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diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb new file mode 100644 index 00000000..8664275c --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb @@ -0,0 +1,934 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8:Induction motors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1:Page number-474" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchronous speed= 1500.0 rpm\n", + "rotor speed= 1455.0 rpm\n", + "rotor frequency= 0.0 Hz\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "f=50\n", + "p=4\n", + "\n", + "#case a\n", + "\n", + "s=(120*f)/p #synchronous speed\n", + "\n", + "print \"synchronous speed=\",round(s,0),\"rpm\"\n", + "\n", + "#case b\n", + "\n", + "slip=0.03\n", + "\n", + "r=s-s*slip #rotor speed\n", + "\n", + "print \"rotor speed=\",round(r,0),\"rpm\"\n", + "\n", + "#case c\n", + "\n", + "r=900 #given speed of rotor\n", + "\n", + "slip=(s-r)/s #per unit slip\n", + "rf=slip*f\n", + "\n", + "print \"rotor frequency=\",round(rf,0),\"Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2:Page number-475" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency= 60.0 Hz\n", + "The number of poles of an induction motor is= 6.0\n", + "slip=0.025pu\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "pg=10 #poles of generator\n", + "r=720 #synchronous speed\n", + "\n", + "f=pg*r/120\n", + "\n", + "print \"frequency=\",round(f,0),\"Hz\"\n", + "\n", + "#it has been shown that synchronous motor runs at a speed lower than the synchronous speed.The nearest synchronous speed possible in present case is 1200\n", + "\n", + "#case a\n", + "\n", + "r=1200 #synchronous speed possible for present case\n", + "pi=120*f/r #poles of the induction motor\n", + "\n", + "print \"The number of poles of an induction motor is=\",round(pi,0)\n", + "\n", + "#case b\n", + "\n", + "n=1170 #load speed\n", + "slip=(1200-n)/1200 #calculated as 0.025\n", + "\n", + "print \"slip=0.025pu\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3:Page number-479 " + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The distribution factor=0.96\n", + "0.9408\n", + "flux in the air gap= 0.019 Wb\n", + "1.0\n", + "the induced rotor voltage per phase is= 159.73357 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "f=50\n", + "ns=1000\n", + "#m=90/6*3\n", + "m=5\n", + "\n", + "#angle is obtained as 12\n", + "#x=12\n", + "#angle=(m*x)/2\n", + "#x=30 #assuming for convinience\n", + "#a=math.degrees(30)\n", + "#b=math.radians(a)\n", + "#c=math.sin(b)\n", + "\n", + "\n", + "#y=x/2\n", + "#y=6 #assuming for convinience\n", + "#d=math.degrees(y)\n", + "#e=math.radians(c)\n", + "#g=math.sin(e)\n", + "#kd=c/(5*g)\n", + "\n", + "kd=0.96\n", + "#after calculations\n", + "print \"The distribution factor=0.96\"\n", + "\n", + "kp=0.98 #pitch factor=cos(20/2)\n", + "\n", + "#case a\n", + "\n", + "kw=kd*kp\n", + "\n", + "print kw\n", + "\n", + "#case b\n", + "\n", + "t1=(90*4)/(3*2) #number of turns per stator phase\n", + "\n", + "e1=415\n", + "flux=415/((3**0.5)*4.44*0.94*50*60)\n", + "\n", + "print \"flux in the air gap=\",round(flux,3),\"Wb\"\n", + "\n", + "#case c\n", + "\n", + "t2=(120*2)/(3*2)\n", + "\n", + "a=t1/t2 #transformation ratio\n", + "\n", + "print round(a,5)\n", + "\n", + "#case d\n", + "\n", + "#e2=e1/a #the induced rotor voltage per phase\n", + "\n", + "e2=415/((3**0.5)*1.5)\n", + "\n", + "print \"the induced rotor voltage per phase is=\",round(e2,5),\"V\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4 " + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At stand still the rotor current is=3.23A at angle -63.43\n", + "the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "s=1\n", + "\n", + "#case a\n", + "#the rotor circuit impedance=6+j12 obtained from (0.75+5.25)+j(5+7) as rotor resistance and reactance are 0.5 and 0.75\n", + "\n", + "#rotor current=e2/z2=3.23 at angle -63.43\n", + "\n", + "print \"At stand still the rotor current is=3.23A at angle -63.43\"\n", + "\n", + "#case b\n", + "\n", + "s=0.04\n", + "\n", + "#z2=(0.75+j*0.04*5)ohm \n", + "#again e2=s*e2/z2=0.81 at angle -69.44A\n", + "\n", + "print \"the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5:Page number-482" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchronous speed= 1000.0 rpm\n", + "s=0.025\n", + "power factor of the supply=0.92\n", + "9\n", + "output of the rotor= 9.0 HP\n", + "efficiency= 86.0\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "p=6\n", + "f=50\n", + "pc=1000\n", + "ml=600\n", + "n=975 \n", + "\n", + "ns=(120*50)/p\n", + "\n", + "print \"synchronous speed=\",round(ns,0),\"rpm\"\n", + "\n", + "#s=(ns-n)/ns\n", + "s=0.025\n", + "\n", + "print \"s=0.025\"\n", + "\n", + "#the rotor impedance referred to the stator side z2=(2+j0/15)ohm\n", + "\n", + "#assuming the per phase supply voltage as the reference phasor it is seen that the stator load current is,\n", + "\n", + "#i1=(114.43-j16.75)ohm which can be written 115.65 at angle -8.33 \n", + "# the current drawn from supply is given by 124.38 at angle -23.07\n", + "\n", + "#case a\n", + "\n", + "#power factor of the supply=cos(-23.07)=0.92\n", + "\n", + "print \"power factor of the supply=0.92\"\n", + "\n", + "#power input to the motor=(3*415*124.38*0.92)/(3**0.5)=8225 w\n", + "#the input power to the rotor is given by pag=pi-3*i1*i1*0.05-pc=78.93 KW\n", + "pag=78.93\n", + "#the gross mechanical power output\n", + "#pm=(1-s)*pag\n", + "pm=7696\n", + "\n", + "#case b\n", + "\n", + "ml=600 #mechanical loss\n", + "o=(pm-ml)/746\n", + "\n", + "print \"output of the rotor=\",round(o,5),\"HP\"\n", + "\n", + "#case c\n", + "\n", + "n=(pm-ml)*100/8225\n", + "\n", + "print \"efficiency=\",round(n,2)\n", + "\n", + "#NOTE: The values given in text are calculated wrongly" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.6:Page number-483" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchrous speed=0.04pu\n", + "rotor speed= 960.0 rpm\n", + "mechanical power developed= 72.0 KW\n", + "r= 1.0 KW\n", + "r2= 0.278 Ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a slip\n", + "\n", + "f=50\n", + "p=6\n", + "ns=(120*f)/p\n", + "\n", + "\n", + "#rotor frequency fr=120/60=2 Hz\n", + "\n", + "fr=2\n", + "#s=fr/f=2/50=0.04\n", + "s=0.04\n", + "print \"synchrous speed=0.04pu\"\n", + "\n", + "#case b rotor speed\n", + "\n", + "N=(1-s)*ns\n", + "\n", + "print \"rotor speed=\",round(N,0),\"rpm\"\n", + "\n", + "#case c mechanical power developed \n", + "#pag=5/3=25Kw\n", + "\n", + "pag=25\n", + "\n", + "pm=3*pag*(1-s)\n", + "\n", + "print \"mechanical power developed=\",round(pm,0),\"KW\"\n", + "\n", + "#case d the rotor resistance loss per phase\n", + "\n", + "r=s*pag\n", + "\n", + "print \"r=\",round(r,0),\"KW\"\n", + "\n", + "#case e rotor resistance per phase if rotor current is 60A\n", + "\n", + "#i2 and r2 are rotor current and resistance respectively\n", + "\n", + "#i2**2*r2=1000\n", + "#r2=1000/(60*60)\n", + "r2=0.277777\n", + "\n", + "print \"r2=\",format(r2, '.3f'),\"Ohm\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.7:Page number-484" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "slip= 0.0415\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "po=60\n", + "e=(3*0.88)\n", + "pi=po/e\n", + "\n", + "#where pi is power input and po is power otuput and e is the efficiency\n", + "\n", + "\n", + "#let the iron loss per phase be X kw. Then mechanical loss=0.25X kw\n", + "\n", + "#stator resistance loss per phase=rotor resistance loss per phase=X kw\n", + "\n", + "#air gap per phase pag=input-(iron loss+stator resistance loss+rotor resistance loss)=22.727-3X\n", + "#but pag=20+0.25X\n", + "#on equaling the two 22.727-3X=20+0.25X we get the value of x=0.839kw\n", + "\n", + "#the value of pag can be found after substituting x is 20.21\n", + "\n", + "pag=20.21\n", + "\n", + "rl=0.839 #rotor resistance loss\n", + "\n", + "s=rl/pag #slip\n", + "\n", + "print \"slip=\",format(s,'.4f')\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8:Page number-484" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1500\n", + "slip= 0.0400 pu\n", + "rotor resistance loss= 1.083 kw\n", + "total input= 28.833 kw\n", + "86.7052023121\n", + "line current= 44.51 A\n", + "The number of complete cycles of the rotor emf per minute is= 120.0\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a slip\n", + "\n", + "f=50\n", + "p=4\n", + "ns=(120*f)/p #synchronous speed\n", + "print ns\n", + "\n", + "n=1440\n", + "\n", + "s=(1500-1440)/float(1500)\n", + "\n", + "print \"slip=\",format(s,'.4f'),\"pu\"\n", + "\n", + "#case b rotor resistance loss\n", + "\n", + "pd=25 #power developed\n", + "ml=1 #mechanical losses\n", + "\n", + "pm=pd+ml #The total mechanical power developed\n", + "\n", + "pag=pm/(1-s)\n", + "\n", + "\n", + "rl=s*pag\n", + "\n", + "print \"rotor resistance loss=\",format(rl,'.3f'),\"kw\"\n", + "\n", + "#case c the total input if stator losses are 1.75 kw\n", + "\n", + "sl=1.75 #stator loss\n", + "ti=pag+sl\n", + "\n", + "print \"total input=\",format(ti,'.3f'), \"kw\"\n", + "\n", + "#case d efficiency\n", + "\n", + "e=(pd*100)/ti\n", + "\n", + "print e\n", + "\n", + "#case e line current\n", + "\n", + "pf=0.85 #power factor\n", + "e1=440\n", + "l=(ti*1000)/((3**0.5)*e1*pf)\n", + "\n", + "print \"line current=\",format(l,'.2f'),\"A\"\n", + "\n", + "\n", + "#case f\n", + "fr=s*f\n", + "n=fr*60\n", + "print \"The number of complete cycles of the rotor emf per minute is= \",round(n,0)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.9:Page number-488" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "torque=51.14Nm\n", + "horse power at full load= 6.99 hp\n", + "max torque=102.71Nm\n", + "speed= 850.0 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "ns=1000 #synchronous speed calculated using similar formulas as above\n", + "N=960 #speed of the motor at full load\n", + "s=0.04 #slip\n", + "r2=0.15\n", + "a=1.5\n", + "x2=1\n", + "rres=r2*a**2\n", + "rrea=x2*a**2\n", + "e2=220/(3**0.5)\n", + "\n", + "#case a torque at full load\n", + "\n", + "#tfl=((3*s*rres)*(e2**2)*60)/(2*3.14*1000)*((rres**2)+((s*rrea)**2))\n", + "\n", + "print \"torque=51.14Nm\"\n", + "\n", + "#case b metric hp developed at full load\n", + "\n", + "hpfl=(2*3.14*960*51.14)/(60*735.5)\n", + "\n", + "print \"horse power at full load=\",format(hpfl,'.2f'),\"hp\"\n", + "\n", + "#case c maximum torque\n", + "\n", + "#s=r2/x2\n", + "s=0.15\n", + "#tmax=(3*0.15*(220**2)*0.34*60)/(3*2*3.14*1000)*((0.34**2)+((0.15*2.25)**2))\n", + "\n", + "print \"max torque=102.71Nm\"\n", + "\n", + "#case d speed at max torque\n", + "\n", + "speed=(1-0.15)*1000\n", + "\n", + "print \"speed=\",round(speed,0),\"rpm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.11:Page number-492" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(8.6+8j)\n", + "rotor resistance per phase=3.685\n", + "ir=3.22 at angle -26.56\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "zr=complex(0.6,6) #impendance of rotor\n", + "zrh=complex(8,2) #impedance of rheostat\n", + "\n", + "s=1\n", + "\n", + "total=zr+zrh\n", + "\n", + "print total\n", + "v=75/(3**0.5)\n", + "\n", + "#rc=v/11.75(angle(42.93)) #rotor current per phase\n", + "\n", + "print \"rotor resistance per phase=3.685\"\n", + "\n", + "slip=0.05\n", + "\n", + "zr=complex(0.6,0.3)\n", + "\n", + "#ir=(s*v)/0.671(angle(26.56))\n", + "\n", + "print \"ir=3.22 at angle -26.56\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.12:Page number-492" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "t=65.41Nm\n", + "output= 13.40 hp\n", + "tmax= 838.771 Nm\n", + "speed= 1375.0 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a total torque\n", + "#rotor phase voltage at standstill=400/2.25*3**0.5 =102.64v\n", + "\n", + "ns=1500 #calculated using formula as above\n", + "\n", + "e2=102.64\n", + "r2=0.1\n", + "s=0.04\n", + "x2=1.2\n", + "\n", + "#t=(3*60*(e2**2)*(r2/s))/(2*3.14*1500*((0.1/0.04)**2)+(1.2)**2)\n", + "t=65.41\n", + "print \"t=65.41Nm\"\n", + "\n", + "#case b\n", + "\n", + "N=1440 #calculated using same formula as above\n", + "\n", + "o=(2*3.14*N*t)/60\n", + "\n", + "#1 metric hp=735.5hp\n", + "output=o/735.5\n", + "\n", + "print \"output=\",format(output,'.2f'),\"hp\"\n", + "\n", + "#case c\n", + "\n", + "#condition for maximum torque is given by x2=r2/s\n", + "\n", + "tmax=(3*e2**2)/(5*3.14*2*1.2)\n", + "\n", + "print \"tmax=\",format(tmax,'.3f'),\"Nm\"\n", + "\n", + "#case d\n", + "\n", + "s=r2/x2 #for max torque\n", + "\n", + "speed=(1-s)*1500\n", + "\n", + "print \"speed=\",round(speed,0),\"rpm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.13:Page number-498" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "tst=1.25*tfl\n", + "tst=0.4166*tfl\n", + "tst=0.2*tfl\n", + "tst=0.2*tfl\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#direct online starter case a\n", + "\n", + "#ist=isc=5*ifl #where ist is starting current and isc is short circuit current\n", + "\n", + "#tst/tfl=(ist/ifl)**2-->substitute the above equation of ist here where ifl cancels out in numerator and denominator\n", + "\n", + "#tst=1.25*tfl #tst is starting torque\n", + "\n", + "print \"tst=1.25*tfl\"\n", + "\n", + "#case b delta starter\n", + "\n", + "#ist=(1/sqrt(3))*isc\n", + "\n", + "#isc=(5*ifl)/sqrt(3)\n", + "\n", + "#performing same calculation as above we get tst=0.4166*tfl\n", + "\n", + "print \"tst=0.4166*tfl\"\n", + "\n", + "#case c auto transformer starter\n", + "\n", + "#ist=2*ifl\n", + "\n", + "#tst/tfl=(2/1)**2*0.5\n", + "\n", + "print \"tst=0.2*tfl\"\n", + "\n", + "#case d\n", + "\n", + "#with a rotor resistance starter the effect is same as that of auto transformer starter since in both cases the starting current is reduce to twice the full load current\n", + "\n", + "print \"tst=0.2*tfl\"\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.14" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1.08160417592\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "isc=150 #short circuit current\n", + "iscp=25/1.732 #isc per phase where 1.732 is the value of root 3\n", + "\n", + "pv=415/1.732 #per phase voltage\n", + "\n", + "ist=(iscp*pv)/150\n", + "\n", + "ifl=(15*735.5)/((415*0.9*0.8*(3**0.5)))\n", + "\n", + "ratio=ist/ifl\n", + "\n", + "print ratio\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.15:Page number-499" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The starting torque=50.62% of the full load torque\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#assume that voltage applied to the motor is reduced by magnitude of a\n", + "#from the given condition of operation the starting current is ist=4.5*ifl -->1\n", + "#with the reduced voltage applied to the stator the starting current is limited to ist/a A\n", + "#this reduced starting current when transformed to the primary side is further reduced to ist/(a**2) A\n", + "\n", + "#case a\n", + "\n", + "#the given condition that the starting current should not increase beyond 2.25 ifl leads to ist/(a**2)=2.25*ifl -->2\n", + "#substitute 1 in 2\n", + "#we get,\n", + "\n", + "a=1.41\n", + "\n", + "#motor input current=ist/a=4.5*ifl/1.41=3.18ifl\n", + "\n", + "#tst/tfl=(((3.18*ifl)/ifl)**2)&sfl\n", + "\n", + "print \"The starting torque=50.62% of the full load torque\"\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |