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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 17 : Two Phase Gas Liquid Systems Saturation Condensation and Vaporization"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.1 Page no. 511\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 1. ;\t\t\t#H2C2O4- [mol]\n",
"ex_O2 = 248. ;\t\t#Excess air- [%]\n",
"f_C = 65/100. ;\t\t# Fraction of Carbon which convert to CO2\n",
"P = 101.3 ;\t\t\t# Atmospheric pressure-[kPa]\n",
"\n",
"# Calculations\n",
"O2_req = F*0.5 ;\t\t\t# O2 required by the above reaction-[mol]\n",
"O2_in = (1. + ex_O2*F/100)*0.5 ;\t\t\t# Mol. of O2 entering\n",
"\n",
"# Use Elemental balance moles of species in output \n",
"n_CO2 = f_C*2 ;\t\t\t# [mol]\n",
"n_H2O = (2*F)/2. ;\t\t# From 2H balance-[mol]\n",
"n_N2 = ((O2_in*0.79)/(0.21)) ;\t\t\t# From 2N balance-[mol]\n",
"n_CO = 2-n_CO2 ;\t\t\t # From C balance-[mol]\n",
"n_O2 = ((4 + O2_in*2)-(n_H2O + n_CO + 2*n_CO2))/2 ;\t\t\t# From O2 balance-[mol]\n",
"total_mol = n_CO2 + n_H2O + n_N2 + n_CO + n_O2 ;\t\t\t# Total moles in output stream-[mol]\n",
"y_H2O = n_H2O/total_mol ;\t# Mole fraction of H2O\n",
"pp_H2O = y_H2O*P ;\t\t\t# Partial pressure of H2O-[kPa]\n",
"\n",
"\n",
"# Results\n",
"print 'Partial pressure of H2O %.2f kPa.'%pp_H2O\n",
"print 'Use partial pressure of H2O to get dew point temperature T from steam table: T = 316.5 K'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Partial pressure of H2O 9.10 kPa.\n",
"Use partial pressure of H2O to get dew point temperature T from steam table: T = 316.5 K\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.2 Page no. 517\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"gas = 1. ;\t\t\t# Entering gas-[g mol]\n",
"T = 26. ;\t\t\t# Temperature (for isothermal process)-[degree C]\n",
"vp = 99.7 ;\t\t\t# vapour pressure of benzene at 26 C-[mm of Hg]\n",
"\n",
"# Analysis of entering gas \n",
"f_C6H6 = 0.018 ;\t\t\t# Mol fraction of benzene\n",
"f_air = 0.982 ;\t\t\t# Mol fraction of air\n",
"mol_C6H6 = 0.018*gas ;\t\t# Moles of benzene-[g mol]\n",
"mol_air = 0.982*gas ;\t\t# Moles of air-[g mol]\n",
"\n",
"# Calculations\n",
"# Analysis of exit gas\n",
"C6H6_rec = 95./100 ;\t\t\t# Fraction of benzene recovered\n",
"C6H6_out = 1-C6H6_rec ;\t\t#Fraction of benzene in exit stream\n",
"C6H6_out = mol_C6H6*C6H6_out ;\t#Moles of benzene in exit stream-[g mol]\n",
"air_out = mol_air ;\t\t\t #Moles of air in exit stream-[g mol]\n",
"total_mol = C6H6_out+air_out ;\t# Total moles in exit stream\n",
"y_C6H6_out = C6H6_out/total_mol ;\t\t\t# Mole fraction of benzene in exit\n",
"P = vp/y_C6H6_out ;\t\t\t # Pressure total of exit\n",
"\n",
"# Results\n",
"print ' Pressure total at exit of compressor %.2e mm of Hg.'%P\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Pressure total at exit of compressor 1.09e+05 mm of Hg.\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.3 Page no. 519\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"ex_air = .4 ;\t\t\t# Fraction of excess air required\n",
"w_C = 12. ;\t\t\t # Mol. wt. of C-[g]\n",
"mol_C = 71./w_C ;\t\t#[kg mol]\n",
"w_H2 = 2.016 ;\t\t\t# Mol. wt. of H2 - [g] \n",
"mol_H2 = 5.6/w_H2;\n",
"air_O2 = 0.21;\t\t\t# Fraction of O2 in air\n",
"air_N2 = 0.79;\t\t\t# Fraction of N2 in air\n",
"\n",
"# Calculations\n",
"CO2_1 = 1. ;\t\t\t# By Eqn. (a) CO2 produced -[kg mol]\n",
"H2O_1 = 2. ;\t\t\t# By Eqn. (a) H2O produced -[kg mol]\n",
"Req_O2_1 = 2. ;\t\t\t# By Eqn. (a) -[kg mol]\n",
"ex_O2_1 = Req_O2_1*ex_air ;\t\t\t# Excess O2 required -[kg mol]\n",
"O2_1 = Req_O2_1 + ex_O2_1 ;\t\t\t # Total O2 required - [kg mol]\n",
"N2_1 = O2_1*(air_N2/air_O2) ;\t\t\t#Total N2 required - [kg mol]\n",
"Total_1 = CO2_1 + H2O_1 + N2_1 + ex_O2_1 ;\t\t\t# Total gas produced- [kg mol]\n",
"\n",
"CO2_2 = 1 ;\t\t\t# By Eqn. (a) CO2 produced -[kg mol]\n",
"H2O_2 = mol_H2/mol_C ;\t\t\t# By Eqn. (a) H2O produced -[kg mol]\n",
"Req_O2_2 = 1 + (mol_H2/mol_C)*(1./2) ;\t\t\t# By Eqn. (b) and (c) -[kg mol]\n",
"ex_O2_2 = Req_O2_2*ex_air ;\t\t\t# Excess O2 required -[kg mol]\n",
"O2_2 = Req_O2_2 + ex_O2_2; \t\t\t# Total O2 required - [kg mol]\n",
"N2_2 = O2_2*(air_N2/air_O2); \t\t\t#Total N2 required - [kg mol]\n",
"Total_2 = CO2_2 + H2O_2 + N2_2 + ex_O2_2 ;\t\t\t# Total gas produced- [kg mol]\n",
"\n",
"P = 100. ;\t\t\t# Total pressure -[kPa]\n",
"p1 = P*(H2O_1/Total_1) ;\t\t\t# Partial pressure of water vapour in natural gas - [kPa]\n",
"Eq_T1 = 52.5 ;\t\t\t# Equivalent temperature -[degree C]\n",
"p2 = P*(H2O_2/Total_2) ;\t\t\t# Partial pressure of water vapour in coal - [kPa]\n",
"Eq_T2 = 35 ;\t\t\t# Equivalent temperature -[degree C]\n",
"\n",
"# Results\n",
"print ' Natural gas Coal'\n",
"print ' ---------------------- --------------------'\n",
"print 'Partial pressure: %.1f kPa %.1f kPa'%(p1,p2 )\n",
"print 'Equivalent temperature: %.1f C %.1f C'%(Eq_T1,Eq_T2 );\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Natural gas Coal\n",
" ---------------------- --------------------\n",
"Partial pressure: 14.0 kPa 5.5 kPa\n",
"Equivalent temperature: 52.5 C 35.0 C\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.4 Page no. 522\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 30. ;\t\t\t# Volume of initial gas-[m**3]\n",
"P_F = 98.6 ;\t\t\t# Pressure of gas-[kPa]\n",
"T_F = 273.+100 ;\t\t\t# Temperature of gas-[K]\n",
"P_p = 109. ;\t\t\t#[kPa]\n",
"T_p = 14.+273 ;\t\t\t# Temperature of gas-[K]\n",
"R = 8.314 ;\t\t\t# [(kPa*m**3)/(k mol*K)] \n",
"# Additional condition\n",
"vpW_30 = 4.24 ;\t\t\t#Vapour pressure-[kPa]\n",
"vpW_14 = 1.60 ;\t\t\t#Vapour pressure-[kPa]\n",
"n_F = (P_F*F)/(R*T_F) ;\t\t\t# Number of moles in F\n",
"\n",
"# Calculations\n",
"# Material balance to calculate P & W\n",
"P = (n_F*((P_F-vpW_30)/P_F))/((P_p-vpW_14)/P_p) ;\t\t\t# P from mat. bal. of air -[kg mol]\n",
"W = (n_F*(vpW_30/P_F))- P*(vpW_14/P_p); \t\t\t# W from mat. bal. of water -[kg mol]\n",
"iW = n_F*(vpW_30/P_F) ;\t\t\t# Initial amount of water -[kg mol]\n",
"fr_con = W/iW ;\t\t\t#Fraction of water condenseed \n",
"\n",
"# Results\n",
"print ' Fraction of water condenseed %.3f.'%fr_con\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Fraction of water condenseed 0.668.\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.5 Page no. 527\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"P = 100. ;\t\t\t# Pressure of air-[kPa]\n",
"T = 20. + 273 ;\t\t\t# Temperature of air-[K]\n",
"R = 8.314 ;\t\t\t# [(kPa*m**3)/(k mol*K)] \n",
"EOH = 6 ;\t\t\t# Amount of ethyl alcohol to evaporate-[kg]\n",
"mw_EOH = 46.07 ;\t\t\t# Mol.wt. of 1 k mol ethyl alcohol-[kg]\n",
"\t\t\t\n",
"# Calculations\n",
"# Additional data needed\n",
"vp_EOH = 5.93 ;\t\t\t# Partial pressure of alcohol at 20 C-[kPa]\n",
"vp_air = P-vp_EOH ;\t\t\t# Partial pressure of air at 20 C-[kPa]\n",
"n_EOH = EOH/mw_EOH ;\t\t\t#Moles of ethyl alcohol -[kg mol]\n",
"n_air = (n_EOH*vp_air)/vp_EOH ;\t\t\t# Moles of air -[kg mol]\n",
"V_air = n_air*R*T/P ;\t\t\t# Volume of air required\n",
"\n",
"# Results\n",
"print ' Volume of air required to evaporate 6 kg of ethyl alcohol is %.1f cubic metre . '%V_air\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Volume of air required to evaporate 6 kg of ethyl alcohol is 50.3 cubic metre . \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 17.6 Page no. 529\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Variables\n",
"P = 760. ;\t\t\t# Pressure -[ mm of Hg]\n",
"vp = 40. ;\t\t\t# vapour pressure of n-heptane-[mm of Hg]\n",
"\n",
"# Calculations\n",
"K = 10**((math.log10(vp/P)-0.16)/1.25) ;\n",
"x = 0.5 ;\t\t\t# mole fraction after t_half\n",
"x0 = 1. ;\t\t\t# initial mole fraction \n",
"t_half = (math.log(x/x0))/(-K);\t\t\t# Time required to reduce the concentration to one-half-[min]\n",
"\n",
"# Results\n",
"print 'Time required to reduce the concentration to one-half is %.1f min. '%t_half\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required to reduce the concentration to one-half is 9.8 min. \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}