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-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 44 : INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.1 , PAGE NO :- 1769"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Total annual cost in group drive = RS. 11000.0\n",
-      "Total annual cost in individual drive = RS. 11400.0\n",
-      "Hence,Individual drive is costlier than group drive.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A motor costing Rs. 10,000 is used for group drive in a certain installation.How will its total annual cost compare with\n",
-    "the case where four individuals motors each costing Rs. 4000 were used? With group drive,the energy consumption is\n",
-    "50MWh whereas it is 45MWh for individual drive.The cost of electric energy is 20 paise/kWh.Assume depriciation,\n",
-    "maintenance and other fixed charges at 10% in the case of group drive and 15 percent in the case of individual drive.'''\n",
-    "\n",
-    "\n",
-    "#Group drive\n",
-    "cost_g  = 10000.0               #Rs      (Capital cost)\n",
-    "other_g = 0.1*cost_g            #Rs      (Annual depriciation,maintenance and other charges)\n",
-    "enrgy_g = 50.0*1000*20/100.0         #Rs      (Annual cost of energy)\n",
-    "total_g = enrgy_g + other_g     #Rs      (total annual cost)\n",
-    "\n",
-    "#Individual drive\n",
-    "cost_i  = 4*4000.0               #Rs      (Capital cost)\n",
-    "other_i = 0.15*cost_i            #Rs      (Annual depriciation,maintenance and other charges)\n",
-    "enrgy_i = 45.0*1000*20/100.0          #Rs      (Annual cost of energy)\n",
-    "total_i = enrgy_i + other_i      #Rs      (total annual cost)\n",
-    "print \"Total annual cost in group drive = RS.\",total_g\n",
-    "print \"Total annual cost in individual drive = RS.\",total_i\n",
-    "print \"Hence,Individual drive is costlier than group drive.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.2 , PAGE NO :- 1775 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Additional resistance =  5.7 ohm.\n",
-      "Initial braking Torque =  955.41 N-m.\n",
-      "Torque at 360 rpm =  766.53 N-m.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 40-kW,440-V,d.c shunt motor is braked by plugging.Calculate(i)the value of resistance that must be placed in series with the\n",
-    "armature circuit to limit the initial braking current to 150A (ii)the braking torque and(iii)the torque when motor speed falls\n",
-    "to 360 rpm. Armature resistance Ra = 0.1 ohm,ful-load Ia=100A,full-load speed=600 rpm.'''\n",
-    "\n",
-    "V = 440.0          #V    (applied voltage)\n",
-    "Ib = 150.0         #A    (initial braking current)\n",
-    "Ra = 0.1           #ohm  (armature resistance)\n",
-    "Ia = 100.0         #A    (armature current)\n",
-    "N1 = 600.0         #rpm  (full load speed)\n",
-    "N2 = 360.0         #rpm  (decreased speed)\n",
-    "Eb = V - Ia*Ra     #V     (back emf)\n",
-    "#Voltage across the armature at the start of braking\n",
-    "V2 = V + Eb        #V\n",
-    "#(i)Since initial braking current is limited to 150A,total armature circuit resistance required is\n",
-    "Rt = V2/Ib        #ohm\n",
-    "#Therefore additional resistance R is\n",
-    "R = Rt - Ra        #ohm\n",
-    "#(ii)For a shunt motor,Torque(Tb) is propotional to Ia\n",
-    "Tb = 40*1000/(2*3.14*600/60)             #N/m\n",
-    "# .'. Initial braking torque/full-load torque = initial braking current/full-load current\n",
-    "\n",
-    "T_ini = Tb*(Ib/Ia)                  #N/m\n",
-    "#(iii)The decrease in Eb is directly propotional to decrease in motor speed\n",
-    "Eb_360 = Eb*(N2/N1)            #V\n",
-    "Ia_360 = (V+Eb_360)/Rt             #A\n",
