{ "metadata": { "name": "", "signature": "sha256:d44af4aca076d4b10368c1a053109550134d8acdf7cc735b45d4d83101e80a0e" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 6 : Synchronous Machines" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.1 Page No : 297" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# GIVEN DATA\n", "\n", "f = 50; # Generating Frequency in Hertz\n", "\n", "\n", "# CALCULATIONS\n", "# For Case(a)\n", "\n", "Ns_a = 3000; # Synchronous speed in RPM\n", "p_a = (120*f)/Ns_a; # Number of poles\n", "\n", "# For Case(b)\n", "\n", "Ns_b = 1000; # Synchronous speed in RPM\n", "p_b = (120*f)/Ns_b; # Number of poles\n", "\n", "# For Case(c)\n", "\n", "Ns_c = 300; # Synchronous speed in RPM\n", "p_c = (120*f)/Ns_c; # Number of poles\n", "\n", "# For Case(d)\n", "\n", "Ns_d = 40; # Synchronous speed in RPM\n", "p_d = (120*f)/Ns_d; # Number of poles\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.1 : SOLUTION :-\") ;\n", "print \" For Casea) Ns = %.f, p = %.f \"%(Ns_a,p_a);\n", "print \" For Caseb) Ns = %.f, p = %.f \"%(Ns_b,p_b);\n", "print \" For Casec) Ns = %.f, p = %.f \"%(Ns_c,p_c);\n", "print \" For Cased) Ns = %.f, p = %.f \"%(Ns_d,p_d);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.1 : SOLUTION :-\n", " For Casea) Ns = 3000, p = 2 \n", " For Caseb) Ns = 1000, p = 6 \n", " For Casec) Ns = 300, p = 20 \n", " For Cased) Ns = 40, p = 150 \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.2 Page No : 301" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# GIVEN DATA\n", "\n", "f = 60.; # Generating Frequency in Hertz\n", "Ns = 1200.; # Synchronous speed in RPM\n", "Ns_r = 1000.; # Alternator running speed in RPM\n", "\n", "\n", "# CALCULATIONS\n", "\n", "p = (120*f)/Ns; # Total number of poles\n", "f_r = (p*Ns_r)/120; # Alternator running frequency in Hertz\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.2 : SOLUTION :-\") ;\n", "print \" Alternator running frequency, f = %.f Hz \"%(f_r);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.2 : SOLUTION :-\n", " Alternator running frequency, f = 50 Hz \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.3 Page No : 303" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# GIVEN DATA\n", "\n", "p = 2.; # Total number of poles\n", "\n", "\n", "# CALCULATIONS\n", "# For Case(a)\n", "\n", "f_a = 10; # Frequency in Hertz\n", "Ns_a = (120*f_a)/p; # Synchronous speed in RPM \n", "\n", "# For Case(b)\n", "\n", "f_b = 50; # Frequency in Hertz\n", "Ns_b = (120*f_b)/p; # Synchronous speed in RPM \n", "\n", "# For Case(c)\n", "\n", "f_c = 60; # Frequency in Hertz\n", "Ns_c = (120*f_c)/p; # Synchronous speed in RPM \n", "\n", "# For Case(d)\n", "\n", "f_d = 100; # Frequency in Hertz\n", "Ns_d = (120*f_d)/p; # Synchronous speed in RPM \n", "\n", "# For Case(e)\n", "\n", "f_e = 400; # Frequency in Hertz\n", "Ns_e = (120*f_e)/p; # Synchronous speed in RPM \n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.3 : SOLUTION :-\");\n", "print \" For Case a) When f = %.f, Synchronous speed, Ns = %.f RPM \"%(f_a,Ns_a)\n", "print \" For Case b) When f = %.f, Synchronous speed, Ns = %.f RPM \"%(f_b,Ns_b)\n", "print \" For Case c) When f = %.f, Synchronous speed, Ns = %.f RPM \"%(f_c,Ns_c)\n", "print \" For Case d) When f = %.f, Synchronous speed, Ns = %.f RPM \"%(f_d,Ns_d)\n", "print \" For Case e) When f = %.f, Synchronous speed, Ns = %.f RPM \"%(f_e,Ns_e)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.3 : SOLUTION :-\n", " For Case a) When f = 10, Synchronous speed, Ns = 600 RPM \n", " For Case b) When f = 50, Synchronous speed, Ns = 3000 RPM \n", " For Case c) When f = 60, Synchronous speed, Ns = 3600 RPM \n", " For Case d) When f = 100, Synchronous speed, Ns = 6000 RPM \n", " For Case e) When f = 400, Synchronous speed, Ns = 24000 RPM \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.4 Page No : 304" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "import cmath\n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.4 : Given Data \";\n", "print \" Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \";\n", "print \" IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \";\n", "m = 3; # Total Number of phase \n", "p = 6; # Total number of Poles \n", "V = 400; # Operating voltage in Volts\n", "I = 13.5; # Operating current in Amphere\n", "N = 1000; # Speed in RPM\n", "Ia_scc = 13.5; # SCC test Armature current in Amphere at If = 9.5 A\n", "If_scc = 9.5; # SCC test field Rated current in Amphere\n", "Ia_zpf = 13.5; # ZPF test Armature current in Amphere at If = 24 A\n", "If_zpf = 24; # ZPF test field Rated current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386\n", "Ra = 1.0; # Armature resistance in Ohms\n", "\n", "# For case (a)\n", "\n", "BC = 120; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386\n", "Xl = BC/(math.sqrt(3)*Ia_scc); # Per phase leakage reactance in Ohms\n", "\n", "\n", "# For Case (b.1) 0.8 pf Lagging\n", "\n", "pfa_b1 = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "Er_b1 = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b1))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b1))))*(Ra+1j*Xl))); # Induced Voltage in Volts \n", "R_b1 = 10; A_b1 = 9.5; #From OCC the field current required for Er_b1 (Should be in Line-line Voltage) Er_b1 = 379.12V will get R_b1 & A_b1 value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b1 = 136.35; # Angle between R_b1 & A_b1 (figure 6.16(a) & Page no:-388) = 90'+9.48'+36.87' = 136.35'\n", "F_b1 = math.sqrt((R_b1**2)+(A_b1**2)-(2*R_b1*A_b1*math.degrees(math.degrees(math.cos(math.radians(angle_b1)))))); # From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere\n", "Eo_b1 = 525; # Corresponding to field current F_b1 = 18.12 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)\n", "r_b1 = 100*((Eo_b1-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b.2) 0.8 pf Leading\n", "\n", "pfa_b2 = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "Er_b2 = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b2))))+1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b2))))*(Ra+1j*Xl))); # Induced Voltage in Volts \n", "R_b2 = 8.3; A_b2 = 9.5; #From OCC the field current required for Er_b2 (Should be in Line-line Voltage) Er_b1 = 363.71 V will get R_b2 & A_b2 value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b2 = 70.61; # Angle between R_b2 & A_b2 (figure 6.16(b) & Page no:-388) = 90'+17.48'-36.87' = 70.61'\n", "F_b2 = cmath.sqrt((R_b2**2)+(A_b2**2)-(2*R_b2*A_b2*math.degrees(math.degrees(math.cos(math.radians(angle_b2)))))); # From phasor diagram in figure 6.16(b) & Page no:-388 the neccessary field excitation in Amphere\n", "Eo_b2 = 338; # Corresponding to field current F_b2 = 10.36 A the open circuit EMF from OCC is 338 V (Figure6.15 & Page no:-386)\n", "r_b2 = 100*((Eo_b2-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b.3) Unity pf Leading\n", "\n", "pfa_b3 = math.acos(math.radians(1.0)); # Power factor angle in degree\n", "Er_b3 = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b3))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b3))))*(Ra+1j*Xl))); # Induced Voltage in Volts \n", "R_b3 = 13; A_b3 = 9.5; #From OCC the field current required for Er_b3 (Should be in Line-line Voltage) Er_b1 = 440.30 V will get R_b3 & A_b3 value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b3 = 105.81; # Angle between R_b3 & A_b3 (figure 6.16(c) & Page no:-388) = 90'+15.81' = 105.81'\n", "F_b3 = cmath.sqrt((R_b3**2)+(A_b3**2)-(2*R_b3*A_b3*math.degrees(math.degrees(math.cos(math.radians(angle_b3)))))); # From phasor diagram in figure 6.16(c) & Page no:-388 the neccessary field excitation in Amphere\n", "Eo_b3 = 520; # Corresponding to field current F_b2 = 18.10 A the open circuit EMF from OCC is 520 V (Figure6.15 & Page no:-386)\n", "r_b3 = 100*((Eo_b3-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b.4) ZPF Lagging\n", "\n", "pfa_b4 = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "Er_b4 = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b4))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b4))))*(Ra+1j*Xl))); # Induced Voltage in Volts \n", "R_b4 = 18; A_b4 = 9.5; #From OCC the field current required for Er_b4 (Should be in Line-line Voltage) Er_b4 = 521 V will get R_b4 & A_b4 value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b4 = 177.57; # Angle between R_b4 & A_b4 = 90'-2.43'+90' = 177.57'\n", "F_b4 = cmath.sqrt((R_b4**2)+(A_b4**2)-(2*R_b4*A_b4*math.degrees(math.degrees(math.cos(math.radians(angle_b4)))))); # The neccessary field excitation in Amphere\n", "Eo_b4 = 570; # Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)\n", "r_b4 = 100*((Eo_b4-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b.4) ZPF Lagging\n", "\n", "pfa_b5 = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "Er_b5 = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b5))))+1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b5))))*(Ra+1j*Xl))); # Induced Voltage in Volts \n", "R_b5 = 6.0; A_b5 = 9.50; #From OCC the field current required for Er_b5 (Should be in Line-line Voltage) Er_b5 = 280.70 V will get R_b5 & A_b5 value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b5 = 4.77; # Angle between R_b5 & A_b5 = 90'-4.77'-90' = 4.77'\n", "F_b5 = cmath.sqrt((R_b5**2)+(A_b5**2)-(2*R_b5*A_b5*math.degrees(math.degrees(math.cos(math.radians(angle_b5)))))); # The neccessary field excitation in Amphere\n", "Eo_b5 = 135; # Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 135 V (Figure6.15 & Page no:-386)\n", "r_b5 = 100*((Eo_b5-V)/V); # Percentage regulation \n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.4 : SOLUTION :-\") ;\n", "print \" a) Per phase leakage reactance, Xl = %.2f Ohms \"%(Xl)\n", "print \" For Case b.1) 0.8 pf Lagging Open circuit EMF EMF = %.f V \"%(Eo_b1)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b1) \n", "print \" For Case b.2) 0.8 pf Leading Open circuit EMF, EMF = %.f V \"%(Eo_b2)\n", "print \" Percenatge Regulation, R = %.1f Percenatge \"%(r_b2) \n", "print \" For Case b.3) Unity pf Lagging Open circuit EMF, EMF = %.f V \"%(Eo_b3)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_b3) \n", "print \" For Case b.4) ZPF Lagging Open circuit EMF, EMF = %.f V\"%(Eo_b4)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b4) \n", "print \" For Case b.5) ZPF Leading Open circuit EMF, EMF = %.f V \"%(Eo_b5)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b5) \n", "print (\" Calculated Answer in Tabular Column\")\n", "print \" Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \"\n", "print \" Open circuit EMF V) %.f %.f %.f %.f %.f \"%(Eo_b1,Eo_b2,Eo_b3,Eo_b4,Eo_b5)\n", "print \" Percenatge Regulation %.2f %.1f %.f. %.2f %.2f \"%(r_b1,r_b2,r_b3,r_b4,r_b5)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.4 : Given Data \n", " Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n", " IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n", "EXAMPLE : 6.4 : SOLUTION :-\n", " a) Per phase leakage reactance, Xl = 5.13 Ohms \n", " For Case b.1) 0.8 pf Lagging Open circuit EMF EMF = 525 V \n", " Percenatge Regulation, R = 0.00 Percenatge \n", " For Case b.2) 0.8 pf Leading Open circuit EMF, EMF = 338 V \n", " Percenatge Regulation, R = -100.0 Percenatge \n", " For Case b.3) Unity pf Lagging Open circuit EMF, EMF = 520 V \n", " Percenatge Regulation, R = 0 Percenatge \n", " For Case b.4) ZPF Lagging Open circuit EMF, EMF = 570 V\n", " Percenatge Regulation, R = 0.00 Percenatge \n", " For Case b.5) ZPF Leading Open circuit EMF, EMF = 135 V \n", " Percenatge Regulation, R = -100.00 Percenatge \n", " Calculated Answer in Tabular Column\n", " Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n", " Open circuit EMF V) 525 338 520 570 135 \n", " Percenatge Regulation 0.00 -100.0 0. 0.00 -100.00 \n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.5 Page No : 307" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.5a Data is same as Example 6.4: Given Data \";\n", "print \" Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \";\n", "print \" IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \";\n", "m = 3; # Total Number of phase in Induction Motor\n", "p = 6; # Total number of Poles of Induction Motor\n", "V = 400; # Operating voltage of the Induction motor in Volts\n", "I = 13.5; # Operating current of the Induction motor in Amphere\n", "N = 1000; # speed of the Induction motor in RPM\n", "Ia_scc = 13.5; # SCC test Armature current in Amphere at If = 9.5 A\n", "If_scc = 9.5; # SCC test field Rated current in Amphere\n", "Ia_zpf = 13.5; # ZPF test Armature current in Amphere at If = 24 A\n", "If_zpf = 24; # ZPF test field Rated current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386\n", "Ra = 1.0; # Armature resistance in Ohms\n", "v = V/math.sqrt(3); # Rated phase Voltage in Volts\n", "\n", "# For case (a)\n", "\n", "EMF_a1 = 345; # From OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386 open-circuit line-line voltage per phase is 345vVfor If = 9.50A in Volts\n", "Zs_a1 = EMF_a1/(Ia_zpf*math.sqrt(3)); # Unsaturated synchronous impedance at If=9.50A in Ohms\n", "Xs_a1 = math.sqrt((Zs_a1**2)-(Ra**2)); # Synchronous reactance at If =9.50A in Ohms\n", "Ia_a2 = 15.75; # Current from SCC in Figure6.15 & Page no:-386 is 15.75A for correspounding to the rated Voltage in Volts\n", "Zs_a2 = V/(Ia_a2*math.sqrt(3)); # Unsaturated synchronous impedance at If=9.50A in Ohms\n", "Xs_a2 = math.sqrt((Zs_a2**2)-(Ra**2)); # Synchronous reactance at If =9.50A in Ohms\n", "\n", "# For Case (b.1) 0.8 pf Lagging\n", "\n", "pfa_b1 = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "real_b1 = (v+Ia_zpf*Ra*math.degrees(math.degrees(math.cos(math.radians(pfa_b1))))+Ia_zpf*Xs_a1*math.degrees(math.degrees(math.sin(math.radians(pfa_b1)))));\n", "imag_b1 = (Ia_zpf*Xs_a1*math.degrees(math.degrees(math.cos(math.radians(pfa_b1))))-Ia_zpf*Ra*math.degrees(math.degrees(math.sin(math.radians(pfa_b1)))));\n", "E_b1 = math.sqrt(real_b1**2+imag_b1**2); # Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for lagging pf\n", "del_b1 = math.degrees(math.atan(imag_b1/real_b1)); # Power angle in degree\n", "r_b1 = 100*(E_b1-v)/v; # Percantage regulation \n", "\n", "\n", "# For Case (b.2) 0.8 pf Leading\n", "\n", "pfa_b2 = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "real_b2 = (v+Ia_zpf*Ra*math.degrees(math.degrees(math.cos(math.radians(pfa_b2))))-Ia_zpf*Xs_a1*math.degrees(math.degrees(math.sin(math.radians(pfa_b2)))));\n", "imag_b2 = (Ia_zpf*Xs_a1*math.degrees(math.degrees(math.cos(math.radians(pfa_b2))))+Ia_zpf*Ra*math.degrees(math.degrees(math.sin(math.radians(pfa_b2)))));\n", "E_b2 = math.sqrt(real_b2**2+imag_b2**2); # Induced Voltage pr phase in Volts from Figure6.19 (b) & Page no:-394 shows the phasor diagram for leading pf\n", "del_b2 = math.degrees(math.atan(imag_b2/real_b2)); # Power angle in degree\n", "r_b2 = 100*(E_b2-v)/v; # Percantage regulation \n", "\n", "\n", "# For Case (b.3) Unity pf\n", "\n", "pfa_b3 = math.acos(math.radians(1.0)); # Power factor angle in degree\n", "real_b3 = (v+Ia_zpf*Ra);\n", "imag_b3 = (Ia_zpf*Xs_a1);\n", "E_b3 = math.sqrt(real_b3**2+imag_b3**2); # Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for unity pf\n", "del_b3 = math.degrees(math.atan(imag_b3/real_b3)); # Power angle in degree\n", "r_b3 = 100*(E_b3-v)/v; # Percantage regulation \n", "\n", "# For Case (b.4) ZPF pf Lagging\n", "\n", "pfa_b4 = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "real_b4 = (v+Ia_zpf*Xs_a1);\n", "imag_b4 = (-Ia_zpf*Ra);\n", "E_b4 = math.sqrt(real_b4**2+imag_b4**2); # Induced Voltage pr phase in Volts ZPF for lagging pf\n", "del_b4 = math.degrees(math.atan(imag_b4/real_b4)); # Power angle in degree\n", "r_b4 = 100*(E_b4-v)/v; # Percantage regulation \n", "\n", "# For Case (b.5) ZPF pf Leading\n", "\n", "pfa_b5 = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "real_b5 = (v-Ia_zpf*Xs_a1);\n", "imag_b5 = (Ia_zpf*Ra);\n", "E_b5 = math.sqrt(real_b5**2+imag_b5**2); # Induced Voltage pr phase in Volts ZPF for lagging pf\n", "del_b5 = math.degrees(math.atan(imag_b5/real_b5)); # Power angle in degree\n", "r_b5 = 100*(E_b5-v)/v; # Percantage regulation \n", "\n", "\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.5a : SOLUTION :-\") ;\n", "print \" a.1) Synchronous reactance for rated current at If = %.2f, Xs = %.2f Ohms \"%(If_scc,Xs_a1)\n", "print \" a.2) Synchronous reactance for rated per phase Voltage at v = %.f, Xs = %.2f Ohms \"%(v,Xs_a2)\n", "print \" For Case b.1) 0.8 pf Lagging Induced EMF per phase , EMF = %.2f V \"%(E_b1)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b1) \n", "print \" Power angle = %.2f degree \"%(del_b1) \n", "print \" For Case b.2) 0.8 pf Leading Induced EMF per phase, EMF = %.2f V \"%(E_b2)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b2) \n", "print \" Power angle = %.2f degree \"%(del_b2) \n", "print \" For Case b.3) Unity pf Lagging Induced EMF per phase, EMF = %.2f V \"%(E_b3)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b3)\n", "print \" Power angle = %.2f degree \"%(del_b3) \n", "print \" For Case b.4) ZPF Lagging Induced EMF per phase, EMF = %.2f V\"%(E_b4)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b4) \n", "print \" Power angle = %.1f degree \"%(del_b4) \n", "print \" For Case b.5) ZPF Leading Induced EMF per phase, EMF = %.2f V \"%(E_b5)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b5) \n", "print \" Power angle = %.2f degree \"%(del_b5) \n", "print (\" Calculated Answer in Tabular Column\")\n", "print \" Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \"\n", "print \" Open circuit EMF V) %.2f %.2f %.2f %.2f %.2f \"%(E_b1,E_b2,E_b3,E_b4,E_b5)\n", "print \" Percenatge Regulation %.2f %.2f %.2f. %.2f %.2f \"%(r_b1,r_b2,r_b3,r_b4,r_b5)\n", "print \" Power angle %.2f %.2f %.2f. %.1f %.2f \"%(del_b1,del_b2,del_b3,del_b4,del_b5)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.5a Data is same as Example 6.4: Given Data \n", " Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n", " IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n", "EXAMPLE : 6.5a : SOLUTION :-\n", " a.1) Synchronous reactance for rated current at If = 9.50, Xs = 14.72 Ohms \n", " a.2) Synchronous reactance for rated per phase Voltage at v = 231, Xs = 14.63 Ohms \n", " For Case b.1) 0.8 pf Lagging Induced EMF per phase , EMF = 653910.50 V \n", " Percenatge Regulation, R = 283051.55 Percenatge \n", " Power angle = 84.54 degree \n", " For Case b.2) 0.8 pf Leading Induced EMF per phase, EMF = 653897.98 V \n", " Percenatge Regulation, R = 283046.13 Percenatge \n", " Power angle = 87.65 degree \n", " For Case b.3) Unity pf Lagging Induced EMF per phase, EMF = 315.03 V \n", " Percenatge Regulation, R = 36.41 Percenatge \n", " Power angle = 39.11 degree \n", " For Case b.4) ZPF Lagging Induced EMF per phase, EMF = 429.88 V\n", " Percenatge Regulation, R = 86.14 Percenatge \n", " Power angle = -1.8 degree \n", " For Case b.5) ZPF Leading Induced EMF per phase, EMF = 34.93 V \n", " Percenatge Regulation, R = -84.88 Percenatge \n", " Power angle = 22.74 degree \n", " Calculated Answer in Tabular Column\n", " Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n", " Open circuit EMF V) 653910.50 653897.98 315.03 429.88 34.93 \n", " Percenatge Regulation 283051.55 283046.13 36.41. 86.14 -84.88 \n", " Power angle 84.54 87.65 39.11. -1.8 22.74 \n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.6 Page No : 309" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "import cmath \n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.4 Data is same as Exaple 6.4 : Given Data \";\n", "print \" Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \";\n", "print \" IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \";\n", "m = 3; # Total Number of phase in Induction Motor\n", "p = 6; # Total number of Poles of Induction Motor\n", "V = 400; # Operating voltage of the Induction motor in Volts\n", "I = 13.5; # Operating current of the Induction motor in Amphere\n", "N = 1000; # speed of the Induction motor in RPM\n", "Ia_scc = 13.5; # SCC test Armature current in Amphere at If = 9.5 A\n", "If_scc = 9.5; # SCC test field Rated current in Amphere\n", "Ia_zpf = 13.5; # ZPF test Armature current in Amphere at If = 24 A\n", "If_zpf = 24; # ZPF test field Rated current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386\n", "Ra = 1.0; # Armature resistance in Ohms\n", "v = V/math.sqrt(3); # Rated phase voltage in Volts\n", "\n", "\n", "# For Case (a) 0.8 pf Lagging\n", "\n", "pfa_a = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "E_a = v+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_a))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_a)))))*Ra); # Induced Voltage in Volts \n", "# E_a = v+(Ia_scc*(cosd(pfa_a)-%i*sind(pfa_a))*Ra);\n", "R1_a = 11.8; A_a = 9.50; #From OCC the field current required for E_a (Should be in Line-line Voltage) E_a = 419.05V will get R1_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_a = 124.95; # Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'-1.92'+36.87' = 124.95'\n", "F_a = math.sqrt((R1_a**2)+(A_a**2)-(2*R1_a*A_a*math.degrees(math.cos(math.radians(angle_a))))); # From phasor diagram in figure 6.21(a) & Page no:-400 the neccessary field excitation in Amphere\n", "Eo_a = 538; # Corresponding to field current F_a = 18.94 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)\n", "r_a = 100*((Eo_a-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b) 0.8 pf Leading\n", "\n", "pfa_b = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "E_b = v+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_b))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_b)))))*Ra); # Induced Voltage in Volts \n", "R1_b = 11.80; A_b = 9.50; #From OCC the field current required for E_b (Should be in Line-line Voltage) E_b = 419.10V will get R1_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b = 55.07; # Angle between R1_b & A_b (Figure6.21b & Page no:-400) = 90'-1.92'-36.87' = 55.07'\n", "F_b = cmath.sqrt((R1_b**2)+(A_b**2)-(2*R1_b*A_b*math.degrees(math.cos(math.radians(angle_b))))); # From phasor diagram in figure 6.21(b) & Page no:-400 the neccessary field excitation in Amphere\n", "Eo_b = 382; # Corresponding to field current F_b = 10.10 A the open circuit EMF from OCC is 382 V (Figure6.15 & Page no:-386)\n", "r_b = 100*((Eo_b-V)/V); # Percentage regulation \n", "\n", "# For Case (c) Unity pf\n", "\n", "pfa_c = math.degrees(math.acos(math.radians(1))); # Power factor angle in degree\n", "E_c = v+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_c))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_c)))))*Ra); # Induced Voltage in Volts \n", "R1_c = 12.10; A_c = 9.50; #From OCC the field current required for E_c (Should be in Line-line Voltage) E_c = 423.50V will get R1_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_c = 90; # Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'\n", "F_c = math.sqrt((R1_c**2)+(A_c**2)-(2*R1_c*A_c*math.degrees(math.cos(math.radians(angle_c))))); # From phasor diagram in figure 6.21(c) & Page no:-400 the neccessary field excitation in Amphere\n", "Eo_c = 480; # Corresponding to field current F_c = 15.38 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)\n", "r_c = 100*((Eo_c-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (d) ZPF Lagging\n", "\n", "pfa_d = math.acos(math.radians(0.0)); # Power factor angle in degree\n", "E_d = v+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_d))))-1j*math.degrees(math.degrees(math.sin(math.radians(pfa_d)))))*Ra); # Induced Voltage in Volts \n", "R1_d = 11.20; A_d = 9.50; #From OCC the field current required for E_d (Should be in Line-line Voltage) E_d = 400.80V will get R1_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_d = 179.40; # Angle between R1_d & A_d = 90'-0.6'+90' = 179.40'\n", "F_d = math.sqrt((R1_d**2)+(A_d**2)-(2*R1_d*A_d*math.degrees(math.cos(math.radians(angle_d))))); # From phasor diagram the neccessary field excitation in Amphere\n", "Eo_d = 545; # Corresponding to field current F_d = 18.12 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)\n", "r_d = 100*((Eo_d-V)/V); # Percentage regulation \n", "\n", "# For Case (d) ZPF Lagging\n", "\n", "pfa_e = math.acos(math.radians(0.0)); # Power factor angle in degree\n", "E_e = v+(Ia_scc*(math.degrees(math.degrees(math.cos(math.radians(pfa_e))))+1j*math.degrees(math.degrees(math.sin(math.radians(pfa_e)))))*Ra); # Induced Voltage in Volts \n", "R1_e = 11.20; A_e = 9.50; #From OCC the field current required for E_e (Should be in Line-line Voltage) E_d = 400.80V will get R1_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_e = 0.60; # Angle between R1_e & A_e = 90'+0.6'-90' = 0.60'\n", "F_e = cmath.sqrt((R1_e**2)+(A_e**2)-(2*R1_e*A_e*math.degrees(math.cos(math.radians(angle_e))))); # From phasor diagram the neccessary field excitation in Amphere\n", "Eo_e = 63; # Corresponding to field current F_e = 1.70 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)\n", "r_e = 100*((Eo_e-V)/V); # Percentage regulation \n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.6 : SOLUTION :-\") ;\n", "print \" For Case a) 0.8 pf Lagging Open circuit EMF, EMF = %.f V \"%(Eo_a)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_a) \n", "print \" For Case b) 0.8 pf Leading Open circuit EMF, EMF = %.f V \"%(Eo_b)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b) \n", "print \" For Case c) Unity pf Lagging Open circuit EMF, EMF = %.f V \"%(Eo_c)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_c) \n", "print \" For Case d) ZPF Lagging Open circuit EMF, EMF = %.f V\"%(Eo_d)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_d) \n", "print \" For Case e) ZPF Leading Open circuit EMF, EMF = %.f V \"%(Eo_e)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_e) \n", "print (\" Calculated Answer