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-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b0605e72f5da8476b3859cf3dc90196dcbe80cabc58b8c8d37cf5b28482c7ee1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15 : Circulation Systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 1, Page 369\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "thalf=1;   #Half life of catalyst in s\n",
-      "F=960;     #Feed rate of oil in tons/day\n",
-      "W=50;      #Weight of the bed in tons\n",
-      "a=0.5;     #Activity after time equal to half life\n",
-      "abar=0.01; #Average activity of the catalyst\n",
-      "\n",
-      "#CALCULATION\n",
-      "Ka=-math.log(a)/thalf;#Rate constant is s**-1, assuming I order kinetics from Eqn.(12)\n",
-      "Fs=Ka*W*abar/(1-abar);#Circulation rate of solids from Eqn.(16)\n",
-      "x=(Fs*60*60*24.0)/F;  #Circulation rate per feed of oil\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nSolid recirculation per feed of oil =%ftons of solid circulated/ton feed oil'%x\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "Solid recirculation per feed of oil =31.506690tons of solid circulated/ton feed oil\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2, Page 370\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "deltaHr1=1260.;                               #Enthalpy change during endothermic reaction in kJ/kg\n",
-      "deltaHr2=-33900.;                             #Enthal[y change during exothermic reaction in kJ/kg\n",
-      "H1=703.;                                      #Enthalpy of feed oil in kJ/kg\n",
-      "T1=260.;                                      #Temperature of feed oil in degree celcius\n",
-      "H3=1419.;                                     #Enthalpy of cracked product in kJ/kg\n",
-      "T3=500.;                                      #Temperature of cracked product in degree celcius\n",
-      "Ta=20.;                                       #Temperature of entering air in degree celcius\n",
-      "Cpa=1.09;                                     #Specific heat of entering air in kJ/kg K\n",
-      "Cpf=1.05;                                     #Specific heat of flue gases in kJ/kg K\n",
-      "Cps=1.01;                                     #Specific heat of solids in kJ/kg K\n",
-      "Cpv=3.01;                                     #Specific heat of vaporized feed in kJ/kg K\n",
-      "T4=[520.,540.,560.,580.,600.,620.,640.,660.]; #Temperature of flue gas in degree celcius\n",
-      "V=22.4;                                       #Volume of 1 mole of Carbon dioxide gas in N-m**3\n",
-      "M=12.;                                        #Molecular weight of carbon in kg\n",
-      "rho=1.293;                                    #Density of carbon dioxide gas in kg/N-m**3\n",
-      "xa=0.21;                                      #Mass fraction of oxygen in air\n",
-      "betac=0.07;                                   #Mass fraction of carbon\n",
-      "\n",
-      "#CALCULATION\n",
-      "n=len(T4);\n",
-      "i=0;\n",
-      "x1 = [0,0,0,0,0,0,0,0]\n",
-      "x2 = [0,0,0,0,0,0,0,0]\n",
-      "excess_air = [0,0,0,0,0,0,0,0]\n",
-      "\n",
-      "x2min=betac*(V*rho/(M*xa));#Minimum amount of air required for complete combustion\n",
-      "while i<n:\n",
-      "    x1[i]=(deltaHr1+0.93*H3-H1)/(Cps*(T4[i]-T3));#Fs/F1 by simplifying the overall energy balance\n",
-      "    x2[i]=((0.07*(-deltaHr2)-(deltaHr1+0.93*H3-H1))/(Cpf*(T4[i]-Ta)))-0.07;#F2/F1 by simplifying the energy balance for regenerator\n",
-      "    if x2[i]>x2min:\n",
-      "        excess_air[i]=(x2[i]-x2min)/x2min; #Excess air used\n",
-      "    else:\n",
-      "        excess_air[i]=0;\n",
-      "    i=i+1;\n",
-      "\n",
-      "#OUTPUT    \n",
-      "print 'T4(degree celcius)',\n",
-      "print '\\tFs/F1',\n",
-      "print '\\t\\tF2/F1',\n",
-      "print '\\t\\tExcess air(percentage)'\n",
-      "i=0;\n",
-      "while i<n:\n",
-      "    print '%f'%T4[i],\n",
-      "    print '\\t\\t%f'%x1[i],\n",
-      "    print '\\t%f'%x2[i],\n",
-      "    print '\\t%f'%(excess_air[i]*100);\n",
-      "    i=i+1;\n",
-      "\n",
-      "#Disclaimer: The values of F2/F1 obtained by manual calculation has close correspondance to the ones obtained as the output, whereas it deviates largely from the values given in textbook.\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "T4(degree celcius) \tFs/F1 \t\tF2/F1 \t\tExcess air(percentage)\n",
-        "520.000000 \t\t92.904455 \t0.875390 \t8.807235\n",
-        "540.000000 \t\t46.452228 \t0.839029 \t4.287699\n",
-        "560.000000 \t\t30.968152 \t0.805362 \t0.102944\n",
-        "580.000000 \t\t23.226114 \t0.774099 \t0.000000\n",
-        "600.000000 \t\t18.580891 \t0.744992 \t0.000000\n",
-        "620.000000 \t\t15.484076 \t0.717825 \t0.000000\n",
-        "640.000000 \t\t13.272065 \t0.692412 \t0.000000\n",
-        "660.000000 \t\t11.613057 \t0.668586 \t0.000000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3, Page 379\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "Fs=100.;        #Solid flowrate in kg/s\n",
-      "ephsilon1=0.55;\n",
-      "ephsilon2=0.5;\n",
-      "p1=120.;         #Pressure at upper level in kPa\n",
-      "rhos=1000.;      #Density of solid in kg/m**3\n",
-      "rhog=1.;         #Density of gas in kg/m**3\n",
-      "gc=1.;           #Conversion factor\n",
-      "g=9.81;          #Acceleration due to gravity in m/s**2\n",
-      "di=0.34;         #Diameter of downcomer in m\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "x=(ephsilon1/ephsilon2)*((1-ephsilon2)/(1-ephsilon1));#To find pressure at lower level using Eqn.(30)\n",
-      "p2=x*p1;#Pressure at lower level using Eqn.(30)\n",
-      "deltap=p2-p1;\n",
-      "ephsilonbar=0.5*(ephsilon1+ephsilon2);\n",
-      "deltah=(deltap*10**3*gc)/(rhos*(1-ephsilonbar)*g);#Static head height from Eqn.(28)\n",
-      "At=0.25*pi*di**2;#Area of downcomer\n",
-      "Gs=Fs/At;#Flux of solids in downcomer\n",
-      "Gg=Gs*(ephsilon1/(1-ephsilon1))*(rhog/rhos)*(x-1);#Required gas aeration rate from Eqn.(31)\n",
-      "Fg=Gg*At;#Flow rate of gas required\n",
-      "\n",
-      "#OUTPUT\n",
-      "print '\\nThe required flow rate of gas required for location of %fm below downcomer is %.4fkg/s'%(deltah,Fg)\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "The required flow rate of gas required for location of 5.722768m below downcomer is 0.0272kg/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4, Page 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "Fs=600;#Solid circulation rate in kg/s\n",
-      "dpbar=60;#Mean size of solids in micrometer\n",
-      "pA=120;#Pressure in vessel A in kPa\n",
-      "pB=180;#Pressure in vessel B in kPa\n",
-      "LfA=8;#Bed height in vessel A in m\n",
-      "LfB=8;#Bed height in vessel B i m\n",
-      "#Bulk densities in kg/m**3\n",
-      "rho12=100;\n",
-      "rho34=400;\n",
-      "rho45=550;\n",
-      "rho67=200;\n",
-      "rho78=200;\n",
-      "rho910=400;\n",
-      "rho1011=400;\n",
-      "rho1112=550;\n",
-      "rho13=100;\n",
-      "deltapdA=7;#Pressure drop across the distributor in regenerator in kPa\n",
-      "deltapdB=8;#Pressure drop across the distributor in reactor in kPa\n",
-      "deltap12=(9+4);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
-      "deltap78=(15+3);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa\n",
-      "deltap45=20;#Friction loss across the reactor's stripper downcomer in kPa\n",
-      "deltap1112=4;#Friction loss across the regenerator's downcomer in kPa\n",
-      "deltapvA=5;#Pressure drop assigned for the control valve in regenerator in kPa\n",
-      "deltapvB=15;#Pressure drop assigned for the control valve in reactor in kPa\n",
-      "deltah12=15;#Height of the riser in m\n",
-      "deltah86=30;#Height of the riser in m\n",
-      "deltah1011=7;#Height difference h10-h11 in m\n",
-      "g=9.81;#Acceleration due to gravity in m/s**2\n",
-      "gc=1;#Conversion factor\n",
-      "pi=3.14;\n",
-      "\n",
-      "#CALCULATION\n",
-      "Gs=900;#From Fig.(8), to find dt\n",
-      "dt=math.sqrt((4/math.pi)*Fs/Gs);#Diameter of the downcomer\n",
-      "#Height of downcomer A from Eqn.(7)\n",
-      "deltahA=(1/(rho1112*g))*((pB-pA)*gc*(10**3)+(deltap12+deltapdB+deltap1112+deltapvA)*gc*10**3-rho12*g*(-deltah12)-rho34*g*(-LfB)-rho1011*g*deltah1011);\n",
-      "#Height of downcomer B from Eqn.(8)\n",
-      "deltahB=(1/(rho45*g))*(-(pB-pA)*gc*10**3+(deltap45+deltapvB+deltap78+deltapdA)*gc*10**3+rho78*g*deltah86+rho910*g*LfA)\n",
-      "\n",
-      "#OUTPUT\n",
-      "print 'Height of downcomer for:'\n",
-      "print '\\tRegenerator:%d m'%deltahA\n",
-      "print '\\tReactor:%.1f m'%deltahB\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Height of downcomer for:\n",
-        "\tRegenerator:20 m\n",
-        "\tReactor:16.7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
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