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{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 19 : The Phase Rule and Vapor Liquid Equilibria"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 19.1 Page No. 563\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"N1 = 1.;\n",
"P1 = 1. ;\t\t\t# Number of phases present\n",
"C1 = 1. ;\t\t\t#Number of components present\n",
"F1 = 2.-P1+C1 ;\t\t\t#Number of degree of freedom\n",
"print ' (a) Number of degree of freedom of pure benzene is %i.\\n Therefore %i additional \\\n",
"intensive variables must be specified to fix to fix the system.'%(F1,F1)\n",
"\n",
"\t\t\t# (b)\n",
"N2 = 1.;\n",
"P2 = 2. ;\t\t\t# Number of phases present\n",
"C2 = 1. ;\t\t\t#Number of components present\n",
"F2 = 2.-P2+C2 ;\t\t\t#Number of degree of freedom\n",
"print '(b) Number of degree of freedom of a mixture of ice and water only is %i.\\\n",
" \\nTherefore %i additional intensive variables must be specified to fix the system. '%(F2,F2)\n",
"\n",
"\t\t\t# (c)\n",
"N3 = 2.;\n",
"P3 = 2. ;\t\t\t# Number of phases present\n",
"C3 = 2. ;\t\t\t#Number of components present\n",
"F3 = 2.-P3+C3 ;\t\t\t#Number of degree of freedom\n",
"print '(c) Number of degree of freedom of a mixture of liquid benzene,benzene vapour and\\\n",
" helium gas is %i. \\nTherefore %i additional intensive variables must be specified to fix the system. '%(F3,F3)\n",
"\n",
"\t\t\t# (d)\n",
"N4 = 2.;\n",
"P4 = 2. ;\t\t\t# Number of phases present\n",
"C4 = 2. ;\t\t\t#Number of components present\n",
"F4 = 2.-P4+C4 ;\t\t\t#Number of degree of freedom\n",
"print '(d) Number of degree of freedom of a mixture of salt and water designed to achieve\\\n",
" a specific vapour pressure is %i. \\nTherefore %i additional intensive variables must be\\\n",
" specified to fix the system. '%(F4,F4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" (a) Number of degree of freedom of pure benzene is 2.\n",
" Therefore 2 additional intensive variables must be specified to fix to fix the system.\n",
"(b) Number of degree of freedom of a mixture of ice and water only is 1. \n",
"Therefore 1 additional intensive variables must be specified to fix the system. \n",
"(c) Number of degree of freedom of a mixture of liquid benzene,benzene vapour and helium gas is 2. \n",
"Therefore 2 additional intensive variables must be specified to fix the system. \n",
"(d) Number of degree of freedom of a mixture of salt and water designed to achieve a specific vapour pressure is 2. \n",
"Therefore 2 additional intensive variables must be specified to fix the system. \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 19.2 Page No.564\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"N1 = 5.;\n",
"P1 = 1.; \t\t\t# Number of phases present,here 1 gas \n",
"C1 = 3. ;\t\t\t#Number of independent components present,here 3 because 3 elements(C,O and H)\n",
"F1 = 2-P1+C1 ;\t\t\t#Number of degree of freedom\n",
"print ' (a) Number of degree of gas composed of CO,CO2,H2,H2O and CH4 is %i. \\n \\\n",
"Therefore %i additional intensive variables must be specified to fix the system. '%(F1,F1)\n",
"\n",
"# (b)\n",
"N2 = 4.;\n",
"P2 = 4. ;\t\t\t# Number of phases present,here 3 different solid phases and 1 gas phase\n",
"C2 = 3. ;\t\t\t#Number of components present, here 3 because 3 elements(Zn,O and C) ,you can also use method explained \n",
" #in Appendix L1\n",
"F2 = 2.-P2+C2 ;\t\t#Number of degree of freedom\n",
"print '(b) Number of degree of freedom of a mixture of ZnO(s), C(s) ,CO(g) and Zn(s) is %i. \\n \\\n",
"Therefore %i additional intensive variables must be specified to fix the system. '%(F2,F2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" (a) Number of degree of gas composed of CO,CO2,H2,H2O and CH4 is 4. \n",
" Therefore 4 additional intensive variables must be specified to fix the system. \n",
"(b) Number of degree of freedom of a mixture of ZnO(s), C(s) ,CO(g) and Zn(s) is 1. \n",
" Therefore 1 additional intensive variables must be specified to fix the system. \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 19.3 Page No :576"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from scipy.optimize import fsolve\n",
