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{
"metadata": {
"name": "",
"signature": "sha256:a951bca59afca6e246f5da2ffc1465ed3d7457760d6d4747a778b92b4eb41d0e"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10 : Absorption"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.1 pg : 313"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"c = 0.92\n",
"F = 93 \t# ft**-1\n",
"nu = 2 \t# cs\n",
"dl = 63 \t# lb/ft**3\n",
"dg = 2.8 \t# lb/ft**3\n",
"G = 23 \t#lb/sex\n",
"\t\n",
"#Calculations\n",
"G11 = c*((dl-dg)**0.5)/(((F)**0.5)*(nu**0.05))\t# lb/ft**2-sec\n",
"A = G/G11\t# ft**2\n",
"d = math.sqrt(4*A/math.pi)\t#ft\n",
"\t\n",
"#Results\n",
"print \"The diameter of the tower is %.1f ft\"%(d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diameter of the tower is 6.4 ft\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.3.1 pg : 318"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"G = 2.3 \t # Gas flow in gmol/sec\n",
"L = 4.8 \t # Liquid flow in gmol/sec\n",
"y0 = 0.0126 \t# entering gas Mole fraction of CO2\n",
"yl = 0.0004 \t# Exiting gas mole fraction of CO2 \n",
"xl = 0. \t# Exiting liquid mole fraction of CO2\n",
"d = 40. \t# Diameter of the tower in cm\n",
"x0star = 0.0080\t# if the amine left in equilibrium with the entering gas would contain 0.80 percent C02\n",
"Kya = 5.*10**-5 \t# Overall M.T.C and the product times the area per volume in gmol/cm**3-sec\n",
"\t\n",
"#Calculations\n",
"A =math.pi*(d**2)/4\n",
"x0 = ((G*(y0-yl))/(L)) + xl \t# Entering liquid mole fraction of CO2\n",
"m = y0/x0star \t# Equilibirum consmath.tant\n",
"c1 = G/(A*Kya)\n",
"c2 = 1/(1-(m*G/L))\n",
"c3 = math.log((y0-m*x0)/(yl-m*xl))\n",
"l = (G/(A*Kya))*(1/(1-((m*G)/L)))*(math.log((y0-m*x0)/(yl-m*xl)))/100 \t#length of the tower in metres\n",
"\t\n",
"#Results\n",
"print \"The length of the tower is %.1f m\"%(l)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The length of the tower is 3.2 m\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.3.2 pg : 319"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"\t\n",
"#initialization of variables\n",
"l = 200. \t# Length of the tower in cm\n",
"d = 60. \t# diameter of the tower\n",
"Lf = 300. \t# Liquid flow in cc/sec\n",
"Kx = 2.2*10**-3 \t# dominant transfer co efficient in liquid in cm/sec\n",
"\t\n",
"#Calculations\n",
"A = math.pi*60*60/4 \t# Area of the cross section in sq cm\n",
"L = Lf/A \t# Liquid flux in cm**2/sec\n",
"ratio = 1/(math.exp((l*Kx)/L))\n",
"percentage = (1-ratio)*100 \t# Percentage removal of Oxygen\n",
"\t\n",
"#Results\n",
"print \"the percentage of oxygen we can remove is %.1f\"%(percentage)\n",
"\n",
"# Rounding of error in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the percentage of oxygen we can remove is 98.4\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.4.1 pg : 324"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\t\n",
"#initialization of variables\n",
"y1in = 0.37 \t# mole fraction of Ammonia in gas mixture entering\n",
"y2in =0.16 \t# mole fraction of nitrogen in gas mixture entering\n",
"y3in = 0.47 \t# mole fraction of hydrogen in gas mixture entering\n",
"x1out = 0.23 \t# mole fraction of Ammonia in liquid coming out\n",
"y1out = 0.01 \t# mole fraction of ammonia in gas coming out\n",
"G0 = 1.20 \t # Gas glow entering in m**3/sec\n",
"Mu = 1.787*0.01*0.3048/2.23 \t# liquid viscousity in american units\n",
"dl = 62.4 \t# Density of liquid in lb/ft**3\n",
"KG = 0.032 \t# Overall m.t.c in gas phase in gas side m/sec\n",
"a = 105 \t# surface area in m**2/m**3\n",
"gc = 32.2 \t# acceleration due to gravity in ft/sec**2\n",
"dg = 0.0326 \t# Density of gas in lb/ft**3\n",
"#Molecular weights of Ammonia , N2 , H2\n",
"M1 = 17.\n",
"M2 = 28.\n",
"M3 = 2.\n",
"Nu = 1. \t# Viscousity \n",
"\t\n",
"#Calculations\n",
"AG0 = (y2in+y3in)*G0/22.4 \t# Total flow of non absorbed gases in kgmol/sec\n",
"ANH3 = y1in*G0/22.4- (y1out*AG0)/(1-y1out) \t# Ammonia absorbed kgmol/sec\n",
"AL0 = ((1-x1out)/x1out)*ANH3 \t# the desired water flow in kgmol/sec\n",
"avg1 = 11.7 \t# Average mol wt of gas\n",
"avg2 = 17.8 \t# avg mol wt of liquid\n",
"TFG = avg1*AG0/(y2in+y3in)\t#Total flow of gas in kg/sec\n",
"TFL = avg2*AL0/(1-x1out)\t#total flow of liquid in kg/sec\n",
"F = 45 \t# Packing factor\n",
"GFF = 1.3*((dl-dg)**0.5)/((F**0.5)*(Nu**0.05))\t# Flux we require in lb/ft**2-sec\n",
"GFF1 = GFF*0.45/(0.3**2) \t# in kg/m**2-sec (answer wrong in textbook)\n",
"Area = TFG/GFF1 \t# Area of the cross section of tower\n",
"dia = (math.sqrt(4*Area/math.pi)) \t# diameter in metres\n",
"HTU = (22.4*AG0/math.pi*dia**2)/(KG*a*4)\n",
"NTU = 5555\n",
"l = HTU*NTU \t# Length of the tower\n",
"\t\n",
"#Results\n",
"print \"The flow of pure water into the top of the tower %.4f kgmol/sec\"%(AL0)\n",
"print \" The diameter of the tower is %.1f m\"%(dia)\n",
"print \" The length of the tower is %.f m\"%(l)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flow of pure water into the top of the tower 0.0652 kgmol/sec\n",
" The diameter of the tower is 0.3 m\n",
" The length of the tower is 10 m\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
