{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "# Chapter 11:Radiation heat transfer"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.3:pg-445"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\n",
      "The view factors F13 and F31 between the surfaces 1 and 3 are \n",
      "F13= 0.04\n",
      "F31= 0.032\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 3\"\n",
    "print\"The view factors F13 and F31 between the surfaces 1 and 3 are \"\n",
    "#Determine the view factors F13 and F31 between the surfaces 1 and 3.\n",
    "#F1-2,3=F12+F13\n",
    "#So F13=F1-2,3-F12\n",
    "#Let F1-2,3=F123\n",
    "#From Radiation Shape factor b/w two perpendicular rectangles with a commom edge table we get F12=.027,F1-2,3=0.31\n",
    "F123=0.31;#View factor\n",
    "F12=.27;#View factor\n",
    "F13=F123-F12#View factor\n",
    "print\"F13=\",F13\n",
    "#A1,A2 and A3 are the emitting surface areas\n",
    "#From reciprocity relation F31=(A1/A3)/F13\n",
    "A1=2;\n",
    "A3=2.5;\n",
    "F31=(A1/A3)*F13\n",
    "print\"F31=\",F31"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.4:pg-447"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\n",
      "F124= 0.04\n",
      "F24= 0.06\n",
      "The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
      "F14= 0.02\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 4\"\n",
    "#Determine the view factors F14 for the composite surface .\n",
    "#From the table of radiation shape factor b/w two perpendicular surfaces F1,2-3,4=0.14 and F1,2-3=0.1\n",
    "#By subdivision of the recieving surfaces we get F1,2-4=F1,2-3,4-F1,2-3\n",
    "#Let F1,2-4=F124 , F1,2-3,4=F1234 , F1,2-3=F123\n",
    "F1234=0.14;#View factor\n",
    "F123=0.1;#View factor\n",
    "F124=F1234-F123;#View factor\n",
    "print\"F124=\",F124\n",
    "#Again from the table of radiation shape factor b/w two perpendicular surfaces F2-3,4=0.24 , F23=0.18\n",
    "#Let F2-3,4=F234\n",
    "F234=0.24;#View factor\n",
    "F23=0.18;#View factor\n",
    "#By subdivision of the recieving surfaces we get F24=F2-3,4-F23\n",
    "F24=F234-F23;#view factor\n",
    "print\"F24=\",F24\n",
    "#A1 and A2 are the emitting surface areas.\n",
    "A1=12;\n",
    "A2=12;\n",
    "#Now by subdivision of emitting surfaces F1,2-4=(1/(A1+A2))*(A1*F14+A2*F24)\n",
    "#This implies F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\n",
    "print\"The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2\"\n",
    "F14=((F124*(A1+A2))-(A2*F24))/A2\n",
    "print\"F14=\",F14"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.5:pg-453"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\n",
      "The view factors of cylindrical surface with respect to the base are\n",
      "F13= 0.84\n",
      "F31= 0.0\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 5\"\n",
    "#Consider a cylinder having length,L=2r determine the view factor of cylindrical surface with respect to the base.\n",
    "#From the graph of radiation shape factor b/w parallel coaxial disks of equal diameter F12=0.16\n",
    "F12=0.16;#View factor\n",
    "#By the summation rule of an enclosure F11+F12+F13=1\n",
    "#But F11=0(since the base surface is flat)\n",
    "F11=0;#View factor\n",
    "print\"The view factors of cylindrical surface with respect to the base are\"\n",
    "F13=1-F12-F11#view factor\n",
    "print\"F13=\",F13\n",
    "#By making use of reciprocity theorem we have F31=(A1/A2)*F13\n",
    "#A1 and A2 are emitting surface areas\n",
    "#A1/A2=(pi*r**2)/(2*pi*r*2*r)=1/4\n",
    "#Let A1/A2=A\n",
    "A=1/4;\n",
    "F31=(A)*F13\n",
    "print\"F31=\",F31"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.6:pg-456"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\n",
      "The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4) in W\n",
      "Q= -2918.916\n",
      "Here minus sign indicates that the net heat transfer is from surface2 to surface1\n",
      "The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 0K in W\n",
      "Q1= 2894.4216\n",
      "The view factor of surface 1 with respect to surrounding is\n",
      "F1s= 0.89\n",
      "The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \n",
      "Q1= 1055.04525\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 6\"\n",
    "#Two rectangles length,L=1.5m by breadth,B=3.0m are parallel and directly opposed.\n",
    "L=1.5;\n",
    "B=3;\n",
    "#They are 3m apart\n",
    "#Temprature(T1) of surface 1 is 127°C or 400K and temprature(T2) of surface 2 is 327°C or 600K \n",
    "T1=400;\n",
    "T2=600;\n",
    "#Area (A) is the product of L and B\n",
    "A1=L*B;\n",
    "#Stefan -Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
    "sigma=5.67*10**-8;\n",
    "#From the graph of radiation shape factor b/w parallel rectangles F12=0.11\n",
    "F12=0.11;#View factor\n",
    "#The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4)\n",
    "print\"The rate of heat transfer is given by Q=A1*F12*sigma*(T1**4-T2**4) in W\"\n",
    "Q=A1*F12*sigma*(T1**4-T2**4)\n",
    "print\"Q=\",Q\n",
    "print\"Here minus sign indicates that the net heat transfer is from surface2 to surface1\"\n",
    "#Surface1 recieves energy only from surface 2,since the surrounding is at 0K.\n",
    "#Therefore Q1=A1*Eb1-A2*F21*Eb2\n",
    "#This implies Q1 can also be written as A1*sigma*(T1**4-F12*T2**4)\n",
    "#From reciprocity theorem F21=F12 (since A1=A2)\n",
    "F21=F12;#view factor\n",
    "print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 0K in W\" \n",
    "Q1=A1*sigma*(T1**4-F12*T2**4)\n",
    "print\"Q1=\",Q1\n",
    "#In the case when surrounding is at temprature, Ts=300K ,the energy recieved from the surrounding by the surface 1 has to be considered.\n",
    "Ts=300;\n",
    "#Applying summation rule of view factors F11+F12+F1s=1\n",
    "F11=0;#view factor\n",
    "print\"The view factor of surface 1 with respect to surrounding is\"\n",
    "F1s=1-F11-F12\n",
    "print\"F1s=\",F1s\n",
    "#subscript s denotes the surroundings\n",
    "#Q1=A1*Eb1-A2*F21*Eb2-As*Fs1*Ebs\n",
    "#With the help of reciprocity theorem A2*F21=A1*F12 , As*Fs1=A1*F1s\n",
    "#Therefore we can write Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
    "print\"The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W \"\n",
    "Q1=A1*sigma*(T1**4-F12*T2**4-F1s*Ts**4)\n",
    "print\"Q1=\",Q1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.7:pg-470"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\n",
      "The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1) in W\n",
      "H= 1764.51561476\n",
      "The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\n",
      "H= 3276.95757027\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 7\"\n",
    "#Two parallel infinite surafces are maintained at tempratures T2=200°C or 473.15K and T1=300°C or 573.15K\n",
    "T1=573.15;\n",
    "T2=473.15;\n",
    "#The emissivity(emi) is 0.7 for both the surfaces which are gray.\n",
    "emi1=0.7;\n",
    "emi2=0.7;\n",
    "#stefan=boltzman constant(sigma)=5.67*10**-8W/(m**2*K**4)\n",
    "sigma=5.67*10**-8;\n",
    "#The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1)\n",
    "#Let Q/A=H\n",
    "print\"The net rate of heat transfer per unit area is given Q/A=(sigma*(T1**4-T2**4))/((1/Ɛ1)+(1/Ɛ2)-1) in W\"\n",
    "H=(sigma*(T1**4-T2**4))/((1/emi1)+(1/emi2)-1)\n",
    "print\"H=\",H\n",
    "#When the two surfaces are black\n",
    "#This implies emiisivity(emi)=1 for both surfaces\n",
    "#So,The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4)\n",
    "print\"The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1**4-T2**4) in W\"\n",
    "H=sigma*(T1**4-T2**4)\n",
    "print\"H=\",H"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.8:pg-482"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\n",
      "The net rate of heat exchange between the spheres when the surfaces are black is Q=A1*sigma*(T1**4-T2**4) in W \n",
      "Q= 779.311327631\n",
      "The net rate of radiation exchange when one surface is gray and other is diffuse is given by Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) in W\n",
      "Q1= 346.360590058\n",
      "The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1) in W\n",
      "Q2= 389.655663816\n",
      "Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\n",
      "E= 12.5\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 8\"\n",
    "#Two concentric spheres of diameters D1=0.5m and D2=1m are separated by an air space.\n",
    "#The surface tempratures are T1=400K and T2=300K\n",
    "T1=400;\n",
    "T2=300;\n",
    "D1=0.5;\n",
    "D2=1;\n",
    "#A1 and A2 are the areas in m**2 of surface 1 and surface 2 respectively\n",
    "A1=(math.pi*D1**2);\n",
    "A2=(math.pi*D2**2);\n",
    "#Stefan-Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
    "sigma=5.67*10**-8;\n",
    "#The emissivity is represented by emi \n",
    "#The radiation heat exchange in case of two concentric sphere is given by Q=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) \n",
    "#When the spheres are black emi1=emi2=1\n",
    "emi1=1;\n",
    "emi2=1;\n",
    "#Hence Q=A1*sigma*(T1**4-T2**4)\n",
    "print\"The net rate of heat exchange between the spheres when the surfaces are black is Q=A1*sigma*(T1**4-T2**4) in W \"\n",
    "Q=A1*sigma*(T1**4-T2**4)\n",
    "print\"Q=\",Q\n",
    "#The net rate of radiation exchange when one surface is gray and other is diffuse having emi1=0.5 and emi2=0.5  \n",
    "emi1=0.5;\n",
    "emi2=0.5;\n",
    "print\"The net rate of radiation exchange when one surface is gray and other is diffuse is given by Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1)) in W\"  \n",
    "Q1=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1))\n",
    "print\"Q1=\",Q1\n",
    "#The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1)\n",
    "emi2=1;#emissivity of outer surface\n",
    "print\"The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1) in W\"\n",
    "Q2=(A1*sigma*(T1**4-T2**4))/((1/emi1)+(A1/A2)*(1/emi2-1))\n",
    "print\"Q2=\",Q2\n",
    "print\"Error(E) is given By ((Q2-Q1)/Q1)*100 in percentage\"\n",
    "E=((Q2-Q1)/Q1)*100\n",
    "print\"E=\",E"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex11.10:pg-484"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\n",
      "F12= 0.5\n",
      "Equivalent resistance of thermal network (R) is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
      "R= 2.09523809524\n",
      "The radiation flux leaving the hot wall is Q/A=[sigma*(T1**4-T2**4)]/(A*R) in W/m**2\n",
      "H= 26655.2740483\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 11, Example 10\"\n",
    "#Given a furnace which can be approximated as an equuilateral triangle duct\n",
    "#The hot wall is maintained at temprature (T1)=1000K and has emmisivity(emi1)=0.75\n",
    "#The cold wall is at temprature(T2)=350K and has emmisivity(emi2)=0.7\n",
    "T1=1000;\n",
    "T2=350;\n",
    "emi1=0.75;\n",
    "emi2=0.7;\n",
    "#Stefan-Boltzman constant(sigma)=5.67*10**-8 W/(m**2*K**4)\n",
    "sigma=5.67*10**-8;\n",
    "#The third wall is reradiating zone having Q3=0\n",
    "#The radiation flux leaving the hot wall is Q/A=(sigma*(T1**4-T2**4))/(A*R)\n",
    "#By summation rule F33+F31+F32=1\n",
    "#F33=0(in consideration of surface to be plane)\n",
    "#From symmetry F31=F32\n",
    "F31=0.5;#View factors\n",
    "F32=F31;#View factors\n",
    "F33=0;#View factors\n",
    "#From reciprocity theorem F13=F31 and F23=F32=0.5 (since A1=A2=A3=A)\n",
    "F13=F31;#View factors\n",
    "F23=F32;#View factors\n",
    "#Again F11+F12+F13=1 from summation rule\n",
    "F11=0;#View factors\n",
    "F12=1-F13-F11;#View factors\n",
    "print\"F12=\",F12\n",
    "#R1,R2,R12,R13,R23 are the resistances\n",
    "#R is equivalent resistance of thermal network is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
    "R1=(1-emi1)/(emi1);\n",
    "R2=(1-emi2)/(emi2);\n",
    "R12=1/(F12);\n",
    "R13=1/(F13);\n",
    "R23=1/(F23);\n",
    "#R is equivalent resistance of thermal network \n",
    "print\"Equivalent resistance of thermal network (R) is given by R1+((1/R12)+(1/(R13+R23)))**-1+R2\"\n",
    "R=R1+((1/R12)+(1/(R13+R23)))**-1+R2\n",
    "print\"R=\",R\n",
    "#The radiation flux leaving the hot wall is Q/A.\n",
    "print\"The radiation flux leaving the hot wall is Q/A=[sigma*(T1**4-T2**4)]/(A*R) in W/m**2\"\n",
    "#Since A gets cancelled in the factor (A*R)\n",
    "#So Q/A=(sigma*(T1**4-T2**4))/(R)\n",
    "#Let Q/A=H\n",
    "H=(sigma*(T1**4-T2**4))/(R)\n",
    "print\"H=\",H"
   ]
  }
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