Added(A)/Deleted(D) following books

 A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter40.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter41.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter43.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter44.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter45.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter46.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter49.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_1.png
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_2.png
A  A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_3.png

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# CHAPTER 49 : ILLUMINATION"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.1 , PAGE NO :- 1899"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Illumination at point = 0.76 lm/m^2 .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A lamp giving out 1200 lm in all directions is suspended 8 m above the working plane. Calculate the illumination at a point on\n",
+    "the working plane 6 m away from the foot of the lamp.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 1200/(4*3.14)      #Cd      (luminous intensity of lamp)\n",
+    "h = 8.0                #m       (height)\n",
+    "b = 6.0                #m       (breadth)  \n",
+    "length = m.sqrt(h**2 + b**2)  #m\n",
+    "\n",
+    "cosQ = h/length\n",
+    "E = I*cosQ/length**2    #lm/m^2\n",
+    "\n",
+    "print \"Illumination at point =\",round(E,2),\"lm/m^2 .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.2 , PAGE NO :- 1899"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Distance between A and B is = 19.08 m.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A small light source with intensity uniform in all directions is mounted at a height of 10 metres above a horizontal surface.\n",
+    "Two points A and B both lie on the surface with point A directly beneath the source. How far is B from A if the illumination at B\n",
+    "is only 1/10 as great as at A ?'''\n",
+    "\n",
+    "from sympy import Eq,solve,Symbol\n",
+    "\n",
+    "#let the intensity of lamp be I and distance between A and B be x metres\n",
+    "x = Symbol('x')\n",
+    "\n",
+    "l = 10.0        #m     (vertical distance)\n",
+    "#Illumination at point A\n",
+    "Ea = I/l**2        #lux\n",
+    "#Illumination at point B\n",
+    "\n",
+    "Eb = I/(l**2)*(l/(l**2 + x**2)**0.5)**3\n",
+    "\n",
+    "I = 10.0           #lm     (assumed value as the equation does not depend on I)\n",
+    "#As Eb = 1/10*Ea\n",
+    "eq = Eq(Eb,Ea/10.0)\n",
+    "x = solve(eq)\n",
+    "\n",
+    "x1 = x[1]\n",
+    "\n",
+    "print \"Distance between A and B is =\",round(x1,2),\"m.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## EXAMPLE 49.3 , PAGE NO :- 1900"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Illumination due to 4 lamps =  6.16 lm/m^2 .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A corridor is lighted by 4 lamps spaced 10 m apart and suspended at a height of 5 m above the centre line of the floor.\n",
+    "If each lamp gives 200 C.P. in all directions below the horizontal,find the illumination at the point on the floor mid-way\n",
+    "between the second and third lamps.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 200.0    #C.P       (luminous intensity)\n",
+    "h = 5.0      #m         (height between lamps and ground)\n",
+    "l1 = 15.0    #m         (horizantal distance 1)\n",
+    "l2 = 5.0    #m         (horizantal distance 2)\n",
+    "\n",
+    "d1 = m.sqrt(h**2 + l1**2)    #m   (Dist btwn L1 and mid-pt)\n",
+    "d2 = m.sqrt(h**2 + l2**2)    #m   (Dist btwn L2 and mid-pt)\n",
+    "\n",
+    "#(i)Illumination due to L1\n",
+    "#L = (I/r^2)*cosQ\n",
+    "L1 = (I/d1**2)*(h/d1)      #lm/m^2\n",
+    "\n",
+    "#(ii)Illumination due to L2\n",
+    "L2 = (I/d2**2)*(h/d2)      #lm/m^2\n",
+    "\n",
+    "#Illumination at mid-pt due to 4-lamps\n",
+    "Lt = 2*(L1+L2)             #lm/m^2\n",
+    "\n",
+    "print \"Illumination due to 4 lamps = \",round(Lt,2),\"lm/m^2 .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.4 , PAGE NO :- 1901"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Reading of photometer = 0.066 lm/m^2.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Two lamps A and B of 200 candela and 400 candela respectively are situated 100 m apart. The height of A above the ground \n",
+    "level is 10 m and that of B is 20 m. If a photometer is placed at the centre of the line joining the two lamp posts,\n",
+    "calculate its reading.'''\n",
+    "\n",
+    "import math as m\n",
+    "I1 = 200.0     #Cd    (lamp 1 intensity)\n",
+    "I2 = 400.0     #Cd    (lamp 2 intensity)\n",
+    "h1 = 10.0      #m     (height between lamp 1 and ground)\n",
+    "h2 = 20.0      #m     (height between lamp 2 and ground)\n",
+    "d1 = 50.0      #m     (horizontal distance from 1)\n",
+    "d2 = 50.0      #m     (horizontal distance from 2)\n",
+    "\n",
+    "l1 = m.sqrt(h1**2 + d1**2)\n",
+    "l2 = m.sqrt(h2**2 + d2**2)\n",
+    "\n",
+    "cosQ1 = h1/l1\n",
+    "cosQ2 = h2/l2\n",
+    "\n",
+    "#Illumination at point C = Illumination due to 1 + Illumination due to 2\n",
+    "I_tot = (I1/l1**2)*cosQ1 + (I2/l2**2)*cosQ2      #lm/m^2\n",
+    "\n",
+    "print \"Reading of photometer =\",round(I_tot,3),\"lm/m^2.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.5 , PAGE NO :- 1901"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Average Brightness = 6375.22 cd/m^2.\n",
+      "cost of running = Rs 150.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''The average luminous output of an 80-W fluorescent lamp 1.5 metre in length and 3.5 cm diameter is 3300 lumens. Calculate its\n",
+    "average brightness.If the auxiliary gear associated with the lamp consumes a load equivalent to 25 percent of the lamp,\n",
+    "calculate the cost of running a twin unit for 2500 hours at 30 paise per kWh.'''\n",
+    "\n",
+    "\n",
+    "length  = 1.5    #m       (lamp output length)\n",
+    "dia = 3.5e-2     #m       (lamp output diameter)\n",
+    "l_flux = 3300.0  #lumens  (luminous flux)\n",
+    "P = 80.0         #W       (Power output)\n",
+    "\n",
+    "#Surface area of lamp\n",
+    "sa = 3.14*dia*length    #m^2\n",
+    "\n",
+    "#Flux emmited per unit area\n",
+    "fluxA = l_flux/sa          #lm/m^2\n",
+    "#Therefore,\n",
+    "B = fluxA/3.14             #cd/m^2\n",
+    "print \"Average Brightness =\",round(B,2),\"cd/m^2.\"\n",
+    "#Total load of twin fitting\n",
+    "load = 2*(P +0.25*P)      #W\n",
+    "time = 2500.0             #hr\n",
+    "enrgy = load*time/1000    #kWh   (Energy consumed)\n",
+    "\n",
+    "#Total cost\n",
+    "cost = enrgy*0.3     #Rs\n",
+    "print \"cost of running = Rs\",round(cost)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.6 , PAGE NO :- 1901"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lamp wattage = 4415.63 W.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A small area 7.5 m in diameter is to be illuminated by a lamp suspended at a height of 4.5 m over the centre of the area.\n",
+    "The lamp having an efficiency of 20 lm/w is fitted with a reflector which directs the light output only over the surface to be\n",
+    "illuminated,giving uniform candle power over this angle. Utilisation coefficient = 0.40. Find out the wattage of the lamp.\n",
+    "Assume 800 lux of illumination level from the lamp.'''\n",
+    "\n",
+    "dia =  7.5     #m      (diameter)\n",
+    "h = 4.5        #m      (height)\n",
+    "E = 800.0      #lux    (illumination) \n",
+    "eff = 20.0     #lm/w   (lamp efficiency)\n",
+    "A = 3.14*(dia**2)/4\n",
+    "#Luminous flux reaching the surface\n",
+    "flux = A*E      #lm\n",
+    "\n",
+    "#Total flux emmited is\n",
+    "f_out = flux/0.4 #lm\n",
+    "\n",
+    "#Lamp in watts\n",
+    "watt = f_out/eff   #W\n",
+    "print \"lamp wattage =\",round(watt,2),\"W.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.7 , PAGE NO :- 1902"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Illumination =  5.0 lux.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A lamp of 100 candela is placed 1 m below a plane mirror which reflects 90% of light falling on it. The lamp is hung 4 m above\n",
+    "ground.Find the illumination at a point on the ground 3 m away from the point vertically below the lamp.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "h1 = 4.0     #m   (height of 1 from ground)\n",
+    "d1 = 3.0     #m   (horizontal distance 1)\n",
+    "I1 = 100.0   #cd  (intenstity)\n",
+    "\n",
+    "l1 = m.sqrt(h1**2 + d1**2)  #m\n",
+    "cosQ1 = h1/l1\n",
+    "#The lamp L1 will produce the image L2 1m behind the mirror.Therefore,\n",
+    "\n",
+    "h2 = h1+1+1  #m   (height of 2 from ground)\n",
+    "d2 = 3.0     #m   (horizontal distance 2)\n",
+    "I2 = 0.9*I1  #cd  (intensity)\n",
+    "\n",
+    "l2 = m.sqrt(h2**2 + d2**2)  #m\n",
+    "cosQ2 = h2/l2\n",
+    "\n",
+    "#Illumination at the required point is\n",
+    "\n",
+    "E = I1/l1**2*cosQ1 + I2/l2**2*cosQ2   #lux\n",
+    "\n",
+    "print \"Illumination = \",round(E),\"lux.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.8 , PAGE NO :- 1902"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Average illumination = 924.08 lux .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A light source having an intensity of 500 candle in all directions is fitted with a reflector so that it directs 80% of its\n",
+    "light along a beam having a divergence of 15º. What is the total light flux emitted along the beam? What will be the average\n",
+    "illumination produced on a surface normal to the beam direction at a distance of 10 m? '''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 500.0      #cd        (intensity)\n",
+    "Q = 15.0       #degrees   (Beam angle)\n",
+    "h = 10.0       #m         (height)\n",
+    "\n",
+    "#Total flux emmited is\n",
+    "flux = 0.8*(4*3.14*I)        #lm\n",
+    "#radius of circle to be illuminated\n",
+    "r = h*m.tan(Q/2*(3.14/180))       #m\n",
+    "\n",
+    "#Area of surface to be illuminated is\n",
+    "A = 3.14*(r*r)         #m^2\n",
+    "\n",
+    "#Avg illumination\n",
+    "avg = flux/A           #lux\n",
+    "\n",
+    "print \"Average illumination =\",round(avg,2),\"lux .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.9  , PAGE NO :- 1902"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "----Without reflector ----\n",
+      "Illumination at centre = 0.75 lm/m^2.\n",
+      "Illumination at edge = 0.54 lm/m^2.\n",
+      "----With reflector ----\n",
+      "Illumination at every point = 6.0 lm/m^2\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A lamp has a uniform candle power of 300 in all directions and is fitted with a reflector which directs 50% of the total\n",
+    "emitted light uniformly on to a flat circular disc of 20 m diameter placed 20 m vertically below the lamp. Calculate the\n",
+    "illumination (a) at the centre and (b)at the edge of the surface without the reflector. Repeat these two calculations with\n",
+    "the reflector provided.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 300.0       #Cd   (intensity)\n",
+    "h = 20.0        #m    (height)\n",
+    "dia = 20.0      #m    (diameter of luminous area)\n",
+    "\n",
+    "#(i)Without reflector\n",
+    "Ec = I/h**2      #lm/m^2  (illumination at centre)\n",
+    "\n",
+    "theta = m.atan((dia/2)/h)\n",
+    "l = m.sqrt(h**2 + (dia/2)**2)     #m    (distance between edge and source lamp)\n",
+    "\n",
+    "Eb = I/l**2*m.cos(theta)     #lm/m^2   (illuminaton at edge)\n",
+    "print \"----Without reflector ----\"\n",
+    "print \"Illumination at centre =\",round(Ec,2),\"lm/m^2.\"\n",
+    "print \"Illumination at edge =\",round(Eb,2),\"lm/m^2.\"\n",
+    "\n",
+    "#(ii)With reflector\n",
+    "#Luminous output of lamp\n",
+    "lflux = I*4*3.14       #lm\n",
+    "\n",
+    "#flux directed by reflector\n",
+    "reflux = 0.5*lflux     #lm\n",
+    "\n",
+    "#Area of disc\n",
+    "A = 3.14*(dia*dia)/4   #m^2\n",
+    "\n",
+    "#Illumination at every point will be same and will be equal to\n",
+    "Et = reflux/A          #lm/m^2\n",
+    "\n",
+    "print \"----With reflector ----\"\n",
+    "print \"Illumination at every point =\",round(Et,2),\"lm/m^2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.10 , PAGE NO :- 1903"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Iluminaton at P =  11.11 cd/m^2\n",
+      "Iluminaton at Q =  3.93 cd/m^2\n",
+      "Total radiations sent =  314.0 lumens.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A light is placed 3 m above the ground and its candle power is 100 cos θ in any downward direction making an angle q with the\n",
+    "vertical. If P and Q are two points on the grond, P being vertically under the light and the distance PQ being 3 m, calculate.\n",
+    "(a) the illumination of the ground at P and also at Q.\n",
+    "(b) the total radiations sent down by the lamp.'''\n",
+    "\n",
+    "import math as m\n",
+    "from scipy.integrate import quad\n",
+    "\n",
+    "r1  = 3.0                                   #m\n",
+    "r2 = m.sqrt(3**2 + 3**2)                    #m\n",
+    "#(a)\n",
+    "#Candela Power along LP\n",
+    "CP1 = 100.0*m.cos(0)                        #cd\n",
+    "#Illumination at P is\n",
+    "Ep = CP1/(r1**2)                            #cd/m^2\n",
+    "\n",
+    "#Candela Power along LQ\n",
+    "CP2 = 100.0*m.cos(45*3.14/180)              #cd\n",
+    "\n",
+    "#Illumination at Q is\n",
+    "Eq = CP2/(r2**2)                            #cd/m^2\n",
+    "\n",
+    "print \"Iluminaton at P = \",round(Ep,2),\"cd/m^2\"\n",
+    "print \"Iluminaton at Q = \",round(Eq,2),\"cd/m^2\"\n",
+    "\n",
+    "#After working out , total flux = integral (100*pi*sin2Q*dQ) 0->pi/2\n",
+    "\n",
+    "def integrand(Q):\n",
+    "    return 100*3.14*m.sin(2*Q)\n",
+    "\n",
+    "ans, err = quad(integrand, 0,3.14/2)\n",
+    "print \"Total radiations sent = \",round(ans),\"lumens.\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.11 , PAGE NO :- 1903"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of lamps = 21.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A drawing office containing a number of boards and having a total effective area of 70 m2 is lit by a number of 40 W\n",
+    "incandescent lamps giving 11 lm/W. An illumination of 80 lux is required on the drawing boards. Assuming that 60% of the\n",
+    "total light emitted by the lamps is available for illuminating the drawing boards, estimate the number of lamps required.'''\n",
+    "\n",
+    "A = 70.0        #m^2    (area)\n",
+    "watt = 40.0     #W      (each bulb wattage)\n",
+    "eff = 11.0      #lm/W   (luminous efficacy)\n",
+    "E = 80.0        #lux    (Illumination)\n",
+    "\n",
+    "#Output per lamp is\n",
+    "oplamp = watt*eff    #lm\n",
+    "\n",
+    "#Flux actually used per lamp is\n",
+    "flux = 0.6*oplamp    #lm\n",
+    "\n",
+    "#Now, Total flux required is   Illumination*Area\n",
+    "flux_tot = E*A       #lm\n",
+    "\n",
+    "#Therefore number of lamps required are\n",
+    "N = flux_tot/flux\n",
+    "\n",
+    "print \"Number of lamps =\",round(N)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.12 , PAGE NO :- 1904"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total flux radiated = 62.74 lm.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A perfectly diffusing surface has a luminous intensity of 10 candles at an angle of 60º to the normal. If the area of the \n",
+    "surface is 100 cm2, determine the brightness and total flux radiated.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "I = 10.0        #Cd         (Intensity)\n",
+    "theta = 60.0    #degrees    (angle to normal)\n",
+    "A = 100.0       #cm^2       (Area)\n",
+    "\n",
+    "proA = A*m.cos(theta*3.14/180)     #cm^2\n",
+    "\n",
+    "B = I/proA*(10000)       #cd/m^2    (Brightness)\n",
+    "B = B*3.14               #lm/m^2    (Brightness)\n",
+    "\n",
+    "flux = B*A*10e-5     #lm        (Total flux radiated)\n",
+    "print \"Total flux radiated =\",round(flux,2),\"lm.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.13 , PAGE NO :- 1904"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Brightness 1 = 1.191083e+04 cd/m^2.\n",
+      "Brightness 2 = 0.06 cd/m^2.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Calculate the brightness (or luminance) of snow under an illumination of (a) 44,000 lux and (b) 0.22 lux. Assume that snow\n",
+    "behaves like a perfect diffusor having a reflection factor of 85 per cent.'''\n",
+    "\n",
+    "E1 = 44000.0         #lux      (illumination 1)\n",
+    "E2 = 0.22            #lux      (illumination 2)\n",
+    "rf = 0.85            #         (reflection factor)\n",
+    "\n",
+    "L1 = (E1*rf/3.14)    #cd/m^2    (Brightness 1)\n",
+    "L2 = (E2*rf/3.14)    #cd/m^2    (Brightness 2)\n",
+    "\n",
+    "print \"Brightness 1 = %e cd/m^2.\" %round(L1,2)\n",
+    "print \"Brightness 2 =\",round(L2,2),\"cd/m^2.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.14 , PAGE NO :- 1904"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "luminous intensity of globe is = 44.0 Cd.\n",
+      "percentage absorption =  44.61 %.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 21 cm diameter globe of dense opal glass encloses a lamp emitting 1000 lumens and has uniform brightness of 4e+3 lumen/m^2\n",
+    "when viewed in any direction. What would be the luminous intensity of the globe in any direction? Find what percentage of the\n",
+    "flux emitted by the lamp is absorbed by the globe.'''\n",
+    "\n",
+    "d = 21.0           #cm      (diameter)\n",
