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diff --git a/Introduction_To_Chemical_Engineering/ch1.ipynb b/Introduction_To_Chemical_Engineering/ch1.ipynb new file mode 100644 index 00000000..a09b75a5 --- /dev/null +++ b/Introduction_To_Chemical_Engineering/ch1.ipynb @@ -0,0 +1,900 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Introduction" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.1 page number 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find composition of air by weight\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "y_oxygen = 0.21 #mole fraction of oxygen\n", + "y_nitrogen = 0.79 #mole fraction of nitrogen\n", + "molar_mass_oxygen = 32.\n", + "molar_mass_nitrogen = 28.\n", + "\n", + "# Calculations and Results\n", + "molar_mass_air = y_oxygen*molar_mass_oxygen+y_nitrogen*molar_mass_nitrogen;\n", + "mass_fraction_oxygen =y_oxygen*molar_mass_oxygen/molar_mass_air;\n", + "mass_fraction_nitrogen = y_nitrogen*molar_mass_nitrogen/molar_mass_air;\n", + "\n", + "print \"mass fraction of oxygen = %f \"%(mass_fraction_oxygen)\n", + "print \"mass fraction of nitrogen = %f \"%(mass_fraction_nitrogen)\n", + "\n", + "V1 = 22.4 #in liters\n", + "P1 = 760. #in mm Hg\n", + "P2= 735.56 #in mm Hg\n", + "T1= 273. #in K\n", + "T2 = 298. #in K\n", + "\n", + "V2= (P1*T2*V1)/(P2*T1);\n", + "density = molar_mass_air/V2;\n", + "\n", + "print \"density = %f gm/l\"%(density)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass fraction of oxygen = 0.233010 \n", + "mass fraction of nitrogen = 0.766990 \n", + "density = 1.141558 gm/l\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.2 page number 20\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#find the volume occupied by propane\n", + "\n", + "import math \n", + "# Variables\n", + "mass_propane=14.2 #in kg\n", + "molar_mass=44 #in kg\n", + "\n", + "# Calculations\n", + "moles=(mass_propane*1000)/molar_mass;\n", + "volume=22.4*moles; #in liters\n", + "\n", + "# Results\n", + "print \"volume = %d liters\"%(volume)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "volume = 7229 liters\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.3 page number 20\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the average weight, weight composition, gas volume in absence of SO2\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "y_CO2 = 0.25;\n", + "y_CO = 0.002;\n", + "y_SO2 = 0.012;\n", + "y_N2 = 0.680;\n", + "y_O2 = 0.056;\n", + "\n", + "# Calculations and Results\n", + "Mm = y_CO2*44+y_CO*28+y_SO2*64+y_N2*28+y_O2*32;\n", + "print \" molar mass = %d \"%(Mm)\n", + "\n", + "print \" finding weight composition \"\n", + "w_CO2 = y_CO2*44*100/Mm;\n", + "print \" weight_CO2 = %f \"%(w_CO2)\n", + "w_CO = y_CO*28*100/Mm;\n", + "print \"weight_CO = %f \"%(w_CO)\n", + "w_SO2 = y_SO2*64*100/Mm;\n", + "print \"weight_SO2 = %f \"%( w_SO2)\n", + "w_N2 = y_N2*28*100/Mm;\n", + "print \"weight_N2 = %f \"%( w_N2)\n", + "w_O2 = y_O2*32*100/Mm;\n", + "print \"weight_O2 = %f \"%( w_O2)\n", + "\n", + "print \"if SO2 is removed \"\n", + "v_CO2 = 25;\n", + "v_CO = 0.2;\n", + "v_N2 = 68.0;\n", + "v_O2 = 5.6;\n", + "v = v_CO2+v_CO+v_N2+v_O2;\n", + "v1_CO2 = (v_CO2*100/98.8);\n", + "\n", + "print \"volume_CO2 = %f \"%( v1_CO2)\n", + "v1_CO = (v_CO*100/98.8);\n", + "print \"volume_CO = %f \"%(v1_CO)\n", + "v1_N2 = (v_N2*100/98.8);\n", + "print \"volume_N2 = %f \"%(v1_N2)\n", + "v1_O2 = (v_O2*100/98.8);\n", + "print \"volume_O2 = %f \"%(v1_O2 )\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molar mass = 32 \n", + " finding weight composition \n", + " weight_CO2 = 33.684468 \n", + "weight_CO = 0.171485 \n", + "weight_SO2 = 2.351788 \n", + "weight_N2 = 58.304753 \n", + "weight_O2 = 5.487506 \n", + "if SO2 is removed \n", + "volume_CO2 = 25.303644 \n", + "volume_CO = 0.202429 \n", + "volume_N2 = 68.825911 \n", + "volume_O2 = 5.668016 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.4 page number 24\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find volume of NH3 dissolvable in water\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "p=1. #atm\n", + "H=2.7 #atm\n", + "\n", + "# Calculations and Results\n", + "x=p/H;\n", + "\n", + "mole_ratio = (x)/(1-x);\n", + "moles_of_water=(100*1000)/18.;\n", + "moles_of_NH3=mole_ratio*moles_of_water;\n", + "\n", + "print \"moles of NH3 dissolved = %f\"%(moles_of_NH3)\n", + "\n", + "volume_NH3=(moles_of_NH3*22.4*293)/273;\n", + "print \"volume of NH3 dissolved = %f liters\"%(volume_NH3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "moles of NH3 dissolved = 3267.973856\n", + "volume of NH3 dissolved = 78565.443271 liters\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.5 page number 24\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to calculate amount of CO2 released by water\n", + "\n", + "import math \n", + "# Variables\n", + "p=746 #in mm Hg\n", + "H=1.08*10**6 #in mm Hg, Henry's constant\n", + "\n", + "# Calculations\n", + "x= p/H; #mole fraction of CO2\n", + "X=x*(44./18); #mass ratio of CO2 in water\n", + "\n", + "initial_CO2 = 0.005; #kg CO2/kg H20\n", + "G=1000*(initial_CO2-X);\n", + "\n", + "# Results\n", + "print \"CO2 given up by 1 cubic meter of water = %f kg CO2/cubic meter H20\"%(G)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "CO2 given up by 1 cubic meter of water = 3.311523 kg CO2/cubic meter H20\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.6 page number 27 \n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find vapor pressre of ethyl alchohal\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "pa1 = 23.6; #VP of ethyl alchohal at 10 degree C\n", + "pa3=760. #VP of ethyl alchohal at 78.3 degree C in mm Hg\n", + "pb1 = 9.2 #VP of ethyl water at 10 degree C in mm Hg\n", + "pb3=332. #VP of ethyl water at 78.3 degree C in mm Hg\n", + "\n", + "# Calculations\n", + "C=(math.log10(pa1/pa3)/(math.log10(pb1/pb3)));\n", + "\n", + "pb2=149. #VP of water at 60 degree C in mm Hg\n", + "\n", + "pas=(pb3/pb2);\n", + "pa=C*math.log10(pas);\n", + "pa2=pa3/(10**pa);\n", + "\n", + "# Results\n", + "print \"vapor pressure of ethyl alcholoh at 60 degree C = %f mm Hg\"%(pa2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "vapor pressure of ethyl alcholoh at 60 degree C = 349.872551 mm Hg\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.7 page number 28 \n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find vapor pressure using duhring plot\n", + "\n", + "import math \n", + "# Variables\n", + "t1 = 41. #in degree C\n", + "t2=59. #in degree C\n", + "theta_1 =83. #in degree C\n", + "theta_2=100. #in degree C\n", + "\n", + "# Calculations\n", + "K = (t1-t2)/(theta_1-theta_2);\n", + "t=59+(K*(104.2-100));\n", + "\n", + "# Results\n", + "print \"boiling point of SCl2 at 880 Torr = %f degree celcius\"%(t)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "boiling point of SCl2 at 880 Torr = 63.447059 degree celcius\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.8 page number 29\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the amount of steam released\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "vp_C6H6 = 520. #in torr\n", + "vp_H2O = 225. #in torr\n", + "mass_water=18.\n", + "mass_benzene=78.\n", + "\n", + "# Calculations\n", + "amount_of_steam = (vp_H2O/vp_C6H6)/(mass_benzene/mass_water);\n", + "\n", + "# Results\n", + "print \"amount of steam = %f\"%( amount_of_steam)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "amount of steam = 0.099852\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.9 page number 30\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find equilibrium vapor liquid composition\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "p0b = 385. #vapor pressue of benzene at 60 degree C in torr\n", + "p0t=140. #vapor pressue of toluene at 60 degree C in torr\n", + "xb=0.4;\n", + "xt=0.6;\n", + "\n", + "# Calculations and Results\n", + "pb=p0b*xb;\n", + "pt=p0t*xt;\n", + "P=pb+pt;\n", + "\n", + "print \"total pressure = %.0f torr\"%(P)\n", + "\n", + "yb=pb/P;\n", + "yt=pt/P;\n", + "print \"vapor composition of benzene = %f vapor composition of toluene = %f\"%(yb,yt)\n", + "\n", + "#for liquid boiling at 90 degree C and 760 torr, liquid phase composition\n", + "x=(760.-408)/(1013-408);\n", + "#(1013*x)+(408*(1-x))==760;\n", + "print \"mole fraction of benzene in liquid mixture = %.3f mole fraction of toluene in liquid mixture= %.3f\"%(x,1-x)\n", + "print \"Thus, the liquid mixture contained %.1f mole %% benzene and %.1f mole %% toluene\"%(x*100,(1-x)*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "total pressure = 238 torr\n", + "vapor composition of benzene = 0.647059 vapor composition of toluene = 0.352941\n", + "mole fraction of benzene in liquid mixture = 0.582 mole fraction of toluene in liquid mixture= 0.418\n", + "Thus, the liquid mixture contained 58.2 mole % benzene and 41.8 mole % toluene\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.10 page number 33\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find relation between friction factor and reynold's number\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "#math.log f=y, math.log Re=x, math.log a=c\n", + "sigma_x=23.393;\n", + "sigma_y=-12.437;\n", + "sigma_x2=91.456\n", + "sigma_xy=-48.554;\n", + "\n", + "# Calculations and Results\n", + "m=((6*sigma_xy)-(sigma_x*sigma_y))/(6*sigma_x2-(sigma_x)**2);\n", + "print \"m = %f \"%(m)\n", + "\n", + "c=((sigma_x2*sigma_y)-(sigma_xy*sigma_x))/(6*sigma_x2-(sigma_x)**2);\n", + "print \"c = %f \"%(c)\n", + "\n", + "print \"f=0.084*Re**-0.256\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m = -0.256233 \n", + "c = -1.073825 \n", + "f=0.084*Re**-0.256\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.11 page number 35\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the average velocity\n", + "\n", + "import math\n", + "from numpy import *\n", + "from matplotlib.pyplot import * \n", + "\n", + "%pylab inline\n", + "\n", + "# Variables\n", + "u = array([2,1.92,1.68,1.28,0.72,0]);\n", + "r = array([0,1,2,3,4,5]);\n", + "\n", + "# Calculations\n", + "z = u*r;\n", + "plot(r,z)\n", + "suptitle(\"variation of ur with r\")\n", + "xlabel(\"r\")\n", + "ylabel(\"ur\")\n", + "show()\n", + "#by graphical integration, we get\n", + "u_avg = (2./25)*12.4\n", + "\n", + "# Results\n", + "print \"average velocity = %f cm/s\"%(u_avg)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'new_figure_manager']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x2575250>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "average velocity = 0.992000 cm/s\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.12 page number 37\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the average velocity\n", + "\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "n = 6.;\n", + "h = (3. - 0)/n;\n", + "\n", + "# Calculations and Results\n", + "I = (h/2.)*(0+2*0.97+2*1.78+2*2.25+2*2.22+2*1.52+0);\n", + "u_avg = (2./3**2)*I;\n", + "\n", + "print \"average velocity = %f cm/s\"%(u_avg)\n", + "\n", + "print ('Simpsons rule')\n", + "\n", + "n = 6.;\n", + "h = 3./n;\n", + "I = (h/3)*(0+4*(0.97+2.25+1.52)+2*(1.78+2.22)+0);\n", + "u_avg = (2./3**2)*I;\n", + "\n", + "print \"average velocity = %f cm/s\"%(u_avg)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "average velocity = 0.971111 cm/s\n", + "Simpsons rule\n", + "average velocity = 0.998519 cm/s\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.13 page number 38\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the settling velocity as a function of time\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "z0 = 30.84;\n", + "z1 = 29.89;\n", + "z2 = 29.10;\n", + "h = 4;\n", + "\n", + "# Calculations\n", + "u1_t0 = (-3*z0+4*z1-z2)/(2*h);\n", + "u1_t4 = (-z0+z2)/(2*h);\n", + "u1_t8 = (z0-4*z1+3*z2)/(2*h);\n", + "\n", + "#considering data set for t = 4,8,12 min\n", + "z0 = 29.89;\n", + "z1 = 29.10;\n", + "z2 = 28.30;\n", + "u2_t4 = (-3*z0+4*z1-z2)/(2*h);\n", + "u2_t8 = (-z0+z2)/(2*h);\n", + "u2_t12 = (z0-4*z1+3*z2)/(2*h);\n", + "\n", + "#considering data set for t = 8,12,16 min\n", + "z0 = 29.10;\n", + "z1 = 28.30;\n", + "z2 = 27.50;\n", + "u3_t8 = (-3*z0+4*z1-z2)/(2*h);\n", + "u3_t12 = (-z0+z2)/(2*h);\n", + "u3_t16 = (z0-4*z1+3*z2)/(2*h);\n", + "\n", + "#taking average\n", + "u_t4 = (u1_t4+u2_t4)/2;\n", + "u_t8 = (u1_t8+u2_t8+u3_t8)/3;\n", + "u_t12 = (u2_t12+u3_t12)/2;\n", + "\n", + "# Results\n", + "print \"u_t0 = %f cm/min u_t4 = %f cm/min u_t8 = %f cm/min u_t12 = %f/n cm/min u_t16 =%f/n cm/min \"%(u1_t0,u_t4,u_t8,u_t12,u3_t16)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "u_t0 = -0.257500 cm/min u_t4 = -0.206875 cm/min u_t8 = -0.192083 cm/min u_t12 = -0.200625/n cm/min u_t16 =-0.200000/n cm/min \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.16 page number 49\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the flow rate and pressure drop\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "density_water=988. #in kg/m3\n", + "viscosity_water=55.*10**-5 #in Ns/m2\n", + "density_air=1.21 #in kg/m3\n", + "viscosity_air=1.83*10**-5 #in Ns/m2\n", + "L=1 #length in m\n", + "\n", + "\n", + "# Calculations and Results\n", + "L1=10.*L #length in m\n", + "Q=0.0133;\n", + "\n", + "Q1=((Q*density_water*viscosity_air*L)/(L1*viscosity_water*density_air))\n", + "\n", + "print \"flow rate = %f cubic meter/s\"%(Q1)\n", + "\n", + "#equating euler number\n", + "\n", + "p=9.8067*10**4; #pressure in pascal\n", + "p1=(p*density_water*Q**2*L**4)/(density_air*Q1**2*L1**4);\n", + "\n", + "print \"pressure drop corresponding to 1kp/square cm = %f kP/square cm\"%(p1/p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flow rate = 0.036134 cubic meter/s\n", + "pressure drop corresponding to 1kp/square cm = 0.011062 kP/square cm\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.17 page number 50\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the specific gravity of plasstic\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "L=1. #length of prototype in m\n", + "L1=10*L #length of model in m\n", + "density_prototype=2.65 #gm/cc\n", + "density_water=1. #gm/cc\n", + "\n", + "# Calculations\n", + "density_model=(L**3*(density_prototype-density_water))/(L1**3)+1;\n", + "\n", + "# Results\n", + "print \"specific gravity of plastic = %f\"%(density_model)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "specific gravity of plastic = 1.001650\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.18 page number 53\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find error in actual data and nomographic chat value\n", + "\n", + "import math \n", + "from numpy import linspace\n", + "from matplotlib.pyplot import *\n", + "\n", + "\n", + "# Variables\n", + "#for my\n", + "ly = 8 #in cm\n", + "my = ly/((1/0.25) - (1/0.5));\n", + "lz = 10.15 #in cm\n", + "\n", + "# Calculations and Results\n", + "mz = lz/((1./2.85) - (1/6.76));\n", + "mx = (my*mz)/(my+mz);\n", + "print \"mx = %f cm\"%(mx)\n", + "err = ((1-0.9945)/0.9945)*100;\n", + "print \"error = %f \"%(err)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mx = 3.703774 cm\n", + "error = 0.553042 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "example 1.19 page number 54\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#to find the economic pipe diameter from nomograph\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "#from the nomograph,we get the values of w and density\n", + "w=450. #in kg/hr\n", + "density=1000. #in kg/m3\n", + "d=16. #in mm\n", + "\n", + "# Calculations\n", + "u=(w/density)/(3.14*d**2/4);\n", + "Re=u*density*d/0.001;\n", + "\n", + "# Results\n", + "if Re>2100:\n", + " print \"flow is turbulent and d= %f mm\"%(d)\n", + "else:\n", + " print (\"flow is laminar and this nomograph is not valid\")\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flow is turbulent and d= 16.000000 mm\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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