b = %i*Z_3_b\n", +"X_b = X_1_b+X_2_b+(X_1_b*X_2_b/X_3_b) // Transfer reactance(p.u)\n", +"P_max_b = E_1*E_2/abs(X_b) // Maximum steady state power if static capacitor is replaced by an inductive reactor(p.u)\n", +"P_b = P_max_b*sind(delta) // Value of P(p.u)\n", +"Q_b = (E_1*E_2/abs(X_b))*cosd(delta)-(E_2**2/abs(X_b)) // Value of Q(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.10 : SOLUTION :-')\n", +"printf('\nCase(a): Maximum steady state power if static capacitor is connected, P_max = %.3f p.u', P_max_a)\n", +"printf('\n Value of P = %.3f p.u', P_a)\n", +"printf('\n Value of Q = %.3f p.u', Q_a)\n", +"printf('\nCase(b): Maximum steady state power if static capacitor is replaced by an inductive reactor, P_max = %.3f p.u', P_max_b)\n", +"printf('\n Value of P = %.3f p.u', P_b)\n", +"printf('\n Value of Q = %.4f p.u', Q_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.11: Kinetic_energy_stored_in_the_rotor_at_synchronous_speed_and_Acceleration.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.11 :\n", +"// Page number 303\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"G = 100.0 // Rating of generator(MVA)\n", +"H = 5.0 // Inertia constant(MJ/MVA)\n", +"P_a = 20.0 // Acceleration power(MVA)\n", +"\n", +"// Calculations\n", +"GH = G*H // Energy stored in rotor at synchronous speed(MJ)\n", +"M = GH/(180*f) // Angular momentum\n", +"acceleration = P_a/M // Acceleration(°/sec^2)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.11 : SOLUTION :-')\n", +"printf('\nKinetic energy stored in the rotor at synchronous speed, GH = %.f MJ', GH)\n", +"printf('\nAcceleration = %.f°/sec^2', acceleration)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.12: Kinetic_energy_stored_in_the_rotor_at_synchronous_speed_and_Acceleration.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.12 :\n", +"// Page number 303-304\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"P = 4.0 // Number of poles\n", +"G = 20.0 // Rating of generator(MVA)\n", +"H = 9.0 // Inertia constant(kWsec/MVA)\n", +"P_m = 26800.0 // Rotational loss(hp)\n", +"P_e = 16000.0 // Electric power developed(kW)\n", +"\n", +"// Calculations\n", +"GH = G*H // Energy stored in rotor at synchronous speed(MJ)\n", +"P_m_kW = P_m*0.746 // Rotational loss(kW)\n", +"P_a = P_m_kW-P_e // Acceleration power(kW)\n", +"P_a1 = P_a/1000.0 // Acceleration power(MW)\n", +"M = GH/(180*f) // Angular momentum\n", +"acceleration = P_a1/M // Acceleration(°/sec^2)\n", +"acceleration_1 = acceleration*%pi/180.0 // Acceleration(rad/sec^2)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.12 : SOLUTION :-')\n", +"printf('\nKinetic energy stored in the rotor at synchronous speed, GH = %.f MJ', GH)\n", +"printf('\nAcceleration = %.f°/sec^2 = %.2f rad/sec^2 \n', acceleration,acceleration_1)\n", +"printf('\nNOTE: ERROR: H = 9 kW-sec/MVA, not 9 kW-sec/kVA as mentioned in the textbook statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.13: Change_in_torque_angle_in_that_period_and_RPM_at_the_end_of_10_cycles.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.13 :\n", +"// Page number 304\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"P = 4.0 // Number of poles\n", +"alpha = 200.0 // Acceleration(°/sec^2)\n", +"alpha_rad = 3.49 // Acceleration(rad/sec^2)\n", +"n = 10.0 // Number of cycle\n", +"\n", +"// Calculations\n", +"t = 1/f*n // Time(sec)\n", +"delta_rel = ((alpha_rad*2)**0.5*0.5)**2 // Relation of change in rotor angle with time(rad)\n", +"delta = delta_rel*t**2 // Change in torque angle(rad)\n", +"delta_deg = delta*180/%pi // Change in torque angle in that period(°)\n", +"rpm_rad = (alpha_rad*2*delta)**0.5 // r.p.m(rad/sec)\n", +"rpm = rpm_rad*60.0/(%pi*P) // r.p.m\n", +"speed_rotor = (120*f/P)+rpm // Rotor speed at the end of 10 cycles(r.p.m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.13 : SOLUTION :-')\n", +"printf('\nChange in torque angle in that period, δ = %.4f rad = %.f elect degree', delta,delta_deg)\n", +"printf('\nRotor speed at the end of 10 cycles = %.2f r.p.m', speed_rotor)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.14: Accelerating_torque_at_the_time_the_fault_occurs.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.14 :\n", +"// Page number 304\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Power = 20.0*10**3 // Rating of generator(kVA)\n", +"PF = 0.8 // Lagging power factor\n", +"fault = 0.5 // Reduction in output under fault\n", +"P = 4.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"P_m = Power*PF // Output power before fault(kW)\n", +"P_e = fault*P_m // Output after fault(kW)\n", +"P_a = P_m-P_e // Accelerating power(kW)\n", +"w_s = 4.0*%pi*f/P // Speed\n", +"T_a = P_a*10**3/w_s // Accelerating torque at the time the fault occurs(N-m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.14 : SOLUTION :-')\n", +"printf('\nAccelerating torque at the time the fault occurs, T_a = %.2f N-m', T_a)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.16: Value_of_H_and_in_100_MVA_base.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.16 :\n", +"// Page number 304-305\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"S = 1000.0 // Rating of generator(MVA)\n", +"N = 1500.0 // Speed of alternator(r.p.m)\n", +"WR_sq = 5.0*10**6 // WR^2(lb.ft^2)\n", +"\n", +"// Calculations\n", +"H = 2.31*10**-10*WR_sq*N**2/S // Inertia constant(MJ/MVA)\n", +"H_100 = H*1000.0/100 // Inertia constant on 100 MVA(MJ/MVA)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.16 : SOLUTION :-')\n", +"printf('\nValue of inertia constant, H = %.1f MJ/MVA', H)\n", +"printf('\nValue of inertia constant in 100 MVA base, H = %.f MJ/MVA', H_100)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.17: Equivalent_H_for_the_two_to_common_100_MVA_base.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.17 :\n", +"// Page number 305\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_1 = 500.0 // Rating of generator(MVA)\n", +"H_1 = 4.0 // Inertia constant(MJ/VA)\n", +"MVA_2 = 1000.0 // Rating of generator(MVA)\n", +"H_2 = 3.5 // Inertia constant(MJ/VA)\n", +"MVA = 100.0 // Base MVA\n", +"\n", +"// Calculations\n", +"KE_T = H_1*MVA_1+H_2*MVA_2 // Total KE of the system(MJ)\n", +"H_total = KE_T/MVA // Equivalent H for the two to common 100MVA base(MJ/MVA)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.17 : SOLUTION :-')\n", +"printf('\nEquivalent H for the two to common 100 MVA base, H = %.f MJ/MVA', H_total)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.18: Energy_stored_in_the_rotor_at_the_rated_speed_Value_of_H_and_Angular_momentum.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.18 :\n", +"// Page number 305\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 210.0 // Rating of generator(MVA)\n", +"P = 2.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"MI = 60.0*10**3 // Moment of inertia(kg-mt^2)\n", +"\n", +"// Calculations\n", +"N = 120.0*f/P // Speed(r.p.m)\n", +"KE = 1.0/2*MI*(2*%pi*N/f)**2/10**6 // Energy stored in the rotor at rated speed(MJ)\n", +"H = KE/MVA // Inertia constant(MJ/MVA)\n", +"G = MVA\n", +"M = G*H/(180*f) // Angular momentum(MJ-sec/elect.degree)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.18 : SOLUTION :-')\n", +"printf('\nEnergy stored in the rotor at the rated speed, KE = %.2e MJ', KE)\n", +"printf('\nValue of inertia constant, H = %.2f MJ/MVA', H)\n", +"printf('\nAngular momentum, M = %.3f MJ-sec/elect.degree', M)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.19: Acceleration_of_the_rotor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.19 :\n", +"// Page number 305\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P_accl = 30.0 // Acceleration power(MVA)\n", +"M = 0.474 // Angular momentum(MJ-sec/elect.degree). From Example 10.18\n", +"\n", +"// Calculations\n", +"acceleration = P_accl/M // Acceleration of the rotor(elect.degree/sec^2)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.19 : SOLUTION :-')\n", +"printf('\nAcceleration of the rotor = %.2f elect.degree/sec^2', acceleration)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.1: Operating_power_angle_and_Magnitude_of_P0.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.1 :\n", +"// Page number 270\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Z = 0.1 // Impedance of transmission line(p.u)\n", +"M = 0.3 // Stability margin\n", +"X = 1.0 // Constant(p.u)\n", +"\n", +"// Calculations\n", +"sin_delta_0 = 1-M // Sin(δ_0)\n", +"delta_0 = asind(sin_delta_0) // δ_0(°)\n", +"P_0 = X/Z*sin_delta_0 // Magnitude of P_0(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.1 : SOLUTION :-')\n", +"printf('\nOperating power angle, δ_0 = %.2f° ', delta_0)\n", +"printf('\nP_0 = %.2f p.u', P_0)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.20: Accelerating_power_and_New_power_angle_after_10_cycles.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.20 :\n", +"// Page number 305\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 50.0 // Rating of alternator(MVA)\n", +"P = 4.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"KE = 150.0 // Kinetic energy stored in rotor(MJ)\n", +"P_m = 25.0 // Machine input(MW)\n", +"P_e = 22.5 // Developed power(MW)\n", +"n = 10.0 // Number of cycles\n", +"\n", +"// Calculations\n", +"P_a = P_m-P_e // Accelerating power(MW)\n", +"H = KE/MVA // Inertia constant(MJ/MVA)\n", +"G = MVA\n", +"M_deg = G*H/(180*f) // Angular momentum(MJ-sec/elect.degree)\n", +"M = G*H/(%pi*f) // Angular momentum(MJ-sec/rad)\n", +"acceleration = P_a/M // Accelerating power(rad/sec^2)\n", +"t = 1/f*n // Time(sec)\n", +"delta = 1.309*t**2 // Term in δ\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.20 : SOLUTION :-')\n", +"printf('\nAccelerating power = %.3f rad/sec^2', acceleration)\n", +"printf('\nNew power angle after 10 cycles, δ = (%.3f + δ_0) rad', delta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.21: Kinetic_energy_stored_by_rotor_at_synchronous_speed_and_Acceleration_in.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.21 :\n", +"// Page number 305-306\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"P = 4.0 // Number of poles\n", +"G = 20.0 // Rating of turbo-generator(MVA)\n", +"V = 13.2 // Voltage(kV)\n", +"H = 9.0 // Inertia constant(kW-sec/kVA)\n", +"P_s = 20.0 // Input power less rotational loss(MW)\n", +"P_e = 15.0 // Output power(MW)\n", +"\n", +"// Calculations\n", +"KE = G*H // Kinetic energy stored(MJ)\n", +"M = G*H/(180*f) // Angular momentum(MJ-sec/elect.degree)\n", +"P_a = P_s-P_e // Accelerating power(MW)\n", +"alpha = P_a/M // Acceleration(elect.degree/sec^2)\n", +"alpha_deg = alpha/2.0 // Acceleration(degree/sec^2)\n", +"alpha_rpm = 60.0*alpha_deg/360 // Acceleration(rpm/sec)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.21 : SOLUTION :-')\n", +"printf('\nCase(a): Kinetic energy stored by rotor at synchronous speed, GH = %.f MJ', KE)\n", +"printf('\nCase(b): Acceleration, α = %.f degree/sec^2', alpha_deg)\n", +"printf('\n Acceleration, α = %.2f rpm/sec', alpha_rpm)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.22: Change_in_torque_angle_and_Speed_in_rpm_at_the_end_of_10_cycles.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.22 :\n", +"// Page number 306\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"P = 4.0 // Number of poles\n", +"G = 20.0 // Rating of turbo-generator(MVA)\n", +"V = 13.2 // Voltage(kV)\n", +"H = 9.0 // Inertia constant(kW-sec/kVA)\n", +"P_s = 20.0 // Input power less rotational loss(MW)\n", +"P_e = 15.0 // Output power(MW)\n", +"n = 10.0 // Number of cycles\n", +"\n", +"// Calculations\n", +"KE = G*H // Kinetic energy stored(MJ)\n", +"M = G*H/(180*f) // Angular momentum(MJ-sec/elect.degree)\n", +"P_a = P_s-P_e // Accelerating power(MW)\n", +"alpha = P_a/M // Acceleration(elect.degree/sec^2)\n", +"alpha_deg = alpha/2.0 // Acceleration(degree/sec^2)\n", +"alpha_rpm = 60.0*alpha_deg/360 // Acceleration(rpm/sec)\n", +"t = 1.0/f*n // Time(sec)\n", +"delta = 1.0/2*alpha*t**2 // Change in torque angle(elect.degree)\n", +"N_s = 120*f/P // Synchronous speed(rpm)\n", +"speed = N_s+alpha_rpm*t // Speed at the end of 10 cycles(rpm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.22 : SOLUTION :-')\n", +"printf('\nChange in torque angle in that period, δ = %.f elect degrees.', delta)\n", +"printf('\nSpeed in rpm at the end of 10 cycles = %.2f rpm', speed)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.23: Accelerating_torque_at_the_time_of_fault_occurrence.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.23 :\n", +"// Page number 306\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"G = 20.0 // Rating of turbo-generator(MVA)\n", +"PF = 0.75 // Lagging power factor\n", +"fault = 0.5 // Fault reduces output power\n", +"N_s = 1500.0 // Synchronous speed(rpm). From Example 10.22\n", +"\n", +"// Calculations\n", +"P_prefault = PF*G // Pre-fault output power(MW)\n", +"P_a = P_prefault*fault // Post-fault output power(MW)\n", +"w = 2.0*%pi*N_s/60 // ω(rad/sec)\n", +"T_a = P_a*10**6/w // Accelerating torque at the time of fault occurrence(N-m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.23 : SOLUTION :-')\n", +"printf('\nAccelerating torque at the time of fault occurrence, T_a = %.f N-m', T_a)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.24: Swing_equation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.24 :\n", +"// Page number 306-307\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"x_d = %i*0.2 // Transient reactance of generator(p.u)\n", +"P_e = 0.8 // Power delivered(p.u)\n", +"V_t = 1.05 // Terminal voltage(p.u)\n", +"H = 4.0 // Inertia constant(kW-sec/kVA)\n", +"x_t = %i*0.1 // Transformer reactance(p.u)\n", +"x_l = %i*0.4 // Transmission line reactance(p.u)\n", +"V = 1.0 // Infinite bus voltage(p.u)\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"x_12 = x_d+x_t+(x_l/2) // Reactance b/w bus 1 & 2(p.u)\n", +"y_12 = 1/x_12 // Admittance b/w bus 1 & 2(p.u)\n", +"y_21 = y_12 // Admittance b/w bus 2 & 1(p.u)\n", +"y_10 = 0.0 // Admittance b/w bus 1 & 0(p.u)\n", +"y_20 = 0.0 // Admittance b/w bus 2 & 0(p.u)\n", +"Y_11 = y_12+y_10 // Admittance at bus 1(p.u)\n", +"Y_12 = -y_12 // Admittance b/w bus 1 & 2(p.u)\n", +"Y_21 = -y_12 // Admittance b/w bus 2 & 1(p.u)\n", +"Y_22 = y_21+y_20 // Admittance at bus 2(p.u)\n", +"x_32 = x_t+(x_l/2) // Reactance b/w bus 3 & 1(p.u)\n", +"theta_t = asind(P_e*abs(x_32)/V_t) // Angle(°)\n", +"V_t1 = V_t*exp(%i*theta_t*%pi/180) // Terminal voltage(p.u)\n", +"I = (V_t1-V)/x_32 // Current(p.u)\n", +"E = V_t1+I*x_d // Alternator voltage(p.u)\n", +"sine = poly(0,'sin')\n", +"P_e1 = 2.0*abs(E) // Developed power(p.u) in terms of sin δ\n", +"P_m_P_e = P_e-P_e1*sine\n", +"M = 2*H/(2*%pi*f) // Angular momentum\n", +"acc = (P_e-P_e1*sine)*2*%pi*f/(2*H) // Acceleration = α (rad/sec^2)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.24 : SOLUTION :-')\n", +"printf('\nSwing equation is, %.4f*α = %.1f - %.3fsin δ\n', M,P_e,P_e1)\n", +"printf('\nNOTE: Swing equation is simplified and represented here')\n", +"printf('\n ERROR: x_d = 0.2 p.u, not 0.1 p.u as mentioned in textbook statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.26: Critical_clearing_angle.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.26 :\n", +"// Page number 308-309\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X_d = 0.25 // Transient reactance of generator(p.u)\n", +"X_t1 = 0.15 // Reactance of transformer(p.u)\n", +"X_t2 = 0.15 // Reactance of transformer(p.u)\n", +"X_t3 = 0.15 // Reactance of transformer(p.u)\n", +"X_t4 = 0.15 // Reactance of transformer(p.u)\n", +"X_l1 = 0.20 // Reactance of line(p.u)\n", +"X_l2 = 0.20 // Reactance of line(p.u)\n", +"X_tr = 0.15 // Reactance of transformer(p.u)\n", +"P_m = 1.0 // Power delivered(p.u)\n", +"E = 1.20 // Voltage behind transient reactance(p.u)\n", +"V = 1.0 // Infinite bus voltage(p.u)\n", +"\n", +"// Calculations\n", +"X_14 = X_d+((X_t1+X_t2+X_l1)/2)+X_tr // Reactance before fault(p.u)\n", +"x_1_b = X_t1+X_t2+X_l1 // Reactance(p.u). From figure (b)\n", +"x_2_b = X_l2+X_t4 // Reactance(p.u). From figure (b)\n", +"x_1 = x_1_b*X_t3/(x_1_b+x_2_b+X_t3) // Reactance(p.u). From figure (c)\n", +"x_2 = x_1_b*x_2_b/(x_1_b+x_2_b+X_t3) // Reactance(p.u). From figure (c)\n", +"x_3 = X_t3*x_2_b/(x_1_b+x_2_b+X_t3) // Reactance(p.u). From figure (c)\n", +"X_14_fault = x_1+X_d+x_2+X_tr+((x_1+X_d)*(x_2+X_tr)/x_3) // Reactance under fault(p.u)\n", +"X_14_after_fault = X_d+X_t1+X_l1+X_t2+X_tr // Reactance after fault is cleared(p.u)\n", +"P_max = V*E/X_14 // Maximum power transfer(p.u)\n", +"gamma_1 = (V*E/X_14_fault)/P_max // γ_1\n", +"gamma_2 = (V*E/X_14_after_fault)/P_max // γ_2\n", +"delta_0 = asin(P_m/P_max) // δ_0(radians)\n", +"delta_0_degree = delta_0*180/%pi // δ_0(°)\n", +"delta_m = %pi-asin(P_m/(gamma_2*P_max)) // δ_1(radians)\n", +"delta_m_degree = delta_m*180/%pi // δ_1(°)\n", +"delta_c = acosd((P_m/P_max*(delta_m-delta_0)+gamma_2*cos(delta_m)-gamma_1*cos(delta_0))/(gamma_2-gamma_1)) // Clearing angle(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.26 : SOLUTION :-')\n", +"printf('\nCritical clearing angle, δ_c = %.2f° ', delta_c)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.27: Critical_angle_using_equal_area_criterion.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.27 :\n", +"// Page number 309-310\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Frequency(Hz)\n", +"P_m = 1.0 // Power delivered(p.u)\n", +"P_max = 1.8 // Maximum power(p.u)\n", +"gamma_1_P_max = 0.4 // Reduced maximum power after fault(p.u)\n", +"gamma_2_P_max = 1.30 // Maximum power after fault clearance(p.u)\n", +"\n", +"// Calculations\n", +"delta_0 = asin(P_m/P_max) // δ_0(radians)\n", +"delta_0_degree = delta_0*180/%pi // δ_0(°)\n", +"delta_f = %pi-asin(P_m/(gamma_2_P_max)) // δ_1(radians)\n", +"delta_f_degree = delta_f*180/%pi // δ_1(°)\n", +"gamma_1 = gamma_1_P_max/P_max // γ_1\n", +"gamma_2 = gamma_2_P_max/P_max // γ_2\n", +"delta_c = acosd(1.0/(gamma_2-gamma_1)*((delta_f-delta_0)*sin(delta_0)+(gamma_2*cos(delta_f)-gamma_1*cos(delta_0)))) // Clearing angle(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.27 : SOLUTION :-')\n", +"printf('\nCritical angle, δ_c = %.2f° ', delta_c)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.28: Critical_clearing_angle.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.28 :\n", +"// Page number 310\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"sin_delta_0 = 0.45 // Supplying percent of peak power capacity before fault\n", +"x = 4.0 // Reactance under fault increased\n", +"gamma_2 = 0.7 // Peak power delivered after fault clearance\n", +"\n", +"// Calculations\n", +"delta_0 = asin(sin_delta_0) // δ_0(radians)\n", +"delta_0_degree = delta_0*180/%pi // δ_0(°)\n", +"gamma_1 = 1.0/x // γ_1\n", +"delta_m = %pi-asin(sin_delta_0/(gamma_2)) // δ_m(radians)\n", +"delta_m_degree = delta_m*180/%pi // δ_m(°)\n", +"delta_c = acosd(1.0/(gamma_2-gamma_1)*((delta_m-delta_0)*sin(delta_0)+(gamma_2*cos(delta_m)-gamma_1*cos(delta_0)))) // Clearing angle(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.28 : SOLUTION :-')\n", +"printf('\nCritical clearing angle, δ_c = %.f° ', delta_c)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.2: Minimum_value_of_E_and_VL_Maximum_power_limit_and_Steady_state_stability_margin.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.2 :\n", +"// Page number 270\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"x_s = 0.85 // Reactance(p.u)\n", +"x_T1 = 0.157 // Reactance(p.u)\n", +"x_T2 = 0.157 // Reactance(p.u)\n", +"x_l1 = 0.35 // Reactance(p.u)\n", +"x_l2 = 0.35 // Reactance(p.u)\n", +"E = 1.50 // Sending end voltage(p.u)\n", +"V_L = 1.0 // Load voltage(p.u)\n", +"P_0 = 1.0 // Stable power output(p.u)\n", +"\n", +"// Calculations\n", +"x = x_s+x_T1+x_T2+(x_l1/2) // Total reactance(p.u)\n", +"P_max = E*V_L/x // Maximum power limit(p.u)\n", +"M = (P_max-P_0)/P_max*100 // Steady state stability margin(%)\n", +"V_Lmin = P_0*x/E // Minimum value of V_L(p.u)\n", +"E_min = P_0*x/V_L // Minimum value of E(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.2 : SOLUTION :-')\n", +"printf('\nMinimum value of |E|, |E_min| = %.3f p.u', E_min)\n", +"printf('\nMinimum value of |V_L|, |V_Lmin| = %.3f p.u', V_Lmin)\n", +"printf('\nMaximum power limit, P_0 = %.2f p.u', P_max)\n", +"printf('\nSteady state stability margin, M = %.1f percent', M)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.30: Power_angle_and_Swing_curve_data.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.30 :\n", +"// Page number 310-311\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 60.0 // Frequency(Hz)\n", +"P = 6.0 // Number of poles\n", +"H = 4.0 // Inertia constant(p.u)\n", +"P_e = 1.0 // Power supplied by generator(p.u)\n", +"E = 1.2 // Internal voltage(p.u)\n", +"V = 1.0 // Infinite bus voltage(p.u)\n", +"X = 0.3 // Line reactance(p.u)\n", +"del_t = 0.05 // Δt = Interval step size(sec)\n", +"\n", +"// Calculations\n", +"P_max = E*V/X // Maximum power(p.u)\n", +"delta_0 = asind(P_e/P_max) // δ_0(°)\n", +"G = P_e\n", +"M = G*H/(180*f) // Angular momentum(p.u)\n", +"P_a_0 = 1.0/2*(P_e-0) // (p.u)\n", +"alpha_0 = P_a_0/M // α_0(°/sec^2)\n", +"del_w_r_1 = alpha_0*del_t // Δω_r_1(°/sec)\n", +"w_r_1 = 0+del_w_r_1 // ω_r_1(°/sec)\n", +"del_delta_1 = w_r_1*del_t // Δδ_1(°)\n", +"delta_1 = delta_0+del_delta_1 // δ_1(°)\n", +"P_a_1 = 1.0*(P_e-0) // (p.u)\n", +"alpha_1 = P_a_1/M // α_1(°/sec^2)\n", +"del_w_r_2 = alpha_1*del_t // Δω_r_2(°/sec)\n", +"w_r_2 = del_w_r_1+del_w_r_2 // ω_r_2(°/sec)\n", +"del_delta_2 = w_r_2*del_t // Δδ_2(°)\n", +"delta_2 = delta_1+del_delta_2 // δ_2(°)\n", +"del_w_r_3 = del_w_r_2 // Δω_r_3(°/sec)\n", +"w_r_3 = w_r_2+del_w_r_3 // ω_r_3(°/sec)\n", +"del_delta_3 = w_r_3*del_t // Δδ_3(°)\n", +"delta_3 = delta_2+del_delta_3 // δ_3(°)\n", +"del_w_r_4 = del_w_r_2 // Δω_r_4(°/sec)\n", +"w_r_4 = w_r_3+del_w_r_4 // ω_r_4(°/sec)\n", +"del_delta_4 = w_r_4*del_t // Δδ_4(°)\n", +"delta_4 = delta_3+del_delta_4 // δ_4(°)\n", +"del_w_r_5 = del_w_r_2 // Δω_r_5(°/sec)\n", +"w_r_5 = w_r_4+del_w_r_5 // ω_r_5(°/sec)\n", +"del_delta_5 = w_r_5*del_t // Δδ_5(°)\n", +"delta_5 = delta_4+del_delta_5 // δ_5(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.30 : SOLUTION :-')\n", +"printf('\nPower angle, δ_0 = %.2f° ', delta_0)\n", +"printf('\nValue of δ vs t are:')\n", +"printf('\n_________________________')\n", +"printf('\n t(Sec) : δ(degree)')\n", +"printf('\n_________________________')\n", +"printf('\n %.1f : %.2f°', 0,delta_0)\n", +"printf('\n %.2f : %.2f°', (del_t),delta_1)\n", +"printf('\n %.2f : %.2f°', (del_t+del_t),delta_2)\n", +"printf('\n %.2f : %.2f°', (del_t*3),delta_3)\n", +"printf('\n %.2f : %.2f°', (del_t*4),delta_4)\n", +"printf('\n %.2f : %.2f°', (del_t*5),delta_5)\n", +"printf('\n_________________________')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.3: EX17_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.3 :\n", +"// Page number 270-271\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"E_1 = 1.25 // Sending end voltage(p.u)\n", +"x_d = 1.0 // Reactance(p.u)\n", +"x_T1 = 0.2 // Reactance(p.u)\n", +"x_l1 = 1.0 // Reactance(p.u)\n", +"x_l2 = 1.0 // Reactance(p.u)\n", +"x_T2 = 0.2 // Reactance(p.u)\n", +"E_2 = 1.0 // Receiving end voltage(p.u)\n", +"x_L = 1.0 // Shunt inductor reactance(p.u)\n", +"x_C = 1.0 // Shunt capacitor reactance(p.u)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"Z_1_a = x_d+x_T1+(x_l1/2.0) // Reactance(p.u)\n", +"Z_2_a = x_T2+x_d // Reactance(p.u)\n", +"Z_3_a = x_L // Reactance(p.u)\n", +"Z_a = Z_1_a+Z_2_a+(Z_1_a*Z_2_a/Z_3_a) // Transfer reactance(p.u)\n", +"P_max_1 = E_1*E_2/Z_a // Maximum power transfer if shunt inductor is connected at bus 2(p.u)\n", +"// Case(b)\n", +"Z_1_b = x_d+x_T1+(x_l1/2.0) // Reactance(p.u)\n", +"Z_2_b = x_T2+x_d // Reactance(p.u)\n", +"Z_3_b = -x_C // Reactance(p.u)\n", +"Z_b = Z_1_b+Z_2_b+(Z_1_b*Z_2_b/Z_3_b) // Transfer reactance(p.u)\n", +"P_max_2 = E_1*E_2/Z_b // Maximum power transfer if shunt capacitor is connected at bus 2(p.u)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.3 : SOLUTION :-')\n", +"printf('\nCase(a): Maximum power transfer if shunt inductor is connected at bus 2, P_max1 = %.3f p.u', P_max_1)\n", +"printf('\nCase(b): Maximum power transfer if shunt capacitor is connected at bus 2, P_max2 = %.2f p.u', P_max_2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.4: Maximum_power_transfer_and_Stability_margin.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.4 :\n", +"// Page number 271\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // Voltage(kV)\n", +"L = 220.0 // Line length(km)\n", +"P = 0.58 // Initial real power transfer(p.u)\n", +"PF = 0.85 // Lagging power factor\n", +"V_L = 1.00 // Load bus voltage(p.u)\n", +"x_d = 0.460 // Reactance(p.u)\n", +"x_T1 = 0.200 // Reactance(p.u)\n", +"x_T2 = 0.15 // Reactance(p.u)\n", +"x_line = 0.7 // Reactance(p.u)\n", +"\n", +"// Calculations\n", +"x = x_d+x_T1+x_T2+(x_line/2) // Net reactance(p.u)\n", +"phi = acosd(PF) // Φ(°)\n", +"Q = P*tand(phi) // Reactive power(p.u)\n", +"E = ((V_L+(Q*x/V_L))**2+(P*x/V_L)**2)**0.5 // Excitation voltage of generator(p.u)\n", +"P_max = E*V_L/x // Maximum power transfer(p.u)\n", +"M = (P_max-P)/P_max*100 // Steady state stability margin(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.4 : SOLUTION :-')\n", +"printf('\nMaximum power transfer, P_max = %.2f p.u', P_max)\n", +"printf('\nStability margin, M = %.f percent', M)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.5: QgB_Phase_angle_of_VB_and_What_happens_if_QgB_is_made_zero.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.5 :\n", +"// Page number 271-272\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_A = 1.0 // Voltage at bus A(p.u)\n", +"Z_AB = %i*0.5 // Impedance(p.u)\n", +"S_DA = 1.0 // p.u\n", +"S_DB = 1.0 // p.u\n", +"V_B = 1.0 // Voltage at bus B(p.u)\n", +"\n", +"// Calculations\n", +"// Case(i) & (ii)\n", +"X = abs(Z_AB) // Reactance(p.u)\n", +"sin_delta = 1.0*X/(V_A*V_B) // Sin δ\n", +"delta = asind(sin_delta) // δ(°)\n", +"V_2 = V_B\n", +"V_1 = V_A\n", +"Q_gB = (V_2**2/X)-(V_2*V_1*cosd(delta)/X)\n", +"// Case(iii)\n", +"V_2_3 = 1/2.0**0.5 // Solving quadratic equation from textbook\n", +"delta_3 = acosd(V_2_3) // δ(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.5 : SOLUTION :-')\n", +"printf('\nCase(i) : Q_gB = %.3f', Q_gB)\n", +"printf('\nCase(ii) : Phase angle of V_B, δ = %.f° ', delta)\n", +"printf('\nCase(iii): If Q_gB is equal to zero then amount of power transmitted is, V_2 = %.3f∠%.f° ', V_2_3,delta_3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.6: EX17_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.6 :\n", +"// Page number 272\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"A = 0.98*exp(%i*0.3*%pi/180) // Constant\n", +"B = 82.5*exp(%i*76.0*%pi/180) // Constant(ohm)\n", +"C = 0.0005*exp(%i*90.0*%pi/180) // Constant(mho)\n", +"D = A // Constant\n", +"V_S = 110.0 // Sending end voltage(kV)\n", +"V_R = 110.0 // Receiving end voltage(kV)\n", +"\n", +"// Calculations\n", +"alpha = phasemag(A) // α(°)\n", +"beta = phasemag(B) // β(°)\n", +"P_max = (V_S*V_R/abs(B))-(abs(A)*V_R**2/abs(B)*cosd((beta-alpha))) // Maximum power transfer(MW)\n", +"B_new = abs(B)*sind(beta) // Constant(ohm)\n", +"beta_new = 90.0 // β(°)\n", +"P_max_new = (V_S*V_R/B_new)-(V_R**2/B_new*cosd(beta_new)) // Maximum power transfer(MW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.6 : SOLUTION :-')\n", +"printf('\nSteady state stability limit, P_max = %.2f MW', P_max)\n", +"printf('\nSteady state stability limit if shunt admittance is zero & series resistance neglected, P_max = %.2f MW \n', P_max_new)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.8: EX17_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.8 :\n", +"// Page number 273-275\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"V = 33.0*10**3 // Line voltage(V)\n", +"R = 6.0 // Resistance per phase(ohm)\n", +"X = 15.0 // Reactance per phase(ohm)\n", +"\n", +"// Calculations\n", +"V_S = V/3**0.5 // Sending end phase voltage(V)\n", +"V_R = V/3**0.5 // Receiving end phase voltage(V)\n", +"beta = atand(X/R) // β(°)\n", +"Z = (R**2+X**2)**0.5 // Impedance(ohm)\n", +"delta_0 = 0.0 // δ(°)\n", +"P_0 = (V_R/Z**2)*(V_S*Z*cosd((delta_0-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_1 = 30.0 // δ(°)\n", +"P_1 = (V_R/Z**2)*(V_S*Z*cosd((delta_1-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_2 = 60.0 // δ(°)\n", +"P_2 = (V_R/Z**2)*(V_S*Z*cosd((delta_2-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_3 = beta // δ(°)\n", +"P_3 = (V_R/Z**2)*(V_S*Z*cosd((delta_3-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_4 = 90.0 // δ(°)\n", +"P_4 = (V_R/Z**2)*(V_S*Z*cosd((delta_4-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_5 = 120.0 // δ(°)\n", +"P_5 = (V_R/Z**2)*(V_S*Z*cosd((delta_5-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"delta_6 = (acosd(R/Z))+beta // δ(°)\n", +"P_6 = (V_R/Z**2)*(V_S*Z*cosd((delta_6-beta))-V_R*R)/10**6 // Power received(MW/phase)\n", +"\n", +"\n", +"delta = [delta_0,delta_1,delta_2,delta_3,delta_4,delta_5,delta_6]\n", +"P = [P_0,P_1,P_2,P_3,P_4,P_5,P_6]\n", +"a = gca() ;\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot(delta,P,'ro-')\n", +"a.x_label.text = 'Electrical degree' // labels x-axis\n", +"a.y_label.text = 'Power in MW/phase' // labels y-axis\n", +"xtitle('Fig E10.7 . Power angle diagram') \n", +"xset('thickness',2) // sets thickness of axes\n", +"xstring(70,14.12,'P_max = 14.12 MW/phase(approximately)')\n", +"P_max = V_R/Z**2*(V_S*Z-V_R*R)/10**6 // Maximum power transmitted(MW/phase)\n", +"delta_equal = 0.0 // δ With no phase shift(°)\n", +"P_no_shift = (V_R/Z**2)*(V_S*Z*cosd((delta_equal-beta))-V_R*R)/10**6 // Power transmitted with no phase shift(MW/phase)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.8 : SOLUTION :-')\n", +"printf('\nPower angle diagram is plotted and is shown in the Figure 1')\n", +"printf('\nMaximum power the line is capable of transmitting, P_max = %.2f MW/phase', P_max)\n", +"printf('\nWith equal voltage at both ends power transmitted = %.f MW/phase', abs(P_no_shift))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.9: Maximum_steady_state_power_that_can_be_transmitted_over_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 10: POWER SYSTEM STABILITY\n", +"\n", +"// EXAMPLE : 10.9 :\n", +"// Page number 275\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 132.0*10**3 // Sending end voltage(V)\n", +"Z_line = complex(4,6) // Line impedance per phase(ohm)\n", +"\n", +"// Calculations\n", +"V_S = V/3**0.5 // Sending end phase voltage(V)\n", +"V_R = V/3**0.5 // Receiving end phase voltage(V)\n", +"Z = abs(Z_line) // Impedance(ohm)\n", +"R = real(Z_line) // Resistance per phase(ohm)\n", +"P_max_phase = ((V_S*V_R/Z)-(R*V_R**2/Z**2))/10**6 // Maximum steady state power that can be transmitted over the line(MW/phase)\n", +"P_max_total = 3.0*P_max_phase // Maximum steady state power that can be transmitted over the line(MW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 10.9 : SOLUTION :-')\n", +"printf('\nMaximum steady state power that can be transmitted over the line, P_max = %.f MW (total 3-phase)', P_max_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/Power_System_Engineering_by_S_Chakraborthy/18-LOAD_FREQUENCY_CONTROL_AND_LOAD_SHARING_OF_POWER_GENERATING_SOURCES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/18-LOAD_FREQUENCY_CONTROL_AND_LOAD_SHARING_OF_POWER_GENERATING_SOURCES.ipynb new file mode 100644 index 0000000..5c6c297 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/18-LOAD_FREQUENCY_CONTROL_AND_LOAD_SHARING_OF_POWER_GENERATING_SOURCES.ipynb @@ -0,0 +1,786 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.10: Constants_of_the_second_feeder.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.10 :\n", +"// Page number 336\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_1 = 10000.0 // Total balanced load(kW)\n", +"V = 33000.0 // Voltage(V)\n", +"PF_1 = 0.8 // Lagging power factor\n", +"R = 1.6 // Resistance of feeder(ohm/phase)\n", +"X = 2.5 // Reactance of feeder(ohm/phase)\n", +"load_2 = 4460.0 // Load delivered by feeder(kW)\n", +"PF_2 = 0.72 // Lagging power factor\n", +"\n", +"// Calculations\n", +"I = load_1*1000/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Total line current(A)\n", +"I_1 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Line current of first feeder(A)\n", +"I_2 = I-I_1 // Line current of first feeder(A)\n", +"Z_1 = complex(R,X) // Impedance of first feeder(ohm)\n", +"Z_2 = I_1*Z_1/I_2 // Impedance of second feeder(ohm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.10 : SOLUTION :-')\n", +"printf('\nImpedance of second feeder, Z_2 = %.2f∠%.1f° ohm \n', abs(Z_2),phasemag(Z_2))\n", +"printf('\nNOTE: ERROR: Changes in the obtained answer from that of textbook is due to wrong values of substitution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.11: Necessary_booster_voltages.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.11 :\n", +"// Page number 337\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P = 9.0 // Load supplied from substation(MW)\n", +"V = 33000.0 // Voltage(V)\n", +"PF_1 = 1.0 // Unity power factor\n", +"Z_A = complex(2.0,8.0) // Impedance of circuit A(ohm)\n", +"Z_B = complex(4.0,4.0) // Impedance of circuit B(ohm)\n", +"\n", +"// Calculations\n", +"V_ph = V/3**0.5 // Voltage at receiving end per phase(V)\n", +"P_A = 1.0/3*P // Power supplied by line A(MW)\n", +"P_B = 2.0/3*P // Power supplied by line B(MW)\n", +"I_A = P_A*10**6/(3**0.5*V) // Current through line A(A)\n", +"I_B = P_B*10**6/(3**0.5*V) // Current through line B(A)\n", +"IA_ZA_drop = I_A*Z_A // I_A Z_A drop(V/phase)\n", +"IB_ZB_drop = I_B*Z_B // I_B Z_B drop(V/phase)\n", +"phase_boost = real(IB_ZB_drop)-real(IA_ZA_drop) // Voltage in phase boost(V/phase)\n", +"quad_boost = imag(IB_ZB_drop)-imag(IA_ZA_drop) // Voltage in quadrature boost(V/phase)\n", +"constant_P = V_ph+IA_ZA_drop // Assumed that sending end voltage at P is kept constant(V/phase)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.11 : SOLUTION :-')\n", +"printf('\nVoltage in-phase boost = %.2f V/phase', phase_boost)\n", +"printf('\nVoltage in quadrature boost = %.f V/phase', quad_boost)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.12: Load_on_C_at_two_different_conditions_of_load_in_A_and_B.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.12 :\n", +"// Page number 337\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_A = 15000.0 // Capacity of station A(kW)\n", +"cap_B = 10000.0 // Capacity of station B(kW)\n", +"cap_C = 2000.0 // Capacity of station C(kW)\n", +"speed_reg_A = 2.4/100 // Speed regulation of A\n", +"speed_reg_B = 3.2/100 // Speed regulation of B\n", +"slip_C = 4.5/100 // Full load slip\n", +"local_load_B_a = 10000.0 // Local load on station B(kW)\n", +"local_load_A_a = 0 // Local load on station A(kW)\n", +"local_load_both = 10000.0 // Local load on both station(kW)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"speed_A = speed_reg_A/cap_A // % of speed drop for A\n", +"speed_C = slip_C/cap_C // % of speed drop for C\n", +"speed_B = speed_reg_B/cap_B // % of speed drop for B\n", +"X = local_load_B_a*speed_B/(speed_A+speed_B+speed_C) // Load on C when local load of B is 10000 kW and A has no load(kW)\n", +"// Case(b)\n", +"Y = local_load_both*(speed_B-speed_A)/(speed_A+speed_B+speed_C) // Load on C when both station have local loads of 10000 kW(kW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.12 : SOLUTION :-')\n", +"printf('\nCase(a): Load on C when local load of B is 10000 kW and A has no load, X = %.f kW', X)\n", +"printf('\nCase(b): Load on C when both station have local loads of 10000 kW, Y = %.f kW', Y)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.13: EX18_13.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.13 :\n", +"// Page number 337-338\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 20.0 // Length of cable(km)\n", +"r = 0.248 // Resistance(ohm/km)\n", +"x = 0.50*10**-3 // Inductance(H/m)\n", +"V_gen = 6600.0 // Generation voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"V = 33000.0 // Transmission voltage(V)\n", +"rating = 10.0 // Transformer rating(MVA)\n", +"loss_cu = 100.0 // Copper loss at full load(kW)\n", +"x_tr = 2.5/100 // Transformer reactance\n", +"load = 7.5 // Load to be transmitted(MW)\n", +"PF = 0.71 // Lagging power factor\n", +"\n", +"// Calculations\n", +"R = l*r // Resistance of the cable(ohm)\n", +"I_fl = rating*10**6/(3**0.5*V) // Transformer current at full load(A)\n", +"R_eq = loss_cu*1000/(3*I_fl**2) // Equivalent resistance per phase of transformer(ohm)\n", +"R_total_hv = R+2.0*R_eq // Total resistance per conductor in terms of hv side(ohm)\n", +"X = 2.0*%pi*f*l*x // Reactance of cable per conductor(ohm)\n", +"per_X_tr = V/3**0.5*x_tr/I_fl // % reactance of transformer(ohm)\n", +"X_total_hv = X+2.0*per_X_tr // Total reactance per conductor in terms of hv side(ohm)\n", +"I = load*10**6/(3**0.5*V*PF) // Line current at receiving end(A)\n", +"IR = I*R_total_hv // IR drop(V)\n", +"IX = I*X_total_hv // IX drop(V)\n", +"E_r = V/3**0.5 // Phase voltage at station B(V)\n", +"cos_phi_r = PF\n", +"sin_phi_r = (1-PF**2)**0.5\n", +"E_s = ((E_r*cos_phi_r+IR)**2+(E_r*sin_phi_r+IX)**2)**0.5/1000 // Sending end voltage(kV)\n", +"E_s_ll = 3**0.5*E_s // Sending end line voltage(kV)\n", +"V_booster = 3**0.5*(E_s-E_r/1000) // Booster voltage between lines(kV)\n", +"tan_phi_s = (E_r*sin_phi_r+IX)/(E_r*cos_phi_r+IR) // tanΦ_s\n", +"phi_s = atand(tan_phi_s) // Φ_s(°)\n", +"cos_phi_s = cosd(phi_s) // cosΦ_s\n", +"P_s = 3.0*E_s*I*cos_phi_s // Power at sending end(kW)\n", +"loss = P_s-load*1000 // Loss(kW)\n", +"loss_per = loss/(load*1000)*100 // loss percentage\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.13 : SOLUTION :-')\n", +"printf('\nLoss in the interconnector as a percentage of power received = %.3f percent', loss_per)\n", +"printf('\nRequired voltage of the booster = %.3f kV (in terms of H.V) \n', V_booster)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')\n", +"printf('\n kVA rating of booster is not calculated in textbook and here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1: EX18_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.1 :\n", +"// Page number 330\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"rating = 1000.0 // Rating of alternator(kW)\n", +"load = 1600.0 // Total load(kW)\n", +"X_fl = 100.0 // Full load speed regulation of alernator X(%)\n", +"Y_fl = 104.0 // Full load speed regulation of alernator Y(%)\n", +"X_nl = 100.0 // No load speed regulation of alernator X(%)\n", +"Y_nl = 105.0 // No load speed regulation of alernator Y(%)\n", +"\n", +"// Calculations\n", +"h = poly(0,'h')\n", +"PB = (Y_nl-X_nl)-h\n", +"PR = rating/(Y_nl-X_nl)*PB // Load shared by machine X(kW) in terms of h\n", +"QQ = (Y_fl-X_fl)-h\n", +"RQ = rating/(Y_fl-X_fl)*QQ // Load shared by machine Y(kW) in terms of h\n", +"h_1 = roots(PR+RQ-load)\n", +"PB_1 = (Y_nl-X_nl)-h_1\n", +"PR_1 = rating/(Y_nl-X_nl)*PB_1 // Load shared by machine X(kW)\n", +"QQ_1 = (Y_fl-X_fl)-h_1\n", +"RQ_1 = rating/(Y_fl-X_fl)*QQ_1 // Load shared by machine Y(kW)\n", +"load_cease = rating/(Y_nl-X_nl) // Y cease supply load(kW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.1 : SOLUTION :-')\n", +"printf('\nLoad shared by machine X, PR = %.f kW', PR_1)\n", +"printf('\nLoad shared by machine Y, RQ = %.f kW', RQ_1)\n", +"printf('\nLoad at which machine Y ceases to supply any portion of load = %.f kW', load_cease)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.2: Synchronizing_power_and_Synchronizing_torque_for_no_load_and_full_load.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.2 :\n", +"// Page number 330-331\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 5000.0 // Rating of alternator(kVA)\n", +"N = 1500.0 // Speed(rpm)\n", +"V = 6600.0 // Voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"PF = 0.8 // Lagging power factor\n", +"x = 0.15 // Short circuit reactance\n", +"\n", +"// Calculations\n", +"E = V/3**0.5 // Phase voltage(V)\n", +"I = kVA*1000/(3**0.5*V) // Full load current of alternator(A)\n", +"V_drop = E*x // Synchronous reactance drop(V)\n", +"X = V_drop/I // Synchronous reactance per phase(ohm)\n", +"P = 120*f/N // Number of poles\n", +"n = N/60 // Speed(rps)\n", +"phi = acosd(PF) // Φ(°)\n", +"// Case(a)\n", +"theta_a = 2.0 // For a 4 pole m/c. 1 mech degree = 2 elect degree\n", +"E_s_a = E*sind(theta_a) // Synchronizing voltage(V)\n", +"I_s_a = E_s_a/X // Synchronizing current(A)\n", +"P_s_a = E*I_s_a // Synchronizing power per phase(W)\n", +"P_s_a_total = 3.0*P_s_a // Total synchronizing power(W)\n", +"P_s_a_total_kw = P_s_a_total/1000.0 // Total synchronizing power(kW)\n", +"T_s_a = P_s_a_total/(2*%pi*n) // Synchronizing torque(N-m)\n", +"// Case(b)\n", +"sin_phi = sind(phi)\n", +"OB = ((E*PF)**2+(E*sin_phi+V_drop)**2)**0.5 // Voltage(V)\n", +"E_b = OB // Voltage(V)\n", +"alpha_phi = atand((E*sin_phi+V_drop)/(E*PF)) // α+Φ(°)\n", +"alpha = alpha_phi-phi // α(°)\n", +"E_s_b = 2.0*E_b*sind(2.0/2) // Synchronizing voltage(V)\n", +"I_s_b = E_s_b/X // Synchronizing current(A)\n", +"P_s_b = E*I_s_b*cosd((alpha+1.0)) // Synchronizing power per phase(W)\n", +"P_s_b_total = 3.0*P_s_b // Total synchronizing power(W)\n", +"P_s_b_total_kw = P_s_b_total/1000.0 // Total synchronizing power(kW)\n", +"T_s_b = P_s_b_total/(2*%pi*n) // Synchronizing torque(N-m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.2 : SOLUTION :-')\n", +"printf('\nCase(a): Synchronizing power for no-load, P_s = %.1f kW', P_s_a_total_kw)\n", +"printf('\n Synchronizing torque for no-load, T_s = %.f N-m', T_s_a)\n", +"printf('\nCase(b): Synchronizing power at full-load, P_s = %.1f kW', P_s_b_total_kw)\n", +"printf('\n Synchronizing torque at full-load, T_s = %.f N-m \n', T_s_b)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3: Armature_current_EMF_and_PF_of_the_other_alternator.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.3 :\n", +"// Page number 331-332\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 6600.0 // Voltage(V)\n", +"R = 0.045 // Resistance(ohm)\n", +"X = 0.45 // Reactance(ohm)\n", +"Load = 10000.0*10**3 // Total load(W)\n", +"PF = 0.8 // Lagging power factor\n", +"I_a = 437.5 // Armature current(A)\n", +"\n", +"// Calculations\n", +"I = Load/(3**0.5*V*PF) // Load current(A)\n", +"I_working = PF*I // Working component of current(A)\n", +"I_watless = (1-PF**2)**0.5*I // Watless component of current(A)\n", +"I_second = (I_a**2+I_watless**2)**0.5 // Load current supplied by second alternator(A)\n", +"PF_second = I_a/I_second // Lagging power factor of second alternator\n", +"V_ph = V/3**0.5 // Terminal voltage per phase(V)\n", +"I_R = I_second*R // Voltage drop due to resistance(V)\n", +"I_X = I_second*X // Voltage drop due to reactance(V)\n", +"sin_phi_second = (1-PF_second**2)**0.5\n", +"E = ((V_ph+I_R*PF_second+I_X*sin_phi_second)**2+(I_X*PF_second-I_R*sin_phi_second)**2)**0.5 // EMF of the alternator(V/phase)\n", +"E_ll = 3**0.5*E // Line-to-line EMF of the alternator(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.3 : SOLUTION :-')\n", +"printf('\nArmature current of other alternator = %.1f A', I_second)\n", +"printf('\ne.m.f of other alternator = %.f V (line-to-line)', E_ll)\n", +"printf('\nPower factor of other alternator = %.3f (lagging)', PF_second)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4: EX18_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.4 :\n", +"// Page number 332-333\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X = 10.0 // Reactance(ohm)\n", +"I_a = 220.0 // Armature current(A)\n", +"PF = 1.0 // Unity power factor\n", +"V = 11000.0 // Phase voltage(V)\n", +"emf_raised = 0.2 // EMF rasied by 20%\n", +"\n", +"// Calculations\n", +"I_X = I_a*X // Reactance drop(V)\n", +"E_0 = (V**2+I_X**2)**0.5 // EMF(V)\n", +"E_00 = (1+emf_raised)*E_0 // New value of induced emf(V)\n", +"U = ((E_00**2-I_X**2)**0.5-V)/X // Current(A)\n", +"I_1 = (I_a**2+U**2)**0.5 // Current(A)\n", +"PF_1 = I_a/I_1 // Lagging power factor\n", +"I_X_2 = (E_00**2+V**2)**0.5 // Reactance drop(V)\n", +"I_2 = I_X_2/X // Current corresponding to this drop(A)\n", +"PF_2 = E_00/I_X_2 // Leading power factor\n", +"P_max = V*I_2*PF_2/1000 // Maximum power output(kW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.4 : SOLUTION :-')\n", +"printf('\nNew value of machine current = %.1f A', I_1)\n", +"printf('\nNew vaue of power factor, p.f = %.4f (lagging)', PF_1)\n", +"printf('\nPower output at which alternator break from synchronism = %.f kW', P_max)\n", +"printf('\nCurrent corresponding to maximum load = %.f A', I_2)\n", +"printf('\nPower factor corresponding to maximum load = %.4f (leading) \n', PF_2)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.5: Phase_angle_between_busbar_sections.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.5 :\n", +"// Page number 333\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 10000.0 // Voltage(V)\n", +"rating = 10000.0 // Full load rating(kW)\n", +"V_drop_per = 0.2 // Voltage drop of 20% for 10000 kW\n", +"\n", +"// Calculations\n", +"V_drop = V_drop_per*rating // Voltage drop(V)\n", +"sin_theta_2 = (V_drop/2)/V // Sin(θ/2)\n", +"theta_2 = asind(sin_theta_2) // θ/2(°)\n", +"theta = 2.0*theta_2 // Phase angle between busbar sections, θ(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.5 : SOLUTION :-')\n", +"printf('\nPhase angle between busbar sections, θ = %.2f° \n', theta)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6: Voltage_and_Power_factor_at_this_latter_station.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.6 :\n", +"// Page number 334\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_1 = 20000.0 // Total load(kW)\n", +"V = 11000.0 // Voltage(V)\n", +"PF_1 = 1.0 // Unity power factor\n", +"load_2 = 8000.0 // Load supplied(kW)\n", +"PF_2 = 0.8 // Lagging power factor\n", +"R = 0.5 // Resistance(ohm/phase)\n", +"X = 0.8 // Reactance(ohm/phase)\n", +"\n", +"// Calculations\n", +"I_1 = load_1*1000/(3**0.5*V*PF_1) // Load current(A)\n", +"I_2 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current supplied by local generators(A)\n", +"I_3 = I_1-I_2 // Current through interconnector(A)\n", +"angle_I_3 = phasemag(I_3) // Current through interconnector leads reference phasor by angle(°)\n", +"V_drop = (R+%i*X)*I_3 // Voltage drop across interconnector(V)\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"V_S = V_ph+V_drop // Sending end voltage(V/phase)\n", +"V_S_ll = 3**0.5*V_S // Sending end voltage(V)\n", +"angle_V_S_ll = phasemag(V_S_ll) // Angle of sending end voltage(°)\n", +"PF_S = cosd(angle_I_3-angle_V_S_ll) // Power factor at sending station\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.6 : SOLUTION :-')\n", +"printf('\nVoltage at this latter station = %.f∠%.2f° V (line-to-line)', abs(V_S_ll),angle_V_S_ll)\n", +"printf('\nPower factor at this latter station = %.4f (leading)', PF_S)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.7: Load_received_Power_factor_and_Phase_difference_between_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.7 :\n", +"// Page number 334\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 33000.0 // Voltage(V)\n", +"R = 0.7 // Resistance(ohm/phase)\n", +"X = 3.5 // Reactance(ohm/phase)\n", +"load_1 = 60.0 // Load on generator at station X(MW)\n", +"PF_1 = 0.8 // Lagging power factor\n", +"load_2 = 40.0 // Local load taken by consumer(MW)\n", +"PF_2 = 0.707 // Lagging power factor\n", +"\n", +"// Calculations\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"I_1 = load_1*10**6/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Load current on generator at X(A)\n", +"I_2 = load_2*10**6/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current due to local load(A)\n", +"I_3 = I_1-I_2 // Current through interconnector(A)\n", +"angle_I_3 = phasemag(I_3) // Current through interconnector leads reference phasor by angle(°)\n", +"V_drop = (R+%i*X)*I_3 // Voltage drop across interconnector(V)\n", +"V_Y = V_ph-V_drop // Voltage at Y(V)\n", +"angle_V_Y = phasemag(V_Y) // Angle of voltage at Y(°)\n", +"phase_diff = angle_I_3-angle_V_Y // Phase difference b/w Y_Y and I_3(°)\n", +"PF_Y = cosd(phase_diff) // Power factor of current received by Y\n", +"P_Y = 3*abs(V_Y*I_3)*PF_Y/1000.0 // Power received by station Y(kW)\n", +"phase_XY = abs(angle_V_Y) // Phase angle b/w voltages of X & Y(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.7 : SOLUTION :-')\n", +"printf('\nLoad received from station X to station Y = %.f kW', P_Y)\n", +"printf('\nPower factor of load received by Y = %.4f (lagging)', PF_Y)\n", +"printf('\nPhase difference between voltage of X & Y = %.2f° (lagging) \n', phase_XY)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.8: EX18_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.8 :\n", +"// Page number 335\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_tie = 11000.0 // Tie line Voltage(V)\n", +"Z = (3.5+%i*7.0) // Impedance of tie line(ohm/conductor)\n", +"V = 6600.0 // Bus bar voltage(V)\n", +"Z_per = (2.5+%i*7.5) // Percentage impedance on 1000kVA rating\n", +"kVA = 2500.0 // Load receieved by other(kVA)\n", +"\n", +"// Calculations\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"I_fl_LV = 100.0*V_tie/V_ph // LV side Full load current of each transformer(A)\n", +"R_eq = V_ph*real(Z_per)/(100*I_fl_LV) // Equivalent resistance of transformer(ohm/phase)\n", +"X_eq = 3.0*R_eq // Equivalent reactance of transformer(ohm/phase)\n", +"R_phase = real(Z)*(V/V_tie)**2 // Resistance of line per phase(ohm)\n", +"X_phase = imag(Z)*(V/V_tie)**2 // Resistance of line per phase(ohm)\n", +"R_total = 2.0*R_eq+R_phase // Total resistance per phase(ohm)\n", +"X_total = 2.0*X_eq+X_phase // Total resistance per phase(ohm)\n", +"Z_total = R_total+%i*X_total // Total impedance(ohm/phase)\n", +"I = kVA*1000/(3**0.5*V) // Load current(A)\n", +"V_drop = I*Z_total // Voltage drop per phase(V)\n", +"V_A = V_ph\n", +"V_AA = V_A+V_drop // Sending end voltage per phase(V)\n", +"V_increase = abs(V_AA)-V_A // Increase in voltage required(V/phase)\n", +"percentage_increase = V_increase/V_A*100 // Percentage increase required(%)\n", +"phase_diff = phasemag(V_AA) // Angle at which V_A & V_B are displaced(°)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.8 : SOLUTION :-')\n", +"printf('\nCase(a): Percentage increase in voltage = %.2f percent', percentage_increase)\n", +"printf('\nCase(b): Phase angle difference between the two busbar voltages = %.2f° \n', phase_diff)\n", +"printf('\nNOTE: ERROR: Several calculation mistakes in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.9: Station_power_factors_and_Phase_angle_between_two_busbar_voltages.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES\n", +"\n", +"// EXAMPLE : 11.9 :\n", +"// Page number 335-336\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X = 2.80 // Combined reactance(ohm/phase)\n", +"load_1 = 7000.0 // Consumer load at station A(kW)\n", +"PF_1 = 0.9 // Lagging power factor\n", +"V = 11000.0 // Voltage(V)\n", +"load_2 = 10000.0 // Load supplied by station B(kW)\n", +"PF_2 = 0.75 // Lagging power factor\n", +"\n", +"// Calculations\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"I_1 = load_1*10**3/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Current at A due to local load(A)\n", +"I_2 = load_2*10**3/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Current at B due to local load(A)\n", +"IA_X = 0.5*(load_1+load_2)*1000/(3**0.5*V) // Current(A)\n", +"Y_1 = 220.443/V_ph // Solved manually referring textbook\n", +"X_1 = (1-Y_1**2)**0.5\n", +"angle_1 = atand(Y_1/X_1) // Phasor lags by an angle(°)\n", +"IA_Y = (6849.09119318-V_ph*X_1)/X // Current(A)\n", +"Y_X = IA_Y/IA_X\n", +"angle_2 = atand(Y_X) // Angle by which I_A lags behind V_A(°)\n", +"PF_A = cosd(angle_2) // Power factor of station A\n", +"angle_3 = acosd(PF_2)+angle_1 // Angle by which I_2 lags V_A(°)\n", +"I_22 = load_2*10**3/(3**0.5*V*PF_2)*exp(%i*-angle_3*%pi/180) // Current(A)\n", +"I = 78.7295821622-%i*(IA_Y-177.942225747) // Current(A)\n", +"I_B = I_22-I // Current(A)\n", +"angle_4 = abs(phasemag(I_B))-angle_1 // Angle by which I_B lags behind V_B(°)\n", +"PF_B = cosd(angle_4) // Power factor of station B\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 11.9 : SOLUTION :-')\n", +"printf('\nPower factor of station A = %.4f (lagging)', PF_A)\n", +"printf('\nPower factor of station B = %.4f (lagging)', PF_B)\n", +"printf('\nPhase angle between two bus bar voltages = %.f° (V_B lagging V_A)', angle_1)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/2-THERMAL_STATIONS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/2-THERMAL_STATIONS.ipynb new file mode 100644 index 0000000..df3559a --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/2-THERMAL_STATIONS.ipynb @@ -0,0 +1,187 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: THERMAL STATIONS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1: Limiting_value_and_Coal_per_hour.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 2: THERMAL STATIONS\n", +"\n", +"// EXAMPLE : 2.1 :\n", +"// Page number 25-26\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"//Given data\n", +"M = 15000.0+10.0 // Water evaporated(kg)\n", +"C = 5000.0+5.0 // Coal consumption(kg)\n", +"time = 8.0 // Generation shift time(hours)\n", +"\n", +"//Calculations\n", +"//Case(a)\n", +"M1 = M-15000.0 \n", +"C1 = C-5000.0 \n", +"M_C = M1/C1 // Limiting value of water evaporation(kg)\n", +"//Case(b)\n", +"kWh = 0 // Station output at no load\n", +"consumption_noload = 5000+5*kWh // Coal consumption at no load(kg)\n", +"consumption_noload_hr = consumption_noload/time // Coal consumption per hour(kg)\n", +"\n", +"//Results\n", +"disp('PART I - EXAMPLE : 2.1 : SOLUTION :-')\n", +"printf('\nCase(a): Limiting value of water evaporation per kg of coal consumed, M/C = %.f kg', M_C)\n", +"printf('\nCase(b): Coal per hour for running station at no load = %.f kg\n', consumption_noload_hr)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2: Average_load_on_power_plant.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 2: THERMAL STATIONS\n", +"\n", +"// EXAMPLE : 2.2 :\n", +"// Page number 26\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"//Given data\n", +"amount = 25.0*10**5 // Amount spent in 1 year(Rs)\n", +"value_heat = 5000.0 // Heating value(kcal/kg)\n", +"cost = 500.0 // Cost of coal per ton(Rs)\n", +"n_ther = 0.35 // Thermal efficiency\n", +"n_elec = 0.9 // Electrical efficiency\n", +"\n", +"//Calculations\n", +"n = n_ther*n_elec // Overall efficiency\n", +"consumption = amount/cost*1000 // Coal consumption in 1 year(kg)\n", +"combustion = consumption*value_heat // Heat of combustion(kcal)\n", +"output = n*combustion // Heat output(kcal)\n", +"unit_gen = output/860.0 // Annual heat generated(kWh). 1 kWh = 860 kcal\n", +"hours_year = 365*24.0 // Total time in a year(hour)\n", +"load_average = unit_gen/hours_year // Average load on the power plant(kW)\n", +"\n", +"//Result\n", +"disp('PART I - EXAMPLE : 2.2 : SOLUTION :-')\n", +"printf('\nAverage load on power plant = %.2f kW\n', load_average)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the final answer in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3: Heat_balance_sheet.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 2: THERMAL STATIONS\n", +"\n", +"// EXAMPLE : 2.3 :\n", +"// Page number 26\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"//Given data\n", +"consumption = 0.5 // Coal consumption per kWh output(kg)\n", +"cal_value = 5000.0 // Calorific value(kcal/kg)\n", +"n_boiler = 0.8 // Boiler efficiency\n", +"n_elec = 0.9 // Electrical efficiency\n", +"\n", +"//Calculations\n", +"input_heat = consumption*cal_value // Heat input(kcal)\n", +"input_elec = input_heat/860.0 // Equivalent electrical energy(kWh). 1 kWh = 860 kcal\n", +"loss_boiler = input_elec*(1-n_boiler) // Boiler loss(kWh)\n", +"input_steam = input_elec-loss_boiler // Heat input to steam(kWh)\n", +"input_alter = 1/n_elec // Alternator input(kWh)\n", +"loss_alter = input_alter*(1-n_elec) // Alternate loss(kWh)\n", +"loss_turbine = input_steam-input_alter // Loss in turbine(kWh)\n", +"loss_total = loss_boiler+loss_alter+loss_turbine // Total loss(kWh)\n", +"output = 1.0 // Output(kWh)\n", +"Input = output+loss_total // Input(kWh)\n", +"\n", +"//Results\n", +"disp('PART I - EXAMPLE : 2.3 : SOLUTION :-')\n", +"printf('\nHeat Balance Sheet')\n", +"printf('\nLOSSES: Boiler loss = %.3f kWh', loss_boiler)\n", +"printf('\n Alternator loss = %.2f kWh', loss_alter)\n", +"printf('\n Turbine loss = %.3f kWh', loss_turbine)\n", +"printf('\n Total loss = %.2f kWh', loss_total)\n", +"printf('\nOUTPUT: %.1f kWh', output)\n", +"printf('\nINPUT: %.2f kWh\n', Input)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/20-WAVE_PROPAGATION_ON_TRANSMISSION_LINES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/20-WAVE_PROPAGATION_ON_TRANSMISSION_LINES.ipynb new file mode 100644 index 0000000..0743b4b --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/20-WAVE_PROPAGATION_ON_TRANSMISSION_LINES.ipynb @@ -0,0 +1,182 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20: WAVE PROPAGATION ON TRANSMISSION LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.4: Reflected_and_Transmitted_wave_of_Voltage_and_Current_at_the_junction.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 13: WAVE PROPAGATION ON TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 13.4 :\n", +"// Page number 366\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R_1 = 60.0 // Surge impedance of underground cable(ohm)\n", +"R_2 = 400.0 // Surge impedance of overhead line(ohm)\n", +"e = 100.0 // Maximum value of surge(kV)\n", +"\n", +"// Calculations\n", +"i = e*1000/R_1 // Current(A)\n", +"k = (R_2-R_1)/(R_2+R_1)\n", +"e_ref = k*e // Reflected voltage(kV)\n", +"e_trans = e+e_ref // Transmitted voltage(kV)\n", +"e_trans_alt = (1+k)*e // Transmitted voltage(kV). Alternative method\n", +"i_ref = -k*i // Reflected current(A)\n", +"i_trans = e_trans*1000/R_2 // Transmitted current(A)\n", +"i_trans_alt = (1-k)*i // Transmitted current(A). Alternative method\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 13.4 : SOLUTION :-')\n", +"printf('\nReflected voltage at the junction = %.f kV', e_ref)\n", +"printf('\nTransmitted voltage at the junction = %.f kV', e_trans)\n", +"printf('\nReflected current at the junction = %.f A', i_ref)\n", +"printf('\nTransmitted current at the junction = %.f A\n', i_trans)\n", +"printf('\nNOTE: ERROR: Calculation mistake in textbook in finding Reflected current')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.5: First_and_Second_voltages_impressed_on_C.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 13: WAVE PROPAGATION ON TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 13.5 :\n", +"// Page number 366\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R_A = 500.0 // Surge impedance of line A(ohm)\n", +"R_B = 70.0 // Surge impedance of line B(ohm)\n", +"R_C = 600.0 // Surge impedance of line C(ohm)\n", +"e = 20.0 // Rectangular voltage wave(kV)\n", +"\n", +"// Calculations\n", +"E_2 = e*(1+((R_B-R_A)/(R_B+R_A))) // Transmitted wave(kV)\n", +"E_4 = E_2*(1+((R_C-R_B)/(R_C+R_B))) // First voltage impressed on C(kV)\n", +"E_3 = E_2*(R_C-R_B)/(R_C+R_B) // Reflected wave(kV)\n", +"E_5 = E_3*(R_A-R_B)/(R_A+R_B) // Reflected wave(kV)\n", +"E_6 = E_5*(1+((R_C-R_B)/(R_C+R_B))) // Transmitted wave(kV)\n", +"second = E_4+E_6 // Second voltage impressed on C(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 13.5 : SOLUTION :-')\n", +"printf('\nFirst voltage impressed on C = %.1f kV', E_4)\n", +"printf('\nSecond voltage impressed on C = %.1f kV', second)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.6: Voltage_and_Current_in_the_cable_and_Open_wire_lines.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 13: WAVE PROPAGATION ON TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 13.6 :\n", +"// Page number 367\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Z = 100.0 // Surge impedance of cable(ohm)\n", +"Z_1 = 600.0 // Surge impedance of open wire(ohm)\n", +"Z_2 = 1000.0 // Surge impedance of open wire(ohm)\n", +"e = 2.0 // Steep fronted voltage(kV)\n", +"\n", +"// Calculations\n", +"Z_t = Z_1*Z_2/(Z_1+Z_2) // Resultant surge impedance(ohm)\n", +"E = e*(1+((Z_t-Z)/(Z_t+Z))) // Transmitted voltage(kV)\n", +"I_1 = E*1000/Z_1 // Current(A)\n", +"I_2 = E*1000/Z_2 // Current(A)\n", +"E_ref = e*(Z_t-Z)/(Z_t+Z) // Reflected voltage(kV)\n", +"I_ref = -E_ref*1000/Z // Reflected current(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 13.6 : SOLUTION :-')\n", +"printf('\nVoltage in the cable = %.3f kV', E)\n", +"printf('\nCurrent in the cable, I_1 = %.2f A', I_1)\n", +"printf('\nCurrent in the cable, I_2 = %.3f A', I_2)\n", +"printf('\nVoltage in the open-wire lines i.e Reflected voltage = %.3f kV', E_ref)\n", +"printf('\nCurrent in the open-wire lines i.e Reflected current = %.2f A', I_ref)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/21-LIGHTNING_AND_PROTECTION_AGAINST_OVERVOLTAGES_DUE_TO_LIGHTNING.ipynb b/Power_System_Engineering_by_S_Chakraborthy/21-LIGHTNING_AND_PROTECTION_AGAINST_OVERVOLTAGES_DUE_TO_LIGHTNING.ipynb new file mode 100644 index 0000000..6fd6ae3 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/21-LIGHTNING_AND_PROTECTION_AGAINST_OVERVOLTAGES_DUE_TO_LIGHTNING.ipynb @@ -0,0 +1,151 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21: LIGHTNING AND PROTECTION AGAINST OVERVOLTAGES DUE TO LIGHTNING" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.1: EX21_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 14: LIGHTNING AND PROTECTION AGAINST OVERVOLTAGES DUE TO LIGHTNING\n", +"\n", +"// EXAMPLE : 14.1 :\n", +"// Page number 382\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"RI_072 = 72000.0 // Charactersistic of lightning arrester\n", +"Z_c = 500.0 // Surge impedance(ohm)\n", +"V = 500.0 // Surge voltage(kV)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"V_a = 2.0*V // Voltage at the end of line at open-circuit(kV)\n", +"ratio_a = V_a/V // Ratio of voltage when line in open-circuited\n", +"// Case(b)\n", +"I = V*1000/Z_c // Surge current(A)\n", +"R = RI_072/(I)**0.72 // Resistance of LA(ohm)\n", +"ratio_b = R/Z_c // Ratio of voltage when line is terminated by arrester\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 14.1 : SOLUTION :-')\n", +"printf('\nCase(a): Ratio of voltages appearing at the end of a line when line is open-circuited = %.f', ratio_a)\n", +"printf('\nCase(b): Ratio of voltages appearing at the end of a line when line is terminated by arrester = %.f', ratio_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.2: Choosing_suitable_arrester_rating.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 14: LIGHTNING AND PROTECTION AGAINST OVERVOLTAGES DUE TO LIGHTNING\n", +"\n", +"// EXAMPLE : 14.2 :\n", +"// Page number 383\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"rating = 5000.0 // Rating of transformer(kVA)\n", +"V_hv = 66.0 // HV voltage(kV)\n", +"V_lv = 11.0 // LV voltage(kV)\n", +"V = 66.0 // System voltage(kV)\n", +"fluctuation = 0.1 // Voltage fluctuations\n", +"BIL = 350.0 // BIL for 66kV(kV)\n", +"dynamic_ov = 1.3 // Dynamic over-voltage = 1.3*system operating voltage\n", +"V_power_freq = 1.5 // Power frequency breakdown voltage of arrester = 1.5*arrester rating(kV)\n", +"lower_limit = 0.05 // Margin of lower limit of arrester rating\n", +"\n", +"// Calculation & Result\n", +"disp('PART II - EXAMPLE : 14.2 : SOLUTION :-')\n", +"V_rating = V*(1+fluctuation)*0.8*(1+lower_limit) // Voltage rating of arrester(kV)\n", +"if(round(V_rating)==51) then\n", +" V_rating_choosen = 50.0 // Arrester rating choosen(kV)\n", +" V_discharge = 176.0 // Discharge voltage for 50kV arrester(kV)\n", +" protective_margin = BIL-V_discharge // Protective margin available(kV)\n", +" V_power_frequency_bd = V_rating_choosen*V_power_freq // Power frequency breakdown voltage(kV)\n", +" Over_voltage_dynamic = dynamic_ov*V/3**0.5 // Dynamic overvoltage(kV)\n", +" if(V_power_frequency_bd>Over_voltage_dynamic) then\n", +" printf('\nFirst arrester with rating 50 kV (rms) & discharge voltage 176 kV chosen is suitable')\n", +" end\n", +"elseif(round(V_rating)==61) then\n", +" V_rating_choosen = 60.0 // Arrester rating choosen(kV)\n", +" V_discharge = 220.0 // Discharge voltage for 50kV arrester(kV)\n", +" protective_margin = BIL-V_discharge // Protective margin available(kV)\n", +" V_power_frequency_bd = V_rating_choosen*V_power_freq // Power frequency breakdown voltage(kV)\n", +" Over_voltage_dynamic = dynamic_ov*V/3**0.5 // Dynamic overvoltage(kV)\n", +" if(V_power_frequency_bd>Over_voltage_dynamic)\n", +" printf('\nSecond arrester with rating 60 kV (rms) & discharge voltage 220 kV chosen is suitable')\n", +" end\n", +"else(round(V_rating)==74) then\n", +" V_rating_choosen = 73.0 // Arrester rating choosen(kV)\n", +" V_discharge = 264.0 // Discharge voltage for 50kV arrester(kV)\n", +" protective_margin = BIL-V_discharge // Protective margin available(kV)\n", +" V_power_frequency_bd = V_rating_choosen*V_power_freq // Power frequency breakdown voltage(kV)\n", +" Over_voltage_dynamic = dynamic_ov*V/3**0.5 // Dynamic overvoltage(kV)\n", +" if(V_power_frequency_bd>Over_voltage_dynamic) then\n", +" printf('\nThird arrester with rating 73 kV (rms) & discharge voltage 264 kV chosen is suitable')\n", +" end\n", +"end" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/22-INSULATION_COORDINATION.ipynb b/Power_System_Engineering_by_S_Chakraborthy/22-INSULATION_COORDINATION.ipynb new file mode 100644 index 0000000..d784da7 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/22-INSULATION_COORDINATION.ipynb @@ -0,0 +1,134 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22: INSULATION COORDINATION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.1: Highest_voltage_to_which_the_transformer_is_subjected.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 15: INSULATION CO-ORDINATION\n", +"\n", +"// EXAMPLE : 15.1 :\n", +"// Page number 398-399\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 30.0 // Height of arrester located(m)\n", +"BIL = 650.0 // BIL(kV)\n", +"de_dt = 1000.0 // Rate of rising surge wave front(kV/µ-sec)\n", +"V = 132.0 // Transformer voltage at HV side(kV)\n", +"E_a = 400.0 // Discharge voltage of arrester(kV)\n", +"v = 3.0*10**8 // Velocity of surge propagation(m/sec)\n", +"\n", +"// Calculations\n", +"E_t = E_a+(2.0*de_dt*L/300) // Highest voltage the transformer is subjected(kV)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 15.1 : SOLUTION :-')\n", +"printf('\nHighest voltage to which the transformer is subjected, E_t = %.f kV', E_t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.2: Rating_of_LA_and_Location_with_respect_to_transformer.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 15: INSULATION CO-ORDINATION\n", +"\n", +"// EXAMPLE : 15.2 :\n", +"// Page number 399\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_hv = 132.0 // Voltage at the HV side of transformer(kV)\n", +"V_lv = 33.0 // Voltage at the LV side of transformer(kV)\n", +"V = 860.0 // Insulator allowable voltage(kV)\n", +"Z = 400.0 // Line surge impedance(ohm)\n", +"BIL = 550.0 // BIL(kV)\n", +"\n", +"// Calculations\n", +"V_rating_LA = V_hv*1.1*0.8 // Voltage rating of LA(kV)\n", +"E_a = 351.0 // Discharge voltage at 5 kA(kV)\n", +"I_disc = (2*V-E_a)*1000/Z // Discharge current(A)\n", +"L_1 = 37.7 // Separation distance in current b/w arrester tap and power transformer tap(m)\n", +"dist = 11.0 // Lead length from tap point to ground level(m)\n", +"de_dt = 500.0 // Maximum rate of rise of surge(kV/µ-sec)\n", +"Inductance = 1.2 // Inductance(µH/metre)\n", +"di_dt = 5000.0 // di/dt(A/µ-sec)\n", +"lead_drop = Inductance*dist*di_dt/1000 // Drop in the lead(kV)\n", +"E_d = E_a+lead_drop // (kV)\n", +"V_tr_terminal = E_d+2*de_dt*L_1/300 // Voltage at transformer terminals(kV)\n", +"E_t = BIL/1.2 // Highest voltage the transformer is subjected(kV)\n", +"L = (E_t-E_a)/(2*de_dt)*300 // Distance at which lightning arrester located from transformer(m)\n", +"L_lead = (E_t-E_a*1.1)/(2*de_dt)*300 // Distance at which lightning arrester located from transformer taken 10% lead drop(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 15.2 : SOLUTION :-')\n", +"printf('\nRating of L.A = %.1f kV', V_rating_LA)\n", +"printf('\nLocation of L.A, L = %.f m', L)\n", +"printf('\nLocation of L.A if 10 percent lead drop is considered, L = %.1f m', L_lead)\n", +"printf('\nMaximum distance at which a ligtning arrester is usually connected from transformer is %.f-%.f m', L-2,L+3)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/23-POWER_SYSTEM_GROUNDING.ipynb b/Power_System_Engineering_by_S_Chakraborthy/23-POWER_SYSTEM_GROUNDING.ipynb new file mode 100644 index 0000000..184b2df --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/23-POWER_SYSTEM_GROUNDING.ipynb @@ -0,0 +1,79 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23: POWER SYSTEM GROUNDING" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.1: Inductance_and_Rating_of_arc_suppression_coil.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 16: POWER SYSTEM GROUNDING\n", +"\n", +"// EXAMPLE : 16.1 :\n", +"// Page number 409\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 132.0*10**3 // Voltage(V)\n", +"n = 3.0 // Number of phase\n", +"f = 50.0 // Frequency(Hz)\n", +"l = 50.0 // Line length(km)\n", +"C = 0.0157*10**-6 // Capacitance to earth(F/km)\n", +"\n", +"// Calculations\n", +"L = 1/(n*(2*%pi*f)**2*C*l) // Inductance(H)\n", +"X_L = 2*%pi*f*L // Reactance(ohm)\n", +"I_F = V/(3**0.5*X_L) // Current(A)\n", +"rating = I_F*V/(3**0.5*1000) // Rating of arc suppression coil(kVA)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 16.1 : SOLUTION :-')\n", +"printf('\nInductance, L = %.1f Henry', L)\n", +"printf('\nRating of arc suppression coil = %.f kVA \n', rating)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/24-ELECTRIC_POWER_SUPPLY_SYSTEMS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/24-ELECTRIC_POWER_SUPPLY_SYSTEMS.ipynb new file mode 100644 index 0000000..a53952c --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/24-ELECTRIC_POWER_SUPPLY_SYSTEMS.ipynb @@ -0,0 +1,481 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24: ELECTRIC POWER SUPPLY SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.1: EX24_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.1 :\n", +"// Page number 422-423\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"no_phase = 3.0 // Number of phases in ac transmission system\n", +"V = 380.0*10**3 // Voltage b/w lines(V)\n", +"load = 100.0 // Load(MW)\n", +"PF = 0.9 // Power factor\n", +"l = 150.0 // Line length(km)\n", +"n = 0.92 // Efficiency\n", +"r = 0.045 // Resistance(ohm/km/sq.cm)\n", +"w_cu_1 = 0.01 // Weight of 1 cm^3 copper(kg)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"P_loss = (1-n)*load // Power loss in the line(MW)\n", +"I_L = load*10**6/(3**0.5*V*PF) // Line current(A)\n", +"loss_cu = P_loss/no_phase*10**6 // I^2*R loss per conductor(W)\n", +"R = loss_cu/I_L**2 // Resistance per conductor(ohm)\n", +"R_km = R/l // Resistance per conductor per km(ohm)\n", +"area = r/R_km // Conductor area(Sq.cm)\n", +"volume = area*100.0 // Volume of copper per km run(cm^3)\n", +"W_cu_km = volume*w_cu_1 // Weight of copper per km run(kg)\n", +"W_cu = no_phase*l*1000*W_cu_km // Weight of copper for 3 conductors of 150 km(kg)\n", +"// Case(ii)\n", +"W_cu_dc = 1.0/2*PF**2*W_cu // Weight of copper conductor in dc(kg)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.1 : SOLUTION :-')\n", +"printf('\nWeight of copper required for a three-phase transmission system = %.f kg', W_cu)\n", +"printf('\nWeight of copper required for the d-c transmission system = %.f kg \n', W_cu_dc)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.2: Percentage_increase_in_power_transmitted.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.2 :\n", +"// Page number 423\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P_1 = 1.0 // Assume P1 to be 1\n", +"\n", +"// Calculations\n", +"P_2 = (3.0*2)**0.5 // 3-phase power transmitted in terms of P_1\n", +"inc_per = (P_2-P_1)/P_1*100 // Increase in power transmitted(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.2 : SOLUTION :-')\n", +"printf('\nPercentage increase in power transmitted = %.f percent', inc_per)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.3: Percentage_additional_balanced_load.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.3 :\n", +"// Page number 424\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"PF = 0.95 // Lagging power factor\n", +"\n", +"// Calculations\n", +"P_1 = 1.0 // Power in terms of V*I_1\n", +"P_2 = 2.0*PF**2 // Power in terms of V*I_1\n", +"P_additional_percentage = (P_2-P_1)/P_1*100 // Percentage additional power transmitted in a 3-phase 3-wire system\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.3 : SOLUTION :-')\n", +"printf('\nPercentage additional power transmitted in a 3-phase 3-wire system = %.f percent', P_additional_percentage)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.4: EX24_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.4 :\n", +"// Page number 424-425\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 3.0 // 3-phase 4 wire ac system\n", +"\n", +"// Calculations\n", +"a2_a1 = 1.0/6 // Ratio of cross-sectional area of 2 wire dc to 3-phase 4-wire system\n", +"ratio_cu = 3.5/2*a2_a1 // Copper for 3 phase 4 wire system to copper for 2 wire dc system\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.4 : SOLUTION :-')\n", +"printf('\nCopper for 3-phase 4-wire system/Copper for 2-wire dc system = %.3f : 1', ratio_cu)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.5: Weight_of_copper_required_and_Reduction_of_weight_of_copper_possible.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.5 :\n", +"// Page number 425\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 60.0 // Line length(km)\n", +"P = 5.0 // Load(MW)\n", +"PF = 0.8 // Lagging power factor\n", +"V = 33.0*10**3 // Voltage(V)\n", +"n = 0.85 // Transmission efficiency\n", +"rho = 1.73*10**-8 // Specific resistance of copper(ohm-mt)\n", +"density = 8900.0 // Density(kg/mt^3)\n", +"\n", +"// Calculations\n", +"I = P*10**6/(3**0.5*V*PF) // Line current(A)\n", +"line_loss = (1-n)*P*1000/n // Line loss(kW)\n", +"line_loss_phase = line_loss/3.0 // Line loss/phase(kW)\n", +"R = line_loss_phase*1000/I**2 // Resistance/phase(ohm)\n", +"a = rho*L*1000/R // Area of cross section of conductor(m^2)\n", +"volume = 3.0*a*L*1000 // Volume of copper(m^3)\n", +"W_cu = volume*density // Weight of copper in 3-phase system(kg)\n", +"I_1 = P*10**6/V // Current in single phase system(A)\n", +"R_1 = line_loss*1000/(2*I_1**2) // Resistance in single phase system(ohm)\n", +"a_1 = rho*L*1000/R_1 // Area of cross section of conductor in single phase system(m^2)\n", +"volume_1 = 2.0*a_1*L*1000 // Volume of copper(m^3)\n", +"W_cu_1 = volume_1*density // Weight of copper in 1-phase system(kg)\n", +"reduction_cu = (W_cu-W_cu_1)/W_cu*100 // Reduction in copper(%)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.5 : SOLUTION :-')\n", +"printf('\nWeight of copper required for 3-phase 2-wire system = %.2e kg', W_cu)\n", +"printf('\nReduction of weight of copper possible = %.1f percent \n', reduction_cu)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.6: Economical_cross_section_of_a_3_core_distributor_cable.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.6 :\n", +"// Page number 427-428\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 250.0 // Cable length(m)\n", +"P = 80.0*10**3 // Load(W)\n", +"V = 400.0 // Voltage(V)\n", +"PF = 0.8 // Lagging power factor\n", +"time = 4000.0 // Time of operation(hours/annum)\n", +"a = poly(0,'a') // Area of each conductor(Sq.cm)\n", +"cost_instal = 15.0*a+25 // Cost of cable including installation(Rs/m)\n", +"interest_per = 0.1 // Interest & depreciation\n", +"cost_waste_per = 0.1 // Cost of energy wasted(Rs/unit)\n", +"r = 0.173 // Resistance per km of 1 cm^2(ohm)\n", +"\n", +"// Calculations\n", +"I = P/(3**0.5*V*PF) // Line current(A)\n", +"energy_waste = 3.0*I**2*r/a*L*10**-3*time*10**-3 // Energy wasted per annum(kWh)\n", +"cost_energy_waste = cost_waste_per*energy_waste // Annual cost of energy wasted as losses(Rs)\n", +"capitaL_cost_cable = cost_instal*L // Capital cost of cable(Rs)\n", +"annual_cost_cable = capitaL_cost_cable*cost_waste_per // Annual cost on cable(Rs)\n", +"area = (1081.25/375)**0.5 // Area = a(Sq.cm). Simplified and taken final answer\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.6 : SOLUTION :-')\n", +"printf('\nEconomical cross-section of a 3-core distributor cable, a = %.1f cm^2', area)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.7: Most_economical_cross_section.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.7 :\n", +"// Page number 428\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 110.0*10**3 // Voltage(V)\n", +"l_1 = 24.0*10**6 // Load(MW)\n", +"t_1 = 6.0 // Time(hours)\n", +"l_2 = 8.0*10**6 // Load(MW)\n", +"t_2 = 6.0 // Time(hours)\n", +"l_3 = 4.0*10**6 // Load(MW)\n", +"t_3 = 12.0 // Time(hours)\n", +"PF = 0.8 // Lagging power factor\n", +"a = poly(0,'a') // Cross-section of each conductor(Sq.cm)\n", +"cost_line = 12000.0+8000*a // Cost of line including erection(Rs/km)\n", +"R = 0.19/a // Resistance per km of each conductor(ohm)\n", +"cost_energy = 8.0/100 // Energy cost(Rs/unit)\n", +"interest_per = 0.1 // Interest & depreciation. Assumption\n", +"\n", +"// Calculations\n", +"annual_charge = interest_per*cost_line // Total annual charge(Rs)\n", +"I_1 = l_1/(3**0.5*V*PF) // Line current for load 1(A)\n", +"I_2 = l_2/(3**0.5*V*PF) // Line current for load 2(A)\n", +"I_3 = l_3/(3**0.5*V*PF) // Line current for load 3(A)\n", +"I_2_t = I_1**2*t_1+I_2**2*t_2+I_3**2*t_3 // I^2*t\n", +"annual_energy = 3.0*R*365/1000*I_2_t // Annual energy consumption on account of losses(kWh)\n", +"cost_waste = annual_energy*cost_energy // Cost of energy wasted per annum(Rs)\n", +"area = (2888.62809917355/800.0)**0.5 // Economical cross-section = a(Sq.cm). Simplified and taken final answer\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.7 : SOLUTION :-')\n", +"printf('\nMost economical cross-section, a = %.2f cm^2', area)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.8: Most_economical_current_density_for_the_transmission_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.8 :\n", +"// Page number 428-429\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cost_km_cu = 2800.0 // Cost per km for each copper conductor of sq.cm(Rs)\n", +"LF_I = 80.0/100 // Load factor of load current\n", +"LF_loss = 65.0/100 // Load factor of losses\n", +"interest_per = 10.0/100 // Rate of interest and depreciation\n", +"cost_energy = 5.0/100 // Cost of energy(Rs/kWh)\n", +"rho = 1.78*10**-8 // Resistivity(ohm-m)\n", +"\n", +"// Calculations\n", +"P_2 = cost_km_cu*interest_per // Cost in terms of L(Rs)\n", +"time_year = 365.0*24 // Total hours in a year\n", +"P_3 = cost_energy*rho*10**4*time_year*LF_loss // Cost in terms of I^2 & L(Rs)\n", +"delta = (P_2/P_3)**0.5 // Economical current density for the transmission line(A/sq.cm)\n", +" \n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.8 : SOLUTION :-')\n", +"printf('\nMost economical current density for the transmission line, δ = %.f A/sq.cm', delta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.9: Most_economical_cross_section_of_the_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS\n", +"\n", +"// EXAMPLE : 17.9 :\n", +"// Page number 429\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 1000.0 // Maximum demand(kW)\n", +"energy_cons = 5.0*10**6 // Annual energy consumption(kWh)\n", +"PF = 0.85 // Power factor\n", +"capital_cost = 80000.0 // Capital cost of cable(Rs/km)\n", +"cost_energy = 5.0/100 // Energy cost(Rs/kWh)\n", +"interest_per = 10.0/100 // Rate of interest and depreciation\n", +"r_specific = 1.72*10**-6 // Specific resistance of copper(ohm/cubic.cm)\n", +"V = 11.0 // Voltage(kV)\n", +"\n", +"// Calculations\n", +"I = MD/(3**0.5*V*PF) // Line current corresponding to maximum demand(A)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"LF = energy_cons/(MD*hours_year) // Load factor\n", +"loss_LF = 0.25*LF+0.75*LF**2 // Loss load factor\n", +"P_2 = capital_cost*interest_per // Cost in terms of L(Rs)\n", +"P_3 = 3.0*I**2*r_specific*10**4*hours_year*loss_LF*cost_energy // Cost in terms of I^2 & L(Rs)\n", +"a = (P_3/P_2)**0.5 // Most economical cross-section of conductor(sq.cm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 17.9 : SOLUTION :-')\n", +"printf('\nMost economical cross-section of the conductor, a = %.2f cm^2 \n', a)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook solution')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/25-POWER_DISTRIBUTION_SYSTEMS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/25-POWER_DISTRIBUTION_SYSTEMS.ipynb new file mode 100644 index 0000000..32f6c43 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/25-POWER_DISTRIBUTION_SYSTEMS.ipynb @@ -0,0 +1,534 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25: POWER DISTRIBUTION SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.1: Potential_of_O_and_Current_leaving_each_supply_point.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.1 :\n", +"// Page number 437\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_A = 225.0 // Potential at point A(V)\n", +"R_A = 5.0 // Resistance of line A(ohm)\n", +"V_B = 210.0 // Potential at point B(V)\n", +"R_B = 1.0 // Resistance of line B(ohm)\n", +"V_C = 230.0 // Potential at point C(V)\n", +"R_C = 1.0 // Resistance of line C(ohm)\n", +"V_D = 230.0 // Potential at point D(V)\n", +"R_D = 2.0 // Resistance of line D(ohm)\n", +"V_E = 240.0 // Potential at point E(V)\n", +"R_E = 2.0 // Resistance of line E(ohm)\n", +"\n", +"// Calculations\n", +"V_0 = ((V_A/R_A)+(V_B/R_B)+(V_C/R_C)+(V_D/R_D)+(V_E/R_E))/((1/R_A)+(1/R_B)+(1/R_C)+(1/R_D)+(1/R_E)) // Potential at point O(V)\n", +"I_A = (V_A-V_0)/R_A // Current leaving supply point A(A)\n", +"I_B = (V_B-V_0)/R_B // Current leaving supply point B(A)\n", +"I_C = (V_C-V_0)/R_C // Current leaving supply point C(A)\n", +"I_D = (V_D-V_0)/R_D // Current leaving supply point D(A)\n", +"I_E = (V_E-V_0)/R_E // Current leaving supply point E(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.1 : SOLUTION :-')\n", +"printf('\nPotential of point O, V_0 = %.f V', V_0)\n", +"printf('\nCurrent leaving supply point A, I_A = %.f A', I_A)\n", +"printf('\nCurrent leaving supply point B, I_B = %.f A', I_B)\n", +"printf('\nCurrent leaving supply point C, I_C = %.f A', I_C)\n", +"printf('\nCurrent leaving supply point D, I_D = %.2f A', I_D)\n", +"printf('\nCurrent leaving supply point E, I_E = %.2f A', I_E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.2: EX25_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.2 :\n", +"// Page number 437-438\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I = 600.0 // Constant current drawn(A)\n", +"D = 8.0 // Distance b/w two sub-stations(km)\n", +"V_A = 575.0 // Potential at point A(V)\n", +"V_B = 590.0 // Potential at point B(V)\n", +"R = 0.04 // Track resistance(ohm/km)\n", +"\n", +"// Calculations\n", +"x = poly(0,'x') // x(km)\n", +"I_A = ((-V_B+R*I*D+V_A)-(R*I)*x)/(D*R) // Simplifying\n", +"V_P = V_A-I_A*R*x // Potential at P in terms of x(V)\n", +"dVP_dx = derivat(V_P) // dV_P/dx\n", +"x_sol = roots(dVP_dx) // Value of x(km)\n", +"I_A_1 = ((-V_B+R*I*D+V_A)-(R*I)*x_sol)/(D*R) // Current drawn from end A(A)\n", +"I_B = I-I_A_1 // Current drawn from end B(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.2 : SOLUTION :-')\n", +"printf('\nPoint of minimum potential along the track, x = %.2f km', x_sol)\n", +"printf('\nCurrent supplied by station A, I_A = %.f A', I_A_1)\n", +"printf('\nCurrent supplied by station B, I_B = %.f A \n', I_B)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.3: Position_of_lowest_run_lamp_and_its_Voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.3 :\n", +"// Page number 438-439\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 400.0 // Length of cable(m)\n", +"i = 1.0 // Load(A/m)\n", +"I_1 = 120.0 // Current at 40m from end A(A)\n", +"l_1 = 40.0 // Distance from end A(A)\n", +"I_2 = 72.0 // Current at 72m from end A(A)\n", +"l_2 = 120.0 // Distance from end A(A)\n", +"I_3 = 48.0 // Current at 200m from end A(A)\n", +"l_3 = 200.0 // Distance from end A(A)\n", +"I_4 = 120.0 // Current at 320m from end A(A)\n", +"l_4 = 320.0 // Distance from end A(A)\n", +"r = 0.15 // Cable resistance(ohm/km)\n", +"V_A = 250.0 // Voltage at end A(A)\n", +"V_B = 250.0 // Voltage at end A(A)\n", +"\n", +"// Calculations\n", +"I = poly(0,'I') // Current from end A(A)\n", +"A_A1 = l_1*r*(I-(1.0/2)*i*l_1) // Drop over length(V)\n", +"I_d_1 = 40.0 // Distributed tapped off current(A)\n", +"I_A1_A2 = I-l_1-l_2 // Current fed in over length(A)\n", +"A1_A2 = (l_2-l_1)*r*(I_A1_A2-(1.0/2)*i*(l_2-l_1)) // Drop over length(V)\n", +"I_d_2 = 80.0 // Distributed tapped off current(A)\n", +"I_A2_A3 = I_A1_A2-(I_2+I_d_2) // Current fed in over length(A)\n", +"A2_A3 = (l_3-l_2)*r*(I_A2_A3-(1.0/2)*i*(l_3-l_2)) // Drop over length(V)\n", +"I_d_3 = 80.0 // Distributed tapped off current(A)\n", +"I_A3_A4 = I_A2_A3-(I_3+I_d_3) // Current fed in over length(A)\n", +"A3_A4 = (l_4-l_3)*r*(I_A3_A4-(1.0/2)*i*(l_4-l_3)) // Drop over length(V)\n", +"I_d_4 = 120.0 // Distributed tapped off current(A)\n", +"I_A4_B = I_A3_A4-(I_4+I_d_4) // Current fed in over length(A)\n", +"A4_B = (l-l_4)*r*(I_A4_B-(1.0/2)*i*(l-l_4)) // Drop over length(V)\n", +"V_drop = A_A1+A1_A2+A2_A3+A3_A4+A4_B // Total voltage drop in terms of I\n", +"I = roots(V_drop) // Current(A)\n", +"I_total = 760.0 // Total load current(A)\n", +"I_B = I_total-I // Current from B(A)\n", +"A_A3 = 2.0*r/1000*(l_1*(I-20)+(l_2-l_1)*(I-200)+(l_3-l_2)*(I-352)) // Potential drop over length A_A3(V)\n", +"V_A3 = V_A-A_A3 // Voltage at the lowest run lamp(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.3 : SOLUTION :-')\n", +"printf('\nPosition of lowest-run lamp, A_3 = %.f m', l_3)\n", +"printf('\nVoltage at the lowest-run lamp = %.1f V', V_A3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.4: Point_of_minimum_potential_and_its_Potential.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.4 :\n", +"// Page number 439\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 450.0 // Length of wire(m)\n", +"V_A = 250.0 // Voltage at end A(V)\n", +"V_B = 250.0 // Voltage at end A(V)\n", +"r = 0.05 // Conductor resistance(ohm/km)\n", +"i = 1.5 // Load(A/m)\n", +"I_C = 20.0 // Current at C(A)\n", +"l_C = 60.0 // Distance to C from A(m)\n", +"I_D = 40.0 // Current at D(A)\n", +"l_D = 100.0 // Distance to D from A(m)\n", +"l_E = 200.0 // Distance to E from A(m)\n", +"\n", +"// Calculations\n", +"x = poly(0,'x') // Current to point D from end A(A)\n", +"AD = (I_C+x)*r*l_C+x*r*(l_D-l_C) // Drop in length AD\n", +"BD = (i*r*V_A**2/2)+(I_D-x)*r*(450-l_D) // Drop in length BD\n", +"x_sol = roots(AD-BD) // Current(A)\n", +"I_F = x_sol-I_D // Current supplied to load from end A(A)\n", +"l_F = l_E+(I_F/i) // Point of minimum potential at F from A(m)\n", +"V_F = V_B-(375.0-I_F)*(250-(l_F-200))*r/1000 // Potential at F from end B(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.4 : SOLUTION :-')\n", +"printf('\nPoint of minimum potential occurs at F from A = %.2f metres', l_F)\n", +"printf('\nPotential at point F = %.2f V', V_F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.6: Ratio_of_weight_of_copper_with_and_without_interconnector.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.6 :\n", +"// Page number 440-441\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l_AB = 100.0 // Length between A & B(m)\n", +"l_BC = 150.0 // Length between B & C(m)\n", +"l_CD = 200.0 // Length between C & D(m)\n", +"l_AD = 350.0 // Length between A & D(m)\n", +"l_AE = 200.0 // Length between A & E(m)\n", +"l_ED = 250.0 // Length between E & D(m)\n", +"I_B = 10.0 // Current at B(A)\n", +"I_C = 20.0 // Current at C(A)\n", +"I_D = 50.0 // Current at D(A)\n", +"I_E = 39.0 // Current at E(A)\n", +"\n", +"// Calculations\n", +"x = poly(0,'x') // Current in section AB(A)\n", +"ABCDEA = x*l_AB+(x-I_B)*l_BC+(x-I_B-I_C)*l_CD+(x-I_B-I_C-I_D)*l_ED+(x-I_B-I_C-I_D-I_E)*l_AE // KVL around loop ABCDEA\n", +"x_sol = roots(ABCDEA) // Current in section AB(A)\n", +"V_AD = x_sol*l_AB+(x_sol-I_B)*l_BC+(x_sol-I_B-I_C)*l_CD // Voltage drop from A to D in terms of ρ/a_1(V)\n", +"R_AD = (l_AB+l_BC+l_CD)*(l_AE+l_ED)/(l_AB+l_BC+l_CD+l_AE+l_ED) // Resistance of n/w across terminals AD in terms of ρ/a\n", +"I_AD = V_AD/(R_AD+l_AD) // Current in interconnector AD(A)\n", +"V_A_D = I_AD*l_AD // Voltage drop between A & D in terms of ρ/a_2\n", +"a2_a1 = V_A_D/V_AD\n", +"length_with = (l_AB+l_BC+l_CD+l_AE+l_ED+l_AD) // Length of conductor with interconnector(m)\n", +"length_without = (l_AB+l_BC+l_CD+l_AE+l_ED) // Length of conductor without interconnector(m)\n", +"volume_with = a2_a1*length_with/length_without // Weight of copper with interconnector\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.6 : SOLUTION :-')\n", +"printf('\nRatio of weight of copper with & without interconnector = %.3f : 1 (or) 1 : %.2f', volume_with,1/volume_with)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.7: Potential_difference_at_each_load_point.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.7 :\n", +"// Page number 441-442\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"r_out = 0.05 // Resistance of each outer per 100 metre length(ohm)\n", +"r_neutral = 0.10 // Resistance of each neutral per 100 metre length(ohm)\n", +"V_A = 200.0 // Potential at point A(V)\n", +"V_B = 200.0 // Potential at point B(V)\n", +"l_AC = 100.0 // Length between A & C(m)\n", +"l_CD = 150.0 // Length between C & D(m)\n", +"l_DB = 200.0 // Length between D & B(m)\n", +"l_AF = 200.0 // Length between A & F(m)\n", +"l_FE = 100.0 // Length between F & E(m)\n", +"l_EB = 150.0 // Length between E & B(m)\n", +"I_C = 20.0 // Current at point C(A)\n", +"I_D = 30.0 // Current at point D(A)\n", +"I_F = 60.0 // Current at point F(A)\n", +"I_E = 40.0 // Current at point E(A)\n", +"\n", +"// Calculations\n", +"x = poly(0,'x') // Current in positive outer alone(A)\n", +"equ_1 = r_out*(l_DB*(I_D-x))-r_out*(l_AC*(I_C+x)+l_CD*x)\n", +"x_sol = roots(equ_1) // Current in positive outer alone(A)\n", +"y = poly(0,'y') // Current in negative outer alone(A)\n", +"equ_2 = r_out*((I_E-y)*l_FE+(I_E+I_F-y)*l_AF)-r_out*(l_EB*y)\n", +"y_sol = roots(equ_2) // Current in negative outer alone(A)\n", +"I_pos_out = I_C+x_sol // Current entering positive outer(A)\n", +"I_neg_out = I_E+I_F-y_sol // Current returning via negative outer(A)\n", +"I_middle = I_neg_out-I_pos_out // Current in the middle wire towards G(A)\n", +"r_CD = r_out*l_CD/100.0 // Resistance between C & D(ohm)\n", +"r_D = r_out*l_DB/100.0 // Resistance between D & B(ohm)\n", +"r_IH = r_neutral*l_FE*0.5/100.0 // Resistance between I & H(ohm)\n", +"r_IJ = r_neutral*l_FE*0.5/100.0 // Resistance between I & J(ohm)\n", +"r_GH = r_neutral*l_AF*0.5/100.0 // Resistance between G & H(ohm)\n", +"r_AF = r_out*l_AF/100.0 // Resistance between A & F(ohm)\n", +"I_CD = x_sol // Current flowing into D from C(A)\n", +"I_out_D = I_D-x_sol // Current flowing into D from outer side(A)\n", +"I_GH = I_C+I_middle // Current flowing into H from G(A)\n", +"I_IH = I_F-I_GH // Current flowing into H from I(A)\n", +"I_BJ = I_E-(I_D-I_IH) // Current flowing into J from B(A)\n", +"I_FE = y_sol-I_E // Current flowing into E from F(A)\n", +"I_IJ = I_D-I_IH // Current flowing into J from I(A)\n", +"V_C = V_A-(I_pos_out*r_out-I_middle*r_neutral) // Potential at load point C(A)\n", +"V_D = V_C-(I_CD*r_CD+I_IH*r_IH-I_GH*r_GH) // Potential at load point D(A)\n", +"V_F = V_A-(I_middle*r_neutral+I_GH*r_neutral+I_neg_out*r_AF) // Potential at load point F(A)\n", +"V_E = V_F-(-I_IH*r_IH+I_IJ*r_IJ-I_FE*r_out) // Potential at load point E(A)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.7 : SOLUTION :-')\n", +"printf('\nPotential difference at load point C = %.3f V', V_C)\n", +"printf('\nPotential difference at load point D = %.3f V', V_D)\n", +"printf('\nPotential difference at load point E = %.3f V', V_E)\n", +"printf('\nPotential difference at load point F = %.3f V', V_F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.8: Load_on_the_main_generators_and_On_each_balancer_machine.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.8 :\n", +"// Page number 442-443\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 440.0 // Voltage between outer(V)\n", +"I_pos = 210.0 // Ligting load current on positive side(A)\n", +"I_neg = 337.0 // Ligting load current on negative side(A)\n", +"I_power = 400.0 // Power load current(A)\n", +"P_loss = 1.5 // Loss in each balancer machine(kW)\n", +"\n", +"// Calculations\n", +"P = I_power*V/1000.0 // Power(kW)\n", +"load_pos = I_pos*V*0.5/1000.0 // Load on positive side(kW)\n", +"load_neg = I_neg*V*0.5/1000.0 // Load on negative side(kW)\n", +"loss_total = 2*P_loss // Total loss on rotary balancer set(kW)\n", +"load_main = P+load_pos+load_neg+loss_total // Load on main machine(kW)\n", +"I = load_main*1000/V // Current(A)\n", +"I_M = I-610.0 // Current through balancer machine(A)\n", +"I_G = 127.0-I_M // Current through generator(A)\n", +"output_G = I_G*V*0.5/1000.0 // Output of generator(kW)\n", +"input_M = I_M*V*0.5/1000.0 // Input to balancer machine(kW)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.8 : SOLUTION :-')\n", +"printf('\nLoad on the main machine = %.2f kW', load_main)\n", +"printf('\nOutput of generator = %.2f kW', output_G)\n", +"printf('\nInput to balancer machine = %.2f kW', input_M)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.9: Currents_in_various_sections_and_Voltage_at_load_point_C.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 18: POWER DISTRIBUTION SYSTEMS\n", +"\n", +"// EXAMPLE : 18.9 :\n", +"// Page number 444\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_a = 11.0*10**3 // Line voltage at A(V)\n", +"Z_AB = complex(1.0,0.8) // Impedance between A & B(ohm)\n", +"Z_AC = complex(3.0,2.0) // Impedance between A & C(ohm)\n", +"Z_BD = complex(3.0,4.0) // Impedance between B & D(ohm)\n", +"Z_CD = complex(1.0,0.7) // Impedance between C & D(ohm)\n", +"I_B = 60.0 // Current at B(A)\n", +"I_C = 30.0 // Current at C(A)\n", +"I_D = 50.0 // Current at D(A)\n", +"pf_B = 0.8 // Power factor at B\n", +"pf_C = 0.9 // Power factor at C\n", +"pf_D = 0.707 // Power factor at D\n", +"\n", +"// Calculations\n", +"sin_phi_B = (1-pf_B**2)**0.5\n", +"I_B1 = I_B*(pf_B-%i*sin_phi_B) // Load current(A)\n", +"sin_phi_C = (1-pf_C**2)**0.5\n", +"I_C1 = I_C*(pf_C-%i*sin_phi_C) // Load current(A)\n", +"sin_phi_D = (1-pf_D**2)**0.5\n", +"I_D1 = I_D*(pf_D-%i*sin_phi_D) // Load current(A)\n", +"V_A = V_a/3**0.5 // Phase voltage at A(V)\n", +"I_AC = I_C1 // Current in section AC when C & D is removed(A)\n", +"I_BD = I_D1 // Current in section BD when C & D is removed(A)\n", +"I_AB = I_B1+I_D1 // Current in section AB when C & D is removed(A)\n", +"V_AC_drop = I_AC*Z_AC // Voltage drop at section AC(V)\n", +"V_AB_drop = I_AB*Z_AB // Voltage drop at section AB(V)\n", +"V_BD_drop = I_BD*Z_BD // Voltage drop at section BD(V)\n", +"V_drop_D = V_BD_drop+V_AB_drop // Total drop upto D(V)\n", +"pd_CD = V_drop_D-V_AC_drop // Potential difference between C & D(V)\n", +"Z_C_D = Z_AB+Z_BD+Z_AC // Impedance of network looking from terminal C & D(ohm)\n", +"I_CD = pd_CD/(Z_C_D+Z_CD) // Current flowing in section CD(A)\n", +"I_AC = I_CD+I_C1 // Current flowing in section AC(A)\n", +"I_BD = I_D1-I_CD // Current flowing in section BD(A)\n", +"I_AB = I_BD+I_B1 // Current flowing in section AB(A)\n", +"V_drop_AC = I_AC*Z_AC // Drop caused by current flowing in section AC(V/phase)\n", +"V_drop_AC_line = V_drop_AC*3**0.5 // Drop caused by current flowing in section AC(V)\n", +"V_C = V_a-V_drop_AC_line // Voltage at C(V)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 18.9 : SOLUTION :-')\n", +"printf('\nCurrent in section CD, I_CD = (%.2f%.2fj) A', real(I_CD),imag(I_CD))\n", +"printf('\nCurrent in section AC, I_AC = (%.2f%.2fj) A', real(I_AC),imag(I_AC))\n", +"printf('\nCurrent in section BD, I_BD = (%.2f%.2fj) A', real(I_BD),imag(I_BD))\n", +"printf('\nCurrent in section AB, I_AB = (%.2f%.2fj) A', real(I_AB),imag(I_AB))\n", +"printf('\nVoltage at load point C = %.2f∠%.2f° kV', abs(V_C)/1000,phasemag(V_C))" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/27-SYMMETRICAL_SHORT_CIRCUIT_CAPACITY_CALCULATIONS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/27-SYMMETRICAL_SHORT_CIRCUIT_CAPACITY_CALCULATIONS.ipynb new file mode 100644 index 0000000..b7fbb90 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/27-SYMMETRICAL_SHORT_CIRCUIT_CAPACITY_CALCULATIONS.ipynb @@ -0,0 +1,782 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.10: Initial_symmetrical_rms_current_at_the_generator_terminal.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.10 :\n", +"// Page number 472\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_alt = 625.0 // Alternator rating(kVA)\n", +"V_alt = 480.0 // Voltage rating of alternator(V)\n", +"load = 500.0 // Load(kW)\n", +"V_load = 480.0 // Load voltage(V)\n", +"X_st = 8.0/100 // Sub-transient reactance\n", +"\n", +"// Calculations\n", +"kVA_base = 625.0 // Base kVA\n", +"V_base = 480.0 // Base voltage(V)\n", +"I_load = load/kVA_base // Load cuurent(A)\n", +"V = 1.0 // Terminal voltage(p.u)\n", +"E_st = V+%i*I_load*X_st // Sub-transient voltage(p.u)\n", +"I_st = E_st/(%i*X_st) // Sub-transient current(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.10 : SOLUTION :-')\n", +"printf('\nInitial symmetrical rms current at the generator terminal = (%.1f%.1fj) p.u', real(I_st),imag(I_st))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.11: Sub_transient_current_in_the_fault_in_generator_and_Motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.11 :\n", +"// Page number 472-473\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X_d_st_G = 0.15 // Sub-transient reactance of generator(p.u)\n", +"X_d_st_M = 0.45 // Sub-transient reactance of motor(p.u)\n", +"X = 0.10 // Leakage reactance of transformer(p.u)\n", +"V = 0.9 // Terminal voltage of the generator(p.u)\n", +"I_G = 1.0 // Output current of the generator(p.u)\n", +"PF = 0.8 // Power factor of the load\n", +"\n", +"// Calculations\n", +"sin_phi = (1-PF**2)**0.5\n", +"I = I_G*(PF+%i*sin_phi) // Load current(p.u)\n", +"E_st_G = V+%i*I*X_d_st_G // Sub-transient voltage of the generator(p.u)\n", +"E_st_M = V-%i*I*X_d_st_M // Sub-transient voltage of the motor(p.u)\n", +"I_st_g = E_st_G/(%i*(X_d_st_G+X)) // Sub-transient current in the generator at fault(p.u)\n", +"I_st_m = E_st_M/(%i*(X_d_st_M-X)) // Sub-transient current in the motor at fault(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.11 : SOLUTION :-')\n", +"printf('\nCase(a): Sub-transient current in the fault in generator = %.3f∠%.3f° p.u', abs(I_st_g),phasemag(I_st_g))\n", +"printf('\nCase(b): Sub-transient current in the fault in motor = %.3f∠%.2f° p.u \n', abs(I_st_m),180+phasemag(I_st_m))\n", +"printf('\nNOTE: ERROR: Sub-transient reactance of motor is 0.45 p.u & not 0.35 p.u as mentioned in textbook statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.12: EX27_12.