+    "flux = 1000.0      #lumens  (luminous flux) \n",
+    "B = 4e+3           #lm/m^2  (uniform Brightness)\n",
+    "\n",
+    "#Surface Area of the globe\n",
+    "sa = 3.14*(d*d)*10e-5       #m^2\n",
+    "#Flux emitted by globe is\n",
+    "fluxe = sa*B       #lm\n",
+    "\n",
+    "#luminous intensity of globe is\n",
+    "lint = fluxe/(4*3.14)      #Cd\n",
+    "print \"luminous intensity of globe is =\",round(lint),\"Cd.\"\n",
+    "#Flux absorbed by globe is\n",
+    "fluxab = flux - fluxe        #lm\n",
+    "\n",
+    "#% absorption is\n",
+    "absrp = fluxab/flux*100      #% absorption\n",
+    "print \"percentage absorption = \",round(absrp,2),\"%.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.15 , PAGE NO :- 1904"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Luminous intensity is = 1005440.0 Cd\n",
+      "The beam spread is = 14.25 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 2.5 cm diameter disc source of luminance 1000 cd/cm2 is placed at the focus of a specular parabolic reflector\n",
+    "normal to the axis. The focal length of the reflector is 10 cm, diameter 40 cm and reflectance 0.8. Calculate the axial\n",
+    "intensity and beam-spread. Also show diagrammatically what will happen if the source were moved away from the reflector\n",
+    "along the axis in either direction.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "dia = 0.025         #m         (diameter of disc)\n",
+    "d = 0.4           #m         (diameter of relector) \n",
+    "L = 1000.0e+4     #Cd/m^2    (luminance)\n",
+    "\n",
+    "#Surface area is\n",
+    "A = 3.142*d*d/4    #m^2       (Area)\n",
+    "\n",
+    "#Luminous intensity is\n",
+    "I = 0.8*A*L      #Cd\n",
+    "print \"Luminous intensity is =\",round(I,2),\"Cd\"\n",
+    "\n",
+    "#Let us assume 'theta' as the beam-spread .Then\n",
+    "r = dia/2           #m    (radius)\n",
+    "f = 0.1           #m    (focal length) \n",
+    "theta = 2*m.degrees(m.atan((r/f)))\n",
+    "\n",
+    "print \"The beam spread is =\",round(theta,2),\"degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.16 , PAGE NO :- 1905"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Flux emmited by source is = 6942.15 lm/m^2 .\n",
+      "C.P of globe is =  84.0 Cd.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 22cm diameter globe of opal glass encloses a lamp of uniform luminous intensity 120 C.P. Thirty per cent of light emitted \n",
+    "by the lamp is absorbed by globe. Determine (a) luminance of globe (b) C.P. of globe in any direction.'''\n",
+    "\n",
+    "\n",
+    "d = 0.22       #m     (diameter)\n",
+    "I = 120.0      #Cd    (luminous intensity)\n",
+    "\n",
+    "#surface area is\n",
+    "sa = 3.14*d*d      #m^2\n",
+    "\n",
+    "#Flux emmited by source is\n",
+    "flux = I*(4*3.14)       #lm\n",
+    "#Flux emmited by globe is\n",
+    "reflux = 0.7*flux       #lm\n",
+    "\n",
+    "#(a)Luminance of globe is\n",
+    "L = reflux/sa           #lm/m^2\n",
+    "print \"Flux emmited by source is =\",round(L,2),\"lm/m^2 .\"\n",
+    "#(b) C.P of globe is\n",
+    "cp = reflux/(4*3.14)     #Cd\n",
+    "print \"C.P of globe is = \",round(cp,2),\"Cd.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.17 , PAGE NO :- 1905"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Average luminance = 7722.93 lm/m^2 .\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 0.4 m diameter diffusing sphere of opal glass (20 percent absorption) encloses an incandescent lamp with a luminous flux of\n",
+    "4850 lumens. Calculate the average luminance of the sphere.'''\n",
+    "\n",
+    "\n",
+    "d = 0.4              #m       (diameter)\n",
+    "lflux = 4850.0       #lm      (luminous flux)\n",
+    "reflux = 0.8*lflux   #lm      (flux emmited by globe)\n",
+    "\n",
+    "sa = 3.14*d*d        #m^2     (surface area)\n",
+    "\n",
+    "#Brightness B = flux emmited/surface area . i.e\n",
+    "B = reflux/sa        #lm/m^2  (brightness)\n",
+    "\n",
+    "print \"Average luminance =\",round(B,2),\"lm/m^2 .\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.18 , PAGE NO :- 1907"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Illumination produced is = 122.55 lm/m^2.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A show case is lighted by 4 metre of architectural tubular lamps arranged in a continuous line and placed along the top of the\n",
+    "case.Determine the illumination produced on a horizontal surface 2 metres below the lamps in a position directly underneath the\n",
+    "centre of the 4 m length of the lamps on the assumption that in tubular lamps emit 1,880 lm per metre run.\n",
+    "Neglect the effect of any reflectors which may be used.'''\n",
+    "\n",
+    "import math as m\n",
+    "\n",
+    "L = 4.0       #m        (length of source of light)\n",
+    "d = 2.0       #m        (height)\n",
+    "flux = 1880.0 #lumens      (flux) \n",
+    "#Now\n",
+    "theta = m.atan(L/(2*d))\n",
+    "\n",
+    "#As I = flux/(3.14*3.14*L)\n",
+    "I = 4*flux/(3.14*3.14*L)     #cd/m\n",
+    "\n",
+    "#Illumination produced is\n",
+    "E = I/(2*d)*(m.sin(2*theta) + 2*theta)    #lm/m^2\n",
+    "\n",
+    "print \"Illumination produced is =\",round(E,2),\"lm/m^2.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## EXAMPLE 49.19 , PAGE NO :- 1913"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Luminous flux yield of the source = 1109.2 lm.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''If an integrating sphere 0.6 m in diameter whose inner surface has a reflection coefficient of 0.8 contains a lamp producing\n",
+    "on the portion of the sphere, screened from direct radiation,a luminance of 1000 cd/m2, what is the luminous flux yield of\n",
+    "the source ?'''\n",
+    "\n",
+    "from sympy import Eq,solve,Symbol\n",
+    "\n",
+    "coef = 0.8      #          (reflection coefficient)\n",
+    "L = 1000.0      #cd/m^2    (luminance)\n",
