path: root/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter49.ipynb

{
 "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
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   "source": []
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   "cell_type": "code",
   "execution_count": null,
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