diff options
Diffstat (limited to 'Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb')
| -rw-r--r-- | Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb | 543 |
1 files changed, 543 insertions, 0 deletions
diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb new file mode 100644 index 00000000..a46cced0 --- /dev/null +++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter11.ipynb @@ -0,0 +1,543 @@ +{ + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n" + ] + }, + { + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n", + "\n", + "\n", + "\n", + " \n" + ] + }, + { + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n" + ] + }, + { + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "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/Æ1)+(1/Æ2)-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": [ + " \n", + " \n", + " \n", + " \n", + "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/Æ1)+(1/Æ2)-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/Æ1)+(1/Æ2)-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\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "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": [ + " \n", + " \n", + " \n", + " \n", + "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\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + } + ], + "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 +} |