+    "d = 0.6         #m         (diameter)\n",
+    "\n",
+    "Fl = Symbol('Fl')\n",
+    "E = coef*Fl/(3.14*d*d*(1-coef))     #lm/m^2\n",
+    "L1 = coef*E/3.14           #cd/m^2\n",
+    "#As L is equal to L1\n",
+    "eq = Eq(L,L1)\n",
+    "Fl = solve(eq)\n",
+    "Fl1 = Fl[0]     #lumens\n",
+    "\n",
+    "print \"Luminous flux yield of the source =\",round(Fl1,2),\"lm.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## EXAMPLE 49.20 , PAGE NO :- 1919 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Utilization coefficient =  0.4\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A room 8 m * 12 m is lighted by 15 lamps to a fairly uniform illumination of 100 lm/m^2. Calculate the utilization coefficient\n",
+    "of the room given that the output of each lamp is 1600 lumens.'''\n",
+    "\n",
+    "Area = 8*12      #m^2      (area of room)\n",
+    "num = 15.0       #         (number of lamps)\n",
+    "I = 1600.0       #lumens   (output of each lamp)\n",
+    "E = 100.0        #lm/m^2   (illumination)\n",
+    "\n",
+    "#Lumens emmited by lamp\n",
+    "I_tot = num*I    #lumens\n",
+    "\n",
+    "#Lumens recieved by working plane\n",
+    "I1 = Area*E      #lumens\n",
+    "\n",
+    "#Utilization coefficient is\n",
+    "coef = I1/I_tot\n",
+    "\n",
+    "print \"Utilization coefficient = \",round(coef,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.21 , PAGE NO :- 1919"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of lamps required = 50.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''The illumination in a drawing office 30 m*10 m is to have a value of 250 lux and is to be provided by a number of 300 W\n",
+    "filament lamps. If the coefficient of utilization is 0.4 and the depreciation factor 0.9, determine the number of lamps\n",
+    "required. The luminous efficiency of each lamp is 14 lm/W.'''\n",
+    "\n",
+    "A = 30*10.0     #m^2     (area)\n",
+    "E = 250.0       #lm/m^2  (illumination)\n",
+    "coef = 0.4      #        (coefficient of utilization)\n",
+    "p = 0.9         #        (depriciation factor)\n",
+    "eff = 14.0      #lm/W    (luminous efficiency)\n",
+    "watt = 300.0    #W       (wattage of eacch lamp)\n",
+    "#Now, flux = E*A/coef*p\n",
+    "flux = E*A/(coef*p)   #lm     (output in lumens)\n",
+    "\n",
+    "#Flux emmited per lamp is\n",
+    "Fl2 = watt*eff      #lm\n",
+    "\n",
+    "#No. of lamps required are\n",
+    "num = flux/Fl2\n",
+    "\n",
+    "print \"Number of lamps required =\",round(num)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.22 , PAGE NO :- 1919"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Net saving in load = 600.0 W.\n",
+      "Increase in illumination =  42.22 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Find the total saving in electrical load and percentage increase in illumination if instead of using twelve 150 W tungsten-\n",
+    "filament lamps,we use twelve 80 W fluorescent tubes. It may be assumed that (i) there is a choke loss of 25 per cent of rated\n",
+    "lamp wattage (ii) average luminous efficiency throughout life for each lamp is 15 lm/W and for each tube 40 lm/W and \n",
+    "(iii) coefficient of utilization remains the same in both cases.'''\n",
+    "\n",
+    "#Luminous efficiency\n",
+    "eff1 = 15.0      #lm/W\n",
+    "eff2 = 40.0      #lm/W\n",
+    "\n",
+    "#Total load in filament-lamps\n",
+    "flamp = 12*150.0      #W\n",
+    "#Total load in fluoroscent tubes\n",
+    "tube = 12*(80 + 0.25*80) #W\n",
+    "#Net saving\n",
+    "load = flamp - tube     #W\n",
+    "print \"Net saving in load =\",round(load,2),\"W.\"\n",
+    "\n",
+    "#Let us assume that\n",
+    "#E1 -> illumination with lamps\n",
+    "#E2 -> illumination with tubes\n",
+    "#Now E1/E2 = (O/P in lumens 1)/(O/P in lumens 2)\n",
+    "\n",
+    "tube2 = 12*80.0         #W\n",
+    "\n",
+    "E1_E2 = flamp*eff1/(tube2*eff2)\n",
+    "\n",
+    "#Increase in illumination is given by  %increase = (E2/E1 - 1)*100\n",
+    "increase = (1/E1_E2 - 1)*100.0\n",
+    "\n",
+    "print \"Increase in illumination = \",round(increase,2),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.23 , PAGE NO :- 1919"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of lamps on each tower is = 50.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A football pitch 120 m * 60 m is to be illuminated for night play by similar banks of equal 1000 W lamps supported on twelve \n",
+    "towers which are distributed around the ground to provide approximately uniform illumination of the pitch.Assuming that 40% of\n",
+    "the total light emitted reaches the playing pitch and that an illumination of 1000 lm/m2 is necessary for television purposes,\n",
+    "calculate the number of lamps on each tower. The overall efficiency of the lamp is to be taken as 30 lm/W.'''\n",
+    " \n",
+    "Area = 120.0*60.0       #m^2         (Area of pitch)\n",
+    "E = 1000.0              #lm/m^2      (Illumination of pitch)\n",
+    "\n",
+    "#Flux required is\n",
+    "flux = Area*E           #lm         \n",
+    "\n",
+    "#Since only 40% reaches the ground.Total flux required is\n",
+    "lflux = flux/0.4        #lm\n",
+    "\n",
+    "#There are 12 tower banks . Therefore flux by each tower bank is\n",
+    "flux_each = lflux/12    #lm\n",
+    "\n",
+    "#Output of each 1000 W lamp is\n",
+    "I = 30.0*1000           #lm\n",
+    "\n",
+    "#Therefore, number of each lamps is\n",
+    "num = flux_each/I\n",
+    "\n",
+    "print \"Number of lamps on each tower is =\",round(num)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.24 , PAGE NO :- 1920"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of flouroscent tubes = 384.0\n",
+      "number of twin fittings = 192.0\n",
+      "These can be arranged in 8 rows and 24 columns with space/height ratio = 1.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Design a suitable lighting scheme for a factory 120 m * 40 m with a height of 7 m. Illumination required is 60 lux.\n",
+    "State the number, location and mounting height of 40 W fluorescent tubes giving 45 lm/W. Depreciation factor = 1.2;\n",
+    "utilization factor = 0.5 .'''\n",
+    "\n",
+    "A = 120.0*40.0        #m^2     (Area)\n",
+    "h = 7.0               #m       (Height)\n",
+    "E = 60.0              #lm/m^2  (Illumination)\n",
+    "watt = 40.0           #W       (Wattage of bulb)\n",
+    "eff = 45.0            #lm/W    (Luminous efficiency)\n",
+    "dep = 1.2             #        (Depriciation factor)\n",
+    "uti = 0.5             #        (Utilization factor)\n",
+    "\n",
+    "#Total output flux is\n",
+    "flux = E*A/(uti*1/dep)     #lm\n",
+    "#Flux per tube is\n",
+    "flux_tube = eff*watt       #lm\n",
+    "\n",
+    "#Therefore,number of flouroscent tubes required\n",
+    "num = flux/flux_tube\n",
+    "print \"number of flouroscent tubes =\",round(num)\n",
+    "#For twin fittings\n",
+    "num = num/2\n",
+    "print \"number of twin fittings =\",round(num)\n",
+    "print \"These can be arranged in 8 rows and 24 columns with space/height ratio = 1.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.25 , PAGE NO :- 1920"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of 500-W lamps required = 24.0\n",
+      "Number of 300-W lamps required = 49.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A drawing hall in an engineering college is to be provided with a lighting installation. The hall is 30 m * 20 m * 8 m (high).\n",
+    "The mounting height is 5 m and the required level of illumination is 144 lm/m^2. Using metal filament lamps, estimate\n",
+    "the size and number of single lamp luminaries and also draw their spacing layout. Assume :\n",
+    "Utilization coefficient = 0.6; maintenance factor = 0.75; space/height ratio=1 lumens/watt for 300-W lamp = 13,\n",
+    "lumens/watt for 500-W lamp = 16.'''\n",
+    "\n",
+    "\n",
+    "A = 30.0*20      #m^2      (Area)\n",
+    "E = 144.0        #lm/m^2   (Illumination)\n",
+    "coef = 0.6       #         (Utilization coefficient)\n",
+    "mfac = 0.75      #         (maintenance factor)\n",
+    "\n",
+    "\n",
+    "#The flux is given by\n",
+    "flux = E*A/(coef*mfac)       #lm\n",
+    "\n",
+    "#Lumen output for 500-W lamp\n",
+    "I5 = 500.0*16        #lm\n",
+    "\n",
+    "#Lumen output for 500-W lamp\n",
+    "I3 = 300.0*13        #lm\n",
+    "\n",
+    "#No. of 500 W lamps required is\n",
+    "num5 = flux/I5\n",
+    "print \"Number of 500-W lamps required =\",round(num5)\n",
+    "\n",
+    "#No. of 300 W lamps required is\n",
+    "num3 = flux/I3\n",
+    "print \"Number of 300-W lamps required =\",round(num3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.26 , PAGE NO :- 1920"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total number of lamps = 36.0\n",
+      "Wattage of each lamp is = 500.0 W.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Estimate the number and wattage of lamps which would be required to illuminate a workshop space 60 * 15 metres by means of\n",
+    "lamps mounted 5 metres above the working plane. The average illumination required is about 100 lux.Coefficient of\n",
+    "utilization=0.4 ; Luminous efficiency=16 lm/W.Assume a spacing/height ratio of unity and a candle power depreciation of 20%.'''\n",
+    "\n",
+    "A = 60.0*15.0        #m^2\n",
+    "E = 100.0            #lm/m^2\n",
+    "coef = 0.4           #            (coefficient of utilization)\n",
+    "lum = 16.0           #lm/W        (luminous efficiency)\n",
+    "dep = 1+0.2          #            (depriciation factor)\n",
+    "\n",
+    "#Total flux is given by\n",
+    "flux = E*A/(coef*1/dep)    #lm\n",
+    "\n",
+    "#Total wattage required is\n",
+    "watt = flux/lum            #W\n",
+    "\n",
+    "#Now,space/height ratio is 1.\n",
+    "h = 5.0                #m\n",
+    "\n",
+    "#Therefore along breadth , lamps are\n",
+    "num_b = 15.0/h\n",
+    "\n",
+    "#Therefore along length , lamps are\n",
+    "num_l = 60.0/h\n",
+    "\n",
+    "#Total number of lamps are\n",
+    "num_tot = num_b*num_l\n",
+    "\n",
+    "#Wattage of each lamp is\n",
+    "watt_each = watt/num_tot\n",
+    "\n",
+    "print \"Total number of lamps =\",num_tot\n",
+    "print \"Wattage of each lamp is =\",round(watt_each,-2),\"W.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## EXAMPLE 49.27 , PAGE NO :- 1921"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of 200-W lamps required are = 68.0\n",
+      "Number of 300-W lamps required are = 40.0\n",
+      "Number of 500-W lamps required are = 22.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A drawing hall 40 m * 25 m * 6 high is to be illuminated with metal-filament gas-filled lamps to an average illumination of\n",
+    "90 lm/m^2 on a working plane 1 metre above the floor.Estimate suitable number, size and mounting height of lamps. Sketch the\n",
+    "spacing layout.Assume coefficient of utilization of 0.5, depreciation factor of 1.2 and spacing/height ratio of 1.2\n",
+    "\n",
+    "Size of lamps                 :     200 W        300 W         500 W\n",
+    "Luminous efficiency (in lm/W) :      16           18            20                                '''\n",
+    "\n",
+    "A = 40.0*25.0       #m^2        (Area)\n",
+    "E = 90.0            #lm/m^2     (Illumination)\n",
+    "coef = 0.5          #           (Coefficient of utilization)\n",
+    "dep = 1.2           #           (Depreciation factor)\n",
+    "\n",
+    "#Total flux required is\n",
+    "flux = E*A/(coef*1/1.2)          #lumens\n",
+    "\n",
+    "#Lumen output of each 200-W lamp is\n",
+    "flux_200 = 200.0*16              #lumens\n",
+    "\n",
+    "#Lumen output of each 200-W lamp is\n",
+    "flux_300 = 300.0*18             #lumens\n",
+    "\n",
+    "#Lumen output of each 200-W lamp is\n",
+    "flux_500 = 500.0*20              #lumens\n",
+    "\n",
+    "#Number of 200-W lamps required is\n",
+    "num_200 = flux/flux_200\n",
+    "#Number of 200-W lamps required is\n",
+    "num_300 = flux/flux_300\n",
+    "#Number of 200-W lamps required is\n",
+    "num_500 = flux/flux_500\n",
+    "\n",
+    "print \"Number of 200-W lamps required are =\",round(num_200)\n",
+    "print \"Number of 300-W lamps required are =\",round(num_300)\n",
+    "print \"Number of 500-W lamps required are =\",round(num_500)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.28 , PAGE NO :- 1922"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total wattage required is = 2019.23 W.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A school classroom, 7 m * 10 m * 4 m high is to be illuminated to 135 lm/m^2 on the working plane. If the coefficient of\n",
+    "utilization is 0.45 and the sources give 13 lumens per watt,work out the total wattage required, assuming a depreciation factor\n",
+    "of 0.8 .Sketch roughly the plan of the room, showing suitable positions for fittings, giving reasons for the positions chosen.'''\n",
+    "\n",
+    "A = 7.0*10.0       #m^2     (Area)\n",
+    "E = 135.0          #lm/m^2  (Illumination)\n",
+    "coef = 0.45        #        (Coefficient of utilization)\n",
+    "dep = 0.8          #        (Depreciation factor)\n",
+    "eff = 13.0         #lm/W        (luminous efficiency) \n",
+    "\n",
+    "#Total flux is\n",
+    "flux = E*A/(coef*dep)      #lumens\n",
+    "\n",
+    "#Therefore,total wattage required is\n",
+    "watt = flux/eff              #W\n",
+    "\n",
+    "print \"Total wattage required is =\",round(watt,2),\"W.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.29 , PAGE NO :- 1922"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of 100-W lamps required = 29.0\n",
+      "Number of 200-W lamps required = 13.0\n",
+      "Number of 300-W lamps required = 10.0\n",
+      "Number of 500-W lamps required = 5.0\n",
+      "Number of 1000-W lamps required = 2.0\n",
+      "If we take the mounting height of 5 m, then 300 W lamps would be suitable.\n",
+      "The No.of lamps required would be 10, arranged in two rows, each row having 5 lamps thus giving space/height ratio of 6/5\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A hall 30 m long and 12 m wide is to be illuminated and the illumination required is 50 lm/m2. Calculate the number,\n",
+    "the wattage of each unit and the location and mounting height of the units, taking a depreciation factor of 1.3 and\n",
+    "utilization factor of 0.5, given that the outputs of the different types of lamp are as under :\n",
+    "Watts  : 100       200       300       500        1000\n",
+    "Lumens : 1615      3650      4700      9950       21500                                              '''\n",
+    "\n",
+    "A = 30.0*12.0       #m^2         (Area)\n",
+    "E = 50.0            #lm/m^2      (Illumination)\n",
+    "coef = 0.5          #            (coefficient of utilization)\n",
+    "dep = 1/1.3         #            (depreciation factor)\n",
+    "\n",
+    "#Total flux required is\n",
+    "flux = E*A/(coef*dep)       #lumens\n",
+    "\n",
+    "#For 100-W lamps are used ,Number required\n",
+    "num_100 = flux/1615.0\n",
+    "#For 200-W lamps are used ,Number required\n",
+    "num_200 = flux/3650.0\n",
+    "#For 300-W lamps are used ,Number required\n",
+    "num_300 = flux/4700.0\n",
+    "#For 500-W lamps are used ,Number required\n",
+    "num_500 = flux/9950.0\n",
+    "#For 1000-W lamps are used ,Number required\n",
+    "num_1000 = flux/21500.0\n",
+    "\n",
+    "print \"Number of 100-W lamps required =\",round(num_100)\n",
+    "print \"Number of 200-W lamps required =\",round(num_200)\n",
+    "print \"Number of 300-W lamps required =\",round(num_300)\n",
+    "print \"Number of 500-W lamps required =\",round(num_500)\n",
+    "print \"Number of 1000-W lamps required =\",round(num_1000)\n",
+    "print \"If we take the mounting height of 5 m, then 300 W lamps would be suitable.\"\n",
+    "print \"The No.of lamps required would be 10, arranged in two rows, each row having 5 lamps thus giving space/height ratio of 6/5\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.30 , PAGE NO :- 1924"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of lamps required are = 10.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''It is desired to floodlight the front of a building 42 m wide and 16 m high.Projectors of 30° beam spread and 1000-W lamps\n",
+    "giving 20 lumen/watt are available. If the desired level of illumination is 75 lm/m2 and if the projectors are to be located\n",
+    "at ground level 17 m away,design and show a suitable scheme. Assume the following :\n",
+    "Coefficient of utilization = 0.4     ;      Depreciation factor = 1.3    ;   Waste-light factor = 1.2.   '''\n",
+    "\n",
+    "A = 42.0*16.0           #m^2          (Area)\n",
+    "E = 75.0                #lm/m^2       (Illumination)\n",
+    "W = 1.2                 #             (Waste-light factor)\n",
+    "coef = 0.4              #             (Coefficient of utilization)\n",
+    "dep = 1/1.3             #             (Depreciation factor)\n",
+    "eff = 20.0              #lm/W         (luminous efficiency)\n",
+    "\n",
+    "#Total flux is\n",
+    "flux = E*A*W/(coef*dep)       #lm/m^2\n",
+    "\n",
+    "#Lumen output of each 1000-W lamp is\n",
+    "flux_each = 1000.0*eff\n",
+    "\n",
+    "#Number of lamps required are\n",
+    "num = flux/flux_each\n",
+    "\n",
+    "print \"Number of lamps required are =\",round(num)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.31 , PAGE NO :- 1925"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The number of floodlight projectors required = 24.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Estimate the number of 1000-W floodlight projectors required to illuminate the up per 75 m of one face of a 96 m tower of \n",
+    "width 13 m if approximate initial average luminance is to be 6.85 cd/m2. The projectors are mounted at ground level 51m from base\n",
+    "of the tower.Utilization factor is = 0.2; reflection factor of wall = 25% and efficiency of each lamp = 18 lm/W.'''\n",
+    "\n",
+    "\n",
+    "A = 13.0*75.0      #m^2      (Area to be flood-lighted)\n",
+    "B = 6.85           #cd/m^2   (Average luminance)\n",
+    "watt = 1000.0      #W        (Wattage floodlight projectors)\n",
+    "coef = 0.2         #         (Utilization factor)\n",
+    "ref = 0.25         #         (Reflection factor)\n",
+    "\n",
+    "#Illumination E = pi*B/reflection factor\n",
+    "E = 3.14*B/ref     #lm/m^2\n",
+    "\n",
+    "#Therefore, total flux required is\n",
+    "flux = E*A         #lm\n",
+    "\n",
+    "#Flux to be emmited by lamp is\n",
+    "lflux = flux/coef  #lm\n",
+    "\n",
+    "#Flux from each lamp is\n",
+    "flux_each = 18.0*watt   #lm\n",
+    "\n",
+    "#The number of floodlight projectors required are\n",
+    "num = lflux/flux_each\n",
+    "\n",
+    "print \"The number of floodlight projectors required =\",round(num)+1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.32 , PAGE NO :- 1928"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Diameter d2 =  0.4 *d1.\n",
+      "Length l2 =  1.26 *l1.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''If the filament of a 32 candela, 100-V lamp has a length l and diameter d,calculate the length and diameter of the filament\n",
+    "of a 16 candela 200-V lamp,assuming that the two lamps run at the same intrinsic brilliance.'''\n",
+    "\n",
+    "\n",
+    "#As l*d is directly propotional luminous intensity .\n",
+    "#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 32/16\n",
+    "\n",
+    "a_b = 32.0/16         # (= a/b = l1*d1/l2*d2)\n",
+    "\n",
+    "#As luminous intensity is directly propotional to power output\n",
+    "# 32 o< 100*I1    & 16 o< 200*I2\n",
+    "\n",
+    "I1_I2 = 32*200.0/(16*100.0)     #( = I1/I2)\n",
+    "\n",
+    "\n",
+    "#Also , I o< d^3/2 \n",
+    "d1_d2 = (I1_I2)**(2.0/3)         #( = d1/d2)\n",
+    "\n",
+    "print \"Diameter d2 = \",round(1/d1_d2,2),\"*d1.\"\n",
+    "\n",
+    "#As , d1_d2 = d1/d2\n",
+    "l1_l2 = a_b/d1_d2               #( = l1/l2)\n",
+    "print \"Length l2 = \",round(1/l1_l2,2),\"*l1.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.33 , PAGE NO :- 1928"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Diameter d2 =  0.00198 cm.\n",
+      "Length l2 =  126.0 cm.\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''An incandescent lamp has a filament of 0.005 cm diameter and one metre length. It is required to construct another lamp of\n",
+    "similar type to work at double the supply voltage and give half the candle power. Assuming that the new lamp operates at the same\n",
+    "brilliancy,determine suitable dimensions for its filament.'''\n",
+    "\n",
+    "d1 = 0.005       #cm    (diameter)\n",
+    "l1 = 100.0       #cm    (length) \n",
+    "#As l*d is directly propotional luminous intensity .\n",
+    "#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 2.0/1.0\n",
+    "\n",
+    "a_b = 2.0/1        # (= a/b = l1*d1/l2*d2)\n",
+    "\n",
+    "#As luminous intensity is directly propotional to power output\n",
+    "# I1 o< V1*i1    & I2 o< V2*i2\n",
+    "\n",
+    "I1_I2 = (2.0/1)*(2.0/1)     #( = I1/I2)\n",
+    "\n",
+    "\n",
+    "#Also , I o< d^3/2 \n",
+    "d1_d2 = (I1_I2)**(2.0/3)         #( = d1/d2)\n",
+    "d2 = d1/d1_d2                    #cm\n",
+    "print \"Diameter d2 = \",round(d2,5),\"cm.\"\n",
+    "\n",
+    "#As , d1_d2 = d1/d2\n",
+    "l1_l2 = a_b/d1_d2               #( = l1/l2)\n",
+    "l2 = l1/l1_l2                   #cm \n",
+    "print \"Length l2 = \",round(l2),\"cm.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 49.34 , PAGE NO :- 1928"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 52,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Hence expression for candle power is C = 1.155816e-09 *V^ 4.47\n",
+      "Change of candle power per volt =  268.1 cd/m.\n",
+      "% change in candle power for increase = 19.16\n",
+      "% change in candle power for decrease = -16.67\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''A 60 candle power, 250-V metal filament lamp has a measured candle power of 71.5 candela at 260 V and 50 candela at 240 V.\n",
+    "(a) Find the constant for the lamp in the expression C = aV^b where C = candle power and V = voltage.\n",
+    "(b) Calculate the change of candle power per volt at 250 V. Determine the percentage variation of candle power due to a voltage\n",
+    "variation of æ 4% from the normal value.   '''\n",
+    "\n",
+    "from sympy import Symbol,solve,Eq\n",
+    "\n",
+    "#Given expression is C = a*V^b\n",
+    "b = Symbol('b')\n",
+    "# 71.5/50 = (260/240)^b\n",
+    "lhs = 71.5/50.0\n",
+    "rhs = (260.0/240)**b\n",
+    "eq = Eq(lhs,rhs)\n",
+    "b = solve(eq)\n",
+    "b1 = b[0]          #constant\n",
+    "\n",
+    "a = 71.5/(260.0)**b1\n",
+    "print \"Hence expression for candle power is C = %e\" %a,\"*V^\",round(b1,2)\n",
+    "\n",
+    "#Change of candle power per volt\n",
+    "# dC/dV = b*a*V^b\n",
+    "change = b1*a*((250.0)**(b1))    #cd/V\n",
+    "print \"Change of candle power per volt = \",round(change,1),\"cd/m.\"\n",
+    "\n",
+    "#When voltage is increases by 4% C2/C1 = (1.04)^b\n",
+    "per_change = ( (1.04)**b1  - 1 ) * 100\n",
+    "\n",
+    "print \"% change in candle power for increase =\",round(per_change,2)\n",
+    "\n",
+    "#When voltage is decreases by 4% C2/C1 = (0.96)^b\n",
+    "per_change = ( (0.96)**b1 - 1)* 100\n",
+    "\n",
+    "print \"% change in candle power for decrease =\",round(per_change,2)\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "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
+}