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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 18 : Two Phase Gas Liquid Systems Partial Saturation and Humidity"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 18.1 Page no.539\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"V = 1. ;\t\t\t# Volume of water vapour-[cubic metre]\n",
"rel_h = 43. ;\t\t\t# relative humidity -[%]\n",
"vp_H2O = 1.61 ;\t\t\t# vapour pressure of water at 94 F-[in. of Hg]\n",
"P_H2O = vp_H2O*(rel_h/100) ;\t\t\t# Pressure of water vapour in air-[in. of Hg]\n",
"P = 29.92 ;\t\t\t# [in of Hg]\n",
"T = 94+460. ;\t\t\t# Temperature -[Rankine]\n",
"Ts = 492. ;\t\t\t#Temperature std. -[Rankine]\n",
"mw_H2O = 18. ;\t\t\t# molecular mass of water -[lb]\n",
"\n",
"# Calculations\n",
"H2O = (5280**3*Ts*P_H2O*mw_H2O)/(T*P*359) ;\t\t\t#mass of H2O-[lb]\n",
"# The dew point is temperature at which water vapour in air first condense ,i.e at realative humidity 100 %, therefore\n",
"psat_H2O = P_H2O ;\t\t\t# Saturation pressure of H2O -[in. of Hg]\n",
"\n",
"# Results\n",
"print 'Saturation pressure of H2O %.3f in. of Hg'%psat_H2O\n",
"print 'Use saturation pressure of H2O to get dew point temperature T from steam table: T is about 68-69 F.'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Saturation pressure of H2O 0.692 in. of Hg\n",
"Use saturation pressure of H2O to get dew point temperature T from steam table: T is about 68-69 F.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 18.2 Page no. 541\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"# Data from steam table\n",
"psat_H2O = 31.8 ;\t\t\t# Saturation pressure -[mm of Hg]\n",
"\n",
"#(c)\n",
"H = .0055 ;\t\t\t# Humidity\n",
"mw_H2O = 18. ;\t\t\t# Molecular wt. of water-[lb]\n",
"mw_air = 29. ;\t\t\t# Molecular wt. of air -[lb]\n",
"P = 750. ;\t\t\t# Pressure total -[mm of Hg]\n",
"\n",
"# Calculations\n",
"p_H2O = ((H*mw_air*P)/mw_H2O)/(1+(H*mw_air/mw_H2O)) ;\t\t\t# Partial pressure of water vapour in air-[mm of Hg]\n",
"#(a)\n",
"rel_H = (p_H2O/psat_H2O)*100 ;\t\t\t# relative humidity -[%]\n",
"#(b)\n",
"mol_H = (p_H2O)/(P-p_H2O) ;\t\t\t# Molal humidity\n",
"\n",
"# Results\n",
"print '(a)Relative humidity is %.0f%% .'%rel_H\n",
"print '(b)Molal humidity is %.1e '%mol_H\n",
"print '(c)Partial pressure of water vapour in air is %.1f mm of Hg.'%p_H2O\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Relative humidity is 21% .\n",
"(b)Molal humidity is 8.9e-03 \n",
"(c)Partial pressure of water vapour in air is 6.6 mm of Hg.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 18.3 Page No. 544\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"V_BDA = 1000. ;\t\t\t# Volume of bone dry air(BDA) at 20 C & 108.0 kPa\n",
"mol_V = 22.4 ;\t\t\t# Molar volume of gas at standard condition-[m**3]\n",
"T = 20+273. ;\t\t\t# Temperature of BDA-[K]\n",
"P = 108.0 ;\t\t\t#Pressure of BDA-[kPa]\n",
"Ts = 273. ;\t\t\t# Standard temperature-[K]\n",
"Ps = 101.3 ;\t\t\t#Standard pressure-[kPa]\n",
"W = 0.93 ;\t\t\t# [kg]\n",
"mw_W = 18. ;\t\t\t# mol. wt. of 1kmol water -[kg]\n",
"\n",
"# Calculations\n",
"mol_W = W/mw_W ;\t\t\t# amount of water vapour(W)-[kg mol]\n",
"mol_BDA = (V_BDA*Ts*P)/(T*Ps*mol_V) ;\t\t\t# amount of BDA-[kg mol]\n",
"p_H2O = (mol_W/(mol_W+mol_BDA))*P ;\t\t\t# Partial pressure of H2O-[kPa]\n",
"\n",
"# Get vapour pressure for water at 15 C , namely 1.70 kPa\n",
"psat_H2O = 1.70 ;\t\t\t#vapour pressure for water at 15 C-[kPa]\n",
"rel_H = (p_H2O/psat_H2O) ;\t\t\t#Fractional relative humidity-[]\n",
"\n",
"# Results\n",
"print '(a)Fractional relative humidity of original air was %.3f .'%rel_H\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Fractional relative humidity of original air was 0.074 .\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 18.4 Page no.545\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 1000. ;\t\t\t# Volume of entering moist air at 22 C & 101.0 kPa\n",
"mol_V = 22.4 ;\t\t\t# Molar volume of gas at standard condition-[m**3]\n",
"T_in = 22.+273 ;\t\t\t# Temperature of entering moist air-[K]\n",
"P_in = 101.0 ;\t\t\t#Pressure of entering moist air -[kPa]\n",
"dp_in = 11.+273 ;\t\t\t# Dew point of entering air-[K]\n",
"Ts = 273. ;\t\t\t# Standard temperature-[K]\n",
"Ps = 101.3 ;\t\t\t#Standard pressure-[kPa]\n",
"T_out = 58.+273 ;\t\t\t# Temperature of exiting moist air-[K]\n",
"P_out = 98. ;\t\t\t#Pressure of exiting moist air -[kPa]\n",
"\n",
"# Additional vapour pressure data\n",
"psat_in = 1.31 ;\t\t\t#Vapour pressure of entering moist air -[kPa]\n",
"psat_out = 18.14 ;\t\t\t# Vapour pressure of exiting moist air -[kPa]\n",
"pBDA_in = P_in-psat_in ;\t\t\t# Pressure of entering dry air - [kPa]\n",
"pBDA_out = P_out - psat_out ;\t\t\t# Pressure of exiting dry air - [kPa]\n",
"\n",
"# Calculations\n",
"mol_F = (F*P_in*Ts)/(Ps*T_in*mol_V) ;\t\t\t# Moles of moist air entering-[kg mol]\n",
"mol_P = (mol_F*(pBDA_in/P_in))/(pBDA_out/P_out); \t\t\t#BDA balance- [kg mol]\n",
"mol_W = mol_P-mol_F ;\t\t\t# Total balance -[kg mol]\n",
"mw_BDA = 29. ;\t\t\t# Mol. wt. of dry air\n",
"mw_H2O = 18. ;\t\t\t# Mol. wt. of water vapour\n",
"m_BDA = (mol_F*pBDA_in/P_in)*mw_BDA ;\t\t\t# Mass of dry air entering-[kg]\n",
"m_H2O = (mol_F*psat_in/P_in)*mw_H2O ;\t\t\t# Mass of water vapour entering-[kg]\n",
"wa_in = m_BDA+m_H2O ;\t\t\t#Total wet air entering -[kg]\n",
"H2O_ad = mol_W*mw_H2O/wa_in ;\t\t\t#Water added to each kg of wet air entering the process-[kg]\n",
"\n",
"# Results\n",
"print 'Water added to each kg of wet air entering the process is %.3f kg.'%H2O_ad\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Water added to each kg of wet air entering the process is 0.132 kg.\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 18.5 Page No.547\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"# Given data\n",
"#Basis: F = 29.76 lb mol\n",
"F = 29.76 ;\t\t\t# amount of entering moist air -[lb mol]\n",
"F_rh = 90/100. ;\t\t\t# Relative humidity\n",
"T_in = 100 + 460. ;\t\t\t# Temperature of entering moist air-[Rankine]\n",
"P_in = 29.76 ;\t\t\t#Pressure of entering moist air -[in. of Hg]\n",
"psat_in = 1.93 ;\t\t\t# Saturation pressure from steam table-[in. of Hg]\n",
"T_out = 120 + 460. ;\t\t\t# Temperature of exiting dry air-[Rankine]\n",
"P_out = 131.7 ;\t\t\t#Pressure of exiting dry air -[in. of Hg]\n",
"psat_out = 3.45 ;\t\t\t# Saturation pressure from steam table-[in. of Hg]\n",
"mol_V = 22.4 ;\t\t\t# Molar volume of gas at standard condition-[m**3]\n",
"mw_H2O = 18.02 ;\t\t\t# Mol. wt. of water -[lb]\n",
"mw_air = 29. ;\t\t\t# Mol. wt. of air -[lb]\n",
"p_H2O_in = F_rh*psat_in ;\t\t\t# Partial pressure of water vapour at inlet--[in. of Hg]\n",
"p_air_in = P_in-p_H2O_in ;\t\t\t# Partial pressure of air at inlet--[in. of Hg]\n",
"\n",
"# Calculations\n",
"# Assume condensation takes place , therefore output gas P is saturated,\n",
"P_rh = 1;\t\t\t# Relative humidity of output gas\n",
"p_H2O_out = P_rh*psat_out ;\t\t\t# Partial pressure of water vapour at outlet--[in. of Hg]\n",
"p_air_out = P_out-p_H2O_out ;\t\t\t# Partial pressure of air at outlet--[in. of Hg]\n",
"\n",
"# Get W and P from balance of air and water\n",
"P = (p_air_in*F/P_in)/(p_air_out/P_out) ;\t\t\t# From air balance-[ lb mol]\n",
"W = (p_H2O_in*F/P_in)-(P*p_H2O_out/P_out);\t\t\t# From water balance -[lb mol]\n",
"W_ton = (W*mw_H2O*2000)/(p_air_in*mw_air) ;\t\t\t# Moles of water condenses per ton dry air-[lb mol]\n",
"W_m = mw_H2O*W_ton ;\t\t\t# Mass of water condenses per ton dry air-[lb]\n",
"\n",
"# Results\n",
"# Since W is positive our assumption(condensation takes place ) is right .\n",
"print '(a) Yes water condense out during compression ,since W(%.3f lb mol) is positive our assumption(condensation takes place ) \\nis right .'%W\n",
"print '(b) Amount of water condenses per ton dry air is %.1f lb mol i.e %.0f lb water.'%(W_ton,W_m)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Yes water condense out during compression ,since W(0.983 lb mol) is positive our assumption(condensation takes place ) \n",
"is right .\n",
"(b) Amount of water condenses per ton dry air is 43.6 lb mol i.e 786 lb water.\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}
|
