{ "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": {} } ] }