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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 3 Steady State Conduction Multiple Dimension"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exa 3.1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "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"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.6"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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