diff options
Diffstat (limited to 'Process_Heat_Transfer_by_D_Q_Kern/4-Radiation.ipynb')
-rw-r--r-- | Process_Heat_Transfer_by_D_Q_Kern/4-Radiation.ipynb | 155 |
1 files changed, 155 insertions, 0 deletions
diff --git a/Process_Heat_Transfer_by_D_Q_Kern/4-Radiation.ipynb b/Process_Heat_Transfer_by_D_Q_Kern/4-Radiation.ipynb new file mode 100644 index 0000000..8a19823 --- /dev/null +++ b/Process_Heat_Transfer_by_D_Q_Kern/4-Radiation.ipynb @@ -0,0 +1,155 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Radiation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1: Radiation_between_large_2_planes.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//page 75\n", +"printf('\t example 4.1 \n');\n", +"printf('\t approximate values are mentioned in the book \n');\n", +"T1=1000+460; // R\n", +"T2=800+460; // R\n", +"Q=((0.173)*((14.6)^4-(12.6)^4)); // using eq.4.24,Btu/(hr)*(ft^2)\n", +"printf('\t heat removed from colder wall per unit area is : %.0f Btu/(hr)*(ft^2) \n',Q);\n", +"// end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2: Radiation_between_Planes_with_Different_Emissivities.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"printf('\t example 4.2 \n');\n", +"printf('\t approximate values are mentioned in the book \n');\n", +"T1=1000+460; // R\n", +"T2=800+460; // R\n", +"e1=0.6; // emissivity of hotter wall\n", +"e2=0.8; // emissivity of colder wall\n", +"Q=(((0.173)/((1/0.6)+(1/0.8)-1))*((14.6)^4-(12.6)^4)); // using eq.4.26,heat loss per unit area,Btu/(hr)*(ft^2)\n", +"printf('\t heat removed from colder wall per unit area is : %.0f Btu/(hr)*(ft^2) \n',Q);\n", +"printf('\t For perfect black bodies the value was 3500 Btu/(hr)(ft^2) \n');\n", +"// end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3: Calculation_of_Radiation_from_a_Pipe.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//page 78\n", +"printf('\t example 4.3 \n');\n", +"printf('\t approximate values are mentioned in the book \n');\n", +"T1=125+460; // R\n", +"T2=70+460; // R\n", +"e=0.9; // emissivity,using table 4.1B\n", +"A=(%pi)*(3.375/12)*(1); // area,ft^2/lin ft\n", +"printf('\t area is : %.2f ft^2/lin ft \n',A);\n", +"Q=(0.9)*(0.88)*(0.173)*((T1/100)^4-(T2/100)^4); // heat loss using eq.4.32,Btu/(hr)*(lin ft)\n", +"printf('\t heat loss is : %.1f Btu/(hr)*(lin ft) \n',Q);\n", +"hr=(Q)/((A)*(T1-T2)); // fictitious film coefficient,using eq 4.33,Btu/(hr)(ft^2)(F)\n", +"printf('\t fictitious film coefficient is : %.2f Btu/(hr)(ft^2)(F) \n',hr);\n", +"//end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: Radiation_from_a_Pipe_to_a_Duct.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//page 82\n", +"printf('\t example 4.4 \n');\n", +"printf('\t approximate values are mentioned in the book \n');\n", +"T1=300+460; // R\n", +"T2=75+460; //R\n", +"A1=0.622; // area from table 11 in the appendix A,ft^2/lin ft\n", +"A2=4*(1*1); // surface area of duct,ft^2/lin ft\n", +"e1=0.79; // emissivity of oxidized steel from table 4.1\n", +"e2=0.276; // emissivity of oxidized zinc from table 4.1\n", +"printf('\t surface area of pipe is : %.3f ft^2/lin ft \n',A1);\n", +"printf('\t surface area of duct is : %.0f ft^2/lin ft \n',A2);\n", +"printf('\t The surface of the pipe is not negligible by comparison with that of the duct, and(f) of Table 4.2 applies most nearly \n');\n", +"Fa=1; // from table 4.2\n", +"Fe=((1)/((1/e1)+((A1/A2)*((1/e2)-1)))); // from table 4.2\n", +"printf('\t Fe is : %.2f \n',Fe);\n", +"Q=(0.173*10^-8)*(Fa)*(Fe)*(A1)*((T1)^4-(T2)^4); // heat loss due to radiation,Btu/(hr)*(lin ft)\n", +"printf('\t heat loss due to radiation is : %.0f Btu/(hr)*(lin ft) \n',Q);\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 +} |