Added(A)/Deleted(D) following books

 R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter40.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter40.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter41.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter43.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter43.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter44.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter44.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter45.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter45.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter46.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter46.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter47.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter49.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter49.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter50.ipynb
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_1.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_1.png
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_2.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_2.png
R  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_3.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_3.png
A  BSc_1st_Year_Physics_by_P._Balabhaskar,_N._Srinivasa_Rao,_B._Sanjeeva_Rao/README.txt
A  Electric_Drives_Concepts_And_Applications_by_Vedam_Subrahmanyam/README.txt

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# CHAPTER 47 : ELECTRIC HEATING\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.1 , PAGE NO :- 1841"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of wire = 16.07 m.\n",
+      "Diameter of wire = 2.72 mm.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A resistance oven employing nichrome wire is to be operated from 220 V single-phase supply and is to be rated at 16 kW.\n",
+    "If the temperature of the element is to be limited to 1,170°C and average temperature of the charge is 500°C, find the\n",
+    "diameter and length of the element wire.\n",
+    "Radiating efficiency = 0.57, Emmissivity=0.9, Specific resistance of nichrome=(109e–8)ohm-m.'''\n",
+    "\n",
+    "\n",
+    "P = 16000.0         #W     (output power)\n",
+    "V = 220.0           #V     (applied voltage)\n",
+    "rho = 109.0e-8      #ohm-m (resistivity)\n",
+    "e = 0.9             #      (Emmisivity)\n",
+    "K = 0.57            #      (Radiating efficiency)\n",
+    "T1 = 1170.0 + 273.0 #K     (Temp of hot body)\n",
+    "T2 = 500.0 + 273.0  #K     (Temp of cold body)\n",
+    "\n",
+    "#Now , l/d^2 = pi*V^2/4*rho*P = a .Therefore a is\n",
+    "a = 3.14*(V**2)/(4*rho*P)        # (a = l/d^2) ------ 1\n",
+    " \n",
+    "#Using Stefan's law of radiation \n",
+    "H = 5.72*e*K*((T1/100)**4-(T2/100)**4)    #W/m^2\n",
+    "\n",
+    "#Total heat dissipated = electrical power input\n",
+    "# (pi*d)*l*H = P . Therefore ,let b = l*d. So,\n",
+    "b = P/(H*3.14)\n",
+    "b2 = b**2      #(b2 = l^2*d^2)------------------------ 2\n",
+    "\n",
+    "#Multiplying 1 and 2.\n",
+    "l3 = a*b2    # ( = l^3)\n",
+    "l = l3**(1/3.0)    #m   (length)\n",
+    "d = b/l*1000            #mm   (diameter)\n",
+    "\n",
+    "print \"Length of wire =\",round(l,2),\"m.\"\n",
+    "print \"Diameter of wire =\",round(d,2),\"mm.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.2 , PAGE NO :- 1841"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Width of strip = 7.4 mm.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 30-kW, 3-phase, 400-V resistance oven is to employ nickel-chrome strip 0.254 mm thick for the three star-connected heating\n",
+    "elements.If the wire temperature is to be 1,100°C and that of the charge to be 700°C, estimate a suitable width for the strip.\n",
+    "Assume emissivity = 0.9 and radiating efficiency to be 0.5 and resistivity of the strip material is 101.6e-8 ohm- m.What would be\n",
+    "the temperature of the wire if the charge were cold ?'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "P = 30.0*1000 *(1/3.0)     #W      (Power/phase)\n",
+    "V = 400.0/m.sqrt(3)        #V      (Phase voltage)\n",
+    "rho = 101.6e-8             #ohm-m  (resistivity)\n",
+    "e = 0.9                    #       (emmisivity)\n",
+    "K = 0.5                    #       (radiating efficiency)\n",
+    "t = 0.254e-3               #m      (thickness of strip)\n",
+    "T1 = 1100.0 + 273          #K      (Temp. of hot wire)\n",
+    "T2 = 700.0 + 273           #K      (Temp. of charge)   \n",
+    "R = V*V/P     #ohm       (Resistance  =>    P = V^2/R)\n",
+    "\n",
+    "# R = rho*l/(w*t)   l/w = R*t/rho = a.Therefore a is\n",
+    "a = R*t/rho      #(=l/w)------------------------------------ 1\n",
+    "\n",
+    "#Using stefan's law\n",
+    "H = 5.72*e*K*((T1/100)**4-(T2/100)**4)    #W/m^2\n",
+    "\n",
+    "#Surface area of strip = 2*w*l .\n",
+    "#Total Heat dissipated = electrical power => wl*2H = P . Let b = wl\n",
+    "b = P/(2*H)   #(=wl)----------------------------------------- 2\n",
+    "\n",
+    "#Dividing 1 by 2\n",
+    "w2 = b/a      #(=w*w)\n",
+    "w = m.sqrt(w2)*1000    #mm     (width)\n",
+    "print \"Width of strip =\",round(w,2),\"mm.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.3 , PAGE NO :- 1842"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Loading in kW =  13.51 kW.\n",
+      "Efficency of Tank = 87.41 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A cubic water tank has surface area of 6.0 m^2 and is filled to 90% capacity six times daily. The water is heated from 20°C \n",
+    "to 65°C.The losses per square metre of tank surface per 1°C temperature difference are 6.3 W. Find the loading in kW and the\n",
+    "efficiency of the tank.Assume specific heat of water = 4,200 J/kg/°C and one kWh = 3.6 MJ.'''\n",
+    "\n",
+    "import math as m\n",
+    "sa = 6.0           #m^2      (surface area)\n",
+    "T2 = 65.0          #*C       (final temp)\n",
+    "T1 = 20.0          #*C       (initial temp)\n",
+    "loss = 6.3         #W/*C/m^2 (loss per square metre per 1*C)\n",
+    "s = 4200.0         #J/Kg/*C  (Specific heat)\n",
+    "\n",
+    "#Now, sa = 6*l^2\n",
+    "l = m.sqrt(sa/6)   #m\n",
+    "\n",
+    "#Volume = l^3\n",
+    "V = l**3           #m^3\n",
+    "\n",
+    "#Volume of water to be heated daily is\n",
+    "V2 = 6*V*0.9       #m^3\n",
+    "\n",
+    "#Since 1m^3 = 1000 kg  =>   mass of water to be heated is\n",
+    "mass = V2*1000.0   #kg\n",
+    "\n",
+    "#Heat required to raise temp =\n",
+    "H = mass*s*(T2-T1)  #MJ\n",
+    "H = H/(3.6*10e+5)           #kWh\n",
+    "\n",
+    "#Daily loss from surface\n",
+    "L = 6*loss*(T2-T1)*(24.0/1000)    #kWh\n",
+    "\n",
+    "#Total energy required\n",
+    "Tot = L + H        #kWh    \n",
+    "#(i) Loading in KW is\n",
+    "load = Tot/24      #kW\n",
+    "\n",
+    "#(ii)Efficiency of tank is\n",
+    "eff = (H/Tot)*100.0    #     (% efficency)\n",
+    "print \"Loading in kW = \",round(load,2),\"kW.\"\n",
+    "print \"Efficency of Tank =\",round(eff,2),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.4 , PAGE NO :- 1844"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Power factor = 0.9487\n",
+      "Power drawn from supply = 900.0 kW.\n",
+      "Time required for melting steel = 40.0 minutes. and  46.0 seconds.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 4-phase electric arc furnace has the following data :\n",
+    "Current drawn = 5000 A ; Arc voltage = 50 V\n",
+    "Resistance of transformer referred to secondary = 0.002 ohm\n",
+    "Resistance of transformer referred to secondary = 0.004 ohm\n",
+    "(i) Calculate the power factor and kW drawn from the supply.\n",
+    "(ii) If the overall efficiency of the furnace is 65%, find the time required to melt 2 tonnes of steel if\n",
+    "latent heat of steel = 8.89 kcal/kg, specific heat of steel = 0.12, melting point of steel = 1370°C and\n",
+    "initial temperature of steel = 20°C.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 5000.0        #A       (current drawn)\n",
+    "V = 50.0          #V       (Arc Voltage) \n",
+    "Rs = 0.002        #ohm     (transformer resistance on secondary)\n",
+    "Xs = 0.004        #ohm     (transformer reactance on secondary)\n",
+    "T2 = 1370.0       #*C      (final temp)\n",
+    "T1 = 20.0         #*C      (initial temp)\n",
+    "\n",
+    "# Voltage drop due to resistance =\n",
+    "Vr = I*Rs         #V\n",
+    "\n",
+    "# Voltage drop due to reactance =\n",
+    "Vx = I*Xs         #V\n",
+    "\n",
+    "#Total Voltage is (Using vector sum)\n",
+    "V_tot = m.sqrt((V+Vr)**2 + Vx**2)       #V\n",
+    "\n",
+    "#(i) Supply power factor is\n",
+    "pf = (V+Vr)/V_tot\n",
+    "\n",
+    "#Total Power drawn =>  P = 3*VI*(power factor)\n",
+    "P = 3*V_tot*I*pf/1000        #kW\n",
+    "\n",
+    "print \"Power factor =\",round(pf,4)\n",
+    "print \"Power drawn from supply =\",round(P,2),\"kW.\"\n",
+    "#Energy required to melt 2 tonnes of steel\n",
+    "m = 2000.0     #kg\n",
+    "s = 0.12       #           (specific heat of steel)\n",
+    "L = 8.89       #kcal/kg    (latent heat of steel)\n",
+    "\n",
+    "enrgy = m*s*(T2-T1) + m*L      #kcal\n",
+    "enrgy = enrgy/860.0            #kWh\n",
+    "\n",
+    "#Utilised power\n",
+    "P = 0.65*P               #kW\n",
+    "\n",
+    "#Time required for melting steel\n",
+    "Time = enrgy/P           #hr\n",
+    "Time = Time*60           #min\n",
+    "Sec = (round(Time,2) - round(Time,-1) )*60  #sec\n",
+    "\n",
+    "print \"Time required for melting steel =\",round(Time,-1),\"minutes. and \",round(Sec),\"seconds.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.5 , PAGE  NO :- 1845"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "#Total KVA taken from supply line = 2145.63 KVA .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''If a 3-phase arc furnace is to melt 10 tonne steel in 2 hours, estimate the average input to the furnace if overall\n",
+    "efficiency is 50%. If the current input is 9,000 A with the above kW input and the resistance and reactance of furnace leads\n",
+    "(including transformer) are 0.003 ohm and 0.005 ohm respectively, estimate the arc voltage and total kVA taken from the supply\n",
+    "Specific heat of steel = 444 J /kg/°C ,Latent heat of fusion of steel = 37.25 kJ/kg , Melting point of steel = 1,370 °C.'''\n",
+    "\n",
+    "from sympy import Symbol,solve,Eq,sqrt\n",
+    "import math as m\n",
+    "\n",
+    "mass = 10000.0        #kg       (mass in kg)\n",
+    "t = 2.0               #hr       (time taken to melt)\n",
+    "eff = 50.0            #%        (overall efficiency)\n",
+    "I = 9000.0            #A        (current input)\n",
+    "Rs = 0.003            #ohm      (secondary resistance)\n",
+    "Xs = 0.005            #ohm      (secondary reactance)\n",
+    "s = 444.0             #J/kg/*C  (specific heat of steel)\n",
+    "L = 37250             #J/kg    (latent heat of fusion)\n",
+    "T2 = 1370.0           #*C       (final temp)\n",
+    "T1 = 20.0             #*C       (initial temp)\n",
+    "\n",
+    "#Energy required to melt 10 tonnes of steel\n",
+    "\n",
+    "enrgy = mass*s*(T2-T1) + mass*L   #J\n",
+    "enrgy = enrgy/(1000*3600)     #kWh\n",
+    "\n",
+    "#Avg output power = energy/time\n",
+    "P = enrgy/t        #kW\n",
+    "#Avg input power =\n",
+    "Pin = P/eff*100\n",
+    "\n",
+    "#Voltage drop due to resistance\n",
+    "Vr = I*Rs          #V\n",
+    "#Voltage drop due to reactance\n",
+    "Vx = I*Xs          #V\n",
+    "\n",
+    "#Now,Let Va is arc drop voltage\n",
+    "Va = Symbol('Va')    #V\n",
+    "Vt = sqrt((Va+Vr)**2 + Vx**2)  \n",
+    "pf = (Va+Vr)/Vt     \n",
+    "#Total power input  = 3*(Vt*It*pf)\n",
+    "\n",
+    "eq = Eq(Pin*1000,3*Vt*I*pf)\n",
+    "Va = solve(eq)     #V\n",
+    "\n",
+    "Va1 = Va[0]\n",
+    "Vt = sqrt((Va1+Vr)**2 + Vx**2)\n",
+    "\n",
+    "#Total KVA taken from supply line =\n",
+    "power = 3*Vt*I/1000\n",
+    "\n",
+    "print \"#Total KVA taken from supply line =\",round(power,2),\"KVA .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## EXAMPLE 47.6 , PAGE NO :- 1850"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Efficiency of induction furnace =  59.35 %.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Determine the efficiency of a high-frequency induction furnace which takes 10 minutes to melt 2 kg of a aluminium initially \n",
+    "at a temperature of 20°C. The power drawn by the furnace is 5 kW, specific heat of aluminium = 0.212, melting point of \n",
+    "aluminium = 660° C and latent heat of fusion of aluminium. = 77 kcal/kg.'''\n",
+    "\n",
+    "m = 2.0     #kg                (mass of alluminium)\n",
+    "L = 77.0    #kcal/kg           (Latent heat of fusion)\n",
+    "T2 = 660.0  #*C              (final temp)\n",
+    "T1 = 20.0   #*C              (initial temp)\n",
+    "s = 0.212   #                (specific heat of alluminium)\n",
+    "Pin = 5.0   #kW              (input power)\n",
+    "#Heat required to melt alluminium\n",
+    "H1 = m*L     #kcal\n",
+    "#Heat required to raise the temperature\n",
+    "H2 = m*s*(T2 - T1)   #kcal\n",
+    "#Total heat\n",
+    "heat_tot = H1 + H2   #kcal\n",
+    "#Heat required per hour\n",
+    "enrgy = heat_tot/(10.0/60)  #kcal\n",
+    "#Power delivered to alluminium\n",
+    "Power = enrgy/860     #kW\n",
+    "\n",
+    "eff = Power/Pin*100   #(% efficiency)\n",
+    "\n",
+    "print \"Efficiency of induction furnace = \",round(eff,2),\"%.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.7 , PAGE NO :- 1850"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Power absorbed = 577.0 kW.\n",
+      "power factor = 0.83\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A low-frequency induction furnace has a secondary voltage of 20V and takes 600 kW at 0.6 p.f. when the hearth is full. If the\n",
+    "secondary voltage is kept constant, determine the power absorbed and the p.f. when the hearth is half-full. Assume that the\n",
+    "resistance of the secondary circuit is doubled but the reactance remains the same.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "V = 20.0      #V   (secondary voltage)\n",
+    "P = 600*1000  #W   (Input Power)\n",
+    "pf = 0.6      #    (power factor)\n",
+    "\n",
+    "#Inital secondary current using P = VI*pf\n",
+    "I = P/(V*pf)  #A   (secondary current)\n",
+    "\n",
+    "#Now, pf = cosQ , .'. sinQ = sqrt(1-pf^2)\n",
+    "Vr = V*pf              #V   (Voltage across resistance)\n",
+    "Vx = V*m.sqrt(1-pf**2) #V   (Voltage across reactance)\n",
+    "\n",
+    "#As, Vr = I*R and Vx = I*X\n",
+    "\n",
+    "R = Vr/I    #ohm\n",
+    "X = Vx/I    #ohm\n",
+    "\n",
+    "#When hearth is half-full\n",
+    "R2 = 2*R\n",
+    "X2 = X\n",
+    "pf = R2/m.sqrt(R2**2 + X2**2)  #(new power factor)\n",
+    "\n",
+    "Vr = V*pf      #V   (Voltage across resistance)\n",
+    "#As, Vr = I*R\n",
+    "I = Vr/R2       #A\n",
+    "\n",
+    "power = V*I*pf/1000       #kW\n",
+    "\n",
+    "print \"Power absorbed =\",round(power),\"kW.\"\n",
+    "print \"power factor =\",round(pf,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.8 , PAGE NO :- 1851"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total furnace input = 119.37 kWh .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Estimate the energy required to melt 0.5 tonne of brass in a single-phase induction furnace. If the melt is to be carried out\n",
+    "in 0.5 hour, what must be the average power input to the furnace?\n",
+    "Specific heat of brass = 0.094\n",
+    "Latent heat of fusion of brass = 39 kilocal/kg\n",
+    "Melting point of brass = 920°C\n",
+    "Furnace efficiency = 60.2%\n",
+    "The temperature of the cold charge may be taken as 20°C.'''\n",
+    "\n",
+    "m = 0.5*1000      #kg    (mass of brass)\n",
+    "s = 0.094    #      (specific heat)\n",
+    "T2 = 920.0   #*C    (final temp)\n",
+    "T1 = 20.0    #*C    (initial temp)\n",
+    "L = 39.0     #kcal/kg(Latent heat of fusion)\n",
+    "eff = 60.2   #       (% efficiency)\n",
+    "\n",
+    "#Total amount of heat req to melt 0.5 kg brass\n",
+    "H = m*L + m*s*(T2-T1)    #kcal\n",
+    "H = H/860    #kWh\n",
+    "H_tot = H/(eff)*100    #kWh   (input energy required)\n",
+    "\n",
+    "print \"Total furnace input =\",round(H_tot,2),\"kWh .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.9 , PAGE NO :- 1851"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "height for maximum heat =  0.75 *H.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A low-frequency induction furnace whose secondary voltage is maintained constant at 10 V, takes 400 kW at 0.6 p.f. when the\n",
+    "hearth is full.Assuming the resistance of the secondary circuit to vary inversely as the height of the charge and reactance to \n",
+    "remain constant,find the height upto which the hearth should be filled to obtain maximum heat.'''\n",
+    "\n",
+    "\n",
+    "import math as m\n",
+    "V2 = 10.0    #V       (secondary voltage)\n",
+    "P = 400.0*1000 #kW    (power)\n",
+    "pf = 0.6       #      (power factor)\n",
+    "\n",
+    "\n",
+    "#Secondary current is (Using P = VI*cosQ)\n",
+    "I = P/(V2*pf)    #A\n",
+    "\n",
+    "#Impedance of secondary circuit is\n",
+    "Z = V2/I        #ohm\n",
+    "#Now, R = Z*cosQ    X = Z*sinQ\n",
+    "R = Z*pf                    #ohm   (resistance)\n",
+    "X = Z*m.sqrt(1-pf**2)       #ohm   (reactance)\n",
+    "\n",
+    "#Let height of charge be 'x' times of the full hearth h = x*H\n",
+    "#Resistance varies inersely as height .Therefore,\n",
+    "# R' = R/x\n",
+    "\n",
+    "#Now ,for max heat resistance should be equal to reactance.Therefore,\n",
+    "x = R/X\n",
+    "\n",
+    "print \"height for maximum heat = \",round(x,2),\"*H.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.10 , PAGE NO :- 1853"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Voltage = 798.04 V.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A slab of insulating material 150 cm^2 in area and 1 cm thick is to be heated by dielectric heating. The power required is \n",
+    "400 W at 30 MHz.Material has relative permittivity of 5 and p.f. of 0.05. Determine the necessary voltage. Absolute\n",
+    "permittivity = 8.854e - 12 F/m.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "P = 400.0    #W         (power)\n",
+    "f = 30.0e+6  #Hz        (frequency)\n",
+    "A = 150.0e-4 #m^2       (area)\n",
+    "d = 1.0e-2   #m         (thickness)\n",
+    "er = 5.0     #          (relative permitivity)\n",
+    "e0 = 8.89e-12#F/m       (absolute permitivity)\n",
+    "pf = 0.05    #          (power factor) \n",
+    "#Capacitance\n",
+    "C = (A/d)*(er*e0)     #F\n",
+    "\n",
+    "#Now,   P = (2*pi*f)*(C*V^2)*pf .Therefore V is\n",
+    "V = m.sqrt(P/(2*3.14*f*C*pf))    #V\n",
+    "print \"Voltage =\",round(V,2),\"V.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.11 , PAGE NO :- 1853"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Voltage =  846.45 V.\n",
+      "Current through material = 9.45 A.\n",
+      "Frequency =  58.49 MHz.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''An insulating material 2 cm thick and 200 cm^2 in area is to be heated by dielectric heating. The material has relative \n",
+    "permitivity of 5 and power factor of 0.05.Power required is 400 W and frequency of 40 MHz is to be used. Determine the necessary\n",
+    "voltage and the current that will flow through the material.If the voltage were to be limited to 700 V, what will be the \n",
+    "frequency to get the same loss? '''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "d = 2.0e-2      #m      (Thickness)\n",
+    "A = 200e-4      #m^2    (Area)\n",
+    "er = 5          #       (relative permitivity)\n",
+    "pf = 0.05       #       (power factor)\n",
+    "f = 40.0e+6     #Hz     (frequency)\n",
+    "P = 400.0       #W      (power)\n",
+    "e0 = 8.89e-12   #       (absolute permitivity)\n",
+    "\n",
+    "C = (A/d)*(e0*er)   #F   (Capacitance)\n",
+    "\n",
+    "#Now,  P = 2*pi*f*C*V^2*pf\n",
+    "V = m.sqrt(P/(2*3.14*f*C*pf))     #V\n",
+    "\n",
+    "#Also, P = VI*cosQ .Therefore,current through material\n",
+    "I = P/(V*pf)           #A\n",
+    "\n",
+    "#Heat produced is propotional to V^2*f.  (V2/V1)^2 = (f1/f2)\n",
+    "f2 = f*(V/700)**2   #Hz\n",
+    "f2 = f2/(10e+5)     #MHz \n",
+    "\n",
+    "print \"Voltage = \",round(V,2),\"V.\"\n",
+    "print \"Current through material =\",round(I,2),\"A.\"\n",
+    "print \"Frequency = \",round(f2,2),\"MHz.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 47.12 , PAGE NO :- 1853"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Power input = 750.0 W .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A plywood board of 0.5*0.25*0.02 metre is to be heated from 25 to 125°C in 10 minutes by dielectric heating employing a\n",
+    "frequency of 30 MHz. Determine the power required in this heating process. Assume specific heat of wood 1500/J/kg/°C; \n",
+    "weight of wood 600 kg/m3 and efficiency of process 50%.'''\n",
+    "\n",
+    "\n",
+    "Vol = 0.5*0.25*0.02    #m^3      (Volume of plywood to be heated)\n",
+    "f = 30.0e+6            #Hz       (frequency)\n",
+    "t = 10.0               #minutes  (time)\n",
+    "T2 = 125.0             #*C       (final temperature)\n",
+    "T1 = 25.0              #*C       (initial temperature)\n",
+    "s = 1500.0             #J/kg/*C  (specific heat of wood)\n",
+    "den = 600.0            #kg/m^3   (weight of wood)\n",
+    "eff = 50.0             #%         (efficiency of process)\n",
+    "\n",
+    "wt = den*Vol          #kg   (weight of plywood)\n",
+    "#Heat required to raise the temp is\n",
+    "H = wt*s*(T2-T1)     #J\n",
+    "H = H/3600           #Wh\n",
+    "\n",
+    "#As  P = H/t .Therfore power required for heating\n",
+    "P = H/(10.0/60)      #W\n",
+    "\n",
+    "#As efficiency is 50%\n",
+    "Pin = P/eff*100      #W\n",
+    "\n",
+    "print \"Power input =\",round(Pin,2),\"W .\""
+   ]
+  },
+  {
+   "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",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}