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.12 :\n", +"// Page number 473-474\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_G = 625.0 // Generator rating(kVA)\n", +"V_G = 2.4 // Voltage rating of generator(kV)\n", +"X_st_G = 8.0/100 // Sub-transient reactance of generator\n", +"rating_M = 250.0 // Motor rating(HP)\n", +"V_M = 2.4 // Voltage rating of motor(kV)\n", +"n = 90.0/100 // Efficiency of motor\n", +"X_st_M = 20.0/100 // Sub-transient reactance of motor\n", +"\n", +"// Calculations\n", +"kVA_base = 625.0 // Base kVA\n", +"input_M = rating_M*0.746/n // Each motor input(kVA)\n", +"X_st_m_pu = X_st_M*kVA_base/input_M // Sub-transient reactance of motor(p.u)\n", +"I_base = kVA_base/(3**0.5*V_M) // Base current(A)\n", +"Z_th = %i*X_st_m_pu/3*X_st_G/(X_st_m_pu/3+X_st_G) // Thevenin impedance(p.u)\n", +"I_st = 1.0/Z_th // Initial symmetrical current at F(p.u)\n", +"I_st_g = I_st*(X_st_m_pu/3/(X_st_m_pu/3+X_st_G)) // Fault current rating of generator breaker(p.u)\n", +"I_st_m = (I_st-I_st_g)/3 // Fault current rating of each motor breaker(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.12 : SOLUTION :-')\n", +"printf('\nSub-transient fault current at F = %.2fj p.u', imag(I_st))\n", +"printf('\nFault current rating of generator breaker = %.1fj p.u', imag(I_st_g))\n", +"printf('\nFault current rating of each motor breaker = %.2fj p.u', imag(I_st_m))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.1: Per_unit_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.1 :\n", +"// Page number 466-467\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 500.0 // Generator voltage(V)\n", +"rating = 10.0 // Rating of the generator(kVA)\n", +"n_up = 1.0/2 // Turns ratio of step-up transformer\n", +"Z_line = complex(1.0,2.0) // Transmission line impedance(ohm)\n", +"n_down = 10.0/1 // Turns ratio of step-down transformer\n", +"load = complex(2.0,4.0) // Load(ohm)\n", +"\n", +"// Calculations\n", +"V_base_gen = V // Base voltage(V)\n", +"kVA_base_gen = rating // Base rating(kVA)\n", +"I_base_gen = kVA_base_gen*1000/V_base_gen // Base current(A)\n", +"Z_base_gen = V_base_gen/I_base_gen // Base impedance(ohm)\n", +"V_base_line = V_base_gen/n_up // Voltage base of the transmission line(V)\n", +"kVA_base_line = rating // Base rating of transmission line(kVA)\n", +"I_base_line = kVA_base_line*1000/V_base_line // Base current of transmission line(A)\n", +"Z_base_line = V_base_line/I_base_line // Base impedance of transmission line(ohm)\n", +"Z_line_1 = Z_line/Z_base_line // Impedance of transmission line(p.u)\n", +"V_base_load = V_base_line/n_down // Base voltage at the load(V)\n", +"kVA_base_load = rating // Base rating of load(kVA)\n", +"I_base_load = kVA_base_load*1000/V_base_load // Base current of load(A)\n", +"Z_base_load = V_base_load/I_base_load // Base impedance of load(ohm)\n", +"Z_load = load/Z_base_load // Load impedance(p.u)\n", +"Z_total = Z_line_1+Z_load // Total impedance(p.u)\n", +"I = 1.0/Z_total // Current(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.1 : SOLUTION :-')\n", +"printf('\nCurrent, I = %.3f∠%.2f° p.u', abs(I),phasemag(I))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.2: EX27_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.2 :\n", +"// Page number 467-468\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 33.0 // Transmission line operating voltage(kV)\n", +"R = 5.0 // Transmission line resistance(ohm)\n", +"X = 20.0 // Transmission line reactance(ohm)\n", +"kVA_tr = 5000.0 // Rating of step-up transformer(kVA)\n", +"X_tr = 6.0 // Reactance of transformer(%)\n", +"kVA_A = 10000.0 // Rating of alternator A(kVA)\n", +"X_A = 10.0 // Reactance of alternator A(%)\n", +"kVA_B = 5000.0 // Rating of alternator B(kVA)\n", +"X_B = 7.5 // Reactance of alternator B(%)\n", +"\n", +"// Calculations\n", +"kVA_base = kVA_A // Base rating(kVA)\n", +"X_gen_A = X_A*kVA_base/kVA_A // Reactance of generator A(%)\n", +"X_gen_B = X_B*kVA_base/kVA_B // Reactance of generator B(%)\n", +"X_trans = X_tr*kVA_base/kVA_tr // Reactance of transformer(%)\n", +"X_per = kVA_base*X/(10*kV**2) // X(%)\n", +"R_per = kVA_base*R/(10*kV**2) // R(%)\n", +"Z_F1 = (X_gen_A*X_gen_B/(X_gen_A+X_gen_B))+X_trans // Impedance upto fault(%)\n", +"kVA_F1 = kVA_base*(100/Z_F1) // Short-circuit kVA fed into the fault(kVA)\n", +"R_per_F2 = R_per // R(%)\n", +"X_per_F2 = X_per+Z_F1 // X(%)\n", +"Z_F2 = (R_per_F2**2+X_per_F2**2)**0.5 // Total impedance upto F2(%)\n", +"kVA_F2 = kVA_base*(100/Z_F2) // Short-circuit kVA fed into the fault at F2(kVA)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.2 : SOLUTION :-')\n", +"printf('\nCase(a): kVA at a short-circuit fault between phases at the HV terminal of transformers = %.f kVA', kVA_F1)\n", +"printf('\nCase(b): kVA at a short-circuit fault between phases at load end of transmission line = %.f kVA \n', kVA_F2)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.3: Transient_short_circuit_current_and_Sustained_short_circuit_current_at_X.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.3 :\n", +"// Page number 468-469\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_a = 40000.0 // Capacity of transmission line(kVA)\n", +"x_a = 10.0 // Reactance of transmission line(%)\n", +"kVA_b = 20000.0 // Capacity of transmission line(kVA)\n", +"x_b = 5.0 // Reactance of transmission line(%)\n", +"kVA_c = 50000.0 // Capacity of transmission line(kVA)\n", +"x_c = 20.0 // Reactance of transmission line(%)\n", +"kVA_d = 30000.0 // Capacity of transmission line(kVA)\n", +"x_d = 15.0 // Reactance of transmission line(%)\n", +"kVA_e = 10000.0 // Capacity of transmission line(kVA)\n", +"x_e = 6.0 // Reactance of transmission line(%)\n", +"kVA_T1 = 150000.0 // Capacity of transformer(kVA)\n", +"x_T1 = 10.0 // Reactance of transformer(%)\n", +"kVA_T2 = 50000.0 // Capacity of transformer(kVA)\n", +"x_T2 = 8.0 // Reactance of transformer(%)\n", +"kVA_T3 = 20000.0 // Capacity of transformer(kVA)\n", +"x_T3 = 5.0 // Reactance of transformer(%)\n", +"kVA_GA = 150000.0 // Capacity of generator(kVA)\n", +"x_sA = 90.0 // Synchronous reactance of generator(%)\n", +"x_tA = 30.0 // Transient reactance of generator(%)\n", +"kVA_GB = 50000.0 // Capacity of generator(kVA)\n", +"x_sB = 50.0 // Synchronous reactance of generator(%)\n", +"x_tB = 17.5 // Transient reactance of generator(%)\n", +"V = 33.0 // Feeder voltage(kV)\n", +"\n", +"// Calculations\n", +"kVA_base = 200000.0 // Base rating(kVA)\n", +"X_a = kVA_base/kVA_a*x_a // Reactance(%)\n", +"X_b = kVA_base/kVA_b*x_b // Reactance(%)\n", +"X_c = kVA_base/kVA_c*x_c // Reactance(%)\n", +"X_d = kVA_base/kVA_d*x_d // Reactance(%)\n", +"X_e = kVA_base/kVA_e*x_e // Reactance(%)\n", +"X_T1 = kVA_base/kVA_T1*x_T1 // Reactance(%)\n", +"X_T2 = kVA_base/kVA_T2*x_T2 // Reactance(%)\n", +"X_T3 = kVA_base/kVA_T3*x_T3 // Reactance(%)\n", +"X_sA = kVA_base/kVA_GA*x_sA // Synchronous reactance(%)\n", +"X_tA = kVA_base/kVA_GA*x_tA // Transient reactance(%)\n", +"X_sB = kVA_base/kVA_GB*x_sB // Synchronous reactance(%)\n", +"X_tB = kVA_base/kVA_GB*x_tB // Transient reactance(%)\n", +"X_eq_ab = X_a+X_b // Equivalent reactance of transmission lines a & b(%)\n", +"X_eq_abc = X_eq_ab*X_c/(X_eq_ab+X_c) // Equivalent reactance of transmission line c with series combination of a & b(%)\n", +"X_CF = (X_eq_abc+X_sA)*X_d/(X_eq_abc+X_sA+X_d) // Total reactance b/w sub-station C & F(%)\n", +"// Case(i)\n", +"X_tr_genA = kVA_base/kVA_GA*x_tA // Reactance in transient state of generator A(%)\n", +"X_T1_tr = kVA_base/kVA_T1*x_T1 // Reactance in transient state of transformer T1(%)\n", +"X_CF_tr = X_CF // Total reactance in transient state b/w sub-station C & F(%)\n", +"X_eq_tAF = X_tr_genA+X_T1_tr+X_CF_tr // Equivalent transient reactance from generator A to substation F(%)\n", +"X_tr_genB = kVA_base/kVA_GB*x_tB // Reactance in transient state of generator B(%)\n", +"X_T2_tr = kVA_base/kVA_T2*x_T2 // Reactance in transient state of transformer T2(%)\n", +"X_eq_tBF = X_tr_genB+X_T2_tr // Equivalent transient reactance from generator B to substation F(%)\n", +"X_eq_tF = X_eq_tAF*X_eq_tBF/(X_eq_tAF+X_eq_tBF) // Equivalent transient reactance upto substation F(%)\n", +"X_eq_tfault = X_eq_tF+X_T3 // Equivalent transient reactance upto fault point(%)\n", +"kVA_t_sc = kVA_base/X_eq_tfault*100 // Transient short circuit kVA(kVA)\n", +"I_t_sc = kVA_t_sc/(3**0.5*V) // Transient short circuit rms current(A)\n", +"I_t_sc_peak = 2**0.5*I_t_sc // Peak value of transient short circuit current(A)\n", +"// Case(ii)\n", +"X_S_genA = kVA_base/kVA_GA*x_sA // Reactance in steady state of generator A(%)\n", +"X_eq_SAF = X_S_genA+X_T1+X_CF // Equivalent steady state reactance from generator A to substation F(%)\n", +"X_eq_SBF = X_sB+X_T2 // Equivalent steady state reactance from generator B to substation F(%)\n", +"X_eq_SF = X_eq_SAF*X_eq_SBF/(X_eq_SAF+X_eq_SBF) // Equivalent steady state reactance upto substation F(%)\n", +"X_eq_Sfault = X_eq_SF+X_T3 // Equivalent steady state reactance upto fault point(%)\n", +"kVA_S_sc = kVA_base/X_eq_Sfault*100 // Steady state short circuit kVA(kVA)\n", +"I_S_sc = kVA_S_sc/(3**0.5*V) // Sustained short circuit rms current(A)\n", +"I_S_sc_peak = 2**0.5*I_S_sc // Peak value of sustained short circuit current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.3 : SOLUTION :-')\n", +"printf('\nCase(i) : Transient short circuit current at X = %.f A (peak value)', I_t_sc_peak)\n", +"printf('\nCase(ii): Sustained short circuit current at X = %.f A (peak value) \n', I_S_sc_peak)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.4: Current_in_the_short_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.4 :\n", +"// Page number 469-470\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_gen = 21000.0 // Generator rating(kVA)\n", +"kV_gen = 13.8 // Voltage rating of generator(kV)\n", +"X_tr_gen = 30.0 // Transient reactance of generator(%)\n", +"kVA_trans = 7000.0 // Transformer rating(kVA)\n", +"kV_trans_lv = 13.8 // LV voltage rating of transformer(kV)\n", +"kV_trans_hv = 66.0 // HV voltage rating of transformer(kV)\n", +"X_trans = 8.4 // Reactance of transformer(%)\n", +"l = 50.0 // Tie line length(miles)\n", +"x = 0.848 // Reactance of tie line(ohm/mile)\n", +"l_fault = 20.0 // Location of fault from station A(miles)\n", +"\n", +"// Calculations\n", +"kVA_base = kVA_gen // Base rating(kVA)\n", +"X_A = X_tr_gen // Reactance of generator A(%)\n", +"X_B = X_tr_gen // Reactance of generator B(%)\n", +"X_T1 = 3.0*X_trans // Reactance of transformer T1(%)\n", +"X_T2 = 3.0*X_trans // Reactance of transformer T2(%)\n", +"X_1 = kVA_base/(10*kV_trans_hv**2)*x*l_fault // Reactance(%)\n", +"X_2 = X_1*(l-l_fault)/l_fault // Reactance(%)\n", +"X_AF = X_A+X_T1+X_1 // Resultant reactance A to F(%)\n", +"X_BF = X_B+X_T2+X_2 // Resultant reactance B to F(%)\n", +"X_eq_fault = X_AF*X_BF/(X_AF+X_BF) // Equivalent reactance upto fault(%)\n", +"kVA_SC = kVA_base/X_eq_fault*100 // Short circuit kVA((kVA)\n", +"I_SC = kVA_SC/(3**0.5*kV_trans_hv) // Short circuit current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.4 : SOLUTION :-')\n", +"printf('\nShort circuit current = %.f A \n', I_SC)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.5: Per_unit_values_of_the_single_line_diagram.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.5 :\n", +"// Page number 470-471\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_G1 = 100.0 // Generator rating(MVA)\n", +"X_G1 = 30.0 // Reactance of generator(%)\n", +"MVA_G2 = 150.0 // Generator rating(MVA)\n", +"X_G2 = 20.0 // Reactance of generator(%)\n", +"MVA_G3 = 200.0 // Generator rating(MVA)\n", +"X_G3 = 15.0 // Reactance of generator(%)\n", +"MVA_T1 = 150.0 // Transformer rating(MVA)\n", +"X_T1 = 10.0 // Reactance of transformer(%)\n", +"MVA_T2 = 175.0 // Transformer rating(MVA)\n", +"X_T2 = 8.0 // Reactance of transformer(%)\n", +"MVA_T3 = 200.0 // Transformer rating(MVA)\n", +"X_T3 = 6.0 // Reactance of transformer(%)\n", +"MVA_T4 = 100.0 // Transformer rating(MVA)\n", +"X_T4 = 5.0 // Reactance of transformer(%)\n", +"MVA_T5 = 150.0 // Transformer rating(MVA)\n", +"X_T5 = 5.0 // Reactance of transformer(%)\n", +"Z_L1 = complex(0.5,1.0) // Line impedance(ohm/km)\n", +"L1 = 100.0 // Line length(km)\n", +"Z_L2 = complex(0.4,1.2) // Line impedance(ohm/km)\n", +"L2 = 50.0 // Line length(km)\n", +"Z_L3 = complex(0.4,1.2) // Line impedance(ohm/km)\n", +"L3 = 50.0 // Line length(km)\n", +"Z_L4 = complex(0.3,1.0) // Line impedance(ohm/km)\n", +"L4 = 60.0 // Line length(km)\n", +"kV_L1 = 220.0 // Voltage towards line(kV)\n", +"kV_L2 = 220.0 // Voltage towards line(kV)\n", +"kV_L3 = 132.0 // Voltage towards line(kV)\n", +"kV_L4 = 132.0 // Voltage towards line(kV)\n", +"\n", +"// Calculations\n", +"MVA_base = 200.0 // Base rating(MVA)\n", +"X_d_G1 = (MVA_base/MVA_G1)*(X_G1/100) // Reactance of generator(p.u)\n", +"X_d_G2 = (MVA_base/MVA_G2)*(X_G2/100) // Reactance of generator(p.u)\n", +"X_d_G3 = (MVA_base/MVA_G3)*(X_G3/100) // Reactance of generator(p.u)\n", +"X_T_1 = (MVA_base/MVA_T1)*(X_T1/100) // Reactance of transformer(p.u)\n", +"X_T_2 = (MVA_base/MVA_T2)*(X_T2/100) // Reactance of transformer(p.u)\n", +"X_T_3 = (MVA_base/MVA_T3)*(X_T3/100) // Reactance of transformer(p.u)\n", +"X_T_4 = (MVA_base/MVA_T4)*(X_T4/100) // Reactance of transformer(p.u)\n", +"X_T_5 = (MVA_base/MVA_T5)*(X_T5/100) // Reactance of transformer(p.u)\n", +"Z_L1_base = kV_L1**2/MVA_base // L1 base impedance(ohm)\n", +"Z_L_1 = Z_L1*L1/Z_L1_base // Line impedance(p.u)\n", +"Z_L2_base = kV_L2**2/MVA_base // L2 base impedance(ohm)\n", +"Z_L_2 = Z_L2*L2/Z_L2_base // Line impedance(p.u)\n", +"Z_L3_base = kV_L3**2/MVA_base // L3 base impedance(ohm)\n", +"Z_L_3 = Z_L3*L3/Z_L3_base // Line impedance(p.u)\n", +"Z_L4_base = kV_L4**2/MVA_base // L4 base impedance(ohm)\n", +"Z_L_4 = Z_L4*L4/Z_L4_base // Line impedance(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.5 : SOLUTION :-')\n", +"printf('\np.u values of the single line diagram are as below')\n", +"printf('\nGenerators p.u reactances :')\n", +"printf('\n X_d_G1 = %.1f p.u', X_d_G1)\n", +"printf('\n X_d_G2 = %.3f p.u', X_d_G2)\n", +"printf('\n X_d_G3 = %.2f p.u', X_d_G3)\n", +"printf('\nTransformers p.u reactances :')\n", +"printf('\n X_T1 = %.3f p.u', X_T_1)\n", +"printf('\n X_T2 = %.4f p.u', X_T_2)\n", +"printf('\n X_T3 = %.2f p.u', X_T_3)\n", +"printf('\n X_T4 = %.1f p.u', X_T_4)\n", +"printf('\n X_T5 = %.3f p.u', X_T_5)\n", +"printf('\nLines p.u impedances :')\n", +"printf('\n Z_L1 = (%.3f + %.3fj) p.u', real(Z_L_1),imag(Z_L_1))\n", +"printf('\n Z_L2 = (%.3f + %.3fj) p.u', real(Z_L_2),imag(Z_L_2))\n", +"printf('\n Z_L3 = (%.3f + %.3fj) p.u', real(Z_L_3),imag(Z_L_3))\n", +"printf('\n Z_L4 = (%.3f + %.3fj) p.u \n', real(Z_L_4),imag(Z_L_4))\n", +"printf('\nNOTE: ERROR: (1). Reactance of T2 is 8 percent & not 1 percent as mentioned in the textbook problem statement')\n", +"printf('\n (2). Several calculation mistakes in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.6: Actual_fault_current_using_per_unit_method.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.6 :\n", +"// Page number 471\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_gen = 21000.0 // Generator rating(kVA)\n", +"kV_gen = 13.8 // Voltage rating of generator(kV)\n", +"X_tr_gen = 30.0 // Transient reactance of generator(%)\n", +"kVA_trans = 7000.0 // Transformer rating(kVA)\n", +"kV_trans_lv = 13.8 // LV voltage rating of transformer(kV)\n", +"kV_trans_hv = 66.0 // HV voltage rating of transformer(kV)\n", +"X_trans = 8.4 // Reactance of transformer(%)\n", +"l = 50.0 // Tie line length(miles)\n", +"x = 0.848 // Reactance of tie line(ohm/mile)\n", +"l_fault = 20.0 // Location of fault from station A(miles)\n", +"\n", +"// Calculations\n", +"kVA_base = kVA_gen // Base rating(kVA)\n", +"kV_base_lv = kV_trans_lv // Base voltage on L.V side(kV)\n", +"kV_base_hv = kV_trans_hv // Base voltage on H.V side(kV)\n", +"Z_gen_pu = %i*X_tr_gen/100 // Impedance of generator(p.u)\n", +"Z_trans_pu = %i*X_trans*3/100 // Impedance of transformer(p.u)\n", +"Z_F_left = %i*x*l_fault*kVA_base/(kV_base_hv**2*1000) // Impedance of line to left of fault F(p.u)\n", +"Z_F_right = Z_F_left*(l-l_fault)/l_fault // Impedance of line to right of fault(p.u)\n", +"Z_AF = Z_gen_pu+Z_trans_pu+Z_F_left // Impedance(p.u)\n", +"Z_BF = Z_gen_pu+Z_trans_pu+Z_F_right // Impedance(p.u)\n", +"Z_eq = Z_AF*Z_BF/(Z_AF+Z_BF) // Equivalent impedance(p.u)\n", +"I_F = 1.0/abs(Z_eq) // Fault current(p.u)\n", +"I_base = kVA_base/(3**0.5*kV_base_hv) // Base current(A)\n", +"I_F_actual = I_F*I_base // Actual fault current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.6 : SOLUTION :-')\n", +"printf('\nActual fault current = %.f A \n', I_F_actual)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.7: Sub_transient_fault_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.7 :\n", +"// Page number 471-472\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_G1 = 50.0 // Generator rating(MVA)\n", +"kV_G1 = 15.0 // Voltage rating of generator(kV)\n", +"X_G1 = 0.2 // Reactance of generator(p.u)\n", +"MVA_G2 = 25.0 // Generator rating(MVA)\n", +"kV_G2 = 15.0 // Voltage rating of generator(kV)\n", +"X_G2 = 0.2 // Reactance of generator(p.u)\n", +"kV_T = 66.0 // Voltage rating of transformer(kV)\n", +"X_T = 0.1 // Reactance of transformer(p.u)\n", +"kV_fault = 66.0 // Voltage at fault occurence(kV)\n", +"kv_base = 69.0 // Base voltage(kV)\n", +"MVA_base = 100.0 // Base MVA\n", +"\n", +"// Calculations\n", +"X_d_G1 = X_G1*MVA_base/MVA_G1 // Sub-transient reactance referred to 100 MVA(p.u)\n", +"E_G1 = kV_fault/kv_base // Voltage(p.u)\n", +"X_d_G2 = X_G2*MVA_base/MVA_G2 // Sub-transient reactance referred to 100 MVA(p.u)\n", +"E_G2 = kV_fault/kv_base // Voltage(p.u)\n", +"X_net = X_d_G1*X_d_G2/(X_d_G1+X_d_G2) // Net sub-transient reactance(p.u)\n", +"E_g = (E_G1+E_G2)/2 // Net voltage(p.u). NOTE: Not sure how this comes\n", +"I_fault = E_g/(%i*(X_net+X_T)) // Sub-transient fault current(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.7 : SOLUTION :-')\n", +"printf('\nSub-transient fault current = %.3fj p.u \n', imag(I_fault))\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.8: Voltage_behind_the_respective_reactances.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.8 :\n", +"// Page number 472\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X_d_st = 0.2 // Sub-transient reactance(p.u)\n", +"X_d_t = 0.4 // Transient reactance(p.u)\n", +"X_d = 1.0 // Direct axis reactance(p.u)\n", +"I_pu = 1.0 // Load current(p.u)\n", +"PF = 0.80 // Lagging power factor\n", +"\n", +"// Calculations\n", +"V = 1.0 // Terminal voltage(p.u)\n", +"sin_phi = (1-PF**2)**0.5\n", +"I = I_pu*(PF-%i*sin_phi) // Load current(p.u)\n", +"E_st = V+%i*I*X_d_st // Voltage behind sub-transient reactance(p.u)\n", +"E_t = V+%i*I*X_d_t // Voltage behind transient reactance(p.u)\n", +"E = V+%i*I*X_d // Voltage behind direct axis reactance(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.8 : SOLUTION :-')\n", +"printf('\nVoltage behind sub-transient reactance = %.2f∠%.2f° p.u', abs(E_st),phasemag(E_st))\n", +"printf('\nVoltage behind transient reactance = %.2f∠%.2f° p.u', abs(E_t),phasemag(E_t))\n", +"printf('\nVoltage behind direct axis reactance, E = %.2f∠%.2f° p.u', abs(E),phasemag(E))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.9: Initial_symmetrical_rms_current_in_the_hv_side_and_lv_side.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS\n", +"\n", +"// EXAMPLE : 1.9 :\n", +"// Page number 472\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_G = 7500.0 // Generator rating(kVA)\n", +"kV_G = 6.9 // Voltage rating of generator(kV)\n", +"X_d_st = 9.0/100 // Sub-transient reactance of generator\n", +"X_d_t = 15.0/100 // Transient reactance of generator\n", +"X_d = 100.0 // Synchronous reactance of generator(%)\n", +"kVA_T = 7500.0 // Transformer rating(kVA)\n", +"kV_T_delta = 6.9 // Voltage rating of transformer delta side(kV)\n", +"kV_T_wye = 115.0 // Voltage rating of transformer wye side(kV)\n", +"X = 10.0/100 // Transformer reactance\n", +"\n", +"// Calculations\n", +"I_base_ht = kVA_T/(3**0.5*kV_T_wye) // Base current at ht side(A)\n", +"I_base_lt = kVA_T/(3**0.5*kV_T_delta) // Base current at lt side(A)\n", +"I_f_st = 1.0/(%i*(X_d_st+X)) // Sub-transient current after fault(p.u)\n", +"I_f_ht = abs(I_f_st)*I_base_ht // Initial fault current in h.t side(A)\n", +"I_f_lt = abs(I_f_st)*I_base_lt // Initial fault current in l.t side(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 1.9 : SOLUTION :-')\n", +"printf('\nInitial symmetrical rms current in the h.v side = %.f A', I_f_ht)\n", +"printf('\nInitial symmetrical rms current in the l.v side = %.f A \n', I_f_lt)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/28-FAULT_LIMITING_REACTORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/28-FAULT_LIMITING_REACTORS.ipynb new file mode 100644 index 0000000..6482ab9 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/28-FAULT_LIMITING_REACTORS.ipynb @@ -0,0 +1,247 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28: FAULT LIMITING REACTORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.1: Reactance_necessary_to_protect_the_switchgear.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 2: FAULT LIMITING REACTORS\n", +"\n", +"// EXAMPLE : 2.1 :\n", +"// Page number 479-480\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_A = 2500.0 // Rating of alternator A(kVA)\n", +"x_A = 8.0 // Reactance of alternator A(%)\n", +"kVA_B = 5000.0 // Rating of alternator B(kVA)\n", +"x_B = 6.0 // Reactance of alternator B(%)\n", +"kVA_CB = 150000.0 // Rating of circuit breaker(kVA)\n", +"kVA_T = 10000.0 // Rating of transformer(kVA)\n", +"x_T = 7.5 // Reactance of transformer(%)\n", +"V = 3300.0 // System voltage(V)\n", +"\n", +"// Calculations\n", +"kVA_base = 10000.0 // Base kVA\n", +"X_A = kVA_base/kVA_A*x_A // Reactance of generator A(%)\n", +"X_B = kVA_base/kVA_B*x_B // Reactance of generator B(%)\n", +"X_eq = X_A*X_B/(X_A+X_B) // Combined reactance of A & B(%)\n", +"kVA_SC_G = kVA_base/X_eq*100 // Short-circuit kVA due to generators(kVA)\n", +"kVA_SC_T = kVA_base/x_T*100 // Short-circuit kVA due to grid supply(kVA)\n", +"X = (kVA_base*100/(kVA_CB-kVA_SC_G))-x_T // Reactance necessary to protect switchgear(%)\n", +"I_fl = kVA_base*1000/(3**0.5*V) // Full load current corresponding to 10000 kVA(A)\n", +"X_phase = X*V/(3**0.5*I_fl*100) // Actual value of reactance per phase(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 2.1 : SOLUTION :-')\n", +"printf('\nReactance necessary to protect the switchgear = %.3f ohm/phase', X_phase)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.2: EX28_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 2: FAULT LIMITING REACTORS\n", +"\n", +"// EXAMPLE : 2.2 :\n", +"// Page number 480\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"X = 10.0 // Reactance of reactor(%)\n", +"kVA = 30000.0 // Rating of generator(kVA)\n", +"X_sc = 20.0 // Short-circuit reactance(%)\n", +"\n", +"// Calculations\n", +"X_1 = 1.0/3*(X_sc+X) // Combined reactance of generator A,B,C & associated reactors(%)\n", +"X_2 = X_1+X // Combined reactance upto fault(%)\n", +"X_total_a = X_2/2.0 // Total reactance upto fault(%)\n", +"kVA_SC_a = 100/X_total_a*kVA // Short-circuit kVA(kVA)\n", +"X_total_b = 1.0/4*X_sc // Total reactance upto fault when E,F,G & H are short-circuited(%)\n", +"kVA_SC_b = 100/X_total_b*kVA // Short-circuit kVA(kVA)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 2.2 : SOLUTION :-')\n", +"printf('\nCase(a): kVA developed under short-circuit when reactors are in circuit = %.f kVA', kVA_SC_a)\n", +"printf('\nCase(b): kVA developed under short-circuit when reactors are short-circuited = %.f kVA', kVA_SC_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.4: Reactance_of_each_reactor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 2: FAULT LIMITING REACTORS\n", +"\n", +"// EXAMPLE : 2.4 :\n", +"// Page number 481\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 20000.0 // Rating of generator(kVA)\n", +"f = 50.0 // Frequency(Hz)\n", +"V = 11.0*10**3 // Voltage of generator(V)\n", +"X_G = 20.0 // Generator short-circuit reactance(%)\n", +"x = 60.0 // Reactance falls to 60% normal value\n", +"\n", +"// Calculations\n", +"kVA_base = 20000.0 // Base kVA\n", +"X = poly(0,'X') // Reactance of each reactors E,F,G & H(%)\n", +"X_AE = X+X_G // Reactances of A & E in series(%)\n", +"X_BF = X+X_G // Reactances of B & F in series(%)\n", +"X_CD = X+X_G // Reactances of C & D in series(%)\n", +"X_eq = X_AE/3 // X_eq = X_AE*X_BF*X_CD/(X_BF*X_CD+X_AE*X_CD+X_AE*X_BF). Combined reactances of 3 groups in parallel(%)\n", +"X_f = X_eq+X // Reactances of these groups to fault via tie-bar(%)\n", +"X_sol = roots(6.66666666666667-(100-x)/100*(X_f)) // Value of reactance of each reactors E,F,G & H(%)\n", +"I_fl = kVA_base*1000/(3**0.5*V) // Full load current corresponding to 20000 kVA & 11 kV(A)\n", +"X_ohm = X_sol*V/(3**0.5*100*I_fl) // Ohmic value of reactance X(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 2.4 : SOLUTION :-')\n", +"printf('\nReactance of each reactor = %.4f ohm \n', X_ohm)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.5: Instantaneous_symmetrical_short_circuit_MVA_for_a_fault_at_X.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 2: FAULT LIMITING REACTORS\n", +"\n", +"// EXAMPLE : 2.5 :\n", +"// Page number 481-482\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_base = 10000.0 // Base kVA\n", +"V = 6.6*10**3 // Voltage of generator(V)\n", +"X_A = 7.5 // Reactance of generator A(%)\n", +"X_B = 7.5 // Reactance of generator B(%)\n", +"X_C = 10.0 // Reactance of generator C(%)\n", +"X_D = 10.0 // Reactance of generator D(%)\n", +"X_E = 8.0 // Reactance of reactor E(%)\n", +"X_F = 8.0 // Reactance of reactor F(%)\n", +"X_G = 6.5 // Reactance of reactor G(%)\n", +"X_H = 6.5 // Reactance of reactor H(%)\n", +"\n", +"// Calculations\n", +"Z_1 = X_B*X_C/(X_H+X_B+X_C) // Impedance(%). Fig E2.7\n", +"Z_2 = X_H*X_C/(X_H+X_B+X_C) // Impedance(%). Fig E2.7\n", +"Z_3 = X_B*X_H/(X_H+X_B+X_C) // Impedance(%). Fig E2.7\n", +"Z_4 = Z_2+X_F // Impedance(%). Fig E2.8 & Fig 2.9\n", +"Z_5 = Z_3+X_E // Impedance(%). Fig E2.8 & Fig 2.9\n", +"Z_6 = X_D*Z_1/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10\n", +"Z_7 = X_D*Z_4/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10\n", +"Z_8 = Z_1*Z_4/(X_D+Z_1+Z_4) // Impedance(%). Fig E2.10\n", +"Z_9 = Z_7+X_G // Impedance(%). Fig E2.11 & Fig 2.12\n", +"Z_10 = Z_8+Z_5 // Impedance(%). Fig E2.11 & Fig 2.12\n", +"Z_11 = Z_9*Z_10/(Z_9+Z_10) // Impedance(%). Fig 2.12 & Fig 2.13\n", +"Z_12 = Z_6+Z_11 // Impedance(%). Fig 2.13\n", +"Z_eq = X_A*Z_12/(X_A+Z_12) // Final Impedance(%). Fig 2.13 & Fig 2.14\n", +"MVA_SC = kVA_base*100/(Z_eq*1000) // Instantaneous symmetrical short-circuit MVA for a fault at X(MVA)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 2.5 : SOLUTION :-')\n", +"printf('\nInstantaneous symmetrical short-circuit MVA for a fault at X = %.f MVA \n', MVA_SC)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/29-SYMMETRICAL_COMPONENTS_ANALYSIS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/29-SYMMETRICAL_COMPONENTS_ANALYSIS.ipynb new file mode 100644 index 0000000..38edf6d --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/29-SYMMETRICAL_COMPONENTS_ANALYSIS.ipynb @@ -0,0 +1,562 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29: SYMMETRICAL COMPONENTS ANALYSIS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.10: EX29_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.10 :\n", +"// Page number 494\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R = 20000.0 // Resistance of voltmeter(ohm)\n", +"E_R = 100.0 // Line-to-neutral voltage(A)\n", +"E_Y = 200.0*exp(%i*270.0*%pi/180) // Line-to-neutral voltage(A)\n", +"E_B = 100.0*exp(%i*120.0*%pi/180) // Line-to-neutral voltage(A)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"V_R0 = 1.0/3*(E_R+E_Y+E_B) // Zero sequence voltage(V)\n", +"V_R1 = 1.0/3*(E_R+a*E_Y+a**2*E_B) // Positive sequence voltage(V)\n", +"V_R2 = 1.0/3*(E_R+a**2*E_Y+a*E_B) // Negative sequence voltage(V)\n", +"I_R1 = V_R1/R // Positive sequence current(A)\n", +"I_R2 = V_R2/R // Negative sequence current(A)\n", +"V_Y1 = a**2*V_R1 // Positive sequence voltage of line Y(V)\n", +"V_Y2 = a*V_R2 // Negative sequence voltage of line Y(V)\n", +"V_Y = V_Y1+V_Y2 // Voltmeter reading connected to the yellow line(V)\n", +"I_Y = abs(V_Y)/R*1000 // Current through voltmeter(mA)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.10 : SOLUTION :-')\n", +"printf('\nVoltmeter reading connected to the yellow line, |V_Y| = %.1f V', abs(V_Y))\n", +"printf('\nCurrent through voltmeter, I_Y = %.3f mA \n', I_Y)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.11: Three_line_currents_and_Wattmeter_reading.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.11 :\n", +"// Page number 495\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // Voltage(V)\n", +"Z_ab = 20.0 // Resistor load(ohm)\n", +"Z_bc = -%i*40.0 // Capacitor load(ohm)\n", +"Z_ca = 5.0+%i*10.0 // Inductor and resistance load(ohm)\n", +"\n", +"// Calculations\n", +"V_ab = V // Line voltage(V)\n", +"V_bc = V*exp(%i*-120.0*%pi/180) // Line voltage(V)\n", +"V_ca = V*exp(%i*120.0*%pi/180) // Line voltage(V)\n", +"I_ab = V_ab/Z_ab // Current(A)\n", +"I_bc = V_bc/Z_bc // Current(A)\n", +"I_ca = V_ca/Z_ca // Current(A)\n", +"I_a = I_ab-I_ca // Line current(A)\n", +"I_b = I_bc-I_ab // Line current(A)\n", +"I_c = I_ca-I_bc // Line current(A)\n", +"phi = -120.0-phasemag(I_a) // φ(°)\n", +"P = abs(I_a*V_bc)*cosd(phi)/1000 // Wattmeter reading(kW)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.11 : SOLUTION :-')\n", +"printf('\nLine currents are:')\n", +"printf('\n I_a = %.1f∠%.1f° A', abs(I_a),phasemag(I_a))\n", +"printf('\n I_b = %.1f∠%.2f° A', abs(I_b),phasemag(I_b))\n", +"printf('\n I_c = %.2f∠%.f° A', abs(I_c),phasemag(I_c))\n", +"printf('\nWattmeter reading, P = %.2f kW \n', P)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.1: Positive_Negative_and_Zero_sequence_currents.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.1 :\n", +"// Page number 487-488\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_R = complex(12.0,24.0) // Line current(A)\n", +"I_Y = complex(16.0,-2.0) // Line current(A)\n", +"I_B = complex(-4.0,-6.0) // Line current(A)\n", +"\n", +"// Calculations\n", +"alpha = exp(%i*120.0*%pi/180) // Operator\n", +"I_R0 = 1.0/3*(I_R+I_Y+I_B) // Zero sequence component(A)\n", +"I_R1 = 1.0/3*(I_R+alpha*I_Y+alpha**2*I_B) // Positive sequence component(A)\n", +"I_R2 = 1.0/3*(I_R+alpha**2*I_Y+alpha*I_B) // Negative sequence component(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.1 : SOLUTION :-')\n", +"printf('\nPositive sequence current, I_R1 = (%.3f + %.1fj) A', real(I_R1),imag(I_R1))\n", +"printf('\nNegative sequence current, I_R2 = (%.3f + %.2fj) A', real(I_R2),imag(I_R2))\n", +"printf('\nZero sequence current, I_R0 = (%.1f + %.2fj) A', real(I_R0),imag(I_R0))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.4: Sequence_components_of_currents_in_the_resistors_and_Supply_lines.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.4 :\n", +"// Page number 489-490\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R_bc = 5.0 // Resistance of resistor connected b/w b & c(ohm)\n", +"R_ca = 10.0 // Resistance of resistor connected b/w c & a(ohm)\n", +"R_ab = 20.0 // Resistance of resistor connected b/w a & b(ohm)\n", +"V = 100.0 // Voltage of balanced system(V)\n", +"\n", +"// Calculations\n", +"E_A = -V // Voltage across resistor connected b/w b & c(V)\n", +"angle = 60.0 // Angle in delta system(°)\n", +"E_B = V*exp(%i*60.0*%pi/180) // Voltage across resistor connected b/w c & a(V)\n", +"E_C = V*exp(%i*-60.0*%pi/180) // Voltage across resistor connected b/w a & b(V)\n", +"I_A = E_A/R_bc // Current flowing across resistor connected b/w b & c(A)\n", +"I_B = E_B/R_ca // Current flowing across resistor connected b/w c & a(A)\n", +"I_C = E_C/R_ab // Current flowing across resistor connected b/w a & b(A)\n", +"alpha = exp(%i*120.0*%pi/180) // Operator\n", +"I_A0 = 1.0/3*(I_A+I_B+I_C) // Zero sequence delta current(A)\n", +"I_A1 = 1.0/3*(I_A+alpha*I_B+alpha**2*I_C) // Positive sequence delta current(A)\n", +"I_A2 = 1.0/3*(I_A+alpha**2*I_B+alpha*I_C) // Negative sequence delta current(A)\n", +"I_a0 = 0.0 // Zero sequence star current(A)\n", +"I_a1 = (alpha-alpha**2)*I_A1 // Positive sequence star current(A)\n", +"I_a2 = (alpha**2-alpha)*I_A2 // Negative sequence star current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.4 : SOLUTION :-')\n", +"printf('\nCurrent in the resistors are:')\n", +"printf('\n I_A = (%.f+%.fj) A', real(I_A),imag(I_A))\n", +"printf('\n I_B = (%.f+%.2fj) A', real(I_B),imag(I_B))\n", +"printf('\n I_C = (%.1f%.2fj) A', real(I_C),imag(I_C))\n", +"printf('\nSequence components of currents in the resistors:')\n", +"printf('\n Zero-sequence current, I_A0 = (%.3f+%.2fj) A', real(I_A0),imag(I_A0))\n", +"printf('\n Positive-sequence current, I_A1 = (%.2f+%.fj) A', real(I_A1),imag(I_A1))\n", +"printf('\n Negative-sequence current, I_A2 = (%.2f%.2fj) A', real(I_A2),imag(I_A2))\n", +"printf('\nSequence components of currents in the supply lines:')\n", +"printf('\n Zero-sequence current, I_a0 = %.f A', I_a0)\n", +"printf('\n Positive-sequence current, I_a1 = %.1fj A', imag(I_a1))\n", +"printf('\n Negative-sequence current, I_a2 = (%.1f+%.2fj) A', real(I_a2),imag(I_a2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.5: EX29_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.5 :\n", +"// Page number 490-491\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"E_a = 100.0 // Line to line voltage(V)\n", +"E_b = 150.0 // Line to line voltage(V)\n", +"E_c = 200.0 // Line to line voltage(V)\n", +"\n", +"// Calculations\n", +"e_A = 1.0 // 100 V = 1 unit\n", +"e_B = 1.5 // 150 V = 1 unit\n", +"e_C = 2.0 // 200 V = 1 unit\n", +"cos_alpha = (e_C**2-e_A-e_B**2)/(2*e_B)\n", +"alpha = acosd(cos_alpha) // angle(°)\n", +"cos_beta = (e_A+e_B*cos_alpha)/e_C\n", +"beta = acosd(cos_beta) // angle(°)\n", +"E_A = E_a*exp(%i*180.0*%pi/180) // Voltage(V)\n", +"E_B = E_b*exp(%i*(180.0-alpha)*%pi/180) // Voltage(V)\n", +"E_C = E_c*exp(%i*-beta*%pi/180) // Voltage(V)\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"E_A0 = 1.0/3*(E_A+E_B+E_C) // Zero sequence voltage(V)\n", +"E_A1 = 1.0/3*(E_A+a*E_B+a**2*E_C) // Positive sequence delta voltage(V)\n", +"E_A1_mag = abs(E_A1) // Magnitude of positive sequence delta voltage(V)\n", +"E_a1 = -%i/3**0.5*E_A1 // Positive sequence star voltage(V)\n", +"E_a1_mag = abs(E_a1) // Magnitude of positive sequence star voltage(V)\n", +"E_A2 = 1.0/3*(E_A+a**2*E_B+a*E_C) // Negative sequence delta voltage(V)\n", +"E_A2_mag = abs(E_A2) // Magnitude of negative sequence delta voltage(V)\n", +"E_a2 = %i/3**0.5*E_A2 // Negative sequence star voltage(V)\n", +"E_a2_mag = abs(E_a2) // Magnitude of negative sequence star voltage(V)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.5 : SOLUTION :-')\n", +"printf('\nMagnitude of positive sequence delta voltage, |E_A1| = %.f V', E_A1_mag)\n", +"printf('\nMagnitude of positive sequence star voltage, |E_a1| = %.1f V', E_a1_mag)\n", +"printf('\nMagnitude of negative sequence delta voltage, |E_A2| = %.f V', E_A2_mag)\n", +"printf('\nMagnitude of negative sequence star voltage, |E_a2| = %.f V', E_a2_mag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.6: Current_in_each_line_by_the_method_of_symmetrical_components.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.6 :\n", +"// Page number 491-492\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 2300.0 // Rated voltage(V)\n", +"kVA = 500.0 // kVA rating\n", +"E_A = 2760.0*exp(%i*0*%pi/180) // Line voltage(V)\n", +"E_B = 2300.0*exp(%i*-138.6*%pi/180) // Line voltage(V)\n", +"E_C = 1840.0*exp(%i*124.2*%pi/180) // Line voltage(V)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"E_A1 = 1.0/3*(E_A+a*E_B+a**2*E_C) // Positive sequence voltage(V)\n", +"E_A2 = 1.0/3*(E_A+a**2*E_B+a*E_C) // Negative sequence voltage(V)\n", +"E_a1 = -%i/3**0.5*E_A1 // Positive sequence star voltage(V)\n", +"E_a2 = %i/3**0.5*E_A2 // Negative sequence star voltage(V)\n", +"E_a0 = 0.0 // Zero sequence voltage(V)\n", +"E_a = E_a1+E_a2+E_a0 // Symmetrical voltage component(V)\n", +"R = V**2/(kVA*1000) // Resistance(ohm)\n", +"I_a = abs(E_a)/R // Current in line a(A)\n", +"E_b = a**2*E_a1+a*E_a2+E_a0 // Symmetrical voltage component(V)\n", +"I_b = abs(E_b)/R // Current in line b(A)\n", +"E_c = a*E_a1+a**2*E_a2+E_a0 // Symmetrical voltage component(V)\n", +"I_c = abs(E_c)/R // Current in line c(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.6 : SOLUTION :-')\n", +"printf('\nCurrent in line a, |I_a| = %.1f A', I_a)\n", +"printf('\nCurrent in line b, |I_b| = %.f A', I_b)\n", +"printf('\nCurrent in line c, |I_c| = %.1f A \n', I_c)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.7: Symmetrical_components_of_line_current_if_phase_3_is_only_switched_off.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.7 :\n", +"// Page number 492-493\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 2300.0 // Rated voltage(V)\n", +"kVA = 500.0 // kVA rating\n", +"I_1 = 100.0 // Line current(A)\n", +"I_2 = 100.0*exp(%i*180*%pi/180) // Line current(A)\n", +"I_3 = 0 // Line current(A)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"I_10 = 1.0/3*(I_1+I_2+I_3) // Symmetrical component of line current for phase 1(A)\n", +"I_11 = 1.0/3*(I_1+a*I_2+a**2*I_3) // Symmetrical component of line current for phase 1(A)\n", +"I_12 = 1.0/3*(I_1+a**2*I_2+a*I_3) // Symmetrical component of line current for phase 1(A)\n", +"I_20 = I_10 // Symmetrical component of line current for phase 2(A)\n", +"I_21 = a**2*I_11 // Symmetrical component of line current for phase 2(A)\n", +"I_22 = a*I_12 // Symmetrical component of line current for phase 2(A)\n", +"I_30 = I_10 // Symmetrical component of line current for phase 3(A)\n", +"I_31 = a*I_11 // Symmetrical component of line current for phase 3(A)\n", +"I_32 = a**2*I_12 // Symmetrical component of line current for phase 3(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.7 : SOLUTION :-')\n", +"printf('\nSymmetrical component of line current for phase 1:')\n", +"printf('\n I_10 = %.1f A', abs(I_10))\n", +"printf('\n I_11 = %.2f∠%.f° A', abs(I_11),phasemag(I_11))\n", +"printf('\n I_12 = %.2f∠%.f° A', abs(I_12),phasemag(I_12))\n", +"printf('\nSymmetrical component of line current for phase 2:')\n", +"printf('\n I_20 = %.1f A', abs(I_20))\n", +"printf('\n I_21 = %.2f∠%.f° A', abs(I_21),phasemag(I_21))\n", +"printf('\n I_22 = %.2f∠%.f° A', abs(I_22),phasemag(I_22))\n", +"printf('\nSymmetrical component of line current for phase 3:')\n", +"printf('\n I_30 = %.1f A', abs(I_30))\n", +"printf('\n I_31 = %.2f∠%.f° A', abs(I_31),phasemag(I_31))\n", +"printf('\n I_32 = %.2f∠%.f° A', abs(I_32),phasemag(I_32))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.8: Positive_Negative_and_Zero_sequence_components_of_currents_for_all_phases.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.8 :\n", +"// Page number 493\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_a = 1000.0 // Current to earth(A)\n", +"I_b = 0 // Current(A)\n", +"I_c = 0 // Current(A)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"I_a0 = 1.0/3*(I_a+I_b+I_c) // Zero sequence component of current(A)\n", +"I_b0 = I_a0 // Zero sequence component of current(A)\n", +"I_c0 = I_a0 // Zero sequence component of current(A)\n", +"I_a1 = 1.0/3*(I_a+a*I_b+a**2*I_c) // Positive sequence component of current(A)\n", +"I_b1 = a**2*I_a1 // Positive sequence component of current(A)\n", +"I_c1 = a*I_a1 // Positive sequence component of current(A)\n", +"I_a2 = 1.0/3*(I_a+a**2*I_b+a*I_c) // Negative sequence component of current(A)\n", +"I_b2 = a*I_a2 // Negative sequence component of current(A)\n", +"I_c2 = a**2*I_a2 // Negative sequence component of current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.8 : SOLUTION :-')\n", +"printf('\nZero sequence component of current for all phases are')\n", +"printf('\n I_a0 = %.1f∠%.f° A', abs(I_a0),phasemag(I_a0))\n", +"printf('\n I_b0 = %.1f∠%.f° A', abs(I_b0),phasemag(I_b0))\n", +"printf('\n I_c0 = %.1f∠%.f° A', abs(I_c0),phasemag(I_c0))\n", +"printf('\nPositive sequence component of current for all phases are')\n", +"printf('\n I_a1 = %.1f∠%.f° A', abs(I_a1),phasemag(I_a1))\n", +"printf('\n I_b1 = %.1f∠%.f° A', abs(I_b1),360+phasemag(I_b1))\n", +"printf('\n I_c1 = %.1f∠%.f° A', abs(I_c1),phasemag(I_c1))\n", +"printf('\nNegative sequence component of current for all phases are')\n", +"printf('\n I_a2 = %.1f∠%.f° A', abs(I_a2),phasemag(I_a2))\n", +"printf('\n I_b2 = %.1f∠%.f° A', abs(I_b2),phasemag(I_b2))\n", +"printf('\n I_c2 = %.1f∠%.f° A', abs(I_c2),360+phasemag(I_c2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.9: Currents_in_all_the_lines_and_their_symmetrical_components.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 3: SYMMETRICAL COMPONENTS' ANALYSIS\n", +"\n", +"// EXAMPLE : 3.9 :\n", +"// Page number 493-494\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_A = 1000.0 // Current through line A(A)\n", +"I_C = 0 // Current through line C(A)\n", +"\n", +"// Calculations\n", +"I_B = 1000.0*exp(%i*180.0*%pi/180) // Current through line B(A)\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"I_a0 = 1.0/3*(I_A+I_B+I_C) // Zero sequence component of current(A)\n", +"I_b0 = I_a0 // Zero sequence component of current(A)\n", +"I_c0 = I_a0 // Zero sequence component of current(A)\n", +"I_a1 = 1.0/3*(I_A+a*I_B+a**2*I_C) // Positive sequence component of current(A)\n", +"I_b1 = a**2*I_a1 // Positive sequence component of current(A)\n", +"I_c1 = a*I_a1 // Positive sequence component of current(A)\n", +"I_a2 = 1.0/3*(I_A+a**2*I_B+a*I_C) // Negative sequence component of current(A)\n", +"I_b2 = a*I_a2 // Negative sequence component of current(A)\n", +"I_c2 = a**2*I_a2 // Negative sequence component of current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 3.9 : SOLUTION :-')\n", +"printf('\nCurrent in line A, I_A = %.f∠%.f° A', abs(I_A),phasemag(I_A))\n", +"printf('\nCurrent in line B, I_B = %.f∠%.f° A', abs(I_B),phasemag(I_B))\n", +"printf('\nCurrent in line C, I_C = %.f A', I_C)\n", +"printf('\nSymmetrical current components of line A are:')\n", +"printf('\n I_a0 = %.f A', abs(I_a0))\n", +"printf('\n I_a1 = %.1f∠%.f° A', abs(I_a1),phasemag(I_a1))\n", +"printf('\n I_a2 = %.1f∠%.f° A', abs(I_a2),phasemag(I_a2))\n", +"printf('\nSymmetrical current components of line B are:')\n", +"printf('\n I_b0 = %.f A', abs(I_b0))\n", +"printf('\n I_b1 = %.1f∠%.f° A', abs(I_b1),phasemag(I_b1))\n", +"printf('\n I_b2 = %.1f∠%.f° A', abs(I_b2),phasemag(I_b2))\n", +"printf('\nSymmetrical current components of line C are:')\n", +"printf('\n I_c0 = %.f A', abs(I_c0))\n", +"printf('\n I_c1 = %.1f∠%.f° A', abs(I_c1),phasemag(I_c1))\n", +"printf('\n I_c2 = %.1f∠%.f° A', abs(I_c2),phasemag(I_c2))" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/3-HYDRO_ELECTRIC_STATIONS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/3-HYDRO_ELECTRIC_STATIONS.ipynb new file mode 100644 index 0000000..42b78d5 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/3-HYDRO_ELECTRIC_STATIONS.ipynb @@ -0,0 +1,165 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: HYDRO ELECTRIC STATIONS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1: Firm_capacity_and_Yearly_gross_output.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n", +"\n", +"// EXAMPLE : 3.1 :\n", +"// Page number 41\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Q = 95.0 // Minimum run-off(m^3/sec)\n", +"h = 40.0 // Head(m)\n", +"\n", +"// Calculations\n", +"w = 1000.0 // Density of water(kg/m^3)\n", +"weight = Q*w // Weight of water per sec(kg)\n", +"work_done = weight*h // Work done in one second(kg-mt)\n", +"kW_1 = 75.0/0.746 // 1 kW(kg-mt/sec)\n", +"power = work_done/kW_1 // Power production(kW)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"output = power*365*24.0 // Yearly gross output(kWhr)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 3.1 : SOLUTION :-')\n", +"printf('\nFirm capacity = %.f kW', power)\n", +"printf('\nYearly gross output = %.2e kWhr.', output)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3: Available_continuous_power.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n", +"\n", +"// EXAMPLE : 3.3 :\n", +"// Page number 41\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"A = 200.0 // Catchment area(Sq.km)\n", +"F = 1000.0 // Annual rainfall(mm)\n", +"H = 200.0 // Effective head(m)\n", +"K = 0.5 // Yield factor\n", +"n = 0.8 // Plant efficiency\n", +"\n", +"// Calculations\n", +"P = 3.14*n*K*A*F*H*10**-4 // Available continuous power(kW)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 3.3 : SOLUTION :-')\n", +"printf('\nAvailable continuous power of hydro-electric station , P = %.f kW', P)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4: Minimum_flow_of_river_water_to_operate_the_plant.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n", +"\n", +"// EXAMPLE : 3.4 :\n", +"// Page number 41-42\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_factor = 0.15 // Load factor\n", +"P = 10.0*10**3 // Rated installed capacity(kW)\n", +"H = 50.0 // Head of plant(m)\n", +"n = 0.8 // Efficiency of plant\n", +"\n", +"//Calculation\n", +"units_day = P*load_factor // Total units generated daily on basis of load factor(kWhr)\n", +"units_week = units_day*24.0*7 // Total units generated for one week(kWhr)\n", +"Q = units_week/(9.81*H*n*24*7) // Minimum flow of water(cubic mt/sec)\n", +"\n", +"//Result\n", +"disp('PART I - EXAMPLE : 3.4 : SOLUTION :-')\n", +"printf('\nMinimum flow of river water to operate the plant, Q = %.3f cubic mt/sec', Q)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/30-UNSYMMETRICAL_FAULTS_IN_POWER_SYSTEMS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/30-UNSYMMETRICAL_FAULTS_IN_POWER_SYSTEMS.ipynb new file mode 100644 index 0000000..2b93761 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/30-UNSYMMETRICAL_FAULTS_IN_POWER_SYSTEMS.ipynb @@ -0,0 +1,1023 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30: UNSYMMETRICAL FAULTS IN POWER SYSTEMS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.10: EX30_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.10 :\n", +"// Page number 519-520\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_A = 30.0 // Alternator rating(MVA)\n", +"kV_A = 11.0 // Alternator rating(kV)\n", +"X_1 = 2.5 // Reactance to positive sequence current(ohm)\n", +"X_2 = 0.8*X_1 // Reactance to negative sequence current(ohm)\n", +"X_0 = 0.3*X_1 // Reactance to zero sequence current(ohm)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"Z_1 = %i*X_1 // Positive sequence impedance(ohm)\n", +"Z_2 = %i*X_2 // Negative sequence impedance(ohm)\n", +"Z_0 = %i*X_0 // Zero sequence impedance(ohm)\n", +"Z_02 = Z_0*Z_2/(Z_0+Z_2) // Impedance(ohm)\n", +"E_a = kV_A*1000/3**0.5 // Phase voltage(V)\n", +"I_a1 = E_a/(Z_1+Z_02) // Positive sequence current(A)\n", +"I_a2 = -Z_0/(Z_0+Z_2)*I_a1 // Negative sequence current(A)\n", +"I_a0 = -Z_2/(Z_0+Z_2)*I_a1 // Zero sequence current(A)\n", +"I_0 = I_a0 // Zero sequence current(A)\n", +"I_a = I_a0+I_a1+I_a2 // Line current(A)\n", +"I_b = I_0+a**2*I_a1+a*I_a2 // Line current(A)\n", +"I_c = I_0+a*I_a1+a**2*I_a2 // Line current(A)\n", +"// Case(b)\n", +"I_n = 3*abs(I_0) // Current through ground(A)\n", +"// Case(c)\n", +"V_a2 = Z_02*I_a1 // Negative sequence voltage(V)\n", +"V_a = 3*abs(V_a2) // Voltage of healthy phase to neutral(V)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.10 : SOLUTION :-')\n", +"printf('\nCase(a): Currents in the faulted phase are')\n", +"printf('\n I_a = %.f A', abs(I_a))\n", +"printf('\n I_b = %.f∠%.1f° A', abs(I_b),phasemag(I_b))\n", +"printf('\n I_c = %.f∠%.1f° A', abs(I_c),phasemag(I_c))\n", +"printf('\nCase(b): Current through ground, I_n = %.f A', I_n)\n", +"printf('\nCase(c): Voltage of healthy phase to neutral, V_a = %.f V\n', V_a)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.11: Fault_currents.