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diff --git a/Transport_Phenomena/ch1.ipynb b/Transport_Phenomena/ch1.ipynb new file mode 100644 index 00000000..0a893e9a --- /dev/null +++ b/Transport_Phenomena/ch1.ipynb @@ -0,0 +1,120 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : introduction to transport phenomena" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 1.1 - Page No : 6\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculate the concentration of nitrogen in the tank in\n", + "\n", + "# Variables\n", + "v=0.01283; \t\t\t #[m**3] - volume of tank in m**3\n", + "v=0.4531; \t\t\t #[ft**3] - volume of tank in ft**3\n", + "p=2; \t\t\t #[atm] - pressure\n", + "T=1.8*300; \t\t\t #[degR] - temperature\n", + "R=0.73; \t\t \t #[(atm*ft**3)/(lbmol*degR)] - gas constant\n", + "\n", + "# Calculations\n", + "# usin the equation of state for an ideal gas pv=nRT\n", + "n=(p*v)/(R*T);\n", + "\n", + "xN2=0.5; \t\t\t # fractiom of N2 in math.tank\n", + "nN2=xN2*n;\n", + "Ca=nN2/v;\n", + "\n", + "# Results\n", + "print \"no. of moles , n = %.3e\"%n\n", + "print \"Ca = %.2e lb*mol/ft**3\"%(Ca);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no. of moles , n = 2.299e-03\n", + "Ca = 2.54e-03 lb*mol/ft**3\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 1.2 - Page No :9\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the mass of each concentrated solution to be added.\n", + "\n", + "from numpy import *\n", + "\n", + "# the three unknowns are x,y,z\n", + "# the three equations are-\n", + "# x+y+z = 1500\n", + "# (1) 0.05*x+0.15*y+0.40*z = 1500*0.25\n", + "# (2) 0.95*x+0.00*y+0.452*z = 1500*0.50\n", + "# Variables\n", + "a = array([[1, 1, 1],[0.05, 0.15, 0.40],[0.95, 0 ,0.452]])\n", + "d = array([[1500.],[1500.*0.25],[1500.*0.50]])\n", + "\n", + "# Calculations\n", + "#ainv = linalg.inv(a);\n", + "#sol = ainv * d;\n", + "sol = linalg.solve(a,d)\n", + "# Results\n", + "print \"the amount of concentrated HNO3 is %.0fkg \\\n", + "\\nthe amount of concentrated H2SO4 is %.0fkg \\\n", + "\\nthe amount of waste acids is %.0fkg\"%(sol[1],sol[0],round(sol[2],-1));\n", + "\n", + "# Answer may be different because of rounding error and inbuilt function solve." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the amount of concentrated HNO3 is 307kg \n", + "the amount of concentrated H2SO4 is 423kg \n", + "the amount of waste acids is 770kg\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
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