From 64d949698432e05f2a372d9edc859c5b9df1f438 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- sample_notebooks/kartiksankhla/Chapter2.ipynb | 217 ++++++++++++++++++-------- 1 file changed, 154 insertions(+), 63 deletions(-) mode change 100755 => 100644 sample_notebooks/kartiksankhla/Chapter2.ipynb (limited to 'sample_notebooks/kartiksankhla/Chapter2.ipynb') diff --git a/sample_notebooks/kartiksankhla/Chapter2.ipynb b/sample_notebooks/kartiksankhla/Chapter2.ipynb old mode 100755 new mode 100644 index 21f2d4c4..f12ee152 --- a/sample_notebooks/kartiksankhla/Chapter2.ipynb +++ b/sample_notebooks/kartiksankhla/Chapter2.ipynb @@ -1,7 +1,7 @@ { "metadata": { "name": "", - "signature": "sha256:e984fee9b841dd6e9b7eedf1533b0a0d297cd9f484c047f051ce48a09b156826" + "signature": "sha256:44c6b2962e60454059ed8ab0f850fa5cf7fde8b83f0146551b8d869bf0ff197f" }, "nbformat": 3, "nbformat_minor": 0, @@ -13,7 +13,9 @@ "level": 1, "metadata": {}, "source": [ - "Chapter2-Nuclear Engineering" + "Chapter2-Basic Thermodynamics, Fluid\n", + "Mechanics: Definitions\n", + "of Efficiency" ] }, { @@ -21,25 +23,32 @@ "level": 2, "metadata": {}, "source": [ - "Ex1-pg54" + "Ex1-pg39" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "## Example 2.1\n", "import math\n", - "#determine atoms in deuterium\n", - "## Given data\n", - "atom_h = 6.6*10**24; ## Number of atoms in Hydrogen\n", - "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n", - "isoab_H2 = 0.015; ## Isotropic abundance of deuterium\n", - "## Calculation\n", - "totatom_d=(isoab_H2*atom_h)/100.;\n", - "## Result\n", - "print\"%s %.2e %s \"%('\\n Number of deuterium atoms = ',totatom_d,'');\n", - "\n" + "#calculate the polyefficency and overall total to total efficiency\n", + "\n", + "##given data\n", + "gamma = 1.4;\n", + "pi = 8.;##pressure ratio\n", + "T01 = 300.;##inlet temperature in K\n", + "T02 = 586.4;##outlet temperature in K\n", + "\n", + "##Calculations\n", + "##Calculation of Overall Total to Total efficiency\n", + "Tot_eff = ((pi**((gamma-1.)/gamma))-1.)/((T02/T01)-1.);\n", + "\n", + "##Calculation of polytropic efficiency\n", + "Poly_eff = ((gamma-1.)/gamma)*((math.log(pi))/math.log(T02/T01));\n", + "\n", + "##Results\n", + "print'%s %.2f %s'%('The Overall total-to-total efficiency is ',Tot_eff,'');\n", + "print'%s %.2f %s'%('The polytropic efficiency is ',Poly_eff,'');\n" ], "language": "python", "metadata": {}, @@ -48,41 +57,42 @@ "output_type": "stream", "stream": "stdout", "text": [ - "\n", - " Number of deuterium atoms = 9.90e+20 \n" + "The Overall total-to-total efficiency is 0.85 \n", + "The polytropic efficiency is 0.89 \n" ] } ], - "prompt_number": 5 + "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ - "Ex2-pg54" + "Ex2-pg44" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "## Example 2.2\n", "import math\n", - "#determine atomic weight of oxygen\n", - "## Given data \n", - "## Using the data given in the example 2.2\n", - "atwt_O16 = 15.99492; ## Atomic weight of O-16 isotope\n", - "isoab_O16 = 99.759; ## Abundance of O-16 isotope\n", - "atwt_O17 = 16.99913; ## Atomic weight of O-17 isotope\n", - "isoab_O17 = 0.037; ## Abundance of O-17 isotope\n", - "atwt_O18 = 17.99916; ## Atomic weight of O-18 isotope\n", - "isoab_O18 = 0.204; ## Abundance of O-18 isotope\n", - "## Calculation\n", - "atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;\n", - "## Result\n", - "print\"%s %.2f %s \"%('\\n Atomic Weight of Oxygen = ',atwt_O,'');\n", - "\n" + "#calculate the\n", + "\n", + "##given data\n", + "T01 = 1200.;##Stagnation temperature at which gas enters in K\n", + "p01 = 4.;##Stagnation pressure at which gas enters in bar\n", + "c2 = 572.;##exit velocity in m/s\n", + "p2 = 2.36;##exit pressure in bar\n", + "Cp = 1.160*1000.;##in J/kgK\n", + "gamma = 1.33\n", + "\n", + "##calculations\n", + "T2 = T01 - 0.5*(c2**2)/Cp;##Calculation of exit temperature in K\n", + "Noz_eff = ((1.-(T2/T01))/(1.-(p2/p01)**((gamma-1.)/gamma)));##Nozzle efficiency\n", + "\n", + "##Results\n", + "print'%s %.2f %s'%('Nozzle efficiency is ',Noz_eff,'');\n" ], "language": "python", "metadata": {}, @@ -91,8 +101,7 @@ "output_type": "stream", "stream": "stdout", "text": [ - "\n", - " Atomic Weight of Oxygen = 16.00 \n" + "Nozzle efficiency is 0.96 \n" ] } ], @@ -103,31 +112,31 @@ "level": 2, "metadata": {}, "source": [ - "Ex3-pg55" + "Ex3-pg51" ] }, { "cell_type": "code", "collapsed": false, "input": [ - "## Example 2.3\n", "import math\n", - "#determine rest mass energy of electron\n", - "## Given data\n", - "me = 9.1095*10**(-28); ## Mass of electron in grams\n", - "c = 2.9979*10**10; ## Speed of light in vacuum in cm/sec\n", - "## Calculation\n", - "rest_mass = me*c**2;\n", - "## Result\n", - "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass,' ergs\\n');\n", - "print('Expressing the result in joules')\n", - "## 1 Joule = 10^(-7)ergs\n", - "rest_mass_j = rest_mass*10**(-7);\n", - "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass_j,' joules\\n');\n", - "print('Expressing the result in MeV')\n", - "## 1 MeV = 1.6022*10^(-13)joules\n", - "rest_mass_mev = rest_mass_j/(1.6022*10**(-13));\n", - "print\"%s %.2f %s \"%('\\n Rest mass energy of electron = ',rest_mass_mev,' MeV\\n');\n" + "#calculate the\n", + "\n", + "##given data\n", + "cp = 0.6;##coefficient of pressure\n", + "AR = 2.13;##Area ratio\n", + "N_R1 = 4.66;\n", + "\n", + "##calculations\n", + "cpi = 1. - (1./(AR**2));\n", + "Diff_eff = cp/cpi;##diffuser efficiency\n", + "theta = 2.*(180./math.pi)*math.atan((AR**0.5 - 1.)/(N_R1));##included cone angle\n", + "\n", + "##Results\n", + "print'%s %.2f %s'%('cpi = \\n',cpi,'');\n", + "print'%s %.2f %s'%('The included cone angle can be found = ',theta,' deg.');\n", + "\n", + "\n" ], "language": "python", "metadata": {}, @@ -136,21 +145,103 @@ "output_type": "stream", "stream": "stdout", "text": [ + "cpi = \n", + " 0.78 \n", + "The included cone angle can be found = 11.26 deg.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4-pg52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate the\n", + "\n", + "##given data\n", + "AR = 1.8;##Area ratio\n", + "cp = 0.6;##coefficient of pressure\n", + "N_R1 = 7.85;\n", + "\n", + "##calculations\n", + "Theta = 2.*(180./math.pi)*math.atan((AR**0.5 - 1.)/(N_R1));##included cone angle\n", + "cpi = 1.-(1./(AR**2));\n", + "Diff_eff = cp/cpi;##diffuser efficeincy\n", + "\n", + "##Results\n", + "print'%s %.2f %s'%('The included cone angle can be found = ',Theta,' deg.\\n');\n", + "print'%s %.2f %s'%('cpi = \\n',cpi,'');\n", + "print'%s %.2f %s'%('Diffuser efficiency = ',Diff_eff,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The included cone angle can be found = 4.98 deg.\n", "\n", - " Rest mass energy of electron = 8.19e-07 ergs\n", - " \n", - "Expressing the result in joules\n", - "\n", - " Rest mass energy of electron = 8.19e-14 joules\n", - " \n", - "Expressing the result in MeV\n", - "\n", - " Rest mass energy of electron = 0.51 MeV\n", - " \n" + "cpi = \n", + " 0.69 \n", + "Diffuser efficiency = 0.87 \n" ] } ], "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5-pg53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate the\n", + "\n", + "##given data\n", + "AR = 2.0;##Area ratio\n", + "alpha1 = 1.059;\n", + "B1 = 0.109;\n", + "alpha2 = 1.543;\n", + "B2 = 0.364;\n", + "cp = 0.577;##coefficient of pressure\n", + "\n", + "##calculations\n", + "cp = (alpha1 - (alpha2/(AR**2))) - 0.09;\n", + "Diff_eff = cp/(1.-(1./(AR**2)));##Diffuser efficiency\n", + "\n", + "##Results\n", + "print'%s %.2f %s'%('The diffuser efficiency = ',Diff_eff,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diffuser efficiency = 0.78 \n" + ] + } + ], + "prompt_number": 5 } ], "metadata": {} -- cgit