"import math\n",
"\n",
"# Variables\n",
"P_atm = 1. ;\t\t\t#[atm]\n",
"P = 760. ;\t\t\t#[mm of Hg]\n",
"x_1 = 4./100 ;\t\t\t# Mole fraction of hexane in liquid phase\n",
"# Constant A,B and C for Antoine eqn. of n_hexane \n",
"A1 = 15.8366;\n",
"B1 = 2697.55 ;\n",
"C1 = -48.784;\n",
"# Constant A,B and C for Antoine eqn. of n_octane\n",
"A2 = 15.9798;\n",
"B2 = 3127.60 ;\n",
"C2 = -63.633;\n",
"\n",
"# Calculations\n",
"# Solve for bubble point temperature by eqn. obtained by using Antoine equation\n",
"def f(T):\n",
" return math.exp(A1-(B1/(C1+T)))*x_1 + math.exp(A2-(B2/(C2+T)))*(1-x_1) - P\n",
"T = fsolve(f,390)[0] ;\t\t\t# Bubble point temperature \n",
"\n",
"print 'Bubble point temperature is %.1f K'%T\n",
"\n",
"# Composition of first vapour\n",
"# Get vapour pressure of hexane and octane from Perry, it is\n",
"vp_1 = 3114. ;\t\t\t# vapour pressure of hexane-[mm of Hg]\n",
"vp_2 = 661. ;\t\t\t# vapour pressure of octane-[mm of Hg]\n",
"y_1 = vp_1*x_1/P ;\t\t\t# Mole fraction of hexane in vapour phase\n",
"y_2 = 1- y_1 ;\t\t\t#Mole fraction of octane in vapour phase\n",
"\n",
"# Results\n",
"print ' Composition of first vapour. '\n",
"print 'Component Mole fraction. '\n",
"print 'n_hexane %.3f'%y_1\n",
"print ' n_octane %.3f'%y_2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Bubble point temperature is 393.6 K\n",
" Composition of first vapour. \n",
"Component Mole fraction. \n",
"n_hexane 0.164\n",
" n_octane 0.836\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 19.4 Page no. 577"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# Basis : 100 g solution\n",
"F = 100. ;\t\t\t# Amount of solution-[g]\n",
"P_atm = 1. ;\t\t\t#[atm]\n",
"P = 760. ;\t\t\t# Total pressure -[mm of Hg]\n",
"wf_hex = 68.6/100 ;\t\t\t#Weight fraction of hexane in mixture\n",
"wf_tol = 31.4/100 ;\t\t\t#Weight fraction of toluene in mixture\n",
"mw_hex = 86.17 ;\t\t\t# Mol.wt. of hexane-[g]\n",
"mw_tol = 92.13 ;\t\t\t# Mol.wt. of toluene-[g]\n",
"\n",
"# Calculations\n",
"mol_hex = wf_hex *F/mw_hex ;\t\t\t# moles of hexane-[g mol]\n",
"mol_tol = wf_tol*F/mw_tol ;\t\t\t # moles of toluene-[g mol]\n",
"mol_total = mol_hex + mol_tol ;\t\t\t# Total moles in mixture-[g mol]\n",
"molf_hex = mol_hex/mol_total ;\t\t\t# Mole fraction of hexane \n",
"molf_tol = mol_tol/mol_total ;\t\t\t# Mole fraction of toluene \n",
"# Get vapour pressure of hexane and toluene at 80 deg. C from Perry, it is\n",
"vp_hex = 1020. ;\t\t\t# vapour pressure of hexane-[mm of Hg]\n",
"vp_tol = 290. ;\t\t\t# vapour pressure of toluene-[mm of Hg]\n",
"K_hex = vp_hex/P ;\t\t\t# K-value of hexane\n",
"K_tol = vp_tol/P ;\t\t\t# K-value of toluene\n",
"rec_K_hex = 1/K_hex ;\t\t\t# Reciprocal of K-value of hexane\n",
"rec_K_tol = 1/K_tol ;\t\t\t# Reciprocal of K-value of toluene\n",
"\n",
"# Let L/F = x, then use eqn. 19.11 to find x(L/F) \n",
"def g(x):\n",
" return (molf_hex)/(1-x*(1-rec_K_hex)) + (molf_tol)/(1-x*(1-rec_K_tol))-1\n",
"\n",
"x = fsolve(g,1)[0] ;\t\t\t# L/F value\n",
"\n",
"# Results\n",
"print ' Fraction of liquid(L/F) that will remain at equilibrium after vaporization is %.3f. '%x\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Fraction of liquid(L/F) that will remain at equilibrium after vaporization is 0.744. \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 19.5 Page no. 578\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Vo = 3.0 ;\t\t\t# Initial volume of the solution containing the culture and virus-[L]\n",
"Vp = 0.1 ;\t\t\t# Volume of the polymer solution added to the vessel -[L]\n",
"Kpc = 100. ;\t\t\t# Partition coefficient for virus(cp/cc) between two phases\n",
"\n",
"# Calculations\n",
"Vc = Vo ;\t\t\t# At equilibrium -[L]\n",
"cp_by_co = Vo/(Vp+(Vo/Kpc)) ;\t\t\t\n",
"Fr_rec = cp_by_co*(Vp/Vo) ;\t\t\n",
"\n",
"# Results\n",
"print ' Fraction of the initial virus in the culture phase that is recovered in the polymer phase is %.2f . '%Fr_rec\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Fraction of the initial virus in the culture phase that is recovered in the polymer phase is 0.77 . \n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
