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{
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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14 : Extraction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.3.1 pg : 412"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"Rat1 = (6.5/3)*(1-0.47)\t# as Rat = x0/y0\n",
"m = 0.14 \n",
"H = (6.5*10**3)/3600 \t# Extract flow in g/sec\n",
"L = (3*10**3)/3600. \t# Solvent flow in g/sec\n",
"d= 10. \t# cm\n",
"A = 0.25*math.pi*d**2 \t# cm**2\n",
"l = 65 \t# cm\n",
"\t\n",
"#Calculations and Results\n",
"Kya = ((H/(l*A))*(1/(1-((m*H)/L)))*(math.log((1-0.14*Rat1)/(0.47))))*10**3\t# kg/m**3-sec\n",
"print \"The value of Kya is %.1f kg/m**3-sec\"%(Kya)\n",
"Rat2 = (6.5/3)*(1-0.1)\t#For case B\n",
"l2 = l*(math.log(1/((1-0.14*Rat2)/(0.1))))/(math.log(1/((1-0.14*Rat1)/(0.47))))/100\t# m\n",
"print \"The length for 90 percent recovery is %.1f m\"%(l2)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Kya is 0.3 kg/m**3-sec\n",
"The length for 90 percent recovery is 2.2 m\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.4.1 pg : 415"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialization of variables\n",
"m = 0.018 \n",
"H = 450. \t# litres/hr\n",
"L = 37. \t# litres/hr\n",
"Ynplus1byY1 = 100. \n",
"\t\n",
"#Calculations\n",
"E =m*H/L\n",
"nplus1 = math.log((Ynplus1byY1*((1/E)-1))+1)/math.log(1/E)\n",
"n = nplus1 -1\n",
"print \"The number of ideal stages are %.1f\"%(n)\n",
"N = 0.60 \t#Murphree efficienct\n",
"E1 = (m*H/L) + (1/N) - 1\n",
"nplus1 = math.log((Ynplus1byY1*((1/E1)-1))+1)/math.log(1/E1)\n",
"n=nplus1-1\n",
"print \"The number of stages required if Murphree efficiency is 60 percent is %.f\"%(n)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of ideal stages are 2.9\n",
"The number of stages required if Murphree efficiency is 60 percent is 21\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.5.1 pg : 419"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"F = 5. \t#kg feed\n",
"S = 2. \t# kg solvent\n",
"E = F-S \t# kg extract\n",
"W = 1. \t # kg waste\n",
"EG = 80. \t # ppm\n",
"y0 = (100-99)/100. \t# mole fraction of gold left\n",
"y1 = y0*EG*W/S \t # concentration in raffinate\n",
"\t\n",
"#Calculations\n",
"xN = (EG*W - y1*S)/E \t# solvent concentration\n",
"xNminus1 = ((xN*(E+S)) - EG*W)/F\t#feed stage balance\n",
"N = 1 + ((math.log((xN-xNminus1)/(y1))/math.log(F/S)))\t#numner of stages including feed stage\n",
"\t\n",
"#Results\n",
"print \"The number of stages including feed stage is %d\"%(N)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of stages including feed stage is 5\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
