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| author | Trupti Kini | 2016-06-13 23:30:27 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-06-13 23:30:27 +0600 |
| commit | 200c141d0c5797a4033c448aaf9c73b81e2b819c (patch) | |
| tree | df88ac2ce6febb537f4aa4e1f55b726d960c9162 | |
| parent | 6803fb66aaaaea61e0fffbddf6656dff22d5fe10 (diff) | |
| download | Python-Textbook-Companions-200c141d0c5797a4033c448aaf9c73b81e2b819c.tar.gz Python-Textbook-Companions-200c141d0c5797a4033c448aaf9c73b81e2b819c.tar.bz2 Python-Textbook-Companions-200c141d0c5797a4033c448aaf9c73b81e2b819c.zip | |
Added(A)/Deleted(D) following books
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_10_Tabulated_properties_Steam_Tables.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_11_Properties_of_Gases.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_16_Fluid_Flow_Nozzles_and_Turbines.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_17_Gas_compression.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_18_Refrigeration.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_19_Heat_Transmission.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_2_Work.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_3_Temperature_and_Heat.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_5_First_Law_of_Thermodynamics.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_6_Flow_Procesess_First_law_analysis.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_8_Basic_applications_of_the_second_law.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Combustion_Processes_First_law_analysis.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Gas_cycles.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Properties_of_Gaseous_Mixtures.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/README.txt
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Vapor_cycles.ipynb
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/screenshots/2.png
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/screenshots/3.png
A Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/screenshots/6.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter1_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter2.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter2_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter3.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter3_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter4.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter4_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter5.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter5_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter6.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter6_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter7.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter7_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter8.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter8_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter9.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9_1.ipynb -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/Chapter9_1.ipynb
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/README.txt -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/README.txt
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/1.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_1.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/1_1.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_2.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/1_2.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/2.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2_1.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/2_1.png
R Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/3.png -> Power_Electronics:_Principles_&_Applications_by_J_M_Jacob_/screenshots/3.png
A sample_notebooks/nishumittal/chapter1.ipynb
45 files changed, 4486 insertions, 0 deletions
diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_10_Tabulated_properties_Steam_Tables.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_10_Tabulated_properties_Steam_Tables.ipynb new file mode 100644 index 00000000..716ec20d --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_10_Tabulated_properties_Steam_Tables.ipynb @@ -0,0 +1,463 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 10: Tabulated properties Steam Tables" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat of vaporisation (Btu/lb) = 889.11\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#What is the heat of vaporization of water at 100 psia?\n", + "#initialisation of variables\n", + "p= 100\t\t\t\t\t\t\t\t\t#psia\n", + "vg= 4.432 \t\t\t\t\t\t\t\t#cu ft/lb\n", + "vf= 0.01744 \t\t\t\t\t\t\t#cu ft/lb\n", + "T= 327.8 \t\t\t\t\t\t\t\t#F\n", + "sfg= 1.1286 \t\t\t\t\t\t\t#Bu/lb R\n", + "#CALCULATIONS\n", + "Q=(T+460)*sfg\t\t\t\t\t\t\t#Heat of vaporisation\n", + "#RESULTS\n", + "print '%s %.2f' %('Heat of vaporisation (Btu/lb) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "All the values are obtained from the steam tables\n", + "Pressure (Psia) = 70.00\n", + " \n", + " Temperature (F) = 302.92\n", + " \n", + " Enthalpy (Btu/lb) = 1180.60\n" + ] + } + ], + "source": [ + "#Saturated Steam has entropy of 1.6315 Btu/lb R; What are its pressure,\n", + "#temperature and enthalpy?\n", + "#initialisation of variables\n", + "S= 1.6315 \t\t\t\t\t\t#Btu/lb R\n", + "#CALCULATIONS\n", + "print '%s' %('All the values are obtained from the steam tables')\n", + "P= 70 \t\t\t\t\t\t\t#psia\n", + "t= 302.92 \t\t\t\t\t\t#F\n", + "h= 1180.6 \t\t\t\t\t\t#Btu/lb\n", + "#RESULTS\n", + "print '%s %.2f' %('Pressure (Psia) = ',P)\n", + "print '%s %.2f' % (' \\n Temperature (F) = ',t)\n", + "print '%s %.2f' % (' \\n Enthalpy (Btu/lb) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Internal energy (Btu/lb) = 1087.70\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#What is the internal energy of saturated water vapor at 250F?\n", + "#initialisation of variables\n", + "T= 250 \t\t\t\t\t\t#F\n", + "hg= 1164.0 \t\t\t\t\t#Btu/lb\n", + "P= 29.825 \t\t\t\t\t#Psia\n", + "Vg= 13.821 \t\t\t\t\t#cu ft/lb\n", + "#CALCULATIONS\n", + "ug= hg-(P*Vg*144./778.) #Internal energy\n", + "#RESULTS\n", + "print '%s %.2f' %('Internal energy (Btu/lb) = ',ug)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.4" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume (cu ft/lb) = 2.67\n", + " \n", + " Entropy (Btu/lb R) = 1.15\n", + " \n", + " Enthalpy (Btu/lb) = 831.68\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the properties of a mixture of steam and liquid water at 100 psia, containing 40 percent liquid.\n", + "#initialisation of variables\n", + "P= 100 \t\t\t\t\t\t\t#psia\n", + "n= 40\n", + "vf= 0.01774 \t\t\t\t\t#cu ft/lb\n", + "vg= 4.432 \t\t\t\t\t\t#cu ft/lb\n", + "hf= 298.4 \t\t\t\t\t\t#Btu/lb\n", + "hfg= 888.8 \t\t\t\t\t\t#Btu/lb\n", + "sg= 1.6026 \t\t\t\t\t\t#Btu/lb R\n", + "sfg= 1.1286 \t\t\t\t\t#Btu/lb R\n", + "#CALCULATIONS\n", + "vx= (n/100.)*vf+(1-(n/100.))*vg \t#Volume of mixture\n", + "hx= hf+(1-(n/100.))*hfg\t\t\t#Enthalpy of mixture\n", + "sx= sg-(n/100.)*sfg\t\t\t\t#Entropy of mixture\n", + "#RESULTS\n", + "print '%s %.2f' %('Volume (cu ft/lb) = ',vx)\n", + "print '%s %.2f' %(' \\n Entropy (Btu/lb R) = ',sx)\n", + "print '%s %.2f' %(' \\n Enthalpy (Btu/lb) = ',hx)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.5" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Enthalpy (Btu/lb) = 1160.54\n", + " \n", + " Precise Enthalpy (Btu/lb) = 1160.54\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the enthalpy of wet steam, 0.97 quality, at 100 psia.\n", + "#initialisation of variables\n", + "P= 100 \t\t\t\t\t#psia\n", + "n= 0.97\n", + "hf= 298.4 \t\t\t\t#Btu/lb\n", + "hfg= 888.8 \t\t\t\t#Btu/lb\n", + "hg= 1187.2 \t\t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "hx= hf+n*hfg\t\t\t#Enthalpy\n", + "hx1= hg-(1-n)*hfg\t\t#Precise Enthalpy\n", + "#RESULTS\n", + "print '%s %.2f' % ('Enthalpy (Btu/lb) = ',hx)\n", + "print '%s %.2f' % (' \\n Precise Enthalpy (Btu/lb) = ',hx1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume (cu ft/lb) = 25.38\n", + " \n", + " Entropy (Btu/lb R) = 1.71\n", + " \n", + " Enthalpy (Btu/lb) = 1117.23\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Water at 15 psia has entropy of 1.7050 Btu/lb R. Find its enthalpy and volume.\n", + "#initialisation of variables\n", + "P= 15 \t\t\t\t\t#psia\n", + "S= 1.7050 \t\t\t\t#Btu/lb R\n", + "sg= 1.7549 \t\t\t\t#btu/lb R\n", + "sfg= 1.4415 \t\t\t#Bru/lb R\n", + "hg= 1150.8 \t\t\t\t#btu/lb\n", + "hfg= 969.7 \t\t\t\t#Btu/lb\n", + "vg= 26.29 \t\t\t\t#cu ft/lb\n", + "vfg= 26.27 \t\t\t\t#cu ft/lb\n", + "#CALCULATIONS\n", + "n= (sg-S)/sfg \t\t\t#moisture fraction\n", + "sx= sg-n*sfg\t\t\t#Entropy\n", + "hx= hg-n*hfg\t\t\t#Enthalpy\n", + "vx= vg-n*vfg\t\t\t#Volume\n", + "#RESULTS\n", + "print '%s %.2f' % ('Volume (cu ft/lb) = ',vx)\n", + "print '%s %.2f' % (' \\n Entropy (Btu/lb R) = ',sx)\n", + "print '%s %.2f' % (' \\n Enthalpy (Btu/lb) = ',hx)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.10" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "All the values are obtained from the steam tables\n", + "Volume (cu ft/lb) = 0.01608\n", + " \n", + " Enthalpy (Btu/lb) = 70.67\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the volume and enthalpy of liquid water at 100 F and 1000 psia.\n", + "#initialisation of variables\n", + "T= 100 \t\t\t\t\t\t#F\n", + "P= 1000 \t\t\t\t\t#psia\n", + "dv= -5.1/100000. \t\t\t#cu ft/lb\n", + "dh= 2.70 \t\t\t\t\t#Btu/lb\n", + "vf= 0.01613 \t\t\t\t#cu ft/lb\n", + "hf= 67.97 \t\t\t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "print '%s' %(\"All the values are obtained from the steam tables\")\n", + "h= dh+hf\t\t\t\t\t#Enthalpy\n", + "v= dv+vf\t\t\t\t\t#Volume\n", + "#RESULTS\n", + "print '%s %.5f' %('Volume (cu ft/lb) = ',v)\n", + "print '%s %.2f' %(' \\n Enthalpy (Btu/lb) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 10.11" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "quality= 0.982\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A sample of steam at 200 psia flows to a throttling calorimeter in which the \n", + "#pressure is 15 psia and the temperature 280 F. Find the quality of the sample. \n", + "#At 15 psia and 280 F the enthalpy is found in table 3 to be 1183.2 btu/lb. \t\n", + "#initialisation of variables\n", + "h1= 1183.2 \t\t\t\t#Btu/lb\n", + "hg= 1198.4 \t\t\t\t#Btu/lb\n", + "hfg= 843.0 \t\t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "n= 1-((hg-h1)/hfg)\t\t#Quality\n", + "#RESULTS\n", + "print '%s %.3f' %('quality= ',n)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_11_Properties_of_Gases.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_11_Properties_of_Gases.ipynb new file mode 100644 index 00000000..e62ea4c5 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_11_Properties_of_Gases.ipynb @@ -0,0 +1,289 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 11: Properties of Gases" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 11.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume of the apparatus (cu ft) = 60.64\n" + ] + } + ], + "source": [ + "#A certain piece of apparatus of constant volume is filled with nitrogen \n", + "#at 15 psia, 80 F. from a nitrogen bottle on a weighing scale exactly 3 lb \n", + "#of nitrogen is added to the apparatus. The final pressure and temperature \n", + "#are 25 psia, 75 F. Find the volume of the apparatus.\n", + "import math\n", + "#initialisation of variables\n", + "P= 15. \t\t\t\t\t\t\t\t#psia\n", + "T= 80. \t\t\t\t\t\t\t\t#F\n", + "m= 3. \t\t\t\t\t\t\t\t#lb\n", + "P1= 25.\t\t\t\t\t\t\t\t#psia\n", + "T1= 75.\t\t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "r= (P*(460+T1))/(P1*(T+460))\t\t#ratio\n", + "m2= m/(1-r) \t\t\t\t\t\t#Mass 2\n", + "V2= (m2*55.16*(460+T1))/(P1*144.) \t#Volume 2\n", + "#RESULTS\n", + "print '%s %.2f' %('Volume of the apparatus (cu ft) = ',V2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 11.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "#Specific heat at constant pressure (Btu/lb R) = 0.217\n", + " \n", + "Specific heat at constant volume (Btu/lb R) = 0.155\n", + " \n", + "Specific heat at constant pressure 2 (Btu/lb R) = 0.235\n", + " \n", + "Specific heat at constant volume 2 (Btu/lb R) = 0.160\n", + " \n", + "Specific heat at constant pressure 3 (Btu/lb R) = 0.262\n", + " \n", + "Specific heat at constant volume 3 (Btu/lb R) = 0.200\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Given that O2 has a gas constant of 48.3 and is a diatomic gas, compare its\n", + "#actual Cp at 100, 500, 1500 F with the values computed from the simple KTG\n", + "#initialisation of variables\n", + "R= 48.3 \t\t\t\t\t#ft lb/lb R\n", + "T= 100 \t\t\t\t\t\t#F\n", + "T1= 500 \t\t\t\t\t#F\n", + "T2= 1500 \t\t\t\t\t#F\n", + "k= 1.4\n", + "k1= 1.36\n", + "k2= 1.31\n", + "#CALCULATIONS\n", + "dc= R/778. \t\t\t\t\t#Cp-Cv \n", + "cp= (k/(k-1))*dc \t\t\t#Specific heat at constant pressure\n", + "cv= cp/k \t\t\t\t\t#Specific heat at constant volume\n", + "cp1= (k1/(k1-1))*dc \t\t#Specific heat at constant pressure 2\n", + "cv1= cp/k1 \t\t\t\t\t#Specific heat at constant volume 2 \n", + "cp2= (k2/(k2-1))*dc \t\t#Specific heat at constant pressure 3\n", + "cv2= cp2/k2 \t\t\t\t#Specific heat at constant volume 3\n", + "#RESULTS\n", + "print '%s %.3f' %('#Specific heat at constant pressure (Btu/lb R) = ',cp)\n", + "print '%s %.3f' %(' \\nSpecific heat at constant volume (Btu/lb R) = ',cv)\n", + "print '%s %.3f' %(' \\nSpecific heat at constant pressure 2 (Btu/lb R) = ',cp1)\n", + "print '%s %.3f' %(' \\nSpecific heat at constant volume 2 (Btu/lb R) = ',cv1)\n", + "print '%s %.3f' %(' \\nSpecific heat at constant pressure 3 (Btu/lb R) =',cp2)\n", + "print '%s %.3f' %(' \\nSpecific heat at constant volume 3 (Btu/lb R) = ',cv2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 11.4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Enthalpy change (Btu/lb) = -71.79\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Nitrogen in a frictionless adiabatic process, expands from an initial state\n", + "#of 100 psia, 140 F to a final pressure of 10 psia. How much does the enthalpy\n", + "#change?\n", + "import math\n", + "#initialisation of variables\n", + "P= 10. \t\t\t\t\t\t#psia\n", + "P1= 100. \t\t\t\t\t#psia\n", + "T= 140. \t\t\t\t\t#F\n", + "k= 1.4\n", + "R= 55.16 \t\t\t\t\t#ft lb/lb R\n", + "#CALCULATIONS\n", + "dh= (k*R*(T+460)/(k-1))*(math.pow((P/P1),((k-1)/k))-1)*(72./56000.)\n", + "#RESULTS\n", + "print '%s %.2f' %('Enthalpy change (Btu/lb) = ',dh)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 11.5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "case 1\n", + "\n", + " change in volume = 23.66\n", + "\n", + " case 2\n", + "\n", + " change in volume (cu ft/lb) = 28.70\n", + "At T1=2460 R, from table 1, case 3\n", + "\n", + " change in volume (cu ft/lb) = 28.79\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air initially at 100 psia, 2000 F, expands reversibly and adiabatically\n", + "#to 15 psia. Find the change of enthalpy and of V by perfect gas laws, and\n", + "#by variable specific heats using the gas tables\n", + "import math\n", + "#initialisation of variables\n", + "P= 100 \t\t\t\t\t\t\t\t\t\t\t#psia\n", + "P1= 15 \t\t\t\t\t\t\t\t\t\t\t#psia\n", + "T= 2000 \t\t\t\t\t\t\t\t\t\t#F\n", + "k= 1.4\n", + "R= 53.34 \t\t\t\t\t\t\t\t\t\t#ft lb/lb R\n", + "cp= 0.240 \t\t\t\t\t\t\t\t\t\t#Btu/lb R\n", + "#CALCULATIONS\n", + "v1= (R*(T+460)/(P*144))*math.pow((P/P1),(1/k))\t#Initial volume\n", + "T1= (T+460)*(P1*v1/(P*(R*(T+460)/(P*144)))) \t#Initial temperature\n", + "dh= cp*(T1-T) \t\t\t\t\t\t\t\t\t#Change in enthalpy\n", + "dv= v1-(R*(T+460)/(P*144)) \t\t\t\t\t\t#Change in volume\n", + "print('case 1')\n", + "print '%s %.2f' %('\\n change in volume =', dv)\t\n", + "print('\\n case 2')\n", + "T2=1500 \t\t\t\t\t\t\t\t\t\t#F\n", + "v2=R*(T+460)/(P*144)/0.241\t\t\t\t\t\t#Volume in case 2\n", + "T2=(T2+460)*(P1*v2/(P*(R*(T2+460)/(P*144)))) \t#Temperature in case 2\n", + "deltah=0.276*(T2-460-T) \t\t\t\t\t\t#Change in enthalpy\n", + "dv2=v2-(R*(T+460)/(P*144)) \t\t\t\t\t\t#Change in volume\n", + "print '%s %.2f' %('\\n change in volume (cu ft/lb) = ', dv2)\n", + "print('At T1=2460 R, from table 1, case 3')\n", + "h1=634.34\n", + "pr1=407.3\n", + "pr2=pr1*P1/P \t\t\t\t\t\t\t\t\t#pressure 2\n", + "T2=1075 \t\t\t\t\t\t\t\t\t\t#F\n", + "h2=378.44\n", + "deltah=h2-h1 \t\t\t\t\t\t\t\t\t#Change in enthalpy\n", + "v2=53.34*(T2+460)/(P1*144) \t\t\t\t\t\t#Final volume\n", + "dv3=v2-(R*(T+460)/(P*144)) \t\t\t\t\t\t#Change in volume\n", + "print '%s %.2f' %('\\n change in volume (cu ft/lb) = ',dv3)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_16_Fluid_Flow_Nozzles_and_Turbines.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_16_Fluid_Flow_Nozzles_and_Turbines.ipynb new file mode 100644 index 00000000..37fd32ab --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_16_Fluid_Flow_Nozzles_and_Turbines.ipynb @@ -0,0 +1,238 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 16: Fluid Flow Nozzles and Turbines" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 16.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit area (sq. ft) = 0.01190\n", + "\n", + " Exit area in case 2 (sq. ft) = 0.00521\n" + ] + } + ], + "source": [ + "#(a)Find the exit area of a reversible nozzle to pass 1 lb/sec of steam if the\n", + "#inlet state of steam is 100 psia, 500F and 100 fps velocity and the exit pressure\n", + "#is 10 psia. (b) Will the nozzle be convering or divering and if the latter, what\n", + "#will be the throat area?\n", + "import math\n", + "#initialisation of variables\n", + "h1=1279.1 \t\t\t\t\t\t\t\t\t\t\t#Btu/lb\n", + "s1=1.7085 \t\t\t\t\t\t\t\t\t\t\t#Btu/lb R\n", + "p1= 100 \t\t\t\t\t\t\t\t\t\t\t#psia\n", + "p2=10 \t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "h2=1091.7 \t\t\t\t\t\t\t\t\t\t\t#Btu/lb\n", + "s2=s1\n", + "V1=100 \t\t\t\t\t\t\t\t\t\t\t\t#fps\n", + "v2=36.41 \t\t\t\t\t\t\t\t\t\t\t#cu ft/lb\n", + "w=1 \t\t\t\t\t\t\t\t\t\t\t\t#lb/sec\n", + "#Calculations\n", + "a2=w*v2/(math.sqrt(V1*V1 + 2*24956.243*(h1-h2)))\t#Exit area\n", + "print '%s %.5f' %('Exit area (sq. ft) = ',a2)\n", + "pt=0.55*p1 \t\t\t\t\t\t\t\t\t\t\t#Thorat pressure\n", + "ht=1221.5 \t\t\t\t\t\t\t\t\t\t\t#Btu/lb \n", + "vt=8.841 \t\t\t\t\t\t\t\t\t\t\t#cu ft/lb\n", + "at=w*vt/(math.sqrt(V1*V1 + 2*24956.243*(h1-ht))) \t#Exit area in case 2\n", + "print '%s %.5f' %('\\n Exit area in case 2 (sq. ft) = ',at)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 16.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "from tables\n", + "Throat area (ft^2) = 0.00571\n", + "\n", + " Exit area (ft^2) = 0.17756\n", + "\n", + " Final throat area (ft^2)= 0.00582\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the throat and exit areas for a nozzle to pass 10000 of steam from an initial\n", + "#state of 250 psia, 500F to a final pressure 1 psia, vel coeff. 0.949 and the \n", + "#discharge coeff. is unity. Find the exit jet velocity\n", + "#initialisation of variables\n", + "import math\n", + "w=10000. \t\t\t\t\t\t\t\t\t#lb/hr\n", + "p0=250. \t\t\t\t\t\t\t\t\t#psia\n", + "T1=500. \t\t\t\t\t\t\t\t\t#F\n", + "Pf=1. \t\t\t\t\t\t\t\t\t\t#psia\n", + "vc=0.949\n", + "dc=1\n", + "h0=1263.4 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "s0=1.5949 \t\t\t\t\t\t\t\t\t#btu/lb R\n", + "v2=276.\t\t\t\t\t\t\t\t\t\t#cu ft/lb\n", + "#Calculations\n", + "pt=0.55*p0\t\t\t\t\t\t\t\t\t#Throat pressure\n", + "print '%s' %('from tables')\n", + "hts=1208.2 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "vts=3.415 \t\t\t\t\t\t\t\t\t#cu ft/lb\n", + "h2s=891. \t\t\t\t\t\t\t\t\t#btu/lb\n", + "Vts=math.sqrt(2*32.174*778*(h0-hts))\t\t#Throat velocity\n", + "w=w/3600. \t\t\t\t\t\t\t\t\t#lb/sec \n", + "cw=1\n", + "at=w*vts/(cw*Vts)\t\t\t\t\t\t\t#Throat area\n", + "print '%s %.5f' %('Throat area (ft^2) = ',at)\n", + "V2=math.sqrt(2*32.174*778*(h0-h2s)) \t\t#Exit velocity\n", + "eta=0.9\n", + "h2=h0-eta*(h0-h2s) \t\t\t\t\t\t\t#Enthalpy\n", + "a2s=w*v2/(cw*V2) \t\t\t\t\t\t\t#Exit area\n", + "print '%s %.5f' %('\\n Exit area (ft^2) = ',a2s)\n", + "at=at/0.98\n", + "print '%s %.5f' %('\\n Final throat area (ft^2)= ',at)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 16.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit velocity case 1 (fps) = 1262.76\n", + "\n", + " Throat Area (ft^2) = 0.00246\n", + "\n", + " Exit velocity (fps) = 2029.48\n", + "\n", + " Exit area (ft^2) = 0.00562\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the exit velocity and the throat and exit areas for a nozzle to pass\n", + "#1 lb/sec of air from an inlet state of 150 psia, 340 F, negligible velocity, \n", + "#to an exhaust pressure of 15 psia if the nozzle efficiency is 88% and Cd=1?\n", + "import math\n", + "#initialisation of variables\n", + "k=1.4\n", + "ptbyp0=0.53\n", + "T0=800.\t\t\t\t\t\t\t\t\t\t#R\n", + "cp=778. \n", + "R=0.0425864\n", + "P0=150.\t\t\t\t\t\t\t\t\t\t#psia\n", + "Pt=15. \t\t\t\t\t\t\t\t\t\t#psia\n", + "w=1. \t\t\t\t\t\t\t\t\t\t#lb/sec\n", + "cw=1.0043782\n", + "#Calculations\n", + "Pt2=ptbyp0*Pt \t\t\t\t\t\t\t\t#Pressure\n", + "Tts=T0*math.pow((ptbyp0),((k-1)/k)) \t\t#Temperature\n", + "Vts=math.sqrt(2*32.174*cp*0.24*(T0-Tts)) \t#exit velocity\n", + "print '%s %.2f' %('Exit velocity case 1 (fps) = ',Vts)\n", + "vts=3.12 \t\t\t\t\t\t\t\t\t#cu ft/lb\n", + "at=w*vts/(cw*Vts)\t\t\t\t\t\t\t#Throat area\n", + "print '%s %.5f' %('\\n Throat Area (ft^2) = ', at)\n", + "T2s=T0*math.pow((Pt/P0),((k-1)/k)) \t\t\t#Final temperature\n", + "eta=0.88 \n", + "T2=T0-eta*(T0-T2s) \t\t\t\t\t\t\t#Temperature\n", + "V2=math.sqrt(2*32.5*cp*0.24*(T0-T2)) \t\t#Exit velocity\n", + "print '%s %.2f' %('\\n Exit velocity (fps) = ', V2)\n", + "v2=11.4 \t\t\t\t\t\t\t\t\t#cu ft/lb \n", + "a2=w*v2/V2\n", + "print '%s %.5f' %('\\n Exit area (ft^2) = ',a2)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_17_Gas_compression.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_17_Gas_compression.ipynb new file mode 100644 index 00000000..530b79bf --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_17_Gas_compression.ipynb @@ -0,0 +1,263 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 17: Gas compression" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 17.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfered from the airin each case (Btu/lb) = -16.01\n" + ] + } + ], + "source": [ + "#tests on a reciproacating air compressers with water-cooled cylinders show\n", + "#that it is practical to cool the air sufficently during compression to \n", + "#correspond to a polytropic exponenent n in the vicinity of 1.3. Compare\n", + "#the work per pound of air compressed from 15 psia, 80 F to 90 psia, according\n", + "#to three processes: reversible adiabatic, reversible isothermal, and reversible\n", + "#pv^1.3=c. Find the heat transferred from the air in each case\n", + "import math\n", + "#initialisation of variables\n", + "R= 53.31\n", + "T= 80 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "P2= 90. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "P1= 15. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "n= 1.4\n", + "n1= 1.3\n", + "cv= 0.171\n", + "#CALCULATIONS\n", + "Wk= (n/(n-1))*R*(T+460)*(math.pow((P2/P1),((n-1)/n))-1)\t\t\t\t#Work in 1\n", + "Wn= (n1/(n1-1))*R*(T+460)*(math.pow((P2/P1),((n1-1)/n1))-1) \t\t#Work in 2 \n", + "Wt= R*(T+460)*math.log(P2/P1)\t\t\t\t\t\t\t\t\t\t#Work in 3\n", + "Q= cv*0.778*((n-n1)/(1-n1))*(T+460)*(math.pow((P2/P1),((n-1)/n))-1) #Heat \n", + "#RESULTS\n", + "print '%s %.2f' %('Heat transfered from the airin each case (Btu/lb) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 17.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal effeciency = 0.81\n", + " \n", + " Isothermal effeciency= 1.05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#For the conditions of example 1 find the adiabatic efficiency and the isothermal\n", + "#efficiency of the reversible polytropic compressor\n", + "import math\n", + "#initialisation of variables\n", + "R= 53.31\n", + "T= 80 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "P2= 90. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "P1= 15. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "n= 1.4\n", + "n1= 1.3\n", + "cv= 0.171\n", + "#CALCULATIONS\n", + "Wk= (n/(n-1))*R*(T+460)*(math.pow((P2/P1),((n-1)/n))-1)\t\t\t#Work in 1\n", + "Wn= (n1/(n1-1))*R*(T+460)*(math.pow((P2/P1),((n1-1)/n1))-1)\t\t#Work in 2 \n", + "Wt= R*(T+460)*math.log(P2/P1) \t\t\t\t\t\t\t\t\t#Work in 3\n", + "nc= Wt/Wn \t\t\t\t\t\t\t\t\t\t\t\t\t\t#Thermal efficiency\n", + "nc1=Wk/Wn \t\t\t\t\t\t\t\t\t\t\t\t\t\t#Isothermal effeciency\n", + "##RESULTS\n", + "print '%s %.2f' %('Thermal effeciency = ',nc)\n", + "print '%s %.2f' %(' \\n Isothermal effeciency= ',nc1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 17.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transferred (Btu/lb) = -24.77\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#For the same initial state and final pressure as in ex.1, a real compressor\n", + "#has an efficiency of 95% on the adiabatic basis. The initial and final states\n", + "#correspond to a polytropic compression with n=1.3. Find the heat transferred\n", + "#per pound of air\n", + "import math\n", + "#initialisation of variables\n", + "R= 53.31\n", + "T= 80 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "P2= 90. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "P1= 15. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "n= 1.4\n", + "cp= 0.240\n", + "nc= 0.95\n", + "n1= 1.3\n", + "#CALCULATIONS\n", + "Wk= (n/(n-1))*(R)*(T+460)*(math.pow((P2/P1),((n-1)/n))-1)\t\t#Work in 1\n", + "Wx= -Wk/nc \t\t\t\t\t\t\t\t\t\t\t\t\t\t#Work in 2\n", + "dh= cp*(T+460)*(math.pow((P2/P1),((n1-1)/n1))-1) \t\t\t\t#Enthalpy transferred \n", + "Q= dh+(Wx/778.) \t\t\t\t\t\t\t\t\t\t\t\t#Heat\n", + "#RESULTS\n", + "print '%s %.2f' %('Heat transferred (Btu/lb) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 17.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "single-stage compression = 0.745\n", + " \n", + " two-stage compression = 0.937\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A gas is to be compressed from 5 psia to 83.5 psia. It is known that cooling\n", + "#correspondin to the polytropic exponent of 1.25 is practical and the clearance \n", + "#of the available compressors is 3%. Compare the volumetric efficencies to be \n", + "#anticipated for (a) single-stage compression, and (b)two-stage compression,with\n", + "#equal pressure ratios in the stages.\n", + "import math\n", + "#initialisation of variables\n", + "P1= 83.5\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "P2= 5. \t\t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "n= 3. \t\t\t\t\t\t\t\t\t\t\t\t\t#percent\n", + "n1= 1.25\n", + "#CALCULATIONS\n", + "nv= 1-(n/100.)*(math.pow((P1/P2),(1/n1))-1)\t\t\t\t#Efficiency of single stage\n", + "nv1= 1-(n/100.)*(math.sqrt(math.pow((P1/P2),(1/n1)))-1) #efficiency of two-stage\n", + "#RESULTS\n", + "print '%s %.3f' %('single-stage compression = ',nv)\n", + "print '%s %.3f' %(' \\n two-stage compression = ',nv1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_18_Refrigeration.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_18_Refrigeration.ipynb new file mode 100644 index 00000000..1370106c --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_18_Refrigeration.ipynb @@ -0,0 +1,92 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 18: Refrigeration" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 18.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "coefficient of performance of carnot = 6.05\n", + " \n", + " coefficient of performance = 5.23\n", + " \n", + " Piston displacement (cu ft/min) = 4.48\n" + ] + } + ], + "source": [ + "#A refrigiration plant is to operate with an evaporator saturation temp. is 0 F\n", + "#while rempvin 10000 From a cold room. The condenser is to be cooled by water so \n", + "#that the saturation temp. can be kept at 76 F. the refrigirant is ammonia\n", + "#(a) Assuming the plant works on a cycle like that, find its coefficient of \n", + "#performance and ompare this with the one of a carnot engine\n", + "#(b) If the volumetric efficiency if the compressor is 70% how much piston\n", + "#displacement per min will be needed?\n", + "import math\n", + "#initialisation of variables\n", + "T2 = 0 \t\t\t\t\t#F\n", + "T1= 76. \t\t\t\t#F\n", + "h1= 611.8 \t\t\t\t#Btu/lb\n", + "h4= 127.4 \t\t\t\t#Btu/lb\n", + "h2= 704.4 \t\t\t\t#Btu/lb\n", + "x= 10000 \t\t\t\t#Btu/hr\n", + "v1= 9.116 \t\t\t\t#cu ft/lb\n", + "n=70.\n", + "#CALCULATIONS\n", + "CP= (T2+460)/(T1-T2) \t#Carnot efficiency\n", + "CP1= (h1-h4)/(h2-h1) \t#coefficient of performance\n", + "w= (x/60.)/(h1-h4) \t\t#Work done\n", + "PD=(w*v1)/(n/100.) \t\t#Piston displacement\n", + "#RESULTS\n", + "print '%s %.2f' %('coefficient of performance of carnot = ',CP)\n", + "print '%s %.2f' %(' \\n coefficient of performance = ',CP1)\n", + "print '%s %.2f' %(' \\n Piston displacement (cu ft/min) = ',PD)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_19_Heat_Transmission.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_19_Heat_Transmission.ipynb new file mode 100644 index 00000000..0ec693cd --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_19_Heat_Transmission.ipynb @@ -0,0 +1,467 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 19: Heat Transmission" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat flow (Btu/hr) = 14.55\n", + " \n", + " Temperature at the interface (F) = 40.61\n" + ] + } + ], + "source": [ + "#A large flat wall consists of two well bonded layers of material 8 in thick\n", + "#and 1 in thick. The 8 in thick layer is concrete having a k=0.50. and the 1- in \n", + "#k=0.02. The surface temp is -20 F. and the surface temp. of concrete is 60 F. \n", + "#Find the rate of heat flow per unit area, the temperature at the interface\n", + "#between the two layers and the resistances per unit area of the two layers\n", + "import math\n", + "#initialisation of variables\n", + "t= 8. \t\t\t\t\t\t\t#in\n", + "t1= 1. \t\t\t\t\t\t\t#in\n", + "k= 0.50 \t\t\t\t\t\t#Btu/hr ft F\n", + "k1= 0.02 \t\t\t\t\t\t#Btu/hr ft F\n", + "A= 1 \t\t\t\t\t\t\t#ft^2\n", + "T= 60 \t\t\t\t\t\t\t#F\n", + "T1= -20 \t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "Rc= (t/12.)/(k*A)\t\t\t\t#Resistance 1\n", + "Rf= (t1/12.)/(k1*A)\t\t\t\t#Resistance 2\n", + "R= Rc+Rf \t\t\t\t\t\t#Total resistance\n", + "q= (T-T1)/R \t\t\t\t\t#Heat\n", + "T2= (T+(Rc/Rf)*T1)/(1+(Rc/Rf)) \t#temp of the interface\n", + "#RESULTS\n", + "print '%s %.2f' %('Rate of heat flow (Btu/hr) = ',q)\n", + "print '%s %.2f' %(' \\n Temperature at the interface (F) = ',T2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at the interface (F) = 133.055520\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A steel pipe of 10 in inside dia. and 0.375 in. wall thickness carries steam \n", + "#at 500 f. The pipe is covered by 2 in of insulation to reduce heat losses to\n", + "#surroundings at 80 F. It is known from tests that the convection coefficients\n", + "#for inside and outside are 2500 and 1.6. to protect personnel it is desired\n", + "#that the outside shouldn't exceed 140 f. If ksteel=26. and insulation is 0.045. \n", + "#will the 2 in thickness of insulation meet the requirement?\n", + "import math\n", + "#initialisation of variables\n", + "h1= 2500. \t\t\t\t\t\t\t#Btu/sq ft hr F\n", + "r= 10. \t\t\t\t\t\t\t\t#in\n", + "t= 0.375 \t\t\t\t\t\t\t#in\n", + "Ts= 500. \t\t\t\t\t\t\t#F\n", + "Ta= 80. \t\t\t\t\t\t\t#F\n", + "r2= 5.375 \t\t\t\t\t\t\t#in\n", + "r1= 5. \t\t\t\t\t\t\t\t#in\n", + "r3= 7.375 \t\t\t\t\t\t\t#in\n", + "kp= 26. \t\t\t\t\t\t\t#Btu ft/hr\n", + "ki= 0.045 \t\t\t\t\t\t\t#Btu ft/hr\n", + "h1= 2500. \t\t\t\t\t\t\t#Btu/sq ft hr F\n", + "h3= 1.6 \t\t\t\t\t\t\t#Btu/sq ft hr F\n", + "r4= 14.750\n", + "#CALCULATIONS\n", + "R1= 1/(h1*math.pi*(r/12.))\t\t\t#Resistance 1\n", + "Rp= math.log(r2/r1)/(2*math.pi*kp) \t#Resistance 2\n", + "Ri= math.log(r3/r2)/(2*math.pi*ki) \t#Resistance 3\n", + "R3= 1/(h3*math.pi*(r4/12.)) \t\t#Resistance 4\n", + "R0= R1+Rp+Ri+R3 \t\t\t\t\t#Total reistance\n", + "T3=Ta+ (Ts-Ta)*R3/R0 \t\t\t\t#Interface temp.\n", + "#RESULTS\n", + "print '%s %.6f' %('Temperature at the interface (F) = ',T3)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1000000.0\n", + "Water flow rate (lb/hr) = 25000.00\n", + " \n", + " Area of heat transfer surface (sq ft) = 337.89\n", + " \n", + " temperature of the oil (F) = 131.11\n", + " \n", + " flow rate (lb/hr) = 50000.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Lubricating oil is to be cooled from 170 F to 120 F by water ssupplied at 100F.\n", + "#Inorder to minimize the water requirement it is desired, if possible to have\n", + "#water rise to 140F, the highest permissible temp. The oil flow rate is 40000.\n", + "#the oil cp=0.5 and U=120. (b) Find the water flow rate for the counter flow\n", + "#operation. (c) Find the heat transfer surface required for the counter flow.