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diff --git a/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb b/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb new file mode 100644 index 0000000..3e3e005 --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_R_Joel/9-Heat_transfer.ipynb @@ -0,0 +1,399 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Heat transfer" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: interface_temperature.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.1');\n", +"\n", +"// aim : To determine \n", +"// the heat loss per hour through the wall and interface temperature\n", +"\n", +"// Given values\n", +"x1 = .25;// thickness of brick,[m]\n", +"x2 = .05;// thickness of concrete,[m]\n", +"t1 = 30;// brick face temperature,[C]\n", +"t3 = 5;// concrete face temperature,[C]\n", +"l = 10;// length of the wall, [m]\n", +"h = 5;// height of the wall, [m]\n", +"k1 = .69;// thermal conductivity of brick,[W/m/K]\n", +"k2 = .93;// thermal conductivity of concrete,[W/m/K]\n", +"\n", +"// solution\n", +"A = l*h;// area of heat transfer,[m^2]\n", +"Q_dot = A*(t1-t3)/(x1/k1+x2/k2);// heat transferred, [J/s]\n", +"\n", +"// so heat loss per hour is\n", +"Q = Q_dot*3600*10^-3;// [kJ]\n", +"mprintf('\n The heat lost per hour is = %f kJ\n',Q);\n", +"\n", +"// interface temperature calculation\n", +"// for the brick wall, Q_dot=k1*A*(t1-t2)/x1;\n", +"// hence\n", +"t2 = t1-Q_dot*x1/k1/A;// [C]\n", +"mprintf('\n The interface temperature is = %f C\n',t2);\n", +"\n", +"// End" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: thickness_of_lagging.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.2');\n", +"\n", +"// aim : To determine\n", +"// the minimum \n", +"// thickness of the lagging required\n", +"\n", +"// Given values\n", +"r1 = 75/2;// external radious of the pipe,[mm]\n", +"L = 80;// length of the pipe,[m]\n", +"m_dot = 1000;// flow of steam, [kg/h]\n", +"P = 2;// pressure, [MN/m^2]\n", +"x1 = .98;// inlet dryness fraction\n", +"x2 = .96;// outlet dryness fraction\n", +"k = .08;// thermal conductivity of of pipe, [W/m/K]\n", +"t2 = 27;// outside temperature,[C]\n", +"\n", +"// solution\n", +"// using steam table at 2 MN/m^2 the enthalpy of evaporation of steam is,\n", +"hfg = 1888.6;// [kJ/kg]\n", +"// so heat loss through the pipe is\n", +"Q_dot = m_dot*(x1-x2)*hfg/3600;// [kJ]\n", +"\n", +"// also from steam table saturation temperature of steam at 2 MN/m^2 is,\n", +"t1 = 212.4;// [C]\n", +"// and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1)\n", +"// hence\n", +"r2 = r1*exp(k*2*%pi*L*(t1-t2)*10^-3/Q_dot);// [mm]\n", +"\n", +"t = r2-r1;// thickness, [mm]\n", +"\n", +"mprintf('\n The minimum thickness of the lagging required is = %f mm\n',t);\n", +"\n", +"// End" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: heat_lost_and_interface_temperature.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.3');\n", +"\n", +"// aim : To determine the\n", +"// (a) heat loss per hour\n", +"// (b) interface temperature og lagging\n", +"\n", +"// Given values\n", +"r1 = 50; // radious of steam main,[mm]\n", +"r2 = 90;// radious with first lagging,[mm]\n", +"r3 = 115;// outside radious os steam main with lagging,[mm]\n", +"k1 = .07;// thermal conductivity of 1st lagging,[W/m/K]\n", +"k2 = .1;// thermal conductivity of 2nd lagging, [W/m/K]\n", +"P = 1.7;// steam pressure,[MN/m^2]\n", +"t_superheat = 30;// superheat of steam, [K]\n", +"t3 = 24;// outside temperature of the lagging,[C]\n", +"L = 20;// length of the steam main,[m]\n", +"\n", +"// solution\n", +"// (a)\n", +"// using steam table saturation temperature of steam at 1.7 MN/m^2 is\n", +"t_sat = 204.3;// [C]\n", +"// hence\n", +"t1 = t_sat+t_superheat;// temperature of steam,[C]\n", +"\n", +"Q_dot = 2*%pi*L*(t1-t3)/(log(r2/r1)/k1+log(r3/r2)/k2);// heat loss,[W]\n", +"// heat loss in hour is\n", +"Q = Q_dot*3600*10^-3;// [kJ]\n", +"\n", +"mprintf('\n (a) The heat lost per hour is = %f kJ\n',Q);\n", +"\n", +"// (b)\n", +"// using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) \n", +"t2 = t1-Q_dot*log(r2/r1)/(2*%pi*k1*L);// interface temperature of lagging,[C]\n", +"\n", +"mprintf('\n (b) The interface temperature of the lagging is = %f C\n',t2);\n", +"\n", +"// There is some calculation mistake in the book so answer is not matching\n", +"\n", +"// End" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: energy_emitted.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.4');\n", +"\n", +"// aim : To determine \n", +"// the energy emetted from the surface\n", +"\n", +"// Given values\n", +"h = 3;// height of surface, [m]\n", +"b = 4;// width of surface, [m]\n", +"epsilon_s = .9;// emissivity of the surface\n", +"T = 273+600;// surface temperature ,[K]\n", +"sigma = 5.67*10^-8;// [W/m^2/K^4]\n", +"\n", +"// solution\n", +"As = h*b;// area of the surface, [m^2]\n", +"\n", +"Q_dot = epsilon_s*sigma*As*T^4*10^-3;// energy emitted, [kW]\n", +"\n", +"mprintf('\n The energy emitted from the surface is = %f kW\n',Q_dot);\n", +"\n", +"// End" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: Rate_of_heat_transfer.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.5');\n", +"\n", +"// aim : To determine \n", +"// the rate of energy transfer between furnace and the sphere and its direction\n", +"\n", +"// Given values\n", +"l = 1.25;// internal side of cubical furnace, [m]\n", +"ti = 800+273;// internal surface temperature of the furnace,[K]\n", +"r = .2;// sphere radious, [m]\n", +"epsilon = .6;// emissivity of sphere\n", +"ts = 300+273;// surface temperature of sphere, [K]\n", +"sigma = 5.67*10^-8;// [W/m^2/K^4]\n", +"\n", +"// Solution\n", +"Af = 6*l^2;// internal surface area of furnace, [m^2]\n", +"As =4 *%pi*r^2;// surface area of sphere, [m^2]\n", +"\n", +"// considering internal furnace to be black\n", +"Qf = sigma*Af*ti^4*10^-3;// [kW]\n", +"\n", +"// radiation emitted by sphere is\n", +"Qs = epsilon*sigma*As*ts^4*10^-3; // [kW]\n", +"\n", +"// Hence transfer of energy is\n", +"Q = Qf-Qs;// [kW]\n", +"\n", +"mprintf('\n The transfer of energy will be from furnace to sphere and transfer rate is = %f kW\n',Q);\n", +"\n", +"// There is some calculation mistake in the book so answer is not matching\n", +"\n", +"// End\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6: overall_heat_transfer_coefficient_and_heat_lost.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.6');\n", +"\n", +"// aim : To determine\n", +"// the overall transfer coefficient and the heat loss per hour\n", +"\n", +"// Given values\n", +"x1 = 25*10^-3;// Thickness of insulating board, [m]\n", +"x2 = 75*10^-3;// Thickness of fibreglass, [m]\n", +"x3 = 110*10^-3;// Thickness of brickwork, [m]\n", +"k1 = .06;// Thermal conductivity of insulating board, [W/m K]\n", +"k2 = .04;// Thermal conductivity of fibreglass, [W/m K]\n", +"k3 = .6;// Thermal conductivity of brickwork, [W/m K]\n", +"Us1 = 2.5;// surface heat transfer coefficient of the inside wall,[W/m^2 K]\n", +"Us2 = 3.1;// surface heat transfer coefficient of the outside wall,[W/m^2 K]\n", +"ta1 = 27;// internal ambient temperature, [C]\n", +"ta2 = 10;// external ambient temperature, [C]\n", +"h = 6;// height of the wall, [m]\n", +"l = 10;// length of the wall, [m]\n", +"\n", +"// solution\n", +"U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);// overall heta transfer coefficient,[W/m^2 K]\n", +"\n", +"A = l*h;// area ,[m^2]\n", +"\n", +"Q_dot = U*A*(ta1-ta2);// heat loss [W]\n", +"\n", +"// so heat loss per hour is\n", +"Q = Q_dot*3600*10^-3;// [kJ]\n", +"mprintf('\n The overall heat transfer coefficient for the wall is = %f W/m^2 K\n',U);\n", +"mprintf('\n The heat loss per hour through the wall is = %f kJ\n',Q);\n", +"\n", +"// End" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7: heat_lost_and_surafce_temperature_of_lagging.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear;\n", +"clc;\n", +"disp('Example 9.7');\n", +"\n", +"// aim : To determine \n", +"// the heat loss per hour and the surface temperature of the lagging\n", +"\n", +"// Given values\n", +"r1 = 75*10^-3;// External radiou of the pipe, [m]\n", +"t_l1 = 40*10^-3;// Thickness of lagging1, [m]\n", +"t_l2 = t_l1;\n", +"k1 = .07;// thermal conductivity of lagging1, [W/m K]\n", +"k2 = .1;// thermal conductivity of lagging2, [W/m K]\n", +"Us = 7;// surface transfer coefficient for outer surface, [W/m^2 K]\n", +"L = 50;// length of the pipe, [m]\n", +"ta = 27;// ambient temperature, [C]\n", +"P = 3.6;// wet steam pressure, [MN/m^2]\n", +"\n", +"// solution\n", +"// from steam table saturation temperature of the steam at given pressure is,\n", +"t1 = 244.2;// [C]\n", +"r2 = r1+t_l1;// radious of pipe with lagging1,[m]\n", +"r3 = r2+t_l2;// radious of pipe with both the lagging, [m]\n", +"\n", +"R1 = log(r2/r1)/(2*%pi*L*k1);// resistance due to lagging1,[C/W]\n", +"R2 = log(r3/r2)/(2*%pi*L*k2);// resistance due to lagging2,[C/W]\n", +"R3 = 1/(Us*2*%pi*r3*L);// ambient resistance, [C/W]\n", +"\n", +"// hence overall resistance is,\n", +"Req = R1+R2+R3;// [C/W]\n", +"tdf = t1-ta;// temperature driving force, [C]\n", +"Q_dot = tdf/Req;// rate of heat loss, [W]\n", +"// so heat loss per hour is,\n", +"Q = Q_dot*3600*10^-3;// heat loss per hour, [kJ]\n", +"\n", +"// using eqn [3]\n", +"t3 = ta+Q_dot*R3;// surface temperature of the lagging, [C]\n", +"\n", +"mprintf('\n The heat loss per hour is = %f kJ\n',Q);\n", +"mprintf('\n The surface temperature of the lagging is = %f C\n',t3);\n", +"\n", +"// there is minor variation in the answer\n", +"\n", +"// End" + ] + } +], +"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 +} |