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diff --git a/Fluidization_Engineering/ch4.ipynb b/Fluidization_Engineering/ch4.ipynb new file mode 100644 index 00000000..3ff36672 --- /dev/null +++ b/Fluidization_Engineering/ch4.ipynb @@ -0,0 +1,309 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : The Dense Bed" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 106\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Design of a Perforated Plate Distributor\n", + "\n", + "#Variable declaration\n", + "dt=4; #Vessel diameter in m\n", + "Lmf=2; #Length of the bed in m\n", + "ephsilonmf=0.48; #Void fraction of bed\n", + "rhos=1500; #Density of solid in kg/m**3\n", + "rhog=3.6; #Density of gas in kg/m**3\n", + "myu=2E-5; #Viscosity of gas in kg/m s\n", + "po=3; #Pressure of inlet gas in bar\n", + "uo=0.4; #Superficial velocity of gas in m/s\n", + "uorm=40; #Maximum allowable jet velocity from holes in m/s\n", + "g=9.80; #Acceleration due to gravity in m/s**2\n", + "gc=1;\n", + "pi=3.1428;\n", + "\n", + "#CALCULATION\n", + "#Computation of minimum allowable pressure drop through the distributor\n", + "deltapb=((1-ephsilonmf)*(rhos-rhog)*g*Lmf)/gc; #Calculation of pressure drop in bed using Eqn.(3.17)\n", + "deltapd=0.3*deltapb; #Calculation of pressure drop in distributor using Eqn.(3)\n", + "\n", + "#Computation of orifice coefficient\n", + "Ret=(dt*uo*rhog)/myu;\n", + "if Ret>=3000:\n", + " Cd=0.60;\n", + "elif Ret>=2000:\n", + " Cd=0.61;\n", + "elif Ret>=1000:\n", + " Cd=0.64;\n", + "elif Ret>=500:\n", + " Cd=0.68;\n", + "elif Ret>=300:\n", + " Cd=0.70;\n", + "elif Ret>=100:\n", + " Cd=0.68;\n", + "\n", + "#Computation of gas velocity through orifice\n", + "uor=Cd*((2*deltapd)/rhog)**0.5; #Calculation of gas velocity through orifice by using Eqn.(12)\n", + "f=(uo/uor)*100; #Calculation of fraction of open area in the perforated plate \n", + "\n", + "\n", + "#Computation of number of orifices per unit area of distributor\n", + "dor=[0.001,0.002,0.004]; #Different orifice diameters in m\n", + "n=len(dor);\n", + "i=0;\n", + "Nor = [0.,0.,0.]\n", + "while i<n:\n", + " Nor[i]=(uo*4)/(pi*uor*(dor[i])**2);#Calculation of number of orifices by using Eqn.(13)\n", + " i=i+1;\n", + " \n", + "#OUTPUT\n", + "print 'The pressure drop in bed:%fPa'%deltapb\n", + "print 'The minimum allowable pressure drop in distributor:%fPa'%deltapd\n", + "if uor<uorm:\n", + " print 'The gas veleocity of %fm/s is satisfactory'%uor\n", + "else:\n", + " print 'The gas veleocity of %fm/s is not satisfactory'%uor\n", + "\n", + "if f<10:\n", + " print 'The fraction of open area of %f percent is allowable'%f\n", + "else:\n", + " print 'The fraction of open area of %f percent is not allowable'%f\n", + "\n", + "print 'Diameter of orifice(m)',\n", + "print '\\tNumber of orifices per unit area(per sq.m)'\n", + "\n", + "j=0;\n", + "while j<n:\n", + " print '%f'%dor[j],\n", + " print '\\t\\t%f'%Nor[j]\n", + " j=j+1;\n", + "\n", + "print 'This number can be rounded off.'\n", + "print 'Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter %fm'%dor[2]\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pressure drop in bed:15251.308800Pa\n", + "The minimum allowable pressure drop in distributor:4575.392640Pa\n", + "The gas veleocity of 30.250265m/s is satisfactory\n", + "The fraction of open area of 1.322302 percent is allowable\n", + "Diameter of orifice(m) \tNumber of orifices per unit area(per sq.m)\n", + "0.001000 \t\t16829.610145\n", + "0.002000 \t\t4207.402536\n", + "0.004000 \t\t1051.850634\n", + "This number can be rounded off.\n", + "Since orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter 0.004000m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 108\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Design of a Tuyere Distributor\n", + "\n", + "#Variable declaration\n", + "lor=0.1; #Minimum allowable tuyere spacing in m\n", + "uorm=30; #Maximum allowable jet velocity from the tuyere in m/s\n", + "uo=0.4; #Superficial velocity of gas in m/s\n", + "uor=30.2; #Gas velocity through orifice,from Exa 1, in m/s\n", + "Cd=0.6; #Dicharge coefficient from Exa 1\n", + "rhog=3.6 #Density of gas in kg/m**3\n", + "pi=3.1428;\n", + "\n", + "#CALCULATION\n", + "Nor=1/(lor**2); #Calculation of number of orifices per unit area by assuming minimum spacing for tuyeres\n", + "dor=((4/pi)*(uo/uor)*(1/Nor))**0.5; #Calculation of diameter of inlet orifiec by using Eqn.(13)\n", + "\n", + "#Computation of diameter of hole for different number of holes per tuyere\n", + "q=(lor**2)*uo; #Volumetric flow rate in m**3/s\n", + "Nh=[8.,6.,4.]; #Different number of holes per tuyere\n", + "n=len(Nh);\n", + "i=0;\n", + "dh = [0.,0.,0.]\n", + "while i<n:\n", + " dh[i]=((((q/Nh[i])*(4./pi))/uorm)**0.5);#Calculation of diameter of holes\n", + " i=i+1;\n", + "\n", + "deltaph=(rhog/2.)*((uor/Cd)**2)\n", + "\n", + "#OUTPUT\n", + "print 'Number of holes(number of holes/tuyeres)',\n", + "print '\\tDiameter of hole(m)'\n", + "j=0;\n", + "while j<n:\n", + " print '%f'%Nh[j],\n", + " print '\\t\\t\\t\\t\\t%f'%dh[j]\n", + " j=j+1;\n", + "\n", + "print 'The design chosen is as follows'\n", + "print '\\tTuyeres are as shown in Fig.2(b),page 97'\n", + "print '\\tNumber of holes = %f(Since rectangular pitch is chosen for tuyeres)'%Nh[1]\n", + "print '\\tDiameter of hole = %fm'%dh[1]\n", + "print '\\tDiameter of incoming high-pressure-drop orifice = %fm ID'%dor\n", + "print 'Checking the pressure drop in tuyeres'\n", + "print 'Since pressure drop of %.1f Pa gives sufficiently high \\\n", + "distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.'%deltaph\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of holes(number of holes/tuyeres) \tDiameter of hole(m)\n", + "8.000000 \t\t\t\t\t0.004606\n", + "6.000000 \t\t\t\t\t0.005318\n", + "4.000000 \t\t\t\t\t0.006513\n", + "The design chosen is as follows\n", + "\tTuyeres are as shown in Fig.2(b),page 97\n", + "\tNumber of holes = 6.000000(Since rectangular pitch is chosen for tuyeres)\n", + "\tDiameter of hole = 0.005318m\n", + "\tDiameter of incoming high-pressure-drop orifice = 0.012984m ID\n", + "Checking the pressure drop in tuyeres\n", + "Since pressure drop of 4560.2 Pa gives sufficiently high distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 110\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Power Requirement for a Fluidized Coal Combustor(FBC)\n", + "\n", + "#Variable declaration\n", + "deltapd=[3,10] #Distributor pressure drop in kPa\n", + "deltapd2=10.0; #Distributor pressure drop in kPa\n", + "po=101.0; #Entering air pressure in kPa\n", + "To=20.0; #Entering air temperature in degree C\n", + "y=1.4; #Fugacity of air\n", + "deltapb=10; #Pressure drop in bed in kPa\n", + "p3=103; #Pressure at the bed exit in kPa\n", + "F=8; #Feed rate of coal in tons/hr\n", + "H=25; #Gross heatig value of coal in MJ/kg\n", + "Fa=10; #Air required at standard condition in nm**3/kg\n", + "etac=0.75; #Efficiency of compressor\n", + "etap=36; #Efficiency of plant in %\n", + "\n", + "#CALCULATION\n", + "#Calculation of volumetric flow rate of air\n", + "vo=((F*1000)*Fa*((To+273)/273.0))/3600.0;\n", + "\n", + "#Case(a) Distributor Pressure drop = 3kPa and Case(b) Distributor Pressure drop = 10kPa\n", + "n=len(deltapd);\n", + "i=0;\n", + "p1= [0,0]\n", + "p2 = [0,0]\n", + "ws = [0.,0.]\n", + "while i<n:\n", + " p2[i]=p3+deltapb; #Calculation of pressure at the entrance of the bed\n", + " p1[i]=p2[i]+deltapd[i];#Calculation of pressure before entering the bed\n", + " ws[i]=(y/(y-1))*po*vo*((p1[i]/po)**((y-1)/y)-1)*(1.0/etac);#Calculation of power required for the compressor by Eqn.(18) & Eqn.(20)\n", + " i=i+1;\n", + "\n", + "#Case(c) 50% of the required bypassed to burn the volatile gases. Distributor Pressure drop = 3kPa\n", + "#No change in pressure drop from case(a)\n", + "v1=vo/2.0; #New volumetric flow rate of air\n", + "ws1=455/2.0; #Power required for blower for primary air\n", + "ws2=(y/(y-1))*po*v1*((p3/po)**((y-1)/y)-1)*(1/etac);#Power required for blower for bypassed air\n", + "wst=ws1+ws2; #Total power required for the two blowers\n", + "p=((ws[1]-wst)/ws[1])*100; #Saving in power when compared to case(a)\n", + "\n", + "#OUTPUT\n", + "print 'Case(a)'\n", + "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n", + "print '\\tPower required for compressor = %.0f kW'%ws[0]\n", + "print 'Case(b)'\n", + "print '\\tVolumetric flow rate of air = %.2f m**3/hr'%vo\n", + "print '\\tPower required for compressor = %.0f kW'%ws[1]\n", + "print 'Case(c)'\n", + "print '\\tVolumetric flow rate of air = %.3f m**3/hr'%v1\n", + "print '\\tPower required for compressor for primary air = %.1f kW'%ws1\n", + "print '\\tPower required for blower for bypassed air = %.1f kW'%ws2\n", + "print '\\tTotal power required for the two blowers = %.0f kW'%wst\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case(a)\n", + "\tVolumetric flow rate of air = 23.85 m**3/hr\n", + "\tPower required for compressor = 454 kW\n", + "Case(b)\n", + "\tVolumetric flow rate of air = 23.85 m**3/hr\n", + "\tPower required for compressor = 651 kW\n", + "Case(c)\n", + "\tVolumetric flow rate of air = 11.925 m**3/hr\n", + "\tPower required for compressor for primary air = 227.5 kW\n", + "\tPower required for blower for bypassed air = 31.6 kW\n", + "\tTotal power required for the two blowers = 259 kW\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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