{
 "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",
    "#given\n",
    "f=50\n",
    "p=4\n",
    "#case a\n",
    "s=(120*f)/p #synchronous speed\n",
    "print \"synchronous speed=\",round(s,0),\"rpm\"\n",
    "#case b\n",
    "slip=0.03\n",
    "r=s-s*slip #rotor speed\n",
    "print \"rotor speed=\",round(r,0),\"rpm\"\n",
    "#case c\n",
    "r=900 #given speed of rotor\n",
    "slip=(s-r)/s #per unit slip\n",
    "rf=slip*f\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",
    "#given\n",
    "pg=10 #poles of generator\n",
    "r=720 #synchronous speed\n",
    "f=pg*r/120\n",
    "print \"frequency=\",round(f,0),\"Hz\"\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",
    "#case a\n",
    "r=1200 #synchronous speed possible for present case\n",
    "pi=120*f/r #poles of the induction motor\n",
    "print \"The number of poles of an induction motor is=\",round(pi,0)\n",
    "#case b\n",
    "n=1170  #load speed\n",
    "slip=(1200-n)/1200 #calculated as 0.025\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",
    "#given\n",
    "f=50\n",
    "ns=1000\n",
    "#m=90/6*3\n",
    "m=5\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",
    "#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",
    "kd=0.96\n",
    "#after calculations\n",
    "print \"The distribution factor=0.96\"\n",
    "kp=0.98 #pitch factor=cos(20/2)\n",
    "#case a\n",
    "kw=kd*kp\n",
    "print kw\n",
    "#case b\n",
    "t1=(90*4)/(3*2) #number of turns per stator phase\n",
    "e1=415\n",
    "flux=415/((3**0.5)*4.44*0.94*50*60)\n",
    "print \"flux in the air gap=\",round(flux,3),\"Wb\"\n",
    "#case c\n",
    "t2=(120*2)/(3*2)\n",
    "a=t1/t2 #transformation ratio\n",
    "print round(a,5)\n",
    "#case d\n",
    "#e2=e1/a #the induced rotor voltage per phase\n",
    "e2=415/((3**0.5)*1.5)\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",
    "#given\n",
    "s=1\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",
    "#rotor current=e2/z2=3.23 at angle -63.43\n",
    "print \"At stand still the rotor current is=3.23A at angle -63.43\"\n",
    "#case b\n",
    "s=0.04\n",
    "#z2=(0.75+j*0.04*5)ohm \n",
    "#again e2=s*e2/z2=0.81 at angle -69.44A\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",
    "#given\n",
    "p=6\n",
    "f=50\n",
    "pc=1000\n",
    "ml=600\n",
    "n=975 \n",
    "ns=(120*50)/p\n",
    "print \"synchronous speed=\",round(ns,0),\"rpm\"\n",
    "#s=(ns-n)/ns\n",
    "s=0.025\n",
    "print \"s=0.025\"\n",
    "#the rotor impedance referred to the stator side z2=(2+j0/15)ohm\n",
    "#assuming the per phase supply voltage as the reference phasor it is seen that the stator load current is,\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",
    "#case a\n",
    "#power factor of the supply=cos(-23.07)=0.92\n",
    "print \"power factor of the supply=0.92\"\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",
    "#case b\n",
    "ml=600 #mechanical loss\n",
    "o=(pm-ml)/746\n",
    "print \"output of the rotor=\",round(o,5),\"HP\"\n",
    "#case c\n",
    "n=(pm-ml)*100/8225\n",
    "print \"efficiency=\",round(n,2)\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",
    "#case a slip\n",
    "f=50\n",
    "p=6\n",
    "ns=(120*f)/p\n",
    "#rotor frequency fr=120/60=2 Hz\n",
    "fr=2\n",
    "#s=fr/f=2/50=0.04\n",
    "s=0.04\n",
    "print \"synchrous speed=0.04pu\"\n",
    "#case b rotor speed\n",
    "N=(1-s)*ns\n",
    "print \"rotor speed=\",round(N,0),\"rpm\"\n",
    "#case c mechanical power developed \n",
    "#pag=5/3=25Kw\n",
    "pag=25\n",
    "pm=3*pag*(1-s)\n",
    "print \"mechanical power developed=\",round(pm,0),\"KW\"\n",
    "#case d the rotor resistance loss per phase\n",
    "r=s*pag\n",
    "print \"r=\",round(r,0),\"KW\"\n",
    "#case e rotor resistance per phase if rotor current is 60A\n",
    "#i2 and r2 are rotor current and resistance respectively\n",
    "#i2**2*r2=1000\n",
    "#r2=1000/(60*60)\n",
    "r2=0.277777\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",
    "#given\n",
    "po=60\n",
    "e=(3*0.88)\n",
    "pi=po/e\n",
    "#where pi is power input and po is power otuput and e is the efficiency\n",
    "#let the iron loss per phase be X kw. Then mechanical loss=0.25X kw\n",
    "#stator resistance loss per phase=rotor resistance loss per phase=X kw\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",
    "#the value of pag can be found after substituting x is 20.21\n",
    "pag=20.21\n",
    "rl=0.839 #rotor resistance loss\n",
    "s=rl/pag #slip\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",
    "#case a slip\n",
    "f=50\n",
    "p=4\n",
    "ns=(120*f)/p #synchronous speed\n",
    "print ns\n",
    "n=1440\n",
    "s=(1500-1440)/float(1500)\n",
    "print \"slip=\",format(s,'.4f'),\"pu\"\n",
    "#case b rotor resistance loss\n",
    "pd=25 #power developed\n",
    "ml=1 #mechanical losses\n",
    "pm=pd+ml #The total mechanical power developed\n",
    "pag=pm/(1-s)\n",
    "rl=s*pag\n",
    "print \"rotor resistance loss=\",format(rl,'.3f'),\"kw\"\n",
    "#case c the total input if stator losses are 1.75 kw\n",
    "sl=1.75 #stator loss\n",
    "ti=pag+sl\n",
    "print \"total input=\",format(ti,'.3f'), \"kw\"\n",
    "#case d efficiency\n",
    "e=(pd*100)/ti\n",
    "print e\n",
    "#case e line current\n",
    "pf=0.85 #power factor\n",
    "e1=440\n",
    "l=(ti*1000)/((3**0.5)*e1*pf)\n",
    "print \"line current=\",format(l,'.2f'),\"A\"\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",
    "#given\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",
    "#case a torque at full load\n",
    "#tfl=((3*s*rres)*(e2**2)*60)/(2*3.14*1000)*((rres**2)+((s*rrea)**2))\n",
    "print \"torque=51.14Nm\"\n",
    "#case b metric hp developed at full load\n",
    "hpfl=(2*3.14*960*51.14)/(60*735.5)\n",
    "print \"horse power at full load=\",format(hpfl,'.2f'),\"hp\"\n",
    "#case c maximum torque\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",
    "print \"max torque=102.71Nm\"\n",
    "#case d speed at max torque\n",
    "speed=(1-0.15)*1000\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",
    "zr=complex(0.6,6) #impendance of rotor\n",
    "zrh=complex(8,2)  #impedance of rheostat\n",
    "s=1\n",
    "total=zr+zrh\n",
    "print total\n",
    "v=75/(3**0.5)\n",
    "#rc=v/11.75(angle(42.93)) #rotor current per phase\n",
    "print \"rotor resistance per phase=3.685\"\n",
    "slip=0.05\n",
    "zr=complex(0.6,0.3)\n",
    "#ir=(s*v)/0.671(angle(26.56))\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",
    "#case a total torque\n",
    "#rotor phase voltage at standstill=400/2.25*3**0.5 =102.64v\n",
    "ns=1500 #calculated using formula as above\n",
    "e2=102.64\n",
    "r2=0.1\n",
    "s=0.04\n",
    "x2=1.2\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",
    "#case b\n",
    "N=1440 #calculated using same formula as above\n",
    "o=(2*3.14*N*t)/60\n",
    "#1 metric hp=735.5hp\n",
    "output=o/735.5\n",
    "print \"output=\",format(output,'.2f'),\"hp\"\n",
    "#case c\n",
    "#condition for maximum torque is given by x2=r2/s\n",
    "tmax=(3*e2**2)/(5*3.14*2*1.2)\n",
    "print \"tmax=\",format(tmax,'.3f'),\"Nm\"\n",
    "#case d\n",
    "s=r2/x2  #for max torque\n",
    "speed=(1-s)*1500\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",
    "#direct online starter case a\n",
    "#ist=isc=5*ifl  #where ist is starting current and isc is short circuit current\n",
    "#tst/tfl=(ist/ifl)**2-->substitute the above equation of ist here where ifl cancels out in numerator and denominator\n",
    "#tst=1.25*tfl #tst is starting torque\n",
    "print \"tst=1.25*tfl\"\n",
    "#case b delta starter\n",
    "#ist=(1/sqrt(3))*isc\n",
    "#isc=(5*ifl)/sqrt(3)\n",
    "#performing same calculation as above we get tst=0.4166*tfl\n",
    "print \"tst=0.4166*tfl\"\n",
    "#case c auto transformer starter\n",
    "#ist=2*ifl\n",
    "#tst/tfl=(2/1)**2*0.5\n",
    "print \"tst=0.2*tfl\"\n",
    "#case d\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",
    "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",
    "isc=150 #short circuit current\n",
    "iscp=25/1.732 #isc per phase where 1.732 is the value of root 3\n",
    "pv=415/1.732 #per phase voltage\n",
    "ist=(iscp*pv)/150\n",
    "ifl=(15*735.5)/((415*0.9*0.8*(3**0.5)))\n",
    "ratio=ist/ifl\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",
    "#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",
    "#case a\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",
    "a=1.41\n",
    "#motor input current=ist/a=4.5*ifl/1.41=3.18ifl\n",
    "#tst/tfl=(((3.18*ifl)/ifl)**2)&sfl\n",
    "print \"The starting torque=50.62% of the full load torque\"\n",
    "\n"
   ]
  }
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