From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001
From: Thomas Stephen Lee
Date: Fri, 4 Sep 2015 22:04:10 +0530
Subject: add/remove/update books

---
 Fluidization_Engineering/ch4.ipynb | 306 -------------------------------------
 1 file changed, 306 deletions(-)
 delete mode 100755 Fluidization_Engineering/ch4.ipynb

(limited to 'Fluidization_Engineering/ch4.ipynb')

diff --git a/Fluidization_Engineering/ch4.ipynb b/Fluidization_Engineering/ch4.ipynb
deleted file mode 100755
index cbd2e1e4..00000000
--- a/Fluidization_Engineering/ch4.ipynb
+++ /dev/null
@@ -1,306 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:66650a76faf5eec3efd8e3c62ecb57addf32e1a6b2cb5d90995ca785777b9427"
- },
- "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": [
-      "\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": [
-      "\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": [
-      "\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": {}
-  }
- ]
-}
\ No newline at end of file
-- 
cgit