{
"cells": [
{
"cell_type": "markdown",
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"collapsed": true
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"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 127Â°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 127Â°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 127Â°C or 400K and temprature(T2) of surface 2 is 327Â°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 127Â°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 127Â°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/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": [
"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=200Â°C or 473.15K and T1=300Â°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"
]
},
{
"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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