{
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"source": [
"# Chapter 2:CONDUCTION"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example2.1 pg:13"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
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{
"name": "stdout",
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"text": [
" heat is Btu/hr 69120.0\n",
"\t approximate values are mentioned in the book \n",
"\n"
]
}
],
"source": [
"\n",
"#given\n",
"Tavg=900; # average temperature of the wall,F\n",
"k=0.15; # Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)\n",
"T1=1500; # hot side temperature,F\n",
"T2=300; # cold side temperature,F\n",
"A=192; # surface area,ft^2\n",
"L=0.5; # thickness,ft\n",
"#solution\n",
"Q=(k)*(A)*(T1-T2)/L; # formula for heat,Btu/hr\n",
"print \" heat is Btu/hr \",Q\n",
"print\"\\t approximate values are mentioned in the book \\n\"\n",
"#end\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example2.2 pg:14"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
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"text": [
"\t resistance offered by firebrick : (hr)*(F)/Btu 0.97\n",
"\t resistance offered by insulating brick : (hr)*(F)/Btu 2.2\n",
"\t resistance offered by buildingbrick : (hr)*(F)/Btu 1.25\n",
"\t total resistance offered by three walls : (hr)*(F)/Btu 4.42\n",
"\t heat loss/ft^2 : Btu/hr 331.0\n",
"\t delta is : F 322.0\n",
"\t temperature at interface of firebrick and insulating brick F 1278.0\n",
"\t deltb is : F 729.0\n",
"\t temperature at interface of insulating brick and building brick F 549.0\n",
"\t approximate values are mentioned in the book \n",
"\n"
]
}
],
"source": [
"\n",
"#given\n",
"La=0.66; # Thickness of firebrick wall,ft\n",
"Lb=0.33; # Thickness of insulating brick wall,ft\n",
"Lc=0.5; # Thickness of building brick wall,ft\n",
"Ka=0.68; # themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)\n",
"Kb=0.15; # themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)\n",
"Kc=0.40; # themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)\n",
"A=1.; # surface area,ft^2\n",
"Ta=1600.; # temperature of inner wall,F\n",
"Tb=125.; # temperature of outer wall.F\n",
"#solution\n",
"Ra=La/(Ka)*(A); # formula for resistance,(hr)*(F)/Btu\n",
"print\"\\t resistance offered by firebrick : (hr)*(F)/Btu \",round(Ra,2)\n",
"Rb=Lb/(Kb)*(A); # formula for resistance,(hr)*(F)/Btu\n",
"print\"\\t resistance offered by insulating brick : (hr)*(F)/Btu \",round(Rb,2)\n",
"Rc=Lc/(Kc)*(A); # formula for resistance,(hr)*(F)/Btu\n",
"print\"\\t resistance offered by buildingbrick : (hr)*(F)/Btu \",round(Rc,2)\n",
"R=Ra+Rb+Rc; # total resistance offered by three walls,(hr)*(F)/Btu\n",
"print\"\\t total resistance offered by three walls : (hr)*(F)/Btu \",round(R,2)\n",
"Q=(1600-125)/4.45; # using formula for heat loss/ft^2,Btu/hr\n",
"print\"\\t heat loss/ft^2 : Btu/hr \",round(Q,0)\n",
"# T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F\n",
"delta=(Q)*(Ra); # formula for temperature difference,F\n",
"print\"\\t delta is : F \",round(delta,0)\n",
"T1=Ta-((Q)*(Ra)); # temperature at interface of firebrick and insulating brick,F\n",
"print\"\\t temperature at interface of firebrick and insulating brick F \",round(T1,0)\n",
"deltb=Q*(Rb);\n",
"print\"\\t deltb is : F \",round(deltb,0)\n",
"T2=T1-((Q)*(Rb)); #temperature at interface of insulating brick and building brick,F\n",
"print\"\\t temperature at interface of insulating brick and building brick F \",round(T2,0)\n",
"print\"\\t approximate values are mentioned in the book \\n\"\n",
"#end\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example2.3 pg:15"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\t approximate values are mentioned in the book \n",
"\n",
"\t resistance offered by air film (hr)(F)/Btu 0.79\n",
"\t total resistance (hr)(F)/Btu 5.24\n",
"\t heat loss Btu/hr 282.0\n"
]
}
],
"source": [
"\n",
"print\"\\t approximate values are mentioned in the book \\n\"\n",
"#given\n",
"Lair=0.25/12; # thickness of air film,ft\n",
"Kair=0.0265; # thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)\n",
"A=1; # surface area,ft^2\n",
"#solution\n",
"Rair=Lair/(Kair*(A)); # resistance offered by air film, (hr)(F)/Btu\n",
"print\"\\t resistance offered by air film (hr)(F)/Btu \",round(Rair,2)\n",
"R=4.45; # resistance from previous example 2.2,(hr)(F)/Btu\n",
"Rt=(R)+Rair; # total resistance,(hr)(F)/Btu\n",
"print\"\\t total resistance (hr)(F)/Btu \",round(Rt,2)\n",
"Ta=1600; # temperature of inner wall,F\n",
"Tb=125; # temperature of outer wall,F\n",
"Q=(1600-125)/Rt; # heat loss, Btu/hr\n",
"print\"\\t heat loss Btu/hr \",round(Q,0)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example2.4 pg 16"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\t heat flow is : Btu/(hr)*(ft) 543.0\n",
"\t approximate values are mentioned in the book \n",
"\n"
]
}
],
"source": [
"\n",
"#given\n",
"k=0.63; # thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)\n",
"Do=6. # in\n",
"Di=5. # in\n",
"Ti=200.;# inner side temperature,F\n",
"To=175.; # outer side temperature,F\n",
"#solution\n",
"import math\n",
"from math import log\n",
"q=(2*(3.14)*(k)*(Ti-To))/(log (Do/Di)); # formula for heat flow,Btu/(hr)*(ft)\n",
"print\"\\t heat flow is : Btu/(hr)*(ft) \",round(q,0)\n",
"print\"\\t approximate values are mentioned in the book \\n\"\n",
"# caculation mistake in book\n",
"# end\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"##Example2.5 pg 19"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\t approximate values are mentioned in the book \n",
"\n",
"\t heat loss for linear foot is : Btu/(hr)*(lin ft) 104.4\n",
"\t Check between ts and t1, since delt/R = deltc/Rc \n",
"\t t1 is : F 122.300238658\n",
"\t heat loss for linear foot is : Btu/(hr)*(lin ft) 102.9\n",
"\t Check between ts and t1, since delt/R = deltc/Rc \n",
"\t t1 is : F 125.4\n"
]
}
],
"source": [
"\n",
"print\"\\t approximate values are mentioned in the book \\n\"\n",
"#given\n",
"t1=150; # assume temperature of outer surface of rockwool,F\n",
"ta=70; # temperature of surrounding air,F\n",
"ha=2.23; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
"#solution\n",
"import math\n",
"from math import log\n",
"q=(3.14)*(300-70)/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.23)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake\n",
"print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft) \",round(q,1)\n",
"print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
"t1=300-(((104.8)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
"print\"\\t t1 is : F \",t1\n",
"t1=125; # assume temperature of outer surface of rockwool,F\n",
"ha=2.10; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
"q=((3.14)*(300-70))/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.10)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft)\n",
"print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft)\",round(q,1)\n",
"print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
"t1=300-(((103)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
"print\"\\t t1 is : F \",round(t1,1)\n",
"# end \n"
]
}
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