{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2 : Units and Dimensions" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.1 page no : 24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "# Variables \n", "V1 = 15. #ft**3/min volumetric flow rate\n", "ft = 0.3048 #m relationship\n", "min = 60. #secs relationship\n", "\n", "# Calculation \n", "V = V1*ft**3/min;\n", "\n", "# Result \n", "print \"Volumetric flowrate = %.3e m**3/s\"%V\n", "D = 1000 #kg/m**3\n", "M = V * D;\n", "print \"mass flowrate = %.3f kg/s\"%M\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Volumetric flowrate = 7.079e-03 m**3/s\n", "mass flowrate = 7.079 kg/s\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.2 page no : 24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables \n", "ft = 0.3048; #m \n", "lb = 0.4536; #kg\n", "\n", "# Calculation \n", "P = ft*lb;\n", "\n", "# Result \n", "print \"1 poundal is 1 ft*lb/s**2 = %.4f N\"%P\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1 poundal is 1 ft*lb/s**2 = 0.1383 N\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.3 page no : 24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables \n", "kgf = 9.80665; #N KGF\n", "\n", "# Calculation and Result \n", "cm = 10**-2; #m\n", "P = kgf/cm**2;\n", "print \"1 kgf/cm**2 = %0.3e N/m**2\"%P\n", "lbf = 32.174; #lb*ft#s**2\n", "lb = 0.4535924; #kg\n", "ft = 0.3048; #m\n", "in_ = 0.0254; #m\n", "P1 = lbf*lb*ft/in_**2;\n", "print \"1 lbf/in**2 = %.2f N/m**2\"%P1\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1 kgf/cm**2 = 9.807e+04 N/m**2\n", "1 lbf/in**2 = 6894.75 N/m**2\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.4 page no : 25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables \n", "Q1 = 10000. #kJ/hr rate of heat transfer\n", "kJ = 1000. #J \n", "hr = 3600. #s \n", "\n", "# Calculation \n", "Q = Q1*kJ/hr; #J/s\n", "print \"Q = %.2f J/s\"%Q\n", "x = 0.1; #m\n", "A = 1. #m**2\n", "T = 800. #K\n", "k = x*Q/(A*T);\n", "\n", "# Result \n", "print \"thermal conductivity = %.3f W/(m*k)\"%k\n", "J = 1/4.1868 #cal\n", "k1 = k*J*hr/1000\n", "print \"thermal conductivity = %.3F kcal/(h*m*C)\"%k1\n", "\n", "# note : answers may vary because of rounding off error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Q = 2777.78 J/s\n", "thermal conductivity = 0.347 W/(m*k)\n", "thermal conductivity = 0.299 kcal/(h*m*C)\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.5 page no : 26" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables \n", "F = 300. #N weight of object\n", "a = 9.81; #m/s**2 gravity\n", "\n", "# Calculation \n", "m = F/a; #kg\n", "\n", "# Result \n", "print \"mass in kg = %.2f kg\"%m\n", "lb = 4.535924/10; #kG\n", "m1 = m/lb\n", "print \"mass in pounds = %.2f LB\"%m1\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass in kg = 30.58 kg\n", "mass in pounds = 67.42 LB\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.6 page no : 26" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables \n", "z = 15. #m height\n", "PE = 2000.; #J potential energy\n", "g = 9.8067; #m/s**2 \n", "\n", "# Calculation \n", "m = PE/(z*g);\n", "\n", "# Result \n", "print \"mass = %.3f kg\"%m\n", "v = 50; #m/s\n", "KE = 1./2*m*(v**2)/1000.;\n", "print \"kinetic energy = %.3f kj\"%KE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass = 13.596 kg\n", "kinetic energy = 16.995 kj\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.7 page no : 26" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables \n", "g = 9.81; #m/s**2 gravity\n", "m = 100 * 0.4536; #kg weight \n", "P = 101325.; #N/m**2 standard atomosphere\n", "D1 = 4.; #inch\n", "\n", "# Calculation \n", "D = D1 * 2.54 * 10**-2; #m\n", "A = 3.1415 * (D**2)/4; #m**2\n", "F1 = P * A; #N\n", "F2 = m * g; #N\n", "F = F1 + F2;\n", "\n", "# Result \n", "print \"Total force acting on the gas = %.2f N\"%F\n", "P1 = F / A; #N/m**2\n", "P2 = P1/100000.; #bar\n", "P3 = P1/(6.894757 * 10**3); #psi\n", "print \"Pressure in N/m**2 = %.3e N/m**2\"%P1\n", "print \"Pressure in bar = %.3f bar\"%P2\n", "print \"Pressure in psi = %.2f psi\"%P3\n", "d = 0.4; #m\n", "W = F * d;\n", "print \"Work done = %.2f J\"%W\n", "PE = m * g * d;\n", "print \"Change in potential energy = %.2f J\"%PE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total force acting on the gas = 1266.43 N\n", "Pressure in N/m**2 = 1.562e+05 N/m**2\n", "Pressure in bar = 1.562 bar\n", "Pressure in psi = 22.66 psi\n", "Work done = 506.57 J\n", "Change in potential energy = 177.99 J\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.8 page no : 28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables \n", "#kG = (6.7 * 10**-4) * (( G * (ds + dt) / ds)**0.8) / ((ds**0.4)*(dG**0.2))\n", "\n", "#kG - lbmol/(h ft**2 atm), G - lb/(ft**2 h), ds, dG, dt - feet\n", "#kG1 - kmol/(m**2 h atm), G1 - kg/(m**2 h), ds1, dG1, dt1 - m\n", "G = 0.2048; #G1 * lb/(ft**2 h) velocity\n", "d = 3.2808; #d1 * ft clearance between grids\n", "ds = d; # clearance between grids\n", "dt = d; # clearance between grids\n", "dG = d; # clearance between grids\n", "kG = 4.885; #kG1 (lbmol/(h ft**2 atm) = 4.885 * kmol/(m**2 h atm))\n", "\n", "# Calculation \n", "C = (6.7 * 10**-4) * (( G * d / ds)**0.8) / ((ds**0.4)*(dG**0.2))* kG;\n", "\n", "# Result \n", "print \"Co-efficient = %.2e (kmol)(kg)**-0.8 (m)**0.2 (h)**-0.2 (atm)**-1\"%C\n", "# this is the constant for the equation\n", "# the equation thus becomes,\n", "# kG1 = C * (( G1 * (ds1 + dt1) / ds1)**0.8) / ((ds1**0.4)*(dG1**0.2))\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Co-efficient = 4.51e-04 (kmol)(kg)**-0.8 (m)**0.2 (h)**-0.2 (atm)**-1\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.9 page no : 29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "from sympy import Symbol\n", "\n", "# Variables \n", "#Cp = 7.13 + 0.577 * (10**-3) * t + 0.0248 * (10**-6) * t**2 \n", "\n", "#Cp - Btu/lb-mol F, t - F\n", "#Cp1 - kJ/kmol K , t1 - K\n", "a = 7.13 # first term\n", "b = 0.577 * 10**-3 # second term\n", "c = 0.0248 * 10**-6 # third term\n", "#t = 1.8 * t1 - 459.67\n", "Cp = 4.1868; #Cp1 (Btu/lb-mol F = 4.1868 * (kJ/kmol K) )\n", "t = Symbol('T')\n", "#substituting the above, we get,\n", "#Cp1 = 28.763 + 4.763 * (10**-3) * t1 + 0.3366 * (10**-6) * t**2\n", "a1 = 28.763\n", "\n", "# Calculation \n", "b1 = 4.763 * (10**-3)\n", "c1 = 0.3366 * (10**-6)\n", "\n", "Cp = a1 + b1*t + c1*t**2\n", "# Result \n", "print \"a1 = \",a1\n", "print \"b1 = \",b1\n", "print \"c1 = \",c1\n", "# this are the co efficents for the following equation;\n", "# Cp1 = a1 + b1 * t1 + c1 * (t1)**2\n", "print \"Equation Cp = \",Cp" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a1 = 28.763\n", "b1 = 0.004763\n", "c1 = 3.366e-07\n", "Equation Cp = 3.366e-7*T**2 + 0.004763*T + 28.763\n" ] } ], "prompt_number": 10 } ], "metadata": {} } ] }