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diff --git a/Fundamentals_of_Electrical_Machines/CH_1.ipynb b/Fundamentals_of_Electrical_Machines/CH_1.ipynb new file mode 100755 index 00000000..801b1ed1 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_1.ipynb @@ -0,0 +1,731 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 1: REVIEW OF ELECTRIC CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1, Page number 2-3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Q = 4.0 #Charge(C) \n",
+ "t = 0.54 #Time(sec) \n",
+ "\n",
+ "#Calculation\n",
+ "I = Q/t #Current(A) \n",
+ "\n",
+ "#Result\n",
+ "print('Value of Current is , I = %.2f A' %I) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Current is , I = 7.41 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2, Page number 4-5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = -24.0 #Voltage(V)\n",
+ "I = 3.0 #Current(A)\n",
+ "\n",
+ "#Calculation \n",
+ "P = V*I #Power supplied by the element A(W) \n",
+ "\n",
+ "#Result\n",
+ "print('Power supplied by the element A is , P = %.1f W' %P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power supplied by the element A is , P = -72.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3, Page number 7-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R1 = 5.0 #Resistance(ohm)\n",
+ "R2 = 4.0 #Resistance(ohm)\n",
+ "R3 = 9.0 #Resistance(ohm)\n",
+ "R4 = 6.0 #Resistance(ohm)\n",
+ "V1 = 10.0 #Resistance(ohm)\n",
+ "V2 = 6.0 #Resistance(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "R_th = (R1*R4/(R1+R4))+R2 #Thevenin resistance(ohm) by removing R3 & short-circuiting voltage sources\n",
+ "I = (V1-V2)/(R1+R4) #Current(A) by applying KVL\n",
+ "V_th = 6*I+V2 #Thevenin voltage(V) by applying KVL\n",
+ "I_9ohm = V_th/(R_th+R3) #Current through 9 ohm resistor(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Current through 9 ohm resistor , I_9\u03a9 = %.2f A' %I_9ohm) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through 9 ohm resistor , I_9\u03a9 = 0.52 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4, Page number 10-11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from scipy.integrate import quad\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_t1 = 30.0 #Magnitudes of voltages(V) 0 < t1 < 2\n",
+ "V_t2 = -10.0 #Magnitudes of voltages(V) 2 < t2 < 4\n",
+ "T = 4.0 #Time period(sec) from figure\n",
+ "\n",
+ "#Calculation\n",
+ "def integrand(V):\n",
+ " return V**0\n",
+ "\n",
+ "a, err = quad(integrand, 0, 2)\n",
+ "\n",
+ "def integrand(V):\n",
+ " return V**0\n",
+ "\n",
+ "b, err = quad(integrand, 2, 4)\n",
+ "\n",
+ "V_rms = ((a*V_t1**2+b*V_t2**2)/4)**0.5 #RMS value of voltage waveform(V)\n",
+ "\n",
+ "#Result\n",
+ "print('RMS value , V_rms = %.2f V' %V_rms)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS value , V_rms = 22.36 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5, Page number 15-16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_P = 200.0 #Magnitude of each phase(V) \n",
+ "\n",
+ "#Calculation\n",
+ "V_an = V_P*cmath.exp(1j*0*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_bn = V_P*cmath.exp(1j*-120*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_cn = V_P*cmath.exp(1j*120*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_L = 3**0.5*V_P #Magnitude of line voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Expression of phase voltages are,')\n",
