diff options
Diffstat (limited to 'Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb')
| -rw-r--r-- | Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb | 456 |
1 files changed, 456 insertions, 0 deletions
diff --git a/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb new file mode 100644 index 00000000..339b7e2c --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb @@ -0,0 +1,456 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 - Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 256" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.1\n", + " The heat lost per hour is (kJ) = 10815.0\n", + " The interface temperature is (C) = 8.2\n" + ] + } + ], + "source": [ + "#pg 256\n", + "print('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", + "print ' The heat lost per hour is (kJ) = ',round(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", + "print ' The interface temperature is (C) = ',round(t2,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 258" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.2\n", + " The minimum thickness of the lagging required is (mm) = 38.8\n" + ] + } + ], + "source": [ + "#pg 258\n", + "print('Example 9.2');\n", + "\n", + "# aim : To determine\n", + "# the minimum \n", + "# thickness of the lagging required\n", + "import math\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*math.exp(k*2*math.pi*L*(t1-t2)*10**-3/Q_dot);# [mm]\n", + "\n", + "t = r2-r1;# thickness, [mm]\n", + "#results\n", + "print ' The minimum thickness of the lagging required is (mm) = ',round(t,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 260" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.3\n", + " (a) The heat lost per hour is (kJ) = 8770.0\n", + " (b) The interface temperature of the lagging is (C) = 71.5\n", + "There is some rounding off error in the book, so answer is not matching\n" + ] + } + ], + "source": [ + "#pg 260\n", + "print('Example 9.3');\n", + "\n", + "# aim : To determine the\n", + "# (a) heat loss per hour\n", + "# (b) interface temperature og lagging\n", + "import math\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*math.pi*L*(t1-t3)/(math.log(r2/r1)/k1+math.log(r3/r2)/k2);# heat loss,[W]\n", + "# heat loss in hour is\n", + "Q = Q_dot*3600*10**-3;# [kJ]\n", + "\n", + "print ' (a) The heat lost per hour is (kJ) = ',round(Q)\n", + "\n", + "# (b)\n", + "# using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) \n", + "t2 = t1-Q_dot*math.log(r2/r1)/(2*math.pi*k1*L);# interface temperature of lagging,[C]\n", + "\n", + "print ' (b) The interface temperature of the lagging is (C) = ',round(t2,1)\n", + "\n", + "print 'There is some rounding off error in the book, so answer is not matching'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 265" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.4\n", + " The energy emitted from the surface is (kW) = 355.7\n" + ] + } + ], + "source": [ + "#pg 265\n", + "print('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", + "#results\n", + "print ' The energy emitted from the surface is (kW) = ',round(Q_dot,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 265" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.5\n", + " The transfer of energy will be from furnace to sphere and transfer rate is (kW) = 703.0\n", + " There is some calculation mistake in the book, so answer is not matching\n" + ] + } + ], + "source": [ + "#pg 265\n", + "print('Example 9.5');\n", + "\n", + "# aim : To determine \n", + "# the rate of energy transfer between furnace and the sphere and its direction\n", + "import math\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 *math.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", + "#results\n", + "print ' The transfer of energy will be from furnace to sphere and transfer rate is (kW) = ',round(Q)\n", + "print' There is some calculation mistake in the book, so answer is not matching'\n", + "\n", + "# End\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 271" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.6\n", + " The overall heat transfer coefficient for the wall is (W/m**2 K) = 0.313\n", + " The heat loss per hour through the wall is (kJ) = 1148.0\n" + ] + } + ], + "source": [ + "#pg 271\n", + "print('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", + "#results\n", + "print ' The overall heat transfer coefficient for the wall is (W/m**2 K) = ',round(U,3)\n", + "print ' The heat loss per hour through the wall is (kJ) = ',round(Q)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 272" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.7\n", + " The heat loss per hour is (kJ) = 24533.0\n", + " The surface temperature of the lagging is (C) = 46.99\n", + "there is minor variation in the answer due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#pg 272\n", + "print('Example 9.7');\n", + "\n", + "# aim : To determine \n", + "# the heat loss per hour and the surface temperature of the lagging\n", + "import math\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 = math.log(r2/r1)/(2*math.pi*L*k1);# resistance due to lagging1,[C/W]\n", + "R2 = math.log(r3/r2)/(2*math.pi*L*k2);# resistance due to lagging2,[C/W]\n", + "R3 = 1/(Us*2*math.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", + "#results\n", + "print ' The heat loss per hour is (kJ) = ',round(Q,0)\n", + "print ' The surface temperature of the lagging is (C) = ',round(t3,2)\n", + "\n", + "print 'there is minor variation in the answer due to rounding off error in textbook'\n", + "\n", + "# End\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |