{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4: Analysis of heat conduction and some steady one-dimensional problems" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.8, Page number: 172" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variables\n", "d=0.02; #diameter of alluminium rod,m\n", "k=205; #thermal conductivity of rod,W/(m.K)\n", "l=0.08; #length of rod, m\n", "T1=423; #wall temperature, K\n", "T2=299; #air temperatutre, K\n", "h=120; #convective coefficient, W/(m**2*K)\n", "\n", "#Calculations\n", "mL=math.sqrt(h*(math.pow(l,2))/(k*d/4)); # formula for mL=((h*Perimeter*l**2)/(k*Area))**0.5\n", "Bi=h*l/k #Biot no.\n", "a1=(math.cosh(0)+(Bi/mL)*math.sinh(0))/(math.cosh(mL)+(Bi/mL)*math.sinh(mL));#formula for temperature difference T-Ttip\n", "Ttip1=T2+a1*(T1-T2); # exact tip temperature, C\n", "Tt1=Ttip1-273; #Exact tip temp., K\n", "a2=(math.cosh(0)+(Bi/mL)*math.sinh(0))/(math.cosh(mL));#dimensionless temp. at tip if heat transfer from the tip is not considered\n", "Ttip2=T2+a2*(T1-T2); #Approximate tip temp., K\n", "Tt2=Ttip2-273; #Approximate tip temp., C\n", "\n", "#Results\n", "print \"The exact tip temperature is :\",round(Tt1,3),\"C\\n\"\n", "print \"Approximate tip temperature is : \",round(Tt2,3),\" C\\n\"\n", "print \"Thus the insulated tip approximation is adequate for the computation in this case.\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The exact tip temperature is : 111.428 C\n", "\n", "Approximate tip temperature is : 114.659 C\n", "\n", "Thus the insulated tip approximation is adequate for the computation in this case.\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.9, Page number: 174" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variables\n", "T1=423; #wall temperature, K\n", "d=0.02; #diameter of alluminium rod,m\n", "k=205; #thermal conductivity of rod,W/(m.K)\n", "l=0.08; #length of rod, m\n", "T2=299; #air temperatutre, K \n", "h=120; #convective coefficient, W/(m**2*K)\n", "mL=0.8656;\n", "\n", "#Calculations\n", "mr=mL*(d/(2*l)); # by looking at graph of 1-Qact/Q(no temp.depression) vs. mr*math.tanh(mL), we can find out the value of Troot. 1-Qact./Q(no temp. depression) = 0.05 so heat flow is reduced by 5 percent\n", "Troot=T1-(T1-T2)*0.05; #Actual temperature of root, K (0.05 is from graph)\n", "Tr=Troot-273; #Actual temperature of root, \u00b0C\n", "\n", "#Results\n", "print \"Actual temperature of root is :\",Tr,\"C , the correction is modest in this \\n\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Actual temperature of root is : 143.8 C , the correction is modest in this \n", "\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.10, Page number: 178" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variables\n", "T1=308; #air temperature, K\n", "Q=0.1; # heat transferred,W\n", "k=16; #thermal conductivity of wires, W/(m*K)\n", "d=0.00062; #diameter of wire,m\n", "Heff=23; #convection coefficient, W/(m**2*K)\n", "A=1.33*math.pow(10,-4); #Aera of resistor surface, m^2 (from example 2.8)\n", "#the wires act actn as very long fins connected to ressistor hence math.tanh(mL)=1\n", "\n", "#Calculations\n", "R1=1/math.sqrt(k*Heff*math.pow(math.pi,2)*math.pow(d,3)/4); #Fin resistance, K/W\n", "Req=math.pow((1/R1+1/R1+7.17*A+13*A),-1); #the 2 thermal ressistances are in parallel to the thermal ressistance for natural...\n", "#convection and thermal radiation from the ressistor's surface found in previous eg.\n", "Tres=T1+Q*Req; #Resistor temperature, K\n", "Trs=Tres-273; #Resistor temperature, \u00b0C\n", "\n", "#Results \n", "print \"Resistor temperature is :\",round(Trs,2),\"C or about 10 C lower than before.\\n\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resistor temperature is : 62.68 C or about 10 C lower than before.\n", "\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.11, Page number: 181" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variables\n", "D1=0.03; # outer diameter, m\n", "T1=358; #hot water temperature, K\n", "t1=0.0008; #thickness of fins, m\n", "D2=0.08; # diameter of fins, m\n", "t2=0.02; # spacing between fins, m\n", "h1=20; # convection coefficient, W/(m**2*K)\n", "h2=15; #convection coefficient with fins, W/(m**2*K)\n", "To=295; #surrounding temperature, K\n", "\n", "#Calculations\n", "Q=math.pi*D1*h1*(T1-To); # if fins are not added.\n", "Q1=math.ceil(Q1); #heat loss without fins,W/m\n", "# we set wall temp.=water temp..since the wall is constantly heated by water, we should not have a root temp. depression problem after the fins are added.hence by looking at the graph, ml(l/Perimeter)**0.5=(h*(D2/2-D1/2)/(125*0.025*t1)) = 0.306, we obtain n(efficiency)=89 percent\n", "Qfin=math.ceil(Q*(t2-t1)/t2 + 0.89*(2*3.14*(math.pow(D2,2)/4-math.pow(D1,2)/4))*50*h2*(T1-To)) #Heat transferred with fins, K/W\n", "\n", "#Results\n", "print \"Heat trnsferred without fins is :\",Q1,\"W/m\\n\"\n", "print \"Heat transferred with fins is :\", round(Qfin,3),\"W/m or 4.02 times heat loss without fins.\\n\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat trnsferred without fins is : 199.0 W/m\n", "\n", "Heat transferred with fins is : 478.0 W/m or 4.02 times heat loss without fins.\n", "\n" ] } ], "prompt_number": 9 } ], "metadata": {} } ] }