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{
"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": {}
}
]
}
|
