path: root/Engineering_Heat_Transfer/CHAPTER11.ipynb

{
 "metadata": {
  "name": "CHAPTER11"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 11: Introduction to Radiation Heat Transfer"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.1 Page NO.554"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Calculation of the value of the solid angle subtended by surfaces dA2 and dA3 with respect to dA1 (b) the intensity of emission from dA, in the direction of the other areas (c) the rate at which radiation emitted by dA, is intercepted by the other areas\n",
      "\n",
      "#Given\n",
      "# solid angle is calculate using the equation dw=dA*cos(Beta)/r**2\n",
      "# Beta is the angle between the surface normal of a receiver surface and the line connecting the two surfaces\n",
      "# For area A2\n",
      "# dimensions are 1X1 in, so\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "dA2=(1*1)/144.0\n",
      "Beta1=40*math.pi/180.0\n",
      "r=4\n",
      "dw2_1=dA2*math.cos(Beta1)/r**2\n",
      "dA3=dA2\n",
      "Beta2=0\n",
      "dw3_1=dA3*cos(Beta2)/r**2\n",
      "\n",
      "theta2=math.pi*50/180.0\n",
      "theta3=math.pi*60/180.0\n",
      "I_theta2=2000*(1-0.4*(sin(theta2))**2)\n",
      "I_theta3=2000*(1-0.4*(sin(theta3))**2)\n",
      "\n",
      "dA1=1/144.0\n",
      "dq1_2=I_theta2*dA1*math.cos(theta2)*dw2_1          #In book calculation mistake\n",
      "dq1_3=I_theta3*dA1*math.cos(theta2)*dw3_1\n",
      "\n",
      "#Result\n",
      "print\"(a)The solid angle subtended by area dA2 with respect to dA1 is \",round(dw2_1,4),\"sr\"\n",
      "print\" The solid angle subtended by area dA3 with respect to dA1 is  \",round(dw3_1,4),\"sr\"\n",
      "print\"(b) The intensity of radiation emitted from dA1 in the direction of dA2 is \",round(I_theta2,0),\"BTU/(hr.sq.ft.sr)\"\n",
      "print\" The intensity of radiation emitted from dA1 in the direction of dA3 is\",round(I_theta3,0),\"BTU/(hr.sq.ft.sr)\"\n",
      "print\"(c)The rate at which radiation emitted by dA1 is intercepted by dA2 is \",round(dq1_2,4),\"BTU/hr\"\n",
      "print\" The rate at which radiation emitted by dA1 is intercepted by dA3 is \",round(dq1_3,4),\"BTU/hr\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The solid angle subtended by area dA2 with respect to dA1 is  0.0003 sr\n",
        " The solid angle subtended by area dA3 with respect to dA1 is   0.0004 sr\n",
        "(b) The intensity of radiation emitted from dA1 in the direction of dA2 is  1531.0 BTU/(hr.sq.ft.sr)\n",
        " The intensity of radiation emitted from dA1 in the direction of dA3 is 1400.0 BTU/(hr.sq.ft.sr)\n",
        "(c)The rate at which radiation emitted by dA1 is intercepted by dA2 is  0.0023 BTU/hr\n",
        " The rate at which radiation emitted by dA1 is intercepted by dA3 is  0.0027 BTU/hr\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.2 Page NO.557"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Calculation of the value of the solid angle subtended by surfaces dA2 with respect to dA1 (b)  the rate at which radiation emitted by dA1 is intercepted by dA2 (c)  the irradiation associated with dA2\n",
      "\n",
      "#Given\n",
      "# solid angle is calculate using the equation dw=dA*cos(Beta)/r**2\n",
      "# The angle Beta is 0 because the surface normal of dA2 is directed at dA1\n",
      "dA2=0.02*0.02\n",
      "Beta=0\n",
      "r=1\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "dw2_1=dA2*math.cos(Beta)/r**2\n",
      "dA1=dA2\n",
      "theta=math.pi*30/180.0\n",
      "I_theta=1000# The intensity of radiation leaving dA1 in any direction is 1 000 W/(m**2.sr\n",
      "dq1_2=I_theta*dA1*cos(theta)*dw2_1\n",
      "# The irradiation associated with dA2 can be found by dividing the incident radiation by the receiver area\n",
      "dQ1_2=dq1_2/dA2\n",
      "\n",
      "#Result\n",
      "print\"(a)The solid angle subtended by area dA2 with respect to dA1 is \",round(dw2_1,4),\"sr\"\n",
      "print\"(b)The rate at which radiation emitted by dA1 is intercepted by dA2 is\",round(dq1_2,5),\"W\"\n",
      "print\"(c)The irradiation associated with dA2 is \",round(dQ1_2,3),\"W/sq.m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The solid angle subtended by area dA2 with respect to dA1 is  0.0004 sr\n",
        "(b)The rate at which radiation emitted by dA1 is intercepted by dA2 is 0.00014 W\n",
        "(c)The irradiation associated with dA2 is  0.346 W/sq.m\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.3 Page NO. 563"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# (a) Calculation of the emissivity of the hole.(b) the rate of radiant emission from the hole\n",
      "\n",
      "#Given\n",
      "D=2.5/12.0          # diameter in ft\n",
      "L=4.5/12.0          # length in ft\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "A=(2*math.pi*D**2/4)+(math.pi*D*L)\n",
      "A_hole=math.pi*(1/(8.0*12.0))**2/4.0\n",
      "f=A_hole/A          # fraction of area removed\n",
      "# for rolled and polished aluminum, that emissivity = 0.039 from appendix table E1\n",
      "emissivity=0.039\n",
      "emissivity_hole=emissivity/(emissivity+(1-emissivity)*f)\n",
      "\n",
      "sigma=0.1714e-8      # stefan Boltzmann constant in BTU/(hr~ft**2 degree R)\n",
      "T=150+460            # temperature in degree R\n",
      "qe=emissivity_hole*sigma*T**4\n",
      "Qe=A_hole*qe\n",
      "\n",
      "#Result\n",
      "print\"(a)The emissivity of the hole is %.4f\",round(emissivity_hole,4)\n",
      "print\"(b)The heat lost by the hole is \",round(Qe,4),\"BTU/hr\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The emissivity of the hole is %.4f 0.9933\n",
        "(b)The heat lost by the hole is  0.0201 BTU/hr\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.4 Page NO.568"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of the percentage of total emitted energy that lies in the visible range. \n",
      "\n",
      "#Given\n",
      "T=2800              #Temprature\n",
      "lambda1=4e-7        #Wavelength\n",
      "lambda2=7e-7\n",
      "hT=lambda1*T\n",
      "\n",
      "#Calculation\n",
      "lambdaT=lambda2*T\n",
      "I1=0.0051           #Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1\n",
      "I2=0.065            #Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1\n",
      "dI=I2-I1\n",
      "\n",
      "#Result\n",
      "print\"The percentage of total emitted energy that lies in the visible range is\",round(dI*100,0),\"percant\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The percentage of total emitted energy that lies in the visible range is 6.0 percant\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.5 Page NO. 570"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Estimation of the surface temperature of the sun and the emitted heat flux\n",
      "\n",
      "#Given\n",
      "lambda_max=0.5e-6   # maximum wavelength in m\n",
      "# From Wien\u2019s Displacement Law we can write lambda_max*T=2.898e-3 m.K\n",
      "T=2.898e-3/lambda_max\n",
      "# The heat flux is given by the Stefan-Boltzmann Equation as q=sigma*T**4\n",
      "sigma=5.675e-8      # value of Stefan-Boltzmann constant in W/(m**2.K**4)\n",
      "q=sigma*T**4\n",
      "\n",
      "#result\n",
      "print\"The Surface Temperature of the Sun is \",round(T,2),\"K\"\n",
      "print\"The heat flux emitted is \",round(q,0),\"W/sq.m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Surface Temperature of the Sun is  5796.0 K\n",
        "The heat flux emitted is  64044136.0 W/sq.m\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 11.6 Page NO.575"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# (a) Calculation of the rate at which the sun\u2019s radiant energy is transmitted through the glass windshield. The interior of the car is considered to be a black body that radiates at 100\u00b0F. (b) Calculation of the rate at which radiant energy from the car interior is transmitted through the glass windshield. \n",
      "\n",
      "#Given\n",
      "lambda1=300e-9         # lower limit of wavelength\n",
      "lambda2=380e-9         # upper limit of wavelength\n",
      "T=5800\n",
      "lambda1_T=lambda1*T\n",
      "lambda2_T=lambda2*T\n",
      "I1=0.101               #Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1\n",
      "I2=0.0334              #Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1\n",
      "dI=abs(I2-I1)\n",
      "t=dI*0.68               # transmissivity\n",
      "q=1100                  # radiation received by car in W/sq.m\n",
      "q_in=t*q                # energy transmitted from the sun through the glass\n",
      "\n",
      "\n",
      "#Calculation\n",
      "Tb=311                 # temperature of black body source in K\n",
      "lambda1_Tb=lambda1*Tb\n",
      "lambda2_Tb=lambda2*Tb\n",
      "dI_b=0                # Table 11.1 gives negligibly small values of the corresponding integrals.\n",
      "t_b=dI_b*0.68         # transmissivity\n",
      "q_out=t_b*q\n",
      "\n",
      "#Result\n",
      "print\"(a)The energy transmitted from the sun through the glass is \",round(q_in,1),\"W/sq.m\"\n",
      "print\"(b)the rate at which radiant energy from the car interior is transmitted through the glass windshield is\",q_out,\"W/sq.m\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The energy transmitted from the sun through the glass is  50.6 W/sq.m\n",
        "(b)the rate at which radiant energy from the car interior is transmitted through the glass windshield is 0.0 W/sq.m\n"
       ]
      }
     ],
     "prompt_number": 32
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}