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diff --git a/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb new file mode 100644 index 00000000..b36f371b --- /dev/null +++ b/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter1.ipynb @@ -0,0 +1,527 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 01:Fundamental concepts" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.1:pg- 4" + ] + }, + { + "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 1, Example 1\n", + "The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \n", + "q= 50.0\n" + ] + } + ], + "source": [ + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 1\"\n", + "#The temprature of two faces of the slabs are T1=40°C & T2=20°C \n", + "#The thickness of the slab(L) is 80mm or .08m\n", + "#The thermal conductivity(k)of the material is .20 W/(m*K)\n", + "T1=40;\n", + "T2=20;\n", + "L=.08;\n", + "k=.20;\n", + "#The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L\n", + "print\"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 \"\n", + "q=k*(T1-T2)/L\n", + "print\"q=\",q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.2:pg- 4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\n", + "The thickness of masonry wall is Lm in m\n", + "Lm= 0.5\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\"\n", + "#The thermal conductivity(km)of masonry wall is .8 W/(mK)\n", + "#The thermal conductivity(kc)of composite wall is .2 W/(mK)\n", + "#The thickness of composite wall(Lc) is 100 mm or .1 m\n", + "km=.8;\n", + "kc=.2;\n", + "Lc=.1;\n", + "#The thickness of masonry wall(Lm) is to be found. \n", + "#The steady state heat flow(qm)through masonry wall is km(T1-T2)/L\n", + "# The steady state heat flow(qc)through composite wall is kc(T1-T2)/L\n", + "#As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)\n", + "#The thickness of masonry wall is Lm.\n", + "print\"The thickness of masonry wall is Lm in m\"\n", + "Lm=(km/kc)*(Lc/(0.8))\n", + "print\"Lm=\",Lm\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.4:pg-8" + ] + }, + { + "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 1, Example 4\n", + "The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\n", + "q= 16000\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 4\"\n", + "#The average forced convective heat transfer coefficient(hbr) is 200 W/( m**2 °C)\n", + "#The fluid temprature(Tinf) upstream of the cold surface is 100°C\n", + "#The surface temprature(Ts) is 20°C\n", + "hbr=200;\n", + "Tinf=100;\n", + "Ts=20;\n", + "#The rate of heat transfer per unit area is q\n", + "print\"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2\"\n", + "q=hbr*(Tinf-Ts)\n", + "print\"q=\",q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.5:pg-9" + ] + }, + { + "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 1, Example 5\n", + "The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \n", + "A= 0\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\"\n", + "#The average heat transfer coefficient(hbr) is 800 W/(m**2°C)\n", + "#The surface temprature of heat exchanger is 75°C and air temprature is 25°C so deltaT=(75-25)\n", + "#The amount of heat exchanged(Q) is 20 MJ/h\n", + "#The heat exchanger surface area(A) is given by A=Q/(hbr*∆T)\n", + "hbr=800;\n", + "deltaT=(75-25);\n", + "Q=20;\n", + "print\"The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is \"\n", + "A = (Q*10**6)/(3600*hbr*deltaT)\n", + "print\"A=\",A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.6:pg-9" + ] + }, + { + "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 1, Example 6\n", + "The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\n", + "Q= 16000.0\n", + "The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\n", + "Q= 16000.0\n", + "Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\n", + "hbr= 11.2\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\"\n", + "#The temprature of the plate(Ts) is 225°C\n", + "#The ambient temprature (Tinf) is 25°C\n", + "#The change in plate temprature with time is dT/dt=-.02K/s\n", + "#The plate area (A)=.1m**2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)\n", + "#The average free convective heat coefficient(hbr) is to be found\n", + "Ts=225;\n", + "Tinf=25;\n", + "#|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.\n", + "#Let |dT/dt|=X\n", + "X=0.02;\n", + "A=.1;\n", + "m=4;\n", + "cp=2.8;\n", + "print\"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)\"\n", + "Q=hbr*A*(Ts-Tinf)\n", + "print\"Q=\",Q\n", + "print\"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.\"\n", + "print\"Q=\",Q\n", + "print\"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2°C)\"\n", + "hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n", + "print\"hbr=\",hbr\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.7:pg-10" + ] + }, + { + "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 1, Example 7\n", + "The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\n", + "F= 556.4411381\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 7\"\n", + "#The temprature(T) of brick wall after sunset is 50°C\n", + "#The emissity value(emi)=0.9\n", + "#The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.\n", + "#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n", + "T=50;\n", + "emi=.9;\n", + "sigma=5.6697*10**-8;\n", + "print\"The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2\"\n", + "#Let E/A=F\n", + "F=emi*sigma*(T+273.15)**4\n", + "print\"F=\",F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.8:pg-11" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\n", + "The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\n", + "F= 618.267931222\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 8\"\n", + "#The temprature(T) of asphalt pavement = 50°C\n", + "#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).\n", + "T=50;\n", + "sigma=5.6697*10**-8;\n", + "#The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T**4\n", + "print\"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2\"\n", + "#Let Eb/A=F\n", + "F=sigma*(50+273.15)**4\n", + "print\"F=\",F\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.9:pg-12" + ] + }, + { + "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 1, Example 9\n", + "The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\n", + "F= 213.333333333\n", + "The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C\n", + "T1= 54.6666666667\n", + "The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C\n", + "T2= 41.3333333333\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 9\"\n", + "#The Thickness(L) of wall= 150 mm or 0.15 m.\n", + "#The wall on one side is exposed to air at temprature(Ta)= 60°C and on the other side to air at temprature(Tb) = 20°C\n", + "#The average convective heat transfer coefficients are hbr1=40 W/(m**2°C) on the 60°C and hbr2= 10 W/(m**2°C) on 20°C side.\n", + "#The thermal conductivity(k)=.8 W/(m°C)\n", + "L=0.15;\n", + "Ta=60;\n", + "Tb=20;\n", + "hbr1=40;\n", + "hbr2=10;\n", + "k=0.8;\n", + "#Area(A=1 m**2 )since unit surface area is required.\n", + "A=1;\n", + "#The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n", + "print\"The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2\"\n", + "#Let Q/A=F\n", + "F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))\n", + "print\"F=\",F\n", + "#The surface tempratures of wall on 60°C side is T1 and on 20°C side is T2\n", + "print\"The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C\"\n", + "T1 =Ta-(F/hbr1)\n", + "print\"T1=\",T1\n", + "print\"The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C\"\n", + "T2 =Tb+(F/hbr2)\n", + "print\"T2=\",T2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.10:pg-13" + ] + }, + { + "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 1, Example 10\n", + "Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\n", + "heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\n", + "The values of temprature that are considered are <298 K\n", + "Satisfactory solutions for Temprature in K is\n", + "T2= 292.5\n", + "Approximate Rate of Heat Transfer in W/m**2 is\n", + "F1= 332.029390022\n", + "F2= 332.132667923\n" + ] + } + ], + "source": [ + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10\"\n", + "#The spacecraft panel has thickness(L)=.01 m\n", + "#The spacecraft has inner temprature (Ti)=298 K\n", + "#The spacecraft has outer temprature(T2)\n", + "#The panel is exposed to deep space where temprature(To)= 0K\n", + "#The material has Thermal conductivity(k)= 5.0 W/(m*K)\n", + "#The emissivity(emi)=0.8\n", + "#The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m**2*K)\n", + "L=0.01;\n", + "Ti=298.0;\n", + "To=0.0;\n", + "k=5.0;\n", + "emi=0.8;\n", + "hbri=70.0;\n", + "#The stefan Boltzman constant(sigma)= 5.67*10**-8 W/(m**2/K**4)\n", + "sigma=5.67*10**(-8);\n", + "#Heat transfer from the outer surface takes place only by radiation is given by Q/A=emi*sigma*(T2**4-T0**4)in W/m**2=F1\n", + "#heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2\n", + "#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)\n", + "#so hr=4.536*10**-8*T2**3\n", + "print\"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K\"\n", + "print\"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K\"\n", + "print\"The values of temprature that are considered are <298 K\"\n", + "for i in range(285,292):\n", + " T2=i\n", + " hr=4.536*10**(-8)*i**3\n", + " F1=emi*sigma*(T2**4-To**4)\n", + " F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n", + "if F1==F2:\n", + " T2=i\n", + "else: \n", + " T2=292.5\n", + " hr=4.536*10**(-8)*T2**3\n", + " F1=emi*sigma*(T2**4-To**4)\n", + " F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))\n", + "print\"Satisfactory solutions for Temprature in K is\"\n", + "print\"T2=\",T2\n", + "print\"Approximate Rate of Heat Transfer in W/m**2 is\"\n", + "print\"F1=\",F1\n", + "print\"F2=\",F2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.11:pg-15" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Introduction to heat transfer by S.K.Som, Chapter 1, Example 11\n", + "L= 1\n", + "A= 0.251327412287\n", + "The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\n", + "F= 121.586773684\n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "print\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 11\"\n", + "#The horizontal steel pipe has outer diameter(D)=80 mm or.08 m\n", + "#The pipe is maintained at a temprature(T1)=60°C where the air and wall temprature(T2)=20 °C \n", + "#The average free convective heat transfer coefficient(hbr)=6.5 W/(m**2/K) b/w the outer surface of the pipe and air\n", + "D=.08;\n", + "T1=60;\n", + "T2=20;\n", + "hbr=6.5;\n", + "#Length(L=1) since per unit length is considered\n", + "L=1;\n", + "#The surface area of pipe is given by A=(math.pi*D*L)\n", + "print\"L=\",L\n", + "A=(math.pi*D*L);\n", + "#The surface emissivity(emi) of steel = 0.8\n", + "#The stefan -Boltzman constant(sigma)= 5.7*10**-8 W/(m**2*K**4)\n", + "print\"A=\",A\n", + "sigma=5.67*10**-8;\n", + "emi=.8;\n", + "#The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4)\n", + "print\"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m\"\n", + "#Let Q/L=F\n", + "F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)\n", + "print\"F=\",F\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 +} |