-    "Tb_360 = Tb*(Ia_360/Ia)        #N-m\n",
-    "\n",
-    "print \"Additional resistance = \",R,\"ohm.\"\n",
-    "print \"Initial braking Torque = \",round(T_ini,2),\"N-m.\"\n",
-    "print \"Torque at 360 rpm = \",round(Tb_360,2),\"N-m.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.3 , PAGE NO :- 1776"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Plugging torque =  131.92 N-m.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 30-kW,400-V,3-phase,4 pole,50-Hz induction motor has full-load slip of 5%.If the ratio of standstill reactance to resistance\n",
-    "per motor phase is 4 ,estimating the plugging torque at full speed.'''\n",
-    "\n",
-    "Power = 30.0           #kW      (power consumed)\n",
-    "f = 50.0               #Hz      (frequency)\n",
-    "P = 4                  #        (pole)\n",
-    "s1 = 0.05               #        (slip)\n",
-    "Ns = 120*f/P           #rpm        (Synchronus speed)\n",
-    "Nf = Ns*(1-s1)          #rpm        (Full-load speed)\n",
-    "Tf = Power*1000/(2*3.14*Nf/60)  #N-m   (Full-load torque)\n",
-    "R1_X1 = 4.0 \n",
-    "\n",
-    "# As T is propotional to (s*R2*E2^2)/(R2^2 + s^2*X2^2) i.e (T2/T1) = (s2/s1)*(1+s1^2(X2/R2)^2/1+s2^2(X2/R2)^2)\n",
-    "s2 = 2-s1\n",
-    "Tp = (s2/s1)*(1 +(R1_X1)*(R1_X1)*s1*s1)/(1 +(R1_X1)*(R1_X1)*s2*s2)*Tf\n",
-    "print \"Plugging torque = \",round(Tp,2),\"N-m.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.4 , PAGE NO :- 1780"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Hence,energy returned to supply lines = 43.43 kWh.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 500 tonne electric trains travel down a descending gradient of 1 in 80 for 90 seconds during which period its speed is\n",
-    "reduced from 100 km/h to 60 km/h by regenerative braking.Compute the energy returned to the lines of kWh if tractive\n",
-    "resistance = 50N/t;allowance for rotational inertia = 10%;overall efficiency of the system = 75%.'''\n",
-    "\n",
-    "\n",
-    "G = 1/80.0*100          #      (percent gradient)\n",
-    "M = 500.0               #tonne (electric train)\n",
-    "Me_M = 1.1              #      (ratio of rotational mass to stationary mass)\n",
-    "V1 = 100.0              #km/h  (initial speed)\n",
-    "V2 = 60.0               #km/h  (decreased speed)\n",
-    "t = 90.0                #s     (braking period)\n",
-    "r = 50.0                #N/t      (tractive resistance)\n",
-    "eff = 0.75              #      (efficiency)\n",
-    "d = (V1+V2)/2*t/3600\n",
-    "#Hence,energy returned to supply line is\n",
-    "enrgy = 0.75*(0.01072*(Me_M)*(V1*V1 - V2*V2) + d*(27.25*G - 0.2778*r))  #Wh/tonne\n",
-    "enrgy = enrgy*500/1000.0  #kWh\n",
-    "print \"Hence,energy returned to supply lines =\",round(enrgy,2),\"kWh.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.5 , PAGE NO :- 1780"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power returned to line = 509.0 kW.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 350-t electric train has its speed reduced by regenerative braking from 60 to 40km/h over a distance of 2km along down gradient\n",
-    "of 1.5%.Calculate (i)electrical energy and (ii)average power returned to the line.Assume specific train resistance = 50 N/t;rotational\n",
-    "inertia effect = 10% ; conversion efficency of the system = 75%.'''\n",
-    "\n",
-    "\n",
-    "M = 350.0     #tonne                (Mass of train)\n",
-    "eff = 0.75    #                     (efficiency)\n",
-    "V1 = 60.0     #km/h                 (initial speed)\n",
-    "V2 = 40.0     #km/h                 (final speed)\n",
-    "Me = 1.1*M    #                     (rotational mass )\n",
-    "G = 1.5       #                     (percent gradient)\n",
-    "d = 2.0       #km                   (distance)\n",
-    "r = 50.0      #N/t                 (train resistance)\n",
-    "\n",
-    "#Energy returned to line is\n",
-    "enrgy = eff*(0.01072*Me/M*(V1*V1 - V2*V2) + d*(27.25*G - 0.2778*r))   #Wh/t\n",
-    "enrgy = enrgy*M/1000         #kWh\n",
-    "\n",
-    "#(ii)\n",
-    "\n",
-    "speed = (V1+V2)/2     #km/h         (Average speed)\n",
-    "time = d/speed        #h            (time taken)\n",
-    "\n",
-    "power = enrgy/time    #kW           (power returned)\n",
-    "\n",
-    "print \"Power returned to line =\",round(power),\"kW.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.6 , PAGE NO :- 1780"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power fed to line =  424.38 kW.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''If in Example 44.5,regenerative braking is applied in such a way that train speed on down gradient remains constant\n",
-    "at 60 km/h,what would be the power fed into the line?'''\n",
-    "\n",
-    "\n",
-    "M = 350.0      #tonne          (Mass of train)\n",
-    "G = 1.5        #               (percent gradient)\n",
-    "r = 50.0       #N/t            (train resistance)\n",
-    "V = 60.0       #km/h           (speed)\n",
-    "eff = 0.75\n",
-    "#In down-gradient motors act as generators .Force generated\n",
-    "Ft = 98*M*G - M*r    #N\n",
-    "#Power that can be recuperated is\n",
-    "P = Ft*(1000.0/3600)*V                 #W\n",
-    "#Power actually sent to line\n",
-    "P = eff*P/1000                         #kW\n",
-    "print \"Power fed to line = \",round(P,2),\"kW.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.7 , PAGE NO :- 1780 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power fed =  650.0 kW.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A train weighing 500 tonne is going down a gradient of 20 in 1000.It is desired to  maintain train speed at 40 km/h by regenerative\n",
-    "braking.Calculate the power fed into the line.Tractive resistance is 40 N/t and allow rotational inertia of 10% and efficiency\n",
-    "of conversion of 75%.'''\n",
-    "\n",
-    "M = 500.0         #tonne       (Mass of train)\n",
-    "G = 20/1000.0*100 #            (percent gradient)\n",
-    "r = 40.0          #N/t         (Tractive resistance)\n",
-    "V = 40.0          #km/h        (speed)\n",
-    "eff = 0.75        #            (efficiency)     \n",
-    "#Tractive Force when motors are driven as generators is\n",
-    "Ft = 98*M*G - M*r        #N\n",
-    "#Power that can be drawn is\n",
-    "P = 0.2778*Ft*V         #W\n",
-    "#Power actually fed\n",
-    "P = eff*P/1000               #kW\n",
-    "print \"Power fed = \",round(P,0),\"kW.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.8 , PAGE NO :- 1781"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Additional resistance =  1.94 ohm.\n",
-      "Initial braking torque =  1042.27 N-m.\n",
-      "Braking torque at 200rpm =  719.84 N-m.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 250V d.c shunt motor,taking an armature current of 150 A and running at 550 r.p.m is braked by reversing the connections to the\n",
-    "armature and inserting additional resistance in series with it.Calculate:\n",
-    "(a)the value of series resistance required to limit the initial current to 240A.\n",
-    "(b)the initial value of braking torque.\n",
-    "(c)the value of braking torque when the speed has fallen to 200 r.p.m.\n",
-    "The armature resistance is 0.09 ohm.Neglect winding friction and iron losses.'''\n",
-    "\n",
-    "V  = 250.0          #V     (applied voltage)\n",
-    "Ia = 150.0          #A     (armature current)\n",
-    "Ib = 240.0          #A     (initial braking current)\n",
-    "N  = 550.0          #rpm   (speed)\n",
-    "N2 = 200.0          #rpm   (decreased speed)  \n",
-    "Ra = 0.09           #ohm  \n",
-    "#Induced emf at full-load\n",
-    "Eb = V - Ia*Ra      #V\n",
-    "#Voltage across the armature at braking\n",
-    "Vb = V + Eb         #V\n",
-    "#Resistance to limit the current to 240A\n",
-    "Rt = Vb/240          #ohm\n",
-    "#Resistance to be added in the circuit\n",
-    "R = Rt - Ra          #ohm\n",
-    "\n",
-    "#(ii)\n",
-    "Tf = V*Ia/(2*3.14*N/60)     #N-m  (Full load torque)\n",
-    "# Initial braking torque/full-load torque = initial braking current/full-load current\n",
-    "\n",
-    "T_ini = Tf*(Ib/Ia)        #N-m  (initial braking torque)\n",
-    "\n",
-    "Eb_200 = Eb*N2/N          #V    (Back emf at 200 rpm)\n",
-    "Ia_200 = (V + Eb_200)/Rt  #A    (Current drawn at 200 rpm)\n",
-    "Tb_200 = Tf*Ia_200/Ia     #N-m\n",
-    "\n",
-    "print \"Additional resistance = \",round(R,2),\"ohm.\"\n",
-    "print \"Initial braking torque = \",round(T_ini,2),\"N-m.\"\n",
-    "print \"Braking torque at 200rpm = \",round(Tb_200,2),\"N-m.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.9 , PAGE NO :- 1781"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Starting torque Ts =  0.25 *Tfl.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 400V 3-phase squirrel cage induction motor has a full load slip of 4%.A stand-still impedance of 1.54 ohm and the full load current\n",
-    "equal to 30A.The maximum starting current which may be taken from line is 75A.What taping must be provided on an auto-transformer starter\n",
-    "to limit the current to this value and what would be the starting torque available in terms of full-load torque ?'''\n",
-    "\n",
-    "import math as m\n",
-    "from sympy import Eq,solve,Symbol\n",
-    "\n",
-    "#considering Transformer action V2/V1 = I1/I2 = X\n",
-    "V1 = 400.0/(m.sqrt(3))      #V     (applied voltage)\n",
-    "I1 = 75.0                   #A     (max starting current)\n",
-    "Z = 1.54                    #ohm   (impedance)\n",
-    "X = Symbol('X')\n",
-    "I2 = I1/X                   #A\n",
-    "V2 = Z*I2                   #V\n",
-    "eq = Eq(V1*I1,V2*I2)\n",
-    "X = solve(eq)               \n",
-    "X1 = X[1]                   #ohm\n",
-    "\n",
-    "I2 = I1/X1                   #A\n",
-    "sfl = 0.04                   #   (full-load slip)\n",
-    "Ifl = 30.0                   #A   (full-load current)\n",
-    "\n",
-    "#Ts/Tfl = X^2*(Is/Ifl)^2*sfl\n",
-    "\n",
-    "Ts_Tfl = (X1*X1)*(I2/Ifl)*(I2/Ifl)*sfl       # (Ts/Tfl)\n",
-    "print \"Starting torque Ts = \",round(Ts_Tfl,2),\"*Tfl.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.10 , PAGE NO :- 1782"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Additional resistance(constant torque) =  1.02 ohm.\n",
-      "Additional resistance(torque propotional to speed square) =  1.77 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 220V,10 HP shunt motor has field and armature resistances of 122ohm and 0.3ohm respectively.Calculate the resistance to be\n",
-    "inserted in the armature circuit to reduce the speed to 80% assuming motor efficiency at full load to be 80%.\n",
-    "(a)When torque is to remain constant.\n",
-    "(b)When torque is propotional to square of the speed.'''\n",
-    "\n",
-    "\n",
-    "V = 220.0        #V     (applied voltage)\n",
-    "Rf = 122.0       #ohm   (field resistance)\n",
-    "Ra = 0.3         #ohm   (armature resistance)\n",
-    "\n",
-    "If = V/Rf              #A     (field current)\n",
-    "\n",
-    "m_out = 10*746         #W   (motor output)\n",
-    "m_in  = m_out/0.8      #W   (motor input)\n",
-    "Il = m_in/V            #A   (line current)\n",
-    "\n",
-    "Ia = Il - If           #A   (armature current)\n",
-    "\n",
-    "Eb1 = V - Ia*Ra        #V   (back emf)\n",
-    "\n",
-    "#As flux is constant     N2/N1 = Eb2/Eb1 and N2 = N1*0.8\n",
-    "\n",
-    "Eb2 = Eb1*0.8          #V       (back emf at reduced speed)\n",
-    "\n",
-    "#(a) Torque remains constant ,hence Ia remains constant .Using Eb2 = V - Ia*R\n",
-    "Rt = (V-Eb2)/Ia        #ohm       (Total resistance required)\n",
-    "\n",
-    "#Therefore, additional resistance is\n",
-    "R = Rt - Ra            #ohm\n",
-    "print \"Additional resistance(constant torque) = \",round(R,2),\"ohm.\"\n",
-    "\n",
-    "#(b)   As (T2/T1) = (N2/N1)^2 and   (T2/T1) = (Ia2/Ia1)\n",
-    "\n",
-    "T2_T1 =  (0.8)*(0.8)      # (T2/T1)\n",
-    "\n",
-    "Ia2 = Ia*T2_T1           #A        (Changed armature current)\n",
-    "\n",
-    "#(b) Using Eb2 = V - Ia*R\n",
-    "Rt = (V-Eb2)/Ia2        #ohm       (Total resistance required)\n",
-    "\n",
-    "\n",
-    "#Therefore, additional resistance is\n",
-    "R = Rt - Ra            #ohm\n",
-    "print \"Additional resistance(torque propotional to speed square) = \",round(R,2),\"ohm.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.11 , PAGE NO :- 1783"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Initial braking torque =  713.33 N-m.\n",
-      "Electric braking torque at 1/2 speed =  539.96 N-m.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 37.5 HP, 220V DC shunt motor with a full load speed of 535 rpm is to be braked by plugging.Estimate the value of resistance \n",
-    "which should be placed in series with it to limit the initial braking current to 200 A.What would be the initial value of the\n",
-    "electric braking torque and the value when the speed had fallen to half its full load value?Armature resistance of motor is \n",
-    "0.086 ohm and full load armature current is 140A.'''\n",
-    "\n",
-    "V = 220.0          #V          (applied voltage)\n",
-    "Ia = 140.0         #A          (full-load armature current)\n",
-    "Ra = 0.086         #ohm        (armature resistance)\n",
-    "Ib = 200.0         #A          (braking current)\n",
-    "P = 37.5*746       #W          (Power)\n",
-    "N = 535.0          #rpm        (Speed)\n",
-    "Eb = V - Ia*Ra     #V          (Back emf)\n",
-    "#Total voltage during braking\n",
-    "Vb = Eb + V        #V\n",
-    "\n",
-    "#Total resistance required is (using R = V/I)\n",
-    "Rt = Vb/Ib         #ohm\n",
-    "Rt = round(Rt,2)   #ohm\n",
-    "#Therefore,additional resistance required is\n",
-    "R = Rt - Ra        #ohm\n",
-    "#We know that P = Torque*w where w is\n",
-    "w = 2*3.1416*N/60    #rad/s            (angular velocity)\n",
-    "Tfl = P/w          #N-m              (full-load torque)\n",
-    "#As torque is propotional to I\n",
-    "#(Initial braking torque/Initial braking current) = (Full load torque/Full load current)\n",
-    "T_ini = Tfl*(Ib/Ia) #N-m             (Initial braking torque)\n",
-    "\n",
-    "#As speed is propotional to back emf\n",
-    "Eb_2 = Eb/2         #V           (Back emf at 1/2 speed)\n",
-    "Ib_2 = (V+Eb_2)/Rt  #A           (initial braking current at 1/2 speed)\n",
-    "\n",
-    "T_ini2 = Tfl*(Ib_2/Ia) #N-m      (Initial braking torque at 1/2 speed)\n",
-    "print \"Initial braking torque = \",round(T_ini,2),\"N-m.\"\n",
-    "print \"Electric braking torque at 1/2 speed = \",round(T_ini2,2),\"N-m.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## EXAMPLE 44.12 , PAGE NO :- 1783"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Speed when torque is constant is =  600.61 rpm.\n",
-      "Speed when torque is propotional to speed square =  546.14 rpm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 500V series motor having armature and field resistances of 0.2 and 0.3 ohm runs at 500 rpm when taking 70A.Assumimg unsaturated\n",
-    "field find out its speed when field diverter of 0.648 ohm is used for following load whose torque\n",
-    "(a) remains constant\n",
-    "(b) varies as square of speed  '''\n",
-    "\n",
-    "\n",
-    "from sympy import Eq,solve,Symbol\n",
-    "V = 500.0     #V      (applied voltage)\n",
-    "Ra = 0.2      #ohm    (armature resistance)\n",
-    "Rf = 0.3      #ohm    (field resistance)\n",
-    "N1 = 500.0     #rpm    (speed)\n",
-    "Ia = 70.0     #A      (armature current)\n",
-    "Rd = 0.684    #ohm    (diverter resistance)\n",
-    "Eb1 = V - Ia*(Ra + Rf)    #V   (Back emf)\n",
-    "\n",
-    "#(a)Let Ia2 be armature current when diverter is used\n",
-    "\n",
-    "Ia2 = Symbol('Ia2')\n",
-    "\n",
-    "#Now If2 (field current when diverter is used) is\n",
-    "If2 = Ia2*Rd/(Rf+Rd)\n",
-    "\n",
-    "#As Torque is constant Ia1*(Flux1) = Ia2*(Flux2)  Also,Ia1=If1 is propotional to (flux1)\n",
-    "eq = Eq(Ia*Ia/If2-Ia2,0)\n",
-    "Ia2 = solve(eq)      \n",
-    "Ia_2 = Ia2[1]         #A     (Armature current when diverter is used)\n",
-    "\n",
-    "If2 = Ia_2*Rd/(Rf+Rd) #A      (field current when diverter is used)\n",
-    "\n",
-    "#Resistance of field with diverter\n",
-    "Rfd = Rf*Rd/(Rf+Rd)   #ohm\n",
-    "#Total resistance\n",
-    "Rt = Rfd + Ra         #ohm\n",
-    "\n",
-    "Eb2 = V - Ia_2*Rt      #V   (Back emf when diverter is used)\n",
-    "\n",
-    "#Now,  (N1/N2) = (Eb1*flux1/Eb2*flux2)  and flux is propotional to If\n",
-    "N2 = (Eb2/Eb1)*N1*(Ia/If2)     #rpm\n",
-    "print \"Speed when torque is constant is = \",round(N2,2),\"rpm.\"\n",
-    "####################################################################################\n",
-    "\n",
-    "#(B)Let Ia22 be armature current when diverter is used\n",
-    "#Now, (N1/N2)^2 = T1/T2   As (T1/T2) = Ia1*Ia1/(Ia2*If2)\n",
-    "\n",
-    "Ia22 = Symbol('Ia22')\n",
-    "#Now If2 (field current when diverter is used) is\n",
-    "If2 = Ia22*Rd/(Rf+Rd)\n",
-    "\n",
-    "N1_N2a = Ia*Ia/(Ia22*If2)                 #N1_N2a -> (N1/N2)^2\n",
-    "#Also,  N1/N2 = Eb1*flux2/(Eb2*flux1)\n",
-    "N1_N2b = Eb1*If2/((V - Ia22*Rt)*Ia)       #N1_N2b -> (N1/N2)\n",
-    "\n",
-    "eq = Eq(N1_N2a,N1_N2b*N1_N2b)\n",
-    "Ia22 = solve(eq)\n",
-    "Ia_22 = Ia22[1]                       #A     (Armature current when diverter is used)\n",
-    "\n",
-    "If2 = Ia_22*Rd/(Rf+Rd)\n",
-    "N1_N2b = Eb1*If2/((V - Ia_22*Rt)*Ia)  \n",
-    "#Using equation of N1_N2b = N1/N2\n",
-    "N2  = N1/N1_N2b     #rpm\n",
-    "\n",
-    "print \"Speed when torque is propotional to speed square = \",round(N2,2),\"rpm.\"\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "## EXAMPLE 44.13 , PAGE NO :- 1784 "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Additonal resistance required is =  5.9 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 200V series motor runs at 1000 rpm and takes 20A . Armature and field resistance is 0.4 ohm.Calculate the\n",
-    "resistance to be inserted in series so as to reduce the speed to 800 rpm,assuming torque to vary as cube of the\n",
-    "speed and unsaturated field.'''\n",
-    "\n",
-    "from sympy import Eq,Symbol,solve\n",
-    "\n",
-    "V = 200.0        #V      (applied voltage)\n",
-    "N1 = 1000.0      #rpm    (speed 1)\n",
-    "Raf = 0.4        #ohm    (armature and field resistance)\n",
-    "N2 = 800.0       #rpm    (speed 2)\n",
-    "Ia1 = 20.0       #A      (armature current)\n",
-    "\n",
-    "\n",
-    "#Given,   T1/T2 = (N1/N2)^3\n",
-    "\n",
-    "T1_T2a  = (N1/N2)*(N1/N2)*(N1/N2)        # (Ratio   T1/T2)\n",
-    "\n",
-    "#Also T1/T2 = Ia1*flux1/Ia2*flux2   and Ia is propotional to flux\n",
-    "#Let Ia2 be armature current when speed is 800 rpm.\n",
-    "\n",
-    "Ia2 = Symbol('Ia2')\n",
-    "T1_T2b = Ia1*Ia1/(Ia2*Ia2)\n",
-    "eq = Eq(T1_T2a,T1_T2b)\n",
-    "Ia2 = solve(eq)\n",
-    "Ia_2 = Ia2[1]         #A        (armature current 2)\n",
-    "\n",
-    "Eb1 = V - Ia1*Raf      #V         (Back emf 1)\n",
-    "\n",
-    "#As Eb1/Eb2 = N1*I1/(N2*I2).Therefore Back emf 2 is\n",
-    "Eb2 = Eb1*(N2/N1)*(Ia_2/Ia1)       #V   (Back emf 2)\n",
-    "\n",
-    "#Also Eb2 = V - Ia2*Rt .Therefore total resistance required is\n",
-    "Rt = (V - Eb2)/Ia_2         #ohm\n",
-    "#Therefore,additional resistance required is\n",
-    "R = Rt - Raf               #ohm\n",
-    "print \"Additonal resistance required is = \",round(R,2),\"ohm.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "## EXAMPLE 44.14 , PAGE NO :- 1785"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Speed at which torque is 75% of initial value = 242.79 rpm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 220V,500 rpm DC shunt motor with an armature resistance of 0.08 ohm and full load armature current of 150A is to be braked by \n",
-    "plugging.Estimate the value of resistance which is to be placed in series with the armature to limit initial braking current to\n",
-    "200A.What would be the speed at which the electric braking torque is 75% of its initial value.'''\n",
-    "\n",
-    "from sympy import Eq,solve,Symbol\n",
-    "\n",
-    "V = 220.0     #V        (applied voltage)\n",
-    "N1 = 500.0    #rpm      (speed 1)\n",
-    "Ra = 0.08     #ohm      (armature resistance)\n",
-    "Ia = 150.0    #A        (armature current)\n",
-    "Ib = 200.0    #A        (initial braking current)\n",
-    "\n",
-    "\n",
-    "Eb1 = V - Ia*Ra        #V   (Back emf) \n",
-    "#Voltage across armature when braking starts\n",
-    "Vb = V + Eb1           #V\n",
-    "\n",
-    "#Resistance in armature circuit\n",
-    "Rt = Vb/Ib              #ohm \n",
-    "#Additional resistance required\n",
-    "R = Rt - Ra             #ohm \n",
-    "#Since field flux is constant therefore 75% torque is produced when armature current is 75% of Ib.\n",
-    "#As (Eb1/Eb2) = (N1/N2)\n",
-    "N2 = Symbol('N2')          #rpm\n",
-    "Eb2 = Eb1*(N2/N1)          #V\n",
-    "\n",
-    "#Voltage across armature when braking starts is V1=V2 =>\n",
-    "V1 = (0.75*Ib)*Rt           #V\n",
-    "V2 = V + Eb2                #V\n",
-    "eq = Eq(V1,V2)\n",
-    "N2 = solve(eq)              #rpm\n",
-    "N_2 = N2[0]                 #rpm\n",
-    "print \"Speed at which torque is 75% of initial value =\",round(N_2,2),\"rpm.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "## EXAMPLE 44.15 , PAGE NO :- 1785"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Speed of motor with the shunted armature connection = 986.07 rpm.\n",
-      "Series motor can't be started on no-load.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A D.C series motor operating at 250V D.C mains and draws 25A and runs at 1200 rpm Ra = 0.1 ohm and Rs = 0.3 ohm.A resistance of\n",
-    "25 ohm is placed in parallel with the armature of motor.Determine:\n",
-    "(i)The speed of motor with the shunted armature connection,if the magnetic circuit remains unsaturated and the load torque remains\n",
-    "constant.\n",
-    "(ii)No load speed of motor.'''\n",
-    "\n",
-    "from sympy import Eq,Symbol,solve\n",
-    "\n",
-    "V = 250.0       #V     (applied voltage)\n",
-    "Ia = 25.0       #A     (armature current)\n",
-    "N1 = 1200.0     #rpm   (speed)\n",
-    "Ra = 0.1        #ohm   (armature resistance)\n",
-    "Rse = 0.3       #ohm   (series field resistance)\n",
-    "Rd = 25.0       #ohm   (diverter resistance)\n",
-    "\n",
-    "#Let I2 flow from series winding , Ia2 be new armature current and Id be diverter current\n",
-    "\n",
-    "I2 = Symbol('I2')\n",
-    "Vd = V - Rse*I2          #V     (Voltage across diverter)\n",
-    "Id = Vd/Rd               #A     (V=IR) \n",
-    "Ia2 = I2 - Id            #A     (new armature current)\n",
-    "\n",
-    "#AS T is constant,   (flux1)*Ia1 = (flux2)*Ia2\n",
-    "\n",
-    "eq = Eq(Ia*Ia,(I2)*Ia2)   #  As, flux1/flux2 = Ia/I2\n",
-    "I2 = solve(eq)\n",
-    "I_2 = I2[1]               #A   (current through series winding)\n",
-    "\n",
-    "Ia2 = Ia*Ia/I_2            #A\n",
-    "Eb1 = V - Ia*(Ra+Rse)     #V     (Back emf 1) \n",
-    "\n",
-    "Eb2 = V - I_2*Rse - Ia2*Ra  #V    (Back emf 2)\n",
-    "\n",
-    "#N2/N1 = Eb2*flux1/Eb1*flux2\n",
-    "N2 = N1*(Eb2/Eb1)*(Ia/I_2)    #rpm\n",
-    "print \"Speed of motor with the shunted armature connection =\",round(N2,2),\"rpm.\"\n",
-    "print \"Series motor can't be started on no-load.\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {
-    "collapsed": true
-   },
-   "source": [
-    "## EXAMPLE 44.16 , PAGE NO :- 1786"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Additional Resistance required in (i) is = 0.8 ohm.\n",
-      "Additional Resistance required in (ii) is = 1.27 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "'''A 4 pole,50 Hz,slip ring Induction Motor has rotor resistance and stand still reactance referred to stator of 0.2 ohm and 1 ohm\n",
-    "per phase respectively.At full load,it runs at 1440 rpm.Determine the value of resistance to be inserted in rotor in ohm/phase to\n",
-    "operate at a speed of 1200 rpm,if:\n",
-    "(i)Load torque remains constant                       (ii)Load torque varies as square of the speed\n",
-    "Neglect stator resistance and leakage reactance.'''\n",
-    "\n",
-    "from sympy import Eq,solve,Symbol\n",
-    "\n",
-    "p = 4.0      # poles\n",
-    "f = 50.0     #Hz       (frequency)\n",
-    "R2 = 0.2     #ohm      (rotor resistance)\n",
-    "X2 = 1.0     #ohm      (stand still reactance)\n",
-    "N1 = 1440.0  #rpm      (speed)\n",
-    "N2 = 1200.0  #rpm      (new speed)\n",
-    "\n",
-    "Ns = 120*f/p #rpm      (synchronus speed)\n",
-    "s1 = (Ns-N1)/Ns   #rpm (slip 1)\n",
-    "s2 = (Ns-N2)/Ns   #rpm (slip 2)\n",
-    "\n",
-    "#(i)Load torque is constant i.e (T1 = T2)\n",
-    "#T is propotional to (s/R2) , (T1/T2) = (s1/S2)*(R2/Rt).Therefore, new resistance required is \n",
-    "\n",
-    "Rt = (s2/s1)*R2    #ohm    (total resistance)\n",
-    "r = Rt - R2        #ohm    (additional resistance)\n",
-    "\n",
-    "print \"Additional Resistance required in (i) is =\",round(r,2),\"ohm.\"\n",
-    "\n",
-    "#(ii) Load torque varies as square of speed (i.e  T1/T2 = (N1/N2)^2 )\n",
-    "T1_T2a = (N1/N2)*(N1/N2)      #(T1/T2)\n",
-    "\n",
-    "Rt = Symbol('Rt')\n",
-    "T1_T2b = (s1*R2/(R2*R2 + (s1*X2)*(s1*X2)))/(s2*Rt/(Rt*Rt + (s2*X2)*(s2*X2)))\n",
-    "eq = Eq(T1_T2a,T1_T2b)\n",
-    "Rt = solve(eq)\n",
-    "R_t = Rt[1]              #ohm   (total resistance)\n",
-    "\n",
-    "r = R_t - R2             #ohm   (additional resistance)\n",
-    "print \"Additional Resistance required in (ii) is =\",round(r,2),\"ohm.\""
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "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",
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