\n", + "#(d) with the water flow rate found in (b), to what temp. could the oil be \n", + "#cooled in a parallel exchanger of unlimited area (e) In the exchanger of (d)\n", + "#how much water flow would be required to cool oil to 120F?\n", + "import math\n", + "#initialisation of variables\n", + "wh= 40000. \t\t\t\t\t\t\t\t\t#lb.hr\n", + "cph= 0.5 \t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "th1= 170. \t\t\t\t\t\t\t\t\t#F\n", + "th2= 120. \t\t\t\t\t\t\t\t\t#F\n", + "cpc= 1 \t\t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "tc2= 140. \t\t\t\t\t\t\t\t\t#F\n", + "tc1= 100. \t\t\t\t\t\t\t\t\t#F\n", + "t= 140 \t\t\t\t\t\t\t\t\t\t#F\n", + "U= 120 \t\t\t\t\t\t\t\t\t\t#Btu/sq ft hr F\n", + "#CALCULATIONS\n", + "dh= t-th2 \t\t\t\t\t\t\t\t\t#Change in temp. for hot\n", + "dc= tc2-tc1 \t\t\t\t\t\t\t\t#Change in temp. for cold\n", + "wc= (wh*cph*(th1-th2))/(cpc*dc) \t\t\t#weight of cold\n", + "dtm= (-(tc1-th2)-(th1-tc2))/math.log((-tc1+th2)/(th1-tc2)) #lmtd\n", + "q= wh*cph*(th1-th2) \t\t\t\t\t\t#heat\n", + "print(q)\n", + "A= q/(U*dtm) \t\t\t\t\t\t\t\t#area\n", + "th2= ((wh/wc)*(cph/cpc)*th1+tc1)/((wh/wc)*(cph/cpc)+1)#Hot final\n", + "wc1= (wh*cph*(th1-th2))/(cpc*(th2-tc1)) \t#weight of cold final\n", + "#RESULTS\n", + "print '%s %.2f' %('Water flow rate (lb/hr) = ',wc)\n", + "print '%s %.2f' %(' \\n Area of heat transfer surface (sq ft) = ',A)\n", + "print '%s %.2f' %(' \\n temperature of the oil (F) = ',th2)\n", + "print '%s %.2f' %(' \\n flow rate (lb/hr) = ',wc1*2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Film coefficient (Btu/sq ft hr F) = 14.59\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the film coefficient for airr flowing at 100 fps thru a tube of 1 in outside\n", + "#diameter and 18 gage thickness , if the avg. bulk air temp. is 600F, the pressure\n", + "#is 1 atm,a nd the tube wall is 200 F?\n", + "import math\n", + "#initialisation of variables\n", + "Tw= 200. \t\t\t#F\n", + "Ta= 600. \t\t\t#F\n", + "V= 100 \t\t\t\t#fps\n", + "Di= 0.902 \t\t\t#in\n", + "d= 0.0375 \t\t\t#lb/cu ft\n", + "u= 0.000020 \t\t#lbm/sec\n", + "cp= 0.25 \t\t\t#Btu/lb F\n", + "k= 0.027 \t\t\t#Btu/sq ft hr\n", + "#CALCULATIONS\n", + "NRe= (Di*V*d)/(u*12)#Reynolds number \n", + "Npr= 0.66 \t\t\t#prandtl number\n", + "h= k*0.023*math.pow(NRe,0.8)*math.pow(Npr,0.4)*12/Di \n", + "#RESULTS\n", + "print '%s %.2f' %('Film coefficient (Btu/sq ft hr F) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Film coefficient (Btu/sq ft hr F) = 14.38\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the film coefficient for the conditions of example 5, using eq. 17\n", + "import math\n", + "#initialisation of variables\n", + "Tw= 200 \t\t\t\t\t#F\n", + "Ta= 600 \t\t\t\t\t#F\n", + "cpb= 0.25 \t\t\t\t\t# Btu/lb F\n", + "tf= 0.68\n", + "uf= 0.000017 \t\t\t\t#lbm/sec ft\n", + "D= 0.902 \t\t\t\t\t#in\n", + "V= 100. \t\t\t\t\t#fps\n", + "d= 0.0375 \t\t\t\t\t#lb/cu ft\n", + "#CALCULATIONS\n", + "Nre= (D/12.)*V*d/uf\t\t\t#reynolds number\n", + "Npr= 0.68 \t\t\t\t\t#prandtl number\n", + "h= cpb*V*3600*d*0.023/(math.pow(Nre,0.2)*math.pow(Npr,(2./3.)))\n", + "#RESULTS\n", + "print '%s %.2f' %('Film coefficient (Btu/sq ft hr F) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss (Btu/hr) = 140.52\n", + " \n", + " hr (Btu/sq ft hr F) = 1.18\n", + " \n", + " hc (Btu/sq ft hr F) = 1.24\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#An uninsulated 3 in steam pipe passes through a room in which the air and all\n", + "#solid surfaces are at an average temp. of 70 F. If the surface temp. of the \n", + "#steam pipe is 200F, estimate the heat loss per foot of pipe by radiation and\n", + "# compare the relative magnitudes of losses by radiation and by free convection.\n", + "import math\n", + "#initialisation of variables\n", + "A1= 0.916 \t\t\t#ft^2\n", + "e1= 0.8\n", + "s= 0.173 \t\t\t#BTU s^-1 in^-2 R^-4\n", + "T= 200 \t\t\t\t#F\n", + "T1= 70 \t\t\t\t#F\n", + "D= 0.292\n", + "#CALCULATIONS\n", + "q= (A1/math.pow(10,6))*e1*s*((math.pow((T+460),4)/100.)-(math.pow((T1+460),4)/100.))\n", + "hr= q/(A1*(T-T1))\t#Coefficient of convection\n", + "hc= 0.27*math.pow(((T-T1)/D),0.25) #free convection coefficient\n", + "#RESULTS\n", + "print '%s %.2f' %('Heat loss (Btu/hr) = ',q)\n", + "print '%s %.2f' %(' \\n hr (Btu/sq ft hr F) = ',hr)\n", + "print '%s %.2f' %(' \\n hc (Btu/sq ft hr F) = ',hc)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 19.8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss (Btu/hr) = 9683.74\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A peep hole in the furnace wall of 13.5 in thickness is 8 in. square. the inside \n", + "#of the furnace is at uniform temp. of 1500 F and the external surroundings are at\n", + "#120 F. estimate the rate of heat loss by radiation through the open peep hole? \n", + "import math\n", + "#initialisation of variables\n", + "T= 120. \t\t\t\t\t#F\n", + "T1= 1500. \t\t\t\t\t#F\n", + "A= 64./144.\n", + "F= 0.86\n", + "Fe= 1\n", + "s= 0.173 \t\t\t\t\t#BTU s^-1 in^-2 R^-4\n", + "#CALCULATIONS\n", + "q= (A/math.pow(10,6))*F*Fe*s*((math.pow((T1+460),4)/100.)-(math.pow((T+460),4)/100.))\n", + "#RESULTS\n", + "print '%s %.2f' %('Heat loss (Btu/hr) = ',q)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_2_Work.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_2_Work.ipynb new file mode 100644 index 00000000..14873eb2 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_2_Work.ipynb @@ -0,0 +1,168 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 2: Work" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 2.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "P dV work for this process (ft lb) = 39810.42\n" + ] + } + ], + "source": [ + "#A stationary fluid system is subjected to a process in which the pressure and\n", + "#volume change according to the relation pv^1.4=C. The initial pressure and \n", + "#volume are respectively 100 psia and 3 cu ft, the final pressure is 20 psia\n", + "#Find the magnitude and direction of pdv work for this process\n", + "import math\n", + "#initialisation of variables\n", + "P= 100. \t\t\t\t\t\t#psia\n", + "V= 3. \t\t\t\t\t\t\t#cu ft\n", + "P1= 20. \t\t\t\t\t\t#psia\n", + "n= 1.4\n", + "#CALCULATIONS\n", + "V1= V*math.pow((P/P1),(1/n))\t#Final volume\n", + "W= (P1*V1*144-P*V*144)/(1-n) \t#work done\n", + "#144 is the conversion factor to convert in^2 to ft^2\n", + "#RESULTS\n", + "print '%s %.2f' %('P dV work for this process (ft lb) = ',W)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 2.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total system work during the 1 minute period (ft lb) = -13012.80\n" + ] + } + ], + "source": [ + "#In a stationary fluid system, a paddle wheel supplies work at the rate of\n", + "#1 hp. During a certain period of 1 min, the system expands in volume from \n", + "#1 cu.f t to 3 cu. ft. while the pressure remains constant at 69.4 psia.\n", + "#Find the total system work during the 1 min period.\n", + "#initialisation of variables\n", + "W= 1 \t\t\t\t\t\t#hp\n", + "P= 69.4 \t\t\t\t\t#psia\n", + "V2= 3 \t\t\t\t\t\t#cu\n", + "V1= 1 \t\t\t\t\t\t#cu\n", + "#CALCULATIONS\n", + "Wb= -W*33000 \t\t\t\t#Paddle wheel work\n", + "Wa= P*(V2-V1)*144 \t\t\t#Piston work\n", + "Q= Wa+Wb \t\t\t\t\t#Total work\n", + "#RESULTS\n", + "print '%s %.2f' %('Total system work during the 1 minute period (ft lb) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 2.3" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net work done by the fluid in the cylinder (ft lb) = 6335.55\n" + ] + } + ], + "source": [ + "#An engine has gone bore and stroke of 11 by 15 in. An indicator diagram \n", + "#taken from this engine has an area of 1.60 sq. in. and the length of 2.4 in.\n", + "#The k=80 psi per in. How much net work has been done by the fluid in the \n", + "#cylinder upon the engine piston during the engine cycle represented by\n", + "#the diagram.?\n", + "import math\n", + "#initialisation of variables\n", + "r= 11. \t\t\t\t\t#in\n", + "l= 15. \t\t\t\t\t#in\n", + "A= 1.6 \t\t\t\t\t#in\n", + "l1= 2.4 \t\t\t\t#in\n", + "\n", + "#CALCULATIONS\n", + "a= math.pi*r*r/4. \t\t#Piston area \t\t\n", + "L= l/12.\n", + "Pm= (A/l1)*80 \t\t\t#Mean effective pressure\n", + "W= a*L*Pm \t\t\t\t#Net work\n", + "#RESULTS\n", + "print '%s %.2f' %('Net work done by the fluid in the cylinder (ft lb) = ',W)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_3_Temperature_and_Heat.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_3_Temperature_and_Heat.ipynb new file mode 100644 index 00000000..e79cb271 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_3_Temperature_and_Heat.ipynb @@ -0,0 +1,263 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 3: Temperature and Heat" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 3.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pounds of water needed per pound of iron (lb water/lb iron) = 1.92\n" + ] + } + ], + "source": [ + "#It is desired to cool iron parts from 500 F to 100 F by dropping them into \n", + "#water intially at 75 F . The specific heat of the iron is 0.120 and the \n", + "#specific heat of water may be assumed to be 1. Assuming that all the heat from\n", + "#the iron goes to the water and tat none of the water evaporates and that none\n", + "#of the water are needed per pound of iron?\n", + "#initialisation of variables\n", + "T2w= 100. \t\t\t\t\t\t\t\t#F\n", + "T1w= 75. \t\t\t\t\t\t\t\t#F\n", + "cw= 1. \t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "T2i= 100. \t\t\t\t\t\t\t\t#F\n", + "T1i= 500. \t\t\t\t\t\t\t\t#F\n", + "ci= 0.12 \t\t\t\t\t\t\t\t#Btu/lb F\n", + "mi= 1\n", + "#CALCULATIONS\n", + "Mw= -mi*ci*(T2i-T1i)/(cw*(T2w-T1w)) \t#Mass of water required\n", + "#RESULTS\n", + "print '%s %.2f' %('Pounds of water needed per pound of iron (lb water/lb iron) = ',Mw)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 3.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net heat transferred to the system (Btu) = -1292.27\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#The specific heat of a certain gas is given by the function. How much heat is\n", + "#transferred from a system consisting of 5 lb of this gas to cool it at \n", + "#constant pressure from 1540 F to 540 F?\n", + "#initialisation of variables\n", + "import math\n", + "import scipy\n", + "from scipy import integrate\n", + "m=5. \t\t\t\t\t\t\t\t\t#lb\n", + "T1=1540. +460 \t\t\t\t\t\t\t#R\n", + "T2=540+460.\t\t\t\t\t\t\t\t#R\n", + "#CALCULATIONS\n", + "def cp(T):\n", + " cp=0.248+0.448*T*T/math.pow(10,8)\n", + " return cp;\n", + "\n", + "Qdot=scipy.integrate.quad(cp,T1,T2)\t\t#Heat\n", + "Q=m*Qdot[0]\n", + "#Results\n", + "print '%s %.2f' %('Net heat transferred to the system (Btu) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 3.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat required (Btu) = 3057.95\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Ten pounds of solid sulfur at 70 C are to be heated at constant pressre until \n", + "#it is a vapor at 1 atm pressure. How much heat is required?\n", + "#initialisation of variables\n", + "m= 10 \t\t\t\t#lb\n", + "cp= 0.180 \t\t\t#Btu/lb F\n", + "cp1= 0.235 \t\t\t#Btu/lb F\n", + "L= 15.8 \t\t\t#btu/lb\n", + "L1= 120 \t\t\t#btu/lb\n", + "T1= 70 \t\t\t\t#F\n", + "T2= 235 \t\t\t#F\n", + "T3= 832 \t\t\t#F\n", + "#CALCULATIONS\n", + "Qa= m*cp*(T2-T1)\t#Heat to raise solid temperature\n", + "Qb= m*L \t\t\t#Heat to melt solid \n", + "Qc= m*cp1*(T3-T2)\t#Heat to raise liquid temperature\n", + "Qd= m*L1 \t\t\t#Heat to vaporize liquid\n", + "Q= Qa+Qb+Qc+Qd \t\t#Total Heat\n", + "#RESULTS\n", + "print '%s %.2f' %('Heat required (Btu) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 3.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ice weight (lb of ice) = 7.68\n", + " \n", + " Additional ice required (lb of ice) = 0.177\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A piece of ice having aan intital temperature of 22 F is dropped into an \n", + "#insulated tank, which contains 40 lb of water at 70 F. If the temperature \n", + "#of water, after equilibrium is reached, is 40F, how many pounds did the ice\n", + "#weigh? Assume no heat transfer with other bodies hs occured?\n", + "#initialisation of variables\n", + "m= 40 \t\t\t\t#lb\n", + "m1= 10 \t\t\t\t#lb\n", + "cp= 1.00 \t\t\t#Btu/lb F\n", + "cp1= 0.501 \t\t\t#Btu/lb F\n", + "cp2= 0.092 \t\t\t#Btu/lb F\n", + "L= 143.3 \t\t\t#btu/lb\n", + "L1= 120 \t\t\t#btu/lb\n", + "T1= 22 \t\t\t\t#F\n", + "T2= 32 \t\t\t\t#F\n", + "T3= 40 \t\t\t\t#F\n", + "T4= 70 \t\t\t\t#F\n", + "#CALCULATIONS\n", + "Qa= cp1*(T2-T1)\t\t#Heat to raise solid temperature\n", + "Qb= L \t\t\t\t#Heat to melt solid\n", + "Qc= cp*(T3-T2) \t\t#Heat to raise liquid temperature\n", + "Qd= m*cp*(T3-T4) \t#Heat required to lower water from 70 to 40F\n", + "mi= -Qd/(Qa+Qb+Qc)\t#Mass\n", + "Qe= m1*cp2*(T3-T4) \t#Heat in case 2 due to copper\n", + "mi1= -Qe/(Qa+Qb+Qc) #Mass in case 2\n", + "#RESULTS\n", + "print '%s %.2f' %('Ice weight (lb of ice) = ',mi)\n", + "print '%s %.3f' %(' \\n Additional ice required (lb of ice) = ',mi1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_5_First_Law_of_Thermodynamics.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_5_First_Law_of_Thermodynamics.ipynb new file mode 100644 index 00000000..c5127493 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_5_First_Law_of_Thermodynamics.ipynb @@ -0,0 +1,263 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 5: First Law of Thermodynamics" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 5.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in internal energy of the gas (Btu) = -20.82\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A stationary mass of gas is compressed without friction from an intial state\n", + "#of 10 cu ft abd 15 psia to a final state of 5 cu ft and 15 psia, the pressure\n", + "#remaining constant during the process. How much does the internal energy of \n", + "#the gs change?\n", + "#initialisation of variables\n", + "p= 15 \t\t\t\t\t\t\t\t#psia\n", + "V2= 5 \t\t\t\t\t\t\t\t#cu\n", + "V1= 10 \t\t\t\t\t\t\t\t#cu\n", + "E= 34.7 \t\t\t\t\t\t\t#Btu\n", + "#CALCULATIONS\n", + "dE= -E-((p*(V2-V1)*144.)/(778.))\t#Change in internal energy\n", + "#RESULTS\n", + "print '%s %.2f' %('Change in internal energy of the gas (Btu) = ',dE)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 5.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in internal enenrgy (Btu) = -50.00\n" + ] + } + ], + "source": [ + "#A stationary system of conctant volume experience a temperature rise of 35 F\n", + "#when a certain process occurs. The heat transferred in the proccess is 34 btu. \n", + "#The specific heat at constant volume for the pure substance in the system is\n", + "# 1.2, and the system contains 2 lb of substance. Determine the internal \n", + "#energy change and the work done?\n", + "#initialisation of variables\n", + "m= 2 \t\t\t#lb\n", + "T2= 35 \t\t\t#F\n", + "cv= 1.2 \t\t#Btu/lb F\n", + "Q= 34 \t\t\t#Btu\n", + "#CALCULATIONS\n", + "U= m*cv*T2\t\t#Internal energy\n", + "W= Q-U \t\t\t#Work done\n", + "#RESULTS\n", + "print '%s %.2f' %('Change in internal enenrgy (Btu) = ',W)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 5.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From keenan and keyes steam tables\n", + "The second method follows the same procedure hence the same calculations are used\n", + "Change in internal enenrgy (Btu) = 670.96\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#In the steam tables,it is set forth that during the evaporation of 1 lb\n", + "#of water at 500 psia and 467.01 F the V increases from 0.0197 to 0.9278,\n", + "#while the enthalpy increases from 449.4 to 1204.4. How much work is done by\n", + "#a stationary system consisting of 1 lb of water when, because of an inflow \n", + "#heat, the system changes from liquid to vapor at 500 psia. How much does the\n", + "#internal energy change?\n", + "#initialisation of variables\n", + "print '%s' %('From keenan and keyes steam tables')\n", + "p= 500 \t\t\t\t#psia\n", + "V2= 0.9278 \t\t\t#cu ft/lb\n", + "V1= 0.0197 \t\t\t#cu ft/lb\n", + "h= 1204.4 \t\t\t#Btu/lb\n", + "h1= 449.4 \t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "W= p*144*(V2-V1)\t#Work done\n", + "U= h-h1-(W/778.) \t#Internal energy\n", + "#RESULTS\n", + "print '%s' %('The second method follows the same procedure hence the same calculations are used')\n", + "print '%s %.2f' %('Change in internal enenrgy (Btu) = ',U)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 5.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in internal enenrgy (Btu) = -50.82\n", + " \n", + " Work done (Btu) = 529.37\n", + " \n", + " Heat generated (Btu) = 478.55\n", + " \n", + " Work done (Btu) = 80.82\n", + "The answers given in the textbook are wrong.please calculate them personally.\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#The internal energy of a certain susbtance is given by u= 0.48pv + 35.\n", + "#A system composed of 3 lb of this substance expands from an intial pressure\n", + "#of 75 psia and volume of 6 to a final pressure of 15 psia in a process in \n", + "#which pressure and volume are related by pv^1.2=c. \n", + "#(a)If the expression is frictionless, determine Q, U and W for the process.\n", + "#(b) In another proccess the same system again exmapns according to the same \n", + "#initial state to the same final state as in part a but the heat in this case\n", + "#is 30 btu. Find the work for this process\n", + "import math\n", + "#initialisation of variables\n", + "m= 3 \t\t\t\t\t\t\t\t\t\t#lb\n", + "V1= 6 \t\t\t\t\t\t\t\t\t\t#cu ft\n", + "p1= 75. \t\t\t\t\t\t\t\t\t#psia\n", + "p2= 15. \t\t\t\t\t\t\t\t\t#psia\n", + "n= 1.2\n", + "Q1= 30 \t\t\t\t\t\t\t\t\t\t#Btu\n", + "#CALCULATIONS\n", + "V2= V1*math.pow((p1/p2),(1/n)) \t\t\t\t#Final volume\n", + "U= (m/3)*(0.480*p2*V2+35-0.480*p1*V1-35) \t#Internal energy\n", + "W= (p2*V2-p1*V1)/(1-n) \t\t\t\t\t\t#Work done\n", + "Q= U+W \t\t\t\t\t\t\t\t\t\t#Enthalpy\n", + "W1= Q1-U \t\t\t\t\t\t\t\t\t#Work done in case 2\n", + "#RESULTS\n", + "print '%s %.2f' %('Change in internal enenrgy (Btu) = ',U)\n", + "print '%s %.2f' %(' \\n Work done (Btu) = ',W)\n", + "print '%s %.2f' %(' \\n Heat generated (Btu) = ',Q)\n", + "print '%s %.2f' %(' \\n Work done (Btu) = ',W1)\n", + "print '%s' %('The answers given in the textbook are wrong.please calculate them personally.')\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_6_Flow_Procesess_First_law_analysis.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_6_Flow_Procesess_First_law_analysis.ipynb new file mode 100644 index 00000000..021d290f --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_6_Flow_Procesess_First_law_analysis.ipynb @@ -0,0 +1,208 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 6: Flow Procesess First law analysis" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 6.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Poweroutput (hp) = 219.56\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A turbine operates under steady flow conditions, receiving steam at the \n", + "#following state: P=170 psia, T=368.4 F and Cv=2.675, U=1111.9, v=6000 and z=10\n", + "#Steam leaves at the following state: P=3 psia, T=141.5 F and Cv=100.9, \n", + "#U=914.6, v=300 and z=0. heat loss=1000. Rate of flow=2500.what is Power?\n", + "import math\n", + "#initialisation of variables\n", + "u1= 1111.9 \t\t\t#Btu/lb\n", + "p= 170\t\t\t \t#psia\n", + "v1= 2.675 \t\t\t#cu ft/lb\n", + "V1= 6000 \t\t\t#ft/min\n", + "g0= 32.2 \t\t\t#ft/sec^2\n", + "g= 32.2 \t\t\t#ft/sec^2\n", + "z= 10 \t\t\t\t#ft\n", + "Q= 1000\t\t\t\t#Btu/hr\n", + "u2= 914.6 \t\t\t#Btu/lb\n", + "p1= 3 \t\t\t\t#psia\n", + "v2= 100.9 \t\t\t#cu ft/lb\n", + "V2= 300 \t\t\t#ft/sec\n", + "g0= 32.2 \t\t\t#ft/sec^2\n", + "g= 32.2 \t\t\t#ft/sec^2\n", + "z1= 0 \t\t\t\t#ft\n", + "#CALCULATIONS\n", + "#The numbers used are conversion factors for different units\n", + "Wx= (u1+(p*v1*144./778.)+(math.pow((V1/60.),2)/(2*g*778))+(z/778.)-(Q/2500.)-u2-(p1*v2*144./778.)-((V2*V2)/(2*g*778.)))*2500.\n", + "#RESULTS\n", + "print '%s %.2f' %('Poweroutput (hp) = ',Wx*0.000393014779)\n", + "#It is the conversion factor from btu/hr to hp\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 6.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steam supplied to the heater (lb/hr) = 1456.70\n" + ] + } + ], + "source": [ + "#A certain water heater operates under steady flow conditions receiving 500\n", + "#of water at t=165F and enthalpy=132.9. The water is heated by mixing with steam \n", + "#which is supplied to the heater at temp 215 F and enthalpy 1150. The mixture\n", + "#leaves the heater as liquid water at temp 212 F, enthalpy 180. How many pounds\n", + "#per hr of steam must be supplied to the heater? \n", + "#initialisation of variables\n", + "w1= 500 \t\t\t\t\t#lb/min\n", + "h1= 132.9 \t\t\t\t\t#Btu/lb\n", + "h2= 1150 \t\t\t\t\t#Btu/lb\n", + "h3= 180 \t\t\t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "w2= w1*(h3-h1)*60/(h2-h3)\t#Steam supplied\n", + "#RESULTS\n", + "print '%s %.2f' %('Steam supplied to the heater (lb/hr) = ',w2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 6.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average velovity at section1 (fps) = 939.56\n", + " \n", + " Average velovity at section2 (fps) = 1033.51\n", + "\n", + " rate of flow (lb/sec) = 4.15\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Steam is flowing steadily through a pipe 2 in in dia in which there is a \n", + "#pressure drop due to friction. the pipe is thoroughly insulated so that \n", + "#heat loss is neligible. At a certain section in the pipe, steam pressure\n", + "#is 100 psia, the specific volume is 4.937 and enthalpy=1227.6. at another\n", + "# section, downstream first, the corresopnding parameters are 90, 5.434 \n", + "#and 1223.9. Find the average velocity at each of the sections mentioned, \n", + "#the rate of flow\n", + "import math\n", + "#initialisation of variables\n", + "h1= 1227.6 #Btu/lb\n", + "h2= 1223.9 #Btu/lb\n", + "g= 32.2 #ft/sec^2\n", + "v1= 4.937 #cu ft/lb\n", + "d= 2./12. #in\n", + "A1=math.pi*d*d /4.\n", + "#CALCULATIONS\n", + "V1= math.sqrt((2*g*(h1-h2)*778)/((1.1)*1.1-1))\n", + "V2= 1.1*V1\n", + "w=A1*V1/v1\n", + "#RESULTS\n", + "print '%s %.2f' %('Average velovity at section1 (fps) = ',V1)\n", + "print '%s %.2f' %(' \\n Average velovity at section2 (fps) = ',V2)\n", + "print '%s %.2f' %('\\n rate of flow (lb/sec) = ', w)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_8_Basic_applications_of_the_second_law.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_8_Basic_applications_of_the_second_law.ipynb new file mode 100644 index 00000000..65257aa0 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Chapter_8_Basic_applications_of_the_second_law.ipynb @@ -0,0 +1,341 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 8: Basic applications of the second law" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum thermal efficiency (percent) = 6.42\n" + ] + } + ], + "source": [ + "#It is proposed to obtain a power from the hot surface water of tropical seas\n", + "#using the cold water from the depths as a sink for heat rejection. The surface\n", + "#water is 85 F the deep water is at 50 F. In the light of the second law is such\n", + "#scheme possible? If so, what is the max. thermal efficiency possible under law?\n", + "#initialisation of variables\n", + "T1= 85. \t\t\t\t\t#F\n", + "T2= 50. \t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "n= (T1-T2)/(T1+460)\t\t\t#Max. efficiency\n", + "n1= n*100 \t\t\t\t\t#percentage\n", + "#RESULTS\n", + "print '%s %.2f' %('Maximum thermal efficiency (percent) = ',n1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum thermal efficiency (percent) = 63.58\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A proposed steam power plant will provide for supplying heat to steam at temp\n", + "#upto 1050F. The temperature of heat rejection is about 90 F. It is stated that\n", + "#the efficiency of the plant will approach 34%. Does this seem reasonable per law?\n", + "#initialisation of variables\n", + "T1= 1050. \t \t\t\t\t#F\n", + "T2= 90. \t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "n= (T1-T2)/(T1+460)\t\t\t#Efficiency\n", + "n1= n*100 \t\t\t\t\t#Percentage\n", + "#RESULTS\n", + "print '%s %.2f' %('Maximum thermal efficiency (percent) = ',n1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Entropy change (Btu/Fabs) = 0.0430\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#How much does te entropy of a pound if air change when the air is heated \n", + "#irreversibly from 1 atm pressure, 50 F to 1 atm, 150 F if the Cp=0.240?\n", + "import math\n", + "#initialisation of variables\n", + "m= 1 \t\t\t\t\t\t\t\t\t\t\t#lb\n", + "cp= 0.240 \t\t\t\t\t\t\t\t\t\t#btu/lb F\n", + "T2= 150 \t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 50 \t\t\t\t\t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "S= m*cp*(math.log(460.+T2)-math.log(460.+T1))\t#Entropy\n", + "#RESULTS\n", + "print '%s %.4f' %('Entropy change (Btu/Fabs) = ',S)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Entropy change (Btu/Fabs) = 0.0430\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If in the previous example, the process is adiabatic with work against friction\n", + "#How much the entropy change during this process\n", + "import math\n", + "#initialisation of variables\n", + "m= 1 \t\t\t\t\t\t\t\t\t\t\t#lb\n", + "cp= 0.240 \t\t\t\t\t\t\t\t\t\t#btu/lb F\n", + "T2= 150 \t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 50 \t\t\t\t\t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "S= m*cp*(math.log(460.+T2)-math.log(460.+T1))\t#Entropy\n", + "#RESULTS\n", + "print '%s %.4f' %('Entropy change (Btu/Fabs) = ',S)\n", + "#This result is same as the above since change in entropy does not depend on the process involved\n", + "# but only on the initial and final states\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Entropy change (Btu/R) = 0.530\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#In a steam boiler, hot gases from a fire transfer heat to water which evaporates\n", + "#at const. temp. In certain case the gases are cooled from 2000 F to 1000F while\n", + "#the water evaportates to 400 F. the Cp=0.24 and L=826. No wastage. How much does\n", + "#the entropy increase as a result of the irreversible heat transfer.?\n", + "import math\n", + "#initialisation of variables\n", + "Q= 826 \t\t\t\t\t\t\t\t\t\t\t#Btu/lb\n", + "T= 400.\t\t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 1000. \t\t\t\t\t\t\t\t\t\t#F\n", + "T2= 2000. \t\t\t\t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "Sw= Q/(T+460.)\t\t\t\t\t\t\t\t\t#Entropy change for water\n", + "Sg= (Q/T1)*(math.log(T1+460)-math.log(T2+460))\t#Entropy change for gas\n", + "S= Sw+Sg\t\t\t\t\t\t\t\t\t\t#Total entropy change\n", + "#RESULTS\n", + "print '%s %.3f' %('Total Entropy change (Btu/R) = ',S)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 8.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Loss percent = 48\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#For the above example find the increase in unavailable energy due to the \n", + "#irreversible heat transfer. Assume the temperature of the surroundings is 80 F?\n", + "import math\n", + "#initialisation of variables\n", + "Q= 826.\t\t\t\t\t\t\t\t\t\t\t\t#Btu/lb\n", + "T= 400.\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 1000. \t\t\t\t\t\t\t\t\t\t\t#F\n", + "T2= 2000. \t\t\t\t\t\t\t\t\t\t\t#F\n", + "T3= 80.\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "Sw= Q/(T+460.)\t\t\t\t\t\t\t\t\t\t#Entropy change for water\n", + "Sg= (Q/T1)*(math.log(T1+460)-math.log(T2+460))\t\t#Entropy change for gas\n", + "S= Sw+Sg \t\t\t\t\t\t\t\t\t\t\t#Total entropy change\n", + "Q1= (T3+460.)*S \t\t\t\t\t\t\t\t\t#Heat generated\n", + "Q2= Q+(T3+460.)*Sg \t\t\t\t\t\t\t\t\t#Total heat\n", + "n= Q1/Q2 \t\t\t\t\t\t\t\t\t\t\t#Efficiency\n", + "n1= n*100 \t\t\t\t\t\t\t\t\t\t\t#Loss percent\n", + "#RESULTS\n", + "print '%s %d' %('Loss percent = ',n1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Combustion_Processes_First_law_analysis.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Combustion_Processes_First_law_analysis.ipynb new file mode 100644 index 00000000..4c71316d --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Combustion_Processes_First_law_analysis.ipynb @@ -0,0 +1,404 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 14: Combustion Processes First law analysis" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Theoretical air for combustion (lb air per lb C8H18) = 15.06\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the theoretical air for combustion of octane to CO2 and H2O\n", + "#initialisation of variables\n", + "M= 114 \t\t\t\t\t\t\t\t#lb\n", + "Mo= 32 \t\t\t\t\t\t\t\t#lb\n", + "Mn= 28 \t\t\t\t\t\t\t\t#lb\n", + "Mc= 44 \t\t\t\t\t\t\t\t#lb\n", + "Mw= 18 \t\t\t\t\t\t\t\t#lb\n", + "#CALCULATIONS\n", + "Ma= (12.5*Mo+(12.5)*(79./21.)*Mn)/114. #theoretical air\n", + "#RESULTS\n", + "print '%s %.2f' %('Theoretical air for combustion (lb air per lb C8H18) = ',Ma)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part b\n", + "\n", + " From steam tables\n", + "\n", + " Partial pressure of water (psia) = 1.78\n", + "\n", + " Dew point at the pressure (F) = 121.60\n", + "\n", + " Part c\n", + "\n", + " Partial pressure of water (psia) = 0.95\n", + "\n", + " The liquid water formed per pound of the fuel is (lb) 0.55\n" + ] + } + ], + "source": [ + "#Fuel oil containing 86 percent of carbon and 14 percent hydrogen by mass is to\n", + "#be burned with 10% excess air. (b) If the pressure is 15 psia. What is the dew\n", + "# point of the products? (c)If the products are cooled to 100 F at 15 psia how\n", + "#much liquid water will condense per pound of fuel burned?\n", + "#Initialization of variables\n", + "MW=18. \t\t\t\t#gm/mol\n", + "MCO2=44. \t\t\t#gm/mol\n", + "MN2=28. \t\t\t#gm/mol\n", + "MO2=32. \t\t\t#gm/mol\n", + "P=15. \t\t\t\t#psia\n", + "#calculations\n", + "xw=(0.074/MW)/(0.184/MCO2 + 0.074/MW + 0.02/MO2 + 0.722/MN2)\n", + "Pw=xw*P\n", + "print '%s' %(\"Part b\")\n", + "print '%s' %(\"\\n From steam tables\")\n", + "T=121.6 \t\t\t#F\n", + "print '%s %.2f' %('\\n Partial pressure of water (psia) = ',Pw)\n", + "print '%s %.2f' %('\\n Dew point at the pressure (F) = ',T)\n", + "print '%s' %(\"\\n Part c\")\n", + "pw2=0.9492 \t\t\t#Pressure\n", + "xw2=pw2/P\n", + "y = MW*(0.0346*xw2/(1-xw2))\n", + "diff=0.074-y\n", + "lb=17.15 \t\t\t#lb/ lbm of products\n", + "w=lb*diff\n", + "print '%s %.2f' %('\\n Partial pressure of water (psia) = ',pw2)\n", + "print '%s %.2f' %('\\n The liquid water formed per pound of the fuel is (lb) ',w)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.5" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat Transfer from the system (Btu) = -13513.78\n" + ] + } + ], + "source": [ + "#The internal energy of reaction for burning carbon to CO2 at 68F is -14087 . \n", + "#Find the heat trasnferred when a system composed of 1 lb of C and 4 lb of O2\n", + "#at 300 C burns at constant volume at a final temp of 1000 F . Cp=0.17\n", + "#initialisation of variables\n", + "mO2=1.33 \t\t\t\t\t\t\t\t\t\t#lb\n", + "mCO2=3.67 \t\t\t\t\t\t\t\t\t\t#lb\n", + "CvO2=0.155 \t\t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "CvCO2=0.165 \t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "Cc=0.170 \t\t\t\t\t\t\t\t\t\t#Btu/lb F\n", + "t2=1000. \t\t\t\t\t\t\t\t\t\t#F\n", + "tB=68. \t\t\t\t\t\t\t\t\t\t\t#F\n", + "t=300. \t\t\t\t\t\t\t\t\t\t\t#F\n", + "mC=1\n", + "mO=4\n", + "#Calculations\n", + "deltaE1=mO2*CvO2*(t2-tB) + mCO2*CvCO2*(t2-tB)\t#Energy change\n", + "deltaE2=mC*Cc*(tB-t) + mO*CvO2*(tB-t) \t\t\t#Energy change\n", + "E= -14087 \t\t\t\t\t\t\t\t\t\t#Btu \n", + "Q=deltaE1+E+deltaE2\t\t\t\t\t\t\t\t#Heat transfer\n", + "#Results\n", + "print '%s %.2f' %('Heat Transfer from the system (Btu) =',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.6" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Constant pressure heating value (Btu/lb formula wt) = -242211.42\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Given the reaction for burning of CO and the Cp=4344. Find the Cv?\n", + "#initialisation of variables\n", + "HV=4344 \t\t\t#Btu/lb\n", + "m=56 \t\t\t\t#lb\n", + "R=1.986 \t\t\t#Btu/lb mol R\n", + "Tb=530 \t\t\t\t#R\n", + "#Calculations\n", + "HR=m*HV \t\t\t#Heat of reaction\n", + "Eb=-HR-R*Tb*(2-3) \t#Heatin value\n", + "print '%s %.2f' %('Constant pressure heating value (Btu/lb formula wt) = ',Eb)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.7" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In case 1,Final temp (F) = 4940.02\n", + "\n", + " In case 2, Final temp (F) = 3986.47\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Assume carbon burns with air in a steady flow process. If theoretical air\n", + "#is used, calculate the products temp. for adiabatic combustion, assuming\n", + "#the products have constant sp. heats of room temp. magnitude\n", + "#initialisation of variables\n", + "mC=1 \t\t\t\t\t\t\t\t\t#lb\n", + "mO2=2.67\t\t\t\t \t\t\t\t#lb\n", + "mN2=8.78 \t\t\t\t\t\t\t\t#lb\n", + "mCO2=3.67 \t\t\t\t\t\t\t\t#lb\n", + "mN2=8.78 \t\t\t\t\t\t\t\t#lb\n", + "tB=77\t\t \t\t\t\t\t\t\t#F\n", + "deltaH=14087 \t\t\t\t\t\t\t#Btu/lb\n", + "CpCO2=0.196 \t\t\t\t\t\t\t#Btu/lb F\n", + "CpCO2f=0.3 \t\t\t\t\t\t\t\t#Btu/lb F\n", + "CpN2=0.248 \t\t\t\t\t\t\t\t#Btu/lb F\n", + "CpN2f=0.285 \t\t\t\t\t\t\t#Btu/lb F\n", + "#Calculations\n", + "t2= tB+ deltaH/(mCO2*CpCO2 + mN2*CpN2)\t#Final temp in case 1\n", + "t2f=tB+ deltaH/(mCO2*CpCO2f + mN2*CpN2f) #Final temp in case 2\n", + "#Results\n", + "print '%s %.2f' %('In case 1,Final temp (F) = ',t2)\n", + "print '%s %.2f' %('\\n In case 2, Final temp (F) = ',t2f)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.8" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency = 0.914\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Suppose 1 pound of carbon burns at Cp so that 0.9 goes into CO2, 0.05lb goes \n", + "#into CO and 0.05 emerges as unburned carbon; find the efficiency of the \n", + "#combustion process\n", + "#initialisation of variables\n", + "HR=14087 \t\t#Btu\n", + "HRC=3952 \t\t#Btu\n", + "x1=0.9\n", + "x2=0.05\n", + "#Calculations\n", + "HR1=x1*HR \t\t#Heat of reaction\n", + "HR2=x2*HRC \t#Heat of reaction\n", + "e=(HR2+HR1)/HR \t#Efficiency\n", + "#Results\n", + "print '%s %.3f' %('Efficiency = ',e)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 14.9" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "efficiency = 0.75\n" + ] + } + ], + "source": [ + "#A steam generator produces 10000 lb/hr of steam at 150 psia, saturated vapor. \n", + "#In addition, there is drawn off from the boiler 500 lb/hr of saturated liquid\n", + "# at 150 psia as \"blow down\" The feed water is supplied to the boiler at 210 F.\n", + "#Oil fuel having a hv of 19500 is burned in the furnace at the rate of 700 lb/hr.\n", + "#Find the efficiency of the steam generator, as defined above.\n", + "#initialisation of variables\n", + "hvi=19500. \t\t\t\t\t#Btu/hr\n", + "Q=10240000.\t\t\t\t\t#Btu/hr\n", + "rate=700.\t\t\t\t\t#lb/hr\n", + "#calculations\n", + "Hv=rate*hvi\n", + "efficiency=Q/Hv\n", + "#results\n", + "print '%s %.2f' %('efficiency = ',efficiency)" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Gas_cycles.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Gas_cycles.ipynb new file mode 100644 index 00000000..8b184ea7 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Gas_cycles.ipynb @@ -0,0 +1,219 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 15: Gas cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 15.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency = 0.37\n", + "Efficiency in case 2= 0.357\n" + ] + } + ], + "source": [ + "#Find the efficiency, and the work per pound of fluid circulated, for a \n", + "#Baryton cycle working between pressures of 15 psia and 75 psia, if the \n", + "#minimum temp. in the cycle is 550 R and the max. temp is 1700R?\n", + "import math\n", + "#initialisation of variables\n", + "p= 15.\t\t\t\t\t\t\t\t\t#psia\n", + "p1= 75.\t\t\t\t\t\t\t\t\t#psia\n", + "T= 550. \t\t\t\t\t\t\t\t#R\n", + "T1= 1700. \t\t\t\t\t\t\t\t#R\n", + "k= 1.4\n", + "#CALCULATIONS\n", + "Ta= T*math.pow((p1/p),((k-1)/k))\t\t#Temperature at A\n", + "Tc= T1/(math.pow((p1/p),((k-1)/k))) \t#Temperature at C\n", + "cp= 0.24 \n", + "Q1= cp*(T1-Ta) \t\t\t\t\t\t\t#Heat in 1\n", + "Q2= cp*(Tc-T) \t\t\t\t\t\t\t#Heat in 2\n", + "Wnet= Q1-Q2 \t\t\t\t\t\t\t#Work done\n", + "n= Wnet/Q1 \t\t\t\t\t\t\t\t#efficiency\n", + "hb= 422.59 \t\t\t\t\t\t\t\t#Btu/lb\n", + "hc= 269.27 \t\t\t\t\t\t\t\t#Btu/lb\n", + "ha= 208.41\t \t\t\t\t\t\t\t#Btu/lb\n", + "hd= 131.46 \t\t\t\t\t\t\t\t#btu/lb\n", + "Q1i= hb-ha\t\t\t\t\t\t\t\t#Heat in 1 case 2\n", + "Q2i= hc-hd \t\t\t\t\t\t\t\t#Heat in 2 case 2\n", + "Wnet1= Q1i-Q2i \t\t\t\t\t\t\t#work in case 2\n", + "n1= Wnet1/Q1i \t\t\t\t\t\t\t#efficiency 2\n", + "#RESULTS\n", + "print '%s %.2f' %('Efficiency = ',n)\n", + "print '%s %.3f' %( 'Efficiency in case 2= ',n1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 15.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency = 0.45\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Repeat for example 1 for a cycle using a regenerator of 75% effectiveness\n", + "#initialisation of variables\n", + "import math\n", + "p= 15. \t\t\t\t\t\t\t\t#psia\n", + "p1= 75. \t\t\t\t\t\t\t#psia\n", + "T= 550 \t\t\t\t\t\t\t\t#R\n", + "T1= 1700 \t\t\t\t\t\t\t#R\n", + "k= 1.4\n", + "n= 75.\n", + "cp= 0.24\n", + "#CALCULATIONS\n", + "Ta= T*math.pow((p1/p),((k-1)/k))\t#Temperature at A\n", + "Tc= T1/(math.pow((p1/p),((k-1)/k)))\t#Temperature at C\n", + "Ta1= (n/100.)*(Tc-Ta)+Ta \t\t\t#Temperature at A in case 2\n", + "Tc1= Ta+Tc-Ta1 \t\t\t\t\t\t#Temperature at C in case 2\n", + "Q1= cp*(T1-Ta1) \t\t\t\t\t#Heat in 1\n", + "Q2= cp*(Tc1-T) \t\t\t\t\t\t#heat in 2\n", + "Wnet= Q1-Q2 \t\t\t\t\t\t#Net work done\n", + "n1= Wnet/Q1 \t\t\t\t\t\t#Efficiency\n", + "#CALCULATIONS\n", + "print '%s %.2f' %('Efficiency = ',n1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 15.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency = 0.181\n", + " \n", + " air rate (lb air/hphr) = 52.04\n", + " \n", + " back work ratio = 1.87\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Find the efficiency, air rate and back work ratio for a gas turbine plant\n", + "#of the following description. T=60 F, p1=15. p2/p1=6; macchine efficiences\n", + "#of compressor and turbine both 60%. LHV=18500. Turbine inlet temp=1450 F.\n", + "#initialisation of variables\n", + "h1= 124.27 \t\t\t#Btu/lb\n", + "Pr1= 1.2147 \t\t#psia\n", + "r= 6\n", + "p4= 15. \t\t\t#psia\n", + "p1= 15. \t\t\t#psia\n", + "h2s= 197.5 \t\t\t#Btu/lb\n", + "Wnet= 48.9 \t\t\t#Btu/lb air\n", + "hs= 18500 \t\t\t#Btu/lb\n", + "wfbywa= 0.0146 \t\t#lb fuel/lb sir\n", + "W= 2545 \t\t\t#Btu/lb air\n", + "dh=-91.5 \t\t\t#Btu/lb\n", + "Wc= 91.5 \t\t\t#Btu/lb air\n", + "#CALCULATIONS\n", + "n= Wnet/(wfbywa*hs)\t#Efficiency\n", + "n1= W/Wnet\t\t\t#Air rate\n", + "n2= Wc/Wnet \t\t#Back work ratio\n", + "#RESULTS\n", + "print '%s %.3f' %('Efficiency = ',n)\n", + "print '%s %.2f' %(' \\n air rate (lb air/hphr) = ',n1)\n", + "print '%s %.2f' %(' \\n back work ratio = ',n2)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Properties_of_Gaseous_Mixtures.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Properties_of_Gaseous_Mixtures.ipynb new file mode 100644 index 00000000..2178e157 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Properties_of_Gaseous_Mixtures.ipynb @@ -0,0 +1,330 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 12: Properties of Gaseous Mixtures" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 12.1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "relative humidity= 0.59\n", + " \n", + " specific humidity (lb vapour/lb air) = 0.0090\n" + ] + } + ], + "source": [ + "#In a dew point apparatus a metal beaer is cooled by gradually adding ice water\n", + "#to the water initally in the beaker at room temp. The moisture from the room\n", + "#circulating around the beaker beins to condense on the beaker when its temp\n", + "#is 70 F and the pressure is 15 psia, find (1) the partial pressure of water\n", + "#vapor in the room air and (2) the parts by mass of the same.\n", + "#initialisation of variables\n", + "P= 15.0 \t\t\t\t#psia\n", + "T= 55 \t\t\t\t\t#F\n", + "P1= 0.2141 \t\t\t\t#psia\n", + "ma= 29. \t\t\t\t#lb\n", + "mb= 18. \t\t\t\t#lb\n", + "P2= 0.2141 \t\t\t\t#psia\n", + "P3= 0.3631 \t\t\t\t#psia\n", + "#CALCULATIONS\n", + "dp= P-P1 \t\t\t\t#Change in pressure\n", + "r= (dp*ma)/(P1*mb) \t\t#parts by mass\n", + "r1= r/(r+1) \t\t\t#ratio\n", + "r2= 1/(r+1) \t\t\t#ratio\n", + "r4= r2/r1 \t\t\t\t#specific humidity\n", + "P= P2/P3 \t\t\t\t#relative humidity\n", + "#RESULTS\n", + "print '%s %.2f' %('relative humidity= ',P)\n", + "print '%s %.4f' %(' \\n specific humidity (lb vapour/lb air) = ',r4)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 12.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pounds of water vapour enter the furnance per pound of dry air (lb vapour/lb air) = 0.0163\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air supplied to a furnacce has RH 75%, T=80 F and P=10 in of water. The\n", + "#barometer reads 29.5 in of mercury. How many pounds of water vapor enter\n", + "#the furnace per pound of dry air?\n", + "#initialisation of variables\n", + "h= 29.5 \t\t\t\t#in\n", + "n= 75.\n", + "T= 80 \t\t\t\t\t#F\n", + "h1= 10 \t\t\t\t\t#in\n", + "mw= 0.380*18\n", + "ma= 14.47*29\n", + "d= 13.6 \t\t\t\t#kg/m^3\n", + "P= 0.5069 \t\t\t\t#psi\n", + "#CALCULATIONS\n", + "Pw= (n/100.)*P \t\t\t#partial pressure\n", + "P= (h+(h1/d))*(0.491) \t#Total pressure\n", + "pa= P-Pw \t\t\t\t#partial pressure of air\n", + "r= mw/ma \t\t\t\t# pounds of water\n", + "#RESULTS\n", + "print '%s %.4f' %('Pounds of water vapour enter the furnance per pound of dry air (lb vapour/lb air) = ',r)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 12.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "relative humidity (lb water/lb dry air) = 0.00923\n" + ] + } + ], + "source": [ + "#A mixture of air and water vapor at 75 F and 14.7 psia has RH 0.5; find its\n", + "#specific humidity and its dew-point temp.?\n", + "#initialisation of variables\n", + "n= 0.5\n", + "T= 75 \t\t\t#F\n", + "P= 14.7 \t\t#psia\n", + "pg= 0.4298 \t\t#psia\n", + "pw= 0.2149 \t\t#psia\n", + "#CALCULATIONS\n", + "pw1= n*pg \t\t#partial pressure of water\n", + "pa= P-pw1 \t\t#partial pressure of air\n", + "r= 0.622*pw/pa \t#RH\n", + "#RESULTS\n", + "print '%s %.5f' %('relative humidity (lb water/lb dry air) = ',r)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 12.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Method 1\n", + "\n", + " In method 1, Enthalpy (Btu/lb of dry air) = 12.37\n", + "\n", + " Method 2\n", + "In method 2, Enthalpy (btu/lb of dry air) = 12.38\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#In the process shown, the air is received at 1 atm, 40 F and Rh 60% and it is\n", + "#desired to discharge it at 70 F, RH 50%. How much water at 45F must be supplied\n", + "#per pound of dry air passing through the apparatus?\n", + "#initialisation of variables\n", + "r2= 0.0078 \t\t\t\t\t\t#lb water /lb dry air\n", + "r1= 0.0032 \t\t\t\t\t\t#lb water /lb dry air\n", + "h2= 25.33 \t\t\t\t\t\t#Btu/lb\n", + "h1= 12.9 \t\t\t\t\t\t#Btu/lb\n", + "pg= 0.1217 \t\t\t\t\t\t#psia\n", + "p= 14.7 \t\t\t\t\t\t#psia\n", + "h3= 13 \t\t\t\t\t\t\t#Btu/lb\n", + "n= 60.\n", + "t2=70.\n", + "t1=40.\n", + "cpa=0.240\n", + "R2= 0.00788 \t\t\t\t\t#lb/lb of dry air\n", + "w1= 0.00477 \t\t\t\t\t#lb/lb of dry air\n", + "#CALCULATIONS\n", + "print '%s' %('Method 1')\n", + "w= r2-r1 \t\t\t\t\t\t#water to be supplied\n", + "q= h2-h1-w*h3 \t\t\t\t\t#energy supplied\n", + "print '%s %.2f' %('\\n In method 1, Enthalpy (Btu/lb of dry air) = ',q)\n", + "print('\\n Method 2')\n", + "R1= 0.622*(n/100.)