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"metadata": {},
"source": [
"# Chapter 3 Steady State Conduction Multiple Dimension"
]
},
{
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"metadata": {},
"source": [
"## Exa 3.1"
]
},
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"execution_count": 1,
"metadata": {
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"name": "stdout",
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"text": [
"Heat lost by the pipe is 859.9 W\n"
]
}
],
"source": [
"#Example Number 3.1\n",
"# Calculate the heat loss by the pipe\n",
"\n",
"# Variable declaration\n",
"\n",
"d = 0.15 \t\t\t# [m] diameter of pipe\n",
"r = d/2 \t\t\t# [m] radius of pipe\n",
"L = 4 \t\t\t\t# [m] length of pipe\n",
"Tp = 75\t\t\t\t# [degree celsius] pipe wall temperature\n",
"Tes = 5 \t\t\t# [degree celsius] earth surface temperature\n",
"k = 0.8\t\t\t\t# [W/m per deg C] thermal conductivity of earth \n",
"D = 0.20 \t\t\t# [m] depth of pipe inside earth\n",
"\n",
"\t# We may calculate the shape factor for this situation using equation given in \ttable 3-1 \n",
"\t\n",
"\t# since D<3*r\n",
"#Calculation\n",
"import math\n",
"S = (2*math.pi*L)/math.acosh(D/r) \t# [m] shape factor\n",
"\t# the heat flow is calculated from \n",
"q = k*S*(Tp-Tes) \t\t\t# [W]\n",
"\n",
"#Result\n",
"print\"Heat lost by the pipe is\",round(q,1),\"W\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 3.2"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat lost through the walls is: 8.592 kW\n"
]
}
],
"source": [
"#Example Number 3.2 \n",
"# Calculate heat loss through the walls\n",
"\n",
"# VARIABLE DECLARATION\n",
"\n",
"a = 0.5 \t # [m] length of side of cubical furnace\n",
"Ti = 500 \t # [degree celsius] inside furnace temperature\n",
"To = 50 \t # [degree celsius] outside temperature\n",
"k = 1.04 \t # [W/m per degree celsius] thermal conductivity of fireclay brick \n",
"t = 0.10 \t # [m] wall thickness\n",
"A = a*a \t # [square meter] area of one face \n",
"\t\t # we compute the total shape factor by adding the shape factors \t\t \t for the walls, edges and corners\n",
"\n",
"#Calculation\n",
"Sw = A/t\t # [m] shape factor for wall\n",
"Se = 0.54*a \t # [m] shape factor for edges\n",
"Sc = 0.15*t\t # [m] shape factor for corners\n",
"\n",
"\t\t # there are six wall sections, twelve edges and eight corners, so \t\t\tthe total shape factor S is\n",
"\n",
"S = 6*Sw+12*Se+8*Sc \t# [m]\n",
"\t\t \n",
"\t\t# the heat flow is calculated as \n",
"\n",
"q = k*S*(Ti-To) \t# [W]\n",
"\n",
"#Result\n",
"print\"Heat lost through the walls is:\",round(q/1000,3),\"kW\" \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 3.3"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat lost by disk is: 198.46 W\n"
]
}
],
"source": [
"#Example Number 3.3\n",
"# Calculate the heat loss by the disk\n",
"\n",
"# Variable declaration\n",
"\n",
"import math\n",
"d = 0.30 \t# [m] diameter of disk\n",
"r = d/2 \t# [m] radius of disk\n",
"Td = 95 \t# [degree celsius] disk temperature\n",
"Ts = 20 \t# [degree celsius] isothermal surface temperature\n",
"k = 2.1 \t# [W/m per degree celsius] thermal conductivity of medium \n",
"D = 1.0 \t# [m] depth of disk in a semi-infinite medium\n",
"\t# We have to calculate shape factor using relation given in table (3-1) \n",
"\t# We select the relation for the shape factor is for the case D/(2*r)>1\n",
"\n",
"#Calculation\n",
"S = (4*math.pi*r)/((math.pi/2)-math.atan(r/(2*D)))\t # [m] shape factor\n",
"\t# heat lost by the disk is \n",
"q = k*S*(Td-Ts) \t\t\t\t\t # [W]\n",
"\n",
"#Result\n",
"print\"Heat lost by disk is:\",round(q,2),\"W\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 3.4"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat transfer between the disks is: 308.4 W\n"
]
}
],
"source": [
"#Example Number 3.4 \n",
"#Calculate the heat transfer betwwen the disks\n",
"\n",
"# Variable declaration\n",
"\n",
"d = 0.50\t # [m] diameter of both disk\n",
"r = d/2 \t # [m] radius of disk\n",
"Td1 = 80\t # [degree celsius] first disk temperature\n",
"Td2 = 20 \t # [degree celsius] second disk temperature\n",
"k = 2.3 \t # [W/m per degree celsius] thermal conductivity of medium \n",
"D = 1.5\t\t # [m] seperation of disk in a infinite medium\n",
"\t# We have to calculate shape factor using relation given in table (3-1) \n",
"\t# We select the relation for the shape factor is for the case D>5*r\n",
"#Calculation\n",
"import math\n",
"\n",
"S = (4*math.pi*r)/((math.pi/2)-math.atan(r/D)) # [m] shape factor\n",
"q = k*S*(Td1-Td2) # [W]\n",
"\n",
"#Result\n",
"print\"Heat transfer between the disks is:\",round(q,1),\"W\" \n",
"\n"
]
}
],
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