{ "metadata": { "name": "", "signature": "sha256:a4eb855d2f94bc4d3b88cc418db7aa787fc638da4c248a8ac28f23e0760aff9c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 3 : OneDimensional Steady State Heat Conduction" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1 Page No : 45" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "l = 5;\t\t \t#Length of the wall in m\n", "h = 4;\t\t\t #Height of the wall in m\n", "L = 0.25;\t\t\t#Thickness of the wall in m\n", "T = [110,40];\t\t#Temperature on the inner and outer surface in degree C\n", "k = 0.7;\t\t\t#Thermal conductivity in W/m.K\n", "x = 0.20;\t\t\t#Distance from the inner wall in m\n", "\n", "# Calculations\n", "A = l*h;\t\t\t #Arear of the wall in m**2\n", "Q = (k*A*(T[0]-T[1]))/L;\t \t\t#Heat transfer rate in W\n", "T = (((T[1]-T[0])*x)/L)+T[0];\t\t\t#Temperature at interior point of the wall, 20 cm distant from the inner wall in degree C\n", "\n", "# Results\n", "print 'a)Heat transfer rate is %i W \\n \\\n", "b)Temperature at interior point of the wall, 20 cm distant from the inner wall is %i degree C'%(Q, T)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a)Heat transfer rate is 3920 W \n", " b)Temperature at interior point of the wall, 20 cm distant from the inner wall is 54 degree C\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2 Page No : 48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "Di = 0.05;\t\t \t#Inner diameter of hollow cylinder in m\n", "Do = 0.1;\t\t\t #Outer diameter of hollow cylinder in m\n", "T = [200,100];\t\t\t#Inner and outer surface temperature in degree C\n", "k = 70;\t\t\t #Thermal conductivity in W/m.K\n", "\n", "# Calculations\n", "ro = (Do/2);\t\t\t#Outer radius of hollow cylinder in m\n", "ri = (Di/2);\t\t\t#Inner radius of hollow cylinder in m\n", "Q = ((2*3.14*k*(T[0]-T[1]))/(math.log(ro/ri)));\t\t\t#Heat transfer rate in W\n", "r1 = (ro+ri)/2;\t\t\t#Radius at halfway between ro and ri in m\n", "T1 = T[0]-((T[0]-T[1])*(math.log(r1/ri)/(math.log(ro/ri))));\t\t\t#Temperature of the point halfway between the inner and outer surface in degree C\n", "\n", "# Results\n", "print 'Heat transfer rate is %3.1f W /m \\n \\\n", "Temperature of the point halfway between the inner and outer surface is %3.1f degree C'%(Q,T1)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat transfer rate is 63420.9 W /m \n", " Temperature of the point halfway between the inner and outer surface is 141.5 degree C\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3 Page No : 51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "Di = 0.1;\t\t\t#Inner diameter of hollow sphere in m\n", "Do = 0.3;\t\t\t#Outer diameter of hollow sphere in m\n", "k = 50. \t\t\t#Thermal conductivity in W/m.K\n", "T = [300,100];\t\t\t#Inner and outer surface temperature in degree C\n", "\n", "# Calculations\n", "ro = (Do/2);\t\t\t#Outer radius of hollow sphere in m\n", "ri = (Di/2);\t\t\t#Inner radius of hollow sphere in m\n", "Q = ((4*3.14*ro*ri*k*(T[0]-T[1]))/(ro-ri))/1000;\t\t\t#Heat transfer rate in W\n", "r = ri+(0.25*(ro-ri));\t\t\t#The value at one-fourth way of te inner and outer surfaces in m\n", "T = ((ro*(r-ri)*(T[1]-T[0]))/(r*(ro-ri)))+T[0];\t\t\t#Temperature at a point a quarter of the way between the inner and outer surfaces in degree C\n", "\n", "# Results\n", "print 'Heat flow rate through the sphere is %3.2f kW \\n \\\n", " Temperature at a point a quarter of the way between the inner and outer surfaces is %i degree C'%(Q,T)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat flow rate through the sphere is 9.42 kW \n", " Temperature at a point a quarter of the way between the inner and outer surfaces is 200 degree C\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4 Page No : 55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L = 0.4;\t\t\t#Thickness of the furnace in m\n", "T = [300,50];\t\t\t#Surface temperatures in degree C\n", "#k = 0.005T-5*10**-6T**2\n", "\n", "# Calculations\n", "q = ((1./L)*(((0.005/2)*(T[0]**2-T[1]**2))-((5*10**-6*(T[0]**3-T[1]**3))/3)));\t\t\t#Heat loss per square meter surface area in W/m**2\n", "\n", "# Results\n", "print 'Heat loss per square meter surface area is %3.0f W/m**2'%(q)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss per square meter surface area is 435 W/m**2\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.5 Page No : 55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L = 0.2;\t\t\t #Thickness of the wall in m\n", "T = [1000,200];\t\t\t#Surface temperatures in degree C\n", "ko = 0.813;\t\t\t #Value of thermal conductivity at T = 0 in W/m.K\n", "b = 0.0007158;\t\t\t#Temperature coefficient of thermal conductivity in 1./K\n", "\n", "# Calculations\n", "km = ko*(1+((b*(T[0]+T[1]))/2));\t\t\t#Constant thermal conductivity in W/m.K\n", "q = ((km*(T[0]-T[1]))/L);\t\t\t #Rate of heat flow in W/m**2\n", "\n", "# Results\n", "print 'Rate of heat flow is %3.0f W/m**2'%(q)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Rate of heat flow is 4649 W/m**2\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.6 Page No : 58" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# Variables\n", "r = [0.01,0.02];\t\t\t#Inner and outer radius of a copper cylinder in m\n", "T = [310,290];\t\t\t#Inner and Outer surface temperature in degree C\n", "ko = 371.9;\t\t\t#Value of thermal conductivity at T = 0 in W/m.K\n", "b = (9.25*10**-5);\t\t\t#Temperature coefficient of thermal conductivity in 1./K\n", "\n", "# Calculations\n", "Tm = ((T[0]-150)+(T[1]-150))/2;\t\t\t#Mean temperature in degree C\n", "km = ko*(1-(b*Tm));\t\t\t #Constant thermal conductivity in W/m.K\n", "q = ((2*3.14*km*(T[0]-T[1]))/math.log(r[1]/r[0]))/1000;\t\t\t#Heat loss per unit length in kW/m\n", "\n", "# Results\n", "print 'Heat loss per unit length is %3.2f kW/m'%(q)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss per unit length is 66.45 kW/m\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.8 Page No : 63" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L1 = 0.5;\t\t \t#Thickness of the wall in m\n", "k1 = 1.4;\t\t\t #Thermal conductivity in W/m.K\n", "k2 = 0.35;\t\t\t #Thermal conductivity of insulating material in W/m.K\n", "q = 1450.;\t\t\t #Heat loss per square metre in W\n", "T = [1200,15];\t\t\t#Inner and outer surface temperatures in degree C\n", "\n", "# Calculations\n", "L2 = (((T[0]-T[1])/q)-(L1/k1))*k2;\t\t\t#Thickness of the insulation required in m\n", "\n", "# Results\n", "print 'Thickness of the insulation required is %3.3f m'%(L2)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thickness of the insulation required is 0.161 m\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.9 Page No : 64" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L1 = 0.006; \t\t\t#Thickness of each glass sheet in m\n", "L2 = 0.002;\t \t\t#Thickness of air gap in m\n", "Tb = -20;\t\t \t#Temperature of the air inside the room in degree C\n", "Ta = 30;\t\t\t #Ambient temperature of air in degree C\n", "ha = 23.26;\t\t\t #Heat transfer coefficient between glass and air in W/m**2.K\n", "kglass = 0.75;\t\t\t#Thermal conductivity of glass in W/m.K\n", "kair = 0.02;\t\t\t#Thermal conductivity of air in W/m.K\n", "\n", "# Calculations\n", "q = ((Ta-Tb)/((1./ha)+(L1/kglass)+(L2/kair)+(L1/kglass)+(1./ha)));\t\t\t#Rate of heat leaking into the room per unit area of the door in W/m**2\n", "\n", "# Results\n", "print 'Rate of heat leaking into the room per unit area of the door is %3.1f W/m**2'%(q)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Rate of heat leaking into the room per unit area of the door is 247.5 W/m**2\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.10 Page No : 65" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "LA = 0.05;\t\t\t#Length of section A in m\n", "LB = 0.1;\t\t\t#Length of section A in m\n", "LC = 0.1;\t\t\t#Length of section A in m\n", "LD = 0.05;\t\t\t#Length of section A in m\n", "LE = 0.05;\t\t\t#Length of section A in m\n", "kA = 50;\t\t\t#Thermal conductivity of section A in W/m.K\n", "kB = 10;\t\t\t#Thermal conductivity of section B in W/m.K\n", "kC = 6.67;\t\t\t#Thermal conductivity of section C in W/m.K\n", "kD = 20;\t\t\t#Thermal conductivity of section D in W/m.K\n", "kE = 30;\t\t\t#Thermal conductivity of section E in W/m.K\n", "Aa = 1;\t\t\t#Area of section A in m**2\n", "Ab = 0.5;\t\t\t#Area of section B in m**2\n", "Ac = 0.5;\t\t\t#Area of section C in m**2\n", "Ad = 1;\t\t\t#Area of section D in m**2\n", "Ae = 1;\t\t\t#Area of section E in m**2\n", "T = [800,100];\t\t\t#Temperature at inlet and outlet temperatures in degree C\n", "\n", "# Calculations\n", "Ra = (LA/(kA*Aa)); \t\t\t#Thermal Resistance of section A in K/W\n", "Rb = (LB/(kB*Ab));\t \t\t#Thermal Resistance of section B in K/W\n", "Rc = (LC/(kC*Ac));\t\t \t#Thermal Resistance of section C in K/W\n", "Rd = (LD/(kD*Ad));\t\t\t #Thermal Resistance of section D in K/W\n", "Re = (LE/(kE*Ae));\t\t \t#Thermal Resistance of section E in K/W\n", "Rf = ((Rb*Rc)/(Rb+Rc));\t\t\t#Equivalent resistance of section B and section C in K/W\n", "R = Ra+Rf+Rd+Re;\t\t\t #Equivalent resistance of all sections in K/W\n", "Q = ((T[0]-T[1])/R)/1000;\t\t\t#Heat transfer through the composite wall in kW\n", "\n", "# Results\n", "print 'Heat transfer through the composite wall is %3.1f kW'%(Q)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat transfer through the composite wall is 40.8 kW\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.11 Page No : 66" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "T1 = 2000;\t\t\t#Temperature of hot gas in degree C\n", "Ta = 45;\t\t\t#Room air temperature in degree C\n", "Qr1 = 23.260;\t\t\t#Heat flow by radiation from gases to inside surface of the wall in kW/m**2\n", "h = 11.63;\t\t\t#Convective heat transfer coefficient in W/m**2.\n", "C = 58;\t\t\t#Thermal conductance of the wall in W/m**2.K\n", "Q = 9.3;\t\t\t#Heat flow by radiation from external surface to the surrounding in kW.m**2\n", "T2 = 1000;\t\t\t#Interior wall temperature in degree C\n", "\n", "# Calculations\n", "qr1 = Qr1;\t\t\t#Haet by radiation in kW/m**2\n", "qc1 = h*((T1-T2)/1000);\t\t\t#Heat by conduction in kW/m**2\n", "q = qc1+qr1;\t\t\t#Total heat entering the wall in kW/m**2\n", "R = (1./C);\t\t\t#Thermal resistance in m**2.K/W\n", "T3 = T2-(q*R*1000);\t\t\t#External wall temperature in degree C\n", "Ql = q-Q;\t\t\t#Heat loss due to convection kW/m**2\n", "h4 = (Ql*1000)/(T3-Ta);\t\t\t#Convective conductance in W/m**2.K\n", "\n", "# Results\n", "print 'The surface temperature is %i degree C \\n \\\n", "The convective conductance is %3.1f W/m**2.K'%(T3,h4)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The surface temperature is 398 degree C \n", " The convective conductance is 72.4 W/m**2.K\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.12 Page No : 67" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L1 = 0.125;\t\t\t#Thickness of fireclay layer in m\n", "L2 = 0.5;\t\t\t#Thickness of red brick layer in m\n", "T = [1100,50];\t\t\t#Temperatures at inside and outside the furnaces in degree C\n", "k1 = 0.533;\t\t\t#Thermal conductivity of fireclay in W/m.K\n", "k2 = 0.7;\t\t\t#Thermal conductivity of red brick in W/m.K\n", "\n", "# Calculations\n", "R1 = (L1/k1);\t\t\t#Resismath.tance of fireclay per unit area in K/W\n", "R2 = (L2/k2);\t\t\t#Resistance of red brick per unit area in K/W\n", "R = R1+R2;\t\t\t#Total resistance in K/W\n", "q = (T[0]-T[1])/R;\t\t\t#Heat transfer in W/m**2\n", "T2 = T[0]-(q*R1);\t\t\t#Temperature in degree C\n", "T3 = T[1]+(q*R2*0.5);\t\t\t#Temperature at the interface between the two layers in degree C\n", "km = 0.113+(0.00023*((T2+T3)/2));\t\t\t#Mean thermal conductivity in W/m.K\n", "x = ((T2-T3)/q)*km;\t\t\t#Thickness of diatomite in m\n", "\n", "# Results\n", "print 'Amount of heat loss is %3.1f W/m**2 \\n \\\n", "Thickness of diatomite is %3.4f m'%(q,x )\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Amount of heat loss is 1106.7 W/m**2 \n", " Thickness of diatomite is 0.0932 m\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.13 Page No : 70" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# Variables\n", "Di = 0.1;\t\t\t#I.D of the pipe in m\n", "L = 0.01;\t\t\t#Thickness of the wall in m\n", "L1 = 0.03;\t\t\t#Thickness of insulation in m\n", "Ta = 85;\t\t\t#Temperature of hot liquid in degree C\n", "Tb = 25;\t\t\t#Temperature of surroundings in degree C\n", "k1 = 58;\t\t\t#Thermal conductivity of steel in W/m.K\n", "k2 = 0.2;\t\t\t#Thermal conductivity of insulating material in W/m.K\n", "ha = 720;\t\t\t#Inside heat transfer coefficient in W/m**2.K\n", "hb = 9;\t\t\t #Outside heat transfer coefficient in W/m**2.K\n", "D2 = 0.12;\t\t\t#Inner diameter in m\n", "r3 = 0.09;\t\t\t#Radius in m\n", "\n", "# Calculations\n", "q = ((2*3.14*(Ta-Tb))/((1./(ha*(Di/2)))+(1./(hb*r3))+(math.log(D2/Di)/k1)+(math.log(r3/(D2/2))/k2)));\t\t\t#Heat loss fro an insulated pipe in W/m\n", "\n", "# Results\n", "print 'Heat loss from an insulated pipe is %3.2f W/m'%(q)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss from an insulated pipe is 114.43 W/m\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.14 Page No : 71" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "Di = 0.1;\t\t\t#I.D of the pipe in m\n", "Do = 0.11;\t\t\t#O.D of the pipe in m\n", "L = 0.005;\t\t\t#Thickness of the wall in m\n", "k1 = 50;\t\t\t#Thermal conductivity of steel pipe line in W/m.K\n", "k2 = 0.06;\t\t\t#Thermal conductivity of first insulating material in W/m.K\n", "k3 = 0.12;\t\t\t#Thermal conductivity of second insulating material in W/m.K\n", "T = [250,50];\t\t\t#Temperature at inside tube surface and outside surface of insulation in degree C\n", "r3 = 0.105;\t\t\t#Radius of r3 in m as shown in fig.3.14 on page no.71\n", "r4 = 0.155;\t\t\t#Radius of r4 in m as shown in fig.3.14 on page no.71\n", "\n", "# Calculations\n", "r1 = (Di/2);\t\t\t#Radius of the pipe in m\n", "r2 = (Do/2);\t\t\t#Radius of the pipe in m\n", "q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3)));\t\t\t#Loss of heat per metre length of pipe in W/m\n", "T3 = ((q*math.log(r4/r3))/(2*3.14*k3))+T[1];\t\t\t#Interface temperature in degree C\n", "\n", "# Results\n", "print 'Loss of heat per metre length of pipe is %3.1f W/m \\n \\\n", "Interface temperature is %3.1f degree C'%(q,T3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loss of heat per metre length of pipe is 89.6 W/m \n", " Interface temperature is 96.3 degree C\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.15 Page No : 72" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "Di = 0.1;\t\t\t#I.D of the pipe in m\n", "Do = 0.11;\t\t\t#O.D of the pipe in m\n", "L = 0.005;\t\t\t#Thickness of the wall in m\n", "k1 = 50;\t\t\t#Thermal conductivity of steel pipe line in W/m.K\n", "k3 = 0.06;\t\t\t#Thermal conductivity of first insulating material in W/m.K\n", "k2 = 0.12;\t\t\t#Thermal conductivity of second insulating material in W/m.K\n", "T = [250,50];\t\t\t#Temperature at inside tube surface and outside surface of insulation in degree C\n", "r3 = 0.105;\t\t\t#Radius of r3 in m as shown in fig.3.14 on page no.71\n", "r4 = 0.155;\t\t\t#Radius of r4 in m as shown in fig.3.14 on page no.71\n", "\n", "# Calculations\n", "r1 = (Di/2);\t\t\t#Radius of the pipe in m\n", "r2 = (Do/2);\t\t\t#Radius of the pipe in m\n", "q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3)));\t\t\t#Loss of heat per metre length of pipe in W/m\n", "\n", "# Results\n", "print 'Loss of heat per metre length of pipe is %3.2f W/m'%(q)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loss of heat per metre length of pipe is 105.71 W/m\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.16 Page No : 73" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "D1 = 0.1;\t\t\t#O.D of the pipe in m\n", "P = 1373;\t\t\t#Pressure of saturated steam in kPa\n", "D2 = 0.2;\t\t\t#Diameter of magnesia in m\n", "k1 = 0.07;\t\t\t#Thermal conductivity of magnesia in W/m.K\n", "k2 = 0.08;\t\t\t#Thermal conductivity of asbestos in W/m.K\n", "D3 = 0.25;\t\t\t#Diameter of asbestos in m\n", "T3 = 20;\t\t\t#Temerature under the canvas in degree C\n", "t = 12;\t\t\t #Time for condensation in hours\n", "l = 150;\t\t\t#Lemgth of pipe in m\n", "T1 = 194.14;\t\t\t#Saturation temperature of steam in degree C from Table A.6 (Appendix A) at 1373 kPa on page no. 643\n", "hfg = 1963.15;\t\t\t#Latent heat of steam in kJ/kg from Table A.6 (Appendix A) at 1373 kPa on page no. 643\n", "\n", "# Calculations\n", "r1 = (D1/2);\t\t\t#Radius of the pipe in m\n", "r2 = (D2/2);\t\t\t#Radius of magnesia in m\n", "r3 = (D3/2);\t\t\t#Radius of asbestos in m\n", "Q = (((2*3.14*l*(T1-T3))/((math.log(r2/r1)/k1)+(math.log(r3/r2)/k2)))*(3600./1000));\t\t\t#Heat transfer rate in kJ/h\n", "m = (Q/hfg);\t\t\t#Mass of steam condensed per hour\n", "m1 = (m*t);\t\t\t#Mass of steam condensed in 12 hours\n", "\n", "# Results\n", "print 'Mass of steam condensed in 12 hours is %3.2f kg'%(m1)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass of steam condensed in 12 hours is 284.43 kg\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.17 Page No : 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "D1 = 0.1;\t\t\t#I.D of the first pipe in m\n", "D2 = 0.3;\t\t\t#O.D of the first pipe in m\n", "k1 = 70;\t\t\t#Thermal conductivity of first material in W/m.K\n", "D3 = 0.4;\t\t\t#O.D of the second pipe in m\n", "k2 = 15;\t\t\t#Thermal conductivity of second material in W/m.K\n", "T = [300,30];\t\t\t#Inside and outside temperatures in degree C\n", "\n", "# Calculations\n", "r1 = (D1/2);\t\t\t#Inner Radius of first pipe in m\n", "r2 = (D2/2);\t\t\t#Outer Radius of first pipe in m\n", "r3 = (D3/2);\t\t\t#Radius of second pipe in m\n", "Q = ((4*3.14*(T[0]-T[1]))/(((r2-r1)/(k1*r1*r2))+((r3-r2)/(k2*r2*r3))))/1000;\t\t\t#Rate of heat flow through the sphere in kW\n", "\n", "# Results\n", "print 'Rate of heat flow through the sphere is %3.2f kW'%(Q)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Rate of heat flow through the sphere is 11.24 kW\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.18 Page No : 77" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "Di = 0.1;\t\t\t#I.D of a steam pipe in m\n", "Do = 0.25;\t\t\t#I.D of a steam pipe in m\n", "k = 1.;\t\t\t#Thermal conductivity of insulating material in W/m.K\n", "T = [200.,20];\t\t\t#Steam temperature and ambient temperatures in degree C\n", "h = 8.;\t\t\t#Convective heat transfer coefficient between the insulation surface and air in W/m**2.K\n", "\n", "# Calculations\n", "ri = (Di/2);\t\t\t#Inner Radius of steam pipe in m\n", "ro = (Do/2);\t\t\t#Outer Radius of steam pipe in m\n", "rc = (k/h)*100;\t\t\t#Critical radius of insulation in cm\n", "q = ((T[0]-T[1])/((math.log(ro/ri)/(2*3.14*k)+(1./(2*3.14*ro*h)))));\t\t\t#Heat loss per metre of pipe at critical radius in W/m\n", "Ro = (q/(2*3.14*ro*h))+T[1];\t\t\t#Outer surface temperature in degree C\n", "\n", "# Results\n", "print 'Heat loss per metre of pipe at critical radius is %i W/m \\n \\\n", "Outer surface temperature is %3.2f degree C'%(q,Ro)\n", "\n", "# Note : Answer in book is wrong. Please check manually." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss per metre of pipe at critical radius is 589 W/m \n", " Outer surface temperature is 113.93 degree C\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.19 Page No : 78" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "Di = 0.001;\t\t \t#Diameter of copper wire in m\n", "t = 0.001;\t\t \t#Thickness of insulation in m;\n", "To = 20;\t\t \t#Temperature of surrondings in degree C\n", "Ti = 80;\t\t \t#Maximum temperature of the plastic in degree C\n", "kcopper = 400;\t\t\t#Thermal conductivity of copper in W/m.K\n", "kplastic = 0.5;\t\t\t#Thermal conductivity of plastic in W/m.K\n", "h = 8;\t \t\t#Heat transfer coefficient in W/m**2.K\n", "p = (3*10**-8);\t\t\t#Specific electric resistance of copper in Ohm.m\n", "\n", "# Calculations\n", "r = (Di/2);\t\t\t#Radius of copper tube in m\n", "ro = (r+t);\t\t\t#Radius in m \n", "R = (p/(3.14*r*r*0.01));\t\t\t#Electrical resistance per meter length in ohm/m\n", "Rth = ((1./(2*3.14*ro*h))+(math.log(ro/r)/(2*3.14*kplastic)));\t\t\t#Thermal resistance of convection film insulation per metre length \n", "Q = ((Ti-To)/Rth);\t\t\t#Heat transfer in W\n", "I = math.sqrt(Q/R);\t\t\t#Maximum safe current limit in A\n", "rc = ((kplastic*100)/h);\t\t\t#Critical radius in cm\n", "\n", "# Results\n", "print 'The maximum safe current limit is %3.3f A \\n \\\n", "As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto %3.1f cm \\n \\\n", "considerably in increasing the radius of plastic covering'%(I,rc)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The maximum safe current limit is 1.074 A \n", " As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto 6.2 cm \n", " considerably in increasing the radius of plastic covering\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.20 Page No : 83" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "L = 0.1;\t\t\t#Thickness of the wall in m\n", "Q = (4*10**4);\t\t\t#Heat transfer rate in W/m**3\n", "h = 50;\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", "T = 20;\t\t\t#Ambient air temperature in degree C\n", "k = 15;\t\t\t#Thermal conductivity of the material in W/m.K\n", "\n", "# Calculations\n", "Tw = (T+((Q*L)/(2*h)));\t\t\t#Surface temperature in degree C\n", "Tmax = (Tw+((Q*L*L)/(8*k)));\t\t\t#Maximum temperature in the wall in degree C\n", "\n", "# Results\n", "print 'Surface temperature is %i degree C \\nMaximum temperature in the wall is %3.3f degree C'%(Tw,Tmax)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Surface temperature is 60 degree C \n", "Maximum temperature in the wall is 63.333 degree C\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.21 Page No : 85" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "Do = 0.006;\t\t\t#Outer diameter of hallow cylinder in m\n", "Di = 0.004;\t\t\t#Inner diameter of hallow cylinder in m\n", "I = 1000;\t\t\t#Current in A\n", "T = 30;\t\t\t#Temperature of water in degree C\n", "h = 35000;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "k = 18;\t\t\t#Thermal conductivity of the material in W/m.K\n", "R = 0.1;\t\t\t#Electrical reisitivity of the material in ohm.mm**2/m\n", "\n", "# Calculations\n", "ro = (Do/2);\t\t\t#Outer radius of hallow cylinder in m\n", "ri = (Di/2);\t\t\t#Inner radius of hallow cylinder in m\n", "V = ((3.14*(ro**2-ri**2)));\t\t\t#Vol. of wire in m**2\n", "Rth = (R/(3.14*(ro**2-ri**2)*10**6));\t\t\t#Resistivity in ohm/mm**2\n", "q = ((I*I*Rth)/V);\t\t\t#Heat transfer rate in W/m**3\n", "To = T+(((q*ri*ri)/(4*k))*((((2*k)/(h*ri))-1)*((ro/ri)**2-1)+(2*(ro/ri)**2*math.log(ro/ri))));\t\t\t#Temperature at the outer surface in degree C\n", "\n", "# Results\n", "print 'Temperature at the outer surface is %3.2f degree C'%(To)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature at the outer surface is 57.44 degree C\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.22 Page No : 88" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "D = 0.025;\t\t\t#Diameter of annealed copper wire in m\n", "I = 200;\t\t\t#Current in A\n", "R = (0.4*10**-4);\t#Resistance in ohm/cm\n", "T = [200,10];\t\t#Surface temperature and ambient temperature in degree C\n", "k = 160;\t\t\t#Thermal conductivity in W/m.K\n", "\n", "# Calculations\n", "r = (D/2);\t\t\t#Radius of annealed copper wire in m\n", "Q = (I*I*R*100);\t#Heat transfer rate in W/m\n", "V = (3.14*r*r);\t #Vol. of wire in m**2\n", "q = (Q/V);\t\t\t#Heat loss in conductor in W/m**2\n", "Tc = T[0]+((q*r*r)/(4*k));\t\t\t#Maximum temperature in the wire in degree C\n", "h = ((r*q)/(2*(T[0]-T[1])));\t\t#Heat transfer coefficient in W/m**2.K\n", "\n", "# Results\n", "print 'Maximum temperature in the wire is %3.2f degree C \\nHeat transfer coefficient is %3.2f W/m**2.K'%(Tc,h)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Maximum temperature in the wire is 200.08 degree C \n", "Heat transfer coefficient is 10.73 W/m**2.K\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.23 Page No : 89" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "p = 100.;\t\t\t#Resistivity of nichrome in \u00b5 ohm-cm \n", "Q = 10000.;\t\t\t#Heat input of a heater in W\n", "T = 1220.;\t\t\t#Surface temperature of nichrome in degree C\n", "Ta = 20.;\t\t\t#Temperature of surrounding air in degree C\n", "h = 1150.;\t\t\t#Outside surface coeffient in W/m**2.K\n", "k = 17.;\t\t\t#Thermal conductivity of nichrome in W/m.K\n", "L = 1.; \t\t\t#Length of heater in m\n", "\n", "# Calculations\n", "d = (Q/((T-Ta)*3.14*h))*1000;\t#Diameter of nichrome wire in mm\n", "A = (3.14*d*d)/4;\t\t\t #Area of the wire in m**2\n", "R = ((p*10**-8*L)/A);\t\t\t#Resistance of the wire in ohm\n", "I = math.sqrt(Q/R)/1000;\t\t#Rate of current flow in A\n", "\n", "# Results\n", "print 'Diameter of nichrome wire is %3.4f mm \\n \\\n", "Rate of current flow is %i A'%(d,I)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Diameter of nichrome wire is 2.3078 mm \n", " Rate of current flow is 204 A\n" ] } ], "prompt_number": 33 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.24 Page No : 93" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "Do = 0.025;\t\t\t#O.D of the rod in m\n", "k = 20;\t\t\t#Thermal conductivity in W/m.K\n", "Q = (2.5*10**6);\t\t\t#Rate of heat removal in W/m**2\n", "\n", "# Calculations\n", "ro = (Do/2);\t\t\t#Outer radius of the rod in m\n", "q = ((4*Q)/(ro));\t\t\t#Heat transfer rate in W/m**3\n", "T = ((-3*q*ro**2)/(16*k));\t\t\t#Temperature drop from the centre line to the surface in degree C\n", "\n", "# Results\n", "print 'Temperature drop from the centre line to the surface is %3.3f degree C'%(T)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature drop from the centre line to the surface is -1171.875 degree C\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.25 Page No : 95" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "Q = 300;\t\t\t#Heat produced by the oranges in W/m**2\n", "s = 0.08;\t\t\t#Size of the orange in m\n", "k = 0.15;\t\t\t#Thermal conductivity of the sphere in W/m.K\n", "\n", "# Calculations\n", "q = (3*Q)/(s/2);\t\t\t#Heat flux in W/m**2\n", "Tc = 10+((q*(s/2)**2)/(6*k));\t\t\t#Temperature at the centre of the sphere in degree C\n", "\n", "# Results\n", "print 'Temperature at the centre of the orange is %i degree C'%(Tc)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature at the centre of the orange is 50 degree C\n" ] } ], "prompt_number": 35 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.26 Page No : 102" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "To = 140;\t\t\t#Temperature at the junction in degree C\n", "Ti = 15;\t\t\t#Temperature of air in the room in degree C\n", "D = 0.003;\t\t\t#Diameter of the rod in m\n", "h = 300;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "k = 150;\t\t\t#Thermal conductivity in W/m.K\n", "\n", "# Calculations\n", "P = (3.14*D);\t\t\t#Perimeter of the rod in m\n", "A = (3.14*D**2)/4;\t\t\t#Area of the rod in m**2\n", "Q = math.sqrt(h*P*k*A)*(To-Ti);\t\t\t#Total heat dissipated by the rod in W\n", "\n", "# Results\n", "print 'Total heat dissipated by the rod is %3.3f W'%(Q)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total heat dissipated by the rod is 6.841 W\n" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.27 Page No : 103" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "D = 0.025\t\t\t#Diameter of the rod in m\n", "Ti = 22.;\t\t\t#Temperature of air in the room in degree C\n", "x = 0.1;\t\t\t#Dismtance between the points in m\n", "T = [110.,85.];\t\t#Temperature sat two points in degree C\n", "h = 28.4;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "\n", "# Calculations\n", "m = -math.log((T[1]-Ti)/(T[0]-Ti))/x;\t\t\t#Calculation of m for obtaining k\n", "P = (3.14*D);\t\t\t#Perimeter of the rod in m\n", "A = (3.14*D**2)/4;\t\t\t#Area of the rod in m**2\n", "k = ((h*P)/((m)**2*A));\t\t\t#Thermal conductivity of the rod material in W/m.K\n", "\n", "# Results\n", "print 'Thermal conductivity of the rod material is %3.1f W/m.K'%(k)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thermal conductivity of the rod material is 406.8 W/m.K\n" ] } ], "prompt_number": 38 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.28 Page No : 103" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "L = 0.06;\t\t\t#Length of the turbine blade in m\n", "A = (4.65*10**-4);\t#Cross sectional area in m**2\n", "P = 0.12;\t\t\t#Perimeter in m\n", "k = 23.3;\t\t\t#Thermal conductivity of stainless steel in W/m.K\n", "To = 500;\t\t\t#Temperature at the root in degree C\n", "Ti = 870;\t\t\t#Temperature of the hot gas in degree C\n", "h = 442;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "\n", "# Calculations\n", "m = math.sqrt((h*P)/(k*A));\t\t\t#Calculation of m for calculating heat transfer rate\n", "X = (To-Ti)/math.cosh(m*L);\t\t\t#X for calculating tempetarure distribution\n", "Q = math.sqrt(h*P*k*A)*(To-Ti)*math.tanh(m*L);\t\t\t#Heat transfer rate in W\n", "\n", "# Results\n", "print 'Temperature distribution is given by :\\n T-Ti = %i cosh[%3.2f%3.2f-x)] cosh[%3.2f%3.2f)] \\n \\\n", "Heat transfer rate is %3.1f W'%(To-Ti,m,L,m,L,Q)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature distribution is given by :\n", " T-Ti = -370 cosh[69.970.06-x)] cosh[69.970.06)] \n", " Heat transfer rate is -280.4 W\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.29 Page No : 104" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "W = 1;\t\t\t#Length of the cylinder in m\n", "D = 0.05;\t\t\t#Diameter of the cylinder in m\n", "Ta = 45;\t\t\t#Ambient temperature in degree C\n", "n = 10;\t\t\t#Number of fins\n", "k = 120;\t\t\t#Thermal conductivity of the fin material in W/m.K\n", "t = 0.00076;\t\t\t#Thickness of fin in m\n", "L = 0.0127;\t\t\t#Height of fin in m\n", "h = 17;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "Ts = 150;\t\t\t#Surface temperature of cylinder in m\n", "\n", "# Calculations\n", "P = (2*W);\t\t\t#Perimeter of cylinder in m\n", "A = (W*t);\t\t\t#Surface area of cyinder in m**2\n", "m = round(math.sqrt((h*P)/(k*A)),2);\t\t\t#Calculation of m for determining heat transfer rate\n", "Qfin = (math.sqrt(h*P*k*A)*(Ts-Ta)*((math.tanh(m*L)+(h/(m*k)))/(1+((h/(m*k))*math.tanh(m*L)))));\t\t\t#Heat transfer through the fin in kW\n", "Qb = h*((3.14*D)-(n*t))*W*(Ts-Ta);\t\t\t#Heat from unfinned (base) surface in W\n", "Q = ((Qfin*10)+Qb);\t\t\t#Total heat transfer in W\n", "Ti = ((Ts-Ta)/(math.cosh(m*L)+((h*math.sinh(m*L))/(m*k))));\t\t\t#Ti to calculate temperature at the end of the fin in degree C\n", "T = (Ti+Ta);\t\t\t#Temperature at the end of the fin in degree C\n", "\n", "# Results\n", "print 'Rate of heat transfer is %3.2f W \\nTemperature at the end of the fin is %3.2f degree C'%(Q,T )\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Rate of heat transfer is 723.99 W \n", "Temperature at the end of the fin is 146.74 degree C\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.31 Page No : 109" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "t = 0.025;\t\t\t#Thickness of fin in m\n", "L = 0.1;\t\t\t#Length of fin in m\n", "k = 17.7;\t\t\t#Thermal conductivity of the fin material in W/m.K\n", "p = 7850;\t\t\t#Density in kg/m**3\n", "Tw = 600;\t\t\t#Temperature of the wall in degree C\n", "Ta = 40;\t\t\t#Temperature of the air in degree C\n", "h = 20;\t\t\t #Heat transfer coefficient in W/m**2.K\n", "I0 = 2.1782;\t\t\t#Io value taken from table 3.2 on page no.108\n", "I1 = 1.48871;\t\t\t#I1 value taken from table 3.2 on page no. 108\n", "\n", "# Calculations\n", "B = math.sqrt((2*L*h)/(k*t));\t\t\t#Calculation of B for determining temperature distribution \n", "\n", "X = ((Tw-Ta)/2.1782);\t\t\t#Calculation of X for determining temperature distribution \n", "Y = (2*B);\t\t\t#Calculation of Y for determining temperature distribution \n", "Q = (math.sqrt(2*h*k*t)*(Tw-Ta)*((1.48871))/(2.1782));\n", "m = ((p*t*L)/2);\t\t\t#Mass of the fin per meter of width in kg/m\n", "q = (Q/m);\t\t\t#Rate of heat flow per unit mass in W/kg\n", "\n", "# Results\n", "print 'Temperature distribution is T = %i+%3.1f(%3.4f\u221ax) \\n \\\n", "Rate of heat flow per unit \\\n", " mass of the fin is %3.2f W/kg'%(Ta,X,Y,q)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature distribution is T = 40+257.1(6.0132\u221ax) \n", " Rate of heat flow per unit mass of the fin is 164.10 W/kg\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.32 Page No : 116" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "t = 0.002;\t\t\t#Thickness of fin in m\n", "L = 0.015;\t\t\t#Length of fin in m\n", "k1 = 210.;\t\t\t#Thermal conductivity of aluminium in W/m.K\n", "h1 = 285.;\t\t\t#Heat transfer coefficient of aluminium in W/m**2.K\n", "k2 = 40.;\t\t\t#Thermal conductivity of steel in W/m.K\n", "h2 = 510.;\t\t\t#Heat transfer coefficient of steel in W/m**2.K\n", "\n", "# Calculations\n", "Lc = (L+(t/2));\t\t\t#Corrected length of fin in m\n", "mLc1 = Lc*math.sqrt((2*h1)/(k1*t));\t\t\t#Calculation of mLc for efficiency\n", "n1 = math.tanh(mLc1)/mLc1;\t\t\t#Efficiency of fin when aluminium is used\n", "mLc2 = Lc*math.sqrt((2*h2)/(k2*t));\t\t\t#Calculation of mLc for efficiency\n", "n2 = math.tanh(mLc2)/mLc2;\t\t\t#Efficiency of fin when steel is used\n", "\n", "\n", "# Results\n", "print 'Efficiency of fin when aluminium is used is %3.4f \\n \\\n", "Efficiency of fin when steel is used is %3.3f'%(n1,n2)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Efficiency of fin when aluminium is used is 0.8983 \n", " Efficiency of fin when steel is used is 0.524\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.33 Page No : 117" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "k = 200;\t\t\t#Thermal conductivity of aluminium in W/m.K\n", "t = 0.001;\t\t\t#Thickness of fin in m\n", "L = 0.015;\t\t\t#Width of fin in m\n", "D = 0.025;\t\t\t#Diameter of the tube in m\n", "Tb = 170;\t\t\t#Fin base temperature in degree C\n", "Ta = 25;\t\t\t#Ambient fluid temperature in degree C\n", "h = 130;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "\n", "# Calculations\n", "Lc = (L+(t/2));\t\t\t#Corrected length of fin in m\n", "r1 = (D/2);\t\t\t#Radius of tube in m\n", "r2c = (r1+Lc);\t\t\t#Corrected radius in m\n", "Am = t*(r2c-r1);\t\t\t#Corrected area in m**2\n", "x = Lc**(3/2)*math.sqrt(h/(k*Am));\t\t\t#x for calculating efficiency\n", "n = 0.82;\t\t\t#From fig. 3.18 on page no. 112 efficiency is 0.82\n", "qmax = (2*3.14*(r2c**2-r1**2)*h*(Tb-Ta));\t\t\t#Maximum heat transfer in W\n", "qactual = (n*qmax);\t\t\t#Actual heat transfer in W\n", "\n", "# Results\n", "print 'Heat loss per fin is %3.2f W'%(qactual)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss per fin is 60.94 W\n" ] } ], "prompt_number": 51 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.34 Page No : 117" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "k = 16;\t\t\t#Thermal conductivity of fin in W/m.K\n", "L = 0.1;\t\t\t#Length of fin in m\n", "D = 0.01;\t\t\t#Diameter of fin in m\n", "h = 5000;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "\n", "# Calculations\n", "P = (3.14*D);\t\t\t#Perimeter of fin in m\n", "A = (3.14*D**2)/4;\t\t\t#Area of fin in m**2\n", "m = math.sqrt((h*P)/(k*A));\t\t\t#Calculation of m for determining heat transfer rate\n", "n = math.tanh(m*L)/math.sqrt((h*A)/(k*P));\t\t\t#Calculation of n for checking whether installation of fin is desirable or not\n", "x = (n-1)*100;\t\t\t#Conversion into percentage\n", "\n", "# Results\n", "print 'This large fin only produces an increase of %i percent in heat dissipation, \\\n", " so naturally this configuration is undesirable'%(x)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "This large fin only produces an increase of 13 percent in heat dissipation, so naturally this configuration is undesirable\n" ] } ], "prompt_number": 52 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.35 Page No : 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "k = 55.8;\t\t\t#Thermal conductivity of steel in W/m.K\n", "t = 0.0015;\t\t\t#Thickness of steel tube in m\n", "L = 0.12;\t\t\t#Length of steel tube in m\n", "h = 23.3;\t\t\t#Heat transfer coefficient in W/m**2.K\n", "Tl = 84;\t\t\t#Temperature recorded by the thermometer in degree C\n", "Tb = 40;\t\t\t#Temperature at the base of the well in degree C\n", "\n", "# Calculations\n", "m = math.sqrt(h/(k*t));\t\t\t#Calculation of m for determining the temperature distribution\n", "x = 1./math.cosh(m*L);\t\t\t#Calculation of x for determining the temperature distribution\n", "Ti = ((Tl-(x*Tb))/(1-x));\t\t\t#Temperature distribution in degree C\n", "T = (Ti-Tl);\t\t\t#Measurement error in degree C\n", "\n", "# Results\n", "print 'Measurement error is %3.0f degree C'%(T)\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Measurement error is 16 degree C\n" ] } ], "prompt_number": 16 } ], "metadata": {} } ] }