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.11 :\n", +"// Page number 520-521\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 6.0 // Number of alternator\n", +"kV_A = 6.6 // Alternator rating(kV)\n", +"X_1 = 0.9 // Positive sequence reactance(ohm)\n", +"X_2 = 0.72 // Negative sequence reactance(ohm)\n", +"X_0 = 0.3 // Zero sequence reactance(ohm)\n", +"Z_n = 0.2 // Resistance of grounding resistor(ohm)\n", +"\n", +"// Calculations\n", +"E_a = kV_A*1000/3**0.5 // Phase voltage(V)\n", +"// Case(a)\n", +"Z_1_a = %i*X_1/n // Positive sequence impedance when alternators are in parallel(ohm)\n", +"Z_2_a = %i*X_2/n // Negative sequence impedance when alternators are in parallel(ohm)\n", +"Z_0_a = %i*X_0/n // Zero sequence impedance when alternators are in parallel(ohm)\n", +"I_a_a = 3*E_a/(Z_1_a+Z_2_a+Z_0_a) // Fault current assuming 'a' phase to be fault(A)\n", +"// Case(b)\n", +"Z_0_b = 3*Z_n+%i*X_0 // Zero sequence impedance(ohm)\n", +"I_a_b = 3*E_a/(Z_1_a+Z_2_a+Z_0_b) // Fault current(A)\n", +"// Case(c)\n", +"Z_0_c = %i*X_0 // Zero sequence impedance(ohm)\n", +"I_a_c = 3*E_a/(Z_1_a+Z_2_a+Z_0_c) // Fault current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.11 : SOLUTION :-')\n", +"printf('\nCase(a): Fault current if all alternator neutrals are solidly grounded, I_a = %.f A', imag(I_a_a))\n", +"printf('\nCase(b): Fault current if one alternator neutral is grounded & others isolated, I_a = %.1f∠%.1f° A', abs(I_a_b),phasemag(I_a_b))\n", +"printf('\nCase(c): Fault current if one alternator neutral is solidly grounded & others isolated, I_a = %.2fj A\n', imag(I_a_c))\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.12: EX30_12.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.12 :\n", +"// Page number 521-522\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_A = 30.0 // Alternator rating(MVA)\n", +"kV_A = 6.6 // Alternator rating(kV)\n", +"X_G = 10.0 // Reactance of alternator(%)\n", +"kV_lv_T = 6.6 // Transformer lv side rating(kV) \n", +"kV_hv_T = 33.0 // Transformer hv side rating(kV) \n", +"X_T = 6.0 // Reactance of transformer(%)\n", +"kV_line = 33.0 // Transmission line voltage(kV)\n", +"X_line = 4.0 // Transmission line reactance(ohm)\n", +"X_g2 = 70.0 // Negative sequence reactance is 70% of +ve sequence reactance of generator(%)\n", +"\n", +"// Calculations\n", +"MVA_base = 30.0 // Base MVA\n", +"kV_base = 6.6 // Base kV\n", +"Z_base = kV_base**2/MVA_base // Base impedance(ohm)\n", +"Z_g1 = %i*Z_base*X_G/100 // Positive sequence impedance of alternator(ohm)\n", +"Z_T1 = %i*Z_base*X_T/100 // Positive sequence impedance of transformer(ohm)\n", +"Z_L1 = %i*(kV_base/kV_line)**2*X_line // Positive sequence impedance of transmission line(ohm)\n", +"Z_g2 = X_g2/100*Z_g1 // Negative sequence impedance of alternator(ohm)\n", +"Z_T2 = %i*Z_base*X_T/100 // Negative sequence impedance of transformer(ohm)\n", +"Z_T0 = %i*Z_base*X_T/100 // Zero sequence impedance of transformer(ohm)\n", +"Z_L2 = Z_L1 // Negative sequence impedance of transmission line(ohm)\n", +"Z_1 = Z_g1+Z_T1+Z_L1+Z_T1 // Positive sequence impedance(ohm)\n", +"Z_2 = Z_g2+Z_T2+Z_L2+Z_T2 // Negative sequence impedance(ohm)\n", +"Z_0 = Z_T0 // Zero sequence impedance(ohm)\n", +"E_a = kV_base*1000/3**0.5 // Base voltage(V)\n", +"// Case(a)\n", +"I_sc = E_a/Z_1 // Fault current if all 3 phases short circuited(A)\n", +"// Case(b)\n", +"I_a = 3*E_a/(Z_1+Z_2+Z_0) // Fault current if single line is grounded assuming 'a' to be grounded(A)\n", +"// Case(c)\n", +"I_b = %i*3**0.5*E_a/(Z_1+Z_2) // Fault current for a short circuit between two lines(A)\n", +"I_c = -%i*3**0.5*E_a/(Z_1+Z_2) // Fault current for a short circuit between two lines(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.12 : SOLUTION :-')\n", +"printf('\nCase(a): Fault current if all 3 phases short circuited, I_sc = %.f∠%.f° A', abs(I_sc),phasemag(I_sc))\n", +"printf('\nCase(b): Fault current if single line is grounded, I_a = %.fj A', imag(I_a))\n", +"printf('\nCase(c): Fault current for a short circuit between two lines, I_b = %.f A', real(I_b))\n", +"printf('\n Fault current for a short circuit between two lines, I_c = %.f A\n', real(I_c))\n", +"printf('\nNOTE: ERROR: (1).Calculation mistake in Z_2 in the textbook solution')\n", +"printf('\n (2).Transformer reactance is 6 percent, not 5 percent as in problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.13: Sub_transient_current_in_the_faulty_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.13 :\n", +"// Page number 522\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 6.9 // Alternator rating(kV)\n", +"MVA = 10.0 // Alternator rating(MVA)\n", +"X_st = 0.15 // Sub-transient reactance(p.u)\n", +"X_2 = 0.15 // Negative sequence reactance(p.u)\n", +"X_0 = 0.05 // Zero sequence reactance(p.u)\n", +"X = 0.397 // Grounding reactor(ohm)\n", +"\n", +"// Calculations\n", +"MVA_base = 10.0 // Base MVA\n", +"kV_base = 6.9 // Base kV\n", +"Z_base = kV_base**2/MVA_base // Base impedance(ohm)\n", +"Z_n = X/Z_base // Grounding reactor(p.u)\n", +"Z_1 = %i*X_st // Positive sequence impedance(p.u)\n", +"Z_2 = %i*X_2 // Negative sequence impedance(p.u)\n", +"Z_0 = %i*(X_0+3*Z_n) // Zero sequence impedance(p.u)\n", +"E_a = 1.0 // Phase voltage(p.u)\n", +"I_a_pu = 3*E_a/(Z_1+Z_2+Z_0) // Sub-transient current in the faulty phase(p.u)\n", +"I_base = kV_base*1000/(3**0.5*Z_base) // Base current(A)\n", +"I_a = abs(I_a_pu)*I_base // Sub-transient current in the faulty phase(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.13 : SOLUTION :-')\n", +"printf('\nSub-transient current in the faulty phase, I_a = %.f A\n', I_a)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.14: Initial_symmetrical_rms_current_in_all_phases_of_generator.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.14 :\n", +"// Page number 522-523\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 10000.0 // Generator rating(kVA)\n", +"kV = 13.8 // Generator rating(kV)\n", +"X_st = 10.0 // Sub-transient reactance(%)\n", +"X_2 = 10.0 // Negative sequence reactance(%)\n", +"X_0 = 5.0 // Zero sequence reactance(%)\n", +"X = 8.0 // Grounding reactor(%)\n", +"X_con = 6.0 // Reactance of reactor connecting generator & transformer(%)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"Z_1 = %i*(X_st+X_con)/100 // Positive sequence impedance(p.u)\n", +"Z_2 = %i*(X_2+X_con)/100 // Negative sequence impedance(p.u)\n", +"Z_0 = %i*X_con/100 // Zero sequence impedance(p.u)\n", +"E_a = 1.0 // Phase voltage(p.u)\n", +"I_a1 = E_a/(Z_1+Z_2+Z_0) // Sub-transient current in the faulty phase(p.u)\n", +"I_A1 = %i*I_a1 // Positive sequence current(p.u)\n", +"I_A2 = -%i*I_a1 // Negative sequence current(p.u)\n", +"I_A = I_A1+I_A2 // Initial symmetrical r.m.s current in phase a(p.u)\n", +"I_B1 = a**2*I_A1 // Positive sequence current(p.u)\n", +"I_B2 = a*I_A2 // Negative sequence current(p.u)\n", +"I_B = I_B1+I_B2 // Initial symmetrical r.m.s current in phase b(p.u)\n", +"I_C1 = a*I_A1 // Positive sequence current(p.u)\n", +"I_C2 = a**2*I_A2 // Negative sequence current(p.u)\n", +"I_C = I_C1+I_C2 // Initial symmetrical r.m.s current in phase c(p.u)\n", +"I_base = kVA/(3**0.5*kV) // Base current(A)\n", +"I_A_amp = I_A*I_base // Initial symmetrical r.m.s current in phase a(p.u)\n", +"I_B_amp = I_B*I_base // Initial symmetrical r.m.s current in phase b(p.u)\n", +"I_C_amp = I_C*I_base // Initial symmetrical r.m.s current in phase c(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.14 : SOLUTION :-')\n", +"printf('\nInitial symmetrical r.m.s current in all phases of generator are,')\n", +"printf('\n I_A = %.f A', abs(I_A_amp))\n", +"printf('\n I_B = %.f∠%.f° A', abs(I_B_amp),phasemag(I_B_amp))\n", +"printf('\n I_C = %.f∠%.f° A', abs(I_C_amp),phasemag(I_C_amp))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.1: EX30_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.1 :\n", +"// Page number 510-512\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 15.0 // Generator rating(MVA)\n", +"kV = 6.9 // Generator voltage(kV)\n", +"X_1 = 25.0 // Positive sequence reactance(%)\n", +"X_2 = 25.0 // Negative sequence reactance(%)\n", +"X_0 = 8.0 // Zero sequence reactance(%)\n", +"X = 6.0 // Reactor placed in line(%)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"Z_1 = %i*X_1/100 // Positive sequence impedance(p.u)\n", +"Z_2 = %i*X_2/100 // Negative sequence impedance(p.u)\n", +"Z_g0 = %i*X_0/100 // Impedance(p.u)\n", +"Z = %i*X/100 // Impedance(p.u)\n", +"Z_0 = Z_g0+3*Z // Zero sequence impedance(p.u)\n", +"E_a = 1.0 // Voltage(p.u)\n", +"E_b = a**2*E_a // Voltage(p.u)\n", +"// Case(a)\n", +"I_a0_a = 0 // Current(A)\n", +"I_a1_a_pu = 1.0/(Z_1+Z_2) // Current(p.u)\n", +"I_a1_a = I_a1_a_pu*MVA*1000/(3**0.5*kV) // Current(A)\n", +"I_a2_a = -I_a1_a // Current(A)\n", +"I_b0_a = 0 // Current(A)\n", +"I_b1_a = a**2*I_a1_a // Current(A)\n", +"I_b2_a = a*I_a2_a // Current(A)\n", +"I_a_a = I_a1_a+I_a2_a // Line current(A)\n", +"I_b_a = I_b1_a+I_b2_a // Line current(A)\n", +"I_c_a = -I_b_a // Line current(A)\n", +"I_g_a = 0 // Ground wire current(A)\n", +"V_a_a = (E_a-I_a1_a*Z_1-I_a2_a*Z_2-I_a0_a*Z_0)*kV*1000/3**0.5 // Voltage(V)\n", +"V_b_a = (a**2*E_a+%i*3**0.5*I_a1_a_pu*Z_1)*kV*1000/3**0.5 // Voltage(V)\n", +"V_c_a = V_b_a // Voltage(V)\n", +"// Case(b)\n", +"I_a1_b_pu = E_a/(Z_1+(Z_2*Z_0/(Z_2+Z_0))) // Current(p.u)\n", +"I_a1_b = I_a1_b_pu*MVA*1000/(3**0.5*kV) // Current(A)\n", +"I_a2_b_pu = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)\n", +"I_a2_b = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b // Current(A)\n", +"I_a0_b_pu = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)\n", +"I_a0_b = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b // Current(A)\n", +"I_a_b = I_a0_b+I_a1_b+I_a2_b // Line current(A)\n", +"I_b_b = I_a0_b+a**2*I_a1_b+a*I_a2_b // Line current(A)\n", +"I_c_b = I_a0_b+a*I_a1_b+a**2*I_a2_b // Line current(A)\n", +"I_0_b = 3*I_a0_b // Current in the ground resistor(A)\n", +"V_a_b_pu = E_a-I_a1_b_pu*Z_1-I_a2_b_pu*Z_2-I_a0_b_pu*Z_0 // Voltage(p.u)\n", +"V_a_b = abs(V_a_b_pu)*kV*1000/(3**0.5) // Voltage(V)\n", +"V_b_b = 0 // Voltage(V)\n", +"V_c_b = 0 // Voltage(V)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.1 : SOLUTION :-')\n", +"printf('\nCase(a): Initial symmetrical rms line current when ground is not involved in fault, I_a = %.f A', abs(I_a_a))\n", +"printf('\n Initial symmetrical rms line current when ground is not involved in fault, I_b = %.f A', real(I_b_a))\n", +"printf('\n Initial symmetrical rms line current when ground is not involved in fault, I_c = %.f A', real(I_c_a))\n", +"printf('\n Ground wire current = %.f A', I_g_a)\n", +"printf('\n Line to neutral voltage, V_a = %.f V', real(V_a_a))\n", +"printf('\n Line to neutral voltage, V_b = %.f V', real(V_b_a))\n", +"printf('\n Line to neutral voltage, V_c = %.f V', real(V_c_a))\n", +"printf('\nCase(b): Initial symmetrical rms line current when fault is solidly grounded, I_a = %.f A', abs(I_a_b))\n", +"printf('\n Initial symmetrical rms line current when fault is solidly grounded, I_b = (%.f+%.fj) A', real(I_b_b),imag(I_b_b))\n", +"printf('\n Initial symmetrical rms line current when fault is solidly grounded, I_c = (%.f+%.fj) A', real(I_c_b),imag(I_c_b))\n", +"printf('\n Ground wire current = %.fj A', imag(I_0_b))\n", +"printf('\n Line to neutral voltage, V_a = %.f V', V_a_b)\n", +"printf('\n Line to neutral voltage, V_b = %.f V', V_b_b)\n", +"printf('\n Line to neutral voltage, V_c = %.f V\n', V_c_b)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here and approximation in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.2: Current_in_the_line_with_two_lines_short_circuited.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.2 :\n", +"// Page number 512\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 10000.0 // Generator rating(kVA)\n", +"f = 50.0 // Frequency(Hz)\n", +"I_1 = 30.0 // Positive sequence current(%)\n", +"I_2 = 10.0 // Negative sequence current(%)\n", +"I_0 = 5.0 // Zero sequence current(%)\n", +"d = 1.0/100 // Diameter of conductor(m)\n", +"D = 5.0 // Triangular spacing(m)\n", +"kV = 30.0 // Generator voltage on open-circuit(kV)\n", +"l = 20.0 // Distance of line at short circuit occurance(km)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"Z_g1 = kV**2*I_1*I_2/kVA // Positive phase sequence reactance of generator(ohm)\n", +"Z_g2 = Z_g1*I_2/I_1 // Negative phase sequence reactance of generator(ohm)\n", +"Z_g0 = Z_g1*I_0/I_1 // Zero phase sequence reactance of generator(ohm)\n", +"r = d/2 // Radius of conductor(m)\n", +"Z_l1 = 2.0*%pi*f*(0.5+4.606*log10(D/r))*10**-7*l*1000 // Positive phase sequence reactance of line(ohm)\n", +"Z_l2 = 2.0*%pi*f*(0.5+4.606*log10(D/r))*10**-7*l*1000 // Negative phase sequence reactance of line(ohm)\n", +"Z_1 = %i*(Z_g1+Z_l1) // Z1 upto the point of fault(ohm)\n", +"Z_2 = %i*(Z_g2+Z_l2) // Z2 upto the point of fault(ohm)\n", +"E_a = kV*1000/3**0.5 // Phase voltage(V)\n", +"I_a1 = E_a/(Z_1+Z_2) // Positive sequence current in line a(A)\n", +"I_a2 = -I_a1 // Negative sequence current in line a(A)\n", +"I_a0 = 0 // Zero sequence current in line a(A)\n", +"I_b0 = 0 // Zero sequence current in line b(A)\n", +"I_c0 = 0 // Zero sequence current in line c(A)\n", +"I_a = I_a0+I_a1+I_a2 // Current in line a(A)\n", +"I_b = I_b0+a**2*I_a1+a*I_a2 // Current in line b(A)\n", +"I_c = I_c0+a*I_a1+a**2*I_a2 // Current in line c(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.2 : SOLUTION :-')\n", +"printf('\nCurrent in line a, I_a = %.f A', abs(I_a))\n", +"printf('\nCurrent in line b, I_b = %.f A', real(I_b))\n", +"printf('\nCurrent in line c, I_c = %.f A', real(I_c))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.3: EX30_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.3 :\n", +"// Page number 512-513\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 10000.0 // Alternator rating(kVA)\n", +"Z_g1 = complex(0.5,4.7) // Positive sequence impedance(ohm/phase)\n", +"Z_g2 = complex(0.2,0.6) // Negative sequence impedance(ohm/phase)\n", +"Z_g0 = complex(0,0.43) // Zero sequence impedance(ohm/phase)\n", +"Z_l1 = complex(0.36,0.25) // Impedance(ohm)\n", +"Z_l2 = complex(0.36,0.25) // Impedance(ohm)\n", +"Z_l0 = complex(2.9,0.95) // Impedance(ohm)\n", +"V = 6600.0 // Voltage(V)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"// Case(a)\n", +"E_a = V/3**0.5 // Phase voltage(V)\n", +"Z_1 = Z_g1+Z_l1 // Z1 upto the point of fault(ohm)\n", +"Z_2 = Z_g2+Z_l2 // Z2 upto the point of fault(ohm)\n", +"Z_0 = Z_g0+Z_l0 // Z0 upto the point of fault(ohm)\n", +"I_a = 3*E_a/(Z_1+Z_2+Z_0) // Fault current(A)\n", +"// Case(b)\n", +"I_a0 = abs(I_a)/3 // Zero sequence current of line a(A)\n", +"I_a1 = abs(I_a)/3 // Positive sequence current of line a(A)\n", +"I_a2 = abs(I_a)/3 // Negative sequence current of line a(A)\n", +"I_b0 = I_a0 // Zero sequence current of line b(A)\n", +"I_b1 = a**2*I_a1 // Positive sequence current of line b(A)\n", +"I_b2 = a*I_a2 // Negative sequence current of line b(A)\n", +"I_c0 = I_a0 // Zero sequence current of line c(A)\n", +"I_c1 = a*I_a1 // Positive sequence current of line c(A)\n", +"I_c2 = a**2*I_a2 // Negative sequence current of line c(A)\n", +"// Case(c)\n", +"V_b = E_a/(Z_1+Z_2+Z_0)*((a**2-a)*Z_2+(a**2-1)*Z_0) // Voltage of the line b(V)\n", +"V_c = E_a/(Z_1+Z_2+Z_0)*((a-a**2)*Z_2+(a-1)*Z_0) // Voltage of the line c(V)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.3 : SOLUTION :-')\n", +"printf('\nCase(a): Fault current, |I_a| = %.f A', abs(I_a))\n", +"printf('\nCase(b): Zero sequence current of line a, I_a0 = %.f A', I_a0)\n", +"printf('\n Positive sequence current of line a, I_a1 = %.f A', I_a1)\n", +"printf('\n Negative sequence current of line a, I_a2 = %.f A', I_a2)\n", +"printf('\n Zero sequence current of line b, I_b0 = %.f A', I_b0)\n", +"printf('\n Positive sequence current of line b, I_b1 = (%.1f%.1fj) A', real(I_b1),imag(I_b1))\n", +"printf('\n Negative sequence current of line b, I_b2 = (%.1f+%.1fj) A', real(I_b2),imag(I_b2))\n", +"printf('\n Zero sequence current of line c, I_c0 = %.f A', I_c0)\n", +"printf('\n Positive sequence current of line c, I_c1 = (%.1f+%.1fj) A', real(I_c1),imag(I_c1))\n", +"printf('\n Negative sequence current of line c, I_c2 = (%.1f%.1fj) A', real(I_c2),imag(I_c2))\n", +"printf('\nCase(c): Voltage of the sound line to earth at fault, |V_b| = %.f V', abs(V_b))\n", +"printf('\n Voltage of the sound line to earth at fault, |V_c| = %.f V\n', abs(V_c))\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.4: EX30_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.4 :\n", +"// Page number 513-514\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 11000.0 // Alternator voltage(V)\n", +"kVA = 50000.0 // Alternator rating(kVA)\n", +"Z_l1 = complex(0.4,0.7) // Positive sequence impedance of feeder(ohm)\n", +"Z_l2 = complex(0.4,0.7) // Negative sequence impedance of feeder(ohm)\n", +"Z_l0 = complex(0.7,3.0) // Zero sequence impedance of feeder(ohm)\n", +"Z_g1_A = complex(0,0.6) // Positive sequence reactance(ohm)\n", +"Z_g1_B = complex(0,0.6) // Positive sequence reactance(ohm)\n", +"Z_g2_A = complex(0,0.4) // Negative sequence reactance(ohm)\n", +"Z_g2_B = complex(0,0.4) // Negative sequence reactance(ohm)\n", +"Z_g0_A = complex(0,0.2) // Zero sequence reactance(ohm)\n", +"Z_g0_B = complex(0,0.2) // Zero sequence reactance(ohm)\n", +"Z_n_A = complex(0,0.2) // Neutral reactance(ohm)\n", +"Z_n_B = complex(0,0.2) // Neutra reactance(ohm)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"Z_g1 = 1.0/((1/Z_g1_A)+(1/Z_g1_B)) // Equivalent positive sequence impedance(ohm)\n", +"Z_g2 = 1.0/((1/Z_g2_A)+(1/Z_g2_B)) // Equivalent negative sequence impedance(ohm)\n", +"Z_g0 = 1.0/((1/Z_g0_A)+(1/Z_g0_B)) // Equivalent zero sequence impedance(ohm)\n", +"Z_n = 1.0/((1/Z_n_A)+(1/Z_n_B)) // Equivalent neutral impedance(ohm)\n", +"Z_1 = Z_l1+Z_g1 // Positive sequence impedance(ohm)\n", +"Z_2 = Z_l2+Z_g2 // Negative sequence impedance(ohm)\n", +"Z_0 = Z_l0+Z_g0+3*Z_n // Zero sequence impedance(ohm)\n", +"Z = Z_0*Z_2/(Z_0+Z_2) // Impedance(ohm)\n", +"E_R = V/3**0.5 // Phase voltage(V)\n", +"I_R1 = E_R/(Z_1+Z) // Postive sequence current(A)\n", +"I_R2 = -Z*I_R1/Z_2 // Negative sequence current(A)\n", +"I_R0 = -Z*I_R1/Z_0 // Zero sequence current(A)\n", +"I_R = I_R0+I_R1+I_R2 // Fault current in line(A)\n", +"I_Y = I_R0+a**2*I_R1+a*I_R2 // Fault current in line(A)\n", +"I_B = I_R0+a*I_R1+a**2*I_R2 // Fault current in line(A)\n", +"I_earth = 3.0*I_R0 // Current through earth reactance(A)\n", +"V_neutral = abs(I_earth*Z_n) // Magnitude of potential above earth attained by generator neutral(V)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.4 : SOLUTION :-')\n", +"printf('\nFault current in the line R, I_R = %.f A', abs(I_R))\n", +"printf('\nFault current in the line Y, I_Y = (%.f%.fj) A', real(I_Y),imag(I_Y))\n", +"printf('\nFault current in the line B, I_B = (%.f+%.fj) A', real(I_B),imag(I_B))\n", +"printf('\nPotential above earth attained by the alternator neutrals = %.f V\n', V_neutral)\n", +"printf('\nNOTE: ERROR: Voltage is 11000 not 11000 kV as given in textbook statement')\n", +"printf('\n Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.5: Fault_currents.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.5 :\n", +"// Page number 514-515\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 6600.0 // Alternator voltage(V)\n", +"kVA = 10000.0 // Alternator rating(kVA)\n", +"x_1 = 15.0 // Reactance to positive sequence current(%)\n", +"x_2 = 75.0 // Reactance to negative sequence current(%)\n", +"x_0 = 30.0 // Reactance to zero sequence current(%)\n", +"R_earth = 0.3 // Earth resistance(ohm)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"E_g = V/3**0.5 // Phase voltage(V)\n", +"// Case(a)\n", +"I = kVA*1000/(3**0.5*V) // Full load current of each alternator(A)\n", +"X = x_1*V/(100*3**0.5*I) // Positive sequence reactance(ohm)\n", +"Z_g1 = %i*X // Equivalent positive sequence impedance(ohm)\n", +"Z_g2 = Z_g1*x_2/100 // Equivalent negative sequence impedance(ohm)\n", +"Z_g0 = Z_g1*x_0/100 // Equivalent zero sequence impedance(ohm)\n", +"Z_1 = Z_g1/3 // Positive sequence impedance(ohm)\n", +"Z_2 = Z_g2/3 // Negative sequence impedance(ohm)\n", +"Z_0 = Z_g0/3 // Zero sequence impedance(ohm)\n", +"I_a_a = 3*E_g/(Z_1+Z_2+Z_0) // Fault current(A)\n", +"// Case(b)\n", +"Z_0_b = Z_g0 // Impedance(ohm)\n", +"I_a_b = 3*E_g/(Z_1+Z_2+Z_0_b) // Fault current(A)\n", +"// Case(c)\n", +"Z_0_c = R_earth*3+Z_g0 // Impedance(ohm)\n", +"I_a_c = 3*E_g/(Z_1+Z_2+Z_0_c) // Fault current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.5 : SOLUTION :-')\n", +"printf('\nCase(a): Fault current if all the alternator neutrals are solidly earthed, I_a = %.fj A', imag(I_a_a))\n", +"printf('\nCase(b): Fault current if only one of the alternator neutrals is solidly earthed & others isolated = %.fj A', imag(I_a_b))\n", +"printf('\nCase(c): Fault current if one of alternator neutrals is earthed through resistance & others isolated = %.f A\n', abs(I_a_c))\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.6: Fault_current_for_line_fault_and_Line_to_ground_fault.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.6 :\n", +"// Page number 515-516\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_G = 2000.0 // Generator rating(kVA)\n", +"X_G = 10.0 // Generator reactance(%)\n", +"kVA_T1 = 2000.0 // Transformer rating(kVA)\n", +"lv_T1 = 6.6 // LV side voltage(kV)\n", +"hv_T1 = 11.0 // HV side voltage(kV)\n", +"X_T1 = 5.0 // Transformer reactance(%)\n", +"X_cable = 0.5 // Cable reactance(ohm)\n", +"V_cable = 11.0 // Cable voltage(V)\n", +"kVA_T2 = 2000.0 // Transformer rating(kVA)\n", +"lv_T2 = 6.6 // LV side voltage(kV)\n", +"hv_T2 = 11.0 // HV side voltage(kV)\n", +"X_T2 = 5.0 // Transformer reactance(%)\n", +"\n", +"// Calculations\n", +"a = exp(%i*120.0*%pi/180) // Operator\n", +"kVA_base = 2000.0 // Base kVA\n", +"kV = 6.6 // Base voltage(kV)\n", +"X_1 = X_G*kV**2*10/kVA_base // 10% reactance at 6.6 kV(ohm)\n", +"X_2 = X_T1*kV**2*10/kVA_base // 5% reactance at 6.6 kV(ohm)\n", +"X_3 = (kV/hv_T1)**2*X_cable // 0.5 ohm at 11kV when referred to 6.6kV(ohm)\n", +"Z_g1 = %i*X_1 // Positive sequence impedance of generator(ohm)\n", +"Z_g2 = Z_g1*0.7 // Negative sequence impedance of generator equal to 70% of +ve sequence impedance(ohm)\n", +"T1_Z_T1_1 = %i*X_2 // Positive sequence impedance of transformer(ohm)\n", +"T1_Z_T1_2 = %i*X_2 // Negative sequence impedance of transformer(ohm)\n", +"Z_C1 = %i*X_3 // Positive sequence impedance of cable(ohm)\n", +"Z_C2 = %i*X_3 // Negative sequence impedance of cable(ohm)\n", +"T2_Z_T2_1 = %i*X_2 // Positive sequence impedance of transformer(ohm)\n", +"T2_Z_T2_2 = %i*X_2 // Negative sequence impedance of transformer(ohm)\n", +"Z_1 = Z_g1+T1_Z_T1_1+Z_C1+T2_Z_T2_1 // Positive sequence impedance(ohm)\n", +"Z_2 = Z_g2+T1_Z_T1_2+Z_C2+T2_Z_T2_2 // Negative sequence impedance(ohm)\n", +"Z_0 = %i*X_2 // Zero sequence impedance(ohm)\n", +"E_a = kV*1000/3**0.5 // Phase voltage(V)\n", +"// Case(a)\n", +"I_a1 = E_a/(Z_1+Z_2) // Positive sequence current(A)\n", +"I_a2 = -I_a1 // Negative sequence current(A)\n", +"I_a0 = 0 // Zero sequence current(A)\n", +"I_a = I_a1+I_a2+I_a0 // Fault current in line a(A)\n", +"I_b = (a**2-a)*I_a1 // Fault current in line b(A)\n", +"I_c = -I_b // Fault current in line c(A)\n", +"// Case(b)\n", +"I_a_b = 3*E_a/(Z_1+Z_2+Z_0) // Fault current for line to ground fault(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.6 : SOLUTION :-')\n", +"printf('\nCase(a): Fault current for line fault are')\n", +"printf('\n I_a = %.f A', abs(I_a))\n", +"printf('\n I_b = %.f A', abs(I_b))\n", +"printf('\n I_c = %.f A', abs(I_c))\n", +"printf('\nCase(b): Fault current for line to ground fault, |I_a| = %.f A\n', abs(I_a_b))\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.7: Fault_current_for_a_LG_fault_at_C.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.7 :\n", +"// Page number 516-518\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_G1 = 40.0 // Generator rating(MVA)\n", +"kV_G1 = 13.2 // Generator voltage(kV)\n", +"X_st_G1 = 0.15 // Sub-transient reactance(p.u)\n", +"X_2_G1 = 0.15 // Negative sequence reactance(p.u)\n", +"X_0_G1 = 0.08 // Zero sequence reactance(p.u)\n", +"MVA_G3 = 60.0 // Generator rating(MVA)\n", +"kV_G3 = 13.8 // Generator voltage(kV)\n", +"X_st_G3 = 0.20 // Sub-transient reactance(p.u)\n", +"X_2_G3 = 0.20 // Negative sequence reactance(p.u)\n", +"X_0_G3 = 0.08 // Zero sequence reactance(p.u)\n", +"MVA_T1 = 40.0 // Transformer rating(MVA)\n", +"kV_lv_T1 = 13.8 // Transformer low voltage(kV)\n", +"kV_hv_T1 = 138 // Transformer high voltage(kV)\n", +"X_1_T1 = 0.10 // Positive sequence reactance(p.u)\n", +"X_2_T1 = 0.10 // Negative sequence reactance(p.u)\n", +"X_0_T1 = 0.08 // Zero sequence reactance(p.u)\n", +"MVA_T5 = 30.0 // Transformer rating(MVA)\n", +"kV_lv_T5 = 13.8 // Transformer low voltage(kV)\n", +"kV_hv_T5 = 138 // Transformer high voltage(kV)\n", +"X_1_T5 = 0.10 // Positive sequence reactance(p.u)\n", +"X_2_T5 = 0.10 // Negative sequence reactance(p.u)\n", +"X_0_T5 = 0.08 // Zero sequence reactance(p.u)\n", +"X_neutral = 0.05 // Reactance of reactor connected to generator neutral(p.u)\n", +"\n", +"// Calculations\n", +"MVA_base = 100.0 // Base MVA\n", +"kV_line = 138.0 // Base voltage for line(kV)\n", +"kV_G = 13.8 // Base voltage for generator(kV)\n", +"X_st_G1_pu = %i*X_st_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)\n", +"X_2_G1_pu = %i*X_2_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)\n", +"X_g0_G1_pu = %i*X_0_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)\n", +"X_gn_G1_pu = %i*X_neutral*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)\n", +"X_st_G3_pu = %i*X_st_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)\n", +"X_2_G3_pu = %i*X_2_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)\n", +"X_g0_G3_pu = %i*X_0_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)\n", +"X_gn_G3_pu = %i*X_neutral*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)\n", +"X_1_T1_pu = %i*X_1_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)\n", +"X_2_T1_pu = %i*X_2_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)\n", +"X_0_T1_pu = %i*X_0_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)\n", +"X_1_T5_pu = %i*X_1_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)\n", +"X_2_T5_pu = %i*X_2_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)\n", +"X_0_T5_pu = %i*X_0_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)\n", +"X_1_line_20 = %i*20.0*100/kV_line**2 // Impedance of 20 ohm line(p.u)\n", +"X_2_line_20 = %i*20.0*100/kV_line**2 // Impedance of 20 ohm line(p.u)\n", +"X_0_line_20 = 3.0*X_1_line_20 // Impedance of 20 ohm line(p.u)\n", +"X_1_line_10 = %i*10.0*100/kV_line**2 // Impedance of 10 ohm line(p.u)\n", +"X_2_line_10 = %i*10.0*100/kV_line**2 // Impedance of 10 ohm line(p.u)\n", +"X_0_line_10 = 3.0*X_1_line_10 // Impedance of 10 ohm line(p.u)\n", +"// Positive,negative and zero sequence network\n", +"Z_1_1 = X_1_T1_pu+X_1_T1_pu+X_1_line_20 // Impedance(p.u)\n", +"Z_2_1 = X_1_T1_pu+X_1_T5_pu+X_1_line_10 // Impedance(p.u)\n", +"Z_3_1 = X_1_T1_pu+X_1_T5_pu+X_1_line_10 // Impedance(p.u)\n", +"Z_4_1 = Z_1_1*Z_2_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)\n", +"Z_5_1 = Z_3_1*Z_1_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)\n", +"Z_6_1 = Z_3_1*Z_2_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)\n", +"Z_7_1 = X_st_G1_pu+Z_4_1 // Impedance(p.u)\n", +"Z_8_1 = X_st_G1_pu+Z_5_1 // Impedance(p.u)\n", +"Z_9_1 = Z_7_1*Z_8_1/(Z_7_1+Z_8_1) // Impedance in parallel(p.u). Refer Fig E4.14(e) & E4.14(f)\n", +"Z_10_1 = Z_9_1+Z_6_1 // Impedance(p.u). Refer Fig E4.14(f) & E4.14(g)\n", +"Z_11_1 = Z_10_1*X_st_G3_pu/(Z_10_1+X_st_G3_pu) // Impedance in parallel(p.u). Refer Fig E4.14(g) & E4.14(h)\n", +"Z_1 = Z_11_1 // Positive sequence impedance(p.u)\n", +"Z_2 = Z_1 // Negative sequence impedance(p.u)\n", +"Z_0 = X_g0_G3_pu+3.0*X_gn_G3_pu // Zero sequence impedance(p.u)\n", +"E_g = 1.0 // Voltage(p.u)\n", +"I_f_pu = 3*E_g/(Z_1+Z_2+Z_0) // L-G fault current(p.u)\n", +"I_f = abs(I_f_pu)*MVA_base*1000/(3**0.5*kV_G) // Actual fault current(A)\n", +"MVA_fault = abs(I_f_pu)*MVA_base // Fault MVA\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.7 : SOLUTION :-')\n", +"printf('\nFault current for a L-G fault at C = %.f A\n', I_f)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.8: Fault_current_when_a_single_phase_to_earth_fault_occurs.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.8 :\n", +"// Page number 518-519\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV_G = 11.0 // Generator rating(kV)\n", +"X_1_G = %i*0.1 // Positive sequence reactance of generator(p.u)\n", +"X_2_G = %i*0.1 // Negative sequence reactance of generator(p.u)\n", +"X_0_G = %i*0.02 // Zero sequence reactance of generator(p.u)\n", +"Z = 1.0 // Earthing resistor(ohm) \n", +"X_1_T1 = %i*0.1 // Positive sequence reactance of 2-winding transformer(p.u)\n", +"X_2_T1 = %i*0.1 // Negative sequence reactance of 2-winding transformer(p.u)\n", +"X_0_T1 = %i*0.1 // Zero sequence reactanc of 2-winding transformere(p.u)\n", +"X_1_T2_hv = %i*0.05 // Positive sequence reactance of hv 3-winding transformer(p.u)\n", +"X_2_T2_hv = %i*0.05 // Negative sequence reactance of hv 3-winding transformer(p.u)\n", +"X_0_T2_hv = %i*0.05 // Zero sequence reactanc of hv 3-winding transformere(p.u)\n", +"X_1_T2_lv_1 = %i*0.02 // Positive sequence reactance of lv 3-winding transformer(p.u)\n", +"X_2_T2_lv_1 = %i*0.02 // Negative sequence reactance of lv 3-winding transformer(p.u)\n", +"X_0_T2_lv_1 = %i*0.02 // Zero sequence reactanc of lv 3-winding transformere(p.u)\n", +"X_1_T2_lv_2 = %i*0.05 // Positive sequence reactance of lv 3-winding transformer(p.u)\n", +"X_2_T2_lv_2 = %i*0.05 // Negative sequence reactance of lv 3-winding transformer(p.u)\n", +"X_0_T2_lv_2 = %i*0.05 // Zero sequence reactanc of lv 3-winding transformere(p.u)\n", +"\n", +"// Calculations\n", +"MVA_b = 10.0 // Base MVA\n", +"kV_b = 11.0 // Base voltage(kV)\n", +"Z_n = Z*MVA_b/kV_b**2 // Impedance(p.u)\n", +"Z_1 = X_1_G+X_1_T1+X_1_T2_hv+((X_1_T2_lv_1*X_1_T2_lv_2)/(X_1_T2_lv_1+X_1_T2_lv_2)) // Positive sequence impedance(p.u)\n", +"Z_2 = X_2_G+X_2_T1+X_2_T2_hv+((X_2_T2_lv_1*X_2_T2_lv_2)/(X_2_T2_lv_1+X_2_T2_lv_2)) // Negative sequence impedance(p.u)\n", +"Z_0 = ((X_0_T1+X_0_T2_hv)*X_0_T2_lv_2/(X_0_T1+X_0_T2_hv+X_0_T2_lv_2))+X_0_T2_lv_1+3*Z_n // Zero sequence impedance(p.u)\n", +"E = 1.0 // Voltage(p.u)\n", +"I_f_pu = 3*E/(Z_1+Z_2+Z_0) // Fault current(p.u)\n", +"I_f = MVA_b*1000*abs(I_f_pu)/(3**0.5*kV_b) // Fault current(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.8 : SOLUTION :-')\n", +"printf('\nFault current, I_f = %.f A\n', I_f)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.9: Fault_currents_in_the_lines.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS\n", +"\n", +"// EXAMPLE : 4.9 :\n", +"// Page number 519\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA_G = 10.0 // Generator rating(MVA)\n", +"kV_G = 11.0 // Generator rating(kV)\n", +"X_1_G = 27.0 // Positive sequence reactance of generator(p.u)\n", +"X_2_G = 9.0 // Negative sequence reactance of generator(p.u)\n", +"X_0_G = 4.5 // Zero sequence reactance of generator(p.u)\n", +"X_1_L = 9.0 // Positive sequence reactance of line upto fault(p.u)\n", +"X_2_L = 9.0 // Negative sequence reactance of line upto fault(p.u)\n", +"X_0_L = 0 // Zero sequence reactance of line upto fault(p.u)\n", +"\n", +"// Calculations\n", +"E_a = kV_G*1000/3**0.5 // Phase voltage(V)\n", +"Z_1 = %i*(X_1_G+X_1_L) // Positive sequence reactance(p.u)\n", +"Z_2 = %i*(X_2_G+X_2_L) // Negative sequence reactance(p.u)\n", +"I_b = %i*3**0.5*E_a/(Z_1+Z_2) // Fault current in line b(p.u)\n", +"I_c = -I_b // Fault current in line c(p.u)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 4.9 : SOLUTION :-')\n", +"printf('\nFault current in line b, I_b = %.f A', abs(I_b))\n", +"printf('\nFault current in line c, I_c = %.f A', real(I_c))" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/32-CIRCUIT_BREAKER.ipynb b/Power_System_Engineering_by_S_Chakraborthy/32-CIRCUIT_BREAKER.ipynb new file mode 100644 index 0000000..590afcb --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/32-CIRCUIT_BREAKER.ipynb @@ -0,0 +1,280 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 32: CIRCUIT BREAKER" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.1: EX32_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 6: CIRCUIT BREAKER\n", +"\n", +"// EXAMPLE : 6.1 :\n", +"// Page number 545\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"f = 50.0 // Generator frequency(Hz)\n", +"kV = 7.5 // emf to neutral rms voltage(kV)\n", +"X = 4.0 // Reactance of generator & connected system(ohm)\n", +"C = 0.01*10**-6 // Distributed capacitance(F)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"v = 2**0.5*kV // Active recovery voltage i.e phase to neutral(kV)\n", +"V_max_restrike = v*2 // Maximum restriking voltage i.e phase to neutral(kV)\n", +"// Case(b)\n", +"L = X/(2.0*%pi*f) // Inductance(H)\n", +"f_n = 1/(2.0*%pi*(L*C)**0.5*1000) // Frequency of transient oscillation(kHZ)\n", +"// Case(c)\n", +"t = 1.0/(2.0*f_n*1000) // Time(sec)\n", +"avg_rate = V_max_restrike/t // Average rate of rise of voltage upto first peak of oscillation(kV/s)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 6.1 : SOLUTION :-')\n", +"printf('\nCase(a): Maximum re-striking voltage(phase-to-neutral) = %.1f kV', V_max_restrike)\n", +"printf('\nCase(b): Frequency of transient oscillation, f_n = %.1f kHz', f_n)\n", +"printf('\nCase(c): Average rate of rise of voltage upto first peak of oscillation = %.f kV/s \n', avg_rate)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.3: Rate_of_rise_of_restriking_voltage.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 6: CIRCUIT BREAKER\n", +"\n", +"// EXAMPLE : 6.3 :\n", +"// Page number 545-546\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 132.0 // Voltage(kV)\n", +"pf = 0.3 // Power factor of the fault\n", +"K3 = 0.95 // Recovery voltage was 0.95 of full line value\n", +"f_n = 16000.0 // Natural frequency of the restriking transient(Hz)\n", +"\n", +"// Calculations\n", +"kV_phase = kV/3**0.5 // System voltage(kV)\n", +"sin_phi = sind(acosd(pf)) // Sinφ\n", +"K2 = 1.0\n", +"v = K2*K3*kV/3**0.5*2**0.5*sin_phi // Active recovery voltage(kV)\n", +"V_max_restrike = 2*v // Maximum restriking voltage(kV)\n", +"t = 1.0/(2.0*f_n) // Time(sec)\n", +"RRRV = V_max_restrike/(t*10**6) // Rate of rise of restriking voltage(kV/µ-sec)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 6.3 : SOLUTION :-')\n", +"printf('\nRate of rise of restriking voltage, R.R.R.V = %.2f kV/µ-sec', RRRV)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.5: Voltage_across_the_pole_of_a_CB_and_Resistance_to_be_used_across_the_contacts.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 6: CIRCUIT BREAKER\n", +"\n", +"// EXAMPLE : 6.5 :\n", +"// Page number 565\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 132.0 // Voltage(kV)\n", +"C = 0.01*10**-6 // Phase to ground capacitance(F)\n", +"L = 6.0 // Inductance(H)\n", +"i = 5.0 // Magnetizing current(A)\n", +"\n", +"// Calculations\n", +"V_pros = i*(L/C)**0.5/1000 // Prospective value of voltage(kV)\n", +"R = 1.0/2*(L/C)**0.5/1000 // Resistance to be used across the contacts to eliminate the restriking voltage(k-ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 6.5 : SOLUTION :-')\n", +"printf('\nVoltage across the pole of a CB = %.1f kV', V_pros)\n", +"printf('\nResistance to be used across the contacts to eliminate the restriking voltage, R = %.2f k-ohm\n', R)\n", +"printf('\nNOTE: ERROR: Unit of final answer R is k-ohm, not ohm as in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.6: Rated_normal_current_Breaking_current_Making_current_and_Short_time_rating.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 6: CIRCUIT BREAKER\n", +"\n", +"// EXAMPLE : 6.6 :\n", +"// Page number 567\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I = 1200.0 // Rated normal current(A)\n", +"MVA = 1500.0 // Rated MVA\n", +"kV = 33.0 // Voltage(kV)\n", +"\n", +"// Calculations\n", +"I_breaking = MVA/(3**0.5*kV) // Rated symmetrical breaking current(kA)\n", +"I_making = I_breaking*2.55 // Rated making current(kA)\n", +"I_short = I_breaking // Short-time rating(kA)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 6.6 : SOLUTION :-')\n", +"printf('\nRated normal current = %.f A', I)\n", +"printf('\nBreaking current = %.2f kA (rms)', I_breaking)\n", +"printf('\nMaking current = %.f kA', I_making)\n", +"printf('\nShort-time rating = %.2f kA for 3 secs', I_short)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.8: EX32_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 6: CIRCUIT BREAKER\n", +"\n", +"// EXAMPLE : 6.8 :\n", +"// Page number 569\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 7500.0 // Rated kVA\n", +"X_st = 9.0 // Sub-transient reactance(%)\n", +"X_t = 15.0 // Transient reactance(%)\n", +"X_d = 100.0 // Direct-axis reactance(%)\n", +"kV = 13.8 // Voltage(kV). Assumption\n", +"\n", +"// Calculations\n", +"kVA_base = 7500.0 // Base kVA\n", +"kVA_sc_sustained = kVA_base/X_d*100 // Sustained S.C kVA\n", +"I_sc_sustained = kVA_base/(3**0.5*kV) // Sustained S.C current(A). rms\n", +"I_st = kVA*100/(X_st*3**0.5*kV) // Initial symmetrical rms current in the breaker(A)\n", +"I_max_dc = 2**0.5*I_st // Maximum possible dc component of the short-circuit(A)\n", +"I_moment = 1.6*I_st // Momentary current rating of the breaker(A)\n", +"I_interrupt = 1.1*I_st // Current to be interrupted by the breaker(A)\n", +"I_kVA = 3**0.5*I_interrupt*kV // Interrupting kVA\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 6.8 : SOLUTION :-')\n", +"printf('\nCase(a): Sustained short circuit KVA in the breaker = %.f kVA', kVA_sc_sustained)\n", +"printf('\n Sustained short circuit current in the breaker = %.1f A (rms)', I_sc_sustained)\n", +"printf('\nCase(b): Initial symmetrical rms current in the breaker = %.f A (rms)', I_st)\n", +"printf('\nCase(c): Maximum possible dc component of the short-circuit in the breaker = %.f A', I_max_dc)\n", +"printf('\nCase(d): Momentary current rating of the breaker = %.f A (rms)', I_moment)\n", +"printf('\nCase(e): Current to be interrupted by the breaker = %.f A (rms)', I_interrupt)\n", +"printf('\nCase(f): Interrupting kVA = %.f kVA \n', I_kVA)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook due to more approximation in textbook')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/33-PROTECTIVE_RELAYS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/33-PROTECTIVE_RELAYS.ipynb new file mode 100644 index 0000000..c58702d --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/33-PROTECTIVE_RELAYS.ipynb @@ -0,0 +1,286 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 33: PROTECTIVE RELAYS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.1: Time_of_operation_of_the_relay.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 7: PROTECTIVE RELAYS\n", +"\n", +"// EXAMPLE : 7.1 :\n", +"// Page number 595-596\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_setting = 150.0 // Current setting of IDMT(%)\n", +"t_mult = 0.5 // Time multiplier setting\n", +"ratio_CT = 500.0/5 // CT ratio\n", +"CT_sec = 5.0 // Secondary turn\n", +"I_f = 6000.0 // Fault current\n", +"\n", +"// Calculations\n", +"I_sec_fault = I_f/ratio_CT // Secondary fault current(A)\n", +"PSM = I_sec_fault/(CT_sec*I_setting/100) // Plug setting multiplier\n", +"t = 3.15 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"time_oper = t*t_mult // Operating time(sec)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 7.1 : SOLUTION :-')\n", +"printf('\nTime of operation of the relay = %.3f sec', time_oper)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.2: Time_of_operation_of_the_relay.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 7: PROTECTIVE RELAYS\n", +"\n", +"// EXAMPLE : 7.2 :\n", +"// Page number 596\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"ratio = 525.0/1 // CT ratio\n", +"CT_sec = 1.0 // Secondary turn\n", +"t_mult = 0.3 // Time multiplier setting\n", +"I_f = 5250.0 // Fault current(A)\n", +"\n", +"// Calculations\n", +"I_sec_fault = I_f/ratio // Secondary fault current(A)\n", +"PSM = I_sec_fault/(1.25*CT_sec) // Plug setting multiplier\n", +"t = 3.15 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"time_oper = t*t_mult // Operating time(sec)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 7.2 : SOLUTION :-')\n", +"printf('\nTime of operation of the relay = %.3f sec', time_oper)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.3: EX33_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 7: PROTECTIVE RELAYS\n", +"\n", +"// EXAMPLE : 7.3 :\n", +"// Page number 596\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 20.0 // Transformer MVA\n", +"overload = 30.0 // Overload of transformer(%)\n", +"kV = 11.0 // Bus bar rating(kV)\n", +"CT_trans = 1000.0/5 // Transformer CT\n", +"CT_cb = 400.0/5 // Circuit breaker CT\n", +"ps = 125.0 // Plug setting(%)\n", +"ts = 0.3 // Time setting\n", +"I_f = 5000.0 // Fault current(A)\n", +"t_margin = 0.5 // Discriminative time margin(sec)\n", +"\n", +"// Calculations\n", +"I_sec_fault = I_f/CT_cb // Secondary fault current(A)\n", +"CT_cb_sec = 5.0 // Secondary turn\n", +"PSM = I_sec_fault/(ps/100*CT_cb_sec) // Plug setting multiplier\n", +"t = 2.8 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"time_oper = t*ts // Operating time of feeder relay(sec)\n", +"I_ol = (1+(overload/100))*MVA*1000/(3**0.5*kV) // Overload current(A)\n", +"I_sec_T = I_ol/CT_trans // Secondary current(A)\n", +"CT_T_sec = 5.0 // Secondary turn of transformer\n", +"PSM_T = I_sec_T/CT_T_sec // Minimum plug setting multiplier of transformer\n", +"I_sec_T1 = I_f/CT_trans // Secondary fault current(A)\n", +"ps_T1 = 1.5 // Plug setting as per standard value\n", +"PSM_T1 = I_sec_T1/(CT_T_sec*ps) // Plug setting multiplier of transformer\n", +"t_T1 = 7.0 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"time_setting = (time_oper+t_margin)/t_T1 // Time setting of transformer\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 7.3 : SOLUTION :-')\n", +"printf('\nOperating time of feeder relay = %.2f sec', time_oper)\n", +"printf('\nMinimum plug setting of transformer relay, P.S > %.2f ', PSM_T)\n", +"printf('\nTime setting of transformer = %.3f ', time_setting)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.4: Time_of_operation_of_the_two_relays.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 7: PROTECTIVE RELAYS\n", +"\n", +"// EXAMPLE : 7.4 :\n", +"// Page number 596-597\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_f = 2000.0 // Fault current(A)\n", +"ratio_CT = 200.0/1 // CT ratio\n", +"R_1 = 100.0 // Relay 1 set on(%)\n", +"R_2 = 125.0 // Relay 2 set on(%)\n", +"t_margin = 0.5 // Discriminative time margin(sec)\n", +"TSM_1 = 0.2 // Time setting multiplier of relay 1\n", +"\n", +"// Calculations\n", +"CT_sec = 200.0 // CT secondary\n", +"PSM_1 = I_f*100/(CT_sec*R_1) // PSM of relay 1\n", +"t_1 = 2.8 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"time_oper_1 = TSM_1*t_1 // Operating time of relay with TSM of 0.2(Sec)\n", +"PSM_2 = I_f*100/(CT_sec*R_2) // PSM of relay 2\n", +"t_2 = 3.15 // Time against this PSM(sec). From graph E7.1 in textbook page no 595\n", +"actual_time_2 = time_oper_1+t_margin // Actual time of operation of relay 2(sec)\n", +"TSM_2 = actual_time_2/t_2 // Time setting multiplier of relay 2\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 7.4 : SOLUTION :-')\n", +"printf('\nTime of operation of relay 1 = %.2f sec', time_oper_1)\n", +"printf('\nActual time of operation of relay 2 = %.2f sec', actual_time_2)\n", +"printf('\nT.S.M of relay 2 = %.4f', TSM_2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.6: Will_the_relay_operate_the_trip_of_the_breaker.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 7: PROTECTIVE RELAYS\n", +"\n", +"// EXAMPLE : 7.6 :\n", +"// Page number 611\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_min = 0.1 // Relay minimum pick up current(A)\n", +"slope = 10.0 // Slope characteristic(%)\n", +"CT_ratio = 400.0/5 // CT ratio\n", +"I_1 = 320.0 // Current(A)\n", +"I_2 = 304.0 // Current(A)\n", +"\n", +"// Calculations\n", +"I_op_coil = (I_1-I_2)/CT_ratio // Current in operating coil(A)\n", +"I_re_coil = 1.0*(I_1+I_2)/(2*CT_ratio) // Current in restraining coil(A)\n", +"I_re_coil_slope = I_re_coil*slope/100 // Current in restraining coil with slope(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 7.6 : SOLUTION :-')\n", +"if(I_op_coil<I_re_coil_slope) then\n", +" printf('\nRelay will not trip the circuit breaker')\n", +"else then\n", +" print('\nRelay will trip the circuit breaker')\n", +"end" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/34-PROTECTION_OF_ALTERNATORS_AND_AC_MOTORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/34-PROTECTION_OF_ALTERNATORS_AND_AC_MOTORS.ipynb new file mode 100644 index 0000000..caa8cef --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/34-PROTECTION_OF_ALTERNATORS_AND_AC_MOTORS.ipynb @@ -0,0 +1,416 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 34: PROTECTION OF ALTERNATORS AND AC MOTORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.1: Neutral_earthing_reactance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.1 :\n", +"// Page number 624\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 6600.0 // Alternator Voltage(V)\n", +"P = 2000.0*10**3 // Rating of alternator(W)\n", +"PF = 0.8 // Power factor of alternator\n", +"X = 12.5 // Alternator reactance(%)\n", +"I = 200.0 // Current protection(A)\n", +"per = 10.0 // Percentage of winding unprotected(%)\n", +"\n", +"// Calculations\n", +"I_fl = P/(3**0.5*V*PF) // Full load current of alternator(A)\n", +"x = X*V/(3**0.5*100*I_fl) // Reactance per phase of alternator(ohm)\n", +"x_per = per/100*x // Reactance of 10% of the winding(ohm)\n", +"NA = V/(3**0.5*per) // Voltage induced in winding(V)\n", +"r = ((NA/I)**2-x_per**2)**0.5 // Neutral earthing reactance(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.1 : SOLUTION :-')\n", +"printf('\nNeutral earthing reactance, r = %.2f ohm', r)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.2: EX34_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.2 :\n", +"// Page number 624-625\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 20.0 // Generator rating(MVA)\n", +"V = 11.0*10**3 // Generator voltage(V)\n", +"ratio_CT = 1200.0/5 // Ratio of current transformer\n", +"I_min_op = 0.75 // Minimum operating current of relay(A)\n", +"R = 6.0 // Neutral point earthing resistance(ohm)\n", +"\n", +"// Calculations\n", +"I_max_fault = ratio_CT*I_min_op // Maximum fault current to operate relay(A)\n", +"x = I_max_fault*3**0.5*100*R/V // Unprotected portion for R = 6 ohm(%)\n", +"R_1 = 3.0 // Neutral point earthing resistance(ohm)\n", +"x_1 = I_max_fault*3**0.5*100*R_1/V // Unprotected portion for R = 3 ohm(%)\n", +"R_3 = 12.0 // Neutral point earthing resistance(ohm)\n", +"x_3 = I_max_fault*3**0.5*100*R_3/V // Unprotected portion for R = 12 ohm(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.2 : SOLUTION :-')\n", +"printf('\nUnprotected portion of each phase of the stator winding against earth fault, x = %.f percent', x)\n", +"printf('\nEffect of varying neutral earthing resistance keeping relay operating current the same :')\n", +"printf('\n (i) R = 3 ohms')\n", +"printf('\n Unprotected portion = %.1f percent', x_1)\n", +"printf('\n Protected portion = %.1f percent', (100-x_1))\n", +"printf('\n (ii) R = 6 ohms')\n", +"printf('\n Unprotected portion = %.f percent', x)\n", +"printf('\n Protected portion = %.f percent', (100-x))\n", +"printf('\n (iii) R = 12 ohms')\n", +"printf('\n Unprotected portion = %.f percent', x_3)\n", +"printf('\n Protected portion = %.f percent', (100-x_3))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.3: Portion_of_alternator_winding_unprotected.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.3 :\n", +"// Page number 625\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 5000.0 // Alternator rating(kVA)\n", +"V = 6600.0 // Alternator voltage(V)\n", +"X = 2.0 // Synchronous reactance per phase(ohm)\n", +"R = 0.5 // Resistance(ohm)\n", +"ofb = 30.0 // Out-of-balance current(%)\n", +"R_n = 6.5 // Resistance of resistor earthed to star point(ohm)\n", +"\n", +"// Calculations\n", +"I_fl = kVA*1000/(3**0.5*V) // Full load current(A)\n", +"I_ofb = ofb/100*I_fl // Out-of-balance current(A)\n", +"x = R_n/((V/(3**0.5*100*I_ofb))-(R/100)) // Portion of winding unprotected(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.3 : SOLUTION :-')\n", +"printf('\nPortion of alternator winding unprotected, x = %.1f percent', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.4: Will_the_relay_trip_the_generator_CB.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.4 :\n", +"// Page number 625\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_min = 0.15 // Minimum pick up current of relay(A)\n", +"slope = 12.0 // Slope(%)\n", +"CT_ratio = 400.0/5 // CT ratio\n", +"I_1 = 360.0 // Current(A)\n", +"I_2 = 300.0 // Current(A)\n", +"\n", +"// Calculations\n", +"i_1 = I_1/CT_ratio // Current(A)\n", +"i_2 = I_2/CT_ratio // Current(A)\n", +"percentage = (i_1-i_2)/((i_1+i_2)/2)*100 // Percentage(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.4 : SOLUTION :-')\n", +"if(percentage>slope) then\n", +" printf('\nRelay would trip the circuit breaker, since the point lie on +ve torque regime')\n", +"else then\n", +" printf('\nRelay would not trip the circuit breaker, since the point do not lie on +ve torque regime')\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.5: EX34_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.5 :\n", +"// Page number 625-626\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 50.0 // Alternator rating(MVA)\n", +"V = 33.0*10**3 // Alternator voltage(V)\n", +"CT_ratio = 2000.0/5 // CT ratio\n", +"R = 7.5 // Resistor earthed generator neutral(ohm)\n", +"I = 0.5 // Current above which pick up current(A)\n", +"\n", +"// Calculations\n", +"I_min = CT_ratio*I // Minimum current required to operate relay(A)\n", +"x = I_min*R/(V/3**0.5)*100 // Winding unprotected during normal operation(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.5 : SOLUTION :-')\n", +"printf('\nWinding of each phase unprotected against earth when machine operates at nominal voltage, x = %.2f percent', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.6: Portion_of_winding_unprotected.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.6 :\n", +"// Page number 626\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 50.0 // Alternator rating(MVA)\n", +"kV = 11.0 // Alternator voltage(kV)\n", +"X = 2.0 // Synchronous reactance per phase(ohm)\n", +"R = 0.7 // Resistance per phase(ohm)\n", +"R_n = 5.0 // Resistance through which alternator is earthed(ohm)\n", +"ofb = 25.0 // Out-of-balance current(%)\n", +"\n", +"// Calculations\n", +"I_fl = MVA*1000/(3**0.5*kV) // Full load current(A)\n", +"I_ofb = ofb/100*I_fl // Out-of-balance current(A)\n", +"x = R_n/((kV*1000/(3**0.5*100*I_ofb))-(R/100)) // Portion of winding unprotected(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.6 : SOLUTION :-')\n", +"printf('\nPortion of winding unprotected, x = %.f percent', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.7: Percentage_of_winding_that_is_protected_against_earth_faults.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.7 :\n", +"// Page number 626-627\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 11.0 // Alternator voltage(kV)\n", +"MVA = 5.0 // Alternator rating(MVA)\n", +"X = 2.0 // Reactance per phase(ohm)\n", +"ofb = 35.0 // Out-of-balance current(%)\n", +"R_n = 5.0 // Resistance through which star point is earthed(ohm)\n", +"\n", +"// Calculations\n", +"I_fl = MVA*1000/(3**0.5*kV) // Full load current(A)\n", +"I_ofb = ofb/100*I_fl // Out-of-balance current(A)\n", +"x = I_ofb*R_n*100/(kV*1000/3**0.5) // Portion of winding unprotected(%)\n", +"protected = 100.0-x // Winding that is protected against earth faults(%)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.7 : SOLUTION :-')\n", +"printf('\nPercentage of winding that is protected against earth faults = %.2f percent', protected)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.8: Magnitude_of_neutral_earthing_resistance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 8: PROTECTION OF ALTERNATORS AND AC MOTORS\n", +"\n", +"// EXAMPLE : 8.8 :\n", +"// Page number 627\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kV = 11.0 // Alternator voltage(kV)\n", +"P = 100.0 // Alternator maximum rating(MW)\n", +"PF = 0.8 // Power factor\n", +"X = 0.1 // Reactance of alternator(pu)\n", +"i = 500.0 // Current(A)\n", +"per = 10.0 // Windings unprotected(%)\n", +"\n", +"// Calculations\n", +"I = P*1000/(3**0.5*kV*PF) // Rated current of alternator(A)\n", +"a = i/I // Relay setting\n", +"I_n = a*I*100/per // Current through neutral(A)\n", +"R = kV*1000/(3**0.5*I_n) // Magnitude of neutral earthing resistance(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 8.8 : SOLUTION :-')\n", +"printf('\nMagnitude of neutral earthing resistance, R = %.2f ohm\n', R) \n", +"printf('\nNOTE: ERROR: Unit of resistance is not mentioned in textbook solution')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/35-PROTECTION_OF_TRANSFORMERS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/35-PROTECTION_OF_TRANSFORMERS.ipynb new file mode 100644 index 0000000..320e6e7 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/35-PROTECTION_OF_TRANSFORMERS.ipynb @@ -0,0 +1,259 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 35: PROTECTION OF TRANSFORMERS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.2: Ratio_of_CTs.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 9: PROTECTION OF TRANSFORMERS\n", +"\n", +"// EXAMPLE : 9.2 :\n", +"// Page number 635-636\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_lv = 220.0 // LV side voltage of transformer(V)\n", +"V_hv = 11000.0 // HV side voltage of transformer(V)\n", +"ratio_CT = 600.0/(5/3**0.5) // CT ratio on LV side of transformer\n", +"\n", +"// Calculations\n", +"CT_pri = 600.0 // Primary CT\n", +"CT_sec = 5.0/3**0.5 // Secondary CT\n", +"I_1 = V_lv/V_hv*CT_pri // Line current in secondary of transformer corresponding to primary winding(A)\n", +"I_2 = CT_sec*3**0.5 // Current in secondary of CT(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 9.2 : SOLUTION :-')\n", +"printf('\nRatio of CTs on 11000 V side = %.f : %.f \n', I_1,I_2)\n", +"printf('\nNOTE: ERROR: Mistake in representing the final answer in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.3: Ratio_of_CTs_on_high_voltage_side.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 9: PROTECTION OF TRANSFORMERS\n", +"\n", +"// EXAMPLE : 9.3 :\n", +"// Page number 636\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_lv = 11.0*10**3 // LV side voltage of transformer(V)\n", +"V_hv = 66.0*10**3 // HV side voltage of transformer(V)\n", +"ratio_CT = 250.0/5 // CT ratio on LV side of transformer\n", +"\n", +"// Calculations\n", +"V_hv_phase = V_hv/3**0.5 // HV side phase voltage(V)\n", +"ratio_main_T = V_hv_phase/V_lv // Ratio of main transformer\n", +"I_2 = 250.0 // Primary CT\n", +"I_1 = I_2/(ratio_main_T*3**0.5) // Primary line current(A)\n", +"CT_sec = 5.0 // Secondary CT\n", +"secondary_side = CT_sec/3**0.5 // HV side CT secondary\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 9.3 : SOLUTION :-')\n", +"printf('\nRatio of CTs on high voltage side = %.1f : %.1f = (%.f/%.2f√3) : (%.f/√3) ', I_1,secondary_side,I_2,ratio_main_T,CT_sec)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.4: Ratio_of_protective_CTs.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 9: PROTECTION OF TRANSFORMERS\n", +"\n", +"// EXAMPLE : 9.4 :\n", +"// Page number 636\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_hv = 33.0 // HV side voltage of transformer(kV)\n", +"V_lv = 6.6 // LV side voltage of transformer(kV)\n", +"ratio_CT = 100.0/1 // CT ratio on LV side of transformer\n", +"\n", +"// Calculations\n", +"CT_pri = 100.0 // Primary CT\n", +"CT_sec = 1.0 // Secondary CT\n", +"I_hv = V_lv/V_hv*CT_pri // Line current on HV side(A)\n", +"I_lv = CT_sec/3**0.5 // Line current on LV side(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 9.4 : SOLUTION :-')\n", +"printf('\nRatio of protective CTs on 33 kV side = %.f : %.f/√3 = %.f : %.f ', I_hv,CT_sec,3**0.5*I_hv,I_lv*3**0.5)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.5: CT_ratios_on_high_voltage_side.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 9: PROTECTION OF TRANSFORMERS\n", +"\n", +"// EXAMPLE : 9.5 :\n", +"// Page number 636-637\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA = 200.0 // Transformer rating(kVA)\n", +"E_1 = 11000.0 // HV side voltage of transformer(kV)\n", +"E_2 = 400.0 // LV side voltage of transformer(kV)\n", +"ratio_CT = 500.0/5 // CT ratio on LV side of transformer\n", +"I_f = 750.0 // Fault current(A)\n", +"\n", +"// Calculations\n", +"I_2 = 500.0 // Primary CT\n", +"I_1 = 5.0 // Secondary CT\n", +"I_1_T = E_2*I_2/(3**0.5*E_1) // Primary current in transformer(A)\n", +"I_hv_T = I_1_T*3**0.5 // Equivalent line current on HV side(A)\n", +"I_pilot_lv = I_1*3**0.5 // Pilot current on LV side(A)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 9.5 : SOLUTION :-')\n", +"printf('\nCT ratios on high voltage side = %.2f : %.2f \n', I_hv_T,I_pilot_lv)\n", +"printf('\nNOTE: Circulating current is not calculated')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.6: Suitable_CT_ratios.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 9: PROTECTION OF TRANSFORMERS\n", +"\n", +"// EXAMPLE : 9.6 :\n", +"// Page number 640\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MVA = 50.0 // Transformer rating(MVA)\n", +"V_hv = 132.0 // HV side voltage of transformer(kV)\n", +"V_lv = 33.0 // LV side voltage of transformer(kV)\n", +"CT_sec = 1.0 // Secondary CT rating\n", +"\n", +"// Calculations\n", +"I_FL = MVA*1000/(3**0.5*V_lv) // Full-load current(A)\n", +"CT_ratio_33kV = I_FL/CT_sec // CT ratio on 33 kV side\n", +"CT_ratio_132kV = (I_FL*V_lv/V_hv)/(CT_sec/3**0.5) // CT ratio on 132 kV side\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 9.6 : SOLUTION :-')\n", +"printf('\nCT ratio on 33 kV side = %.f : 1 ', CT_ratio_33kV)\n", +"printf('\nCT ratio on 132 kV side = %.f : 1 = %.f√3 : 1 ', CT_ratio_132kV,CT_ratio_132kV/3**0.5)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/36-PROTECTION_OF_TRANSMISSION_LINE_SHUNT_INDUCTORS_AND_CAPACITORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/36-PROTECTION_OF_TRANSMISSION_LINE_SHUNT_INDUCTORS_AND_CAPACITORS.ipynb new file mode 100644 index 0000000..2111286 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/36-PROTECTION_OF_TRANSMISSION_LINE_SHUNT_INDUCTORS_AND_CAPACITORS.ipynb @@ -0,0 +1,154 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 36: PROTECTION OF TRANSMISSION LINE SHUNT INDUCTORS AND CAPACITORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.1: First_Second_and_Third_zone_relay_setting_Without_infeed_and_With_infeed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 10: PROTECTION OF TRANSMISSION LINE, SHUNT INDUCTORS AND CAPACITORS\n", +"\n", +"// EXAMPLE : 10.1 :\n", +"// Page number 647-648\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"G2_per = 70.0 // G2 is fed at 70% distance from A in section AB(%)\n", +"X_T = 10.0 // Transformer reactance(%)\n", +"zone_1_per = 80.0 // Setting for first zone(%)\n", +"zone_2_per = 50.0 // Setting for second zone(%)\n", +"CT_ratio = 400.0/5 // CT ratio\n", +"PT_ratio = 166000.0/110 // PT ratio\n", +"Z_AB = complex(20.0,60.0) // Section AB impedance(ohm)\n", +"Z_BC = complex(10.0,25.0) // Section BC impedance(ohm)\n", +"MVA = 10.0 // Transformer rating(MVA)\n", +"kV_hv = 166.0 // HV side voltage(kV)\n", +"kV_lv = 33.0 // LV side voltage(kV)\n", +"\n", +"// Calculations\n", +"// Case(i) Without infeed\n", +"Z_sec_1 = zone_1_per/100*Z_AB*CT_ratio/PT_ratio // First zone setting(ohm)\n", +"Z_BC_hv = Z_BC*(kV_hv/kV_lv)**2 // Z_BC on 166 kV base(ohm)\n", +"Z_T = %i*10*X_T*kV_hv**2/(MVA*1000) // Transformer impedance(ohm)\n", +"Z_sec_2 = (Z_AB+zone_2_per/100*Z_BC_hv+Z_T)*CT_ratio/PT_ratio // Second zone setting(ohm)\n", +"Z_sec_3 = (Z_AB+Z_BC_hv+Z_T)*CT_ratio/PT_ratio // Third zone setting(ohm)\n", +"// Case(ii) With infeed\n", +"I_AB = 2.0 // Current ratio\n", +"Z_zone_1 = (G2_per/100*Z_AB)+I_AB*(zone_1_per-G2_per)/100*Z_AB // First zone impedance(ohm)\n", +"Z_1 = Z_zone_1*CT_ratio/PT_ratio // First zone setting(ohm)\n", +"Z_zone_2 = (G2_per/100*Z_AB)+I_AB*(((zone_1_per-zone_2_per)/100*Z_AB)+(zone_2_per/100*Z_BC_hv)+Z_T) // Second zone impedance(ohm)\n", +"Z_2 = Z_zone_2*CT_ratio/PT_ratio // Second zone setting(ohm)\n", +"under_reach = Z_zone_2-(Z_AB+zone_2_per/100*Z_BC_hv+Z_T) // Under-reach due to infeed(ohm)\n", +"Z_zone_3 = (G2_per/100*Z_AB)+I_AB*(((zone_1_per-zone_2_per)/100*Z_AB)+Z_BC_hv+Z_T) // Third zone impedance(ohm)\n", +"Z_3 = Z_zone_3*CT_ratio/PT_ratio // Third zone setting(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 10.1 : SOLUTION :-')\n", +"printf('\nCase(i) Without infeed:')\n", +"printf('\n First zone relay setting = (%.2f + %.2fj) ohm', real(Z_sec_1),imag(Z_sec_1))\n", +"printf('\n Second zone relay setting = (%.1f + %.1fj) ohm', real(Z_sec_2),imag(Z_sec_2))\n", +"printf('\n Third zone relay setting = (%.1f + %.1fj) ohm', real(Z_sec_3),imag(Z_sec_3))\n", +"printf('\nCase(ii) With infeed:')\n", +"printf('\n First zone relay setting = (%.3f + %.2fj) ohm', real(Z_1),imag(Z_1))\n", +"printf('\n Second zone relay setting = (%.1f + %.1fj) ohm', real(Z_2),imag(Z_2))\n", +"printf('\n Third zone relay setting = (%.1f + %.fj) ohm\n', real(Z_3),imag(Z_3))\n", +"printf('\nNOTE: ERROR: Calculation mistake in Z_BC. Hence, changes in the obtained answer from that of textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.2: Impedance_seen_by_relay_and_Relay_setting_for_high_speed_backup_protection.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART III : SWITCHGEAR AND PROTECTION\n", +"// CHAPTER 10: PROTECTION OF TRANSMISSION LINE, SHUNT INDUCTORS AND CAPACITORS\n", +"\n", +"// EXAMPLE : 10.2 :\n", +"// Page number 648\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"CT_ratio = 300.0/5 // CT ratio\n", +"PT_ratio = 166000.0/110 // PT ratio\n", +"Z_AB = complex(40.0,160.0) // Section AB impedance(ohm)\n", +"Z_BC = complex(7.5,15.0) // Section BC impedance(ohm)\n", +"kV_hv = 166.0 // HV side voltage(kV)\n", +"kV_lv = 33.0 // LV side voltage(kV)\n", +"MVA = 5.0 // Transformer rating(MVA)\n", +"X_T = 6.04 // Transformer reactance(%)\n", +"\n", +"// Calculations\n", +"Z_T = %i*10*X_T*kV_hv**2/(MVA*1000) // Tranformer impedance(ohm)\n", +"Z_fault = Z_AB+Z_T // Fault impedance(ohm)\n", +"Z_sec = Z_fault*CT_ratio/PT_ratio // Relay setting for primary protection(ohm)\n", +"Z_BC_hv = Z_BC*(kV_hv/kV_lv)**2 // Z_BC on 166 kV base(ohm)\n", +"Z = Z_AB+Z_T+Z_BC_hv // For backup protection of line BC(ohm)\n", +"Z_sec_set = Z*CT_ratio/PT_ratio // Relay setting(ohm)\n", +"\n", +"// Results\n", +"disp('PART III - EXAMPLE : 10.2 : SOLUTION :-')\n", +"printf('\nImpedance seen by relay = (%.f + %.fj) ohm', real(Z_fault),imag(Z_fault))\n", +"printf('\nRelay setting for high speed & backup protection = (%.1f + %.2fj) ohm', real(Z_sec_set),imag(Z_sec_set))" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/39-INDUSTRIAL_APPLICATIONS_OF_ELECTRIC_MOTORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/39-INDUSTRIAL_APPLICATIONS_OF_ELECTRIC_MOTORS.ipynb new file mode 100644 index 0000000..927dc10 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/39-INDUSTRIAL_APPLICATIONS_OF_ELECTRIC_MOTORS.ipynb @@ -0,0 +1,1665 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 39: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.10: Additional_resistance_to_be_inserted_in_the_field_circuit_to_raise_the_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.10 :\n", +"// Page number 687\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 220.0 // DC shunt motor voltage(V)\n", +"I_a1 = 50.0 // Armature current at 800rpm(A)\n", +"N_1 = 800.0 // Speed of dc shunt motor(rpm)\n", +"N_2 = 1000.0 // Speed of dc shunt motor with additional resistance(rpm)\n", +"I_a2 = 75.0 // Armature current with additional resistance(A)\n", +"R_a = 0.15 // Armature resistance(ohm)\n", +"R_f = 250.0 // Field resistance(ohm)\n", +"\n", +"// Calculations\n", +"E_b1 = V-R_a*I_a1 // Back emf at 800 rpm(V)\n", +"I_f1 = V/R_f // Shunt field current(A)\n", +"E_b2 = V-R_a*I_a2 // Back emf at 1000 rpm(V)\n", +"I_f2 = E_b2*N_1*I_f1/(E_b1*N_2) // Shunt field current at 1000 rpm(A)\n", +"R_f2 = V/I_f2 // Field resistance at 1000 rpm(ohm)\n", +"R_add = R_f2-R_f // Additional resistance required(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.10 : SOLUTION :-')\n", +"printf('\nAdditional resistance to be inserted in the field circuit to raise the speed = %.1f ohm\n', R_add)\n", +"printf('\nNOTE: ERROR: Calculation mistake in E_b2 in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.11: Speed_of_motor_with_a_diverter_connected_in_parallel_with_series_field.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.11 :\n", +"// Page number 687\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 220.0 // DC series motor voltage(V)\n", +"I_1 = 20.0 // Armature current at 800rpm(A)\n", +"N_1 = 800.0 // Speed of dc series motor(rpm)\n", +"R_div = 0.4 // Diverter resistance(ohm)\n", +"R_a = 0.5 // Armature resistance(ohm)\n", +"R_f = 0.2 // Series field resistance(ohm)\n", +"\n", +"// Calculations\n", +"E_b1 = V-(R_a+R_f)*I_1 // Back emf at 800 rpm(V)\n", +"I_2 = I_1*R_div/(R_div+R_f) // Series field current at new speed(A)\n", +"E_b2 = V-(R_a*I_1+R_f*I_2) // Back emf at new speed(V)\n", +"N_2 = I_1*N_1*E_b2/(I_2*E_b1) // New speed with diverter(rpm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.11 : SOLUTION :-')\n", +"printf('\nSpeed of motor with a diverter connected in parallel with series field, N_2 = %.f rpm', N_2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.12: Diverter_resistance_as_a_percentage_of_field_resistance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.12 :\n", +"// Page number 687-688\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"speed_per = 15.0 // Motor speed increased by(%)\n", +"\n", +"// Calculations\n", +"N_2 = (100+speed_per)/100 // New speed N_2(rpm)\n", +"phi_2 = 1/N_2*100 // Flux_2 in terms of full load flux\n", +"I_sc1 = 0.75 // New series field current in terms of I_a1\n", +"I_a2 = N_2 // Armature current in terms of I_a1\n", +"R_d = I_sc1/(I_a2-I_sc1)*100 // Diverter resistance in terms of series field resistance(%)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.12 : SOLUTION :-')\n", +"printf('\nDiverter resistance, R_d = %.1f percent of field resistance', R_d)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.13: Additional_resistance_to_be_placed_in_the_armature_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.13 :\n", +"// Page number 689\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 250.0 // Voltage of DC shunt motor(V)\n", +"N_1 = 400.0 // No load speed(rpm)\n", +"R_a = 0.5 // Armature resistance(ohm)\n", +"N_2 = 200.0 // Speed with additional resistance(rpm)\n", +"I_a = 20.0 // Armature current(A)\n", +"\n", +"// Calculations\n", +"k_phi = (V-I_a*R_a)/N_1 // kΦ\n", +"R = (V-k_phi*N_2)/I_a // Resistance(ohm)\n", +"R_add = R-R_a // Additional resistance to be placed in armature circuit(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.13 : SOLUTION :-')\n", +"printf('\nResistance to be placed in the armature circuit = %.f ohm\n', R_add)\n", +"printf('\nNOTE: ERROR: The given data doesnt match with example 1.7 as mentioned in problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.14: Resistance_to_be_connected_in_series_with_armature_to_reduce_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.14 :\n", +"// Page number 689\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // Voltage of DC shunt motor(V)\n", +"hp = 20.0 // Power of DC shunt motor(hp)\n", +"I = 44.0 // Current drawn by motor(A)\n", +"N_1 = 1000.0 // Speed(rpm)\n", +"N_2 = 800.0 // Speed with additional resistance(rpm)\n", +"R_sh = 200.0 // Shunt field resistance(ohm)\n", +"\n", +"// Calculations\n", +"output = hp*746 // Motor output(W)\n", +"I_f1 = V/R_sh // Shunt field current(A)\n", +"I_a1 = I-I_f1 // Armature current(A)\n", +"E_b1 = output/I_a1 // Back emf(V)\n", +"R_a = (V-E_b1)/I_a1 // Armature resistance(ohm)\n", +"I_a2 = I_a1*(N_2/N_1)**2 // Armature current at N2(A)\n", +"E_b2 = N_2/N_1*E_b1 // Back emf at N2(V)\n", +"r = ((V-E_b2)/I_a2)-R_a // Resistance connected in series with armature(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.14 : SOLUTION :-')\n", +"printf('\nResistance to be connected in series with armature to reduce speed, r = %.2f ohm', r)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.15: Ohmic_value_of_resistor_connected_in_the_armature_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.15 :\n", +"// Page number 690\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 15.0 // Power of DC shunt motor(hp)\n", +"V = 400.0 // Voltage of DC shunt motor(V)\n", +"N_reduce = 20.0 // Speed is to be reduced by(%)\n", +"I_f = 3.0 // Field current(A)\n", +"R_a = 0.5 // Armature resistance(ohm)\n", +"n = 0.85 // Efficiency of motor\n", +"\n", +"// Calculations\n", +"motor_input = hp*746/n // Motor input(W)\n", +"I = motor_input/V // Motor current(A)\n", +"I_a1 = I-I_f // Armature current(A)\n", +"I_a2 = I_a1 // Armature current at new speed(A)\n", +"E_b1 = V-I_a1*R_a // Back emf(V)\n", +"E_b2 = E_b1*(100-N_reduce)/100 // Back emf at new speed(V)\n", +"r = ((V-E_b2)/I_a2)-R_a // Ohmic value of resistor connected in the armature circuit(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.15 : SOLUTION :-')\n", +"printf('\nOhmic value of resistor connected in the armature circuit, r = %.2f ohm', r)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.16: External_resistance_per_phase_added_in_rotor_circuit_to_reduce_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.16 :\n", +"// Page number 697-698\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"p = 6.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"R_2 = 0.3 // Rotor resistance per phase(ohm)\n", +"N_1 = 960.0 // Rotor speed(rpm)\n", +"N_2 = 800.0 // New rotor speed with external resistance(rpm)\n", +"\n", +"// Calculations\n", +"N_s = 120*f/p // Synchronous speed(rpm)\n", +"S_1 = (N_s-N_1)/N_s // Slip at full load\n", +"S_2 = (N_s-N_2)/N_s // New slip\n", +"R = (S_2/S_1*R_2)-R_2 // External resistance per phase added in rotor circuit to reduce speed(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.16 : SOLUTION :-')\n", +"printf('\nExternal resistance per phase added in rotor circuit to reduce speed, R = %.1f ohm', R)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.17: Braking_torque_and_Torque_when_motor_speed_has_fallen.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.17 :\n", +"// Page number 699\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 50.0 // DC shunt motor rating(hp)\n", +"V = 440.0 // Voltage(V)\n", +"I_b = 150.0 // Breaking current(A)\n", +"N_reduce = 40.0 // Speed of motor fallen by(%)\n", +"R_a = 0.1 // Armature resistance(ohm)\n", +"I_a_fl = 100.0 // Full-load armature current(A)\n", +"N_fl = 600.0 // Full-load speed(rpm)\n", +"\n", +"// Calculations\n", +"E_b = V-I_a_fl*R_a // Back emf of motor(V)\n", +"V_a = V+E_b // Voltage across armature when braking starts(V)\n", +"R_b = V_a/I_b // Resistance required(ohm)\n", +"R_extra = R_b-R_a // Extra resistance required(ohm)\n", +"T_fl = hp*746*60/(2*%pi*N_fl) // Full-load torque(N-m)\n", +"T_initial_b = T_fl*I_b/I_a_fl // Initial breaking torque(N-m)\n", +"E_b2 = E_b*(100-N_reduce)/100 // Back emf at new speed(V)\n", +"I = (V+E_b2)/R_b // Current(A)\n", +"EBT = T_fl*I/I_a_fl // Torque when motor speed reduced by 40%(N-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.17 : SOLUTION :-')\n", +"printf('\nBraking torque = %.1f N-m', T_initial_b)\n", +"printf('\nTorque when motor speed has fallen, E.B.T = %.1f N-m\n', EBT)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.18: Initial_plugging_torque_and_Torque_at_standstill.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.18 :\n", +"// Page number 699-700\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // Voltage of IM(V)\n", +"p = 4.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"hp = 25.0 // Power developed(hp)\n", +"S = 0.04 // Slip\n", +"R_X_2 = 1.0/4 // Ratio of rotor resistance to standstill reactance i.e R2/X2\n", +"\n", +"// Calculations\n", +"N_s = 120*f/p // Synchronous speed(rpm)\n", +"N_fl = N_s*(1-S) // Full load speed(rpm)\n", +"T_fl = hp*735.5*60/(2*%pi*N_fl*9.81) // Full-load torque(kg-m)\n", +"S_1 = 1.0 // Slip at standstill\n", +"X_R_2 = 1.0/R_X_2 // Ratio of standstill reactance to rotor resistance\n", +"T_s_fl = S_1/S*((1+(S*X_R_2)**2)/(1+(S_1*X_R_2)**2)) // T_standstill/T_fl\n", +"T_standstill = T_s_fl*T_fl // Standstill torque(kg-m)\n", +"S_instant = (N_s+N_fl)/N_s // Slip at instant of plugging\n", +"T_initial = (S_instant/S)*((1+(S*X_R_2)**2)/(1+(S_instant*X_R_2)**2))*T_fl // Initial plugging torque(kg-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.18 : SOLUTION :-')\n", +"printf('\nInitial plugging torque = %.1f kg-m', T_initial)\n", +"printf('\nTorque at standstill = %.f kg-m\n', T_standstill)\n", +"printf('\nNOTE: ERROR: Calculation mistake from full-load torque onwards. Hence, change in obtained answer from that of textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.19: Value_of_resistance_to_be_connected_in_motor_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.19 :\n", +"// Page number 701\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"T = 312.5 // Load torque(N-m)\n", +"N = 500.0 // Speed limit(rpm)\n", +"R_total = 1.0 // Total resistance of armature & field(ohm)\n", +"\n", +"// Calculations\n", +"input_load = 2*%pi*N*T/60 // Input from load(W)\n", +"E = 345.0 // Voltage from magnetization curve(V). From Fig E1.5 page no 701\n", +"I = 47.5 // Current from magnetization curve(A). From Fig E1.5 page no 701\n", +"R = E/I // Resistance(ohm)\n", +"R_add = R-R_total // Additional resistance required(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.19 : SOLUTION :-')\n", +"printf('\nValue of resistance to be connected in motor circuit = %.2f ohm', R_add)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.1: Total_annual_cost_of_group_drive_and_Individual_drive.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.1 :\n", +"// Page number 676\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"capital_cost_group = 8000.0 // Capital cost of group drive(Rs)\n", +"n_single = 5.0 // Number of individual drive\n", +"capital_cost_single = 2500.0 // Capital cost of individual drive(Rs)\n", +"energy_cons_group = 40000.0 // Annual energy consumption of group drive(kWh)\n", +"energy_cons_single = 30000.0 // Annual energy consumption of group drive(kWh)\n", +"cost_energy = 8.0/100 // Cost of energy per kWh(Rs)\n", +"dmo_group = 12.0 // Depreciation,maintenance & other fixed charges for group drive(%)\n", +"dmo_single = 18.0 // Depreciation,maintenance & other fixed charges for individual drive(%)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"annual_cost_energy_a = energy_cons_group*cost_energy // Annual cost of energy(Rs)\n", +"dmo_cost_a = capital_cost_group*dmo_group/100 // Depreciation,maintenance & other fixed charges per year for group drive(Rs)\n", +"yearly_cost_a = annual_cost_energy_a+dmo_cost_a // Total yearly cost(Rs)\n", +"// Case(b)\n", +"total_cost = capital_cost_single*n_single // Capital cost of individual drive(Rs)\n", +"annual_cost_energy_b = energy_cons_single*cost_energy // Annual cost of energy(Rs)\n", +"dmo_cost_b = total_cost*dmo_single/100 // Depreciation,maintenance & other fixed charges per year for individual drive(Rs)\n", +"yearly_cost_b = annual_cost_energy_b+dmo_cost_b // Total yearly cost(Rs)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.1 : SOLUTION :-')\n", +"printf('\nTotal annual cost of group drive = Rs. %.f ', yearly_cost_a)\n", +"printf('\nTotal annual cost of individual drive = Rs. %.f ', yearly_cost_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.20: EX39_20.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.20 :\n", +"// Page number 702\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"V = 500.0 // Shunt motor voltage(V)\n", +"load = 400.0 // Hoist load(kg)\n", +"speed = 2.5 // Hoist raised speed(m/sec)\n", +"n_motor = 0.85 // Efficiency of motor\n", +"n_hoist = 0.75 // Efficiency of hoist\n", +"\n", +"// Calculations\n", +"P_output = load*speed*9.81 // Power output from motor(W)\n", +"P_input = P_output/(n_motor*n_hoist) // Motor input(W)\n", +"I = P_input/V // Current drawn from supply(A)\n", +"output_G = load*speed*9.81*n_motor*n_hoist // Generator output(W)\n", +"R = V**2/output_G // Resistance required in the armature circuit for rheostatic braking(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.20 : SOLUTION :-')\n", +"printf('\nCurrent drawn by the motor from supply = %.1f A', I)\n", +"printf('\nResistance required in the armature circuit for rheostatic braking, R = %.f ohm', R)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.21: One_hour_rating_of_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.21 :\n", +"// Page number 705\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"t = 1.0 // Time(hour)\n", +"hp = 15.0 // Motor rating(hp)\n", +"T = 2.0 // Time constant(hour)\n", +"theta_f = 40.0 // Temperature rise(°C)\n", +"\n", +"// Calculations\n", +"P = (1.0/(1-exp(-t/T)))**0.5*hp // One-hour rating of motor(hp)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.21 : SOLUTION :-')\n", +"printf('\nOne-hour rating of motor, P = %.f hp\n', P)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.22: Final_temperature_rise_and_Thermal_time_constant_of_the_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.22 :\n", +"// Page number 706\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 10.0 // Motor rating(hp)\n", +"d = 0.7 // Diameter of cylinder(m)\n", +"l = 1.0 // Length of cylinder(m)\n", +"w = 380.0 // Weight of motor(kgm)\n", +"heat_specific = 700.0 // Specific heat(J/kg/1°C)\n", +"heat_dissipation = 15.0 // Outer surface heat dissipation rate(W/sq.cm/°C)\n", +"n = 0.88 // Efficiency\n", +"\n", +"// Calculations\n", +"output = hp*735.5 // Output of motor(W)\n", +"loss = (1-n)/n*output // Losses(W)\n", +"area_cooling = %pi*d*l // Cooling surface area(sq.m)\n", +"theta_m = loss/(area_cooling*heat_dissipation) // Final temperature rise(°C)\n", +"T_sec = w*heat_specific/(area_cooling*heat_dissipation) // Thermal time constant(sec)\n", +"T_hour = T_sec/3600 // Thermal time constant(hours)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.22 : SOLUTION :-')\n", +"printf('\nFinal temperature rise, θ_m = %.1f°C', theta_m)\n", +"printf('\nThermal time constant of the motor = %.2f hours\n', T_hour)\n", +"printf('\nNOTE: ERROR: Mistake in calculating thermal time constant in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.23: Half_hour_rating_of_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.23 :\n", +"// Page number 706\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 25.0 // Motor rating(hp)\n", +"T = 100.0/60 // Heating time constant(hour)\n", +"theta = 40.0 // Temperature rise(°C)\n", +"t = 0.5 // Time(hour)\n", +"n = 0.85 // Motor maximum efficiency\n", +"\n", +"// Calculations\n", +"output = hp*735.5/1000 // Output of motor(kW)\n", +"output_max = output*n // Power at maximum efficiency(kW)\n", +"theta_f2 = theta/(1-exp(-t/T)) // θ_f2(°C)\n", +"loss = 1+(output/output_max)**2 // Losses at 18.4 kW output in terms of W\n", +"P = ((theta_f2/theta*loss)-1)**0.5*output_max // Half-hour rating of motor(kW)\n", +"P_hp = P*1000/735.5 // Half-hour rating of motor(hp)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.23 : SOLUTION :-')\n", +"printf('\nHalf-hour rating of motor, P = %.f kW = %.1f hp (metric)\n', P,P_hp)\n", +"printf('\nNOTE: ERROR: Calculation mistake from final temperature rise onwards in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.24: EX39_24.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.24 :\n", +"// Page number 706\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"theta_f1 = 40.0 // Temperature rise(°C)\n", +"T = 100.0 // Heating time constant(min)\n", +"rated_2 = 2.0 // Motor at twice the continuously rating\n", +"\n", +"// Calculations\n", +"loss_cu = 2.0**2 // Copper loss at twice full load in terms of W\n", +"loss_total = loss_cu+1 // Total losses at full load in terms of W\n", +"theta_f2 = theta_f1*loss_total/rated_2 // θ_f2(°C)\n", +"t = log(1-(theta_f1/theta_f2))*(-T) // Time for which motor can run at twice the continuously rated output without overheating(min)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.24 : SOLUTION :-')\n", +"printf('\nMotor can run at twice the continuously rated output without overheating for time, t = %.f min', t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.25: Maximum_overload_that_can_be_carried_by_the_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.25 :\n", +"// Page number 706-707\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kW = 20.0 // Motor output(kW)\n", +"theta_1 = 50.0 // Temperature rise not to be exceeded on overload(°C)\n", +"t_1 = 1.0 // Time on overload(hour)\n", +"theta_2 = 30.0 // Temperature rise on full-load(°C)\n", +"t_2 = 1.0 // Time on full-load(hour)\n", +"theta_3 = 40.0 // Temperature rise on full-load(°C)\n", +"t_3 = 2.0 // Time on full-load(hour)\n", +"\n", +"// Calculations\n", +"e_lambda = 1.0/3 // Obtained directly from textbook\n", +"theta_f = theta_2/(1-e_lambda) // θ_f(°C)\n", +"theta_f1 = theta_1/(1-e_lambda) // θ'_f(°C)\n", +"P = (theta_f1/theta_f)**0.5*kW // Maximum overload that can be carried by the motor(kW)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.25 : SOLUTION :-')\n", +"printf('\nMaximum overload that can be carried by the motor, P = %.1f kW', P)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.26: Required_size_of_continuously_rated_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.26 :\n", +"// Page number 707-708\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp_1 = 100.0 // Motor load(hp)\n", +"t_1 = 10.0 // Time of operation(min)\n", +"hp_2 = 0 // Motor load(hp)\n", +"t_2 = 5.0 // Time of operation(min)\n", +"hp_3 = 60.0 // Motor load(hp)\n", +"t_3 = 8.0 // Time of operation(min)\n", +"hp_4 = 0 // Motor load(hp)\n", +"t_4 = 4.0 // Time of operation(min)\n", +"\n", +"// Calculations\n", +"t_total = t_1+t_2+t_3+t_4 // Total time of operation(min)\n", +"rms = ((hp_1**2*t_1+hp_2**2*t_2+hp_3**2*t_3+hp_4**2*t_4)/t_total)**0.5 // rms horsepower\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.26 : SOLUTION :-')\n", +"printf('\nRequired size of continuously rated motor = %.f H.P\n', rms)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook')\n", +"printf('\n Actual value is written here instead of standard values')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.27: Suitable_size_of_the_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.27 :\n", +"// Page number 708\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp_1 = 200.0 // Motor load(hp)\n", +"t_1 = 5.0 // Time of operation(min)\n", +"hp_2 = 100.0 // Motor load(hp)\n", +"t_2 = 10.0 // Time of operation(min)\n", +"hp_3 = 0 // Motor load(hp)\n", +"t_3 = 3.0 // Time of operation(min)\n", +"\n", +"// Calculations\n", +"m = hp_1/t_1 // Slope of uniform rise power\n", +"t_total = t_1+t_2+t_3 // Total time of operation(min)\n", +"ans = integrate('(m*x)**2','x', 0, t_1) // Integarted uniform area upto 5 min\n", +"rms = ((ans+hp_2**2*t_2+hp_3**2*t_3)/t_total)**0.5 // rms horsepower\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.27 : SOLUTION :-')\n", +"printf('\nrms horsepower = %.1f HP. Therefore, a motor of %.f H.P should be selected', rms,rms+4)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.28: Time_taken_to_accelerate_the_motor_to_rated_speed_against_full_load_torque.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.28 :\n", +"// Page number 710\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 440.0 // DC shunt motor voltage(V)\n", +"hp = 50.0 // Motor rating(hp)\n", +"N = 600.0 // Speed(rpm)\n", +"I = 80.0 // Current at full-load(A)\n", +"I_1 = 1.1 // Lower current limit in terms of full current\n", +"I_2 = 1.5 // Upper current limit in terms of full current\n", +"J = 20.0 // Moment of inertia(kg-m^2)\n", +"\n", +"// Calculations\n", +"T = hp*746*60/(2*%pi*N) // Full load torque of motor(N-m)\n", +"T_avg_start = (I_1+I_2)/2*T // Average starting torque(N-m)\n", +"T_g = ((I_1+I_2)/2-1)*T // Torque available for acceleration(N-m)\n", +"alpha = T_g/J // Angular acceleration(rad/sec^2)\n", +"t = 2*%pi*N/(60*alpha) // Time taken to accelerate the motor(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.28 : SOLUTION :-')\n", +"printf('\nTime taken to accelerate the motor to rated speed against full load torque, t = %.2f sec\n', t)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.29: Time_taken_to_accelerate_the_motor_to_rated_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.29 :\n", +"// Page number 710\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 50.0 // Motor rating(hp)\n", +"N = 600.0 // Speed(rpm)\n", +"energy = 276.0 // Stored energy(kg-m/hp)\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"T = hp*746*60/(2*%pi*N*g) // Full load torque of motor(kg-m)\n", +"J = hp*energy*2*g/(2*%pi*N/60)**2 // Moment of inertia(kg-m^2)\n", +"alpha = T*g/J // Angular acceleration(rad/sec^2)\n", +"t = 2*%pi*N/(60*alpha) // Time taken to accelerate the motor to rated speed(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.29 : SOLUTION :-')\n", +"printf('\nTime taken to accelerate the motor to rated speed, t = %.2f sec', t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.2: EX39_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.2 :\n", +"// Page number 680\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_sc = 6.0 // Short circuit current = 6 times full load current\n", +"s_fl = 5.0 // Full load slip(%)\n", +"tap = 60.0 // Auto-tranformer tapping(%)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"I_s_fl_a = I_sc/3.0 // I_s/I_fl\n", +"T_s_fl_a = I_s_fl_a**2*s_fl/100 // Starting torque in terms of full-load torque with star-delta starter\n", +"// Case(b)\n", +"I_s_fl_b = tap/100*I_sc // I_s/I_fl\n", +"T_s_fl_b = I_s_fl_b**2*s_fl/100 // Starting torque in terms of full-load torque with auto-transformer starter\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.2 : SOLUTION :-')\n", +"printf('\nCase(a): Starting torque in terms of full-load torque with star-delta starter, I_s/I_fl = %.1f ', T_s_fl_a)\n", +"printf('\nCase(b): Starting torque in terms of full-load torque with auto-transformer starter, I_s/I_fl = %.3f ', T_s_fl_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.30: Time_taken_to_accelerate_a_fly_wheel.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.30 :\n", +"// Page number 710\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"J = 1270.0 // Moment of inertia of fly-wheel(kg-m^2)\n", +"N = 500.0 // Speed(rpm)\n", +"hp = 50.0 // Motor rating(hp)\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"T = hp*746*60/(2*%pi*N*g) // Full load torque of motor(kg-m)\n", +"T_m = 2*T // Accelerating torque(kg-m)\n", +"alpha = T_m*g/J // Angular acceleration(rad/sec^2)\n", +"t = 2*%pi*N/(60*alpha) // Time taken to accelerate a fly-wheel(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.30 : SOLUTION :-')\n", +"printf('\nTime taken to accelerate a fly-wheel, t = %.1f sec', t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.31: EX39_31.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.31 :\n", +"// Page number 710-711\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"N_1 = 1000.0 // Speed of dc shunt motor(rpm)\n", +"N_2 = 400.0 // Speed of dc shunt motor(rpm)\n", +"R = 14.0 // Resistance connected across armature(ohm)\n", +"E_1 = 210.0 // EMF induced in armature at 1000 rpm(V)\n", +"J = 17.0 // Moment of inertia(kg-m^2)\n", +"T_F = 1.0 // Frictional torque(kg-m)\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"output = E_1**2/R // Motor output(W)\n", +"T_E = output*60/(2*%pi*N_1*g) // Electric braking torque(kg-m)\n", +"w_1 = 2*%pi*N_1/60 // ω_1(rad/sec)\n", +"k = T_E/w_1\n", +"t = J/(g*k)*log(N_1/N_2) // Time taken for dc shunt motor to fall in speed with constant excitation(sec)\n", +"kw = T_E*N_2/N_1 // kω\n", +"t_F = J/(g*k)*log((1+T_E)/(1+kw)) // Time for the same fall if frictional torque exists(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.31 : SOLUTION :-')\n", +"printf('\nTime taken for dc shunt motor to fall in speed with constant excitation, t = %.1f sec', t)\n", +"printf('\nTime for the same fall if frictional torque exists, t = %.1f sec', t_F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.32: EX39_32.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.32 :\n", +"// Page number 711\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // Voltage of synchronous motor(V)\n", +"p = 8.0 // Number of poles\n", +"J = 630.0 // Moment of inertia(kg-m^2)\n", +"T_E = 165.0 // Braking torque(kg-m)\n", +"kw_1 = 690.0 // Electric braking torque(kg-m)\n", +"T_F = 1.4 // Frictional torque(kg-m)\n", +"f = 50.0 // Frequency(Hz). Assumed normal supply frequency\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"// Case(a) Plugging\n", +"T_B = T_E+T_F // Torque(kg-m)\n", +"beta = T_B*g/J // Retardation(rad/sec^2)\n", +"N_s = 120*f/p // Synchronous speed(rad/sec)\n", +"w = 2*%pi*N_s/60 // ω(rad/sec)\n", +"t_a = integrate('-1.0/beta','w', w, 0) // Time taken to stop the motor(sec)\n", +"n_a = integrate('-w/(2*%pi*beta)','w', w, 0) // Number of revolutions\n", +"// Case(b) Rheostatic braking\n", +"k = kw_1/w\n", +"t_b = J/(g*k)*log((T_F+kw_1)/T_F) // Time taken to stop the motor(sec)\n", +"n_b = 1.0/(2*%pi*k)*(J/(g*k)*(T_F+kw_1)*(1-exp(-k*g*t_b/J))-T_F*t_b) // Number of revolutions\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.32 : SOLUTION :-')\n", +"printf('\nCase(a): Time taken to come to standstill by plugging, t = %.1f sec', t_a)\n", +"printf('\n Number of revolutions made to come to standstill by plugging, n = %.f revolutions', n_a)\n", +"printf('\nCase(b): Time taken to come to standstill by rheostatic braking, t = %.1f sec', t_b)\n", +"printf('\n Number of revolutions made to come to standstill by rheostatic braking, n = %.f revolutions\n', n_b)\n", +"printf('\nNOTE: ERROR: Calculation mistake in finding number of revolution in case(a) in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.33: Inertia_of_flywheel_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.33 :\n", +"// Page number 712-713\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 500.0 // Rating of IM(hp)\n", +"N_nl = 40.0 // No-load speed(rpm)\n", +"S_fl = 0.12 // Slip at full-load\n", +"T_l = 41500.0 // Load torque(kg-m)\n", +"t = 10.0 // Duration of each rolling period(sec)\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"T_fl = hp*746*60/(2*%pi*N_nl*g*(1-S_fl)) // Torque at full-load(kg-m)\n", +"T_m = 2.0*T_fl // Motor torque at any instant(kg-m)\n", +"slip = S_fl*N_nl // Slip(rpm)\n", +"slip_rad = slip*2*%pi/60 // Slip(rad/sec)\n", +"k = slip_rad/T_fl\n", +"J = -g*t/(k*log(1-(T_m/T_l))) // Inertia of flywheel(kg-m^2)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.33 : SOLUTION :-')\n", +"printf('\nInertia of flywheel required, J = %.3e kg-m^2\n', J)\n", +"printf('\nNOTE: ERROR : J = 2.93*10^6 kg-m^2 and not 2.93*10^5 as mentioned in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.34: Moment_of_inertia_of_the_flywheel.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.34 :\n", +"// Page number 713\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"T_l = 150.0 // Load torque(kg-m)\n", +"t = 15.0 // Duration of load torque(sec)\n", +"T_m = 85.0 // Motor torque(kg-m)\n", +"N = 500.0 // Speed(rpm)\n", +"s_fl = 0.1 // Full-load slip\n", +"\n", +"// Calculations\n", +"g = 9.81\n", +"slip = N*s_fl*2*%pi/60 // Slip(rad/sec)\n", +"k = slip/T_m\n", +"T_0 = 0 // No-load torque(kg-m)\n", +"J = -g*t/(k*log((T_l-T_m)/(T_l-T_0))) // Moment of inertia of flywheel(kg-m^2)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.34 : SOLUTION :-')\n", +"printf('\nInertia of flywheel required, J = %.f kg-m^2\n', J)\n", +"printf('\nNOTE: ERROR : Calculation mistake in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.3: EX39_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.3 :\n", +"// Page number 680-681\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // IM voltage(V)\n", +"s_fl = 5.0 // Full-load slip(%)\n", +"I_fl = 20.0 // Full load current drawn from supply by IM(A)\n", +"Z = 2.5 // Impedance per phase(ohm)\n", +"I_max = 50.0 // Maximum current drawn(A)\n", +"\n", +"// Calculations\n", +"V_phase = V/3**0.5 // Normal phase voltage(V)\n", +"P = (100**2*I_max*Z/V_phase)**0.5 // Tapping to be provided to auto-transformer(%)\n", +"I_s = I_max/(P/100) // Starting current taken by motor(A)\n", +"T_s_fl = (I_s/I_fl)**2*s_fl/100 // Starting torque in terms of full-load torque\n", +"T_s_fl_R = (I_max/I_fl)**2*s_fl/100 // Starting torque in terms of full-load torque when a resistor is used\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.3 : SOLUTION :-')\n", +"printf('\nTapping to be provided on an auto-transformer, P = %.1f percent', P)\n", +"printf('\nStarting torque in terms of full-load torque, T_s = %.3f*T_fl ', T_s_fl)\n", +"printf('\nStarting torque in terms of full-load torque if a resistor were used in series, T_s = %.4f*T_fl ', T_s_fl_R)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.4: EX39_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.4 :\n", +"// Page number 681-682\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 30.0 // Power of cage IM(hp)\n", +"V = 500.0 // Cage IM voltage(V)\n", +"P = 4.0 // Number of poles\n", +"f = 50.0 // Frequency(Hz)\n", +"I_fl = 33.0 // Full load current(A)\n", +"s = 4.0/100 // Slip\n", +"Z = 3.5 // Impedance per phase(ohm)\n", +"tap = 60.0 // Auto-transformer tap setting(%)\n", +"\n", +"// Calculations\n", +"// Case(1)\n", +"I_s_1 = 3**0.5*(V/Z) // Starting current taken from line(A)\n", +"N_s = 120*f/P // Speed(rpm)\n", +"N_fl = N_s-N_s*s // Full load speed of motor(rpm)\n", +"T_fl = hp*746*60/(2*%pi*N_fl) // Full load torque(N-m)\n", +"T_s_1 = (I_s_1/I_fl)**2*s*T_fl // Starting torque(N-m)\n", +"// Case(2)\n", +"V_ph = V/3**0.5 // Phase voltage in star(V)\n", +"I_s_2 = V_ph/Z // Starting current(A/phase)\n", +"T_s_2 = (I_s_2/(I_fl/3**0.5))**2*s*T_fl // Starting torque(N-m)\n", +"// Case(3)\n", +"V_ph_at = V*tap/(3**0.5*100) // Phase voltage of auto-transformer secondary(V)\n", +"V_impressed = V_ph_at*3**0.5 // Volatage impressed on delta-connected stator(V)\n", +"I_s_3 = V_impressed/Z // Starting current(A/phase)\n", +"I_s_line = 3**0.5*I_s_3 // Motor starting line current from auto-transformer secondary(A)\n", +"I_s_line_3 = tap/100*I_s_line // Starting current taken from supply(A)\n", +"T_s_3 = (I_s_3/(I_fl/3**0.5))**2*s*T_fl // Starting torque(N-m)\n", +"// Case(4)\n", +"I_s_4 = 3**0.5*V/Z // Starting current from line(A)\n", +"T_s_4 = T_fl*s*(I_s_4/I_fl)**2 // Starting torque(N-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.4 : SOLUTION :-')\n", +"printf('\nCase(1): Starting torque for direct switching, T_s = %.f N-m', T_s_1)\n", +"printf('\n Starting current taken from supply line for direct switching, I_s = %.f A', I_s_1)\n", +"printf('\nCase(2): Starting torque for star-delta starting, T_s = %.f N-m', T_s_2)\n", +"printf('\n Starting current taken from supply line for star-delta starting, I_s = %.1f A per phase', I_s_2)\n", +"printf('\nCase(3): Starting torque for auto-transformer starting, T_s = %.f N-m', T_s_3)\n", +"printf('\n Starting current taken from supply line for auto-transformer starting, I_s = %.f A', I_s_line_3)\n", +"printf('\nCase(4): Starting torque for series-parallel switch, T_s = %.f N-m', T_s_4)\n", +"printf('\n Starting current taken from supply line for series-parallel switch, I_s = %.f A\n', I_s_4)\n", +"printf('\nNOTE: ERROR: Calculation mistakes and more approximation in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.5: EX39_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.5 :\n", +"// Page number 682\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 400.0 // IM voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"I_s = 5.0 // Full voltage starting current in terms of full load current\n", +"T_s = 2.0 // Full voltage starting torque in terms of full load torque\n", +"tap = 65.0 // Auto-tranformer tapping(%)\n", +"\n", +"// Calculations\n", +"V_ph = V/3**0.5 // Phase voltage(V)\n", +"V_ph_motor = tap/100*V_ph // Motor phase voltage when auto-transformer is used(V)\n", +"I_ph_motor = tap/100*I_s // Motor phase current in terms of full load current\n", +"I_1 = tap/100*I_ph_motor // Line current from supply in terms of full load current\n", +"T = (tap/100)**2*T_s // Starting torque in terms of full load current\n", +"V_applied = V_ph/2**0.5 // Voltage to be applied to develop full-load torque(V)\n", +"I_line = V_applied/V_ph*I_s // Line current in terms of full load current\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.5 : SOLUTION :-')\n", +"printf('\nCase(i): Motor current per phase = %.2f*I_fl ', I_ph_motor)\n", +"printf('\nCase(ii): Current from the supply, I_1 = %.2f*I_fl ', I_1)\n", +"printf('\nCase(iii): Starting torque with auto-transformer starter, T = %.3f*T_fl ', T)\n", +"printf('\nVoltage to be applied if motor has to develop full-load torque at starting, V = %.f V', V_applied)\n", +"printf('\nLine current from the supply to develop full-load torque at starting = %.2f*I_fl ', I_line)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.6: Ratio_of_starting_current_to_full_load_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.6 :\n", +"// Page number 682\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hp = 10.0 // IM rating(hp)\n", +"V = 400.0 // IM voltage(V)\n", +"pf = 0.8 // Lagging power factor\n", +"n = 0.9 // Efficiency of IM\n", +"I_sc = 7.2 // Short-circuit current at 160V(A)\n", +"V_sc = 160.0 // Voltage at short-circuit(V)\n", +"\n", +"// Calculations\n", +"I_fl = hp*746/(3**0.5*V*pf*n) // Full-load line current(A)\n", +"I_sc_fv = V/V_sc*I_sc // Short-circuit current at full voltage(A)\n", +"I_s = I_sc_fv/3.0 // Starting current with star-delta starter(A)\n", +"I_s_fl = I_s/I_fl // Ratio of starting current to full load current\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.6 : SOLUTION :-')\n", +"printf('\nRatio of starting current to full-load current, I_s/I_fl = %.1f \n', I_s_fl)\n", +"printf('\nNOTE: ERROR: Calculation mistake in final answer in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.7: Resistance_to_be_placed_in_series_with_shunt_field.