in Tabular Column\")\n", "print \" Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \"\n", "print \" Open circuit EMF V) %.f %.f %.f %.f %.f \"%(Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)\n", "print \" Percenatge Regulation %.2f %.2f %.f %.2f %.2f \"%(r_a,r_b,r_c,r_d,r_e)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.4 Data is same as Exaple 6.4 : Given Data \n", " Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n", " IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n", "EXAMPLE : 6.6 : SOLUTION :-\n", " For Case a) 0.8 pf Lagging Open circuit EMF, EMF = 538 V \n", " Percenatge Regulation, R = 0.00 Percenatge \n", " For Case b) 0.8 pf Leading Open circuit EMF, EMF = 382 V \n", " Percenatge Regulation, R = -100.00 Percenatge \n", " For Case c) Unity pf Lagging Open circuit EMF, EMF = 480 V \n", " Percenatge Regulation, R = 0 Percenatge \n", " For Case d) ZPF Lagging Open circuit EMF, EMF = 545 V\n", " Percenatge Regulation, R = 0.00 Percenatge \n", " For Case e) ZPF Leading Open circuit EMF, EMF = 63 V \n", " Percenatge Regulation, R = -100.00 Percenatge \n", " Calculated Answer in Tabular Column\n", " Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n", " Open circuit EMF V) 538 382 480 545 63 \n", " Percenatge Regulation 0.00 -100.00 0 0.00 -100.00 \n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.7 Page No : 314" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "import cmath \n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.7 Data as same as Example 6.4 : Given Data \";\n", "print \" Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \";\n", "print \" IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \";\n", "m = 3; # Total Number of phase in Induction Motor\n", "p = 6; # Total number of Poles of Induction Motor\n", "V = 400; # Operating voltage of the Induction motor in Volts\n", "I = 13.5; # Operating current of the Induction motor in Amphere\n", "N = 1000; # speed of the Induction motor in RPM\n", "Ia_scc = 13.5; # SCC test Armature current in Amphere at If = 9.5 A\n", "If_scc = 9.5; # SCC test field Rated current in Amphere\n", "Ia_zpf = 13.5; # ZPF test Armature current in Amphere at If = 24 A\n", "If_zpf = 24; # ZPF test field Rated current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386\n", "Ra = 1.0; # Armature resistance in Ohms\n", "BC = 120; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386\n", "Xl = BC/(math.sqrt(3)*Ia_scc); # Per phase leakage reactance in Ohms for this referring to example6.4 & page no:- 386 \n", "\n", "# For Case (a) 0.8 pf Lagging\n", "\n", "pfa_a = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "Er_a = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_a)))-1j*math.degrees(math.sin(math.radians(pfa_a))))*(Ra+1j*Xl)); # Induced Voltage in Volts \n", "R_a = 9.8; A_a = 9.5; #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 479.60V will get R_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_a = 126.87; # Angle between R_a & A_a (Figure6.22(a) & Page no:-403) = 90'+36.87' = 126.87'\n", "F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a))))); # From phasor diagram in figure 6.22(a) & Page no:-403 the neccessary field excitation in Amphere\n", "Eo_a = 560; # Corresponding to field current ( OF'=OF+FF'), F_a = 17.28 + 6.2 = 23.46 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)\n", "r_a = 100*((Eo_a-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b) 0.8 pf Leading\n", "\n", "pfa_b = math.acos(math.radians(0.8)); # Power factor angle in degree\n", "Er_b = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_b)))+1j*math.degrees(math.sin(math.radians(pfa_b))))*(Ra+1j*Xl)); # Induced Voltage in Volts \n", "R_b = 9.8; A_b = 9.5; #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 363.90 V will get R_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_b = 53.13; # Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'-36.87' = 53.13'\n", "F_b = cmath.sqrt((R_b**2)+(A_b**2)-(2*R_b*A_b*math.degrees(math.cos(math.radians(angle_b))))); # From phasor diagram in figure 6.22(b) & Page no:-403 the neccessary field excitation in Amphere\n", "Eo_b = 380; # Corresponding to field current ( OF'=OF+FF') F_b = 8.62+1.5=10.12 A the open circuit EMF from OCC is 380 V (Figure6.15 & Page no:-386)\n", "r_b = 100*((Eo_b-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (c) Unity pf Leading\n", "\n", "pfa_c = math.acos(math.radians(1.0)); # Power factor angle in degree\n", "Er_c = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_c)))-1j*math.degrees(math.sin(math.radians(pfa_c))))*(Ra+1j*Xl)); # Induced Voltage in Volts \n", "R_c = 9.8; A_c = 9.5; #From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 440.11 V will get R_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_c = 90; # Angle between R_c & A_c (Figure6.22c & Page no:-403) = 90' = 90'\n", "F_c = math.sqrt((R_c**2)+(A_c**2)-(2*R_c*A_c*math.degrees(math.cos(math.radians(angle_c))))); # From phasor diagram in figure 6.22(c) & Page no:-403 the neccessary field excitation in Amphere\n", "Eo_c = 510; # Corresponding to field current ( OF'=OF+FF') F_c = 13.65+3.0=16.65A the open circuit EMF from OCC is 510 V (Figure6.15 & Page no:-386)\n", "r_c = 100*((Eo_c-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (d) ZPF Lagging\n", "\n", "pfa_d = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "Er_d = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_d)))-1j*math.degrees(math.sin(math.radians(pfa_d))))*(Ra+1j*Xl)); # Induced Voltage in Volts \n", "R_d = 9.8; A_d = 9.5; #From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 570.20 V will get R_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_d = 180.0; # Angle between R_d & A_d = 90'+90' = 180'\n", "F_d = math.sqrt((R_d**2)+(A_d**2)-(2*R_d*A_d*math.degrees(math.cos(math.radians(angle_d))))); # The neccessary field excitation in Amphere\n", "Eo_d = 600; # Corresponding to field current ( OF'=OF+FF') F_d = 19.3+16=35.30 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)\n", "r_d = 100*((Eo_d-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (e) ZPF Lagging\n", "\n", "pfa_e = math.degrees(math.acos(math.radians(0))); # Power factor angle in degree\n", "Er_e = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_e)))+1j*math.degrees(math.sin(math.radians(pfa_e))))*(Ra+1j*Xl)); # Induced Voltage in Volts \n", "R_e = 9.8; A_e = 9.50; #From OCC the field current required for Er_e (Should be in Line-line Voltage) Er_e = 281.10 V will get R_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)\n", "angle_e = 0.0; # Angle between R_e & A_e = 90'-90' = 0.0'\n", "F_e = cmath.sqrt((R_e**2)+(A_e**2)-(2*R_e*A_e*math.degrees(math.cos(math.radians(angle_e))))); # The neccessary field excitation in Amphere\n", "Eo_e = 5; # Corresponding to field current ( OF'=OF+FF') F_e = 0.0+0.30=0.30 A the open circuit EMF from OCC is 5 V (Figure6.15 & Page no:-386)\n", "r_e = 100*((Eo_e-V)/V); # Percentage regulation \n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.7 : SOLUTION :-\") ;\n", "print \" Per phase leakage reactance, Xl = %.2f Ohms \"%(Xl)\n", "print \" For Case a) 0.8 pf Lagging Open circuit EMF, EMF = %.f V \"%(Eo_a)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_a) \n", "print \" For Case b) 0.8 pf Leading Open circuit EMF, EMF = %.f V \"%(Eo_b)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_b) \n", "print \" For Case c) Unity pf Lagging Open circuit EMF, EMF = %.f V \"%(Eo_c)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_c) \n", "print \" For Case d) ZPF Lagging Open circuit EMF, EMF = %.f V\"%(Eo_d)\n", "print \" Percenatge Regulation, R = %.f Percenatge \"%(r_d) \n", "print \" For Case e) ZPF Leading Open circuit EMF, EMF = %.f V \"%(Eo_e)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_e) \n", "print (\" Calculated Answer in Tabular Column\")\n", "print \" Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \"\n", "print \" Open circuit EMF V) %.f %.f %.f %.f %.f \"%(Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)\n", "print \" Percenatge Regulation %.f %.f %.1f %.f %.2f \"%(r_a,r_b,r_c,r_d,r_e)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.7 Data as same as Example 6.4 : Given Data \n", " Vocv 215 284 320 380 400 422 452 472 488 508 520 532 540 552 560 \n", " IfA 6.5 8 9 10 11 12 14 15 16 17 18 19 20 22 24 \n", "EXAMPLE : 6.7 : SOLUTION :-\n", " Per phase leakage reactance, Xl = 5.13 Ohms \n", " For Case a) 0.8 pf Lagging Open circuit EMF, EMF = 560 V \n", " Percenatge Regulation, R = 0 Percenatge \n", " For Case b) 0.8 pf Leading Open circuit EMF, EMF = 380 V \n", " Percenatge Regulation, R = -100 Percenatge \n", " For Case c) Unity pf Lagging Open circuit EMF, EMF = 510 V \n", " Percenatge Regulation, R = 0 Percenatge \n", " For Case d) ZPF Lagging Open circuit EMF, EMF = 600 V\n", " Percenatge Regulation, R = 0 Percenatge \n", " For Case e) ZPF Leading Open circuit EMF, EMF = 5 V \n", " Percenatge Regulation, R = -100.00 Percenatge \n", " Calculated Answer in Tabular Column\n", " Power Factor 0.8 Lag 0.8 Lead 1.0 ZPF Lag ZPF Lead \n", " Open circuit EMF V) 560 380 510 600 5 \n", " Percenatge Regulation 0 -100 0.0 0 -100.00 \n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.9 Page No : 315" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import imag,real\n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.9 : Given Data \";\n", "print \" VockV 10 10.80 11.50 12.10 12.60 13 14 14.50 14.80 \";\n", "print \" IfA 175 200 225 250 275 300 400 450 500 \";\n", "p = 6; # Total number of Poles of Alternator\n", "V = 11*10**3; # Operating voltage of the Alternator in Volts\n", "N = 1500; # speed of the Alternator in RPM\n", "Ia_scc = 2099; # SCC test Armature current in Amphere at If = 200 A\n", "If_scc = 200; # SCC test field Rated current in Amphere\n", "Ia_pt = 2099; # Pottier test Armature current in Amphere at If = 450 A\n", "If_pt = 450; # Pottier test field Rated current in Amphere\n", "VA = 40*10**6; # VA rating of the Alternator in Volts-Amphere\n", "f = 50; # Operating Frequency of the Alternator in Hertz\n", "pf = 0.8; # Power factor (lagging)\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407\n", "\n", "v = V/math.sqrt(3); # Rated phase Voltage in Volts\n", "I = VA/(math.sqrt(3)*V); # Full-load phase current in Amphere\n", "Xl = 0.4481; # Leakage reactance in Ohms From OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407 \n", "\n", "\n", "# For Case(a) General Method\n", "\n", "pfa_a = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "Er_a = (V/math.sqrt(3))+(Ia_scc*(math.degrees(math.cos(math.radians(pfa_a)))-1j*math.degrees(math.sin(math.radians(pfa_a))))*Xl); # Induced Voltage in Volts \n", "R_a = 208.4; A_a = 200; #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 11043.66 V will get R_a & A_a value Respectively from SCC (Figure6.24 & Page no:-407)\n", "angle_a = 131.93; # Angle between R_a & A_a (Figure6.25(a) & Page no:-408) = 90'+5.06'+36.87' = 131.93'\n", "F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a))))); # From phasor diagram in figure 6.25(a) & Page no:-408 the neccessary field excitation in Amphere\n", "Eo_a = 13720; # Corresponding to field current, F_a = 373 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)\n", "r_a = 100*((Eo_a-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case (b) ASA Method\n", "\n", "pfa_b = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "Er_b = (V/math.sqrt(3))+Ia_scc*(math.degrees(math.cos(math.radians(pfa_b)))-1j*math.degrees(math.sin(math.radians(pfa_b))))*Xl; # Induced Voltage in Volts \n", "R_b = 160; A_b = 200; #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 11043.66 V will get R_b & A_b value Respectively from SCC (Figure6.24 & Page no:-407)\n", "angle_b = 126.87; # Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'+36.87' = 126.87'\n", "F_b = math.sqrt((R_b**2)+(A_b**2)-(2*R_b*A_b*math.degrees(math.cos(math.radians(angle_b))))); # From phasor diagram in figure 6.25(b) & Page no:-408 the neccessary field excitation in Amphere\n", "Eo_b = 13660; # Corresponding to field current ( OF'=OF+FF') F_b = 337.88+15.38=337.88 A the open circuit EMF from OCC is 13660 V (Figure6.15 & Page no:-386)\n", "r_b = 100*((Eo_b-V)/V); # Percentage regulation \n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.9 : SOLUTION :-\") ;\n", "print \" For Case a) GeneralZPF) Method Induced EMF, EMF = %.f < %.2f V \"%(abs(Er_a),math.degrees(math.atan2(Er_a.imag,Er_a.real)))\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_a) \n", "print \" For Case b) ASA Method Induced EMF, EMF = %.f < %.2f V \"%(abs(Er_b),math.degrees(math.atan2(Er_b.imag,Er_b.real)))\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(r_b)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- For Case a) GeneralZPF) Method a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f \"%(abs(Er_a),math.degrees(math.atan2(Er_a.imag,Er_a.real)))\n", "print \" For Case b) ASA Method a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f \"%(abs(Er_b),math.degrees(math.atan2(Er_b.imag,Er_b.real)))\n", "print \" CALCULATION OF THE POWER ANGLE IS NOT CALCULATED IN THE TEXT BOOK FOR THIS PROBLEM \"\n", "print \" INDUCED EMF AND PERCENTAGE REGULATION IS APPROXIMATED VALUE BECACUSE IN THE TEXT BOOK, CALCULATED INDUCED EMF IS WRONGLY PRINTED\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.9 : Given Data \n", " VockV 10 10.80 11.50 12.10 12.60 13 14 14.50 14.80 \n", " IfA 175 200 225 250 275 300 400 450 500 \n", "EXAMPLE : 6.9 : SOLUTION :-\n", " For Case a) GeneralZPF) Method Induced EMF, EMF = 54351 < -82.49 V \n", " Percenatge Regulation, R = 0.00 Percenatge \n", " For Case b) ASA Method Induced EMF, EMF = 54351 < -82.49 V \n", " Percenatge Regulation, R = 0.00 Percenatge \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- For Case a) GeneralZPF) Method a) Induced EMF = 6376<-5.07 degree instead of 54351 < -82.49 \n", " For Case b) ASA Method a) Induced EMF = 6376<-5.07 degree instead of 54351 < -82.49 \n", " CALCULATION OF THE POWER ANGLE IS NOT CALCULATED IN THE TEXT BOOK FOR THIS PROBLEM \n", " INDUCED EMF AND PERCENTAGE REGULATION IS APPROXIMATED VALUE BECACUSE IN THE TEXT BOOK, CALCULATED INDUCED EMF IS WRONGLY PRINTED\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.10 Page No : 318" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import real,imag\n", "\n", "# GIVEN DATA\n", "\n", "m = 3; # Total Number of Phase in Alternator\n", "p = 2; # Total number of Poles of Alternator\n", "V = 11*10**3; # Operating voltage of the Alternator in Volts\n", "VA = 10*10**6; # VA rating of the Alternator in Volts-Amphere\n", "f = 50; # Operating Frequency of the alternator in Hertz\n", "pf = 0.8; # Power factor (lagging)\n", "Vf = 12*10**3; # Operating field voltage of the Alternator in Volts\n", "If = 160; # Field Current in Amphere\n", "Ra = 0.05; # Armature resistance per phase in Ohms\n", "Xs = 1.5; # Winding leakage reactance per phase in Ohms\n", "A = 150; # The armature MMF at rated current is equivalent to Field Current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "\n", "Vt = V/math.sqrt(3); # Rated per phase Voltage in Volts\n", "Ia = VA/(math.sqrt(3)*V); # Rated Armature Current in Amphere\n", "pfa = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "Er = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*(Ra+1j*Xs); # Induced EMF in Volts\n", "R_a = 90 + math.degrees(math.atan2(Er.imag,Er.real)); # Angle of R in Degree\n", "R = 160 * ( 1j * (R_a) * math.pi/180)**2; # (Line-line Voltage) Er = 11902.40V will get R from Air gap Characteristics\n", "A_n = A * ( 1j * (-pfa) * math.pi/180)**2;\n", "F = R - A_n; # Field Current required to produce the excitation EMF in Amphere\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.10: SOLUTION :-\");\n", "print \" a) Field Current required to produce the excitation EMF, F = %.2f A \"%(abs(F))\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.10: SOLUTION :-\n", " a) Field Current required to produce the excitation EMF, F = 22.74 A \n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.11 Page No : 322" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "print \" EXAMPLE : 6.11 : Given Data \";\n", "print \" Voc V 12 13 13.8 14.5 15.1 \";\n", "print \" IfA 175 200 225 250 275 \";\n", "V = 11*10**3; # Operating voltage of the Synchronous generator in Volts\n", "VA = 50*10**6; # VA rating of the Synchronous generator in Volts-Amphere\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "N = 1500; # Speed of the Synchronous generator in RPM\n", "If_scc = 200; # SCC test field Rated current in Amphere at rated Short circuit current\n", "If_zpf = 400; # ZPF test field Rated current in Amphere at rated voltage and rated current\n", "pf = 0.8; # Power factor (lagging)\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.30 & Page no:-413\n", "\n", "Vt = V/math.sqrt(3); # Rated per phase Voltage in Volts\n", "Ia = VA/(math.sqrt(3)*V); # Rated Armature Current in Amphere\n", "pfa = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "O = 13000; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413\n", "Xs = O/(math.sqrt(3)*Ia); # Synchronous reactance per phase in Ohms\n", "BC = 4000; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413\n", "Xl = BC/(math.sqrt(3)*Ia ); # Per phase leakage reactance in Ohms\n", "\n", "# For Case (a) General (ZPF) Method\n", "\n", "Er_a = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*(1j*Xl); # Induced EMF in Volts\n", "R_a = 220; A_a = 200; #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 13776V will get R_a & A_a value Respectively from SCC (Figure6.30 & Page no:-403)\n", "angle_a = 140.3; # Angle between R_a & A_a = 90'+13.43'+36.87' = 140.3'\n", "F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a))))); # From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere\n", "Eo_a = 20000; # Corresponding to field current F_a = 470.90 A the open circuit EMF from OCC is 20000 V (Figure6.30 & Page no:-413)\n", "r_a = 100*((Eo_a-V)/V); # Percentage regulation \n", "\n", "\n", "# For Case(b) EMF Method \n", "\n", "Er_b = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*(1j*Xs); # Induced Voltage in Volts \n", "F_b = 500; #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 21404 V will get 500A from SCC (Figure6.15 & Page no:-386)\n", "\n", "# For Case (c) MMF Method\n", "\n", "Er_c = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*0; # Induced Voltage in Volts ( Zero is multipied because Armature reismath.tance is zero (not mentioned))\n", "R_c = 160; A_c = 200; #From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 11000 V will get R_c & A_c value Respectively from SCC (Figure6.30 & Page no:-413)\n", "angle_c = 126.27; # Angle between R_c & A_c = 90'-0'+36.87' = 126.27' {can refer figure 6.21a at page no:-400}\n", "F_c = math.sqrt((R_c**2)+(A_c**2)-(2*R_c*A_c*math.degrees(math.cos(math.radians(angle_c))))); # From phasor diagram {can refer figure 6.21a at page no:-400} the neccessary field excitation in Amphere\n", "\n", "\n", "# For Case (d) ASA Method\n", "\n", "Er_d = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*(1j*Xl); # Induced Voltage in Volts \n", "R_d = 220; A_d = 200; #From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 13800 V will get R_d & A_d value Respectively from SCC (Figure6.30 & Page no:-413)\n", "angle_d = 126.87; # Angle between R_d & A_d = 90'+36.87' = 126.87'{can refer figure 6.22a at page no:-40}\n", "F_d1 = math.sqrt((R_d**2)+(A_d**2)-(2*R_d*A_d*math.degrees(math.cos(math.radians(angle_d))))); # from Phasor diagram {can refer figure 6.2a at page no:-400 The neccessary field excitation in Amphere\n", "F_d = F_d1 + 30; # from Phasor diagram {can refer figure 6.2a at page no:-400 The Total neccessary field excitation in Amphere\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.11 : SOLUTION :-\") ;\n", "print \" a) Leakage Reactance, Xl = %.2f Ohms \"%(Xl)\n", "print \" b) Synchronous Reactance, Xs = %.2f Ohms \"%(Xs)\n", "print \" For Case a) General ZPF) Method Field Current required for maintaing the rated terminal voltage\\\n", " for rated kVA rating at %.2f Lagging Power factor , F = %.2f A \"%(pf,F_a)\n", "print \" For Case a) EMF Method Field Current required for maintaing the rated terminal voltage for rated\\\n", " kVA rating at %.2f Lagging Power factor , F = %.f A \"%(pf,F_b)\n", "print \" For Case a) MMF Method Field Current required for maintaing the rated terminal voltage for rated\\\n", " kVA rating at %.2f Lagging Power factor , F = %.2f A \"%(pf,F_c)\n", "print \" For Case a) ASA Method Field Current required for maintaing the rated terminal voltage for rated kVA\\\n", " rating at %.2f Lagging Power factor , F = %.f A \"%(pf,F_d)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- For Case a) General ZPF) Method a) Field Current required for maintaining the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = 470.90 A instead of %.2f A \"%(pf,F_a);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.11 : Given Data \n", " Voc V 12 13 13.8 14.5 15.1 \n", " IfA 175 200 225 250 275 \n", "EXAMPLE : 6.11 : SOLUTION :-\n", " a) Leakage Reactance, Xl = 0.88 Ohms \n", " b) Synchronous Reactance, Xs = 2.86 Ohms \n", " For Case a) General ZPF) Method Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1991.92 A \n", " For Case a) EMF Method Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 500 A \n", " For Case a) MMF Method Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1494.97 A \n", " For Case a) ASA Method Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1795 A \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- For Case a) General ZPF) Method a) Field Current required for maintaining the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 470.90 A instead of 1991.92 A \n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.12 Page No : 324" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "\n", "print \" EXAMPLE : 6.12 : Given Data \";\n", "print \" VocV 175 250 280 300 330 350 370 380 \";\n", "print \" IfA 10 17 20 23 30 38 50 60 \";\n", "print \" VzpfV - - - 0 130 210 265 280 \";\n", "V = 433; # Operating voltage of the Alternator in Volts\n", "N = 3000; # speed of the Alternator in RPM\n", "VA = 20*10**3; # VA rating of the Alternator in Volts-Amphere\n", "f = 50; # Operating Frequency of the Alternator in Hertz\n", "pf = 0.8; # Power factor (lagging)\n", "\n", "\n", "# CALCULATIONS\n", "# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.35 & Page no:-420\n", "\n", "Vt = V/math.sqrt(3); # Rated per phase Voltage in Volts\n", "Ia = VA/(math.sqrt(3)*V); # Rated Armature Current in Amphere\n", "pfa = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "O = 298; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413\n", "Xs = O/(math.sqrt(3)*Ia); # Synchronous reactance per phase in Ohms\n", "BC = 70; # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413\n", "Xl = BC/(math.sqrt(3)*Ia ); # Per phase leakage reactance in Ohms\n", "E = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*(1j*Xs); # Induced EMF in Volts umath.sing EMF Method\n", "c = 380-60; # The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation\n", "y = 694.50; # The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation\n", "# Extrapolation Equation is y = (x*(380-370)/(60-50))*c\n", "x = y - c; # The required field current in Amphere\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.12 : SOLUTION :-\") ;\n", "print \" a) Leakage Reactance, Xl = %.2f Ohms \"%(Xl)\n", "print \" b) Synchronous Reactance, Xs = %.2f Ohms \"%(Xs)\n", "print \" c) Field Current required for maintaing the rated terminal voltage for rated kVA\\\n", " rating at %.2f Lagging Power factor , F = %.2f A \"%(pf,x)\n", "\n", " \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " EXAMPLE : 6.12 : Given Data \n", " VocV 175 250 280 300 330 350 370 380 \n", " IfA 10 17 20 23 30 38 50 60 \n", " VzpfV - - - 0 130 210 265 280 \n", "EXAMPLE : 6.12 : SOLUTION :-\n", " a) Leakage Reactance, Xl = 1.52 Ohms \n", " b) Synchronous Reactance, Xs = 6.45 Ohms \n", " c) Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 374.50 A \n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.13 Page No : 328" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "\n", "V = 400; # Operating voltage of the Synchronous generator in Volts\n", "VA = 60*10**3; # VA rating of the Synchronous generator in Volts-Amphere\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "xd = 1.5; # Direct axis reactances in Ohms\n", "xq = 0.6; # Quadrature axis reactances in Ohms\n", "\n", "\n", "# CALCULATIONS\n", "\n", "I = VA/(math.sqrt(3)*V); # Rated current in Amphere\n", "v = V/math.sqrt(3); # Rated Phase Votage in Volts\n", "\n", "# For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)\n", "\n", "pf_a = 0.8; # Power factor\n", "pfa_a = math.degrees(math.acos(math.radians(pf_a))); # Power factor angle in deg\n", "pa_a = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_a))))/(v+I*xq*math.degrees(math.sin(math.radians(pfa_a)))))); # Power angle in deg\n", "Iq_a = I*math.cos(math.radians(pfa_a+pa_a)); # Current in Amphere\n", "Id_a = I*math.sin(math.radians(pfa_a+pa_a)); # Current in Amphere\n", "Eo_a = math.sqrt((v+Id_a*xd*math.degrees(math.cos(math.radians(pa_a)))-Iq_a*xq*math.degrees(math.sin(math.radians(pa_a))))**2 + (Id_a*xd*math.degrees(math.sin(math.radians(pa_a)))+Iq_a*xq*math.degrees(math.cos(math.radians(pa_a))))**2); # Induced EMF in Volts\n", "pr_a = ((Eo_a-v)/v)*100; # Percentage regulation\n", "\n", "# For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)\n", "\n", "pf_b = 1.0; # Power factor\n", "pfa_b= math.degrees(math.acos(math.radians(pf_b))); # Power factor angle in deg\n", "pa_b = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_b))))/(v+I*xq*math.degrees(math.sin(math.radians(pfa_b)))))); # Power angle in deg\n", "Iq_b = I*math.cos(math.radians(pfa_b+pa_b));\n", "Id_b = I*math.sin(math.radians(pfa_b+pa_b));\n", "Eo_b = math.sqrt((v+Id_b*xd*math.degrees(math.cos(math.radians(pa_b)))-Iq_b*xq*math.degrees(math.sin(math.radians(pa_b))))**2 + (Id_b*xd*math.degrees(math.sin(math.radians(pa_b)))+Iq_b*xq*math.degrees(math.cos(math.radians(pa_b))))**2); # Induced EMF in Volts\n", "pr_b = ((Eo_b-v)/v)*100; # Percentage regulation\n", "\n", "# For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)\n", "\n", "pf_c = 0.8; # Power factor\n", "pfa_c = math.degrees(math.acos(math.radians(pf_c))); # Power factor angle in deg\n", "pa_c = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_c))))/(v-I*xq*math.degrees(math.sin(math.radians(pfa_c)))))); # Power angle in deg\n", "Iq_c = I*math.cos(math.radians(pfa_c-pa_c));\n", "Id_c = I*math.sin(math.radians(pfa_c-pa_c));\n", "Eo_c = math.sqrt((v-Id_c*xd*math.degrees(math.cos(math.radians(pa_c)))-Iq_c*xq*math.degrees(math.sin(math.radians(pa_c))))**2 + (-Id_c*xd*math.degrees(math.sin(math.radians(pa_c)))+Iq_c*xq*math.degrees(math.cos(math.radians(pa_c))))**2); # Induced EMF in Volts\n", "pr_c = ((Eo_c-v)/v)*100; # Percentage regulation\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.13: SOLUTION :-\");\n", "print \" For Case a) 0.80 lagging Power factor Induced EMF, EMF = %.2f V \"%(Eo_a)\n", "print \" Power angle = %.3f degree \"%(pa_a)\n", "print \" Percenatge Regulation, R = %.1f Percenatge \"%(pr_a) \n", "print \" For Case b) Unity Power factor Induced EMF, EMF = %.2f V \"%(Eo_b)\n", "print \" Power angle = %.2f degree \"%(pa_b)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(pr_b) \n", "print \" For Case c) 0.80 leading Power factor Induced EMF, EMF = %.2f V \"%(Eo_c)\n", "print \" Power angle = %.2f degree \"%(pa_c)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(pr_c) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.13: SOLUTION :-\n", " For Case a) 0.80 lagging Power factor Induced EMF, EMF = 7673.86 V \n", " Power angle = 0.742 degree \n", " Percenatge Regulation, R = 3222.9 Percenatge \n", " For Case b) Unity Power factor Induced EMF, EMF = 7673.84 V \n", " Power angle = 0.93 degree \n", " Percenatge Regulation, R = 3222.87 Percenatge \n", " For Case c) 0.80 leading Power factor Induced EMF, EMF = 7212.02 V \n", " Power angle = -0.87 degree \n", " Percenatge Regulation, R = 3022.90 Percenatge \n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.14 Page No : 332" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "\n", "v = 1.0; # Operating voltage of the Synchronous generator in pu\n", "xd = 1.0; # Direct axis reactances in pu\n", "xq = 0.5; # Quadrature axis reactances in pu\n", "I = 1.0; # Rated current in pu\n", "\n", "\n", "# CALCULATIONS\n", "\n", "# For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)\n", "\n", "pf_a = 0.8; # Power factor\n", "pfa_a = math.degrees(math.acos(math.radians(pf_a))); # Power factor angle in deg\n", "pa_a = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_a))))/(v+I*xq*math.degrees(math.sin(math.radians(pfa_a)))))); # Power angle in deg\n", "Iq_a = I*math.cos(math.radians(pfa_a+pa_a));\n", "Id_a = I*math.sin(math.radians(pfa_a+pa_a));\n", "Eo_a = math.sqrt((v+Id_a*xd*math.degrees(math.cos(math.radians(pa_a)))-Iq_a*xq*math.degrees(math.sin(math.radians(pa_a))))**2 + (Id_a*xd*math.degrees(math.sin(math.radians(pa_a)))+Iq_a*xq*math.degrees(math.cos(math.radians(pa_a))))**2); # Induced EMF in Volts\n", "pr_a = ((Eo_a-v)/v)*100; # Percentage regulation\n", "\n", "# For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)\n", "\n", "pf_b = 1.0; # Power factor\n", "pfa_b= math.degrees(math.acos(math.radians(pf_b))); # Power factor angle in deg\n", "pa_b = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_b))))/(v+I*xq*math.degrees(math.sin(math.radians(pfa_b)))))); # Power angle in deg\n", "Iq_b = I*math.cos(math.radians(pfa_b+pa_b));\n", "Id_b = I*math.sin(math.radians(pfa_b+pa_b));\n", "Eo_b = math.sqrt((v+Id_b*xd*math.degrees(math.cos(math.radians(pa_b)))-Iq_b*xq*math.degrees(math.sin(math.radians(pa_b))))**2 + (Id_b*xd*math.degrees(math.sin(math.radians(pa_b)))+Iq_b*xq*math.degrees(math.cos(math.radians(pa_b))))**2); # Induced EMF in Volts\n", "pr_b = ((Eo_b-v)/v)*100; # Percentage regulation\n", "\n", "# For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)\n", "\n", "pf_c = 0.8; # Power factor\n", "pfa_c = math.degrees(math.acos(math.radians(pf_c))); # Power factor angle in deg\n", "pa_c = math.degrees(math.atan((I*xq*math.degrees(math.cos(math.radians(pfa_c))))/(v-I*xq*math.degrees(math.sin(math.radians(pfa_c)))))); # Power angle in deg\n", "Iq_c = I*math.cos(math.radians(pfa_c-pa_c));\n", "Id_c = I*math.sin(math.radians(pfa_c-pa_c));\n", "Eo_c = math.sqrt((v-Id_c*xd*math.degrees(math.cos(math.radians(pa_c)))-Iq_c*xq*math.degrees(math.sin(math.radians(pa_c))))**2 + (-Id_c*xd*math.degrees(math.sin(math.radians(pa_c)))+Iq_c*xq*math.degrees(math.cos(math.radians(pa_c))))**2); # Induced EMF in Volts\n", "pr_c = ((Eo_c-v)/v)*100; # Percentage regulation\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.14: SOLUTION :-\");\n", "print \" For Case a) 0.80 lagging Power factor Induced EMF, EMF = %.4f V \"%(Eo_a)\n", "print \" Power angle = %.1f degree \"%(pa_a)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(pr_a) \n", "print \" For Case b) Unity Power factor Induced EMF, EMF = %.2f V \"%(Eo_b)\n", "print \" Power angle = %.2f degree \"%(pa_b)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(pr_b) \n", "print \" For Case c) 0.80 leading Power factor Induced EMF, EMF = %.4f V \"%(Eo_c)\n", "print \" Power angle = %.2f degree \"%(pa_c)\n", "print \" Percenatge Regulation, R = %.2f Percenatge \"%(pr_c) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.14: SOLUTION :-\n", " For Case a) 0.80 lagging Power factor Induced EMF, EMF = 58.2957 V \n", " Power angle = 0.8 degree \n", " Percenatge Regulation, R = 5729.57 Percenatge \n", " For Case b) Unity Power factor Induced EMF, EMF = 58.30 V \n", " Power angle = 0.97 degree \n", " Percenatge Regulation, R = 5729.56 Percenatge \n", " For Case c) 0.80 leading Power factor Induced EMF, EMF = 56.2959 V \n", " Power angle = -0.83 degree \n", " Percenatge Regulation, R = 5529.59 Percenatge \n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.15 Page No : 337" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import real,imag\n", "\n", "# GIVEN DATA\n", "\n", "\n", "If = 1.25; # Given that rated voltage at air gap line for this field current in pu\n", "IF = 0.75; # Rated current in SC test for this field current in pu\n", "Ia = 1.0; # Rated current in Per unit\n", "pf = 0.8; # Power factor\n", "V = 1.0; # Rated Volatge in pu\n", "\n", "\n", "# CALCULATIONS\n", "\n", "pfa = math.degrees(math.acos(math.radians(pf))); # Power factor angle in deg\n", "Voc = (V*IF)/If; # Open circuit volatge in pu\n", "xs = Voc/Ia; # Syncronous reactance in pu\n", "E = V + Ia*(math.degrees(math.cos(math.radians(pfa)))-(1j)*math.degrees(math.sin(math.radians(pfa))))*(1j*xs); # Induced EMF in pu\n", "a = abs(E)*If;\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.15: SOLUTION :-\");\n", "print \" Induced EMF, E = %.2f < %.2f pu \"%(abs(E),math.degrees(math.atan2(E.imag,E.real)))\n", "print \" The field current required for %.2f pu voltage on air gap line %.1f pu \"%(abs(E),a)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.15: SOLUTION :-\n", " Induced EMF, E = 35.38 < 0.78 pu \n", " The field current required for 35.38 pu voltage on air gap line 44.2 pu \n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.16 Page No : 340" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "\n", "V = 440; # Operating voltage of the alternator in Volts\n", "VA = 20*10**3; # VA rating of the alternator in Volts-Amphere\n", "f = 50; # Operating Frequency of the alternator in Hertz\n", "N = 3000; # Rotation of the alternator in RPM\n", "Ra = 0.0; # Armature reismath.tance in Ohms\n", "xl = 0.6; # Armature reactances in Ohms\n", "pf = 0.8; # Power factor lagging\n", "pfa = math.degrees(math.acos(math.radians(pf))); # ower factor angle in deg\n", "p = (120*f)/N; # Number of poles\n", "w = (2*math.pi*f); # Rotation speed in Radians per second\n", "v = V/math.sqrt(3); # Rated phase voltage in Volts\n", "I = VA/(math.sqrt(3)*V); # Rated curent in Amphere\n", "If = I; # Given field current = armature current from SCC test in Amphere\n", "E = 16*If; # Open-circuit EMF at field current in Volts given from Equation E = 16If refer page no. 431\n", "xs = E/(If*math.sqrt(3)); # Synchronous reactance in Ohms\n", "Eo = math.sqrt((v+I*xs*math.degrees(math.sin(math.radians(pfa))))**2 + (I*xs*math.degrees(math.cos(math.radians(pfa))))**2); # Induced EMF in Volts\n", "pa = math.degrees(math.atan(193.98/399.49)); # From above equation Eo\n", "pr = ((Eo-v)/v)*100; # Percent regulation \n", "P = (3*v*Eo*math.degrees(math.sin(math.radians(pa))))/(xs*1000); # Power inKilo-Watts\n", "T = (P*1000)/w; # Torque devolped in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.16: SOLUTION :-\");\n", "print \" Induced EMF, EMF = %.f V \"%(Eo)\n", "print \" Power angle = %.2f degree \"%(pa)\n", "print \" Power, P = %.3f kW \"%(P) \n", "print \" Counter Torque, T = %.2f N-m \"%(T)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- a) Induced EMF, EMF = 471 V instead of %.f V \"%(Eo)\n", "print \" b) Power angle = 18.05 degree instead of %.2f degree \"%(pa)\n", "print \" c) Power, P = 12.003 kV instead of %.3f kW \"%(P) \n", "print \" d) Counter Torque, T = 38.23 N-m instead of %.2f N-m \"%(T)\n", "print \" From Calculation of the Induced EMFE), rest all the Calculated values in the TEXT BOOK is WRONG because\\\n", " of the Induced EMFE) value is WRONGLY calculated and the same used for the further Calculation part \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.16: SOLUTION :-\n", " Induced EMF, EMF = 14144 V \n", " Power angle = 25.90 degree \n", " Power, P = 29202.939 kW \n", " Counter Torque, T = 92955.84 N-m \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- a) Induced EMF, EMF = 471 V instead of 14144 V \n", " b) Power angle = 18.05 degree instead of 25.90 degree \n", " c) Power, P = 12.003 kV instead of 29202.939 kW \n", " d) Counter Torque, T = 38.23 N-m instead of 92955.84 N-m \n", " From Calculation of the Induced EMFE), rest all the Calculated values in the TEXT BOOK is WRONG because of the Induced EMFE) value is WRONGLY calculated and the same used for the further Calculation part \n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.17 Page No : 342" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "V = 440.; # Operating voltage of the Synchronous Motor in pu\n", "E = 200.; # Induced voltage in Volts\n", "xs = 8.0; # Synchronous reactance in Ohms\n", "f = 50.; # Frequency in Hertz\n", "pa = 36.; # Power angle in degree\n", "\n", "\n", "# CALCULATIONS\n", "\n", "v = V/math.sqrt(3); # Rated phase voltage in Volts\n", "ws = 2*math.pi*f; # Synchronous speed in Radians per second\n", "# To calculate the power factor angle refer page no 438 n figure 6.50\n", "# Since E*math.cos(delta) < v so Power factor is lagging, let power factor angle be theta from ohasor diagram figure 6.50:- page no. 438\n", "# v = E*math.cos(delta) + I*xs*math.sin(theta), I*math.sin(theta) = (254-0.809*200)/8 = 11.525\n", "# Similarly, E*math.sin(delta) = I*xs*math.cos(theta), I*math.cos(theta) = (200*0.59)8 = 14.70\n", "# From above two equations, math.tan(theta) = 0.784\n", "theta = -38.1; # Power factor angle in degree (minus sign because of lagging)\n", "pf = math.degrees(math.cos(math.radians(theta))); # Power factor lagging\n", "I = 14.7/math.degrees(math.cos(math.radians(theta))); # Line current in Amphere (I*math.cos(theta) = 14.7) \n", "p = 3 * v * 14.7; # Input to motor in watts ( p = 3*V*I*math.cos(theta), I*math.cos(theta) = 14.7) \n", "P = (3*E*v*math.degrees(math.sin(math.radians(pa))))/(xs*1000); # Power in Kilo-watts\n", "T = (P*1000)/ws; # Torque in Newton-meter\n", "# For Power factor unity\n", "# let the current will be I2, thus 3*v*I2 = 3*v*I*math.cos(theta) , I2 = I*math.cos(theta) = 14.10 A\n", "# let ecitation will be E2, thus v = E2*math.cos(delta2) and E2*math.sin(delta2) = I2*xs, E2*math.cos(delta2) = 254 and E2*math.sin(delta2) = 117.60, by solving these two equations we get E2 = math.sqrt(254**2+117.6**2) = 279.90 V and delta2 = math.degrees(math.atan(117.6/254) = 24.84 degree\n", "E2 = 279.90;\n", "delta2 = 24.84;\n", "P_1 = (3*v*E2*math.degrees(math.sin(math.radians(delta2))))/(xs*1000); # Power in kilo-watts\n", "T_1 = (P_1*1000)/ws; # Torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.17: SOLUTION :-\");\n", "print \" a) Line current, I = %.2f A \"%(I)\n", "print \" b) Power factor angle = %.1f degree \"%(theta)\n", "print \" c) Power , P = %.3f kW \"%(P)\n", "print \" d) Torque , T = %.2f N-m \"%(T)\n", "print \" e) Power factor = %.2f lagging \"%(pf)\n", "print \" To make the Power factor to UNITY requirements are:- a) Excitation EMF, E = %.2f V \"%(E2)\n", "print \" b) Power angle = %.2f degree \"%(delta2)\n", "print \" c) Power , P = %.3f kW \"%(P_1)\n", "print \" d) Torque , T = %.2f N-m \"%(T_1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.17: SOLUTION :-\n", " a) Line current, I = 0.33 A \n", " b) Power factor angle = -38.1 degree \n", " c) Power , P = 641.645 kW \n", " d) Torque , T = 2042.42 N-m \n", " e) Power factor = 45.09 lagging \n", " To make the Power factor to UNITY requirements are:- a) Excitation EMF, E = 279.90 V \n", " b) Power angle = 24.84 degree \n", " c) Power , P = 641.781 kW \n", " d) Torque , T = 2042.85 N-m \n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.18 Page No : 347" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "\n", "v = 1100; # Operating voltage of the Synchronous Motor in Volts\n", "p = 4; # Total number of Poles\n", "m = 3; # number of phase\n", "xs = 5.0; # Synchrouons reactances in Ohms\n", "f = 50; # Frequency in Hertz\n", "delta = 9; # Power angle in degree\n", "p_hp = 150; # Motor delivering power in HP\n", "eta = 89/100; # Efficiency of motor\n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "ws = (4*math.pi*f)/p; # Synchronous speed in Radians per second\n", "# We have (746*150)/0.89) = 125730.34 W = math.sqrt(3)*1100*I*math.cos(theta) refer page no. 440, thus we get I*math.cos(theta) = 12530.34/(1100*math.sqrt(3)) = 65.99 and E*math.sin(delta) = I*xs*math.cos(theta)\n", "E = (xs*65.99)/math.degrees(math.sin(math.radians(delta))); # Exitation EMF per phase in Volts\n", "# math.since E*math.cos(delta) > V, therefore the machine is over excited and power factor is leading, thus we get V = E*math.cos(delta) + I*xs*math.sin(theta), I*math.sin(theta) = (635.1-2109.2*math.cos(9)/5 = -289.586 and we have I*math.cos(theta) = 65.99 thus by solving these two equations we get theta = math.degrees(math.atan(-286.586/65.99) = 77.16 degre\n", "theta = 77.16; # Power factor angle in degree\n", "I = 65.99/math.degrees(math.cos(math.radians(theta))); # Current in Amphere\n", "pf = math.degrees(math.cos(math.radians(theta))); # Power factor leading\n", "P = (3*V*E*math.degrees(math.sin(math.radians(delta))))/(xs*1000); # Power in kilo-Watts\n", "T = (P*1000)/ws; # torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.18: SOLUTION :-\");\n", "print \" a) Excitation EMF, E = %.1f V \"%(E)\n", "print \" b) Line current, I = %.2f A \"%(I)\n", "print \" c) Power factor = %.3f leading \"%(pf)\n", "print \" d) Power , P = %.4f kW \"%(P)\n", "print \" e) Torque , T = %.2f N-m \"%(T)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- a) Power , P = 13.0667 kW instaed of %.4f kW \"%(P)\n", "print \" b) Torque , T = 83.22 N-m instaed of %.2f N-m \"%(T)\n", "print \" From Calculation of the PowerP), rest all the Calculated values in the TEXT BOOK is WRONG because of the\\\n", " PowerP) value is WRONGLY calculated and the same used for the further Calculation part \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.18: SOLUTION :-\n", " a) Excitation EMF, E = 36.8 V \n", " b) Line current, I = 5.18 A \n", " c) Power factor = 12.733 leading \n", " d) Power , P = 125.7278 kW \n", " e) Torque , T = 800.41 N-m \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- a) Power , P = 13.0667 kW instaed of 125.7278 kW \n", " b) Torque , T = 83.22 N-m instaed of 800.41 N-m \n", " From Calculation of the PowerP), rest all the Calculated values in the TEXT BOOK is WRONG because of the PowerP) value is WRONGLY calculated and the same used for the further Calculation part \n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.19 Page No : 351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "v = 440; # Operating voltage of the Synchronous Motor in Volts\n", "p = 6; # Total number of Poles\n", "m = 3; # Number of phase\n", "xs = 5; # Synchrouons reactances per phase in Ohms\n", "f = 50; # Frequency in Hertz\n", "p_hp = 10; # Motor delivering power in HP\n", "loss = 1000; # Total iron,copper and friction losses in Watts\n", "pf = 0.8; # Power factor lagging\n", "I = 10; # Motor drawing current in Amphere at 0.8 PF lagging\n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "ws = (4*math.pi*f)/p; # Synchronous speed in Radians per second\n", "theta = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "Po = p_hp*746; # Output power in Watts\n", "Pi = Po+loss; # Input power in Watts\n", "eta = (Po/Pi)*100; # Efficiency\n", "# we have V = E*math.cos(delta) - I*xs*math.sin(theta), 254 = E*math.cos(delta) - 5*10*0.6, so E*math.cos(delta) = 254 + 30 = 284 and E*math.sin(delta) = I*xs*math.cos(theta) = 5*10*0.8 = 40 by solving these two equations we get delta = math.degrees(math.atan(40/284) = 8.01 degree\n", "delta = 8.01; # Power angle in degree\n", "E = 40/math.degrees(math.sin(math.radians(delta))); # Induced EMF per phase in Volts\n", "P = (3*V*E*math.degrees(math.sin(math.radians(delta))))/(xs*1000); # Power in Kilo-watts\n", "T = (P*1000)/ws; # Torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.19: SOLUTION :-\");\n", "print \" a) Efficiency, eta = %.2f Percent \"%(eta)\n", "print \" b) Induced EMF, E = %.f V per phase and Power Torque) angle = %.2f degree \"%(E,delta)\n", "print \" c) Power , P = %.4f kW \"%(P)\n", "print \" d) Torque , T = %.2f N-m \"%(T) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.19: SOLUTION :-\n", " a) Efficiency, eta = 0.00 Percent \n", " b) Induced EMF, E = 5 V per phase and Power Torque) angle = 8.01 degree \n", " c) Power , P = 6.0968 kW \n", " d) Torque , T = 58.22 N-m \n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.20 Page No : 352" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "v = 11*10**3; # Operating voltage of the Synchronous Motor in Volts\n", "p = 4; # Total number of Poles\n", "m = 3; # number of phase\n", "xs = 7; # Synchrouons reactances per phase in Ohms\n", "f = 50; # Frequency in Hertz\n", "KVA = 1500; # KVA rating (whole)\n", "kva = 500; # Each case KVA rating \n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "ws = (4*math.pi*f)/p; # Synchronous speed in Radians per second\n", "I = (math.sqrt(3)*kva)/v; # Phase Current in Amphere\n", "\n", "# For Case (a) 0.8 pf lagging\n", "\n", "pf_a = 0.8; # Power factor lagging\n", "pfa_a = math.degrees(math.acos(math.radians(pf_a))); # Power factor angle in degree\n", "# we have E*math.cos(delta) = V - I*xs*math.sin(theta) = 6351-78.73*7*0.6 = 6020.334 and E*math.sin(delta) = I*xs*math.cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6036.46 V and delta = math.degrees(math.atan(440.888/6020.334) = 4.19 degree\n", "E_a = 6036.46; # Induced Voltage in Volts\n", "delta_a = 4.19; # Power angle in degree\n", "P_a = (3*V*E_a*math.degrees(math.sin(math.radians(delta_a))))/(xs*10**6); # Power in Mega-Watts\n", "T_a = (P_a*10**6)/ws; # Torque in Newton-meter\n", "\n", "# For Case (b) 0.8 pf leading\n", "\n", "pf_b = 0.8; # Power factor lagging\n", "pfa_b = math.degrees(math.acos(math.radians(pf_b))); # Power factor angle in degree\n", "# we have E*math.cos(delta) = V + I*xs*math.sin(theta) = 6351+78.73*7*0.6 = 6681.666 and E*math.sin(delta) = I*xs*math.cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6696.2 V and delta = math.degrees(math.atan(440.888/6681.666) = 3.78 degree\n", "E_b = 6696.2; # Induced Voltage in Volts\n", "delta_b = 3.78; # Power angle in degree\n", "P_b = (3*V*E_b*math.degrees(math.sin(math.radians(delta_b))))/(xs*10**6); # Power in Mega-Watts\n", "T_b = (P_b*10**6)/ws; # Torque in Newton-meter\n", "\n", "# For Case (c) UPf\n", "\n", "pf_c = 1.0; # Power factor lagging\n", "pfa_c = math.degrees(math.acos(math.radians(pf_c))); # Power factor angle in degree\n", "# we have E*math.cos(delta) = V = 6351 and E*math.sin(delta) = I*xs = 78.73*7 = 551.11 thus we get by sloving these two equatins E = 6374.9 V and delta = math.degrees(math.atan(551.11/6351) = 4.96 degree\n", "E_c = 6374.9; # Induced Voltage in Volts\n", "delta_c = 4.96; # Power angle in degree\n", "P_c = (3*V*E_c*math.degrees(math.sin(math.radians(delta_c))))/(xs*10**6); # Power in Mega-Watts\n", "T_c = (P_c*10**6)/ws; # Torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.20: SOLUTION :-\");\n", "print \" For Case a) 0.80 pf lagging :- a) Induced EMF, E = %.2f V \"%(E_a)\n", "print \" b) Power , P = %.1f MW \"%(P_a)\n", "print \" c) Torque , T = %.2f N-m \"%(T_a)\n", "print \" For Case b) 0.80 pf leading :- a) Induced EMF, E = %.1f V \"%(E_b)\n", "print \" b) Power , P = %.3f MW \"%(P_b)\n", "print \" c) Torque , T = %.2f N-m \"%(T_b)\n", "print \" For Case a) UPf :- a) Induced EMF, E = %.1f V \"%(E_c)\n", "print \" b) Power , P = %.2f MW \"%(P_c)\n", "print \" c) Torque , T = %.f N-m \"%(T_c)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.20: SOLUTION :-\n", " For Case a) 0.80 pf lagging :- a) Induced EMF, E = 6036.46 V \n", " b) Power , P = 68.8 MW \n", " c) Torque , T = 437869.39 N-m \n", " For Case b) 0.80 pf leading :- a) Induced EMF, E = 6696.2 V \n", " b) Power , P = 68.843 MW \n", " c) Torque , T = 438268.77 N-m \n", " For Case a) UPf :- a) Induced EMF, E = 6374.9 V \n", " b) Power , P = 85.95 MW \n", " c) Torque , T = 547202 N-m \n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.21 Page No : 355" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "v = 440; # Operating voltage of the Synchronous Motor in Volts\n", "f = 50; # Operating Frequency of the Synchronous Motor in Hertz\n", "xd = 10; # Direct axis reactances in Ohms\n", "xq = 7.0; # Quadrature axis reactances in Ohms\n", "p = 6; # Total number of Poles\n", "pf = 0.8; # Power factor lagging\n", "i = 10; # Motor drawing current in Amphere \n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "ws = (4*math.pi*f)/p; # Synchronous speed in Radians per second\n", "theta = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "I = 10*(math.degrees(math.cos(math.radians(theta)))+(1j*math.degrees(math.sin(math.radians(theta))))); # Motor drawing current in Amphere at 0.8 PF leading\n", "delta = math.degrees(math.atan( (i*xq*math.degrees(math.cos(math.radians(theta)))) / (V+i*xq*math.degrees(math.sin(math.radians(theta)))) )); # Power angle for motoring mode in degree\n", "# delta = atand((i*xq*cosd(theta))/(V+i*xq*sind(theta))); // Power angle for motoring mode in degree\n", "Iq = i*math.cos(math.radians(theta+delta)); # Current in Amphere\n", "Id = i*math.sin(math.radians(theta+delta)); # Current in Amphere\n", "Eo = V*math.degrees(math.cos(math.radians(delta))) + Id*xd; # Induced EMF in Volts\n", "P = ((3*V*Eo*math.degrees(math.sin(math.radians(delta))))/xd)+(3*V**2*((1/xq)-(1/xd))*math.sin(math.radians(2*delta)))/2; # Power in Watts\n", "T = ((3*V*Eo*math.degrees(math.sin(math.radians(delta))))/(xd*ws))+(3*V**2*((1/xq)-(1/xd))*math.sin(math.radians(2*delta)))/(2*ws); # Torque in Newton-meter \n", "\n", "# when the machine is running as alternator, the magnitude of induced EMF = 323.38V. Let the new current will be Inew at lagging power factor thetanew. Now torque angle is 10.71 deg from phasor diagram Figure 6.51 and page no. 444 we get V+Id*xd*math.cos(delta)-Iq*xq*math.sin(delta) = Eo*math.cos(delta), 254+9.825*Id-1.3Iq = 317.75, 9.825*Id-1.3*Iq = 63.75, 7.56*Id-Iq = 49 and we have Id*xd*math.sin(delta)+Iq*xq*math.cos(math.radianselta) = Eo*math.sin(delta), 1.85*Id+6.88*Iq = 60.1, 0.27*Id+Iq = 8.74 by solving these two equations we get Idnew = 123.85/10.095 = 12.27A and Iqnew = 5.43A\n", "Iqnew = 5.43; # New current in Amphere\n", "Idnew = 12.27; # New current in Amphere\n", "Inew = math.sqrt(Idnew**2 + Iqnew**2); # New total Current in Amphere\n", "# We know that torque angle, math.tan(delta) = (I*xd*math.cos(theta))/(V+I*Xq*math.sin(theta)) so by calutaion for new power factor angle thetanew we get, math.tan(10.17) = (13.42*7*math.cos(thetanew))/(254+13.42*7*math.sin(thetanew)), 0.189(254+13.42*7*math.sin(thetanew) = 13.42*7*math.cos(thetanew), 48-93.94math.cos(thetanew)+17.75*math.sin(thetanew) = 0 by solving this equatuon we gwt thetanew = 49.5 lagging\n", "thetanew = 49.5; # New power factor angle in degree\n", "pfnew = math.degrees(math.cos(math.radians(thetanew))); # Power factor lagging\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.21: SOLUTION :-\");\n", "print \" a) Induced EMF, E = %.2f V \"%(Eo)\n", "print \" b) Power Torque) angle = %.2f degree \"%(delta)\n", "print \" Power , P = %.2f W \"%(P)\n", "print \" Torque , T = %.2f N-m \"%(T)\n", "print \" c) when the machine is running as alternator requirements are:- New Current = %.2f A\"%(Inew)\n", "print \" Power factor = %.3f lagging \"%(pfnew)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.21: SOLUTION :-\n", " a) Induced EMF, E = 14653.83 V \n", " b) Power Torque) angle = 0.75 degree \n", " Power , P = 840567.62 W \n", " Torque , T = 8026.83 N-m \n", " c) when the machine is running as alternator requirements are:- New Current = 13.42 A\n", " Power factor = 37.211 lagging \n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.22 Page No : 359" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import imag,real\n", "\n", "# GIVEN DATA\n", "\n", "E1 = 1100 + (1j*0); # EMFs of two identicel synchronous Generators in Volts per phase \n", "E2 = 1100*(math.degrees(math.cos(math.radians(5)))-(1j*math.degrees(math.sin(math.radians(5))))); # EMF in Volts per phase \n", "Zl = 1.0 + (1j*1.0); # Load impedance in Ohms per phase\n", "Zs1 = 0.15 + (1j*2.1); # Synchronous impedance in Ohms per phase\n", "Zs2 = 0.2 + (1j*3.3); # Synchronous impedance in Ohms per phase\n", "f = 50; # Frequency in Hertz\n", "\n", "\n", "# CALCULATONS\n", "\n", "Ys1 = 1/Zs1; # Synchronous Admitmath.tance in Ohms per phase\n", "Ys2 = 1/Zs2; # Synchronous Admitmath.tance in Ohms per ohase\n", "Yl = 1/Zl; # Load Admitmath.tance in Ohms per ohase\n", "V = ((E1*Ys1)+(E2*Ys2))/(Yl+Ys2+Ys1); # Terminal Voltage in Volts per phase (From Millman's Theorem)\n", "I1 = (E1-V)/Zs1; # Individual current in Amphere per phase\n", "I2 = (E2-V)/Zs2; # Individual current in Amphere per phase\n", "P1 = abs(V)*abs(I1)*math.cos(math.radians(math.degrees(math.atan2(V.imag,V.real)))-math.degrees(math.atan2(I1.imag,I1.real))); # Per phase actice power in Watts\n", "P2 = abs(V)*abs(I2)*math.cos(math.radians(math.degrees(math.atan2(V.imag,V.real)))-math.degrees(math.atan2(I2.imag,I2.real))); # Per phase actice power in Watts\n", "Ic = (E2-E1)/(Zs1+Zs2); # No-load circulating current in Amphere per phase\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.22 : SOLUTION :-\");\n", "print \" a) Terminal Voltage per phase, V = %.2f < %.1f V \"%(abs(V),math.degrees(math.atan2(V.imag,V.real)))\n", "print \" b) Individual currents per phase, I1 = %.f < %.1f A I2 = %.1f < %.1f A \"%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)),abs(I2),math.degrees(math.atan2(I2.imag,I2.real)))\n", "print \" c) Per phase Active Power , P1 = %.f W P2 = %.1f W \"%(P1,P2)\n", "print \" d) No-load current per phase, Ic = %.2f < %.2f A \"%(abs(Ic),math.degrees(math.atan2(Ic.imag,Ic.real)))\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.22 : SOLUTION :-\n", " a) Terminal Voltage per phase, V = 14088.39 < -24.8 V \n", " b) Individual currents per phase, I1 = 6221 < 67.3 A I2 = 15122.8 < -86.1 A \n", " c) Per phase Active Power , P1 = 14868984 W P2 = -148034989.2 W \n", " d) No-load current per phase, Ic = 11444.43 < -91.38 A \n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.23 Page No : 364" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "p = 4; # Number of the poles in the Alternator\n", "f = 50; # Frequency in Hertz\n", "pkw = 500; # Alternator delivering load in kilo-watts\n", "pkwinc = 1000; # Generator increases its share of the common elictrical in kilo-watts\n", "Kj = 1.5; # Inertia acceleration coefficient for the combined prime mover-alternator in N-m/elec deg/second square\n", "Kd = 12; # Damping torque coefficient in N-m/elec deg/second \n", "delta1 = 9; # Initial value of the Power angle in degree\n", "\n", "\n", "# CALCULATIONS\n", "\n", "delta2 = (pkwinc/pkw)*delta1; # Final value (maximum value) of the Power angle in degree (considering Linear variation)\n", "ws = (4*math.pi*f)/p; # Rotational speed in Radians per second\n", "Ts = (pkw*1000)/ws; # Synchornizing torque at 500kW in N-m\n", "Ks = Ts/delta1; # Synchornizing torque cofficient at 500kW in N-m/elec-deg\n", "# Laplace transform of the swing Equation can be written as :- s**2 + ((Kd/Kj)*s) + (Ks/Kj) = 0, s**2 + (12/1.5)s + (353.86/1.5) = 0 and compring with the smath.degrees(math.atanard equation s**2 + s(2*zeta*Wn) + Wn**2 = 0 we get:- mentined below (refer page no. 454 and 455)\n", "Wn = math.sqrt(Ks/Kj); # Natural frequency of oscillations in Radians per second\n", "fn = Wn/(2*math.pi); # Frequency of natural oscillations in Hertz\n", "zeta = (1*Kd)/(2*Wn*Kj); # Damping ratio\n", "Wd = Wn*(math.sqrt(1-zeta**2)); # Frequency of damped oscillations in radians/s\n", "fd = Wd/(2*math.pi); # Frequency of damped oscillations in Hertz\n", "ts = 5/(zeta*Wn); # Settling time in second\n", "deltamax = delta1 + 1.42*(delta2-delta1); # The maximum overshoot for damping ratio of 0.2604 is about 42% the maximum appoximate value of the overshoot in terms of 1% tolearance band in Electrical degree\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.23: SOLUTION :-\");\n", "print \" a.1) Final value maximum value) of the Power angle considering Linear variation), delta2 = %.f degree \"%(delta2)\n", "print \" a.2) Natural frequency of oscillations, Ns = %.2f radians/s \"%(Wn)\n", "print \" a.3) Damping ratio, zeta = %.4f \"%(zeta)\n", "print \" a.4) Frequency of damped oscillations, Wd = %.2f radians/s \"%(Wd)\n", "print \" a.5) Settling time, ts = %.2f seconds \"%(ts)\n", "print \" b) The maximum overshoot for damping ratio of 0.2604 is about 42 percent the maximum appoximate value of the overshoot in terms of 1 percent tolearance band is, deltamax = %.2f degree \"%(deltamax)\n", "print \" FOR CASE C CANNOT BE DO IT IN THIS BECAUSE AS IT REQUIRES MATLAB SIMULINK \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.23: SOLUTION :-\n", " a.1) Final value maximum value) of the Power angle considering Linear variation), delta2 = 18 degree \n", " a.2) Natural frequency of oscillations, Ns = 15.36 radians/s \n", " a.3) Damping ratio, zeta = 0.2605 \n", " a.4) Frequency of damped oscillations, Wd = 14.83 radians/s \n", " a.5) Settling time, ts = 1.25 seconds \n", " b) The maximum overshoot for damping ratio of 0.2604 is about 42 percent the maximum appoximate value of the overshoot in terms of 1 percent tolearance band is, deltamax = 21.78 degree \n", " FOR CASE C CANNOT BE DO IT IN THIS BECAUSE AS IT REQUIRES MATLAB SIMULINK \n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.25 Page No : 365" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "v = 440; # Operating voltage of the Synchronous generator in Volts\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "m = 3; # Total number of Phase\n", "pf = 0.8; # Power factor lagging\n", "Il = 100; # Motor drawing current in Amphere \n", "xs = 2; # Synchronous reactances in Ohms\n", "delta = 20; # Power angle in degree\n", "P = 50*10**3; # Total Power developed by the motor in Watts\n", "Ppp = (50*10**3)/3; # Power developed by the motor per phase in Watts\n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "Eo = (Ppp*xs)/(3*V*math.degrees(math.sin(math.radians(delta)))); # Per phase Induced voltage in Volts\n", "# Let us assume thetam is Power factor angle in degree and Im is the Motor current now, from phasor diagram figure 6.67 page no. 465 we get, Eo*math.degrees(math.cos(math.radians(delta)) = V+Im*xs*math.degrees(math.sin(math.radians(thetam)), Im*math.degrees(math.sin(math.radians(thetam)) = ((383.84*math.degrees(math.cos(math.radians(20)))-254.03)/2 = 53.35 and Im*xs*math.degrees(math.cos(math.radians(thetam)) = Eo*math.sin(delta), Im*math.degrees(math.cos(math.radians(theta)) = ((383.84*math.degrees(math.sin(math.radians(20)))/2 = 65.60 by sloving these two equations we get Im = math.sqrt(65.60**2 + 53.35**2) = 84.56 A and thetam = math.degrees(math.atan(53.35/65.60) = 39.13 degree\n", "Im = math.sqrt(65.60**2 + 53.35**2); # Motor current in Amphere\n", "thetam = math.degrees(math.atan(53.35/65.60)); # Power factor angle in degree\n", "kVA = (math.sqrt(3)*V*Im*math.degrees(math.sin(math.radians(thetam))))/1000; # Rective kVA of the motor in kVAR\n", "thetal = math.degrees(math.acos(math.radians(pf))); # Load power factor angle in degree\n", "thetaR = math.degrees(math.atan((Im*math.degrees(math.sin(math.radians(thetam))) - Il*math.degrees(math.sin(math.radians(thetal)))) / (Im*math.degrees(math.cos(math.radians(thetam))) + Il*math.degrees(math.cos(math.radians(thetal)))))); # resultant Power factor angle in degree\n", "ovpf = math.degrees(math.cos(math.radians(thetaR))); # Overall Power factor lagging\n", "IR = math.sqrt((Im*math.degrees(math.sin(math.radians(thetam)))-Il*math.degrees(math.sin(math.radians(thetal))))**2 + (Im*math.degrees(math.cos(math.radians(thetam)))+Il*math.degrees(math.cos(math.radians(thetal))))**2); # resultant (magnitude) current in Amphere refer phasor diagram figure 6.69 page no. 467\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.25: SOLUTION :-\");\n", "print \" a) Rective kVA of the motor = %.3f kVAR \"%(kVA)\n", "print \" b) Overall Power factor of the load and motor = %.4f lagging and \"%(ovpf)\n", "print \" resultant magnitude) current = %.2f A \"%(IR)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.25: SOLUTION :-\n", " a) Rective kVA of the motor = 1344.961 kVAR \n", " b) Overall Power factor of the load and motor = 47.0232 lagging and \n", " resultant magnitude) current = 4677.18 A \n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.27 Page No : 367" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import real,imag\n", "\n", "\n", "# GIVEN DATA\n", "\n", "v = 440; # Operating voltage of the Synchronous generator in Volts\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "m = 3; # Total number of Phase\n", "xs = 5; # Synchronous reactances in Ohms\n", "Eo = 500; # Indduced Voltage in Volts per phase\n", "R1 = 0.1; # Circuit Parameter in Ohms\n", "R2 = 0.1; # Circuit Parameter in Ohms\n", "X1 = 1.55; # Circuit Parameter in Ohms\n", "X2 = 1.55; # Circuit Parameter in Ohms\n", "s = 0.03; # Slip\n", "P = 30*10**3; # Total Power developed by the motor in Watts\n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "Ii = V/math.sqrt((R1+R2/s)**2 + (X1+X2)**2); # Per phase induction motor current in Amphere\n", "thetal = math.degrees(math.atan((X1+X2)/(R1+R2/s))); # Power factor angle of the induction motor in degree\n", "pf = math.degrees(math.cos(math.radians(thetal))); # Power factor of the induction motor lagging\n", "# Let us assume thetam is leading Power factor angle in degree and Im is the synchronous Motor current now, from phasor diagram figure 6.70 page no. 469\n", "delta = math.asin(math.radians((xs*P)/(3*V*Eo))); # Power angle in degree\n", "# From phasor diagram figure 6.70 page no. 469 we have, Im*xs*math.cos(thetam) = Eo*math.sin(delta), Im*math.cos(delta) = ((500*math.sin(math.radians(23.18))/5 = 39.37 and Eo*math.degrees(math.cos(math.radians(delta)) = V+Im*xs*math.degrees(math.sin(math.radians(thetam)), Im*math.degrees(math.sin(math.radians(thetam)) = ((500*math.cos(math.radians(23.18))-254.03)/5 = 41.12 by sloving these two equations we get Im = math.sqrt(39.37**2 + 41.12**2) = 56.93 A and thetam = math.degrees(math.atan(41.12/39.37) = 46.25 degree\n", "Im = math.sqrt(39.37**2 + 41.12**2); # Motor current in Amphere\n", "thetam = math.degrees(math.atan(41.12/39.37)); # Power factor angle in degree\n", "kVA = (math.sqrt(3)*V*Im*math.degrees(math.sin(math.radians(thetam))))/1000; # Rective kVA of the motor in kVAR\n", "II = Ii * ( 1j * (-thetal) * math.pi/180)**2; # Induction Motor current in Amphere\n", "Im = Im * ( 1j * thetam * math.pi/180)**2; # Synchronous Motor current in Amphere\n", "It = II + Im; # Total per phase current in Amphere\n", "ovpf = math.degrees(math.cos(math.radians(math.degrees(math.atan2(It.imag,It.real))))); # Overall Power factor leading\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.27: SOLUTION :-\");\n", "print \" a) Reactive kVA of the motor = %.3f kVAR \"%(kVA)\n", "print \" b) Individual currents:- Induction Motor current, II = %.2f + i%.2f) A Synchronous Motor current,\\\n", " Im = %.2f + i%.2f) A \"%(II.real,II.imag,Im.real,Im.imag)\n", "print \" c) resultant overall) current = %.2f < %.2f A \"%(abs(It),math.degrees(math.atan2(It.imag,It.real)))\n", "print \" d) Overall Power factor = %.4f leading \"%(ovpf)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.27: SOLUTION :-\n", " a) Reactive kVA of the motor = 1036.641 kVAR \n", " b) Individual currents:- Induction Motor current, II = -29.62 + i-0.00) A Synchronous Motor current, Im = -37.09 + i0.00) A \n", " c) resultant overall) current = 66.71 < 180.00 A \n", " d) Overall Power factor = -57.2958 leading \n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.28 Page No : 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "\n", "V = 400; # Operating voltage of the Synchronous generator in Volts\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "xd = 12; # Direct axis reactances in Ohms\n", "xq = 5; # Quadrature axis reactances in Ohms\n", "delta = 15; # Power(Torque) angle in degree\n", "p = 2; # Number of the poles \n", "m = 3; # Number of the phase\n", "\n", "\n", "# CALCULATIONS\n", "\n", "v = V/math.sqrt(3); # Rated Phase Votage in Volts\n", "Ns = (120*f)/p; # Operating