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diff --git a/Fundamentals_of_Electrical_Machines/CH_5.ipynb b/Fundamentals_of_Electrical_Machines/CH_5.ipynb new file mode 100755 index 00000000..27fbc256 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_5.ipynb @@ -0,0 +1,435 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 5: DIRECT CURRENT MOTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1, Page number 182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "l = 10.0 #Conductor length(m)\n",
+ "B = 0.56 #Magnetic flux density(T)\n",
+ "I = 2.0 #Current through conductor(A)\n",
+ "\n",
+ "#Calculation\n",
+ "F = B*I*l #Magnitude of force(N)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnitude of force , F = %.1f N' %F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of force , F = 11.2 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2, Page number 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I = 20.0 #Total current(A)\n",
+ "V_t = 250.0 #Supply voltage(V) \n",
+ "R_sh = 200.0 #Shunt field resistance(ohm)\n",
+ "R_a = 0.3 #Armature resistance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I-I_sh #Armature current(A)\n",
+ "#For case(i)\n",
+ "E_b = V_t-R_a*I_a #Back emf(V)\n",
+ "#For case(ii)\n",
+ "P_md = E_b*I_a #Mechanical power developed(W) \n",
+ "\n",
+ "#Result\n",
+ "print('(i) Value of back emf , E_b = %.1f V' %E_b) \n",
+ "print('(ii) Mechanical power developed in the armature motor , P_md = %.1f W' %P_md)\n",
+ "print('\\nNOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of back emf , E_b = 244.4 V\n",
+ "(ii) Mechanical power developed in the armature motor , P_md = 4582.0 W\n",
+ "\n",
+ "NOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3, Page number 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R_a = 0.7 #Armature circuit resistance(ohm)\n",
+ "V_t = 5.0 #Applied voltage(V)\n",
+ "I_anl = 5.0 #No-load armature current(A)\n",
+ "I_afl = 35.0 #Full-load armature current(A)\n",
+ "\n",
+ "#Calculation\n",
+ "E_bnl = V_t-R_a*I_anl #Back emf under no-load(V)\n",
+ "E_bfl = V_t-R_a*I_afl #Back emf under full-load(V)\n",
+ "E_bc = E_bnl-E_bfl #Change in back emf from no-load to full load(V)\n",
+ " \n",
+ "#Result\n",
+ "print('Change in back emf from no-load to full load , E_bc = %.f V' %E_bc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in back emf from no-load to full load , E_bc = 21 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4, Page number 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I = 40.0 #Current(A)\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "N = 1100.0 #Speed(rpm)\n",
+ "R_a = 0.25 #Armature resistance(ohm)\n",
+ "R_sh = 230.0 #Shunt field resistance(ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I-I_sh #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V) \n",
+ "T_a = 9.55*E_b*I_a/N #Armature torque(N-m)\n",
+ " \n",
+ "#Result\n",
+ "print('Torque developed by the armature , T_a = %.2f N-m' %T_a)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque developed by the armature , T_a = 74.57 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5, Page number 191-192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 6.0 #Number of poles\n",
+ "V_t = 230.0 #Supply voltage to shunt motor(V)\n",
+ "Z = 450.0 #Number of conductors\n",
+ "R_a = 0.8 #Armature resistance(ohm)\n",
+ "I = 30.0 #Current drawn from supply(A)\n",
+ "P_0 = 5560.0 #Output power(W)\n",
+ "I_f = 3.0 #Current through field winding(A)\n",
+ "phi = 25*10**-3 #Flux per pole(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "A = P #Number of parallel paths in lap winding\n",
+ "I_a = I-I_f #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V)\n",
+ "N = 60*A*E_b/(P*Z*phi) #Speed(rpm)\n",
+ "T_sh = 9.55*P_0/N #Shaft torque(N-m)\n",
+ "\n",
+ "#Result\n",
+ "print('Speed , N = %.1f rpm' %N)\n",
+ "print('Shaft torque , T_sh = %.1f N-m' %T_sh)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed , N = 1111.5 rpm\n",
+ "Shaft torque , T_sh = 47.8 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6, Page number 193-194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_Lnl = 5.0 #Current drawn at no-load(A)\n",
+ "V_t = 230.0 #Terminal voltage at no-load(V)\n",
+ "N_nl = 1000.0 #Speed at no-load(rpm)\n",
+ "R_a = 0.2 #Armature resistance(ohm)\n",
+ "R_f = 230.0 #Field resistance(ohm)\n",
+ "I_Lfl = 30.0 #Current drawn at full-load(A)\n",
+ "\n",
+ "#Calculation\n",
+ "#Under No-load condition\n",
+ "I_sh = V_t/R_f #Shunt field current(A)\n",
+ "I_a1 = I_Lnl-I_sh #Armature current(A)\n",
+ "E_b1 = V_t-I_a1*R_a #Back emf(V)\n",
+ "#Under Full-load condition\n",
+ "I_a2 = I_Lfl-I_sh #Armature current(A)\n",
+ "E_b2 = V_t-I_a2*R_a #Back emf(V)\n",
+ "N_2 = (E_b2/E_b1)*N_nl #Motor speed under load condition(rpm)\n",
+ "\n",
+ "#Result\n",
+ "print('Motor speed under load condition , N_2 = %.1f rpm' %N_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Motor speed under load condition , N_2 = 978.2 rpm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7, Page number 194-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_a1 = 65.0 #Current drawn(A)\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "N_1 = 900.0 #Speed(rpm)\n",
+ "R_a = 0.2 #Armature resistance(ohm)\n",
+ "R_sh = 0.25 #Field resistance(ohm)\n",
+ "I_a2 = 15.0 #Line current(A)\n",
+ "phi_1 = 1.0 #Assumtion of flux(Wb)\n",
+ "phi_2 = 0.4*phi_1 #Flux(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "E_b1 = V_t-I_a1*(R_a+R_sh) #Initial back emf(V)\n",
+ "E_b2 = V_t-I_a2*(R_a+R_sh) #Final back emf(V)\n",
+ "N_2 = N_1*E_b2*phi_1/(E_b1*phi_2) #Speed of motor(rpm)\n",
+ "\n",
+ "#Result\n",
+ "print('Speed of motor when line current is 15 A , N_2 = %.f rpm' %N_2)\n",
+ "print('\\nNOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of motor when line current is 15 A , N_2 = 2502 rpm\n",
+ "\n",
+ "NOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8, Page number 197-198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_Lnl = 5.0 #Current drawn at no-load(A)\n",
+ "V_t = 230.0 #Supply voltage at no-load(V)\n",
+ "R_a = 0.25 #Armature circuit resistance(ohm)\n",
+ "R_sh = 115 #Field circuit resistance(ohm)\n",
+ "I_L = 40.0 #Current under load condition(A)\n",
+ "\n",
+ "#Calculation\n",
+ "#Under No-load condition\n",
+ "P_in1 = V_t*I_Lnl #Input power(W)\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a1 = I_Lnl-I_sh #Armature current(A)\n",
+ "P_acu1 = I_a1**2*R_a #Armature copper loss(W)\n",
+ "P_shcu = I_sh**2*R_sh #Shunt field copper loss(W)\n",
+ "P_iron_friction = P_in1-(P_acu1+P_shcu) #Iron and friction losses(W)\n",
+ "#Under load condition\n",
+ "I_a2 = I_L-I_sh #Armature current(A)\n",
+ "P_acu2 = I_a2**2*R_a #Armature copper loss(W)\n",
+ "P_loss = P_iron_friction+P_shcu+P_acu2 #Total losses(W)\n",
+ "P_in2 = V_t*I_L #Input power(W)\n",
+ "P_0 = P_in2-P_loss #Output power(W)\n",
+ "n = (P_0/P_in2)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Iron and friction losses , P_iron_friction = %.2f W' %P_iron_friction)\n",
+ "print('Efficiency , \u03b7 = %.f percent' %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Iron and friction losses , P_iron_friction = 687.75 W\n",
+ "Efficiency , \u03b7 = 84 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9, Page number 198-199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_L = 80.0 #Current drawn(A)\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "N = 800.0 #Speed(rpm)\n",
+ "R_a = 0.1 #Armature resistance(ohm)\n",
+ "R_sh = 50.0 #Shunt field resistance(ohm)\n",
+ "P_if = 1600.0 #Iron and friction losses(W)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_L-I_sh #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V)\n",
+ "#For case(i)\n",
+ "P_in = V_t*I_L #Input power(W)\n",
+ "P_md = E_b*I_a #Mechanical power developed in the armature(W)\n",
+ "P_cu = P_in-P_md #Copper loss(W)\n",
+ "#For case(ii)\n",
+ "T_a = 9.55*E_b*I_a/N #Armature torque(N-m)\n",
+ "#For case(iii)\n",
+ "P_0 = P_md-P_if #Output power(W)\n",
+ "T_sh = 9.55*P_0/N #Shaft torque(N-m)\n",
+ "#For case(iv)\n",
+ "n = (P_0/P_in)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Copper losses , P_cu = %.2f W' %P_cu)\n",
+ "print('(ii) Armature torque , T_a = %.2f N-m' %T_a)\n",
+ "print('(iii) Shaft torque , T_sh = %.2f N-m' %T_sh)\n",
+ "print('(iv) Efficiency , \u03b7 = %.f percent' %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Copper losses , P_cu = 1539.54 W\n",
+ "(ii) Armature torque , T_a = 191.72 N-m\n",
+ "(iii) Shaft torque , T_sh = 172.62 N-m\n",
+ "(iv) Efficiency , \u03b7 = 82 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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