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.7 :\n", +"// Page number 685-686\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 230.0 // Voltage of DC shunt motor(V)\n", +"N_1 = 1000.0 // No load speed(rpm)\n", +"R_sh = 40.0 // Shunt resistance(ohm)\n", +"N_2 = 1200.0 // Speed with series resistance(rpm)\n", +"\n", +"// Calculations\n", +"phi_2 = N_1/N_2 // Flux_2 in terms flux_1\n", +"I_N1 = V/R_sh // Exciting current at 1000 rpm(A)\n", +"phi_1 = 11.9 // Flux corresponding to I_N1(mWb)\n", +"phi_N2 = phi_1*phi_2 // Flux at 1200 rpm(mWb)\n", +"I_phi_N2 = 3.25 // Exciting current corresponding to phi_N2(A)\n", +"R = V/I_phi_N2 // Resistance in field circuit(ohm)\n", +"R_extra = R-R_sh // Resistance to be placed in series with shunt field(ohm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.7 : SOLUTION :-')\n", +"printf('\nResistance to be placed in series with shunt field = %.1f ohm', R_extra)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.9: Speed_and_Current_when_field_winding_is_shunted_by_a_diverter.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS\n", +"\n", +"// EXAMPLE : 1.9 :\n", +"// Page number 686-687\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_f1 = 25.0 // Current without diverter(A)\n", +"N_1 = 500.0 // Speed of dc series motor without diverter(rpm)\n", +"\n", +"// Calculations\n", +"I_a2 = ((3.0/2)**0.5*I_f1**2*3/2)**0.5 // Field current with diverter(A)\n", +"N_2 = I_f1*N_1*3/(2*I_a2) // Speed with diverter(rpm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 1.9 : SOLUTION :-')\n", +"printf('\nSpeed when field winding is shunted by a diverter, N_2 = %.f rpm', N_2)\n", +"printf('\nCurrent when field winding is shunted by a diverter, I_a2 = %.1f A', I_a2)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/40-HEATING_AND_WELDING.ipynb b/Power_System_Engineering_by_S_Chakraborthy/40-HEATING_AND_WELDING.ipynb new file mode 100644 index 0000000..7342e4c --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/40-HEATING_AND_WELDING.ipynb @@ -0,0 +1,490 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 40: HEATING AND WELDING" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.1: Diameter_Length_and_Temperature_of_the_wire.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.1 :\n", +"// Page number 724-725\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P = 15.0*10**3 // Power supplied(W)\n", +"V = 220.0 // Voltage(V)\n", +"T_w = 1000.0 // Temperature of wire(°C)\n", +"T_c = 600.0 // Temperature of charges(°C)\n", +"k = 0.6 // Radiatting efficiency\n", +"e = 0.9 // Emissivity\n", +"\n", +"// Calculations\n", +"rho = 1.016/10**6 // Specific resistance(ohm-m)\n", +"d_square = 4*rho*P/(%pi*V**2) // d^2 in terms of l\n", +"T_1 = T_w+273 // Absolute temperature(°C)\n", +"T_2 = T_c+273 // Absolute temperature(°C)\n", +"H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) // Heat produced(watts/sq.m)\n", +"dl = P/(%pi*H)\n", +"l = (dl**2/d_square)**(1.0/3) // Length of wire(m)\n", +"d = dl/l // Diameter of wire(m)\n", +"T_2_cold = 20.0+273 // Absolute temperature at the 20°C normal temperature(°C)\n", +"T_1_cold = (H/(5.72*10**4*k*e)+(T_2_cold/1000)**4)**(1.0/4)*1000 // Absolute temperature when charge is cold(°C)\n", +"T_1_c = T_1_cold-273 // Temperature when charge is cold(°C)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.1 : SOLUTION :-')\n", +"printf('\nDiameter of the wire, d = %.3f cm', d*100)\n", +"printf('\nLength of the wire, l = %.2f m', l)\n", +"printf('\nTemperature of the wire when charge is cold, T_1 = %.f°C absolute = %.f°C \n', T_1_cold,T_1_c)\n", +"printf('\nNOTE: Slight changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.2: Width_and_Length_of_nickel_chrome_strip.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.2 :\n", +"// Page number 725\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"P = 15.0*10**3 // Power supplied(W)\n", +"V = 220.0 // Voltage(V)\n", +"T_w = 1000.0 // Temperature of wire(°C)\n", +"T_c = 600.0 // Temperature of charges(°C)\n", +"k = 0.6 // Radiatting efficiency\n", +"e = 0.9 // Emissivity\n", +"thick = 0.25/1000 // Thickness of nickel-chrome strip(m)\n", +"\n", +"// Calculations\n", +"rho = 1.016/10**6 // Specific resistance(ohm-m)\n", +"R = V**2/P // Resistance(ohm)\n", +"l_w = R*thick/rho // Length of strip in terms of w\n", +"T_1 = T_w+273 // Absolute temperature(°C)\n", +"T_2 = T_c+273 // Absolute temperature(°C)\n", +"H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) // Heat produced(watts/sq.m)\n", +"wl = P/(2*H)\n", +"w = (wl/l_w)**0.5 // Width of nickel-chrome strip(m)\n", +"l = w*l_w // Length of nickel-chrome strip(m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.2 : SOLUTION :-')\n", +"printf('\nWidth of nickel-chrome strip, w = %.3f cm', w*100)\n", +"printf('\nLength of nickel-chrome strip, l = %.1f m', l)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.3: Power_drawn_under_various_connections.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.3 :\n", +"// Page number 726-727\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"R = 50.0 // Resistance of each resistor in oven(ohm)\n", +"n = 6.0 // Number of resistance\n", +"V = 400.0 // Supply voltage(V)\n", +"tap = 50.0 // Auto-transformer tapping(%)\n", +"\n", +"// Calculations\n", +"// Case(a)(i)\n", +"P_a_i = n*V**2/R*10**-3 // Power consumption for 6 elements in parallel(kW)\n", +"// Case(a)(ii)\n", +"P_each_a_ii = V**2/(R+R)*10**-3 // Power consumption in each group of 2 resistances in series(kW)\n", +"P_a_ii = n/2*P_each_a_ii // Power consumption for 3 groups(kW)\n", +"// Case(b)(i)\n", +"V_b_i = V/3**0.5 // Supply voltage against each resistance(V)\n", +"P_each_b_i = 2*V_b_i**2/R*10**-3 // Power consumption in each branch(kW)\n", +"P_b_i = n/2*P_each_b_i // Power consumption for 2 elements in parallel in each phase(kW)\n", +"// Case(b)(ii)\n", +"V_b_ii = V/3**0.5 // Supply voltage to any branch(V)\n", +"P_each_b_ii = V_b_ii**2/(R+R)*10**-3 // Power consumption in each branch(kW)\n", +"P_b_ii = n/2*P_each_b_ii // Power consumption for 2 elements in series in each phase(kW)\n", +"// Case(c)(i)\n", +"P_each_c_i = V**2/(R+R)*10**-3 // Power consumption by each branch(kW)\n", +"P_c_i = n/2*P_each_c_i // Power consumption for 2 elements in series in each branch(kW)\n", +"// Case(c)(ii)\n", +"P_each_c_ii = 2*V**2/R*10**-3 // Power consumption by each branch(kW)\n", +"P_c_ii = n/2*P_each_c_ii // Power consumption for 2 elements in parallel in each branch(kW)\n", +"// Case(d)\n", +"V_d = V*tap/100 // Voltage under tapping(V)\n", +"ratio_V = V_d/V // Ratio of normal voltage to tapped voltage\n", +"loss = ratio_V**2 // Power loss in terms of normal power\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.3 : SOLUTION :-')\n", +"printf('\nCase(a): AC Single phase 400 V supply')\n", +"printf('\n Case(i) : Power consumption for 6 elements in parallel = %.1f kW', P_a_i)\n", +"printf('\n Case(ii): Power consumption for 3 groups in parallel with 2 element in series = %.1f kW', P_a_ii)\n", +"printf('\nCase(b): AC Three phase 400 V supply with star combination')\n", +"printf('\n Case(i) : Power consumption for 2 elements in parallel in each phase = %.1f kW', P_b_i)\n", +"printf('\n Case(ii): Power consumption for 2 elements in series in each phase = %.1f kW', P_b_ii)\n", +"printf('\nCase(c): AC Three phase 400 V supply with delta combination')\n", +"printf('\n Case(i) : Power consumption for 2 elements in series in each branch = %.1f kW', P_c_i)\n", +"printf('\n Case(ii): Power consumption for 2 elements in parallel in each branch = %.1f kW', P_c_ii)\n", +"printf('\nCase(d): Power loss will be %.2f of the values obtained as above with auto-transformer tapping', loss)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.4: Amount_of_energy_required_to_melt_brass.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.4 :\n", +"// Page number 728\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"w_brass = 1000.0 // Weight of brass(kg)\n", +"time = 1.0 // Time(hour)\n", +"heat_sp = 0.094 // Specific heat\n", +"fusion = 40.0 // Latent heat of fusion(kcal/kg)\n", +"T_initial = 24.0 // Initial temperature(°C)\n", +"melt_point = 920.0 // Melting point of brass(°C)\n", +"n = 0.65 // Efficiency\n", +"\n", +"// Calculations\n", +"heat_req = w_brass*heat_sp*(melt_point-T_initial) // Heat required to raise the temperature(kcal)\n", +"heat_mel = w_brass*fusion // Heat required for melting(kcal)\n", +"heat_total = heat_req+heat_mel // Total heat required(kcal)\n", +"energy = heat_total*1000*4.18/(10**3*3600*n) // Energy input(kWh)\n", +"power = energy/time // Power(kW)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.4 : SOLUTION :-')\n", +"printf('\nAmount of energy required to melt brass = %.f kWh', energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.5: EX40_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.5 :\n", +"// Page number 728-729\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_2 = 12.0 // Secondary voltage(V)\n", +"P = 30.0*10**3 // Power(W)\n", +"PF = 0.5 // Power factor\n", +"\n", +"// Calculations\n", +"I_2 = P/(V_2*PF) // Secondary current(A)\n", +"Z_2 = V_2/I_2 // Secondary impedance(ohm)\n", +"R_2 = Z_2*PF // Secondary resistance(ohm)\n", +"sin_phi = (1-PF**2)**0.5\n", +"X_2 = Z_2*sin_phi // Secondary reactance(ohm)\n", +"h = R_2/X_2\n", +"H_m = h // Height up to which the crucible should be filled to obtain maximum heating effect in terms of H_c\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.5 : SOLUTION :-')\n", +"printf('\nHeight up to which the crucible should be filled to obtain maximum heating effect, H_m = %.3f*H_c \n', H_m)\n", +"printf('\nNOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.6: Voltage_necessary_for_heating_and_Current_flowing_in_the_material.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.6 :\n", +"// Page number 732\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 10.0 // Length of material(cm)\n", +"b = 10.0 // Breadth of material(cm)\n", +"t = 3.0 // Thickness of material(cm)\n", +"f = 20.0*10**6 // Frequency(Hz)\n", +"P = 400.0 // Power absorbed(W)\n", +"e_r = 5.0 // Relative permittivity\n", +"PF = 0.05 // Power factor\n", +"\n", +"// Calculations\n", +"e_0 = 8.854*10**-12 // Absolute permittivity\n", +"A = l*b*10**-4 // Area(Sq.m)\n", +"C = e_0*e_r*A/(t/100) // Capacitace of parallel plate condenser(F)\n", +"X_c = 1.0/(2*%pi*f*C) // Reactance of condenser(ohm)\n", +"phi = acosd(PF) // Φ(°)\n", +"R = X_c*tand(phi) // Resistance of condenser(ohm)\n", +"V = (P*R)**0.5 // Voltage necessary for heating(V)\n", +"I_c = V/X_c // Current flowing in the material(A)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.6 : SOLUTION :-')\n", +"printf('\nVoltage necessary for heating, V = %.f V', V)\n", +"printf('\nCurrent flowing in the material, I_c = %.2f A\n', I_c)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.7: Voltage_applied_across_electrodes_and_Current_through_the_material.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.7 :\n", +"// Page number 732-733\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 4.0 // Length of material(cm)\n", +"b = 2.0 // Breadth of material(cm)\n", +"t = 1.0 // Thickness of material(cm)\n", +"l_e = 20.0 // Length of area(cm)\n", +"b_e = 2.0 // Breadth of area(cm)\n", +"dis = 1.6 // Distance of separation of electrode(cm)\n", +"f = 20.0*10**6 // Frequency(Hz)\n", +"P = 80.0 // Power absorbed(W)\n", +"e_r1 = 5.0 // Relative permittivity\n", +"e_r2 = 1.0 // Relative permittivity of air\n", +"PF = 0.05 // Power factor\n", +"\n", +"// Calculations\n", +"e_0 = 8.854*10**-12 // Absolute permittivity\n", +"A_1 = (l_e-l)*b_e*10**-4 // Area of one electrode(sq.m)\n", +"A_2 = l*b*10**-4 // Area of material under electrode(sq.m)\n", +"d = dis*10**-2 // Distance of separation of electrode(m)\n", +"d_1 = t*10**-2 // (m)\n", +"d_2 = (d-d_1) // (m)\n", +"C = e_0*((A_1*e_r2/d)+(A_2/((d_1/e_r1)+(d_2/e_r2)))) // Capacitance(F)\n", +"X_c = 1.0/(2*%pi*f*C) // Reactance(ohm)\n", +"phi = acosd(PF) // Φ(°)\n", +"R = X_c*tand(phi) // Resistance(ohm)\n", +"V = (P*R)**0.5 // Voltage applied across electrodes(V)\n", +"I_c = V/X_c // Current through the material(A)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.7 : SOLUTION :-')\n", +"printf('\nVoltage applied across electrodes, V = %.f V', V)\n", +"printf('\nCurrent through the material, I_c = %.1f A\n', I_c)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.8: Time_taken_to_melt_Power_factor_and_Electrical_efficiency_of_the_furnace.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 2: HEATING AND WELDING\n", +"\n", +"// EXAMPLE : 2.8 :\n", +"// Page number 736-737\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"weight = 3000.0 // Weight of steel(kg)\n", +"I = 5000.0 // Current(A)\n", +"V_arc = 60.0 // Arc voltage(V)\n", +"R_t = 0.003 // Resistance of transformer(ohm)\n", +"X_t = 0.005 // Reactance of transformer(ohm)\n", +"heat_sp = 0.12 // Specific heat of steel\n", +"heat_latent = 8.89 // Latent heat of steel(kilo-cal/kg)\n", +"t_2 = 1370.0 // Melting point of steel(°C)\n", +"t_1 = 18.0 // Initial temperature of steel(°C)\n", +"n = 0.6 // Overall efficiency\n", +"\n", +"// Calculations\n", +"R_arc_phase = V_arc/I // Arc resistance per phase(ohm)\n", +"IR_t = I*R_t // Voltage drop across resistance(V)\n", +"IX_t = I*X_t // Voltage drop across reactance(V)\n", +"V = ((V_arc+IR_t)**2+IX_t**2)**0.5 // Voltage(V)\n", +"PF = (V_arc+IR_t)/V // Power factor\n", +"heat_kg = (t_2-t_1)*heat_sp+heat_latent // Amount of heat required per kg of steel(kcal)\n", +"heat_total = weight*heat_kg // Heat for 3 tonnes(kcal)\n", +"heat_actual_kcal = heat_total/n // Actual heat required(kcal)\n", +"heat_actual = heat_actual_kcal*1.162*10**-3 // Actual heat required(kWh)\n", +"P_input = 3*V*I*PF*10**-3 // Power input(kW)\n", +"time = heat_actual/P_input*60 // Time required(min)\n", +"n_elect = 3*V_arc*I/(P_input*1000)*100 // Electrical efficiency(%)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 2.8 : SOLUTION :-')\n", +"printf('\nTime taken to melt 3 metric tonnes of steel = %.f minutes', time)\n", +"printf('\nPower factor of the furnace = %.2f ', PF)\n", +"printf('\nElectrical efficiency of the furnace = %.f percent\n', n_elect)\n", +"printf('\nNOTE: ERROR: Calculation and substitution mistake in the textbook solution')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/41-ELECTROLYTIC_AND_ELECTRO_METALLURGICAL_PROCESSES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/41-ELECTROLYTIC_AND_ELECTRO_METALLURGICAL_PROCESSES.ipynb new file mode 100644 index 0000000..141dfd5 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/41-ELECTROLYTIC_AND_ELECTRO_METALLURGICAL_PROCESSES.ipynb @@ -0,0 +1,175 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 41: ELECTROLYTIC AND ELECTRO METALLURGICAL PROCESSES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.1: Quantity_of_electricity_and_Time_taken_for_the_process.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 3: ELECTROLYTIC AND ELECTRO-METALLURGICAL PROCESSES\n", +"\n", +"// EXAMPLE : 3.1 :\n", +"// Page number 747-748\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 20.0 // Length of shaft(cm)\n", +"d = 10.0 // Diameter of shaft(cm)\n", +"thick = 1.5 // Layer of nickel(mm)\n", +"J = 195.0 // Current density(A/sq.m)\n", +"n_I = 0.92 // Current efficiency\n", +"g = 8.9 // Specific gravity of nickel\n", +"\n", +"// Calculations\n", +"Wt = %pi*l*d*thick/10*g*10**-3 // Weight of nickel to be deposited(kg)\n", +"ece_nickel = 1.0954 // Electro-chemical equivalent of nickel(kg/1000 Ah)\n", +"Q_I = Wt*1000/(ece_nickel*n_I) // Quantity of electricity required(Ah)\n", +"time = Q_I/(%pi*l*d*10**-4*J) // Time taken(hours)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 3.1 : SOLUTION :-')\n", +"printf('\nQuantity of electricity = %.f Ah', Q_I)\n", +"printf('\nTime taken for the process = %.f hours', time)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.2: Annual_output_of_refined_copper_and_Energy_consumption.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 3: ELECTROLYTIC AND ELECTRO-METALLURGICAL PROCESSES\n", +"\n", +"// EXAMPLE : 3.2 :\n", +"// Page number 748\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"no_cells = 600.0 // Number of cells employed for copper refining\n", +"I = 4000.0 // Current(A)\n", +"V = 0.3 // Voltage per cell(V)\n", +"hour = 90.0 // Time of plant operation(hours)\n", +"ece_cu = 1.1844 // Electro-chemical equivalent of copper(kg/1000 Ah)\n", +"\n", +"// Calculations\n", +"Ah_week = I*hour // Ah per week per cell\n", +"Ah_year = Ah_week*52 // Ah per year per cell\n", +"Wt = no_cells*ece_cu*Ah_year/(1000*10**3) // Weight of copper refined per year(tonnes)\n", +"energy = V*I*no_cells*hour*52/1000 // Energy consumed(kWh)\n", +"consumption = energy/Wt // Consumption(kWh/tonne)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 3.2 : SOLUTION :-')\n", +"printf('\nAnnual output of refined copper = %.f tonnes', Wt)\n", +"printf('\nEnergy consumption = %.1f kWh/tonne\n', consumption)\n", +"printf('\nNOTE: ERROR: Substitution & calculation mistake in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.3: Weight_of_aluminium_produced_from_aluminium_oxide.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 3: ELECTROLYTIC AND ELECTRO-METALLURGICAL PROCESSES\n", +"\n", +"// EXAMPLE : 3.3 :\n", +"// Page number 748\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"hour = 24.0 // Time(hour)\n", +"I = 3500.0 // Average current(A)\n", +"n = 0.9 // Current efficiency\n", +"valency = 3.0 // Aluminium valency\n", +"w = 27.0 // Atomic weight of aluminium\n", +"ece_Ag = 107.98 // Electro-chemical equivalent of silver\n", +"Wt_dep = 0.00111 // Silver deposition by one coulomb(gm)\n", +"\n", +"// Calculations\n", +"chemical_eq_Al = w/valency // Chemical equivalent of aluminium\n", +"eme_Al = Wt_dep/ece_Ag*chemical_eq_Al // Electro-chemical equivalent of aluminium(gm/coulomb)\n", +"Wt_Al_liberated = I*hour*3600*n*eme_Al/1000 // Weight of aluminium liberated(Kg)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 3.3 : SOLUTION :-')\n", +"printf('\nWeight of aluminium produced from aluminium oxide = %.1f kg', Wt_Al_liberated)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/42-ILLUMINATION.ipynb b/Power_System_Engineering_by_S_Chakraborthy/42-ILLUMINATION.ipynb new file mode 100644 index 0000000..ad4e765 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/42-ILLUMINATION.ipynb @@ -0,0 +1,428 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 42: ILLUMINATION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.2: EX42_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.2 :\n", +"// Page number 753\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"lumens = 800.0 // Flux emitted by a lamp(lumens)\n", +"cp = 100.0 // cp of a lamp\n", +"d = 2.0 // Distance b/w plane surface & lamp(m)\n", +"theta_ii = 45.0 // Inclined surface(°)\n", +"theta_iii = 90.0 // Parallel rays(°)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"mscp = lumens/(4.0*%pi) // mscp of lamp\n", +"// Case(b)\n", +"I_i = cp/d**2 // Illumination on the surface when it is normal(lux)\n", +"I_ii = cp/d**2*cosd(theta_ii) // Illumination on the surface when it is inclined to 45°(lux)\n", +"I_iii = cp/d**2*cosd(theta_iii) // Illumination on the surface when it is parallel to rays(lux)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.2 : SOLUTION :-')\n", +"printf('\nCase(a): mscp of the lamp, mscp = %.f ', mscp)\n", +"printf('\nCase(b): Case(i) : Illumination on the surface when it is normal, I = %.f lux', I_i)\n", +"printf('\n Case(ii) : Illumination on the surface when it is inclined to 45°, I = %.3f lux', I_ii)\n", +"printf('\n Case(iii): Illumination on the surface when it is parallel to rays, I = %.f lux\n', abs(I_iii))\n", +"printf('\nNOTE: ERROR: Calculation mistake in case(a) in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.3: EX42_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.3 :\n", +"// Page number 753-754\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cp = 200.0 // cp of a lamp\n", +"per = 0.6 // Reflector directing light\n", +"D = 10.0 // Diameter(m)\n", +"h = 6.0 // Height at which lamp is hung(m)\n", +"\n", +"// Calculations\n", +"flux = cp*4*%pi // Flux(lumens)\n", +"I_i = cp/h**2 // Illumination at the centre without reflector(lux)\n", +"d = (h**2+(D/2)**2)**0.5 // (m)\n", +"I_without = (cp/h**2)*(h/d) // Illumination at the edge without reflector(lux)\n", +"I_with = cp*4*%pi*per/(25*%pi) // Illumination at the edge with reflector(lux)\n", +"theta = acosd(h/d) // θ(°)\n", +"w = 2.0*%pi*(1-cosd(theta/2)) // ω(steradian)\n", +"phi = cp*w // Φ(lumens)\n", +"I_avg = phi/(25*%pi) // Average illumination over the area without reflector(lux)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.3 : SOLUTION :-')\n", +"printf('\nCase(i) : Illumination at the centre without reflector = %.2f lux', I_i)\n", +"printf('\n Illumination at the centre with reflector = %.1f lux', I_with)\n", +"printf('\nCase(ii): Illumination at the edge of the surface without reflector = %.2f lux', I_without)\n", +"printf('\n Illumination at the edge of the surface with reflector = %.1f lux', I_with)\n", +"printf('\nAverage illumination over the area without the reflector, I = %.3f lux\n', I_avg)\n", +"printf('\nNOTE: ERROR: Slight calculation mistake & more approximation in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.5: EX42_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.5 :\n", +"// Page number 754\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"flux = 900.0 // Lamp emitting light(lumens)\n", +"D = 30.5 // Diameter of globe(cm)\n", +"B = 250.0*10**-3 // Uniform brightness(Ambert)\n", +"\n", +"// Calculations\n", +"cp = %pi/4*D**2*(B/%pi) // Candle power\n", +"flux_emit = cp*4*%pi // Flux emitted by globe(lumens)\n", +"flux_abs = flux-flux_emit // Flux absorbed by globe(lumens)\n", +"light_abs_per = flux_abs/flux*100 // Light absorbed(%)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.5 : SOLUTION :-')\n", +"printf('\ncp of the globe = %.f ', cp)\n", +"printf('\nPercentage of light emitted by lamp that is absorbed by the globe = %.1f percent\n', light_abs_per)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.6: Curve_showing_illumination_on_a_horizontal_line_below_lamp.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.6 :\n", +"// Page number 754-755\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cp_0 = 500.0 // Candle power\n", +"theta_0 = 0.0 // θ(°)\n", +"cp_1 = 560.0 // Candle power\n", +"theta_1 = 10.0 // θ(°)\n", +"cp_2 = 600.0 // Candle power\n", +"theta_2 = 20.0 // θ(°)\n", +"cp_3 = 520.0 // Candle power\n", +"theta_3 = 30.0 // θ(°)\n", +"cp_4 = 400.0 // Candle power\n", +"theta_4 = 40.0 // θ(°)\n", +"cp_5 = 300.0 // Candle power\n", +"theta_5 = 50.0 // θ(°)\n", +"cp_6 = 150.0 // Candle power\n", +"theta_6 = 60.0 // θ(°)\n", +"cp_7 = 50.0 // Candle power\n", +"theta_7 = 70.0 // θ(°)\n", +"h = 6.0 // Height of lamp(m)\n", +"\n", +"// Calculations\n", +"I_0 = cp_0/h**2*(cosd(theta_0))**3 // Illumination(lux)\n", +"l_0 = h*tand(theta_0) // Distance(m)\n", +"I_1 = cp_1/h**2*(cosd(theta_1))**3 // Illumination(lux)\n", +"l_1 = h*tand(theta_1) // Distance(m)\n", +"I_2 = cp_2/h**2*(cosd(theta_2))**3 // Illumination(lux)\n", +"l_2 = h*tand(theta_2) // Distance(m)\n", +"I_3 = cp_3/h**2*(cosd(theta_3))**3 // Illumination(lux)\n", +"l_3 = h*tand(theta_3) // Distance(m)\n", +"I_4 = cp_4/h**2*(cosd(theta_4))**3 // Illumination(lux)\n", +"l_4 = h*tand(theta_4) // Distance(m)\n", +"I_5 = cp_5/h**2*(cosd(theta_5))**3 // Illumination(lux)\n", +"l_5 = h*tand(theta_5) // Distance(m)\n", +"I_6 = cp_6/h**2*(cosd(theta_6))**3 // Illumination(lux)\n", +"l_6 = h*tand(theta_6) // Distance(m)\n", +"I_7 = cp_7/h**2*(cosd(theta_7))**3 // Illumination(lux)\n", +"l_7 = h*tand(theta_7) // Distance(m)\n", +"l = [-l_7,-l_6,-l_5,-l_4,-l_3,-l_2,-l_1,l_0,l_0,l_1,l_2,l_3,l_4,l_5,l_6,l_7]\n", +"I = [I_7,I_6,I_5,I_4,I_3,I_2,I_1,I_0,I_0,I_1,I_2,I_3,I_4,I_5,I_6,I_7]\n", +"a = gca() ;\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot(l,I,'ro-') // Plot of illumination curve\n", +"x = [0,0,0,0,0,0]\n", +"y = [0,5,10,11,14,16]\n", +"plot(x,y) // Plot of straight line\n", +"a.x_label.text = 'Distance(metres)' // labels x-axis\n", +"a.y_label.text = 'Illumination(flux)' // labels y-axis\n", +"xtitle('Fig E4.4 . Illumination on a horizontal line below the lamp') \n", +"xset('thickness',2) // sets thickness of axes\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.6 : SOLUTION :-')\n", +"printf('\nThe curve showing illumination on a horizontal line below lamp is represented in Figure E4.4')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.7: Maximum_and_Minimum_illumination_on_the_floor_along_the_centre_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.7 :\n", +"// Page number 755\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 9.15 // Lamp space(m)\n", +"h = 4.575 // Height(m)\n", +"P = 100.0 // Power(candle)\n", +"\n", +"// Calculations\n", +"theta_3_max = 0 // θ(°)\n", +"cos_theta_3_max_cubic = cosd(theta_3_max)**3\n", +"theta_4_max = atand(2) // θ(°)\n", +"cos_theta_4_max_cubic = cosd(theta_4_max)**3\n", +"theta_5_max = atand(4) // θ(°)\n", +"cos_theta_5_max_cubic = cosd(theta_5_max)**3\n", +"theta_6_max = atand(6) // θ(°)\n", +"cos_theta_6_max_cubic = cosd(theta_6_max)**3\n", +"I_max = P/h**2*(cos_theta_3_max_cubic+2*cos_theta_4_max_cubic+2*cos_theta_5_max_cubic+2*cos_theta_6_max_cubic) // Max illumination(lux)\n", +"theta_4_min = atand(1) // θ(°)\n", +"cos_theta_4_min_cubic = cosd(theta_4_min)**3\n", +"theta_5_min = atand(3) // θ(°)\n", +"cos_theta_5_min_cubic = cosd(theta_5_min)**3\n", +"theta_6_min = atand(5) // θ(°)\n", +"cos_theta_6_min_cubic = cosd(theta_6_min)**3\n", +"I_min = P/h**2*2*(cos_theta_4_min_cubic+cos_theta_5_min_cubic+cos_theta_6_min_cubic) // Minimum illumination(lux)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.7 : SOLUTION :-')\n", +"printf('\nMaximum illumination on the floor along the centre line = %.2f lux', I_max)\n", +"printf('\nMinimum illumination on the floor along the centre line = %.2f lux', I_min)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.8: Illumination_on_the_working_plane.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.8 :\n", +"// Page number 758\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"b = 15.25 // Breadth of workshop(m)\n", +"l = 36.6 // Length of workshop(m)\n", +"no = 20.0 // Number of lamps\n", +"P = 500.0 // Power of each lamp(W)\n", +"n = 15.0 // Luminous efficiency of each lamp(lumens/watt)\n", +"df = 0.7 // Depreciation factor\n", +"cou = 0.5 // Co-efficient of utilization\n", +"\n", +"// Calculations\n", +"lumen_lamp = no*P*n // Lamp lumens\n", +"lumen_plane = lumen_lamp*df*cou // Lumens on the working plane\n", +"I = lumen_plane/(l*b) // Illumination(lm/sq.m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.8 : SOLUTION :-')\n", +"printf('\nIllumination on the working plane = %.1f lm per sq.m\n', I)\n", +"printf('\nNOTE: ERROR: The breadth should be 15.25m but mentioned as 5.25m in textbook statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.9: Suitable_scheme_of_illumination_and_Saving_in_power_consumption.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 4: ILLUMINATION\n", +"\n", +"// EXAMPLE : 4.9 :\n", +"// Page number 758-759\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"b = 27.45 // Breadth of hall(m)\n", +"l = 45.75 // Length of hall(m)\n", +"I_avg = 108.0 // Average illumination(lumens/sq.m)\n", +"h = 0.75 // Height(m)\n", +"cou = 0.35 // Co-efficient of utilization\n", +"pf = 0.9 // Pereciation factor\n", +"P_fl = 80.0 // Fluorescent lamp power(W)\n", +"n_100 = 13.4 // Luminous efficiency for 100W filament lamp(lumens/watt)\n", +"n_200 = 14.4 // Luminous efficiency for 200W filament lamp(lumens/watt)\n", +"n_80 = 30.0 // Luminous efficiency for 80W fluorescent lamp(lumens/watt)\n", +"\n", +"// Calculations\n", +"area = b*l // Area to be illuminated(Sq.m)\n", +"I_total = area*I_avg // Total illumination on working plane(lumens)\n", +"gross_lumen = I_total/(cou*pf) // Gross lumens required\n", +"P_required = gross_lumen/n_200 // Power required for illumination(W)\n", +"P_required_kW = P_required/1000 // Power required for illumination(kW)\n", +"no_lamp = P_required/200 // Number of lamps\n", +"P_required_new = gross_lumen/n_80 // Power required when fluorescent lamp used(W)\n", +"P_required_new_kW = P_required_new/1000 // Power required when fluorescent lamp used(kW)\n", +"P_saving = P_required_kW-P_required_new_kW // Saving in power(kW)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 4.9 : SOLUTION :-')\n", +"printf('\nSuitable scheme: Whole area divided into %.f rectangles & 200-watt fitting is suspended at centre of each rectangle', no_lamp)\n", +"printf('\nSaving in power consumption = %.1f kW', P_saving)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/43-ELECTRIC_TRACTION_SPEED_TIME_CURVES_AND_MECHANICS_OF_TRAIN_MOVEMENT.ipynb b/Power_System_Engineering_by_S_Chakraborthy/43-ELECTRIC_TRACTION_SPEED_TIME_CURVES_AND_MECHANICS_OF_TRAIN_MOVEMENT.ipynb new file mode 100644 index 0000000..1cea606 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/43-ELECTRIC_TRACTION_SPEED_TIME_CURVES_AND_MECHANICS_OF_TRAIN_MOVEMENT.ipynb @@ -0,0 +1,636 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 43: ELECTRIC TRACTION SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.10: Minimum_adhesive_weight_of_the_locomotive.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.10 :\n", +"// Page number 784\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 350.0 // Weight of train(tonne)\n", +"G = 1.0 // Gradient\n", +"alpha = 0.8 // Acceleration(km phps)\n", +"u = 0.25 // Co-efficient of adhesion\n", +"r = 44.5 // Train resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"F_t = 277.8*W_e*alpha+W*r+98.1*W*G // Tractive effort(N)\n", +"adhesive_weight = F_t/(u*9.81*1000) // Adhesive weight(tonnes)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.10 : SOLUTION :-')\n", +"printf('\nMinimum adhesive weight of the locomotive = %.1f tonnes\n', adhesive_weight)\n", +"printf('\nNOTE: ERROR: Train resistance is 44.5 N per tonne & not 45 N per tonne as mentioned in textbook problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.11: EX43_11.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.11 :\n", +"// Page number 784\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 400.0 // Weight of train(tonne)\n", +"G = 100.0/75 // Gradient\n", +"alpha = 1.6 // Acceleration(km phps)\n", +"r = 66.75 // Train resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"V = 48.0 // Speed(kmph)\n", +"n = 0.7 // Overall efficiency of equipment\n", +"\n", +"// Calculations\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"F_t = 277.8*W_e*alpha+W*r+98.1*W*G // Tractive effort(N)\n", +"t = V/alpha // Time(sec)\n", +"energy_a = F_t*V*t/(2*3600**2) // Energy usefully employed(kWh)\n", +"G_r = 98.1*G+r // Force(N)\n", +"work_tonne_km = G_r*1000 // Work done per tonne per km(Nw-m)\n", +"energy_b = work_tonne_km/(n*3600) // Energy consumption(Wh per tonne-km)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.11 : SOLUTION :-')\n", +"printf('\nCase(a): Energy usefully employed in attaining speed = %.2f kWh', energy_a)\n", +"printf('\nCase(b): Specific energy consumption at steady state speed = %.1f Wh per tonne-km', energy_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.12: Minimum_adhesive_weight_of_a_locomotive.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.12 :\n", +"// Page number 784-785\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 200.0 // Trailing weight(tonne)\n", +"G = 1.0 // Gradient(%)\n", +"alpha = 1.0 // Acceleration(km phps)\n", +"u = 0.2 // Co-efficient of adhesion\n", +"r = 50.0 // Train resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"W_L = ((277.8*(100+I)/100*alpha)+98.1*G+r)*W/(u*9.81*1000-((277.8*(100+I)/100*alpha)+98.1*G+r)) // Weight of locomotive(tonnes)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.12 : SOLUTION :-')\n", +"printf('\nMinimum adhesive weight of a locomotive, W_L = %.1f tonnes\n', W_L)\n", +"printf('\nNOTE: ERROR: Calculation mistake in textbook solution in calculating W_L')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.1: Maximum_speed_over_the_ru.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.1 :\n", +"// Page number 778\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"speed = 45.0 // Scheduled speed(kmph)\n", +"D = 1.5 // Distance between 2 stops(km)\n", +"t = 20.0 // Time of stop(sec)\n", +"alpha = 2.4 // Acceleration(km phps)\n", +"beta = 3.2 // Retardation(km phps)\n", +"\n", +"// Calculations\n", +"t_total = D*3600/speed // Total time(sec)\n", +"T = t_total-t // Actual time for run(sec)\n", +"k = (alpha+beta)/(alpha*beta) // Constant\n", +"V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5 // Maximum speed over the run(kmph)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.1 : SOLUTION :-')\n", +"printf('\nMaximum speed over the run, V_m = %.f kmph', V_m)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.2: Value_of_retardatio.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.2 :\n", +"// Page number 778\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_m = 65.0 // Maximum speed(kmph)\n", +"t = 30.0 // Time of stop(sec)\n", +"speed = 43.5 // Scheduled speed(kmph)\n", +"alpha = 1.3 // Acceleration(km phps)\n", +"D = 3.0 // Distance between 2 stops(km)\n", +"\n", +"// Calculations\n", +"t_total = D*3600/speed // Total time of run including stop(sec)\n", +"T = t_total-t // Actual time for run(sec)\n", +"V_a = D/T*3600 // Average speed(kmph)\n", +"beta = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/alpha)) // Value of retardation(km phps)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.2 : SOLUTION :-')\n", +"printf('\nValue of retardation, β = %.3f km phps\n', beta)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')\n", +"printf('\n ERROR: β unit is km phps & not km phps as mentioned in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.3: Rate_of_acceleration_required_to_operate_service.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.3 :\n", +"// Page number 778-779\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"speed = 25.0 // Scheduled speed(kmph)\n", +"D = 800.0/1000 // Distance between 2 stations(km)\n", +"t = 20.0 // Time of stop(sec)\n", +"V_m_per = 20.0 // Maximum speed higher than(%)\n", +"beta = 3.0 // Retardation(km phps)\n", +"\n", +"// Calculations\n", +"t_total = D*3600/speed // Total time of run including stop(sec)\n", +"T = t_total-t // Actual time for run(sec)\n", +"V_a = D/T*3600 // Average speed(kmph)\n", +"V_m = (100+V_m_per)*V_a/100 // Maximum speed(kmph)\n", +"alpha = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/beta)) // Value of acceleration(km phps)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.3 : SOLUTION :-')\n", +"printf('\nRate of acceleration required to operate this service, α = %.2f km phps', alpha)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.4: Duration_of_acceleration_Coasting_and_Braking_periods.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.4 :\n", +"// Page number 779\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D = 2.0 // Distance between 2 stations(km)\n", +"V_a = 40.0 // Average speed(kmph)\n", +"V_1 = 60.0 // Maximum speed limitation(kph)\n", +"alpha = 2.0 // Acceleration(km phps)\n", +"beta_c = 0.15 // Coasting retardation(km phps)\n", +"beta = 3.0 // Braking retardation(km phps)\n", +"\n", +"// Calculations\n", +"t_1 = V_1/alpha // Time for acceleration(sec)\n", +"T = 3600*D/V_a // Actual time of run(sec)\n", +"V_2 = (T-t_1-(V_1/beta_c))*beta*beta_c/(beta_c-beta) // Speed at the end of coasting period(kmph)\n", +"t_2 = (V_1-V_2)/beta_c // Coasting period(sec)\n", +"t_3 = V_2/beta // Braking period(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.4 : SOLUTION :-')\n", +"printf('\nDuration of acceleration, t_1 = %.f sec', t_1)\n", +"printf('\nDuration of coasting, t_2 = %.f sec', t_2)\n", +"printf('\nDuration of braking, t_3 = %.f sec', t_3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.5: Tractive_resistance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.5 :\n", +"// Page number 781-782\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"r = 1.0 // Tractive resistance(N/tonne)\n", +"\n", +"// Calculations\n", +"tractive_res_i = 0.278*r // Tractive resistance(N/tonne) = Energy consumption(Wh/tonne-km)\n", +"beta = 1/277.8 // Tractive resistance(N/tonne) = Retardation(km kmps/tonne)\n", +"energy = 98.1*1000/3600 // 1% gradient = energy(Wh per tonne km)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.5 : SOLUTION :-')\n", +"printf('\nCase(i) : Tractive resistance of 1 N per tonne = %.3f Wh per tonne-km', tractive_res_i)\n", +"printf('\nCase(ii) : Tractive resistance of 1 N per tonne = %.5f km phps per tonne', beta)\n", +"printf('\nCase(iii): 1 percent gradient = %.2f Wh per tonne km\n', energy)\n", +"printf('\nNOTE: Slight change in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.6: Torque_developed_by_each_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.6 :\n", +"// Page number 782\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 254.0 // Weight of motor-coach train(tonne)\n", +"no = 4.0 // Number of motor\n", +"t_1 = 20.0 // Time(sec)\n", +"V_m = 40.25 // Maximum speed(kmph)\n", +"G = 1.0 // Gradient(%)\n", +"gamma = 3.5 // Gear ratio\n", +"n = 0.95 // Gear efficiency\n", +"D = 91.5/100 // Wheel diameter(m)\n", +"r = 44.0 // Train resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"alpha = V_m/t_1 // Acceleration(km phps)\n", +"F_t = 277.8*W_e*alpha+W*r+98.1*W*G // Tractive effort(N)\n", +"T = F_t*D/(2*n*gamma) // Torque developed(N-m)\n", +"T_each = T/no // Torque developed by each motor(N-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.6 : SOLUTION :-')\n", +"printf('\nTorque developed by each motor = %.f N-m\n', T_each)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & more approximation in textbook')\n", +"printf('\n ERROR: W = 254 tonne, not 256 tonne as mentioned in textbook problem statement')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.7: Time_taken_by_train_to_attain_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.7 :\n", +"// Page number 782\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 203.0 // Weight of motor-coach train(tonne)\n", +"no = 4.0 // Number of motors\n", +"T = 5130.0 // Shaft torque(N-m)\n", +"V_m = 42.0 // Maximum speed(kmph)\n", +"G = 100.0/250 // Gradient\n", +"gamma = 3.5 // Gear ratio\n", +"n = 0.93 // Gear efficiency\n", +"D = 91.5/100 // Wheel diameter(m)\n", +"r = 45.0 // Train resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"F_t = n*4*T*2*gamma/D // Tractive effort(N)\n", +"alpha = (F_t-W*r-98.1*W*G)/(277.8*W_e) // Acceleration(km phps)\n", +"t_1 = V_m/alpha // Time taken by train to attain speed(sec)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.7 : SOLUTION :-')\n", +"printf('\nTime taken by train to attain speed, t_1 = %.1f sec', t_1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.8: Speed_Time_curve_for_the_run_and_Energy_consumption_at_the_axles_of_train.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.8 :\n", +"// Page number 782-783\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_a = 42.0 // Average speed of train(kmph)\n", +"D = 1400.0/1000 // Distance(km)\n", +"alpha = 1.7 // Acceleration(km phps)\n", +"beta = 3.3 // Retardation(km phps)\n", +"r = 50.0 // Tractive resistance(N/tonne)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"T = D*3600/V_a // Time for run(sec)\n", +"k = (alpha+beta)/(alpha*beta) // Constant\n", +"V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5 // Maximum speed over the run(kmph)\n", +"t_1 = V_m/alpha // Time of acceleration(sec)\n", +"t_3 = V_m/beta // Time(sec)\n", +"t_2 = T-(t_1+t_3) // Time(sec)\n", +"D_1 = D-(V_a*t_1/(2*3600)) // Distance(km)\n", +"We_W = (100+I)/100 // W_e/W\n", +"energy = (0.0107*V_m**2*We_W/D)+(0.278*r*D_1/D) // Energy consumption(Wh per tonne-km)\n", +"a = gca() ;\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot([0,t_1,t_1,(t_1+t_2),(t_1+t_2),(t_1+t_2+t_3)],[0,V_m,V_m,V_m,V_m,0]) // Plotting speed-time curve\n", +"plot([t_1,t_1],[0,V_m],'r--')\n", +"plot([t_1+t_2,t_1+t_2],[0,V_m],'r--')\n", +"a.x_label.text = 'Time(seconds)' // labels x-axis\n", +"a.y_label.text = 'Speed (km/h)' // labels y-axis\n", +"xtitle('Fig E5.1 . Speed-time curve for the run') \n", +"xset('thickness',2) // sets thickness of axes\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.8 : SOLUTION :-')\n", +"printf('\nSpeed-time curve for the run is shown in Figure E5.1')\n", +"printf('\nEnergy consumption at the axles of train = %.1f Wh per tonne-km', energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.9: Acceleration_Coasting_retardation_and_Scheduled_speed.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT\n", +"\n", +"// EXAMPLE : 5.9 :\n", +"// Page number 783\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V_A = 48.0 // Speed(kmph)\n", +"t_1 = 24.0 // Time taken to accelerate from rest to speed(sec)\n", +"t_2 = 69.0 // Coasting time(sec)\n", +"r = 58.0 // Constant resistance(N/tonne)\n", +"beta = 3.3 // Retardation(km phps)\n", +"t_3 = 11.0 // Retardation time(sec)\n", +"t_iii_a = 20.0 // Station stop time(sec)\n", +"t_iii_b = 15.0 // Station stop time(sec)\n", +"I = 10.0 // Rotational inertia(%)\n", +"\n", +"// Calculations\n", +"alpha = V_A/t_1 // Acceleration(km phps)\n", +"V_B = beta*t_3 // Speed at B(km phps)\n", +"beta_c = (V_A-V_B)/t_2 // Retardation during coasting(km phps)\n", +"distance_acc = 1.0/2*t_1*V_A/3600 // Distance covered during acceleration(km)\n", +"distance_coasting = (V_A**2-V_B**2)/(2*beta_c*3600) // Distance covered during coasting(km)\n", +"distance_braking = t_3*V_B/(3600*2) // Distance covered during braking(km)\n", +"distance_total = distance_acc+distance_coasting+distance_braking // Total distance(km)\n", +"speed_iii_a = distance_total*3600/(t_1+t_2+t_3+t_iii_a) // Scheduled speed with a stop of 20 sec(kmph)\n", +"speed_iii_b = distance_total*3600/(t_1+t_2+t_3+t_iii_b) // Scheduled speed with a stop of 15 sec(kmph)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 5.9 : SOLUTION :-')\n", +"printf('\nCase(i) : Acceleration, α = %.f km phps', alpha)\n", +"printf('\nCase(ii) : Coasting retardation, β_c = %.2f km phps', beta_c)\n", +"printf('\nCase(iii): Scheduled speed with a stop of 20 seconds = %.2f kmph', speed_iii_a)\n", +"printf('\n Scheduled speed with a stop of 15 seconds = %.2f kmph\n', speed_iii_b)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/44-MOTORS_FOR_ELECTRIC_TRACTION.ipynb b/Power_System_Engineering_by_S_Chakraborthy/44-MOTORS_FOR_ELECTRIC_TRACTION.ipynb new file mode 100644 index 0000000..5ce578f --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/44-MOTORS_FOR_ELECTRIC_TRACTION.ipynb @@ -0,0 +1,352 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 44: MOTORS FOR ELECTRIC TRACTION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.1: Speed_current_of_the_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 6: MOTORS FOR ELECTRIC TRACTION\n", +"\n", +"// EXAMPLE : 6.1 :\n", +"// Page number 788\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_1 = 10.0 // Current(A)\n", +"T_1 = 54.0 // Torque(N-m)\n", +"I_2 = 20.0 // Current(A)\n", +"T_2 = 142.0 // Torque(N-m)\n", +"I_3 = 30.0 // Current(A)\n", +"T_3 = 250.0 // Torque(N-m)\n", +"I_4 = 40.0 // Current(A)\n", +"T_4 = 365.0 // Torque(N-m)\n", +"I_5 = 50.0 // Current(A)\n", +"T_5 = 480.0 // Torque(N-m)\n", +"I_6 = 60.0 // Current(A)\n", +"T_6 = 620.0 // Torque(N-m)\n", +"I_7 = 70.0 // Current(A)\n", +"T_7 = 810.0 // Torque(N-m)\n", +"E = 500.0 // Operating voltage(V)\n", +"R_a = 0.6 // Armature resistance(ohm)\n", +"\n", +"// Calculations\n", +"N_1 = 9.55*(E-I_1*R_a)*I_1/T_1 // Speed(rpm)\n", +"N_2 = 9.55*(E-I_2*R_a)*I_2/T_2 // Speed(rpm)\n", +"N_3 = 9.55*(E-I_3*R_a)*I_3/T_3 // Speed(rpm)\n", +"N_4 = 9.55*(E-I_4*R_a)*I_4/T_4 // Speed(rpm)\n", +"N_5 = 9.55*(E-I_5*R_a)*I_5/T_5 // Speed(rpm)\n", +"N_6 = 9.55*(E-I_6*R_a)*I_6/T_6 // Speed(rpm)\n", +"N_7 = 9.55*(E-I_7*R_a)*I_7/T_7 // Speed(rpm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 6.1 : SOLUTION :-')\n", +"printf('\nSpeed-current of the motor')\n", +"printf('\n_______________________________________')\n", +"printf('\n Current(A) : Speed(rpm) ')\n", +"printf('\n_______________________________________')\n", +"printf('\n %.f : %.f ', I_1,N_1)\n", +"printf('\n %.f : %.f ', I_2,N_2)\n", +"printf('\n %.f : %.f ', I_3,N_3)\n", +"printf('\n %.f : %.f ', I_4,N_4)\n", +"printf('\n %.f : %.f ', I_5,N_5)\n", +"printf('\n %.f : %.f ', I_6,N_6)\n", +"printf('\n %.f : %.f ', I_7,N_7)\n", +"printf('\n_______________________________________\n')\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.2: Speed_torque_for_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 6: MOTORS FOR ELECTRIC TRACTION\n", +"\n", +"// EXAMPLE : 6.2 :\n", +"// Page number 788-789\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"N_1 = 500.0 // Speed(rpm)\n", +"I_1 = 50.0 // Current(A)\n", +"E_1 = 220.0 // Armature voltage(V)\n", +"I_2 = 100.0 // Current(A)\n", +"E_2 = 350.0 // Armature voltage(V)\n", +"I_3 = 150.0 // Current(A)\n", +"E_3 = 440.0 // Armature voltage(V)\n", +"I_4 = 200.0 // Current(A)\n", +"E_4 = 500.0 // Armature voltage(V)\n", +"I_5 = 250.0 // Current(A)\n", +"E_5 = 540.0 // Armature voltage(V)\n", +"I_6 = 300.0 // Current(A)\n", +"E_6 = 570.0 // Armature voltage(V)\n", +"R_wb = 0.08 // Armature and brush resistance(ohm)\n", +"R_f = 0.05 // Resistance of series field(ohm)\n", +"V = 600.0 // Operating voltage(V)\n", +"\n", +"// Calculations\n", +"R_a = R_wb+R_f // Armature resistance(ohm)\n", +"N_11 = N_1/E_1*(V-I_1*R_a) // Speed(rpm)\n", +"T_1 = 9.55*E_1*I_1/N_1 // Torque(N-m)\n", +"N_2 = N_1/E_2*(V-I_2*R_a) // Speed(rpm)\n", +"T_2 = 9.55*E_2*I_2/N_1 // Torque(N-m)\n", +"N_3 = N_1/E_3*(V-I_3*R_a) // Speed(rpm)\n", +"T_3 = 9.55*E_3*I_3/N_1 // Torque(N-m)\n", +"N_4 = N_1/E_4*(V-I_4*R_a) // Speed(rpm)\n", +"T_4 = 9.55*E_4*I_4/N_1 // Torque(N-m)\n", +"N_5 = N_1/E_5*(V-I_5*R_a) // Speed(rpm)\n", +"T_5 = 9.55*E_5*I_5/N_1 // Torque(N-m)\n", +"N_6 = N_1/E_6*(V-I_6*R_a) // Speed(rpm)\n", +"T_6 = 9.55*E_6*I_6/N_1 // Torque(N-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 6.2 : SOLUTION :-')\n", +"printf('\nSpeed-torque curve for motor')\n", +"printf('\n_______________________________________')\n", +"printf('\n Speed(rpm) : Torque(N-m) ')\n", +"printf('\n_______________________________________')\n", +"printf('\n %.f : %.f ', N_11,T_1)\n", +"printf('\n %.f : %.f ', N_2,T_2)\n", +"printf('\n %.f : %.f ', N_3,T_3)\n", +"printf('\n %.f : %.f ', N_4,T_4)\n", +"printf('\n %.f : %.f ', N_5,T_5)\n", +"printf('\n %.f : %.f ', N_6,T_6)\n", +"printf('\n_______________________________________\n')\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.3: Speed_of_motors_when_connected_in_series.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 6: MOTORS FOR ELECTRIC TRACTION\n", +"\n", +"// EXAMPLE : 6.3 :\n", +"// Page number 790\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 650.0 // Voltage supply(V)\n", +"r_A = 45.0 // Radius of driving wheel(cm)\n", +"r_B = 43.0 // Radius of driving wheel(cm)\n", +"N_A = 400.0 // Speed(rpm)\n", +"drop = 10.0 // Voltage drop(%)\n", +"\n", +"// Calculations\n", +"rho = r_B/r_A\n", +"IR = drop*V/100 // Voltage drop(V)\n", +"V_A = (rho*(V-IR)+IR)/(1+rho) // Voltage(V)\n", +"V_B = V-V_A // Voltage(V)\n", +"N_A_A = N_A*(V_A-IR)/(V-IR) // N'_A(rpm)\n", +"N_B_B = N_A_A*r_A/r_B // N'_B(rpm)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 6.3 : SOLUTION :-')\n", +"printf('\nSpeed of first motor when connected in series, N_A = %.f rpm', N_A_A)\n", +"printf('\nSpeed of second motor when connected in series, N_B = %.f rpm\n', N_B_B)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.4: HP_delivered_by_the_locomotive_when_dc_series_motor_and_Induction_motor_is_used.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 6: MOTORS FOR ELECTRIC TRACTION\n", +"\n", +"// EXAMPLE : 6.4 :\n", +"// Page number 791\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"F_t = 33800.0 // Tractive effort(N)\n", +"V = 48.3 // Velocity(kmph)\n", +"T = 53400.0 // Tractive effort(N)\n", +"\n", +"// Calculations\n", +"HP = F_t*V*1000/(60*60*746) // HP on level track(hp)\n", +"HP_i = HP*(T/F_t)**0.5 // hp delivered by locomotive for dc series motor(hp)\n", +"HP_ii = HP*T/F_t // hp delivered by locomotive for induction motor(hp)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 6.4 : SOLUTION :-')\n", +"printf('\nhp delivered by the locomotive when dc series motor is used = %.f HP', HP_i)\n", +"printf('\nhp delivered by the locomotive when induction motor is used = %.f HP', HP_ii)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.5: New_characteristics_of_motor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 6: MOTORS FOR ELECTRIC TRACTION\n", +"\n", +"// EXAMPLE : 6.5 :\n", +"// Page number 792-793\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"I_1 = 100.0 // Current(A)\n", +"N_1 = 71.0 // Speed(kmph)\n", +"F_t1 = 2225.0 // Tractive effort(N)\n", +"I_2 = 150.0 // Current(A)\n", +"N_2 = 57.0 // Speed(kmph)\n", +"F_t2 = 6675.0 // Tractive effort(N)\n", +"I_3 = 200.0 // Current(A)\n", +"N_3 = 50.0 // Speed(kmph)\n", +"F_t3 = 11600.0 // Tractive effort(N)\n", +"I_4 = 250.0 // Current(A)\n", +"N_4 = 45.0 // Speed(kmph)\n", +"F_t4 = 17350.0 // Tractive effort(N)\n", +"I_5 = 300.0 // Current(A)\n", +"N_5 = 42.0 // Speed(kmph)\n", +"F_t5 = 23200.0 // Tractive effort(N)\n", +"D_A = 101.6 // Size of wheels(cm)\n", +"ratio_gear = 72.0/23 // Gear ratio\n", +"D_B = 106.7 // Size of wheels(cm)\n", +"ratio_gear_new = 75.0/20 // Gear ratio\n", +"\n", +"// Calculations\n", +"N_B = ratio_gear*D_B/(ratio_gear_new*D_A) // Speed in terms of V(kmph)\n", +"F_tB = D_A*ratio_gear_new/(ratio_gear*D_B) // Tractive effort in terms of F_tA(N)\n", +"N_B1 = N_B*N_1 // Speed(kmph)\n", +"F_tB1 = F_tB*F_t1 // Tractive effort(N)\n", +"N_B2 = N_B*N_2 // Speed(kmph)\n", +"F_tB2 = F_tB*F_t2 // Tractive effort(N)\n", +"N_B3 = N_B*N_3 // Speed(kmph)\n", +"F_tB3 = F_tB*F_t3 // Tractive effort(N)\n", +"N_B4 = N_B*N_4 // Speed(kmph)\n", +"F_tB4 = F_tB*F_t4 // Tractive effort(N)\n", +"N_B5 = N_B*N_5 // Speed(kmph)\n", +"F_tB5 = F_tB*F_t5 // Tractive effort(N)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 6.5 : SOLUTION :-')\n", +"printf('\nNew characteristics of motor')\n", +"printf('\n_______________________________________')\n", +"printf('\n Current(A) : Speed(kmph) : F_t(N)')\n", +"printf('\n_______________________________________')\n", +"printf('\n %.f : %.1f : %.f ', I_1,N_B1,F_tB1)\n", +"printf('\n %.f : %.1f : %.f ', I_2,N_B2,F_tB2)\n", +"printf('\n %.f : %.1f : %.f ', I_3,N_B3,F_tB3)\n", +"printf('\n %.f : %.1f : %.f ', I_4,N_B4,F_tB4)\n", +"printf('\n %.f : %.1f : %.f ', I_5,N_B5,F_tB5)\n", +"printf('\n_______________________________________\n')\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/45-CONTROL_OF_MOTORS.ipynb b/Power_System_Engineering_by_S_Chakraborthy/45-CONTROL_OF_MOTORS.ipynb new file mode 100644 index 0000000..3b101d4 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/45-CONTROL_OF_MOTORS.ipynb @@ -0,0 +1,193 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 45: CONTROL OF MOTORS" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.1: Approximate_loss_of_energy_in_starting_rheostats.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 7: CONTROL OF MOTORS\n", +"\n", +"// EXAMPLE : 7.1 :\n", +"// Page number 798\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"no = 2.0 // Number of motors\n", +"V_m = 48.0 // Uniform speed(kmph)\n", +"t = 30.0 // Time(sec)\n", +"F_t_m = 13350.0 // Average tractive effort per motor(N)\n", +"\n", +"// Calculations\n", +"F_t = no*F_t_m // Average tractive effort(N)\n", +"energy = t*F_t*V_m/(2*3600**2) // Useful energy for acceleration(kWh)\n", +"energy_loss = energy/no // Approximate loss of energy in starting rheostats(kWh)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 7.1 : SOLUTION :-')\n", +"printf('\nApproximate loss of energy in starting rheostats = %.3f kWh', energy_loss)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.2: EX45_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 7: CONTROL OF MOTORS\n", +"\n", +"// EXAMPLE : 7.2 :\n", +"// Page number 798\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 175.0 // Weight of multiple unit train(tonnes)\n", +"no = 6.0 // Number of motors\n", +"F_t = 69000.0 // Total tractive effort(N)\n", +"V = 600.0 // Line voltage(V)\n", +"I = 200.0 // Average current(A)\n", +"V_m = 38.6 // Speed(kmph)\n", +"R = 0.15 // Resistance of each motor(ohm)\n", +"\n", +"// Calculations\n", +"alpha = F_t/(277.8*W) // Acceleration(km phps)\n", +"T = V_m/alpha // Time for acceleration(sec)\n", +"t_s = (V-2*I*R)*T/(2*(V-I*R)) // Duration of starting period(sec)\n", +"t_p = T-t_s // (sec)\n", +"energy_total_series = no/2*V*I*t_s // Total energy supplied in series position(watt-sec)\n", +"energy_total_parallel = no*V*I*t_p // Total energy supplied in parallel position(watt-sec)\n", +"total_energy = (energy_total_series+energy_total_parallel)/(1000*3600) // Energy supplied during starting period(kWh)\n", +"energy_waste_series = (no/2)/2*(V-2*I*R)*I*t_s // Energy wasted in starting resistance in series position(watt-sec)\n", +"energy_waste_parallel = no*(V/2)/2*I*t_p // Energy wasted in starting resistance in parallel position(watt-sec)\n", +"total_energy_waste = (energy_waste_series+energy_waste_parallel)/(1000*3600) // Total energy wasted in starting resistance(kWh)\n", +"energy_lost = (no*I**2*R*T)/(1000*3600) // Energy lost in motor resistance(kWh)\n", +"useful_energy = T*F_t*V_m/(2*3600**2) // Useful energy supplied to train(kWh)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 7.2 : SOLUTION :-')\n", +"printf('\nEnergy supplied during the starting period = %.2f kWh', total_energy)\n", +"printf('\nEnergy lost in the starting resistance = %.1f kWh', total_energy_waste)\n", +"printf('\nUseful energy supplied to the train = %.1f kWh', useful_energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.3: EX45_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 7: CONTROL OF MOTORS\n", +"\n", +"// EXAMPLE : 7.3 :\n", +"// Page number 799\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 132.0 // Weight of electric train(tonnes)\n", +"no = 4.0 // Number of motors\n", +"V = 600.0 // Voltage of motor(V)\n", +"I = 400.0 // Current per motor(A)\n", +"F_t_m = 19270.0 // Tractive effort per motor at 400A & 600V(N)\n", +"V_m = 39.0 // Train speed(kmph)\n", +"G = 1.0 // Gradient\n", +"r = 44.5 // Resistance to traction(N/tonne)\n", +"inertia = 10.0 // Rotational inertia(%)\n", +"R = 0.1 // Resistance of each motor(ohm)\n", +"\n", +"// Calculations\n", +"W_e = W*(100+inertia)/100 // Accelerating weight of train(tonne)\n", +"F_t = F_t_m*no // Total tractive effort at 400A & 600V(N)\n", +"alpha = (F_t-W*r-98.1*W*G)/(277.8*W_e) // Acceleration(km phps)\n", +"T = V_m/alpha // Time for acceleration(sec)\n", +"t_s = (V-2*I*R)*T/(2*(V-I*R)) // Duration of starting period(sec)\n", +"V_transition = alpha*t_s // Speed at transition(km phps)\n", +"t_p = T-t_s // (sec)\n", +"loss_series = (no/2*((V-2*I*R)/2)*I*t_s)/(1000*3600) // Energy lost during series period(kWh)\n", +"loss_parallel = (no*(V/2)/2*I*t_p)/(1000*3600) // Energy lost during parallel period(kWh)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 7.3 : SOLUTION :-')\n", +"printf('\nCase(i) : Duration of starting period, t_s = %.1f sec', t_s)\n", +"printf('\nCase(ii) : Speed of train at transition, αt = %.1f sec', V_transition)\n", +"printf('\nCase(iii): Case(a): Rheostatic losses during series starting = %.2f kWh', loss_series)\n", +"printf('\n Case(b): Rheostatic losses during parallel starting = %.2f kWh\n', loss_parallel)\n", +"printf('\nNOTE: ERROR: Calculation mistakes in the textbook solution')" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/46-BRAKING.ipynb b/Power_System_Engineering_by_S_Chakraborthy/46-BRAKING.ipynb new file mode 100644 index 0000000..95ef84d --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/46-BRAKING.ipynb @@ -0,0 +1,288 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 46: BRAKING" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 46.1: Braking_torque.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 8: BRAKING\n", +"\n", +"// EXAMPLE : 8.1 :\n", +"// Page number 806\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 525.0 // Voltage of motor(V)\n", +"I_1 = 50.0 // Current(A)\n", +"T_1 = 216.0 // Torque(N-m)\n", +"I_2 = 70.0 // Current(A)\n", +"T_2 = 344.0 // Torque(N-m)\n", +"I_3 = 80.0 // Current(A)\n", +"T_3 = 422.0 // Torque(N-m)\n", +"I_4 = 90.0 // Current(A)\n", +"T_4 = 500.0 // Torque(N-m)\n", +"V_m = 26.0 // Speed(kmph)\n", +"R_b = 5.5 // Resistance of braking rheostat(ohm)\n", +"R_m = 0.5 // Resistance of motor(ohm)\n", +"\n", +"// Calculations\n", +"I = 75.0 // Current drawn at 26 kmph(A)\n", +"back_emf = V-I*R_m // Back emf of the motor(V)\n", +"R_t = R_b+R_m // Total resistance(ohm)\n", +"I_del = back_emf/R_t // Current delivered(A)\n", +"T_b = T_3*I_del/I_3 // Braking torque(N-m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 8.1 : SOLUTION :-')\n", +"printf('\nBraking torque = %.f N-m', T_b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 46.2: Current_delivered_when_motor_works_as_generator.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 8: BRAKING\n", +"\n", +"// EXAMPLE : 8.2 :\n", +"// Page number 806\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 525.0 // Voltage of motor(V)\n", +"I_1 = 50.0 // Current(A)\n", +"N_1 = 1200.0 // Speed(rpm)\n", +"I_2 = 100.0 // Current(A)\n", +"N_2 = 950.0 // Speed(rpm)\n", +"I_3 = 150.0 // Current(A)\n", +"N_3 = 840.0 // Speed(rpm)\n", +"I_4 = 200.0 // Current(A)\n", +"N_4 = 745.0 // Speed(rpm)\n", +"N = 1000.0 // Speed opearting(rpm)\n", +"R = 3.0 // Resistance(ohm)\n", +"R_m = 0.5 // Resistance of motor(ohm)\n", +"\n", +"// Calculations\n", +"I = 85.0 // Current drawn at 1000 rpm(A)\n", +"back_emf = V-I*R_m // Back emf of the motor(V)\n", +"R_t = R+R_m // Total resistance(ohm)\n", +"I_del = back_emf/R_t // Current delivered(A)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 8.2 : SOLUTION :-')\n", +"printf('\nCurrent delivered when motor works as generator = %.f A', I_del)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 46.3: Energy_returned_to_lines.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 8: BRAKING\n", +"\n", +"// EXAMPLE : 8.3 :\n", +"// Page number 810\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 400.0 // Weight of train(tonne)\n", +"G = 100.0/70 // Gradient(%)\n", +"t = 120.0 // Time(sec)\n", +"V_1 = 80.0 // Speed(km/hr)\n", +"V_2 = 50.0 // Speed(km/hr)\n", +"r_kg = 5.0 // Tractive resistance(kg/tonne)\n", +"I = 7.5 // Rotational inertia(%)\n", +"n = 0.75 // Overall efficiency\n", +"\n", +"// Calculations\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"r = r_kg*9.81 // Tractive resistance(N-m/tonne)\n", +"energy_recuperation = 0.01072*W_e*(V_1**2-V_2**2)/1000 // Energy available for recuperation(kWh)\n", +"F_t = W*(r-98.1*G) // Tractive effort during retardation(N)\n", +"distance = (V_1+V_2)*1000*t/(2*3600) // Distance travelled by train during retardation period(m)\n", +"energy_train = abs(F_t)*distance/(3600*1000) // Energy available during train movement(kWh)\n", +"net_energy = n*(energy_recuperation+energy_train) // Net energy returned to supply system(kWh)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 8.3 : SOLUTION :-')\n", +"printf('\nEnergy returned to lines = %.2f kWh\n', net_energy)\n", +"printf('\nNOTE: ERROR: Calculation mistakes & more approximation in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 46.4: Energy_returned_to_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 8: BRAKING\n", +"\n", +"// EXAMPLE : 8.4 :\n", +"// Page number 810\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"W = 355.0 // Weight of train(tonne)\n", +"V_1 = 80.5 // Speed(km/hr)\n", +"V_2 = 48.3 // Speed(km/hr)\n", +"D = 1.525 // Distance(km)\n", +"G = 100.0/90 // Gradient(%)\n", +"I = 10.0 // Rotational inertia(%)\n", +"r = 53.0 // Tractive resistance(N/tonne)\n", +"n = 0.8 // Overall efficiency\n", +"\n", +"// Calculations\n", +"beta = (V_1**2-V_2**2)/(2*D*3600) // Braking retardation(km phps)\n", +"W_e = W*(100+I)/100 // Accelerating weight of train(tonne)\n", +"F_t = 277.8*W_e*beta+98.1*W*G-W*r // Tractive effort(N)\n", +"work_done = F_t*D*1000 // Work done by this effort(N-m)\n", +"energy = work_done*n/(1000*3600) // Energy returned to line(kWh)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 8.4 : SOLUTION :-')\n", +"printf('\nEnergy returned to the line = %.1f kWh', energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 46.5: Braking_effect_and_Rate_of_retardation_produced_by_this_braking_effect.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 8: BRAKING\n", +"\n", +"// EXAMPLE : 8.5 :\n", +"// Page number 811-812\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"funcprot(0)\n", +"\n", +"// Given data\n", +"area = 16.13 // Area of brakes(sq.cm/pole face)\n", +"phi = 2.5*10**-3 // Flux(Wb)\n", +"u = 0.2 // Co-efficient of friction\n", +"W = 10.0 // Weight of car(tonnes)\n", +"\n", +"// Calculations\n", +"a = area*10**-4 // Area of brakes(sq.m/pole face)\n", +"F = phi**2/(2*%pi*10**-7*a) // Force(N)\n", +"force = F*u // Braking effect considering flux and coefficient of friction(N)\n", +"beta = u*F/(W*1000)*100 // Rate of retardation produced by braking effect(cm/sec^2)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 8.5 : SOLUTION :-')\n", +"printf('\nBraking effect, F = %.f N', force)\n", +"printf('\nRate of retardation produced by this braking effect, β = %.2f cm/sec^2', beta)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/47-ELECTRIC_TRACTION_SYSTEMS_AND_POWER_SUPPLY.ipynb b/Power_System_Engineering_by_S_Chakraborthy/47-ELECTRIC_TRACTION_SYSTEMS_AND_POWER_SUPPLY.ipynb new file mode 100644 index 0000000..26f787e --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/47-ELECTRIC_TRACTION_SYSTEMS_AND_POWER_SUPPLY.ipynb @@ -0,0 +1,124 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 47: ELECTRIC TRACTION SYSTEMS AND POWER SUPPLY" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 47.1: EX47_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 9: ELECTRIC TRACTION SYSTEMS AND POWER SUPPLY\n", +"\n", +"// EXAMPLE : 9.1 :\n", +"// Page number 817-818\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"L = 3.0 // Length of section ACB of rail(km)\n", +"L_B_A = 2.0 // Distance of B from A(km)\n", +"I_load = 350.0 // Loading(A/km)\n", +"r_rail = 0.035 // Resistance of rail(ohm/km)\n", +"r_feed = 0.03 // Resistance of negative feeder(ohm/km)\n", +"\n", +"// Calculations\n", +"x_val = integrate('I_load*(L-x)','x',0,L_B_A)\n", +"I = x_val/(L_B_A-0) // Current in negative feeder(A)\n", +"x = L-(I/I_load) // Distance from feeding point(km)\n", +"C = integrate('r_rail*I_load*x','x',0,x)\n", +"V = r_feed*L_B_A*I // Voltage produced by negative booster(V)\n", +"rating = V*I/1000 // Rating of the booster(kW)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 9.1 : SOLUTION :-')\n", +"printf('\nMaximum potential difference between any two points of the rails, C = %.2f V', C)\n", +"printf('\nRating of the booster = %.1f kW', rating)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 47.2: Maximum_sag_and_Length_of_wire_required.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART IV : UTILIZATION AND TRACTION\n", +"// CHAPTER 9: ELECTRIC TRACTION SYSTEMS AND POWER SUPPLY\n", +"\n", +"// EXAMPLE : 9.2 :\n", +"// Page number 820\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D = 50.0 // Distance between poles(m)\n", +"w = 0.5 // Weight of trolley wire per metre(kg)\n", +"T = 520.0 // Maximum tension(kg)\n", +"\n", +"// Calculations\n", +"l = D/2 // Half distance b/w poles(m)\n", +"d = w*l**2/(2*T) // Sag(m)\n", +"wire_length = 2*(l+(2*d**2/(3*l))) // Length of wire required(m)\n", +"\n", +"// Results\n", +"disp('PART IV - EXAMPLE : 9.2 : SOLUTION :-')\n", +"printf('\nMaximum sag, d = %.4f metres', d)\n", +"printf('\nLength of wire required = %.f metres', wire_length)" + ] + } +], +"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/Power_System_Engineering_by_S_Chakraborthy/7-TARIFFS_AND_ECONOMIC_ASPECTS_IN_POWER_GENERATION.ipynb b/Power_System_Engineering_by_S_Chakraborthy/7-TARIFFS_AND_ECONOMIC_ASPECTS_IN_POWER_GENERATION.ipynb new file mode 100644 index 0000000..f1becc9 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/7-TARIFFS_AND_ECONOMIC_ASPECTS_IN_POWER_GENERATION.ipynb @@ -0,0 +1,1691 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10: Cost_of_generation_per_kWh_at_100_and_50_percent_load_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.10 :\n", +"// Page number 76\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_installed = 210.0*10**3 // Installed capacity of the station(kW)\n", +"capital_cost_kW = 1000.0 // Capital cost of station(Rs/kW)\n", +"fixed_cost_per = 0.13 // Fixed cost = 13% * cost of investment\n", +"variable_cost_per = 1.3 // Variable cost = 1.3*fixed cost\n", +"LF_1 = 1.0 // Load factor\n", +"LF_2 = 0.5 // Load factor\n", +"\n", +"// Calculations\n", +"MD = cap_installed // Maximum demand(kW)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"capital_cost = capital_cost_kW*cap_installed // Capital cost of station(Rs)\n", +"// Case(i) At 100% load factor\n", +"fixed_cost_1 = capital_cost*fixed_cost_per // Fixed cost(Rs)\n", +"variable_cost_1 = variable_cost_per*fixed_cost_1 // Variable cost(Rs)\n", +"operating_cost_1 = fixed_cost_1+variable_cost_1 // Operating cost per annum(Rs)\n", +"units_gen_1 = LF_1*MD*hours_year // Total units generated(kWh)\n", +"cost_gen_1 = operating_cost_1*100/units_gen_1 // Cost of generation per kWh(Paise)\n", +"// Case(ii) At 50% load factor\n", +"fixed_cost_2 = capital_cost*fixed_cost_per // Fixed cost(Rs)\n", +"units_gen_2 = LF_2*MD*hours_year // Total units generated(kWh)\n", +"variable_cost_2 = variable_cost_1*units_gen_2/units_gen_1 // Variable cost(Rs)\n", +"operating_cost_2 = fixed_cost_2+variable_cost_2 // Operating cost per annum(Rs)\n", +"cost_gen_2 = operating_cost_2*100/units_gen_2 // Cost of generation per kWh(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.10 : SOLUTION :-')\n", +"printf('\nCost of generation per kWh at 100 percent load factor = %.2f paise', cost_gen_1)\n", +"printf('\nCost of generation per kWh at 50 percent load factor = %.1f paise', cost_gen_2)\n", +"printf('\nComment: As the load factor is reduced, cost of generation is increased\n')\n", +"printf('\nNOTE: ERROR: (1) In problem statement, Capital cost of station must be Rs. 1000/kW, not Rs. 1000/MW')\n", +"printf('\n (2) Calculation mistake in Total units generated in Case(i) in textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.11: Cost_per_unit_generated.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.11 :\n", +"// Page number 76\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 100.0*10**3 // Maximum demand(kW)\n", +"capital_cost = 200.0*10**6 // Capital cost(Rs)\n", +"LF = 0.4 // Annual load factor\n", +"cost_fueloil = 15.0*10**6 // Annual cost of fuel and oil(Rs)\n", +"cost_tax = 10.0*10**6 // Cost of taxes, wages and salaries(Rs)\n", +"interest = 0.15 // Interest and depreciation\n", +"\n", +"// Calculations\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"units_gen = MD*LF*hours_year // Units generated per annum(kWh)\n", +"fixed_charge = interest*capital_cost // Annual fixed charges(Rs)\n", +"running_charge = cost_fueloil+cost_tax // Annual running charges(Rs)\n", +"annual_charge = fixed_charge+running_charge // Total annual charges(Rs)\n", +"cost_unit = annual_charge*100/units_gen // Cost per unit(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.11 : SOLUTION :-')\n", +"printf('\nCost per unit generated = %.f paise', cost_unit)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.12: Minimum_reserve_capacity_of_station_and_Cost_per_kWh_generated.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.12 :\n", +"// Page number 76-77\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_installed = 500.0 // Installed capacity of the station(MW)\n", +"CF = 0.45 // Capacity factor\n", +"LF = 0.6 // Annual laod factor\n", +"cost_fueloil = 10.0*10**7 // Annual cost of fuel,oil etc(Rs)\n", +"capital_cost = 10**9 // Capital cost(Rs)\n", +"interest = 0.15 // Interest and depreciation\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"MD = cap_installed*CF/LF // Maximum demand(MW)\n", +"cap_reserve = cap_installed-MD // Reserve capacity(MW)\n", +"// Case(ii)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"units_gen = MD*10**3*LF*hours_year // Units generated per annum(kWh)\n", +"fixed_charge = interest*capital_cost // Annual fixed charges(Rs)\n", +"running_charge = cost_fueloil // Annual running charges(Rs)\n", +"annual_charge = fixed_charge+running_charge // Total annual charges(Rs)\n", +"cost_unit = annual_charge*100/units_gen // Cost per kWh generated(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.12 : SOLUTION :-')\n", +"printf('\nCase(i) : Minimum reserve capacity of station = %.f MW', cap_reserve)\n", +"printf('\nCase(ii): Cost per kWh generated = %.f paise', cost_unit)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.13: Two_part_tariff_to_be_charged_from_consumers.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.13 :\n", +"// Page number 77\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"gen_expense = 850000.0 // Annual generation expense(Rs)\n", +"fuel_expense = 2800000.0 // Annual fuel expense(Rs)\n", +"trans_expense = 345000.0 // Annual transmission expense(Rs)\n", +"dist_expense = 2750000.0 // Annual distribution expense(Rs)\n", +"repair_expense = 300000.0 // Annual repairs,etc expense(Rs)\n", +"unit_gen = 600.0*10**6 // Number of units generated per year(kWh)\n", +"MD = 75.0*10**3 // Maximum demand(kW)\n", +"gen = 0.9 // Fixed charges for generation\n", +"fuel = 0.15 // Fixed charges for fuel\n", +"transm = 0.85 // Fixed charges for transmission\n", +"dist = 0.95 // Fixed charges for distribution\n", +"repair = 0.5 // Fixed charges for repairs,etc\n", +"loss_dist = 0.2 // Losses in transmission and distribution\n", +"\n", +"// Calculations\n", +"fixed_gen = gen_expense*gen // Fixed charge on generation(Rs)\n", +"running_gen = gen_expense*(1-gen) // Running charge on generation(Rs)\n", +"fixed_fuel = fuel_expense*fuel // Fixed charge on fuel(Rs)\n", +"running_fuel = fuel_expense*(1-fuel) // Running charge on fuel(Rs)\n", +"fixed_trans = trans_expense*transm // Fixed charge on transmission(Rs)\n", +"running_trans = trans_expense*(1-transm) // Running charge on transmission(Rs)\n", +"fixed_dist = dist_expense*dist // Fixed charge on distribution(Rs)\n", +"running_dist = dist_expense*(1-dist) // Running charge on distribution(Rs)\n", +"fixed_repair = repair_expense*repair // Fixed charge on repairs,etc(Rs)\n", +"running_repair = repair_expense*(1-repair) // Running charge on repairs,etc(Rs)\n", +"fixed_charge = fixed_gen+fixed_fuel+fixed_trans+fixed_dist+fixed_repair // Total fixed charges(Rs)\n", +"running_charge = running_gen+running_fuel+running_trans+running_dist+running_repair // Total running charges(Rs)\n", +"fixed_unit = fixed_charge/MD // Fixed charges per unit(Rs)\n", +"units_dist = unit_gen*(1-loss_dist) // Total number of units distributed(kWh)\n", +"running_unit = running_charge*100/units_dist // Running charges per unit(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.13 : SOLUTION :-')\n", +"printf('\nTwo part tariff is Rs %.3f per kW of maximum demand plus %.3f paise per kWh', fixed_unit,running_unit)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.14: Generation_cost_in_two_part_form.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.14 :\n", +"// Page number 77\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_installed = 100.0*10**3 // Installed capacity of the station(kW)\n", +"capital_cost_kW = 1000.0 // Capital cost(Rs/kW)\n", +"depreciation = 0.15 // Annual depreciation charge\n", +"royalty_kW = 2.0 // Royalty per kW per year(Rs)\n", +"royalty_kWh = 0.03 // Royalty per kWh per year(Rs)\n", +"MD = 70.0*10**3 // Maximum demand(kW)\n", +"LF = 0.6 // Annual load factor\n", +"cost_salary = 1000000.0 // Annual cost of salaries,maintenance charges etc(Rs)\n", +"cost_salary_per = 0.2 // Annual cost of salaries,maintenance charges etc charged as fixed charges\n", +"\n", +"// Calculations\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"unit_gen = MD*LF*hours_year // Units generated/annum(kWh)\n", +"capital_cost = cap_installed*capital_cost_kW // Capital cost of plant(Rs)\n", +"depreciation_charge = depreciation*capital_cost // Depreciation charges(Rs)\n", +"salary_charge = cost_salary_per*cost_salary // Cost on salaries, maintenance etc(Rs)\n", +"fixed_charge = depreciation_charge+salary_charge // Total annual fixed charges(Rs)\n", +"cost_kW_fixed = (fixed_charge/MD)+royalty_kW // Cost per kW(Rs)\n", +"salary_charge_running = (1-cost_salary_per)*cost_salary // Annual running charge on salaries, maintenance etc(Rs)\n", +"cost_kWh_running = (salary_charge_running/unit_gen)+royalty_kWh // Cost per kWh(Rs)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.14 : SOLUTION :-')\n", +"printf('\nGeneration cost in two part form is given by, Rs. (%.2f*kW + %.3f*kWh) ', cost_kW_fixed,cost_kWh_running)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.15: Overall_generating_cost_per_unit_at_50_and_100_percent_capacity_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.15 :\n", +"// Page number 78\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_installed = 100.0*10**3 // Installed capacity of station(kW)\n", +"cost_gen = 30.0 // Generating cost per annum(Rs/kW)\n", +"cost_fixed = 4000000.0 // Fixed cost per annum(Rs)\n", +"cost_fuel = 60.0 // Cost of fuel(Rs/tonne)\n", +"calorific = 5700.0 // Calorific value of fuel(kcal/kg)\n", +"rate_heat_1 = 2900.0 // Plant heat rate at 100% capacity factor(kcal/kWh)\n", +"CF_1 = 1.0 // Capacity factor\n", +"rate_heat_2 = 4050.0 // Plant heat rate at 50% capacity factor(kcal/kWh)\n", +"CF_2 = 0.5 // Capacity factor\n", +"\n", +"// Calculations\n", +"cost_fixed_kW = cost_fixed/cap_installed // Fixed cost per kW(Rs)\n", +"cost_fixed_total = cost_gen+cost_fixed_kW // Fixed cost per kW capacity(Rs)\n", +"average_demand_1 = CF_1*cap_installed // Average demand at 100% capacity factor(kW)\n", +"average_demand_2 = CF_2*cap_installed // Average demand at 50% capacity factor(kW)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"unit_gen_1 = CF_1*hours_year // Energy generated per annum with average demand of 1 kW(kWh)\n", +"unit_gen_2 = CF_2*hours_year // Energy generated per annum with average demand of 0.5 kW(kWh)\n", +"cost_kWh_fixed_1 = cost_fixed_total*100/unit_gen_1 // Cost per kWh due to fixed charge with 100% CF(Paise)\n", +"cost_kWh_fixed_2 = cost_fixed_total*100/unit_gen_2 // Cost per kWh due to fixed charge with 50% CF(Paise)\n", +"kg_kWh_1 = rate_heat_1/calorific // Weight(kg)\n", +"kg_kWh_2 = rate_heat_2/calorific // Weight(kg)\n", +"cost_coal_1 = kg_kWh_1*cost_fuel*100/1000.0 // Cost due to coal at 100% CF(Paise/kWh)\n", +"cost_coal_2 = kg_kWh_2*cost_fuel*100/1000.0 // Cost due to coal at 50% CF(Paise/kWh)\n", +"cost_total_1 = cost_kWh_fixed_1+cost_coal_1 // Total cost per unit with 100% CF(Paise)\n", +"cost_total_2 = cost_kWh_fixed_2+cost_coal_2 // Total cost per unit with 50% CF(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.15 : SOLUTION :-')\n", +"printf('\nOverall generating cost per unit at 100 percent capacity factor = %.3f paise', cost_total_1)\n", +"printf('\nOverall generating cost per unit at 50 percent capacity factor = %.3f paise\n', cost_total_2)\n", +"printf('\nNOTE: Slight changes in obtained answer from that of textbook answer is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.16: EX7_16.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.16 :\n", +"// Page number 78\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 75.0*10**3 // Maximum demand(kW)\n", +"LF = 0.4 // Yearly load factor\n", +"cost_capital = 60.0 // Capital cost(Rs/annum/kW)\n", +"cost_kWh = 1.0 // Cost per kWh transmitted(Paise)\n", +"charge_trans = 2000000.0 // Annual capital charge for transmission(Rs)\n", +"charge_dist = 1500000.0 // Annual capital charge for distribution(Rs)\n", +"diversity_trans = 1.2 // Diversity factor for transmission\n", +"diversity_dist = 1.25 // Diversity factor for distribution\n", +"n_trans = 0.9 // Efficiency of transmission system\n", +"n_dist = 0.85 // Efficiency of distribution system\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"capital_cost = cost_capital*MD // Annual capital cost(Rs)\n", +"fixed_charge_sub = capital_cost+charge_trans // Total fixed charges for supply to substation per annum(Rs)\n", +"sum_MD_sub = MD*diversity_trans // Sum of all maximum demand of substation(kW)\n", +"cost_kW_sub = fixed_charge_sub/sum_MD_sub // Yearly cost per kW demand at substation(Rs)\n", +"running_cost_unit_sub = 1/n_trans // Running cost per unit supplied at substation(Paise)\n", +"// Case(b)\n", +"sum_MD_con = sum_MD_sub*diversity_dist // Sum of all maximum demand of consumer(kW)\n", +"fixed_charge_con = capital_cost+charge_trans+charge_dist // Total fixed charges for supply to cosnumers(Rs)\n", +"cost_kW_con = fixed_charge_con/sum_MD_con // Yearly cost per kW demand on consumer premises(Rs)\n", +"running_cost_unit_con = running_cost_unit_sub/n_dist // Running cost per unit supplied to consumer(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.16 : SOLUTION :-')\n", +"printf('\nCase(a): Yearly cost per kW demand at the substations = Rs. %.2f ', cost_kW_sub)\n", +"printf('\n Cost per kWh supplied at the substations = %.2f paise\n', running_cost_unit_sub)\n", +"printf('\nCase(b): Yearly cost per kW demand at the consumer premises = Rs. %.2f ', cost_kW_con)\n", +"printf('\n Cost per kWh supplied at the consumer premises = %.3f paise', running_cost_unit_con)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.17: Number_of_working_hours_per_week_above_which_the_HV_supply_is_cheaper.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.17 :\n", +"// Page number 79\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"kVA_tariff_hv = 60.0 // HV supply per kVA per annum(Rs)\n", +"kWh_tariff_hv = 3.0/100 // HV supply per kWh annum(Rs)\n", +"kVA_tariff_lv = 65.0 // LV supply per kVA per annum(Rs)\n", +"kWh_tariff_lv = 3.3/100 // LV supply per kWh annum(Rs)\n", +"cost_equip_kVA = 50.0 // Cost of transformers and switchgear per kVA(Rs)\n", +"loss_full_load = 0.02 // Full load transformation loss\n", +"fixed_charge_per = 0.2 // Fixed charges per annum\n", +"no_week = 50.0 // Number of working weeks in a year\n", +"\n", +"// Calculations\n", +"rating_equip = 1000/(1-loss_full_load) // Rating of transformer and switchgear(kVA)\n", +"cost_equip = cost_equip_kVA*rating_equip // Cost of transformers and switchgear(Rs)\n", +"fixed_charge = fixed_charge_per*cost_equip // Fixed charges per annum on HV plant(Rs)\n", +"X = poly(0,'X') // Number of working hours per week\n", +"units_consumed = (no_week*X)*1000.0 // Yearly units consumed by load\n", +"total_units = units_consumed/(1-loss_full_load) // Total units to be paid on HV supply\n", +"// Case(a)\n", +"annual_cost_hv = (kVA_tariff_hv*rating_equip)+(kWh_tariff_hv*cost_equip*X)+fixed_charge // Annual cost(Rs)\n", +"// Case(b)\n", +"annual_cost_lv = (kVA_tariff_lv*1000.0)+(kWh_tariff_lv*units_consumed) // Annual cost(Rs)\n", +"p = annual_cost_hv-annual_cost_lv // Finding unknown value i.e working hours in terms of X\n", +"x = roots(p) // Finding unknown value i.e working hours\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.17 : SOLUTION :-')\n", +"printf('\nAbove %.1f working hours per week the H.V supply is cheaper ', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.18: Cheaper_alternative_to_adopt_and_by_how_much.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.18 :\n", +"// Page number 79-80\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_1 = 10.0*10**3 // Load per annum(kVA)\n", +"time_1 = 1800.0 // Time(hours)\n", +"load_2 = 6.0*10**3 // Load per annum(kVA)\n", +"time_2 = 600.0 // Time(hours)\n", +"load_3 = 0.25*10**3 // Load per annum(kVA)\n", +"time_3 = 400.0 // Time(hours)\n", +"rating_trans = 10.0*10**3 // Transformer rating(kVA)\n", +"pf = 0.8 // Lagging power factor\n", +"n_fl_A = 98.3/100.0 // Full load efficiency of transformer A\n", +"n_fl_B = 98.8/100.0 // Full load efficiency of transformer B\n", +"loss_A = 70.0 // Core loss at rated voltage of transformer A(kW)\n", +"loss_B = 40.0 // Core loss at rated voltage of transformer B(kW)\n", +"cost_A = 250000.0 // Cost of transformer A(Rs)\n", +"cost_B = 280000.0 // Cost of transformer B(Rs)\n", +"interest_per = 0.1 // Interest and depreciation charges\n", +"cost_energy_unit = 3.0 // Energy costs per unit(Paise)\n", +"\n", +"// Calculations\n", +"// Transformer A\n", +"output_A = rating_trans*pf // kW output at full load(kW)\n", +"input_A = output_A/n_fl_A // Input at full load(kW)\n", +"cu_loss_fl_A = input_A-output_A-loss_A // Copper loss at full load(kW)\n", +"cu_loss_2_A = (load_2/load_1)**2*cu_loss_fl_A // Copper loss at 6 MVA output(kW)\n", +"cu_loss_3_A = (load_3/load_1)**2*cu_loss_fl_A // Copper loss at 0.25 MVA output(kW)\n", +"ene_iron_loss_A = loss_A*(time_1+time_2+time_3) // Energy consumed due to iron losses(kWh)\n", +"ene_cu_loss_A = time_1*cu_loss_fl_A+time_2*cu_loss_2_A+time_3*cu_loss_3_A // Energy consumed due to copper losses(kWh)\n", +"total_loss_A = ene_iron_loss_A+ene_cu_loss_A // Total loss per annum(kWh)\n", +"cost_energy_A = cost_energy_unit/100*total_loss_A // Energy cost per annum due to losses(Rs)\n", +"// Transformer B\n", +"output_B = rating_trans*pf // kW output at full load(kW)\n", +"input_B = output_B/n_fl_B // Input at full load(kW)\n", +"cu_loss_fl_B = input_B-output_B-loss_B // Copper loss at full load(kW)\n", +"cu_loss_2_B = (load_2/load_1)**2*cu_loss_fl_B // Copper loss at 6 MVA output(kW)\n", +"cu_loss_3_B = (load_3/load_1)**2*cu_loss_fl_B // Copper loss at 0.25 MVA output(kW)\n", +"ene_iron_loss_B = loss_B*(time_1+time_2+time_3) // Energy consumed due to iron losses(kWh)\n", +"ene_cu_loss_B = time_1*cu_loss_fl_B+time_2*cu_loss_2_B+time_3*cu_loss_3_B // Energy consumed due to copper losses(kWh)\n", +"total_loss_B = ene_iron_loss_B+ene_cu_loss_B // Total loss per annum(kWh)\n", +"cost_energy_B = cost_energy_unit/100*total_loss_B // Energy cost per annum due to losses(Rs)\n", +"diff_capital = cost_B-cost_A // Difference in capital costs(Rs)\n", +"annual_charge = interest_per*diff_capital // Annual charge due to this amount(Rs)\n", +"diff_cost_energy = cost_energy_A-cost_energy_B // Difference in energy cost per annum(Rs)\n", +"cheap = diff_cost_energy-annual_charge // Cheaper in cost(Rs)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.18 : SOLUTION :-')\n", +"printf('\nTransformer B is cheaper by Rs. %.f per year \n', cheap)\n", +"printf('\nNOTE: ERROR: Full load efficiency for transformer B is 98.8 percent, not 98.3 percent as given in problem statement')\n", +"printf('\n Changes in obtained answer from that of textbook answer is due to more precision')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.19: EX7_19.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.19 :\n", +"// Page number 80-81\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"fixed_cost = 4.0*10**4 // Fixed cost of plant(Rs)\n", +"salvage_value = 4.0*10**3 // Salvage value(Rs)\n", +"n = 20.0 // Useful life(years)\n", +"r = 0.06 // Sinking fund depreciation compounded annually\n", +"\n", +"// Calculations\n", +"n_2 = n/2 // Halfway of useful life(years)\n", +"// Case(a)\n", +"total_dep_A = fixed_cost-salvage_value // Total depreciation in 20 years(Rs)\n", +"dep_10_A = total_dep_A/2 // Depreciation in 10 years(Rs)\n", +"value_10_A = fixed_cost-dep_10_A // Value at the end of 10 years(Rs)\n", +"// Case(b)\n", +"P_B = fixed_cost // Capital outlay(Rs)\n", +"q_B = (salvage_value/fixed_cost)**(1/n) // q = (1-p)\n", +"value_10_B = P_B*(q_B)**n_2 // Value at the end of 10 years(Rs)\n", +"// Case(c)\n", +"P_C = fixed_cost // Capital cost of plant(Rs)\n", +"P__C = salvage_value // Scrap value(Rs)\n", +"Q_C = P_C-P__C // Cost of replacement(Rs)\n", +"q_C = Q_C/(((1+r)**n-1)/r) // Yearly charge(Rs)\n", +"amount_dep = q_C*((1+r)**n_2-1)/r // Amount deposited at end of 10 years(Rs)\n", +"value_10_C = P_C-amount_dep // Value at the end of 10 years(Rs)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.19 : SOLUTION :-')\n", +"printf('\nCase(a): Valuation halfway through its life based on Straight line depreciation method = Rs %.1e ', value_10_A)\n", +"printf('\nCase(b): Valuation halfway through its life based on Reducing balance depreciation method = Rs %.2e ', value_10_B)\n", +"printf('\nCase(c): Valuation halfway through its life based on Sinking fund depreciation method = Rs %.2e ', value_10_C)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1: Demand_factor_and_Load_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.1 :\n", +"// Page number 73\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"connected_load = 450.0*10**3 // Connected load(kW)\n", +"maximum_demand = 250.0*10**3 // Maximum demand(kW)\n", +"units_generated = 615.0*10**6 // Units generated per annum(kWh)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"demand_factor = maximum_demand/connected_load // Demand factor\n", +"// Case(ii)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"average_demand = units_generated/hours_year // Average demand(kW)\n", +"load_factor = average_demand/maximum_demand*100 // Load factor(%)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.1 : SOLUTION :-')\n", +"printf('\nCase(i) : Demand factor = %.3f ', demand_factor)\n", +"printf('\nCase(ii): Load factor = %.1f percent', load_factor)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.20: Type_and_hp_ratings_of_two_turbines_for_the_station.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.20 :\n", +"// Page number 81\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"h = 30.0 // Mean head(m)\n", +"area_catch = 250.0 // Catchment area(Square km)\n", +"average_rain = 1.25 // Average rainfall per annum(m)\n", +"utilized_rain = 0.7 // Rainfall utilized\n", +"LF = 0.8 // Expected load factor\n", +"n_turbine = 0.9 // Mechanical efficiency of turbine\n", +"n_gen = 0.95 // Efficiency of generator\n", +"\n", +"// Calculations\n", +"water_avail = utilized_rain*area_catch*10**6*average_rain // Water available(m^3)\n", +"sec_year = 365.0*24*60*60 // Total seconds in a year\n", +"Q = water_avail/sec_year // Quantity available per second(m^3) i.e Discharge(m^3/sec)\n", +"w = 1000.0 // Density of water(kg/m^3)\n", +"n = n_turbine*n_gen // Overall efficiency\n", +"P = 0.736/75*Q*w*h*n // Average output of generator units(kW)\n", +"rating_gen = P/LF // Rating of generator(kW)\n", +"rating_gen_each = rating_gen/2.0 // Rating of each generator(kW)\n", +"rating_turbine = rating_gen/2*(1/(0.736*n_gen)) // Rating of each turbine(metric hp)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.20 : SOLUTION :-')\n", +"printf('\nChoice of units are:')\n", +"printf('\n 2 generators each having maximum rating of %.f kW ', rating_gen_each)\n", +"printf('\n 2 propeller turbines each having maximum rating of %.f metric hp \n', rating_turbine)\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.21: Plot_of_chronological_load_curve_and_Load_duration_curve.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.21 :\n", +"// Page number 81-82\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"t0 = 0.0 // Time 12 morning\n", +"l0 = 4.0 // Load at 12 morning(kW*1000)\n", +"t1 = 1.0 // Time 1 a.m\n", +"l1 = 3.5 // Load at 1 a.m(kW*1000)\n", +"t2 = 2.0 // Time 2 a.m\n", +"l2 = 3.0 // Load at 2 a.m(kW*1000)\n", +"t3 = 3.0 // Time 3 a.m\n", +"l3 = 3.0 // Load at 3 a.m(kW*1000)\n", +"t4 = 4.0 // Time 4 a.m\n", +"l4 = 3.5 // Load at 4 a.m(kW*1000)\n", +"t5 = 5.0 // Time 5 a.m\n", +"l5 = 3.0 // Load at 5 a.m(kW*1000)\n", +"t6 = 6.0 // Time 6 a.m\n", +"l6 = 6.0 // Load at 6 a.m(kW*1000)\n", +"t7 = 7.0 // Time 7 a.m\n", +"l7 = 12.5 // Load at 7 a.m(kW*1000)\n", +"t8 = 8.0 // Time 8 a.m\n", +"l8 = 14.5 // Load at 8 a.m(kW*1000)\n", +"t9 = 9.0 // Time 9 a.m\n", +"l9 = 13.5 // Load at 9 a.m(kW*1000)\n", +"t10 = 10.0 // Time 10 a.m\n", +"l10 = 13.0 // Load at 10 a.m(kW*1000)\n", +"t11 = 11.0 // Time 11 a.m\n", +"l11 = 13.5 // Load at 11 a.m(kW*1000)\n", +"t113 = 11.50 // Time 11.30 a.m\n", +"l113 = 12.0 // Load at 11.30 am(kW*1000)\n", +"t12 = 12.0 // Time 12 noon\n", +"l12 = 11.0 // Load at 12 noon(kW*1000)\n", +"t123 = 12.50 // Time 12.30 noon\n", +"l123 = 5.0 // Load at 12.30 noon(kW*1000)\n", +"t13 = 13.0 // Time 1 p.m\n", +"l13 = 12.5 // Load at 1 p.m(kW*1000)\n", +"t133 = 13.50 // Time 1.30 p.m\n", +"l133 = 13.5 // Load at 1.30 p.m(kW*1000)\n", +"t14 = 14.0 // Time 2 p.m\n", +"l14 = 14.0 // Load at 2 p.m(kW*1000)\n", +"t15 = 15.0 // Time 3 p.m\n", +"l15 = 14.0 // Load at 3 p.m(kW*1000)\n", +"t16 = 16.0 // Time 4 p.m\n", +"l16 = 15.0 // Load at 4 p.m(kW*1000)\n", +"t163 = 16.50 // Time 4.30 p.m\n", +"l163 = 18.0 // Load at 4.30 p.m(kW*1000)\n", +"t17 = 17.0 // Time 5 p.m\n", +"l17 = 20.0 // Load at 5 p.m(kW*1000)\n", +"t173 = 17.50 // Time 5.30 p.m\n", +"l173 = 17.0 // Load at 5.30 p.m(kW*1000)\n", +"t18 = 18.0 // Time 6 p.m\n", +"l18 = 12.5 // Load at 6 p.m(kW*1000)\n", +"t19 = 19.0 // Time 7 p.m\n", +"l19 = 10.0 // Load at 7 p.m(kW*1000)\n", +"t20 = 20.0 // Time 8 p.m\n", +"l20 = 7.5 // Load at 8 p.m(kW*1000)\n", +"t21 = 21.0 // Time 9 p.m\n", +"l21 = 5.0 // Load at 9 p.m(kW*1000)\n", +"t22 = 22.0 // Time 10 p.m\n", +"l22 = 5.0 // Load at 10 p.m(kW*1000)\n", +"t23 = 23.0 // Time 11 p.m\n", +"l23 = 4.0 // Load at 11 p.m(kW*1000)\n", +"t24 = 24.0 // Time 12 morning\n", +"l24 = 4.0 // Load at 12 morning(kW*1000)\n", +"\n", +"// Calculations\n", +"t = [t0,t1,t2,t3,t4,t5,t6,t7,t8,t9,t10,t11,t12,t13,t14,t15,t16,t17,t18,t19,t20,t21,t22,t23,t24]\n", +"l = [l0,l1,l2,l3,l4,l5,l6,l7,l8,l9,l10,l11,l12,l13,l14,l15,l16,l17,l18,l19,l20,l21,l22,l23,l24]\n", +"a = gca() ;\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot(t,l,'ro-') // Plot of Chronological load curve\n", +"T = [0,0.5,1,1.5,2.5,4.5,6,7,9,9.5,10,11,12,13,15.5,18.5,20.5,23.5,24] // Solved time\n", +"L = [20,18,17,15,14.5,14,13.5,13,12.5,12,11,10,7.5,6,5,4,3.5,3,3] // Solved load\n", +"plot(T,L,'--mo') // Plot of load duration curve\n", +"a.x_label.text = 'Time & No. of hours' // labels x-axis\n", +"a.y_label.text = 'Load in 10^3 kW' // labels y-axis\n", +"xtitle('Fig E7.2 . Plot of Chronological load curve and load duration curve') \n", +"xset('thickness',2) // sets thickness of axes\n", +"xstring(17.5,17,'Chronological load curve')\n", +"xstring(1.1,17,'Load duration curve')\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.21 : SOLUTION :-')\n", +"printf('\nThe chronological load curve and the load duration curve is shown in the Figure E7.2\n')\n", +"printf('\nNOTE: The time is plotted in 24 hours format')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.22: EX7_22.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.22 :\n", +"// Page number 82\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 20.0*10**3 // Maximum demand(kW)\n", +"LF = 0.6 // Load factor\n", +"CF = 0.48 // Plant capacity factor\n", +"UF = 0.8 // Plant use factor\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"avg_demand = LF*MD // Average demand(kW)\n", +"ene_daily = avg_demand*24.0 // Daily energy produced(kWh)\n", +"// Case(b)\n", +"cap_installed = avg_demand/CF // Installed capacity(kW)\n", +"cap_reserve = cap_installed-MD // Reserve capacity(kW)\n", +"// Case(c)\n", +"max_ene_C = cap_installed*24.0 // Maximum energy that could be produced daily(kWh)\n", +"// Case(d)\n", +"max_ene_D = ene_daily/UF // Maximum energy that could be produced daily as per schedule(kWh)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.22 : SOLUTION :-')\n", +"printf('\nCase(a): Daily energy produced = %.f kWh', ene_daily)\n", +"printf('\nCase(b): Reserve capacity of plant = %.f kW', cap_reserve)\n", +"printf('\nCase(c): Maximum energy that could be produced daily when plant runs at all time = %.f kWh', max_ene_C)\n", +"printf('\nCase(d): Maximum energy that could be produced daily when plant runs fully loaded = %.f kWh', max_ene_D)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.23: EX7_23.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.23 :\n", +"// Page number 83-84\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_3sets = 600.0 // Capacity of 3 generators(kW)\n", +"no_3 = 3.0 // Number of sets of 600 kW\n", +"cap_4thset = 400.0 // Capacity of 4th generator set(kW)\n", +"no_4 = 1.0 // Number of sets of 400 kW\n", +"MD = 1600.0 // Maximum demand(kW)\n", +"LF = 0.45 // Load factor\n", +"cost_capital_kW = 1000.0 // Capital cost per kW installed capacity(Rs)\n", +"cost_annual_per = 0.15 // Annual cost = 15% of capital cost\n", +"cost_operation = 60000.0 // Annual operation cost(Rs)\n", +"cost_maintenance = 30000.0 // Annual maintenance cost(Rs)\n", +"fixed_maintenance = 1.0/3 // Fixed cost\n", +"variable_maintenance = 2.0/3 // Variable cost\n", +"cost_fuel_kg = 40.0/100 // Cost of fuel oil(Rs/kg)\n", +"cost_oil_kg = 1.25 // Cost of lubricating oil(Rs/kg)\n", +"calorific = 10000.0 // Calorific value of fuel(kcal/kg)\n", +"oil_consum = 1.0/400 // Consumption of lubricating oil. 1kg for every 400kWh generated\n", +"fuel_consum = 1.0/2 // Consumption of fuel. 1kg for every 2kWh generated\n", +"n_gen = 0.92 // Generator efficiency\n", +"heat_lost = 1.0/3 // Heat lost in the fuel to cooling water\n", +"theta = 11.0 // Difference of temperature between inlet and outlet(°C)\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"rating_3set_A = cap_3sets/n_gen // Rating of first 3 sets(kW)\n", +"rating_4th_A = cap_4thset/n_gen // Rating of 4th set(kW)\n", +"// Case(b)\n", +"avg_demand_B = LF*MD // Average demand(kW)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"energy_B = avg_demand_B*hours_year // Annual energy produced(kWh)\n", +"// Case(c)\n", +"total_invest = (no_3*cap_3sets+cap_4thset*no_4)*cost_capital_kW // Total investment(Rs)\n", +"annual_cost = cost_annual_per*total_invest // Annual cost(Rs)\n", +"maintenance_cost = fixed_maintenance*cost_maintenance // Maintenance cost(Rs)\n", +"fixed_cost_total = annual_cost+maintenance_cost // Total fixed cost per annum(Rs)\n", +"fuel_consumption = energy_B*fuel_consum // Fuel consumption(Kg)\n", +"cost_fuel = fuel_consumption*cost_fuel_kg // Cost of fuel(Rs)\n", +"oil_consumption = energy_B*oil_consum // Lubrication oil consumption(Kg)\n", +"cost_oil = oil_consumption*cost_oil_kg // Cost of Lubrication oil(Rs)\n", +"var_maintenance_cost = variable_maintenance*cost_maintenance // Variable part of maintenance cost(Rs)\n", +"variable_cost_total = cost_fuel+cost_oil+var_maintenance_cost+cost_operation // Total variable cost per annum(Rs)\n", +"cost_total_D = fixed_cost_total+variable_cost_total // Total cost per annum(Rs)\n", +"cost_kWh_gen = cost_total_D/energy_B*100 // Cost per kWh generated(Paise)\n", +"// Case(c)\n", +"n_overall = energy_B*860/(fuel_consumption*calorific)*100 // Overall efficiency(%)\n", +"// Case(d)\n", +"weight_water_hr = heat_lost*fuel_consumption/(hours_year*theta)*calorific // Weight of cooling water required(kg/hr)\n", +"weight_water_min = weight_water_hr/60.0 // Weight of cooling water required(kg/min)\n", +"capacity_pump = weight_water_min*MD/avg_demand_B // Capacity of cooling water pump(kg/min)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.23 : SOLUTION :-')\n", +"printf('\nCase(a): Rating of first 3 sets of diesel engine = %.f kW', rating_3set_A)\n", +"printf('\n Rating of 4th set of diesel engine = %.f kW', rating_4th_A)\n", +"printf('\nCase(b): Annual energy produced = %.1e kWh', energy_B)\n", +"printf('\nCase(c): Total fixed cost = Rs %.f ', fixed_cost_total)\n", +"printf('\n Total variable cost = Rs %.f ', variable_cost_total)\n", +"printf('\n Cost per kWh generated = %.f paise', cost_kWh_gen)\n", +"printf('\nCase(d): Overall efficiency of the diesel plant = %.1f percent', n_overall)\n", +"printf('\nCase(e): Quantity of cooling water required per round = %.2e kg/hr = %.f kg/min', weight_water_hr,weight_water_min)\n", +"printf('\n Capacity of cooling-water pumps under maximum load = %.f kg/min \n', capacity_pump)\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.24: EX7_24.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.24 :\n", +"// Page number 84\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_installed = 30.0*10**3 // Rating of each generators(kW)\n", +"no = 4.0 // Number of installed generators\n", +"MD = 100.0*10**3 // Maximum demand(kW)\n", +"LF = 0.8 // Load factor\n", +"cost_capital_kW = 800.0 // Capital cost per kW installed capacity(Rs)\n", +"depreciation_per = 0.125 // Depreciation,etc = 12.5% of capital cost\n", +"cost_operation = 1.2*10**6 // Annual operation cost(Rs)\n", +"cost_maintenance = 600000.0 // Annual maintenance cost(Rs)\n", +"fixed_maintenance = 1.0/3 // Fixed cost\n", +"variable_maintenance = 2.0/3 // Variable cost\n", +"cost_miscellaneous = 100000.0 // Miscellaneous cost(Rs)\n", +"cost_fuel_kg = 32.0/1000 // Cost of fuel oil(Rs/kg)\n", +"calorific = 6400.0 // Calorific value of fuel(kcal/kg)\n", +"n_gen = 0.96 // Generator efficiency\n", +"n_thermal = 0.28 // Thermal efficiency of turbine\n", +"n_boiler = 0.75 // Boiler efficiency\n", +"n_overall = 0.2 // Overall thermal efficiency\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"rating_turbine = cap_installed/(n_gen*0.736) // Rating of each steam turbine(metric hp)\n", +"// Case(b)\n", +"avg_demand_B = LF*MD // Average demand(kW)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"energy_B = avg_demand_B*hours_year // Annual energy produced(kWh)\n", +"// Case(c)\n", +"steam_consumption_C = (0.8+3.5*LF)/LF // Average steam consumption(kg/kWh)\n", +"// Case(d)\n", +"LF_D = 1.0 // Assumption that Load factor for boiler\n", +"steam_consumption_D = (0.8+3.5*LF_D)/LF_D // Steam consumption(kg/kWh)\n", +"energy_D = cap_installed*1.0 // Energy output per hour per set(kWh)\n", +"evaporation_cap = steam_consumption_D*energy_D // Evaporation capacity of boiler(kg/hr)\n", +"// Case(e)\n", +"total_invest = no*cap_installed*cost_capital_kW // Total investment(Rs)\n", +"capital_cost = depreciation_per*total_invest // Capital cost(Rs)\n", +"maintenance_cost = fixed_maintenance*cost_maintenance // Maintenance cost(Rs)\n", +"fixed_cost_total = capital_cost+maintenance_cost // Total fixed cost per annum(Rs)\n", +"var_maintenance_cost = variable_maintenance*cost_maintenance // Variable part of maintenance cost(Rs)\n", +"input_E = energy_B/n_overall // Input into system per annum(kWh)\n", +"weight_fuel = input_E*860/calorific // Weight of fuel(kg)\n", +"cost_fuel = weight_fuel*cost_fuel_kg // Cost of fuel(Rs)\n", +"variable_cost_total = cost_operation+var_maintenance_cost+cost_miscellaneous+cost_fuel // Total variable cost per annum(Rs)\n", +"cost_total_E = fixed_cost_total+variable_cost_total // Total cost per annum(Rs)\n", +"cost_kWh_gen = cost_total_E/energy_B*100 // Cost per kWh generated(Paise)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.24 : SOLUTION :-')\n", +"printf('\nCase(a): Rating of each steam turbine = %.f metric hp', rating_turbine)\n", +"printf('\nCase(b): Energy produced per annum = %.3e kWh', energy_B)\n", +"printf('\nCase(c): Average steam consumption per kWh = %.1f kg/kWh', steam_consumption_C)\n", +"printf('\nCase(d): Evaporation capacity of boiler = %.f kg/hr', evaporation_cap)\n", +"printf('\nCase(e): Total fixed cost = Rs %.2e ', fixed_cost_total)\n", +"printf('\n Total variable cost = Rs %.2e ', variable_cost_total)\n", +"printf('\n Cost per kWh generated = %.2f paise\n', cost_kWh_gen)\n", +"printf('\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.25: Plot_of_hydrograph_and_Average_discharge_available.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.25 :\n", +"// Page number 85\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"w1 = 1.0 // Week 1\n", +"Q1 = 200.0 // Discharge during week 1(m^2/sec)\n", +"w2 = 2.0 // Week 2\n", +"Q2 = 300.0 // Discharge during week 2(m^2/sec)\n", +"w3 = 3.0 // Week 3\n", +"Q3 = 1100.0 // Discharge during week 3(m^2/sec)\n", +"w4 = 4.0 // Week 4\n", +"Q4 = 700.0 // Discharge during week 4(m^2/sec)\n", +"w5 = 5.0 // Week 5\n", +"Q5 = 900.0 // Discharge during week 5(m^2/sec)\n", +"w6 = 6.0 // Week 6\n", +"Q6 = 800.0 // Discharge during week 6(m^2/sec)\n", +"w7 = 7.0 // Week 7\n", +"Q7 = 600.0 // Discharge during week 7(m^2/sec)\n", +"w8 = 8.0 // Week 8\n", +"Q8 = 1000.0 // Discharge during week 8(m^2/sec)\n", +"w9 = 9.0 // Week 9\n", +"Q9 = 500.0 // Discharge during week 9(m^2/sec)\n", +"w10 = 10.0 // Week 10\n", +"Q10 = 400.0 // Discharge during week 10(m^2/sec)\n", +"w11 = 11.0 // Week 11\n", +"Q11 = 500.0 // Discharge during week 11(m^2/sec)\n", +"w12 = 12.0 // Week 12\n", +"Q12 = 700.0 // Discharge during week 12(m^2/sec)\n", +"w13 = 13.0 // Week 13\n", +"Q13 = 100.0 // Discharge during week 13(m^2/sec)\n", +"no_week = 13.0 // Total weeks of discharge\n", +"\n", +"// Calculations\n", +"Q_average = (Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13)/no_week // Average weekly discharge(m^3/sec)\n", +"// Hydrograph\n", +"W = [0,w1,w1,w2,w2,w3,w3,w4,w4,w5,w5,w6,w6,w7,w7,w8,w8,w9,w9,w10,w10,w11,w11,w12,w12,w13,w13,w13]\n", +"Q = [200,Q1,Q2,Q2,Q3,Q3,Q4,Q4,Q5,Q5,Q6,Q6,Q7,Q7,Q8,Q8,Q9,Q9,Q10,Q10,Q11,Q11,Q12,Q12,Q13,Q13,Q13,0]\n", +"a = gca()\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot(W,Q) // Plotting hydrograph\n", +"q = Q_average\n", +"w = [0,w1,w2,w3,w4,w5,w6,w7,w8,w9,w10,w11,w12,w13,14]\n", +"q_dash = [q,q,q,q,q,q,q,q,q,q,q,q,q,q,q] // Plotting average weekly discharge\n", +"plot(w,q_dash,'r--')\n", +"a.x_label.text = 'Time(week)' // labels x-axis\n", +"a.y_label.text = 'Q(m^3/sec)' // labels y-axis\n", +"xtitle('Fig E7.4 . Plot of Hydrograph') \n", +"xset('thickness',2) // sets thickness of axes\n", +"xstring(13,560,'Q_av')\n", +"xstring(12.02,110,'Q_min')\n", +"xstring(2.02,1110,'Q_max')\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.25 : SOLUTION :-')\n", +"printf('\nThe hydrograph is shown in the Figure E7.4')\n", +"printf('\nAverage discharge available for the whole period = %.f m^3/sec', Q_average)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.26: EX7_26.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.26 :\n", +"// Page number 85-86\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"Q1 = 1100.0 // Discharge in descending order(m^3/sec)\n", +"Q2 = 1000.0 // Discharge(m^3/sec)\n", +"Q3 = 900.0 // Discharge(m^3/sec)\n", +"Q4 = 800.0 // Discharge(m^3/sec)\n", +"Q5 = 700.0 // Discharge(m^3/sec)\n", +"Q6 = 600.0 // Discharge(m^3/sec)\n", +"Q7 = 500.0 // Discharge(m^3/sec)\n", +"Q8 = 400.0 // Discharge(m^3/sec)\n", +"Q9 = 300.0 // Discharge(m^3/sec)\n", +"Q10 = 200.0 // Discharge(m^3/sec)\n", +"Q11 = 100.0 // Discharge(m^3/sec)\n", +"no_week = 13.0 // Total weeks of discharge\n", +"h = 200.0 // Head of installation(m)\n", +"n_overall = 0.88 // Overall efficiency of turbine and generator\n", +"w = 1000.0 // Density of water(kg/m^3)\n", +"\n", +"// Calculations\n", +"n1 = 1.0 // Number of weeks for 1100 discharge(m^3/sec)\n", +"n2 = 2.0 // Number of weeks for 1000 and above discharge(m^3/sec)\n", +"n3 = 3.0 // Number of weeks for 900 and above discharge(m^3/sec)\n", +"n4 = 4.0 // Number of weeks for 800 and above discharge(m^3/sec)\n", +"n5 = 6.0 // Number of weeks for 700 and above discharge(m^3/sec)\n", +"n6 = 7.0 // Number of weeks for 600 and above discharge(m^3/sec)\n", +"n7 = 9.0 // Number of weeks for 500 and above discharge(m^3/sec)\n", +"n8 = 10.0 // Number of weeks for 400 and above discharge(m^3/sec)\n", +"n9 = 11.0 // Number of weeks for 300 and above discharge(m^3/sec)\n", +"n10 = 12.0 // Number of weeks for 200 and above discharge(m^3/sec)\n", +"n11 = 13.0 // Number of weeks for 100 and above discharge(m^3/sec)\n", +"P1 = n1/no_week*100 // Percentage of total period for n1\n", +"P2 = n2/no_week*100 // Percentage of total period for n2\n", +"P3 = n3/no_week*100 // Percentage of total period for n3\n", +"P4 = n4/no_week*100 // Percentage of total period for n4\n", +"P5 = n5/no_week*100 // Percentage of total period for n5\n", +"P6 = n6/no_week*100 // Percentage of total period for n6\n", +"P7 = n7/no_week*100 // Percentage of total period for n7\n", +"P8 = n8/no_week*100 // Percentage of total period for n8\n", +"P9 = n9/no_week*100 // Percentage of total period for n9\n", +"P10 = n10/no_week*100 // Percentage of total period for n10\n", +"P11 = n11/no_week*100 // Percentage of total period for n11\n", +"P = [0,P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11]\n", +"Q = [Q1,Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10,Q11] // Plotting flow duration curve\n", +"a = gca() ;\n", +"a.thickness = 2 // sets thickness of plot\n", +"plot(P,Q,'ro-')\n", +"a.x_label.text = 'Percentage of time' // labels x-axis\n", +"a.y_label.text = 'Q(m^3/sec)' // labels y-axis\n", +"xtitle('Fig E7.5 . Plot of Flow-duration curve') \n", +"xset('thickness',2) // sets thickness of axes\n", +"xgrid(4)\n", +"Q_1 = 1.0 // Discharge(m^3/sec)\n", +"P_1 = 0.736/75*w*Q_1*h*n_overall // Power developed for Q_1(kW)\n", +"Q_av = 600.0 // Average discharge(m^3/sec). Obtained from Example 1.7.25\n", +"P_av = P_1*Q_av/1000.0 // Average power developed(MW)\n", +"Q_max = Q1 // Maximum discharge(m^3/sec)\n", +"P_max = P_1*Q_max/1000.0 // Maximum power developed(MW)\n", +"Q_10 = 1070.0 // Discharge for 10% of time(m^3/sec). Value is obtained from graph\n", +"P_10 = P_1*Q_10/1000.0 // Installed capacity(MW)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.26 : SOLUTION :-')\n", +"printf('\nFlow-duration curve is shown in the Figure E7.5')\n", +"printf('\nMaximum power developed = %.f MW', P_max)\n", +"printf('\nAverage power developed = %.f MW', P_av)\n", +"printf('\nCapacity of proposed station = %.f MW \n', P_10)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook solution')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2: Total_energy_generated_annually.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.2 :\n", +"// Page number 73\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"maximum_demand = 480.0*10**3 // Maximum demand(kW)\n", +"LF = 0.4 // Annual load factor\n", +"\n", +"// Calculation\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"energy_gen = maximum_demand*LF*hours_year // Total energy generated annually(kWh)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.2 : SOLUTION :-')\n", +"printf('\nTotal energy generated annually = %.5e kWh', energy_gen)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3: Annual_load_factors_and_Capacity_factors_of_two_power_stations.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.3 :\n", +"// Page number 73\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"cap_baseload = 400.0*10**3 // Installed capacity of base load plant(kW)\n", +"cap_standby = 50.0*10**3 // Installed capacity of standby unit(kW)\n", +"output_baseload = 101.0*10**6 // Annual baseload station output(kWh)\n", +"output_standby = 87.35*10**6 // Annual standby station output(kWh)\n", +"peakload_standby = 120.0*10**3 // Peak load on standby station(kW)\n", +"hours_use = 3000.0 // Hours of standby station use/year(hrs)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"LF_1 = output_standby*100/(peakload_standby*hours_use) // Annual load factor(%)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"CF_1 = output_standby*100/(cap_standby*hours_year) // Annual capacity factor(%)\n", +"// Case(ii)\n", +"peakload_baseload = peakload_standby // Peak load on baseload station(kW)\n", +"LF_2 = output_baseload*100/(peakload_baseload*hours_use) // Annual load factor on baseload station(%)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"CF_2 = output_baseload*100/(cap_baseload*hours_year) // Annual capacity factor on baseload station(%)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.3 : SOLUTION :-')\n", +"printf('\nCase(i) : Standby Station')\n", +"printf('\n Annual load factor = %.2f percent', LF_1)\n", +"printf('\n Annual capacity factor = %.2f percent\n', CF_1)\n", +"printf('\nCase(ii): Base load Station')\n", +"printf('\n Annual load factor = %.2f percent', LF_2)\n", +"printf('\n Annual capacity factor = %.2f percent\n', CF_2)\n", +"printf('\nNOTE: Incomplete solution in the textbook') ;" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4: Reserve_capacity_of_plant.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.4 :\n", +"// Page number 74\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 500.0 // Maximum demand(MW)\n", +"LF = 0.5 // Annual load factor\n", +"CF = 0.4 // Annual capacity factor\n", +"\n", +"// Calculations\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"energy_gen = MD*LF*hours_year // Energy generated/annum(MWh)\n", +"plant_cap = energy_gen/(CF*hours_year) // Plant capacity(MW)\n", +"reserve_cap = plant_cap-MD // Reserve capacity of plant(MW)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.4 : SOLUTION :-')\n", +"printf('\nReserve capacity of plant = %.f MW', reserve_cap)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5: Number_of_units_supplied_annually_Diversity_factor_and_Demand_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.5 :\n", +"// Page number 74\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_1 = 150.0 // Load supplied by station(MW)\n", +"load_2 = 120.0 // Load supplied by station(MW)\n", +"load_3 = 85.0 // Load supplied by station(MW)\n", +"load_4 = 60.0 // Load supplied by station(MW)\n", +"load_5 = 5.0 // Load supplied by station(MW)\n", +"MD = 220.0 // Maximum demand(MW)\n", +"LF = 0.48 // Annual load factor\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"units = LF*MD*hours_year // Number of units supplied annually\n", +"// Case(b)\n", +"sum_demand = load_1+load_2+load_3+load_4+load_5 // Sum of maximum demand of individual consumers(MW)\n", +"diversity_factor = sum_demand/MD // Diversity factor\n", +"// Case(c)\n", +"DF = MD/sum_demand // Demand factor\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.5 : SOLUTION :-')\n", +"printf('\nCase(a): Number of units supplied annually = %.2e units', units)\n", +"printf('\nCase(b): Diversity factor = %.3f ', diversity_factor)\n", +"printf('\nCase(c): Demand factor = %.3f = %.1f percent', DF,DF*100)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6: Annual_load_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.6 :\n", +"// Page number 74\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"power_del_1 = 1000.0 // Power delivered by station(MW)\n", +"time_1 = 2.0 // Time for which power is delivered(hours)\n", +"power_del_2 = 500.0 // Power delivered by station(MW)\n", +"time_2 = 6.0 // Time for which power is delivered(hours)\n", +"days_maint = 60.0 // Maintenance days\n", +"max_gen_cap = 1000.0 // Maximum generating capacity(MW)\n", +"\n", +"// Calculations\n", +"energy_sup_day = (power_del_1*time_1)+(power_del_2*time_2) // Energy supplied for each working day(MWh)\n", +"days_total = 365.0 // Total days in a year\n", +"days_op = days_total-days_maint // Operating days of station in a year\n", +"energy_sup_year = energy_sup_day*days_op // Energy supplied per year(MWh)\n", +"hours_day = 24.0 // Total hours in a day\n", +"working_hours = days_op*hours_day // Hour of working in a year\n", +"LF = energy_sup_year*100/(max_gen_cap*working_hours) // Annual load factor(%)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.6 : SOLUTION :-')\n", +"printf('\nAnnual load factor = %.1f percent', LF)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7: Diversity_factor_and_Annual_load_factor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.7 :\n", +"// Page number 74\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_industry = 750.0 // Industrial consumer load supplied by station(MW)\n", +"load_commercial = 350.0 // Commercial establishment load supplied by station(MW)\n", +"load_power = 10.0 // Domestic power load supplied by station(MW)\n", +"load_light = 50.0 // Domestic light load supplied by station(MW)\n", +"MD = 1000.0 // Maximum demand(MW)\n", +"kWh_gen = 50.0*10**5 // Number of kWh generated per year\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"sum_demand = load_industry+load_commercial+load_power+load_light // Sum of max demand of individual consumers(MW)\n", +"diversity_factor = sum_demand/MD // Diversity factor\n", +"// Case(ii)\n", +"hours_year = 365.0*24 // Total hours in a year\n", +"average_demand = kWh_gen/hours_year // Average demand(MW)\n", +"LF = average_demand/MD*100 // Load factor(%)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.7 : SOLUTION :-')\n", +"printf('\nCase(i) : Diversity factor = %.2f ', diversity_factor)\n", +"printf('\nCase(ii): Annual load factor = %.f percent', LF)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8: Maximum_demand_and_Connected_load_of_each_type.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.8 :\n", +"// Page number 74-75\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"load_domestic = 15000.0 // Domestic load supplied by station(kW)\n", +"diversity_domestic = 1.25 // Diversity factor of domestic load\n", +"DF_domestic = 0.7 // Demand factor of domestic load\n", +"load_commercial = 25000.0 // Commercial load supplied by station(kW)\n", +"diversity_commercial = 1.2 // Diversity factor of commercial load\n", +"DF_commercial = 0.9 // Demand factor of commercial load\n", +"load_industry = 50000.0 // Industrial load supplied by station(kW)\n", +"diversity_industry = 1.3 // Diversity factor of industrial load\n", +"DF_industry = 0.98 // Demand factor of industrial load\n", +"diversity_factor = 1.5 // Overall system diversity factor\n", +"\n", +"// Calculations\n", +"// Case(a)\n", +"sum_demand = load_domestic+load_commercial+load_industry // Sum of max demand of individual consumers(MW)\n", +"MD = sum_demand/diversity_factor // Maximum demand\n", +"// Case(b)\n", +"MD_domestic = load_domestic*diversity_domestic // Maximum domestic load demand(kW)\n", +"connected_domestic = MD_domestic/DF_domestic // Connected domestic load(kW)\n", +"MD_commercial = load_commercial*diversity_commercial // Maximum commercial load demand(kW)\n", +"connected_commercial = MD_commercial/DF_commercial // Connected commercial load(kW)\n", +"MD_industry = load_industry*diversity_industry // Maximum industrial load demand(kW)\n", +"connected_industry = MD_industry/DF_industry // Connected industrial load(kW)\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.8 : SOLUTION :-')\n", +"printf('\nCase(a): Maximum demand = %.f kW', MD)\n", +"printf('\nCase(b): Connected domestic load = %.1f kW', connected_domestic)\n", +"printf('\n Connected commercial load = %.1f kW', connected_commercial)\n", +"printf('\n Connected industrial load = %.1f kW', connected_industry)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9: EX7_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART I : GENERATION\n", +"// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION\n", +"\n", +"// EXAMPLE : 7.9 :\n", +"// Page number 75-76\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"MD = 10000.0 // Maximum demand(kW)\n", +"load_1 = 2000.0 // Load from 11 PM-6 AM(kW)\n", +"t_1 = 7.0 // Time from 11 PM-6 AM(hour)\n", +"load_2 = 3500.0 // Load from 6 AM-8 AM(kW)\n", +"t_2 = 2.0 // Time from 6 AM-8 AM(hour)\n", +"load_3 = 8000.0 // Load from 8 AM-12 Noon(kW)\n", +"t_3 = 4.0 // Time from 8 AM-12 Noon(hour)\n", +"load_4 = 3000.0 // Load from 12 Noon-1 PM(kW)\n", +"t_4 = 1.0 // Time from 12 Noon-1 PM(hour)\n", +"load_5 = 7500.0 // Load from 1 PM-5 PM(kW)\n", +"t_5 = 4.0 // Time from 1 PM-5 PM(hour)\n", +"load_6 = 8500.0 // Load from 5 PM-7 PM(kW)\n", +"t_6 = 2.0 // Time from 5 PM-7 PM(hour)\n", +"load_7 = 10000.0 // Load from 7 PM-9 PM(kW)\n", +"t_7 = 2.0 // Time from 7 PM-9 PM(hour)\n", +"load_8 = 4500.0 // Load from 9 PM-11 PM(kW)\n", +"t_8 = 2.0 // Time from 9 PM-11 PM(hour)\n", +"\n", +"// Calculations\n", +"energy_gen = (load_1*t_1)+(load_2*t_2)+(load_3*t_3)+(load_4*t_4)+(load_5*t_5)+(load_6*t_6)+(load_7*t_7)+(load_8*t_8) // Energy generated during 24 hours(kWh)\n", +"LF = energy_gen/(MD*24.0) // Load factor\n", +"no_units = 3.0 // Number of generating set\n", +"cap_1 = 5000.0 // Capacity of first generating unit(kW)\n", +"cap_2 = 3000.0 // Capacity of second generating unit(kW)\n", +"cap_3 = 2000.0 // Capacity of third generating unit(kW)\n", +"cap_reserve = cap_1 // Reserve capacity(kW) i.e largest size of generating unit\n", +"cap_installed = cap_1+cap_2+cap_3+cap_reserve // Installed capacity(kW)\n", +"cap_factor = energy_gen/(cap_installed*24.0) // Plant capacity factor\n", +"cap_plant = cap_3*t_1+(cap_3+cap_2)*t_2+(cap_2+cap_1)*t_3+cap_2*t_4+(cap_2+cap_1)*t_5+(cap_3+cap_2+cap_1)*t_6+(cap_3+cap_2+cap_1)*t_7+cap_1*t_8 // Capacity of plant running actually(kWh)\n", +"use_factor = energy_gen/cap_plant // Plant use factor\n", +"\n", +"// Results\n", +"disp('PART I - EXAMPLE : 7.9 : SOLUTION :-')\n", +"printf('\nNumber of generator units = %.f', no_units)\n", +"printf('\nSize of generator units required are %.f kW, %.f kW and %.f kW', cap_1,cap_2,cap_3)\n", +"printf('\nReserve plant capacity = %.f kW', cap_reserve)\n", +"printf('\nLoad factor = %.2f = %.f percent', LF,LF*100)\n", +"printf('\nPlant capacity factor = %.4f = %.2f percent', cap_factor,cap_factor*100)\n", +"printf('\nPlant use factor = %.3f = %.1f percent', use_factor,use_factor*100)\n", +"printf('\n\nNOTE: Capacity of plant is directly taken & operating schedule is not displayed here')\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/Power_System_Engineering_by_S_Chakraborthy/9-CONSTANTS_OF_OVERHEAD_TRANSMISSION_LINES.ipynb b/Power_System_Engineering_by_S_Chakraborthy/9-CONSTANTS_OF_OVERHEAD_TRANSMISSION_LINES.ipynb new file mode 100644 index 0000000..168a024 --- /dev/null +++ b/Power_System_Engineering_by_S_Chakraborthy/9-CONSTANTS_OF_OVERHEAD_TRANSMISSION_LINES.ipynb @@ -0,0 +1,1331 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: CONSTANTS OF OVERHEAD TRANSMISSION LINES" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.10: Inductance_per_km_of_3_phase_transmission_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.10 :\n", +"// Page number 109\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 5.0 // Diameter of conductor(cm)\n", +"d_1 = 400.0 // Distance between conductor 1 & 2(cm)\n", +"d_2 = 500.0 // Distance between conductor 2 & 3(cm)\n", +"d_3 = 600.0 // Distance between conductor 1 & 3(cm)\n", +"\n", +"// Calculations\n", +"D_eq = (d_1*d_2*d_3)**(1.0/3) // Equivalent distance(cm)\n", +"r_GMR = 0.7788*d/2.0 // GMR(cm)\n", +"L = 0.2*log(D_eq/r_GMR) // Inductance per phase per km(mH)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.10 : SOLUTION :-')\n", +"printf('\nInductance per km of 3 phase transmission line, L = %.3f mH \n', L)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.11: Inductance_of_each_conductor_per_phase_per_km.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.11 :\n", +"// Page number 109\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 3.0 // Diameter of conductor(cm)\n", +"D_12 = 200.0 // Distance between conductor 1 & 2(cm)\n", +"D_23 = 200.0 // Distance between conductor 2 & 3(cm)\n", +"D_31 = 400.0 // Distance between conductor 1 & 3(cm)\n", +"\n", +"// Calculations\n", +"D_eq = (D_12*D_23*D_31)**(1.0/3) // Equivalent distance(cm)\n", +"r = d/2.0 // Radius of conductor(cm)\n", +"L = (0.5+2*log(D_eq/r))*10**-7 // Inductance/phase/m(H)\n", +"L_mH = L*1000.0*1000.0 // Inductance per phase per km(mH)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.11 : SOLUTION :-')\n", +"printf('\nInductance of each conductor per phase per km, L = %.3f mH \n', L_mH)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.12: Inductance_of_each_conductor_and_Average_inductance_of_each_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.12 :\n", +"// Page number 109-110\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.0 // Diameter of conductor(cm)\n", +"D_ab = 400.0 // Distance between conductor a & b(cm)\n", +"D_bc = 400.0 // Distance between conductor b & c(cm)\n", +"D_ca = 800.0 // Distance between conductor c & a(cm)\n", +"\n", +"// Calculations\n", +"I_ab = 1.0*exp(%i*-240.0*%pi/180) // I_a/I_b\n", +"I_cb = 1.0*exp(%i*-120.0*%pi/180) // I_c/I_b\n", +"r_GMR = 0.7788*d/2.0 // GMR(cm)\n", +"L_a = 2.0*10**-7*complex(log((D_ab*D_ca)**0.5/r_GMR),(3**0.5/2*log(D_ab/D_ca))) // Inductance per phase of A(H/m)\n", +"L_amH = L_a*10.0**6 // Inductance per phase of A(mH/km)\n", +"L_b = 2.0*10**-7*complex(log((D_bc*D_ab)**0.5/r_GMR),(3**0.5/2*log(D_bc/D_ab))) // Inductance per phase of B(H/m)\n", +"L_bmH = L_b*10.0**6 // Inductance per phase of B(mH/km)\n", +"L_c = 2.0*10**-7*complex(log((D_ca*D_bc)**0.5/r_GMR),(3**0.5/2*log(D_ca/D_bc))) // Inductance per phase of C(H/m)\n", +"L_cmH = L_c*10.0**6 // Inductance per phase of C(mH/km)\n", +"D_eq = (D_ab*D_bc*D_ca)**(1.0/3) // Equivalent distance(cm)\n", +"L_avg = 0.2*log(D_eq/r_GMR) // Average inductance per phase(mH/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.12 : SOLUTION :-')\n", +"printf('\nInductance of conductor a, L_a = (%.4f%.2fj) mH/km', real(L_amH),imag(L_amH))\n", +"printf('\nInductance of conductor b, L_b = %.3f mH/km', abs(L_bmH))\n", +"printf('\nInductance of conductor c, L_c = (%.4f+%.2fj) mH/km', real(L_cmH),imag(L_cmH))\n", +"printf('\nAverage inductance of each phase, L_avg = %.3f mH/km', L_avg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.13: Inductance_per_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.13 :\n", +"// Page number 110\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D_a_a = 0.9 // Self GMD of conductor a(cm)\n", +"D_a_aa = 40.0 // Distance between conductor a & a'(cm)\n", +"D_a_b = 1000.0 // Distance between conductor a & b(cm)\n", +"D_a_bb = 1040.0 // Distance between conductor a & b'(cm)\n", +"D_aa_b = 960.0 // Distance between conductor a' & b(cm)\n", +"D_c_a = 2000.0 // Distance between conductor a & c(cm)\n", +"D_c_aa = 1960.0 // Distance between conductor a' & c(cm)\n", +"D_cc_a = 2040.0 // Distance between conductor a & c'(cm)\n", +"\n", +"// Calculations\n", +"D_aa_aa = D_a_a // Self GMD of conductor a'(cm)\n", +"D_aa_a = D_a_aa // Distance between conductor a' & a(cm)\n", +"D_s1 = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD in position 1(cm)\n", +"D_s2 = D_s1 // Self GMD in position 2(cm)\n", +"D_s3 = D_s1 // Self GMD in position 3(cm)\n", +"D_s = (D_s1*D_s2*D_s3)**(1.0/3) // Equivalent self GMD(cm)\n", +"D_aa_bb = D_a_b // Distance between conductor a' & b'(cm)\n", +"D_AB = (D_a_b*D_a_bb*D_aa_b*D_aa_bb)**(1.0/4) // Mutual GMD(cm)\n", +"D_BC = D_AB // Mutual GMD(cm)\n", +"D_cc_aa = D_c_a // Distance between conductor a' & c'(cm)\n", +"D_CA = (D_c_a*D_c_aa*D_cc_a*D_cc_aa)**(1.0/4) // Mutual GMD(cm)\n", +"D_m = (D_AB*D_BC*D_CA)**(1.0/3) // Equivalent Mutual GMD(cm)\n", +"L = 0.2*log(D_m/D_s) // Inductance per phase(mH/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.13 : SOLUTION :-')\n", +"printf('\nInductance per phase, L = %.3f mH/km', L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.14: Inductance_per_phase_of_double_circuit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.14 :\n", +"// Page number 110-111\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"r = 6.0/1000 // Radius of conductor(m)\n", +"D_a_cc = 5.0 // Distance between conductor a & c'(m)\n", +"D_b_bb = 6.0 // Distance between conductor b & b'(m)\n", +"D_c_aa = 5.0 // Distance between conductor c & a'(m)\n", +"D_acc_bbb = 3.0 // Distance between conductor ac' & bb'(m)\n", +"D_bbb_caa = 3.0 // Distance between conductor bb' & ca'(m)\n", +"D_a_c = 6.0 // Distance between conductor a & c(m)\n", +"\n", +"// Calculations\n", +"r_GMR = 0.7788*r // GMR of conductor(m)\n", +"D_a_b = (D_acc_bbb**2+((D_b_bb-D_a_cc)/2)**2)**(1.0/2) // Distance between conductor a & b(m)\n", +"D_a_bb = (D_acc_bbb**2+(D_a_cc+(D_b_bb-D_a_cc)/2)**2)**(1.0/2) // Distance between conductor a & b'(m)\n", +"D_a_aa = ((D_acc_bbb+D_bbb_caa)**2+D_c_aa**2)**(1.0/2) // Distance between conductor a & a'(m)\n", +"D_a_a = r_GMR // Self GMD of conductor a(m)\n", +"D_aa_aa = D_a_a // Self GMD of conductor a'(m)\n", +"D_aa_a = D_a_aa // Distance between conductor a' & a(m)\n", +"D_S1 = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD in position 1(m)\n", +"D_bb_b = D_b_bb // Distance between conductor b' & b(m)\n", +"D_S2 = (D_a_a*D_b_bb*D_aa_aa*D_bb_b)**(1.0/4) // Self GMD in position 2(m)\n", +"D_S3 = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD in position 3(m)\n", +"D_S = (D_S1*D_S2*D_S3)**(1.0/3) // Equivalent self GMD(m)\n", +"D_aa_bb = D_a_b // Distance between conductor a' & b'(m)\n", +"D_aa_b = D_a_bb // Distance between conductor a' & b(m)\n", +"D_AB = (D_a_b*D_a_bb*D_aa_b*D_aa_bb)**(1.0/4) // Mutual GMD(m)\n", +"D_BC = D_AB // Mutual GMD(m)\n", +"D_c_a = D_a_c // Distance between conductor c & a(m)\n", +"D_cc_aa = D_c_a // Distance between conductor a' & c'(m)\n", +"D_cc_a = D_a_cc // Distance between conductor c' & a(m)\n", +"D_CA = (D_c_a*D_c_aa*D_cc_a*D_cc_aa)**(1.0/4) // Mutual GMD(m)\n", +"D_m = (D_AB*D_BC*D_CA)**(1.0/3) // Equivalent Mutual GMD(m)\n", +"L = 0.2*log(D_m/D_S) // Inductance per phase(mH/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.14 : SOLUTION :-')\n", +"printf('\nInductance per phase, L = %.2f mH/km', L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.15: Spacing_between_adjacent_conductor_to_keep_same_inductance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.15 :\n", +"// Page number 111\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D_eq = 2.88 // Equilateral spacing of line(m)\n", +"\n", +"// Calculations\n", +"D = D_eq/2**(1.0/3) // Distance(m)\n", +"D_13 = 2.0*D // Distance between conductor 1 & 3(m)\n", +"D_12 = D // Distance between conductor 1 & 2(m)\n", +"D_23 = D // Distance between conductor 2 & 3(m)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.15 : SOLUTION :-')\n", +"printf('\nSpacing between conductor 1 & 2 to keep inductance same, D_12 = %.1f m', D_12)\n", +"printf('\nSpacing between conductor 2 & 3 to keep inductance same, D_23 = %.1f m', D_23)\n", +"printf('\nSpacing between conductor 1 & 3 to keep inductance same, D_13 = %.1f m', D_13)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.16: Capacitance_of_line_neglecting_and_taking_presence_of_ground.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.16 :\n", +"// Page number 112\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 40.0 // Length of line(km)\n", +"d = 5.0/1000 // Diameter of wire(m)\n", +"D = 1.5 // Spacing between conductor(m)\n", +"h = 7.0 // Height of conductors above ground(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of wire(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"// Neglecting presence of ground\n", +"C_ab_1 = %pi*e/(log(D/r)) // Capacitance(F/m)\n", +"C_ab_12 = C_ab_1*l*1000.0*10**6 // Capacitance(μF)\n", +"// Taking presence of ground\n", +"C_ab_2 = %pi*e/log(D/(r*(1+(D/(2*h))**2)**(1.0/2))) // Capacitance(F/m)\n", +"C_ab_22 = C_ab_2*l*1000.0*10**6 // Capacitance(μF)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.16 : SOLUTION :-')\n", +"printf('\nCapacitance of line neglecting presence of ground, C_ab = %.3f μF', C_ab_12)\n", +"printf('\nCapacitance of line taking presence of ground, C_ab = %.3f μF', C_ab_22)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.17: Capacitance_of_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.17 :\n", +"// Page number 114-115\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.0/100 // Diameter of conductor(m)\n", +"D_AB = 4.0 // Spacing between conductor A & B(m)\n", +"D_BC = 4.0 // Spacing between conductor B & C(m)\n", +"D_CA = 8.0 // Spacing between conductor C & A(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"D = 4.0 // Assuming coomon distance(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"C_A = 2*%pi*e/(log(D/r)-complex(-0.5,0.866)*log(2))*1000.0 // Capacitance of conductor A(F/km)\n", +"C_Au = C_A*10.0**6 // Capacitance of conductor A(μF/km)\n", +"C_B = 2*%pi*e/log(D/r)*1000.0 // Capacitance of conductor B(F/km)\n", +"C_Bu = C_B*10.0**6 // Capacitance of conductor B(μF/km)\n", +"C_C = 2*%pi*e/(log(D/r)-complex(-0.5,-0.866)*log(2))*1000.0 // Capacitance of conductor C(F/km)\n", +"C_Cu = C_C*10.0**6 // Capacitance of conductor C(μF/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.17 : SOLUTION :-')\n", +"printf('\nCapacitance of conductor A, C_A = (%.5f+%.6fj) μF/km', real(C_Au),imag(C_Au))\n", +"printf('\nCapacitance of conductor B, C_B = %.6f μF/km', C_Bu)\n", +"printf('\nCapacitance of conductor C, C_C = (%.5f%.6fj) μF/km', real(C_Cu),imag(C_Cu))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.18: New_value_of_capacitance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.18 :\n", +"// Page number 115\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.0/100 // Diameter of conductor(m)\n", +"D_AB = 4.0 // Spacing between conductor A & B(m)\n", +"D_BC = 4.0 // Spacing between conductor B & C(m)\n", +"D_CA = 8.0 // Spacing between conductor C & A(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"D_eq = (D_AB*D_BC*D_CA)**(1.0/3) // Equivalent distance(m)\n", +"C_n = 2*%pi*e/log(D_eq/r)*1000.0 // Capacitance to neutral(F/km)\n", +"C_nu = C_n*10.0**6 // Capacitance to neutral(μF/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.18 : SOLUTION :-')\n", +"printf('\nNew value of capacitance, C_n = %.5f μF/km \n', C_nu)\n", +"printf('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.19: Capacitance_per_phase_to_neutral_of_a_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.19 :\n", +"// Page number 115\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.6 // Outside diameter of conductor(cm)\n", +"D_RY = 8.0 // Spacing between conductor R & Y(m)\n", +"D_YB = 8.0 // Spacing between conductor Y & B(m)\n", +"D_RB = 16.0 // Spacing between conductor R & B(m)\n", +"h = 13.0 // Height of conductor from ground(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"h_12 = (D_RY**2+(2*h)**2)**(1.0/2) // Height of conductor 1 & 2(m)\n", +"h_23 = h_12 // Height of conductor 2 & 3(m)\n", +"h_31 = (D_RB**2+(2*h)**2)**(1.0/2) // Height of conductor 3 & 1(m)\n", +"h_1 = 2*h // Height of transposed conductor 1(m)\n", +"h_2 = 2*h // Height of transposed conductor 2(m)\n", +"h_3 = 2*h // Height of transposed conductor 3(m)\n", +"D_eq = (D_RY*D_YB*D_RB)**(1.0/3) // Equivalent distance(m)\n", +"h_123 = (h_12*h_23*h_31)**(1.0/3) // Height(m)\n", +"h_1_2_3 = (h_1*h_2*h_3)**(1.0/3) // Height(m)\n", +"C_n = 2*%pi*e/(log(D_eq*100/r)-log(h_123/h_1_2_3))*1000.0 // Capacitance of conductor A(F/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.19 : SOLUTION :-')\n", +"printf('\nCapacitance per phase to neutral of a line, C_n = %.1e F/km', C_n)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: Loop_inductance_and_Reactance_of_transmission_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.1 :\n", +"// Page number 100\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D = 100.0 // Distance between conductors(cm)\n", +"d = 1.25 // Diameter of conductor(cm)\n", +"f = 50.0 // Frequency(Hz)\n", +"\n", +"// Calculations\n", +"r_GMR = 0.7788*d/2.0 // GMR of conductor(cm)\n", +"L = 4.0*10**-4*log(D/r_GMR) // Loop inductance(H/km)\n", +"X_L = 2*%pi*f*L // Reactance of transmission line(ohm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.1 : SOLUTION :-')\n", +"printf('\nLoop inductance of transmission line, L = %.2e H/km', L)\n", +"printf('\nReactance of transmission line, X_L = %.2f ohm', X_L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.20: Phase_to_neutral_capacitance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.20 :\n", +"// Page number 117-118\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.5 // Diameter of conductor(cm)\n", +"D = 200.0 // Distance of separation(cm)\n", +"l = 100.0 // Length of line(km)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(cm)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"D_m = (D*(3**0.5)*D*(3**0.5)*D*D)**(1.0/4) // Mutual GMD(cm)\n", +"D_s = (2*D*r)**(1.0/2) // Self GMD(cm)\n", +"C_n = 2*%pi*e/log(D_m/D_s)*1000.0 // Phase-to-neutral capacitance(F/km)\n", +"C_nu = C_n*l*10.0**6 // Phase-to-neutral capacitance(μF)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.20 : SOLUTION :-')\n", +"printf('\nPhase-to-neutral capacitance, C_n = %.2f μF', C_nu)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.21: Capacitance_per_phase_to_neutral.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.21 :\n", +"// Page number 118\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.5/100 // Diameter of conductor(m)\n", +"D = 5.0 // Distance of separation(m)\n", +"h = 2.0 // Height of separation(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"m = (D**2+h**2)**(1.0/2) // (m)\n", +"n = (D**2+(h*2)**2)**(1.0/2) // (m)\n", +"D_ab = (D*m)**(1.0/2) // Distance between conductor a & b(m)\n", +"D_bc = (D*m)**(1.0/2) // Distance between conductor b & c(m)\n", +"D_ca = (2*D*h)**(1.0/2) // Distance between conductor c & a(m)\n", +"D_eq = (D_ab*D_bc*D_ca)**(1.0/3) // Equivalent GMD(m)\n", +"D_s1 = (r*n)**(1.0/2) // Self GMD in position 1(m)\n", +"D_s2 = (r*h)**(1.0/2) // Self GMD in position 2(m)\n", +"D_s3 = (r*n)**(1.0/2) // Self GMD in position 3(m)\n", +"D_s = (D_s1*D_s2*D_s3)**(1.0/3) // Self GMD(m)\n", +"C_n = 2*%pi*e/log(D_eq/D_s)*1000.0 // Capacitance per phase to neutral(F/km)\n", +"C_nu = C_n*10.0**6 // Capacitance per phase to neutral(μF/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.21 : SOLUTION :-')\n", +"printf('\nCapacitance per phase to neutral, C_n = %.2f μF/km', C_nu)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.22: Capacitive_reactance_to_neutral_and_Charging_current_per_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.22 :\n", +"// Page number 119\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.5/100 // Diameter of conductor(m)\n", +"V = 132.0*10**3 // Line voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"h = 4.0 // Height(m)\n", +"H = 8.0 // Height of separation(m)\n", +"D_1_33 = 7.0 // Distance between conductors 1 & 3'(m)\n", +"D_1_22 = 9.0 // Distance between conductors 1 & 2'(m)\n", +"D_1_11 = 8.0 // Distance between conductors 1 & 1'(m)\n", +"D_1 = 1.0 // Distance(m)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 1.0/(36*%pi)*10**-9 // Constant ε_0\n", +"D_12 = (h**2+D_1**2)**(1.0/2) // Distance between conductors 1 & 2(m)\n", +"D_122 = (h**2+D_1_11**2)**(1.0/2) // Distance between conductors 1 & 2'(m)\n", +"D_111 = (D_1_11**2+D_1_33**2)**(1.0/2) // Distance between conductors 1 & 1'(m)\n", +"D_1_2 = (D_12*D_122)**(1.0/2) // Mutual GMD(m)\n", +"D_2_3 = (D_12*D_122)**(1.0/2) // Mutual GMD(m)\n", +"D_3_1 = (D_1_33*D_1_11)**(1.0/2) // Mutual GMD(m)\n", +"D_eq = (D_1_2*D_2_3*D_3_1)**(1.0/3) // Equivalent GMD(m)\n", +"D_s1 = (r*D_111)**(1.0/2) // Self GMD in position 1(m)\n", +"D_s2 = (r*D_1_22)**(1.0/2) // Self GMD in position 2(m)\n", +"D_s3 = (r*D_111)**(1.0/2) // Self GMD in position 3(m)\n", +"D_s = (D_s1*D_s2*D_s3)**(1.0/3) // Self GMD(m)\n", +"C_n = 2*%pi*e/log(D_eq/D_s) // Capacitance per phase to neutral(F/m)\n", +"X_cn = 1/(2.0*%pi*f*C_n) // Capacitive reactance to neutral(ohms/m)\n", +"V_ph = V/(3**0.5) // Phase voltage(V)\n", +"I_charg = V_ph/X_cn*1000.0 // Charging current per phase(A/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.22 : SOLUTION :-')\n", +"printf('\nCapacitive reactance to neutral, X_cn = %.2e ohms/m', X_cn)\n", +"printf('\nCharging current per phase, I_charg = %.3f A/km', I_charg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.23: Inductive_reactance_Capacitance_and_Capacitive_reactance_of_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.23 :\n", +"// Page number 119\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 0.8/100 // Diameter of conductor(m)\n", +"f = 50.0 // Frequency(Hz)\n", +"D_a_b = 5.0 // Distance between conductors a & b(m)\n", +"D_b_c = 5.0 // Distance between conductors b & c(m)\n", +"D_c_a = 8.0 // Distance between conductors c & a(m)\n", +"l = 25.0 // Length of line(km)\n", +"\n", +"// Calculations\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 8.854*10**-12 // Constant ε_0\n", +"D_e = (D_a_b*D_b_c*D_c_a)**(1.0/3) // Equivalent GMD(m)\n", +"L = 2*((1.0/4)+log(D_e/r))*10**-4 // Inductance(H/km)\n", +"X_L = 2*%pi*f*L // Inductive reactance per km(ohms)\n", +"C = %pi*e/log(D_e/r) // Capacitance(F/m)\n", +"C_l = C*1000.0*l // Capacitance for entire length(F)\n", +"C_lu = C_l*10.0**6 // Capacitance for entire length(μF)\n", +"X_c = 1/(2.0*%pi*f*C_l) // Capacitive reactance to neutral(ohm)\n", +"X_ck = X_c/1000.0 // Capacitive reactance to neutral(kilo-ohm)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.23 : SOLUTION :-')\n", +"printf('\nInductive reactance of the line per kilometer per phase, X_L = %.3f ohm', X_L)\n", +"printf('\nCapacitance of the line, C = %.3f μF', C_lu)\n", +"printf('\nCapacitive reactance of the transmission line, X_c = %.1f kilo-ohm\n', X_ck)\n", +"printf('\nNOTE: ERROR: Change in obtained answer from that of textbook due to wrong substitution in finding Capacitance')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.24: Capacitance_of_the_line_and_Charging_current.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.24 :\n", +"// Page number 119-120\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"V = 250.0 // Line voltage(V)\n", +"f = 50.0 // Frequency(Hz)\n", +"D = 1.5 // Distance of separation(m)\n", +"d = 1.5/100 // Diameter of conductor(m)\n", +"l = 50.0 // Length of line(km)\n", +"\n", +"// Calculations\n", +"// Case(i)\n", +"r = d/2 // Radius of conductor(m)\n", +"e = 8.854*10**-12 // Constant ε_0\n", +"C = %pi*e/log(D/r) // Capacitance(F/m)\n", +"C_l = C*1000.0*l // Capacitance for entire length(F)\n", +"C_lu = C_l*10.0**6 // Capacitance for entire length(μF)\n", +"// Case(ii)\n", +"I_charg = 2.0*%pi*f*C_l*V*1000.0 // Charging current(mA)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.24 : SOLUTION :-')\n", +"printf('\nCase(i) : Capacitance of the line, C = %.3f μF', C_lu)\n", +"printf('\nCase(ii): Charging current, I_charg = %.2f mA', I_charg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.25: Capacitance_of_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.25 :\n", +"// Page number 120\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d_1 = 6.0 // Distance between conductor 1 & 2(m)\n", +"d_2 = 6.0 // Distance between conductor 2 & 3(m)\n", +"d_3 = 12.0 // Distance between conductor 3 & 1(m)\n", +"dia = 1.24/100 // Diameter of conductor(m)\n", +"l = 100.0 // Length of line(km)\n", +"\n", +"// Calculations\n", +"r = dia/2 // Radius of conductor(m)\n", +"e = 8.854*10**-12 // Constant ε_0\n", +"d = (d_1*d_2*d_3)**(1.0/3) // Distance(m)\n", +"C = 2*%pi*e/log(d/r) // Capacitance(F/m)\n", +"C_l = C*1000.0*l // Capacitance for entire length(F)\n", +"C_lu = C_l*10.0**6 // Capacitance for entire length(μF)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.25 : SOLUTION :-')\n", +"printf('\nCapacitance of the line, C = %.3f μF', C_lu)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.26: Capacitance_of_each_line_conductor.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.26 :\n", +"// Page number 120\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 2.0 // Spacing between conductors(m)\n", +"dia = 1.25/100 // Diameter of conductor(m)\n", +"\n", +"// Calculations\n", +"r = dia/2 // Radius of conductor(m)\n", +"e = 8.854*10**-12 // Constant ε_0\n", +"C = 2*%pi*e/log(d/r) // Capacitance(F/m)\n", +"C_u = C*1000*10.0**6 // Capacitance for entire length(μF/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.26 : SOLUTION :-')\n", +"printf('\nCapacitance of each line conductor, C = %.4f μF/km', C_u)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: Inductance_per_phase_of_the_system.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.2 :\n", +"// Page number 101\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 100.0 // Length of 3-phase transmission line(km)\n", +"D = 120.0 // Distance between conductors(cm)\n", +"d = 0.5 // Diameter of conductor(cm)\n", +"\n", +"// Calculations\n", +"r_GMR = 0.7788*d/2.0 // GMR of conductor(cm)\n", +"L = 2.0*10**-4*log(D/r_GMR) // Inductance per phase(H/km)\n", +"L_l = L*l // Inductance per phase for 100km length(H)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.2 : SOLUTION :-')\n", +"printf('\nInductance per phase of the system, L = %.4f H \n', L_l)\n", +"printf('\nNOTE: ERROR: In textbook to calculate L, log10 is used instead of ln i.e natural logarithm. So, there is change in answer')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: Loop_inductance_of_line_per_km.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.3 :\n", +"// Page number 101\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D = 135.0 // Spacing between conductors(cm)\n", +"r = 0.8 // Radius of conductor(cm)\n", +"\n", +"// Calculations\n", +"L = (1+4*log(D/r))*10**-7*1000.0 // Loop inductance per km(H)\n", +"L_mH = L*1000.0 // Loop inductance per km(mH)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.3 : SOLUTION :-')\n", +"printf('\nLoop inductance of line per km, L = %.2f mH', L_mH)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: Inductance_per_phase_of_the_system.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.4 :\n", +"// Page number 101\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"l = 80.0 // Length of 3-phase transmission line(km)\n", +"D = 100.0 // Distance between conductors(cm)\n", +"d = 1.0 // Diameter of conductor(cm)\n", +"\n", +"// Calculations\n", +"r_GMR = 0.7788*d/2.0 // GMR of conductor(cm)\n", +"L = 2.0*10**-7*log(D/r_GMR) // Inductance per phase(H/m)\n", +"L_l = L*l*1000.0 // Inductance per phase for 80km(H)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.4 : SOLUTION :-')\n", +"printf('\nInductance per phase of the system, L = %.4f H \n', L_l)\n", +"printf('\nNOTE: ERROR: Calculation mistake in textbook to find Inductance per phase of the system')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: Total_inductance_of_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.5 :\n", +"// Page number 103-104\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D_a_b = 120.0 // Distance between conductors a & b(cm)\n", +"D_a_bb = 140.0 // Distance between conductors a & b'(cm)\n", +"D_aa_b = 100.0 // Distance between conductors a' & b(cm)\n", +"D_aa_bb = 120.0 // Distance between conductors a' & b'(cm)\n", +"D_a_aa = 20.0 // Distance between conductors a & a'(cm)\n", +"d = 2.0 // Diameter of conductor(cm)\n", +"\n", +"// Calculations\n", +"D_m = (D_a_b*D_a_bb*D_aa_b*D_aa_bb)**(1.0/4) // Mutual GMD(cm)\n", +"D_a_a = 0.7788*d/2.0 // Self GMD of conductor a(cm)\n", +"D_aa_aa = D_a_a // Self GMD of conductor a'(cm)\n", +"D_aa_a = D_a_aa // Distance between conductors a' & a(cm)\n", +"D_s = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD(cm)\n", +"L = 4*10**-4*log(D_m/D_s) // Total inductance of the line(H/km)\n", +"L_mH = L*1000.0 // Total inductance of the line(mH/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.5 : SOLUTION :-')\n", +"printf('\nTotal inductance of the line, L = %.2f mH/km', L_mH)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6: Inductance_of_the_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.6 :\n", +"// Page number 104\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D_a_b = 175.0 // Distance between conductors a & b(cm)\n", +"D_a_aa = 90.0 // Distance between conductors a & a'(cm)\n", +"d = 2.5 // Diameter of conductor(cm)\n", +"\n", +"// Calculations\n", +"GMR = 0.7788*d/2.0 // GMR(cm)\n", +"D_a_a = GMR // Self GMD of conductor a(cm)\n", +"D_aa_aa = D_a_a // Self GMD of conductor a'(cm)\n", +"D_aa_a = 90.0 // Distance between conductors a' & a(cm)\n", +"D_s = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD of conductor A = Self GMD of conductor B(cm)\n", +"D_a_bb = (D_a_aa**2+D_a_b**2)**(1.0/2) // Distance between conductors a & b'(cm)\n", +"D_m = ((D_a_b*D_a_bb)**2)**(1.0/4) // Mutual GMD(cm)\n", +"L = 4*10**-4*log(D_m/D_s) // Inductance of the line(H/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.6 : SOLUTION :-')\n", +"printf('\nInductance of the line, L = %.1e H/km', L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7: Inductance_per_km_of_the_double_circuit_line.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.7 :\n", +"// Page number 104\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"D_a_a = 100.0 // Distance between conductors a & a(cm)\n", +"D_a_b = 25.0 // Distance between conductors a & b(cm)\n", +"d = 2.0 // Diameter of conductor(cm)\n", +"\n", +"// Calculations\n", +"r = d/2.0 // Conductor radius(cm)\n", +"GMR = 0.7788*r // GMR(cm)\n", +"D_a_aa = GMR // GMR of conductors a & a'(cm)\n", +"D_aa_a = D_a_aa // GMR of conductors a' & a(cm)\n", +"D_aa_aa = D_a_a // GMR of conductors a' & a'(cm)\n", +"D_s = (D_a_a*D_a_aa*D_aa_aa*D_aa_a)**(1.0/4) // Self GMD of conductor A = Self GMD of conductor B(cm)\n", +"D_a_bb = (D_a_a**2+D_a_b**2)**(1.0/2) // Distance between conductors a & b'(cm)\n", +"D_aa_b = D_a_bb // Distance between conductors a' & b(cm)\n", +"D_aa_bb = D_a_b // Distance between conductors a' & b'(cm)\n", +"D_m = (D_a_b*D_a_bb*D_aa_b*D_aa_bb)**(1.0/4) // Mutual GMD(cm)\n", +"L = 2*10**-7*log(D_m/D_s) // Inductance/conductor/mt(H)\n", +"L_mH = 2.0*L*1000.0*1000.0 // Loop inductance per km(mH)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.7 : SOLUTION :-')\n", +"printf('\nInductance per km of the double circuit line, L = %.1f mH', L_mH)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8: EX9_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.8 :\n", +"// Page number 104-105\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"n = 7.0 // Number of strands\n", +"r = 1.0 // Radius of each conductor. Assume it 1 for calculation purpose\n", +"\n", +"// Calculations\n", +"D_1_2 = 2.0*r // Distance between conductor 1 & 2\n", +"D_1_6 = 2.0*r // Distance between conductor 1 & 6\n", +"D_1_7 = 2.0*r // Distance between conductor 1 & 7\n", +"D_3_4 = 2.0*r // Distance between conductor 3 & 4\n", +"D_1_4 = 4.0*r // Distance between conductor 1 & 4\n", +"D_1_3 = (D_1_4**2-D_3_4**2)**(1.0/2) // Distance between conductor 1 & 3\n", +"D_1_5 = D_1_3 // Distance between conductor 1 & 5\n", +"GMR = 0.7788*r // GMR\n", +"n_o = n-1 // Number of outside strands\n", +"D_s = (GMR**n*(D_1_2**2*D_1_3**2*D_1_4*D_1_7)**6*(2*r)**n_o)**(1.0/49) // GMR\n", +"overall_radius = 3*r // Overall conductor radius\n", +"ratio = D_s/overall_radius // Ratio of GMR to overall conductor radius\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.8 : SOLUTION :-')\n", +"printf('\nGeometric mean radius of the conductor, D_s = %.3f*r', D_s)\n", +"printf('\nRatio of GMR to overall conductor radius = %.4f ', ratio)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.9: Inductance_of_the_line_per_phase.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// A Texbook on POWER SYSTEM ENGINEERING\n", +"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n", +"// DHANPAT RAI & Co.\n", +"// SECOND EDITION \n", +"\n", +"// PART II : TRANSMISSION AND DISTRIBUTION\n", +"// CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES\n", +"\n", +"// EXAMPLE : 2.9 :\n", +"// Page number 108-109\n", +"clear ; clc ; close ; // Clear the work space and console\n", +"\n", +"// Given data\n", +"d = 1.8 // Diameter of conductor(cm)\n", +"D_A_B = 4.0 // Distance between conductor A & B(cm)\n", +"D_B_C = 9.0 // Distance between conductor B & C(cm)\n", +"D_A_C = 6.0 // Distance between conductor A & C(cm)\n", +"\n", +"// Calculations\n", +"D_eq = (D_A_B*D_B_C*D_A_C)**(1.0/3) // Equivalent distance(cm)\n", +"r_GMR = 0.7788*d/2.0 // GMR(cm)\n", +"L = 2*10**-4*log(D_eq/r_GMR) // Inductance per phase(H/km)\n", +"L_mH = L*1000.0 // Inductance per phase(mH/km)\n", +"\n", +"// Results\n", +"disp('PART II - EXAMPLE : 2.9 : SOLUTION :-')\n", +"printf('\nInductance of the line per phase, L = %.3f mH/km \n', L_mH)\n", +"printf('\nNOTE: ERROR: Calculation mistake in the textbook')" + ] + } +], +"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 +} |