{ "metadata": { "name": "", "signature": "sha256:dfc2bc36e1576488bcd07ee5435a22145b28c649187a0989b470c7cdd7161f42" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 5: Single-Phase Systems" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.1-1, page no. 190" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "wtperct=0.5\n", "Dwater=0.998 #g/cm^3\n", "Dsulfuric=1.834 #g/cm^3\n", "\n", "\n", "#Calculations and printing :\n", "\n", "invPbar=wtperct/Dwater + (1-wtperct)/Dsulfuric\n", "print '%s %.3f' %(\"Density calculated using volume additvity (g/cm^3) =\",1/invPbar)\n", "Pbar=wtperct*Dwater + (1-wtperct)*Dsulfuric\n", "print '%s %.3f' %(\" \\n Density calculated using mass additivity (g/cm^3) =\",Pbar)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Density calculated using volume additvity (g/cm^3) = 1.293\n", " \n", " Density calculated using mass additivity (g/cm^3) = 1.416\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 1, "text": [ "''" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2-1, page no. 192" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "T=23.0+273 #kelvin\n", "P=3.0+14.7 #psi \n", "#conversion of pressure from psig to psi requires addition of 14.67 which is 1 atm\n", "R=0.08206 #lt-atm\n", "MN2=28.0 #molecular wt.\n", "weight=100.0 #grams\n", "\n", "#Calculations and printing :\n", "\n", "n=weight/MN2 #mol\n", "V=n*R*T*14.7/P #lt\n", "print '%s %.3f' %(\"assuming ideal gas behaviour, volume (litres) = \",V)\n", "Vcap=V/n\n", "if(Vcap>5):\n", " print(\"ideal gas equation yields error less than 1 percent for diatomic gas\")\n", "else:\n", " print(\"ideal gas equation yields error greater than 1 percent for diatomic gases\")\n", "\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "assuming ideal gas behaviour, volume (litres) = 72.046\n", "ideal gas equation yields error less than 1 percent for diatomic gas\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 2, "text": [ "''" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2-2, page no. 195" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "T=360.0+273 #Kelvin\n", "P=3.0 #atm\n", "Vdot=1100.0 #kg/h\n", "M=58.1 \n", "\n", "#Calculations and printing :\n", "\n", "ndot=Vdot/M #kmol/h\n", "vdot=ndot*22.4*T/(273*P)\n", "print '%s %.3f' %(\"The volumetric flow rate of the stream (m^3/h) = \",vdot)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The volumetric flow rate of the stream (m^3/h) = 327.781\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 3, "text": [ "''" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2-3, page no. 195" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "V1=10.0 #ft^3\n", "T1=70.0+460 #R\n", "P1=1.0 #atm\n", "P2=2.5 #atm\n", "T2=610.0+460 #R\n", "\n", "#Calculations and printing :\n", "\n", "V2=V1*P1*T2/(P2*T1)\n", "print '%s %.3f' %(\"Volume in final state (ft^3) = \",V2)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Volume in final state (ft^3) = 8.075\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 4, "text": [ "''" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2-4, page no. 196" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "#all the calculations are done in R scale\n", "T2=285.0+460 #R\n", "T1=32.0+460 #R\n", "P2=1.30 #atm\n", "P1=1.0 #atm\n", "V1dot=3.95*100000. #ft^3/h\n", "\n", "\n", "#Calculations and printing :\n", "\n", "#SCFH means ft^3(STP/h)\n", "ndot=3.95*math.pow(10,5)/359.