{
"metadata": {
"name": "Chapter 14"
},
"nbformat": 2,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"source": [
"# Chapter 14 :- Chemical and Phase Equilibrium"
]
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.1 Page no-688"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# The reaction is CO + .5O2 ----> CO2",
"# Part(a)",
"T = 298.0 # in kelvin",
"Rbar = 8.314 # universal gas constant in SI units",
"# From table A-25",
"",
"hfbarCO2 = -393520.0 # in kj/kmol",
"hfbarCO = -110530.0 # in kj/kmol",
"hfbarO2 = 0 # in kj/kmol",
"deltahbarCO2 = 0 # in kj/kmol",
"deltahbarCO = 0 # in kj/kmol",
"deltahbarO2 = 0 # in kj/kmol",
"sbarCO2 = 213.69 # in kj/kmol.K",
"sbarCO = 197.54 # in kj/kmol.K",
"sbarO2 = 205.03 # in kj/kmol.K",
"# From table A-27",
"logKtable = 45.066",
"# Calculations",
"import math",
"deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)",
"lnK = -deltaG/(Rbar*T)",
"logK = (1/math.log(10))*lnK",
"# Results",
"print '-> Part(a) the value of equilibrium constant expressed as log10K is: '",
"print logK",
"print '-> The value of equilibrium constant expressed as log10K from table A-27 is: '",
"print logKtable",
"",
"# Part(b)",
"T = 2000.0 # in kelvin",
"# From table A-23",
"hfbarCO2 = -393520.0 # in kj/kmol",
"hfbarCO = -110530.0 # in kj/kmol",
"hfbarO2 = 0 # in kj/kmol",
"deltahbarCO2 = 100804-9364 # in kj/kmol",
"deltahbarCO = 65408 - 8669 # in kj/kmol",
"deltahbarO2 = 67881 - 8682 # in kj/kmol",
"sbarCO2 = 309.210 # in kj/kmol.K",
"sbarCO = 258.6 # in kj/kmol.K",
"sbarO2 = 268.655 # in kj/kmol.K",
"# Calculations",
"deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)",
"lnK = -deltaG/(Rbar*T)",
"logK = (1/math.log(10))*lnK",
"# From table A-27",
"logKtable = 2.884",
"# Results",
"print '-> Part(b) the value of equilibrium constant expressed as log10K is: '",
"print logK",
"print '-> The value of equilibrium constant expressed as log10K from table A-27 is: '",
"print logKtable"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> Part(a) the value of equilibrium constant expressed as log10K is: ",
"45.094010685",
"-> The value of equilibrium constant expressed as log10K from table A-27 is: ",
"45.066",
"-> Part(b) the value of equilibrium constant expressed as log10K is: ",
"2.88485359375",
"-> The value of equilibrium constant expressed as log10K from table A-27 is: ",
"2.884"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.2 Page no-690"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# Applying conservation of mass, the overall balanced chemical reaction equation is",
"# CO + .5O2 -------> zCO + (z/2)O2 + (1-z)CO2",
"",
"# At 2500 K, Table A-27 gives",
"log10K = -1.44",
"# Part(a)",
"p = 1.0 # in atm",
"# Calculations",
"import math",
"K = (10.0)**(log10K) # equilibrium constant",
"# Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives",
"z = 0.129",
"yCO = 2.0*z/(2.0 + z)",
"yO2 = z/(2.0 + z)",
"yCO2 = 2.0*(1.0 - z)/(2.0 + z)",
"",
"# Results",
"print '-> Part(a) mole fraction of CO is: ',round(yCO,3)",
"print '-> Mole fraction of O2 is: ',round(yO2,3)",
"print '-> Mole fraction of CO2 is: ',round(yCO2,3)",
"",
"# Part(b)",
"p = 10.0 # in atm",
"# Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives",
"z = 0.062",
"yCO = 2.0*z/(2.0 + z)",
"yO2 = z/(2.0 + z)",
"yCO2 = 2.0*(1.0 - z)/(2.0 + z)",
"",
"# Results",
"print '-> Part(b) mole fraction of CO is: ',round(yCO,3)",
"print '-> Mole fraction of O2 is: ',round(yO2,3)",
"print '-> Mole fraction of CO2 is: ',round(yCO2,3)"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> Part(a) mole fraction of CO is: 0.121",
"-> Mole fraction of O2 is: 0.061",
"-> Mole fraction of CO2 is: 0.818",
"-> Part(b) mole fraction of CO is: 0.06",
"-> Mole fraction of O2 is: 0.03",
"-> Mole fraction of CO2 is: 0.91"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.3 Page no-691"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"yCO = 0.298",
"p = 1 # in atm",
"pref = 1 # in atm",
"# With this value of K, table A-27 gives",
"T = 2881",
"",
"# Calculations",
"# Solving yCO = 2z/(2 + z)",
"z = 2*yCO/(2 - yCO)",
"K = (z/(1-z))*(z/(2 + z))**.5*(p/pref)**.5",
"",
"# Result",
"print '-> The temperature T of the mixture in kelvin is: ',T"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The temperature T of the mixture in kelvin is: 2881"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.4 Page no-692"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# For a complete reaction of CO with the theoretical amount of air",
"# CO + .5 O2 + 1.88N2 -----> CO2 + 1.88N2 ",
"# Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is",
"# CO + .5O2 + 1.88N2 ---> zCO + z/2 O2 + (1-z)CO2 + 1.88N2",
"",
"K = 0.0363 # equilibrium constant the solution to Example 14.2",
"p =1.0 # in atm",
"pref = 1.0 # in atm",
"",
"# Calculations",
"# Solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives",
"z = 0.175",
"yCO = 2.0*z/(5.76 + z)",
"yO2 = z/(5.76 + z)",
"yCO2 = 2.0*(1.0-z)/(5.76 + z)",
"yN2 = 3.76/(5.76 + z)",
"",
"# Results",
"print '-> The mole fraction of CO is: ',round(yCO,3)",
"print '-> The mole fraction of O2 is: ',round(yO2,3)",
"print '-> The mole fraction of CO2 is: ',round(yCO2,3)",
"print '-> The mole fraction of N2 is: ',round(yN2,3)"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The mole fraction of CO is: 0.059",
"-> The mole fraction of O2 is: 0.029",
"-> The mole fraction of CO2 is: 0.278",
"-> The mole fraction of N2 is: 0.634"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.5 Page no-692"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# Applying the conservation of mass principle, the overall dissociation reaction is described by",
"# CO2 ----> zCO2 + (1-z)CO + ((1-z)/2)O2",
"",
"p = 1.0 # in atm",
"pref = 1.0 # in atm",
"# At 3200 K, Table A-27 gives",
"log10k = -.189",
"# Solving k = ((1-z)/2)*((1-z)/(3-z))^.5 gives",
"z = 0.422",
"",
"# Calculations",
"k = 10**log10k",
"# From tables A-25 and A-23",
"hfbarCO2 = -393520.0 # in kj/kmol",
"deltahbarCO2 = 174695-9364 # in kj/kmol",
"hfbarCO = -110530.0 # in kj/kmol",
"deltahbarCO = 109667-8669 # in kj/kmol",
"hfbarO2 = 0 # in kj/kmol",
"deltahbarO2 = 114809-8682 # in kj/kmol",
"hfbarCO2r = -393520.0 # in kj/kmol",
"deltahbarCO2r = 0 # in kj/kmol",
"",
"Qcvdot = 0.422*(hfbarCO2 + deltahbarCO2) + 0.578*(hfbarCO + deltahbarCO) + 0.289*(hfbarO2 + deltahbarO2)- (hfbarCO2r + deltahbarCO2r) ",
"",
"# Result",
"print '-> The heat transfer to the reactor, in kJ per kmol of CO2 entering is: '",
"print Qcvdot"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The heat transfer to the reactor, in kJ per kmol of CO2 entering is: ",
"322385.449"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"source": [
"##<heading2>Example 14.8 Page no-701"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"%matplotlib inline"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"",
"Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].",
"For more information, type 'help(pylab)'."
]
}
],
"prompt_number": 15
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# The ionization of cesium to form a mixture of Cs, Cs+, and e-\u0005 is described by",
"# Cs ----> (1-z)Cs + zCs+ + Ze-",
"",
"K = 15.63",
"z = 0.95",
"pref =1 # in atm",
"# Calculation",
"p = pref*K*((1-z**2)/z**2)",
"",
"# Results",
"print '-> The pressure if the ionization of CS is 95 percent complete is: ',p,'atm'",
"",
"x = []",
"y = []",
"from numpy import linspace ",
"from pylab import plot, show ",
"x = linspace(0,10,100)",
"for i in range(0,100):",
" y.append(i)",
" y[i]= 100*((1/(1+x[i]/K))**0.5)",
"",
"plot(x,y)",
"xlabel(\"Pressure (atm)\")",
"ylabel(\"Ionization\")",
"show()"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The pressure if the ionization of CS is 95 percent complete is: 1.68855955679 atm"
]
},
{
"output_type": "display_data",
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16lWyTfU0ZSQiUobk58PSpaYld3q6mUp68EEICrry19aUkYhIGVK5srnV7fPP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\ngg==\n"
}
],
"prompt_number": 16
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.9 Page no-703"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# The overall reaction can be written as",
"# CO2 + .5O2 + .5N2 ----> aCO + bNO + (1-a)CO2 + .5(1+a-b)O2 + .5(1-b)N2",
"",
"# At 3000 K, Table A-27 provides",
"log10K1 = -0.485 # equilibrium constant of the reaction CO2 <--> CO + .5O2",
"log10K2 = -0.913 # equilibrium constant of the reaction .5O2 + .5N2 <-->NO",
"# Solving equations K1 = (a/(1-a))*((1+a-b)/(4+a))^.5 and K2 = 2b/((1+a-b)*(1-b))^.5",
"a = 0.3745",
"b = 0.0675",
"",
"# Calculations",
"K1 = 10**log10K1",
"K2 = 10**log10K2",
"",
"# Result",
"print '-> The composition of the equilibrium mixture, in kmol per kmol of CO2 present initially, is then 0.3745CO, 0.0675NO, 0.6255CO2, 0.6535O2, 0.4663N2.'"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The composition of the equilibrium mixture, in kmol per kmol of CO2 present initially, is then 0.3745CO, 0.0675NO, 0.6255CO2, 0.6535O2, 0.4663N2."
]
}
],
"prompt_number": 17
},
{
"cell_type": "markdown",
"source": [
"##<heading2> Example 14.10 Page no-707"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# Given:-",
"# With data from Table A-2 at 20\u0004C,",
"vf = 1.0018e-3 # in m^3/kg",
"psat = 0.0239 # in bar",
"p = 1.0 # in bar",
"T = 293.15 # in kelvin",
"Rbar = 8.314 # universal gas constant in SI units",
"M = 18.02 # molat mass of water in kg/kmol",
"e=2.715",
"",
"# Calculations",
"pvbypsat = e**(vf*(p-psat)*10**5/((1000*Rbar/M)*T))",
"percent = (pvbypsat-1)*100",
"",
"# Result",
"print '-> The departure, in percent, of the partial pressure of the water vapor from the saturation pressure of water at 20\u0004 is: ',round(percent,3)"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-> The departure, in percent, of the partial pressure of the water vapor from the saturation pressure of water at 20\u0004 is: 0.072"
]
}
],
"prompt_number": 18
}
]
}
]
}