*(pg/(p-pg))\t#Gamma 1 \n", + "R2=0.00788\t\t\t\t\t\t#Gamma 2\n", + "w2=R2-R1 \t\t\t\t\t\t#weight 2\n", + "#All constants are obtained from steam tables\n", + "Q=cpa*(t2-t1)+R2*(1092.6)-R1*(1079.6) -w2*h3\n", + "print '%s %.2f' %('In method 2, Enthalpy (btu/lb of dry air) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 12.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mixture must be cooled to the dew point temperature (F) = 55.30\n", + "Heat removed by the cooling coil (Btu/lb dry air)= -8.87\n", + "Heat supplied by the heating coil (Btu/lb dry air)= 3.63\n", + "Fraction of heat removed in the coil = 0.46\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air at 1 am,75F,70% relative humidity is to be brought to 70 F and 60 % RH\n", + "#. To what temp. must the mixture be cooled?How much heat must be removed by \n", + "#the cooling coil and how much be supplied by the heatin coil per pound of dry\n", + "#air?what fraction of the heat removed in the cooling coil is required to cool\n", + "#and condense the water removed?\n", + "#initialisation of variables\n", + "P= 1. \t\t\t \t#atm\n", + "n= 70.\n", + "T= 75 \t\t\t\t#F\n", + "T1= 70 \t\t\t\t#F\n", + "Td=55.3 \t\t\t#F\n", + "r1= 0.0131 \t\t\t#lb water/lb dry air\n", + "r2= 0.0093 \t\t\t#lb water/lb dry air\n", + "h1= 32.36 \t\t\t#Btu/lb dry air\n", + "h2= 27.03 \t\t\t#Btu/lb dry air\n", + "hd2= 23.40 \t\t\t#Btu/lb dry air\n", + "hf= 23.4 \t\t\t#Btu/lb dry liquid\n", + "hg= 1094.5 \t\t\t#Btu/lb dry liquid\n", + "#CALCULATIONS\n", + "R1= r1-r2\t\t\t#Gamma 1 \n", + "Qc= hd2-h1+R1*hf \t#Cooling\n", + "Qh= h2-hd2 \t\t\t#Heating\n", + "x= R1*(hg-hf) \t\t#Mole fraction\n", + "y= x/(-Qc) \t\t\t#fraction of heat removed\n", + "#RESULTS\n", + "print '%s %.2f' %('The mixture must be cooled to the dew point temperature (F) = ',Td)\n", + "print '%s %.2f' %('Heat removed by the cooling coil (Btu/lb dry air)= ',Qc)\n", + "print '%s %.2f' %('Heat supplied by the heating coil (Btu/lb dry air)= ',Qh)\n", + "print '%s %.2f' %('Fraction of heat removed in the coil = ',y)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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 +} diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/README.txt b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/README.txt new file mode 100644 index 00000000..5d8205be --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/README.txt @@ -0,0 +1,10 @@ +Contributed By: Chaitanya Potti +Course: btech +College/Institute/Organization: IITB +Department/Designation: Chemical engineering +Book Title: Introduction To Thermodynamics And Heat Transfer +Author: D. A. Mooney +Publisher: Longmans Green And Co., London +Year of publication: 1957 +Isbn: 8886332655 +Edition: 1
\ No newline at end of file diff --git a/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Vapor_cycles.ipynb b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Vapor_cycles.ipynb new file mode 100644 index 00000000..308a1861 --- /dev/null +++ b/Introduction_To_Thermodynamics_And_Heat_Transfer_by_D._A._Mooney/Vapor_cycles.ipynb @@ -0,0 +1,303 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 13: Vapor cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 13.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "cycle efficency = 0.318\n", + " \n", + " steam rate (lb steam per hphr) = 6.01\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A rankine cycle operates with steam conditions 200 psia,750 F and exhaust\n", + "#pressure 1 psia. Find the heat supplied, the turbine work, and the pump work\n", + "#per pound of steam. Find the cycle efficiency and steam rate?\n", + "#initialisation of variables\n", + "P= 1 \t\t\t\t\t\t#psia\n", + "P1= 200 \t\t\t\t\t#psia\n", + "T= 750 \t\t\t\t\t\t#F\n", + "v3= 0.01614 \t\t\t\t#cu ft/lb\n", + "h1= 1399.2 \t\t\t\t\t#Bu/lb\n", + "h2= 976 \t\t\t\t\t#Btu/lb\n", + "h3= 69.7 \t\t\t\t\t#Btu/lb\n", + "#CALCULATIONS\n", + "dh= v3*(144./778.)*(P1-P)\t#Change in enthalpy\n", + "h4= h3+dh \t\t\t\t\t#Enthalpy 4 \n", + "Q1= h1-h4\t\t\t\t\t#Heat\n", + "Wt= h1-h2 \t\t\t\t\t#Work\n", + "Wp= h4-h3 \t\t\t\t\t#Work\n", + "n= (Wt-Wp)/Q1 \t\t\t\t#Efficiency\n", + "w= 2545./Wt \t\t\t\t#Steam rate\n", + "#RESULTS\n", + "print '%s %.3f' %('cycle efficency = ',n)\n", + "print '%s %.2f' %(' \\n steam rate (lb steam per hphr) = ',w)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 13.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Engine efficency = 0.752\n", + " \n", + " state of the exhaust steam (Btu/lb) = 1081.075\n" + ] + } + ], + "source": [ + "#A turbine operating under the same steam conditions as given for the cycle\n", + "#of example 1 has a measured steam rate 8. Find the engine efficiency of the\n", + "#tuebine and the state of the exhaust steam.\n", + "#initialisation of variables\n", + "wt= 8. \t\t\t\t#lb/hphr\n", + "h1= 1399.2 \t\t\t#Btu/lb\n", + "h2s= 976. \t\t\t#Btu/lb\n", + "h2= 976. \t\t\t#Btu/lb\n", + "#CACLAULATIONS\n", + "Wt= 2545./wt \t\t#Work per piund\n", + "nt= Wt/(h1-h2s) \t#Efficiency\n", + "h21= h1-Wt \t\t\t#State of echaust steam\n", + "#RESULTS\n", + "print '%s %.3f' %('Engine efficency = ',nt)\n", + "print '%s %.3f' %(' \\n state of the exhaust steam (Btu/lb) = ',h21)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 13.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "All the values have been obtained from steam tables and mollier chart\n", + "h2 (Btu/lb) = 1307.05\n", + "\n", + " h4 (Btu/lb) = 1034.16\n", + "\n", + " For the first rankine cycle\n", + "\n", + " h7 (Btu/lb) = 935.18\n", + "\n", + " For the second rankine cycle\n", + "\n", + " h9 (Btu/lb) = 960.08\n", + "\n", + " Percentage Efficiency of reheat cycle compared to Rankine cycle for the first case = 6.93\n", + "\n", + " Percentage Efficiency of reheat cycle compared to Rankine cycle for the second case = 9.56\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A reheat cycle is to operate with turbines of 85% efficiency, but otherwise\n", + "#with idealized processes. the initial pressure is 2000 psia, and exhaust pressure\n", + "#is 0.5 psia, reheat=400 psia. Max. temp=1000 F. find efficiency and steam rate\n", + "#of this cycle. also of a rankine cycle working between 2000, 1000 and 0.5; also \n", + "#of a rankine cycle between 1400,1000,0.5. this being taken as cycle of max.\n", + "#permissible pressure without reheat. Use turbines oof 85% efficiecy in the rankine cycles\n", + "import math\n", + "print '%s' %('All the values have been obtained from steam tables and mollier chart')\n", + "#initialisation of variables\n", + "h1=1474.5 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "s1=1.5603 \t\t\t\t\t\t\t\t\t#btu/lb R\n", + "h2s=1277.5 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "#Calculations and printfing\n", + "h2=h1-0.85*(h1-h2s)\n", + "print '%s %.2f' %('h2 (Btu/lb) = ',h2)\n", + "h3=1522.4 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "s3=1.7623 \t\t\t\t\t\t\t\t\t#btu/lb R\n", + "h4s=948 #btu/lb\n", + "h4=h3- 0.85*(h3-h4s)\n", + "print '%s %.2f' %('\\n h4 (Btu/lb) = ',h4)\n", + "h5=47.6 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "h6=53.5 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "print ('\\n For the first rankine cycle')\n", + "h7s=840 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "h7=h1-0.85*(h1-h7s)\n", + "print '%s %.2f' %('\\n h7 (Btu/lb) = ',h7)\n", + "print ('\\n For the second rankine cycle')\n", + "h8=1493.2 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "s8=1.6903 \t\t\t\t\t\t\t\t\t#btu/lb R\n", + "h9s=866 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "h9=h8-0.85*(h8-h9s)\n", + "print '%s %.2f' %('\\n h9 (Btu/lb) = ',h9)\n", + "h11=51.5 \t\t\t\t\t\t\t\t\t#btu/lb\n", + "n1=0.401\n", + "n2=0.375\n", + "n3=0.366\n", + "e1=(n1-n2)/n2 \t\t\t\t\t\t\t\t#Efficiency\n", + "print '%s %.2f' %('\\n Percentage Efficiency of reheat cycle compared to Rankine cycle for the first case =',e1*100)\n", + "e2=(n1-n3)/n3 \t\t\t\t\t\t\t\t#Efficiency\n", + "print '%s %.2f' %('\\n Percentage Efficiency of reheat cycle compared to Rankine cycle for the second case =',e2*100)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Exa 13.4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "All the values have been obtained from steam tables and mollier chart\n", + "Fraction of energy supplied = 0.91\n", + " \n", + " Fraction of energy supplied which appears as useful energy= 0.64\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A steam plant operates with initial pressure 250 psia and temp 700 F and exhausts\n", + "#to a heating system at 25 psia. the condensate from the heating system is \n", + "#returned to the boiler plant at 150F, and the heating system utilizes for its\n", + "#intended purpose 90% of the energy transferred from the steam it receives\n", + "#(a) what fraction of energy supplied to the steam plant serves a useful purpose?\n", + "#(b) If two separate steam plants had been setup to produce the same useful energy,\n", + "#one to generate power through a cycle working between 250 psia, 700 F and 1 psia, \n", + "#what fraction of energy supplied would have served a useful purpose?\n", + "#initialisation of variables\n", + "h1= 1371 \t\t#Btu/lb\n", + "h2s= 1149 \t\t#Btu/lb\n", + "h3= 118 \t\t#Btu/lb\n", + "Q1= 1253 \t\t#Btu/lb\n", + "W= 156. \t\t#Btu/lb\n", + "Qw= 680. \t\t#Btu/lb\n", + "#CALCULATIONS\n", + "print '%s' %('All the values have been obtained from steam tables and mollier chart')\n", + "Qh= h1-W-h3 \t#Heat\n", + "y= W+0.9*Qh \t#useful energy\n", + "r= y/Q1 \t\t#fraction \n", + "x= Qh+Qw \t\t#total input\n", + "z= y/x \t\t\t#fraction\n", + "#RESULTS\n", + "print '%s %.2f' %('Fraction of energy supplied = ',r)\n", + "print '%s %.2f' %(' \\n Fraction of energy supplied which appears as useful energy= ',z)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": 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+ "metadata": {
+ "name": "",
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Introduction "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rs=50 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rl=100*Rs\n",
+ "\n",
+ "#Result\n",
+ "print\"Load resistance is\",Rl*10**-3,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load resistance is 5.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rs=10*10**3 #Kohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rl=0.01*Rs\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of load resistance is\",Rl,\"Kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of load resistance is 100.0 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=6\n",
+ "R3=4\n",
+ "\n",
+ "#Calculation\n",
+ "Vth=24\n",
+ "Rth=R3+((R1*R2)/(R1+R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"Thevenin Voltage is\",Vth,\"V\"\n",
+ "print\"Thevenin resistance is\",Rth,\"Kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thevenin Voltage is 24 V\n",
+ "Thevenin resistance is 6 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=10 #V\n",
+ "R=2.0 #Kohm\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Nortan current is\", I,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nortan current is 5.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |
