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{
"metadata": {
"name": "",
"signature": "sha256:e7844fd29e61e44a32fcf2d944211052a5451b3c15cdfd580105fb20e51952a3"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15 : Adsorption"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3.2 pg : 438"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"tE = 33. \t# Time taken for ferric ion to exhaust the bed in min\n",
"tB = 23. \t# Time taken for nickel to break through ferric in min\n",
"l = 120. \t#bed length in cm\n",
"\t\n",
"#Calculations\n",
"Theta = round(2*tB/(tB+tE),1)\n",
"lunused = (1-Theta)*120 \t# cm\n",
"\t\n",
"#Results\n",
"print \"the length of the bed unused is %.1f cm\"%(lunused)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the length of the bed unused is 24.0 cm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3.3 pg : 438"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"tB = 10. \t# min\n",
"tE = 14. \t# min\n",
"l = 0.12 \t#m\n",
"l2 = 10. \t# m\n",
"c = 10000.\n",
"A = 1/10000. \t# m**2\n",
"\t\n",
"#Calculations\n",
"theta = 2*tB/(tB+tE)\n",
"l1 = l*(1-theta)\t# m , length of bed unused in first case\n",
"V1 = c*A*l \t# m**3\n",
"l3 = l2-l1 \t# length of bed unused in second case\n",
"d = math.sqrt(V1*4/(l3*math.pi))\t# m\n",
"V2 = c*(l-l1)*A*l2/l3 \t# volume needed for second case\n",
"\t\n",
"#Results\n",
"print \"The volume of adsorbent needed if the bed is kept 12 cm deep is %.2f m**3\"%(V1)\n",
"print \"The volume of adsorbent needed if the bed length is 10 m long is %.4f m**3\"%(V2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The volume of adsorbent needed if the bed is kept 12 cm deep is 0.12 m**3\n",
"The volume of adsorbent needed if the bed length is 10 m long is 0.1002 m**3\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.4.1 pg : 441"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#intialization of variables\n",
"tB1 = 38. \t# days , breakthrough time\n",
"tE1 = 46. \t# days, exhaustion time\n",
"c = 2. \t# number of times flow doubled\n",
"\t\n",
"#Calculations\n",
"theta1 = 2*tB1/(tB1+tE1)\t# in the first case\n",
"ratio1 = 1-theta1 \t# ratio of unused bed length to total bed length\n",
"ratio2 = ratio1*c\n",
"tB2 = ((1/c)*(tB1 + 0.5*(tE1-tB1)))*ratio2\t#breakthrough time for second case\n",
"tE2 = (c-ratio2)*tB2/ratio2\t#exhaustion time for second case\n",
"\t\n",
"#Results\n",
"print \"The breakthrough time for this case is %.1f days\"%(tB2)\n",
"\n",
"# answer slightly wrong in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The breakthrough time for this case is 4.0 days\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.4.2 pg : 442"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"slope = 0.93/3600 \t# sec**-1\n",
"q0 = 300. \t# 300 times y0 \n",
"E = 0.4 \t # void fraction\n",
"d = 310.*10**-4 \t#cm\n",
"v = 1/60. \t#cm/sec\n",
"Nu = 0.01 \t#cm**2/sec\n",
"D = 5.*10**-6 \t #cm**2/sec\n",
"\t\n",
"#Calculations\n",
"ka1 = slope*q0*(1-E)\t#sec**-1\n",
"k = (D/d)*1.17*((d*v/Nu)**0.58)*((Nu/D)**0.33)\t# cm/sec\n",
"a = (6/d)*(1-E)\t#cm**2/cm**3\n",
"ka2 = k*a\t#sec**-1\n",
"\t\n",
"#Results\n",
"print \"The rate constant is %.3f sec**-1\"%(ka1)\n",
"print \"The rate constant of literature is %.3f sec**-1\"%(ka2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate constant is 0.046 sec**-1\n",
"The rate constant of literature is 0.048 sec**-1\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
