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diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb new file mode 100644 index 00000000..977f0e4b --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb @@ -0,0 +1,308 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 Condensation and Boiling Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of reynolds no. is 49.7 so the laminar assumption was correct\n", + "The heat transfer is 2368.0 w\n", + "Total mass flow condensate is 3.78 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.1\n", + "# condensation on vertical plate\n", + "\n", + "#Variable declaration\n", + "\n", + "\t# we have to check the reynolds no. to that film is laminar or turbulent\n", + "Tf = (100+98)/2 \t\t\t# [degree celsius]\n", + "Tw = 98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t\t# [m/s**(2)]\n", + "L=0.3 \t\t\t\t\t# [m]\n", + "\t# RHOf(RHOf-RHOv)~RHOf**(2)\n", + "\t# let us assume laminar film condensate \n", + "Tsat=100 \t\t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t\t# [J/kg]\n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.943*((RHOf**(2)*g*Hfg*Kf**(3)/(L*MUf*(Tg-Tw)))**(0.25)) \t# [W/sq m deg celsius]\n", + "h=hbar \t\t\t\t\t# [W/square meter degree celsius]\n", + "\t# checking reynolds no. with equation(9-17)\n", + "Ref=4*h*L*(Tsat-Tw)/(Hfg*MUf) \n", + "\n", + "print \"Value of reynolds no. is\",round(Ref,1),\"so the laminar assumption was correct\" \n", + "\t# the heat transfer is now calculated from \n", + "A=0.3*0.3 \t\t\t\t# [square meter]\n", + "q=hbar*A*(Tsat-Tw) \t\t\t# [W]\n", + "mdot=q/Hfg \t\t\t\t# [kg/h]\n", + "\n", + "#Result\n", + "\n", + "print \"The heat transfer is\",round(q),\"w\" \n", + "mdot=mdot*3600 \t\t\t\t# [kg/h]\n", + "print \"Total mass flow condensate is\",round(mdot,2),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total surface area is 3.99 square meter/m\n", + "Heat transfer is 100.1 kW/m\n", + "Total mass flow of condensate is 159.8 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.2\n", + "# condensation on tube tank\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# the condensate properties are obtained from previous example\n", + "\t# replacing L by n*d\n", + "Tw=98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t# [m/s^(2)]\n", + "Tsat=100 \t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t# [J/kg]\n", + "d=0.0127 \t\t\t# [m]\n", + "n=10 \n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.725*((RHOf**(2)*g*Hfg*Kf**(3)/(n*d*MUf*(Tg-Tw)))**(0.25)) # [W/sq m deg C]\n", + "\t# total surface area is \n", + "n=100 \n", + "Al=n*22*d/7 \t\t\t# [square meter]\n", + "print \"Total surface area is\",round(Al,2),\"square meter/m\" \n", + "\t# so the heat transfer is \n", + "Ql=hbar*Al*(Tg-Tw)\t\t # [W]\n", + "\n", + "#Result \n", + "\n", + "print \"Heat transfer is\",round(Ql/1000,2),\"kW/m\" \n", + "\t# total mass flow of condensate is then \n", + "mdotl=Ql/Hfg \t\t\t# [kg/h]\n", + "mdotl=mdotl*3600\t\t# [kg/h]\n", + "print \"Total mass flow of condensate is\",round(mdotl,1),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer in a 1.0 m length of tube is 2810.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 9.4\n", + "# Flow boiling\n", + "# Variable declaration\n", + "\n", + "p =0.5066\t\t\t# [MPa] pressure of water \n", + "d = 0.0254 \t\t\t# [m] diameter of tube \n", + "Tw = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# for calculation we use equation (9-45), noting that \n", + "dT = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# the heat transfer coefficient is calculated as \n", + "\n", + "import math\n", + "h = 2.54*Tw**(3)*math.exp(p/1.551) \t# [W/square meter degree celsius]\n", + "\n", + "\t\t# the surface area for a 1-m length of tube is \n", + "L = 1 \t\t\t\t# [m]\n", + "import math\n", + "A = math.pi*d*L \t\t# [square meter]\n", + "\n", + "\t\t# so the heat transfer is \n", + "q = h*A*dT \t\t\t# [W/m]\n", + "print \"Heat transfer in a 1.0 m length of tube is\",round(q),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flux obtained is 22.8 kW/square meter\n", + "If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is 23.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 9.5\n", + "# water boiling in a pan \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 0.101\t\t\t\t# [MPa] pressure of water \n", + "dT_x = 8 \t\t\t\t# [degree celsius]\n", + "p1 = 0.17 \t\t\t\t# [MPa] given operating pressure\n", + "\t# we will use the simplified relation of table 9-13(page no.-506) for the \t\testimates.we do not know the value of q_by_A and so must choose one of the two \trelation for a horizontal surface from the table\n", + "\t# we anticipate nucleate boiling, so choose\n", + "h = 5.56*dT_x**(3) \t\t\t# [W/square meter degree celsius]\n", + "\t\t\t\t\t# and the heat flux is \n", + "q_by_A = h*dT_x \t\t\t# [W/square meter]\n", + "\t# for operation as a pressure cooker we obtain the value of h from \t\tequation(9-44)\n", + "\n", + "#Calculation\n", + "\n", + "hp = h*(p1/p)**(0.4) \t\t\t# [W/square meter degree celsius]\n", + "\t# the corresponding heat flux is \n", + "q_by_A1 = hp*dT_x \t\t\t# [W/square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat flux obtained is\",round(q_by_A/1000,1),\"kW/square meter\" \n", + "per_inc = 100*(q_by_A1-q_by_A)/q_by_A \n", + "\n", + "print \"If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is\",round(per_inc),\"percent\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thus the heat transfers 14.0 times the heat of a pure copper rod with a substantial temperature gradient\n" + ] + } + ], + "source": [ + "#Example Number 9.6\n", + "# heat-flux comparisons \n", + "\n", + "# Variable declaration\n", + "\n", + "Tw = 200.0 \t\t\t\t# [degree celsius] water temperature \n", + "L = 0.08 \t\t\t\t# [m] length of solid copper bar\n", + "dT = 100.0 \t\t\t\t# [degree C] temp differential in copper bar\n", + "\t#using the data of table 9-4(page no.-508)\n", + "\t# the heat flux per unit area is expressed as q_by_A = -k*del_T/dx\n", + "\t# from table A-2(page no.-) the thermal conductivity of copper is \n", + "k = 374.0 \t\t\t\t# [W/m degree celsius]\n", + "\n", + "#Calculation\n", + "\n", + "q_by_A = -k*(-dT)/L \t\t\t# [W/square meter]\n", + "\n", + "\t# from table 9-4(page no.-508) the typical axial heat flux for a water heat \t\tflux for a water heat pipe is \n", + "q_by_A_axial = 0.67 \t\t\t# [kW/csquare meter]\n", + "q_by_A = q_by_A/(1000*10**(4.0)) \t# [kW/csquare meter]\n", + "time=q_by_A_axial/q_by_A\n", + "\n", + "#Result\n", + "\n", + "print \"Thus the heat transfers \",round(time),\"times the heat of a pure copper rod with a substantial temperature gradient\" " + ] + } + ], + "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 +} |