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diff --git a/A_Heat_Transfer_Text_Book/Chapter4.ipynb b/A_Heat_Transfer_Text_Book/Chapter4.ipynb new file mode 100755 index 00000000..6c6f72bd --- /dev/null +++ b/A_Heat_Transfer_Text_Book/Chapter4.ipynb @@ -0,0 +1,226 @@ +{
+ "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": {}
+ }
+ ]
+}
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