speed in RPM\n", "Ws = (2*math.pi*f)/(p/2); # Synchronous speed in radians/s\n", "T = (3*v**2*math.sin(math.radians(2*delta))/(2*Ws))*((1/xq)-(1/xd)); # Developed Torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.28: SOLUTION :-\");\n", "print \" a) Operating speed, Ns = %.f RPM \"%(Ns)\n", "print \" b) Developed Torque , T = %.2f N-m \"%(T)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.28: SOLUTION :-\n", " a) Operating speed, Ns = 3000 RPM \n", " b) Developed Torque , T = 0.00 N-m \n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.29 Page No : 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "f = 400; # Operating Frequency of the Synchronous generator in Hertz\n", "Ld = 50*10**-3; # Direct axis reactances in Henry\n", "Lq = 15*10**-3; # Quadrature axis reactances in Henry\n", "delta = 15; # Power(Torque) angle in degree\n", "p = 2; # Number of the poles \n", "m = 3; # Number of the phase\n", "I = 10; # Operating current in Amphere\n", "\n", "\n", "# CALCULATIONS\n", "\n", "Ns = (120*f)/p; # Operating speed in RPM\n", "Ws = (2*math.pi*f)/(p/2); # Synchronous speed in radians/s\n", "xd = 2*math.pi*f*Ld; # Direct axis reactances in reactance\n", "xq = 2*math.pi*f*Lq; # Quadrature axis reactances in reactance\n", "E1 = 0; # Induced EMF in Volts (Its ZERO beacuse when field winding current is zero)\n", "v = xq*I; # Applied voltage in Volts\n", "T = (3*v**2*math.sin(math.radians(2*delta))/(2*Ws))*((1/xq)-(1/xd)); # Developed Torque in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.29: SOLUTION :-\");\n", "print \" a) Operating speed, Ns = %.f RPM \"%(Ns)\n", "print \" b) Developed Torque , T = %.5f N-m \"%(T)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- a) xd = 12.56 instead of %.2f Ohms \"%(xd);\n", "print \" b) xq = 3.768 instead of %.2f Ohms \"%(xq);\n", "print \" c) v = 36.68 instead of %.2f V \"%(v);\n", "print \" d) T = 0.07875 instead of %.4f N-m \"%(T);\n", "print \" From Calculation of the d-axis and q-axis reactance xd and xq respectively),\\\n", " rest all the Calculated values in the TEXT BOOK is WRONG because of the d-axis and q-axis reactance xd and\\\n", " xq respectively) value is WRONGLY calculated and the same used for the further Calculation part \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.29: SOLUTION :-\n", " a) Operating speed, Ns = 24000 RPM \n", " b) Developed Torque , T = 0.78750 N-m \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- a) xd = 12.56 instead of 125.66 Ohms \n", " b) xq = 3.768 instead of 37.70 Ohms \n", " c) v = 36.68 instead of 376.99 V \n", " d) T = 0.07875 instead of 0.7875 N-m \n", " From Calculation of the d-axis and q-axis reactance xd and xq respectively), rest all the Calculated values in the TEXT BOOK is WRONG because of the d-axis and q-axis reactance xd and xq respectively) value is WRONGLY calculated and the same used for the further Calculation part \n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.30 Page No : 375" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "f = 50; # Operating Frequency of the Synchronous generator in Hertz\n", "p = 2; # Number of the poles \n", "Pt = 800; # Total loss in Watts\n", "Pr = 10; # Rotational loss in Watts\n", "\n", "\n", "# CALCULATIONS\n", "\n", "Ws = (4*math.pi*f)/p; # Synchronous speed in radians/s\n", "Ph = Pt-Pr; # Hysteresis loss refered to stator in Watts\n", "Th = Ph/Ws; # Torque at the shaft in Newton-meter\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.30: SOLUTION :-\");\n", "print \" a) Power at the shaft, Ph = %.f W \"%(Ph)\n", "print \" )b) Torque at the shaft , Th = %.2f N-m \"%(Th)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.30: SOLUTION :-\n", " a) Power at the shaft, Ph = 790 W \n", " )b) Torque at the shaft , Th = 2.51 N-m \n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.31 Page No : 379" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "# GIVEN DATA\n", "\n", "Pi = 2*10**6; # Power input in Volt-Amphere\n", "v = 6.6*10**3; # Operating voltage in Volts\n", "\n", "\n", "# CALCULATIONS\n", "\n", "I = Pi/(v*math.sqrt(3)); # Rated current in Amphere\n", "V = v/math.sqrt(3); # Phase voltage in Volts\n", "xs = v/(I*math.sqrt(3)); # Synchronous reactance in Ohms\n", "\n", "# For case (a) 0.8 pf lagging\n", "\n", "pf_a = 0.8; # Power factor\n", "pfa_a = math.degrees(math.acos(math.radians(pf_a))); # Power factor angle in degree\n", "a_a = (V + (I*xs*math.degrees(math.sin(math.radians(pfa_a)))));\n", "b_a = (I*xs*math.degrees(math.cos(math.radians(pfa_a))));\n", "E_a = math.sqrt(a_a**2 + b_a**2); # Induced EMF in Volts\n", "delta_a = math.degrees(math.atan(b_a/a_a)); # Torque (power) angle in degree\n", "P_a = (3*V*E_a*math.degrees(math.sin(math.radians(delta_a))))/(xs*10**6); # Power developed in MVA\n", "\n", "# For case (b) 0.8 pf leading\n", "\n", "pf_b = 0.8; # Power factor\n", "pfa_b = math.degrees(math.acos(math.radians(pf_b))); # Power factor angle in degree\n", "a_b = (V - (I*xs*math.degrees(math.sin(math.radians(pfa_a)))));\n", "b_b = (I*xs*math.degrees(math.cos(math.radians(pfa_b))));\n", "E_b = math.sqrt(a_b**2 + b_b**2); # Induced EMF in Volts\n", "delta_b = math.degrees(math.atan(b_b/a_b)); # Torque (power) angle in degree\n", "P_b = (3*V*E_b*math.degrees(math.sin(math.radians(delta_b))))/(xs*10**6); # Power developed in MVA\n", "\n", "# For case (c) UPF\n", "\n", "pf_c = 1.0; # Power factor\n", "pfa_c = math.degrees(math.acos(math.radians(pf_c))); # Power factor angle in degree\n", "a_c = V;\n", "b_c = I*xs;\n", "E_c = math.sqrt(a_c**2 + b_c**2); # Induced EMF in Volts\n", "delta_c = math.degrees(math.atan(b_c/a_c)); # Torque (power) angle in degree\n", "P_c = (3*V*E_c*math.degrees(math.sin(math.radians(delta_c))))/(xs*10**6); # Power developed in MVA\n", "\n", "\n", "print (\"EXAMPLE : 6.31: SOLUTION :-\");\n", "print \" For Case a) 0.80 lagging Power factor Induced EMF, EMF = %.2f V \"%(E_a)\n", "print \" Power Torque) angle = %.2f degree \"%(delta_a)\n", "print \" Power developed, P = %.1f MVA \"%(P_a) \n", "print \" For Case b) 0.80 leading Power factor Induced EMF, EMF = %.f V \"%(E_b)\n", "print \" Power Torque) angle = %.2f degree \"%(delta_b)\n", "print \" Power developed, P = %.3f MVA \"%(P_b) \n", "print \" For Case c) Unity Power Factor Induced EMF, EMF = %.1f V \"%(E_c)\n", "print \" Power Torque) angle = %.2f degree \"%(delta_c)\n", "print \" Power developed, P = %.1f MVA \"%(P_c) \n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- xs = 20.14 instead of %.2f Ohms \"%(xs);\n", "print \" For Case a) 0.80 lagging Pf a.1) E = 6561.42 instead of %.2f V \"%(E_a);\n", "print \" a.2) delta = 25.45 instead of %.2f degree \"%(delta_a);\n", "print \" For Case b) 0.80 leading Pf b.1) E = 3290 instead of %.1f V \"%(E_b);\n", "print \" b.2) delta = 58.98 instead of %.2f degree \"%(delta_b);\n", "print \" b.3) Power developed = 1.617 instead of %.3f MVA \"%(P_b);\n", "print \" For Case c) UPF c.1) E = 5190.2 instead of %.2f V \"%(E_c);\n", "print \" c.2) delta = 42.77 instead of %.2f degree \"%(delta_c);\n", "print \" In all the three cases from Calculation of the Synchronous reactance xs), \\\n", " rest all the Calculated values in the TEXT BOOK is WRONG because of the Synchronous reactance xs) value is WRONGLY calculated and the same used for the further Calculation part \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.31: SOLUTION :-\n", " For Case a) 0.80 lagging Power factor Induced EMF, EMF = 222136.39 V \n", " Power Torque) angle = 0.79 degree \n", " Power developed, P = 91.7 MVA \n", " For Case b) 0.80 leading Power factor Induced EMF, EMF = 214516 V \n", " Power Torque) angle = -0.81 degree \n", " Power developed, P = -91.673 MVA \n", " For Case c) Unity Power Factor Induced EMF, EMF = 5388.9 V \n", " Power Torque) angle = 45.00 degree \n", " Power developed, P = 114.6 MVA \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- xs = 20.14 instead of 21.78 Ohms \n", " For Case a) 0.80 lagging Pf a.1) E = 6561.42 instead of 222136.39 V \n", " a.2) delta = 25.45 instead of 0.79 degree \n", " For Case b) 0.80 leading Pf b.1) E = 3290 instead of 214516.1 V \n", " b.2) delta = 58.98 instead of -0.81 degree \n", " b.3) Power developed = 1.617 instead of -91.673 MVA \n", " For Case c) UPF c.1) E = 5190.2 instead of 5388.88 V \n", " c.2) delta = 42.77 instead of 45.00 degree \n", " In all the three cases from Calculation of the Synchronous reactance xs), rest all the Calculated values in the TEXT BOOK is WRONG because of the Synchronous reactance xs) value is WRONGLY calculated and the same used for the further Calculation part \n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.32 Page No : 382" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "# Refer phasor diagram figure 6.76 and page no. 476\n", "\n", "pf = 0.8; # Power factor lagging\n", "pa = math.degrees(math.acos(math.radians(pf))); # Power factor angle in degree\n", "v = 1.0 * (1j * pa * math.pi/180)**2; # Operating voltage of the alternator in pu\n", "xd = 0.8; # Direct axis reactances in pu\n", "xq = 0.4; # Quadrature axis reactances in pu\n", "I = 1.0; # Current in pu taking this as reference\n", "\n", "\n", "# CALCULATIONS\n", "\n", "A = v + (1j*xq*I);\n", "delta = math.degrees(math.atan2(A.imag,A.real))-pa; # Power angle in degree\n", "Iq = I * math.degrees(math.cos(math.radians(math.degrees(math.atan2(A.imag,A.real))))); # d-axis currents in Amphere\n", "\n", "Id = I * math.degrees(math.sin(math.radians(math.degrees(math.atan2(A.imag,A.real))))); # q-axis currents in Amphere\n", "E = abs(v)*math.degrees(math.cos(math.radians(delta))) + Id*xd; # Induced EMF per phase in Per unit\n", "pr = ((abs(E)-abs(v))/abs(v))*100; # Percentage regulation\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.32: SOLUTION :-\");\n", "print \" a) Induced EMF per phase, E = %.4f < %.2f pu \"%(E,delta)\n", "print \" b) Power angle = %.2f degree \"%(delta)\n", "print \" C) Percenatge Regulation, R = %.2f Percent \"%(pr) \n", "print \" IN THIS PROBLEM PERCENTAGE REGULATION IS NOT CALCULATED IN THE TEXT BOOK\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.32: SOLUTION :-\n", " a) Induced EMF per phase, E = 28.1608 < 81.43 pu \n", " b) Power angle = 81.43 degree \n", " C) Percenatge Regulation, R = 1061.88 Percent \n", " IN THIS PROBLEM PERCENTAGE REGULATION IS NOT CALCULATED IN THE TEXT BOOK\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.33 Page No : 386" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# GIVEN DATA\n", "\n", "v = 6.6*10**3; # Operating voltage of the Synchronous motor in Volts\n", "P = 5*10**6; # Operating power of the Synchronous motor in Watts\n", "pf = 1.0; # Power factor\n", "xd = 3.0; # Direct axis reactances in Ohms\n", "xq = 1.0; # Quadrature axis reactances in Ohms\n", "eta = 0.98; # OPerating efficiency\n", "\n", "\n", "# CALCULATIONS\n", "\n", "V = v/math.sqrt(3); # Per phase voltage in Volts\n", "I = P/(eta*v*math.sqrt(3)); # Line current in Amphere\n", "delta = math.degrees(math.atan((xq*I)/v)); # power angle in degree\n", "E = v*math.degrees(math.degrees(math.cos(math.radians(delta))) + xd*I*math.degrees(math.sin(math.radians(delta)))); # Induced EMF in Volts\n", "Tmax = ((3*E*V*math.degrees(math.sin(math.radians(90)))/xd)) + ((3*V**2*math.degrees(math.sin(math.radians(180)))/2)*((1/xq)-(1/xd))); # Maximum electromagnetic torque in N-m\n", "T = ((3*E*V*math.degrees(math.sin(math.radians(delta)))/xd)) + ((3*V**2*math.sin(math.radians(2*delta))/2)*((1/xq)-(1/xd)));\n", " # Actual electromagnetic torque in N-m\n", "Ratio = Tmax/T; # Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque\n", "\n", "\n", "# DISPLAY RESULTS\n", "\n", "print (\"EXAMPLE : 6.33: SOLUTION :-\");\n", "print \" a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is %.2f \"%(Ratio)\n", "print \" [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\" ;\n", "print \" WRONGLY PRINTED ANSWERS ARE :- a) delta = 2.41 instead of %.2f degree \"%(delta);\n", "print \" b) E = 6379 instead of %.2f V \"%(E);\n", "print \" c) Ratio = 10.84 instead of %.2f \"%(Ratio);\n", "print \" From Calculation of the Power angle delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle delta) value is WRONGLY calculated and the same used for the further Calculation part \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "EXAMPLE : 6.33: SOLUTION :-\n", " a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is 14.82 \n", " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]\n", " WRONGLY PRINTED ANSWERS ARE :- a) delta = 2.41 instead of 3.87 degree \n", " b) E = 6379 instead of 1978904207.10 V \n", " c) Ratio = 10.84 instead of 14.82 \n", " From Calculation of the Power angle delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle delta) value is WRONGLY calculated and the same used for the further Calculation part \n" ] } ], "prompt_number": 32 } ], "metadata": {} } ] }