+ "print('\\t\\t\\t V_an = %.f\u2220%.f\u00b0 V' %(abs(V_an),cmath.phase(V_an)))\n",
+ "print('\\t\\t\\t V_bn = %.f\u2220%.f\u00b0 V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))\n",
+ "print('\\t\\t\\t V_cn = %.f\u2220%.f\u00b0 V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))\n",
+ "print('Magnitude of the line voltage , V_L = %.1f V' %V_L) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Expression of phase voltages are,\n",
+ "\t\t\t V_an = 200\u22200\u00b0 V\n",
+ "\t\t\t V_bn = 200\u2220-120\u00b0 V\n",
+ "\t\t\t V_cn = 200\u2220120\u00b0 V\n",
+ "Magnitude of the line voltage , V_L = 346.4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6, Page number 16-17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "R = 10.0 #Resistance of each coil(ohm)\n",
+ "X = 15.0 #Inductive reactance of each coil(ohm)\n",
+ "V_L = 420.0 #Line voltage(V)\n",
+ "f = 50.0 #Frequency of supply(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V_an = (V_L/3**0.5)*cmath.exp(1j*(0-30)*math.pi/180) #Phase voltage(V)\n",
+ "V_bn = (V_L/3**0.5)*cmath.exp(1j*(-120-30)*math.pi/180) #Phase voltage(V)\n",
+ "V_cn = (V_L/3**0.5)*cmath.exp(1j*(120-30)*math.pi/180) #Phase voltage(V)\n",
+ "Z_P = complex(R,X) #Phase impedance(ohm)\n",
+ "#For case(i)\n",
+ "I_L1 = V_an/Z_P #Line current(A)\n",
+ "I_L2 = V_bn/Z_P #Line current(A)\n",
+ "I_L3 = V_cn/Z_P #Line current(A)\n",
+ "#For case(ii)\n",
+ "pf = R/abs(Z_P) #Power factor\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Values of line currents are,')\n",
+ "print('\\t I_L1 = I_an = %.2f\u2220%.2f\u00b0 A' %(abs(I_L1),cmath.phase(I_L1)*180/math.pi))\n",
+ "print('\\t I_L2 = I_bn = %.2f\u2220%.2f\u00b0 A' %(abs(I_L2),cmath.phase(I_L2)*180/math.pi))\n",
+ "print('\\t I_L3 = I_cn = %.2f\u2220%.2f\u00b0 A' %(abs(I_L3),cmath.phase(I_L3)*180/math.pi))\n",
+ "print('(ii) Power factor is , pf = %.1f lag' %pf)\n",
+ "print('\\nNOTE : I_L2 has an angle -206.31\u00b0 in textbook which is same as 153.69\u00b0 i.e (360-206.31)\u00b0 obtained here')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Values of line currents are,\n",
+ "\t I_L1 = I_an = 13.45\u2220-86.31\u00b0 A\n",
+ "\t I_L2 = I_bn = 13.45\u2220153.69\u00b0 A\n",
+ "\t I_L3 = I_cn = 13.45\u222033.69\u00b0 A\n",
+ "(ii) Power factor is , pf = 0.6 lag\n",
+ "\n",
+ "NOTE : I_L2 has an angle -206.31\u00b0 in textbook which is same as 153.69\u00b0 i.e (360-206.31)\u00b0 obtained here\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7, Page number 19-20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z_P = complex(10,15) #Per phase impedance(ohm)\n",
+ "V_L = 420.0 #Voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "V_ab = V_L*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "V_bc = V_L*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_ca = V_L*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_ab = V_ab/Z_P #Phase current(A)\n",
+ "I_bc = V_bc/Z_P #Phase current(A)\n",
+ "I_ca = V_ca/Z_P #Phase current(A)\n",
+ "#For case(ii)\n",
+ "I_P = abs(I_ab) #Phase current magnitude(A)\n",
+ "I_L = 3**0.5*I_P #Line current magnitude(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Phase currents are,')\n",
+ "print('\\t\\t I_ab = %.2f\u2220%.2f\u00b0 A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))\n",
+ "print('\\t\\t I_bc = %.2f\u2220%.2f\u00b0 A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))\n",
+ "print('\\t\\t I_ca = %.2f\u2220%.2f\u00b0 A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))\n",
+ "print('(ii) Magnitude of line current , I_L = %.2f A' %I_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Phase currents are,\n",
+ "\t\t I_ab = 23.30\u2220-56.31\u00b0 A\n",
+ "\t\t I_bc = 23.30\u2220-176.31\u00b0 A\n",
+ "\t\t I_ca = 23.30\u222063.69\u00b0 A\n",
+ "(ii) Magnitude of line current , I_L = 40.35 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.8, Page number 22-23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_P = 280.0 #Generator Phase voltage(V)\n",
+ "Z_P = complex(2,3) #Line impedance per phase(ohm)\n",
+ "Z_L = complex(4,5) #Load impedance per phase(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "V_An = V_P*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "V_Bn = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_Cn = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "Z_t = Z_P+Z_L #Total impedance(ohm)\n",
+ "I_Aa = V_An/Z_t #Magnitude of line current for phase A(A)\n",
+ "I_Bb = V_Bn/Z_t #Magnitude of line current for phase B(A)\n",
+ "I_Cc = V_Cn/Z_t #Magnitude of line current for phase C(A)\n",
+ "V_an = I_Aa*Z_L #Phase voltage of load(V)\n",
+ "V_bn = I_Bb*Z_L #Phase voltage of load(V)\n",
+ "V_cn = I_Cc*Z_L #Phase voltage of load(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Line currents are,')\n",
+ "print('\\t\\t I_Aa = %.f\u2220%.f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print('\\t\\t I_Bb = %.f\u2220%.f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print('\\t\\t I_Cc = %.f\u2220%.f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('\\nLoad phase voltages are,')\n",
+ "print('\\t\\t V_an = %.1f\u2220%.1f\u00b0 V' %(abs(V_an),cmath.phase(V_an)*180/math.pi))\n",
+ "print('\\t\\t V_bn = %.1f\u2220%.1f\u00b0 V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))\n",
+ "print('\\t\\t V_cn = %.1f\u2220%.1f\u00b0 V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))\n",
+ "print('\\nNOTE : ERROR : Z_L = 6.4\u222038.6\u00b0\u03a9 is taken in textbook solution instead of 6.4\u222051.34\u00b0\u03a9 = (4+j5)\u03a9')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line currents are,\n",
+ "\t\t I_Aa = 28\u2220-53\u00b0 A\n",
+ "\t\t I_Bb = 28\u2220-173\u00b0 A\n",
+ "\t\t I_Cc = 28\u222067\u00b0 A\n",
+ "\n",
+ "Load phase voltages are,\n",
+ "\t\t V_an = 179.3\u2220-1.8\u00b0 V\n",
+ "\t\t V_bn = 179.3\u2220-121.8\u00b0 V\n",
+ "\t\t V_cn = 179.3\u2220118.2\u00b0 V\n",
+ "\n",
+ "NOTE : ERROR : Z_L = 6.4\u222038.6\u00b0\u03a9 is taken in textbook solution instead of 6.4\u222051.34\u00b0\u03a9 = (4+j5)\u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.9, Page number 23-24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(6,8) #Per phase impedance of load(ohm)\n",
+ "V_AN = 340.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_BN = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_CN = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_an = V_AN/Z #Load current(A)\n",
+ "I_bn = V_BN/Z #Load current(A)\n",
+ "I_cn = V_CN/Z #Load current(A)\n",
+ "I_n = I_an+I_bn+I_cn #Neutral current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Phase current in each load = Line current in each load are,')\n",
+ "print('\\t\\t\\t\\t I_an = I_Aa = %.f\u2220%.f\u00b0 A' %(abs(I_an),cmath.phase(I_an)*180/math.pi))\n",
+ "print('\\t\\t\\t\\t I_bn = I_Bb = %.f\u2220%.f\u00b0 A' %(abs(I_bn),cmath.phase(I_bn)*180/math.pi))\n",
+ "print('\\t\\t\\t\\t I_cn = I_Cc = %.f\u2220%.f\u00b0 A' %(abs(I_cn),cmath.phase(I_cn)*180/math.pi))\n",
+ "print('Neutral current is , I_n = %.f A' %abs(I_n))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Phase current in each load = Line current in each load are,\n",
+ "\t\t\t\t I_an = I_Aa = 34\u2220-53\u00b0 A\n",
+ "\t\t\t\t I_bn = I_Bb = 34\u2220-173\u00b0 A\n",
+ "\t\t\t\t I_cn = I_Cc = 34\u222067\u00b0 A\n",
+ "Neutral current is , I_n = 0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.10, Page number 25-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(3,4) #Per phase impedance of load(ohm)\n",
+ "V_AN = 200.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_AB = 3**0.5*V_P*cmath.exp(1j*30*math.pi/180) #Line voltage(V)\n",
+ "V_BC = 3**0.5*V_P*cmath.exp(1j*-90*math.pi/180) #Line voltage(V)\n",
+ "V_CA = 3**0.5*V_P*cmath.exp(1j*150*math.pi/180) #Line voltage(V)\n",
+ "#For case(i)\n",
+ "I_ab = V_AB/Z #Load current(A)\n",
+ "I_bc = V_BC/Z #Load current(A)\n",
+ "I_ca = V_CA/Z #Load current(A)\n",
+ "#For case(ii)\n",
+ "I_Aa = I_ab-I_ca #Line current(A)\n",
+ "I_Bb = I_bc-I_ab #Line current(A)\n",
+ "I_Cc = I_ca-I_bc #Line current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Magnitude of load currents are,')\n",
+ "print('\\t\\t I_ab = %.1f\u2220%.2f\u00b0 A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))\n",
+ "print('\\t\\t I_bc = %.1f\u2220%.2f\u00b0 A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))\n",
+ "print('\\t\\t I_ca = %.1f\u2220%.2f\u00b0 A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))\n",
+ "print('\\n(ii) Magnitude of line currents are,')\n",
+ "print('\\t\\t I_Aa = %.2f\u2220%.2f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print('\\t\\t I_Bb = %.2f\u2220%.2f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print('\\t\\t I_Cc = %.2f\u2220%.2f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('\\nNOTE : ERROR : Calculation mistakes in textbook')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnitude of load currents are,\n",
+ "\t\t I_ab = 69.3\u2220-23.13\u00b0 A\n",
+ "\t\t I_bc = 69.3\u2220-143.13\u00b0 A\n",
+ "\t\t I_ca = 69.3\u222096.87\u00b0 A\n",
+ "\n",
+ "(ii) Magnitude of line currents are,\n",
+ "\t\t I_Aa = 120.00\u2220-53.13\u00b0 A\n",
+ "\t\t I_Bb = 120.00\u2220-173.13\u00b0 A\n",
+ "\t\t I_Cc = 120.00\u222066.87\u00b0 A\n",
+ "\n",
+ "NOTE : ERROR : Calculation mistakes in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11, Page number 28-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(3,4) #Per phase impedance of load(ohm)\n",
+ "V_AN = 150.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_BN = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_CN = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_Aa = V_AN/Z #Line current(A)\n",
+ "I_Bb = V_BN/Z #Line current(A)\n",
+ "I_Cc = V_CN/Z #Line current(A)\n",
+ "pf = Z.real/abs(Z) #Power factor\n",
+ "I = abs(I_Aa) #Magnitude of line current(A)\n",
+ "P = V_P*I*pf*10**-3 #Power supplied to each phase(kW)\n",
+ "P_t = 3*P #Total power supplied(kW)\n",
+ "\n",
+ "#Result\n",
+ "print('Line currents are,')\n",
+ "print(' I_Aa = %.f\u2220%.2f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print(' I_Bb = %.f\u2220%.2f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print(' I_Cc = %.f\u2220%.2f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('Power factor , pf = %.1f ' %pf)\n",
+ "print('Power supplied to each phase , P = %.1f kW' %P)\n",
+ "print('Total Power supplied to the load , P_t = %.1f kW' %P_t)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line currents are,\n",
+ " I_Aa = 30\u2220-53.13\u00b0 A\n",
+ " I_Bb = 30\u2220-173.13\u00b0 A\n",
+ " I_Cc = 30\u222066.87\u00b0 A\n",
+ "Power factor , pf = 0.6 \n",
+ "Power supplied to each phase , P = 2.7 kW\n",
+ "Total Power supplied to the load , P_t = 8.1 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12, Page number 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 120.0 #Total power(kW)\n",
+ "pf = 0.6 #Power factor \n",
+ "\n",
+ "#Calculation\n",
+ "teta = math.acos(pf) #Power factor angle(radians)\n",
+ "teta_deg = teta*180/math.pi #Power factor angle(degree)\n",
+ "P_2 = 1.0/2*((math.tan(teta)*P/3**0.5)+P) #Second wattmeter reading(kW)\n",
+ "\n",
+ "#Result\n",
+ "print('Second wattmeter reading , P_2 = %.1f kW' %P_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Second wattmeter reading , P_2 = 106.2 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.13, Page number 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 5000.0 #Power(W)\n",
+ "pf_1 = 0.8 #Initial Power factor\n",
+ "V = 110.0 #rms Voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "pf_2 = 0.9 #Final Power factor\n",
+ "\n",
+ "#Calculation\n",
+ "phi_1 = math.acos(pf_1) #Initial Power factor angle(radians)\n",
+ "phi_1_deg = phi_1*180/math.pi #Initial Power factor angle(degree)\n",
+ "phi_2 = math.acos(pf_2) #Final Power factor angle(radians)\n",
+ "phi_2_deg = phi_2*180/math.pi #Final Power factor angle(degree)\n",
+ "C = P*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6 #Parallel capacitance(\u00b5F)\n",
+ "\n",
+ "#Result\n",
+ "print('Capacitance , C = %.1f \u00b5F' %C)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance , C = 349.5 \u00b5F\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14, Page number 35-36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "pf_1 = 0.85 #Initial Power factor\n",
+ "kVA = 20.0 #Load(kVA)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "pf_2 = 0.95 #Final Power factor\n",
+ "V = 200.0 #Voltage(V)\n",
+ "R = 0.05 #Resistance(ohm)\n",
+ "X = 0.2 #Inductive reactance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "phi_1 = math.acos(pf_1) #Initial Power factor angle(radians)\n",
+ "phi_1_deg = phi_1*180/math.pi #Initial Power factor angle(degree)\n",
+ "phi_2 = math.acos(pf_2) #Final Power factor angle(radians)\n",
+ "phi_2_deg = phi_2*180/math.pi #Final Power factor angle(degree)\n",
+ "P = kVA*pf_1 #Load power(kW)\n",
+ "C = P*1000*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6 #Parallel capacitance(\u00b5F)\n",
+ "#Before adding capacitor\n",
+ "I_1 = P*1000/(pf_1*V) #Line current(A)\n",
+ "P_1 = I_1**2*R #Power loss in line(W)\n",
+ "#After adding capacitor\\n\",\n",
+ "S = P*1000/pf_2 #Apparent power(VA)\n",
+ "I_2 = S/V #Line current(A)\n",
+ "P_2 = I_2**2*R #Power loss in line(W)\n",
+ "\n",
+ "#Result\n",
+ "print('Capacitance , C = %.1f \u00b5F' %C)\n",
+ "print('Power loss in the line before adding capacitor , P_1 = %.1f W' %P_1)\n",
+ "print('Power loss in the line after adding capacitor , P_2 = %.1f W' %P_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance , C = 393.8 \u00b5F\n",
+ "Power loss in the line before adding capacitor , P_1 = 500.0 W\n",
+ "Power loss in the line after adding capacitor , P_2 = 400.3 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |