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diff --git a/sample_notebooks/UmangAgarwal/UmangAgarwal_version_backup/Sample.ipynb b/sample_notebooks/UmangAgarwal/UmangAgarwal_version_backup/Sample.ipynb new file mode 100755 index 00000000..1eb49726 --- /dev/null +++ b/sample_notebooks/UmangAgarwal/UmangAgarwal_version_backup/Sample.ipynb @@ -0,0 +1,163 @@ +{ + "cells": [ + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "Sample Notebook - Heat and Mass Transfer by R.K. Rajput : Chapter 1 - Basic Concepts\n", + "author: Umang Agarwal\n", + "\n", + "\n", + "# Example 1.1 Page 16-17\n", + "\n", + "L=.045; \t\t \t\t\t#[m] - Thickness of conducting wall\n", + "delT = 350 - 50; \t\t #[C] - Temperature Difference across the Wall\n", + "k=370; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Material\n", + "#calculations\n", + "#Using Fourier's Law eq 1.1\n", + "q = k*delT/(L*10**6); \t\t\t#[MW/m^2] - Heat Flux\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer per unit area =\",q,\" W\");\n", + "#END\n", + "\n", + "# Example 1.2 Page 17\n", + "\n", + "L = .15; \t\t \t\t\t#[m] - Thickness of conducting wall\n", + "delT = 150 - 45; \t\t #[C] - Temperature Difference across the Wall\n", + "A = 4.5; #[m^2] - Wall Area\n", + "k=9.35; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Material\n", + "#calculations\n", + "#Using Fourier's Law eq 1.1\n", + "Q = k*A*delT/L; \t\t\t#[W] - Heat Transfer\n", + "#Temperature gradient using Fourier's Law\n", + "TG = - Q/(k*A); #[C/m] - Temperature Gradient\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer per unit area =\",Q,\" W\");\n", + "print '%s %.2f %s' %(\"\\n \\n The Temperature Gradient in the flow direction =\",TG,\" C/m\");\n", + "#END\n", + "\n", + "# Example 1.3 Page 17-18\n", + "\n", + "x = .0825; \t\t \t\t\t#[m] - Thickness of side wall of the conducting oven\n", + "delT = 175 - 75; \t\t #[C] - Temperature Difference across the Wall\n", + "k=0.044; \t\t\t\t\t#[W/m.C] - Thermal Conductivity of Wall Insulation\n", + "Q = 40.5; #[W] - Energy dissipitated by the electric coil withn the oven \n", + "#calculations\n", + "#Using Fourier's Law eq 1.1\n", + "A = (Q*x)/(k*delT); \t\t#[m^2] - Area of wall\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Area of the wall =\",A,\" m^2\");\n", + "#END\n", + "\n", + "# Example 1.4 Page 18-19\n", + "\n", + "delT = 300-20; \t\t #[C] - Temperature Difference across the Wall\n", + "h = 20; \t\t\t\t\t#[W/m^2.C] - Convective Heat Transfer Coefficient\n", + "A = 1*1.5; #[m^2] - Wall Area\n", + "#calculations\n", + "#Using Newton's Law of cooling eq 1.6\n", + "Q = h*A*delT; \t\t\t#[W] - Heat Transfer\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n", + "#END\n", + "\n", + "# Example 1.5 Page 19\n", + "\n", + "L=.15; \t\t \t\t\t#[m] - Length of conducting wire\n", + "d = 0.0015; #[m] - Diameter of conducting wire\n", + "A = 22*d*L/7; #[m^2] - Surface Area exposed to Convection\n", + "delT = 120 - 100; \t\t #[C] - Temperature Difference across the Wire\n", + "h = 4500; \t\t\t\t\t#[W/m^2.C] - Convective Heat Transfer Coefficient\n", + "print 'Electric Power to be supplied = Convective Heat loss';\n", + "#calculations\n", + "#Using Newton's Law of cooling eq 1.6\n", + "Q = h*A*delT; \t\t\t#[W] - Heat Transfer\n", + "Q = round(Q,1);\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n", + "#END\n", + "\n", + "# Example 1.6 Page 20-21\n", + "\n", + "T1 = 300 + 273; \t\t #[K] - Temperature of 1st surface\n", + "T2 = 40 + 273; #[K] - Temperature of 2nd surface\n", + "A = 1.5; #[m^2] - Surface Area\n", + "F = 0.52; \t\t\t\t #[dimensionless] - The value of Factor due geometric location and emissivity\n", + "sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant\n", + "#calculations\n", + "#Using Stephen-Boltzmann Law eq 1.9\n", + "Q = F*sigma*A*(T1**4 - T2**4) \t #[W] - Heat Transfer\n", + "#Equivalent Thermal Resistance using eq 1.10\n", + "Rth = (T1-T2)/Q; #[C/W] - Equivalent Thermal Resistance\n", + "#Equivalent convectoin coefficient using h*A*(T1-T2) = Q\n", + "h = Q/(A*(T1-T2)); #[W/(m^2*C)] - Equivalent Convection Coefficient\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Q,\" W\");\n", + "print '%s %.2f %s' %(\"\\n The equivalent thermal resistance =\",Rth,\" C/W\");\n", + "print '%s %.2f %s' %(\"\\n The equivalent convection coefficient =\",h,\" W/(m^2 * C)\");\n", + "#END\n", + "\n", + "# Example 1.7 Page 21-22\n", + "\n", + "L = 0.025; #[m] - Thickness of plate\n", + "A = 0.6*0.9; #[m^2] - Area of plate \n", + "Ts = 310; \t\t #[C] - Surface Temperature of plate\n", + "Tf = 15; #[C] - Temperature of fluid(air)\n", + "h = 22; \t\t\t\t\t #[W/m^2.C] - Convective Heat Transfer Coefficient\n", + "Qr = 250; \t\t\t\t #[W] - Heat lost from the plate due to radiation\n", + "k = 45; \t\t\t\t\t #[W/m.C] - Thermal Conductivity of Plate\n", + "#calculations\n", + "# In this problem, heat conducted by the plate is removed by a combination of convection and radiation\n", + "# Heat conducted through the plate = Convection Heat losses + Radiation Losses\n", + "# If Ti is the internal plate temperature, then heat conducted = k*A*(Ts-Ti)/L\n", + "Qc = h*A*(Ts-Tf); #[W] - Convection Heat Loss\n", + "Ti = Ts + L*(Qc + Qr)/(A*k); \t #[C] - Inside plate Temperature\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n Rate of Heat Transfer =\",Ti,\" C\");\n", + "#END\n", + "\n", + "# Example 1.8 Page 22\n", + "\n", + "Ts = 250; \t\t #[C] - Surface Temperature\n", + "Tsurr = 110; #[C] - Temperature of surroundings\n", + "h = 75; \t\t\t\t\t #[W/m^2.C] - Convective Heat Transfer Coefficient\n", + "F = 1; \t\t\t\t #[dimensionless] - The value of Factor due geometric location and emissivity\n", + "sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant\n", + "k = 10; \t\t\t\t\t #[W/m.C] - Thermal Conductivity of Solid\n", + "#calculations\n", + "# Heat conducted through the plate = Convection Heat losses + Radiation Losses\n", + "qr = F*sigma*((Ts+273)**4-(Tsurr+273)**4) #[W/m^2] - #[W] - Heat lost per unit area from the plate due to radiation\n", + "qc = h*(Ts-Tsurr); #[W/m^2] - Convection Heat Loss per unit area\n", + "TG = -(qc+qr)/k; \t #[C/m] - Temperature Gradient\n", + "#results\n", + "print '%s %.2f %s' %(\"\\n \\n The temperature Gradient =\",TG,\" C/m\");\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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |