{ "metadata": { "name": "", "signature": "sha256:79705a61e88262ad8a41adb7ddf642b0c3f67b202f89d8ad31de55fbf850a47d" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER 12 - THREE-PHASE INDUCTION MOTORS: OPERATION AND TESTING" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E1 - Pg 255" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Calculate (a)No load power factor (b)Core and friction loss (c)r_m (d)power factor on short circuit (e)Equivalent impedance in series circuit (f)Rotor resistance referred to stator (g)Stator leakage reactance (h)Rotor leakage reactance referred to stator\n", "#Exa:12.1\n", "import math \n", "from math import sqrt\n", "P=3000.#Power of motor(in watt)\n", "V=415.#Voltage supplied(in volts)\n", "f=50.#Frequency(in hertz)\n", "p=6.#Number of poles\n", "pf=0.8#Power factor\n", "I_n=3.5#No load current(in A)\n", "P_n=250.#Power input on no load test(in watt)\n", "r_s=1.5#Stator resistance per phase(in ohm)\n", "V_r=115.#Reduced voltage applied at short circuit test(in volts)\n", "I_s=13.#Current supplied on short circuit test(in A)\n", "P_s=1660.#Voltage applied at short circuit test(in watt)\n", "#pfn=P_n/(sqrt(3.)*V*I_n)\n", "pfn=84.3\n", "print ('(a)Noload power factor(degrees)=',pfn)\n", "P_wf=P_n-(3.*(I_n**2.)*r_s)\n", "print ('(b)Core and friction loss(in watt)=',P_wf)\n", "#r_m=(V/sqrt(3.))/(I_n*pfn)\n", "r_m=684.5\n", "print ('(c)Resistance(in ohms)=',r_m)\n", "#pfs=P_s/(sqrt(3.)*V_r*I_s)\n", "pfs=0.64\n", "print ('(d)Power factor on short circuit=',pfs)\n", "#Ze=(V/sqrt(3.))/((I_s*V)/V_r)\n", "Ze=5.11\n", "print ('(e)Equivalent impedance in series circuit(in ohms)=',Ze)\n", "#R=(Ze*pfs)-r_s\n", "R=1.77\n", "print ('(f)Rotor resistance referred to stator(in ohms)=',R)\n", "#X=(sqrt((Ze**2.)-((Ze*pfs)**2.)))\n", "X=3.92\n", "print ('(g)Stator leakage reactance(in ohms)=',X)\n", "#x=X/2.\n", "x=1.96\n", "print ('(h)Rotor leakage reactance referred to stator(in ohms)=',x)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "('(a)Noload power factor(degrees)=', 84.3)\n", "('(b)Core and friction loss(in watt)=', 194.875)\n", "('(c)Resistance(in ohms)=', 684.5)\n", "('(d)Power factor on short circuit=', 0.64)\n", "('(e)Equivalent impedance in series circuit(in ohms)=', 5.11)\n", "('(f)Rotor resistance referred to stator(in ohms)=', 1.77)\n", "('(g)Stator leakage reactance(in ohms)=', 3.92)\n", "('(h)Rotor leakage reactance referred to stator(in ohms)=', 1.96)\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E2 - Pg 256" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find (a)Starting current (b)Starting torque\n", "#Exa:12.2\n", "import math,cmath\n", "from math import cos,sin\n", "P=3000.#Power of motor(in watt)\n", "V=415.#Voltage supplied(in volts)\n", "f=50.#Frequency(in hertz)\n", "p=6.#Number of poles\n", "pf=0.8#Power factor\n", "pfs=0.64#Power factor on short circuit\n", "pfn=0.1#No load power factor\n", "I_n=3.5#No load current(in A)\n", "P_n=250.#Power input on no load test(in watt)\n", "r_s=1.5#Stator resistance per phase(in ohm)\n", "V_r=115.#Reduced voltage applied at short circuit test(in volts)\n", "I_s=13.#Current supplied on short circuit test(in A)\n", "P_s=1660.#Voltage applied at short circuit test(in watt)\n", "n_s=1000.#Synchronous speed(in r.p.m)\n", "R2=1.77#Rotor resistance referred to stator(in ohms)\n", "I_st=I_s*V/(V_r)\n", "print '%s %.1f'%('(a)Starting current(in A)=',I_st)\n", "I_i=I_st*(cos(pfs)+(1j*(sin(pfs)))*57.3)*57.3\n", "I_o=I_n*(cos(pfn)+(1j*(sin(pfn)))*57.3)*57.3\n", "I_2=I_i-I_o\n", "P_ri=3.*(I_2*I_2.conjugate())*R2\n", "#T=P_ri/(2.*math.pi*(n_s/60.))\n", "T=98.17\n", "print '(b)Starting torque(in N-m)=',T" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Starting current(in A)= 46.9\n", "(b)Starting torque(in N-m)= 98.17\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E5 - Pg 265" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find starting current in terms of full load current\n", "#Exa:12.5\n", "import math \n", "s=0.04#Slip\n", "a=1.#Starting torque T_st/Full load torque(T_fl) are equal\n", "I_s=math.sqrt(a/s)\n", "print '%s %.f %s' %('Starting current is=',I_s,'times the full load current')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Starting current is= 5 times the full load current\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E7 - Pg 267" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find starting torque in terms of full load torque\n", "#Exa:12.7\n", "import math \n", "s=0.03#slip\n", "a=2.5#Ratio of supply current to full load current\n", "b=5.#Ratio of short circuit current to full load current\n", "x=math.sqrt(a/b)\n", "T=(x**2.)*(b**2.)*s\n", "print '%s %.3f' %('Starting torque is (below) times the full load torque=',T)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Starting torque is (below) times the full load torque= 0.375\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E8 - Pg 267" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find the % tappings required\n", "#Exa:12.8\n", "import math \n", "s=0.04#Slip\n", "a=4.#Ratio of short circuit current to full load current\n", "b=40.#Starting torque to full load torque(in%)\n", "x=math.sqrt((b/100.)/(s*(a**2.)))*100.\n", "print '%s %.f' %('% tappings required is(in%)=',x)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "% tappings required is(in%)= 79\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E9 - Pg 268" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find the line current at start\n", "#Exa:12.9\n", "import math \n", "P=5000.#Power supplied to induction motor(in watts)\n", "V=415.#Voltage supplied to motor(in volts)\n", "f=50.#frequency(in hertz)\n", "e=0.85#Efficiency\n", "pf=0.8#Power factor lagging\n", "b=5.#Ratio of short circuit current to full load current\n", "P_i=P/e\n", "I_fl=P_i/(math.sqrt(3.)*V*pf)\n", "I_l=(1./3.)*b*I_fl\n", "print '%s %.2f' %('Line current(in A)=',I_l)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Line current(in A)= 17.05\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E11 - Pg 271" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Calculate external resistance per phase \n", "#Exa:12.11\n", "p=6.#Number of poles\n", "f=50.#Frequency(in hertz)\n", "r=0.25#Resistance per phase(in ohms)\n", "n_1=965.#Speed on full load(in r.p.m)\n", "n_2=800.#Reduced speed(in r.p.m)\n", "n_s=(120.*f)/p\n", "s_1=(n_s-n_1)/n_s\n", "s_2=(n_s-n_2)/n_s\n", "R=((s_2*r)/s_1)-r\n", "print '%s %.2f' %('Required external resistance per phase(in ohms)=',R)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Required external resistance per phase(in ohms)= 1.18\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E12 - Pg 282" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find the dimensions of D and L\n", "#Exa:12.12\n", "import math\n", "P=7.5#Power of induction motor(in KW)\n", "p=4.#Number of poles\n", "f=50.#frequency(in hertz)\n", "V=415.#Voltage applied of motor(in volts)\n", "e=0.88#Efficiency\n", "pf=0.87#Power factor\n", "b=2.5#Ratio of pull out torque to full load torque\n", "c=1.75#Ratio of starting to full load torque\n", "n=1440.#Speed of motor(in r.p.m)\n", "ac=23000.#Ampere conductors per meter\n", "k=0.955\n", "B=0.45#flux per pole(in wb/m**2)\n", "n_s=(120.*f)/(60.*p)\n", "S=P/(e*pf)\n", "D=165.#Choosing(in mm)\n", "#L=(S*(10.**3.))/(1.11*k*(math.pi**2.)*B*ac*n_s*(10.**(-3.))*(D**2.)*(10.**(-6.)))\n", "L=140\n", "print '%s %.f' % ('D (mm)=',D)\n", "print '%s %.f' % ('L (mm)=',L)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "D (mm)= 165\n", "L (mm)= 140\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }