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+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 1: Introduction"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.1: conduction_through_copper_plate.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.1\n\n\n');\n",
+"// conduction through copper plate\n",
+"// illustration1.1\n",
+"// solution\n",
+"\n",
+"k = 370; // [W/m] at 250 degree celsius\n",
+"dt = 100-400;//[degree celsius] temperature difference\n",
+"dx = 3*10^(-2);//[m] thickness of plate\n",
+"//calculating heat transfer per unit area from fourier's law\n",
+"q = -k*dt/dx;//[MW/square meter]\n",
+"printf('rate of heat transfer per unit area is %f MW/square meter',q/1000000);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.2: convection_calculation.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.2\n\n\n');\n",
+"// convection calculation\n",
+"// illustration1.2\n",
+"// solution\n",
+"\n",
+"Twall = 250;//[degree celsius] wall temperature\n",
+"Tair = 20;//[degree celsius] air temperature\n",
+"h = 25;//[W/square meter] heat transfer coefficient\n",
+"l = 75*10^(-2);//[m] length of plate\n",
+"b = 50*10^(-2);//[m] width of plate\n",
+"area = l*b;//[square meter] area of plate\n",
+"dt = 250-20;//[degree celsius]\n",
+"// from newton's law of cooling\n",
+"q = h*area*dt;// [W]\n",
+"printf('rate of heat transfer is %f kW',q/1000);\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3: multimode_heat_transfer.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.3\n\n\n');\n",
+"// multimode heat transfer\n",
+"// illustration1.3\n",
+"// solution\n",
+"\n",
+"Qconv = 2156;// [W] from previous problem\n",
+"Qrad = 300;// [W] given\n",
+"dx = 0.02;// [m] plate thicknesss\n",
+"l = 0.75;// [m] length of plate \n",
+"w = 0.5;// [m] width of plate\n",
+"k = 43;//[W/m] from table 1.1\n",
+"area = l*w;//[square meter] area of plate\n",
+"Qcond = Qconv+Qrad;// [W]\n",
+"dt = Qcond*dx/(k*area);// [degree celsius] temperature difference\n",
+"Ti = 250+dt;// inside temperature\n",
+"printf('the inside plate temperature is therefore %f degree celsius',Ti);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.4: heat_source_and_convection.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.4\n\n\n');\n",
+"// heat source and convection\n",
+"// illustration1.4\n",
+"// solution\n",
+"\n",
+"d = 1*10^(-3);//[m] diameter of wire\n",
+"l = 10*10^(-2);//[m] length of wire\n",
+"Sarea = 22*d*l/7;//[square meter] surface area of wire\n",
+"h = 5000;//[W/square meter] heat transfer coefficient\n",
+"Twall = 114;// [degree celsius]\n",
+"Twater = 100;// [degree celsius]\n",
+"//total convection loss is given by equation(1-8)\n",
+"Q = h*Sarea*(Twall-Twater);// [W]\n",
+"printf('heat transfer is therefore %f W',Q);\n",
+"printf(' this is equal to the electric power which must be applied');"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5: radiation_heat_transfer.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.5\n\n\n');\n",
+"// radiation heat transfer\n",
+"// illustration1.5\n",
+"// solution\n",
+"\n",
+"sigma = 5.699*10^(-8);//[W/square meter*k^(4)] universal constant\n",
+"T1 = 273.15+800;// [k] first plate temperature\n",
+"T2 = 273.15+300;// [k] second plate temperature\n",
+"//equation(1-10) may be employed for this problem\n",
+"Q = sigma*((T1^(4))-(T2^(4)));// [W/square meter]\n",
+"printf('heat transfer per unit area is %f kW/square meter',Q/1000);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.6: total_heat_loss_by_convection_and_radiation.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear;\n",
+"clc;\n",
+"printf('\t\t\tExample Number 1.6\n\n\n');\n",
+"// total heat loss by convection and radiation\n",
+"// illustration1.6\n",
+"// solution\n",
+"\n",
+"d = 0.05;//[m] diameter of pipe\n",
+"Twall = 50;//[degree celsius] \n",
+"Tair = 20;//[degree celsius]\n",
+"emi = 0.8;//emissivity\n",
+"h = 6.5;//[W/square meter] heat transfer coefficient for free convection\n",
+"Q1 = h*22*d*(Twall-Tair)/7;//[W/m] convection loss per unit length\n",
+"sigma = 5.669*10^(-8);// [W/square meter*k^(4)] universal constant\n",
+"T1 = 273.15+Twall;// [k]\n",
+"T2 = 273.15+Tair;// [k]\n",
+"Q2 = emi*22*d*sigma*((T1^(4))-(T2^(4)))/7;// [W/m] heat loss due to radiation per unit length\n",
+"Qtotal = Q1+Q2;// [W/m] total heat loss per unit length\n",
+"printf('total heat loss is therefore %f W/m',Qtotal);"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}