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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 14: Radiation Heat Transfer"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Wavelength at which the maximum monochromatic emissive power (m) = 7.44e-06\n",
+ " \n",
+ " Coffecient of performnance (W/m^3) = 1.14e+08\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 1,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "#If a blackbody is maintained at 116C, determine (a) the wavelength at which \n",
+ "#the maximum monochromatic emissive power occurs and (b) the maximum \n",
+ "#monochromatic emissive power\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "T= 116. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#C\n",
+ "C1= 3.74*math.pow(10,-16)\n",
+ "C2= 1.44*math.pow(10,-2)\n",
+ "#CALCULATIONS\n",
+ "WLmax= (2893*math.pow(10,-6))/(T+273) \t\t\t\t\t\t\t#Maximum Wavelength \n",
+ "Wb= (C1*math.pow((WLmax),(-5)))/(math.exp(C2/2893*1000000.)-1)\t#Coffecient of performnance\n",
+ "#RESULTS\n",
+ "print '%s %.2e' % ('Wavelength at which the maximum monochromatic emissive power (m) = ',WLmax)\n",
+ "print '%s %.2e' % (' \\n Coffecient of performnance (W/m^3) = ',Wb)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Emissive power for the blackbody (W/m^2) = 1305.19\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 2,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "#Determine the total emissive power for the black body of solved problem 2\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "T= 389 \t\t\t\t\t\t#K\n",
+ "s= 5.7*math.pow(10,-8) \t\t#K^4\n",
+ "#CALCULATIONS\n",
+ "Wb= s*T*T*T*T \t\t\t\t#Emissive power for the blackbody\n",
+ "#RESULTS\n",
+ "print '%s %.2f' % ('Emissive power for the blackbody (W/m^2) = ',Wb)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Average absorptivity of the body at 100 F = 0.51\n",
+ " \n",
+ " Average absorptivity of the body at 2000 F= 0.84\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 3,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "#A gray body at 100F receives radiant energy from a wall at 2000 F at a rate\n",
+ "#of 3.2x 10^4. Simultaneously, the body emits energy at the rate of 140. What\n",
+ "#is the average absorptivity of the body at (a) 100 F and (b) 2000 F\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "T= 100 \t\t\t\t\t\t\t\t#F\n",
+ "T1= 2000 \t\t\t\t\t\t\t#F\n",
+ "W= 3.2*10000. \t\t\t\t\t\t#Btu/hr ft^2\n",
+ "W1= 140. \t\t\t\t\t\t\t#Btu/hr ft^2\n",
+ "s= 0.17*math.pow(10,-8) \t\t\t#Btu/hr ft^2 R^4\n",
+ "#CALCULATIONS\n",
+ "alpha= W/(s*math.pow((T1+460),4)) \t#Average absorptivity at 100\n",
+ "b= W1/(s*math.pow((T+460),4)) \t\t#Average absorptivity at 2000\n",
+ "#RESULTS\n",
+ "print '%s %.2f' % ('Average absorptivity of the body at 100 F = ',alpha)\n",
+ "print '%s %.2f' % (' \\n Average absorptivity of the body at 2000 F= ',b)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Heat loss from the conduit by radiation (Btu/hr per ft) = 1401.66\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 4,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "#A red brick conduit 10 in square has a surface temperature of 300F and is \n",
+ "#mounted inside a large earthen chamber whose walls are at 50 F. estimate\n",
+ "#the heat loss from the conduit by radiation\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "T= 300. \t\t\t\t\t\t\t\t\t\t\t\t#F\n",
+ "T1= 50. \t\t\t\t\t\t\t\t\t\t\t\t#F\n",
+ "s= 0.17*math.pow(10,-8) \t\t\t\t\t\t\t\t#Btu/hr ft^2 R^4\n",
+ "e1= 0.93\n",
+ "A= 10. \t\t\t\t\t\t\t\t\t\t\t\t\t#in\n",
+ "F= 1.\n",
+ "#CALCULATIONS\n",
+ "A1= 10*(40./(12.*10.)) \t\t\t\t\t\t\t\t\t#Area\n",
+ "q= A1*F*e1*s*(math.pow((T+460),4)-math.pow((T1+460),4)) #heat loss\n",
+ "#RESULTS\n",
+ "print '%s %.2f' % ('Heat loss from the conduit by radiation (Btu/hr per ft) = ',q)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Radiation heat transfer coefficient (Btu/hr ft^2 R) = 1.68\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 5,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "#Estimate the radiation heat transfer coefficient for solved problem 5\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "T= 300. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n",
+ "T1= 50. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n",
+ "s= 0.17*math.pow(10,-8) \t\t\t\t\t\t\t\t\t\t#Btu/hr ft^2 R^4\n",
+ "e1= 0.93\n",
+ "F= 1.\n",
+ "#CALCULATIONS\n",
+ "hr= F*e1*s*(math.pow((T+460),4)-math.pow((T1+460),4))/(T-T1)\t#Radiation heat transfer coefficient \n",
+ "#RESULTS\n",
+ "print '%s %.2f' % ('Radiation heat transfer coefficient (Btu/hr ft^2 R) = ',hr)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 14.7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Heat transfer coefficient for nucleate boiling (W/m^2 C) = 6.21e+08\n",
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "''"
+ ]
+ },
+ "execution_count": 8,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "P= 1. #atm\n",
+ "T= 11. #C\n",
+ "Csf= 0.006\n",
+ "Tsat = 170.03\n",
+ "r= 1./3.\n",
+ "s= 1.\n",
+ "dt = Tsat - T\n",
+ "cl= 4.218 #J/gm K\n",
+ "hfg= 2257 #J/gm\n",
+ "Pr= 1.75\n",
+ "ul= 283.1/1000. #gm/m s\n",
+ "s= 57.78/1000. #N/m\n",
+ "pl= 958*1000. #gm/m^3\n",
+ "pv= 598. #gm/m^3\n",
+ "gc= 1000. #gm m/N s^2\n",
+ "g= 9.8 #m/s^2\n",
+ "#CALCULATIONS\n",
+ "p= pl-pv\n",
+ "q= ((math.pow(((cl*dt)/(hfg*Csf*math.pow(Pr,s))),(1/r)))*(ul*hfg))/math.pow(gc/(g*p),(1./2.))\n",
+ "h= q/T\n",
+ "#RESULTS\n",
+ "print '%s %.2e' % ('Heat transfer coefficient for nucleate boiling (W/m^2 C) = ',h)\n",
+ "raw_input('press enter key to exit')"
+ ]
+ }
+ ],
+ "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": "2.7.6"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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