\n", "print '%s %.3f' %(\" Molar flowrate (lb-moles/hr) = \",ndot)\n", "V2dot=V1dot*T2*P1/(T1*P2)\n", "print '%s %.3E' %(\" \\n True volumetric flowrate (ft^3/h) = \",V2dot)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Molar flowrate (lb-moles/hr) = 1100.279\n", " \n", " True volumetric flowrate (ft^3/h) = 4.601E+05\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 5, "text": [ "''" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2-5, page no. 197" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "flowinA=400.0 #L/min\n", "flowinN=419.0 # m^3 STP /min\n", "Pfinal=6.3 #gauge\n", "Tfinal=325.0 # C\n", "Pacetone=501.0 #mm of Hg\n", "Dacetone=791.0 #g/L\n", "Macetone=58.08 #g\n", "T1=300.0 #k\n", "P1=1238.0 #mm Hg original\n", "\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "n2cap=flowinA*Dacetone/Macetone\n", "print '%s %.3f' %(\" \\n Molar flowrate of Acetone (mol Acetone/min) = \",n2cap)\n", "P=Pfinal*760 + 763\n", "y4=Pacetone/P\n", "print '%s %.3f' %(\" \\n Mole fraction of Acetone in the final flow (mol Acetone/mol) = \",y4)\n", "print '%s %.3f' %(\" \\n Mole fraction of Nitrogen in the final flow (mol Nitrogen/mol) = \",1-y4)\n", "n3cap=flowinN/0.0224\n", "n4cap=n2cap/y4\n", "print(\"By using Overall Molar balance,\")\n", "n1cap=n4cap-n2cap-n3cap\n", "V1cap=n1cap*0.0224*T1*760/(1*273*P1)\n", "print '%s %.3f' %(\"Volumetric Flowrate of Nitrogen (Nitrogen/min) = \",V1cap)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", " \n", " Molar flowrate of Acetone (mol Acetone/min) = 5447.658\n", " \n", " Mole fraction of Acetone in the final flow (mol Acetone/mol) = 0.090\n", " \n", " Mole fraction of Nitrogen in the final flow (mol Nitrogen/mol) = 0.910\n", "By using Overall Molar balance,\n", "Volumetric Flowrate of Nitrogen (Nitrogen/min) = 547.119\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 6, "text": [ "''" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.3-1, page no. 202" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "T= -150.8 + 273.2 #k\n", "Vcap= 3./2. #L/mol\n", "Tc=126.2 #k\n", "Pc=33.5 #atm\n", "w=0.040\n", "\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "Pideal=0.08206*T/Vcap\n", "print '%s %.3f' %(\" \\n The value of pressure as per Ideal gas equation (atm) = \",Pideal)\n", "Tr=T/Tc\n", "B0=0.083 - (0.422/math.pow(Tr,1.6))\n", "B1=0.139 - (0.172/math.pow(Tr,4.2))\n", "B=0.08206*Tc*(B0+w*B1)/Pc\n", "Pvirial=0.08206*T*(1+ B/Vcap)/Vcap\n", "print '%s %.3f' %(\"\\n The value of pressure as per Virial gas equation (atm) = \",Pvirial)\n", "e=(Pideal-Pvirial)*100/Pvirial\n", "print '%s %.3f' %(\" \\n Percentage error due to Ideal gas Equation =\",e)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", " \n", " The value of pressure as per Ideal gas equation (atm) = 6.696\n", "\n", " The value of pressure as per Virial gas equation (atm) = 6.196\n", " \n", " Percentage error due to Ideal gas Equation = 8.072\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 7, "text": [ "''" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.3-2, page no. 203" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "V=2.5 #m^3\n", "n=1.00 #Kmol\n", "T= 300 #K\n", "Tc=304.2 #K\n", "Pc=72.9 #atm\n", "w=0.225\n", "R=0.08206\n", "\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "Vcap=V/n\n", "a=0.42747*math.pow(R*Tc,2) /Pc\n", "b=0.08664*R*Tc/Pc\n", "m=0.48508+ 1.5171*w -0.1561*w*w\n", "Tr=T/Tc\n", "alpha=math.pow(1+ m*(1-math.sqrt(Tr)),2)\n", "P=(R*T/(Vcap-b))-(alpha*a/(Vcap*(Vcap+b)))\n", "print '%s %.3f' %(\" \\n Pressure of gas calculated using SRK equation (atm) = \",P)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", " \n", " Pressure of gas calculated using SRK equation (atm) = 9.381\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 9, "text": [ "''" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.4-1, page no. 206" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "Vcap=50.0 #M^3/hr\n", "P=40.0 #bar\n", "T=300.0 #K\n", "R=8.314\n", "M=16.04 #kg/kmol\n", "\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "z=0.934\n", "print '%s %.3f' %(\" \\n From the Table, z= \",z)\n", "ncap=P*Vcap*101.325/(z*R*T*1.01325)\n", "mcap=ncap*M\n", "print '%s %.3f' %(\" \\n Mass flow rate of Methane (Kg/hr) = \",mcap)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", " \n", " From the Table, z= 0.934\n", " \n", " Mass flow rate of Methane (Kg/hr) = 1377.071\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 10, "text": [ "''" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.4-2, page no. 209" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "n=100.0 #gm-moles\n", "V=5.0 #ltr\n", "T= -20.6 + 273.2 #K\n", "Tc=126.2 #K\n", "Pc=33.5 #atm\n", "R=0.08206\n", "\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "Tr=T/Tc\n", "Vrideal=V*Pc/(n*R*Tc)\n", "print '%s %.3f' %(\" \\n Tr= \",Tr)\n", "print '%s %.3f' %(\"\\n Vrideal=\",Vrideal)\n", "z=1.77\n", "print '%s %.3f' %(\"\\n From the graphs, z=\",z)\n", "P=z*R*T*n/V\n", "Pr=P/Pc\n", "print '%s %.3f' %(\" \\n Pr= \",Pr)\n", "print '%s %.3f' %(\" \\n Pressure in the cylinder (atm) = \",P)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", " \n", " Tr= 2.002\n", "\n", " Vrideal= 0.162\n", "\n", " From the graphs, z= 1.770\n", " \n", " Pr= 21.904\n", " \n", " Pressure in the cylinder (atm) = 733.784\n" ] }, { "name": "stdout", "output_type": "stream", "stream": "stdout", "text": [ "press enter key to exit\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 11, "text": [ "''" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.4-3, page no. 212" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Initialization of variables\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "import scipy\n", "from scipy import integrate\n", "\n", "yN2=0.25\n", "yH2=1.0-yN2\n", "P=800.0 #atm\n", "T= -70+273.2 #K\n", "TcH2=33.0 #K\n", "TcN2=126.2 #K\n", "PcH2=12.8 #atm\n", "PcN2=33.5 #atm\n", "R=0.0083\n", "#Calculations and printing :\n", "\n", "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", "print(\"Applying newton corrections for Hydrogen,\")\n", "TcaH2=TcH2+8.\n", "PcaH2=PcH2+8.\n", "Tcbar=yH2*TcaH2 + yN2*TcN2\n", "Pcbar=yH2*PcaH2 + yN2*PcN2\n", "Trbar=T/Tcbar\n", "Prbar=P/Pcbar\n", "print '%s %.3f' %(\" \\n Trbar = \",Trbar)\n", "print '%s %.3f' %(\" \\n Prbar = \",Prbar)\n", "Zm=1.86\n", "print '%s %.4f' %(\" \\n From the graph, Zm = \",Zm)\n", "Vcap=Zm*R*T/P\n", "print '%s %.5f' %(\" \\n Specific Volume of Mixture (L/mol) = \",Vcap)\n", "raw_input('press enter key to exit')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", "Applying newton corrections for Hydrogen,\n", " \n", " Trbar = 3.262\n", " \n", " Prbar = 33.368\n", " \n", " From the graph, Zm = 1.8600\n", " \n", " Specific Volume of Mixture (L/mol) = 0.00392\n" ] } ] } ], "metadata": {} } ] }