{
"metadata": {
"name": "",
"signature": "sha256:70e5d31cbd675202c3287b597ce34b28fa488bc5446177836cd0448444371ac9"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 : Chemical Reaction Equilibria"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.6"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"#Given:\n",
"Go_reac = 97540.; \t\t\t#standard free energy of formation of reactant (J/mol)\n",
"Go_pdt = 51310.; \t\t\t#standard free energy of formation of product (J/mol)\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"T = 298.; \t\t\t#temperature (K)\n",
"#Reaction: N2O4(g) --> 2NO2(g)\n",
"\n",
"# Calculations\n",
"#To calculate equilibrium constant\n",
"#Using eq. 9.50 (Page no.413)\n",
"Go = 2*Go_pdt - Go_reac;\n",
"#Using eq. 9.31 (Page no. 406)\n",
"K = math.e**(-Go/(R*T))\n",
"\n",
"# Results\n",
"print 'The equilbrium constant %f'%K\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The equilbrium constant 0.128684\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.7"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"T1 = 298.; \t\t\t#temperature in K\n",
"Hf = -46100.; \t\t\t#standard heat of formation (J/mol)\n",
"Go = -16500.; \t\t\t#standard free energy change (J/mol)\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"T = 500.; \n",
"#Reaction: N2(g) + 3H2(g) --> 2NH3(g)\n",
"#To calculate the equilibrium constant at 500 K\n",
"#Using eq. 9.50 (Page no. 413)\n",
"\n",
"# Calculations\n",
"del_Go = 2*Go;\n",
"import math\n",
"#Using eq. 9.31 (Page no. 406)\n",
"K1 = math.e**(-del_Go/(R*T1)) \t\t\t#equilibrium const at 298 K\n",
"Ho = 2*Hf; \t\t\t#standard heat of reaction\n",
"\n",
"#Using eq. 9.37 (Page no. 411)\n",
"K = K1*(math.e**((-Ho/R)*(1/T - 1/T1)))\n",
"\n",
"# Results\n",
"print 'The equilibrium constant at 500 K is %f'%K\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The equilibrium constant at 500 K is 0.179981\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.8"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variablesa\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"T2 = 317.; \t\t\t#temperature in K\n",
"T1 = 391.; \t\t\t#(K)\n",
"x2 = 0.31; \t\t\t#mol fraction of n-butane at 317 K\n",
"x1 = 0.43; \t\t\t#mol fraction of iso-butane at 391 K\n",
"\n",
"# Calculations\n",
"#To calculate standard free energy change and heat of reaction\n",
"#At 317 K\n",
"K2 = (1-x2)/x2; \t\t\t#equilibrium constant at 317 K\n",
"K1 = (1-x1)/x1; \t\t\t#equilibrium constant at 391 K\n",
"import math\n",
"#Using eq. 9.31 (Page no. 406)\n",
"#Standard free energy change\n",
"G2 = -R*T2*math.log(K2 )\t\t\t#at 317 K (J/mol)\n",
"G1 = -R*T1*math.log(K1 )\t\t\t#at 391 K (J/mol)\n",
"\n",
"#Using eq. 9.37 (Page no. 411)\n",
"Ho = -math.log(K2/K1)*R/(1/T2 - 1/T1)\n",
"\n",
"# Calculations\n",
"print 'Standard free energy change of the reaction'\n",
"print ' At 317 K is %f J/mol'%G2\n",
"print ' At 391 K is %f J/mol'%G1\n",
"print ' Average value of heat of reaction is %f J/mol'%Ho\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Standard free energy change of the reaction\n",
" At 317 K is -2108.744820 J/mol\n",
" At 391 K is -916.234397 J/mol\n",
" Average value of heat of reaction is -7217.201631 J/mol\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"To = 298.; \t\t\t#temperature in K\n",
"T = 700.; \t\t\t#(K)\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"Hf = -46100.; \t\t\t#standard heat of formation (J/mol)\n",
"Gf = -16500.; \t\t\t#standard free energy of formtion of ammonia (J/mol)\n",
"\n",
"# Calculations\n",
"Ho = 2*Hf;\n",
"Go = 2*Gf;\n",
"alpha = 2*29.75 - 27.27 - 3*27.01;\n",
"betta = (2*25.11 - 4.93 - 3*3.51)*10**-3;\n",
"import math\n",
"#Using eq. 9.46 (Page no. 412)\n",
"del_H = Ho - alpha*To - (betta/2)*To**2;\n",
"#Using eq. 9.48 (Page no. 413)\n",
"A = -(Go - del_H + alpha*To*math.log(To) + (betta/2)*To**2)/(R*To)\n",
"\n",
"#Using eq. 9.47 and 9.48 (Page no. 412)\n",
"K = math.e**((-del_H/(R*T)) + (alpha/R)*math.log(T) + (betta/(2*R))*T + A)\n",
"G = del_H - alpha*T*math.log(T) -(betta/2)*T**2 - A*R*T;\n",
"\n",
"# Results\n",
"print 'At 700 K'\n",
"print ' Equilibrium constant is %3.2e'%K\n",
"print ' Standard free energy change is %f J/mol'%G\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"At 700 K\n",
" Equilibrium constant is 9.99e-05\n",
" Standard free energy change is 53606.315911 J/mol\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.10"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"T = 600.; \t\t\t#temperature in K\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"\n",
"#Gibbs free energy at 600 K (J/mol K)\n",
"Gc = -203.81; \t\t\t#for CO\n",
"Gh = -136.39; \t\t\t#for hydrogen\n",
"Gm = -249.83; \t\t\t#for methanol\n",
"\n",
"#Heats of formation at 298 K (J/mol)\n",
"Hc = -110500.; \t\t\t#for CO\n",
"Hm = -200700.; \t\t\t#for methanol\n",
"import math\n",
"\n",
"# Calculations\n",
"#To calculate equilibrium constant at 600 K\n",
"Go = T*((Gm-Gc-(2*Gh)) + (1/T)*(Hm-Hc))\n",
"\t\t\t#Using eq. 9.31 (Page no. 406)\n",
"K = math.e**(-Go/(R*T))\n",
"\n",
"# Results\n",
"print 'Equilibrium constant is %4.3e'%K\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Equilibrium constant is 1.018e-04\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.11"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"T = 500.; \t\t\t#temperature in K\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"\n",
"#Free energy at 500 K (J/mol K)\n",
"Fn = -177.5; \t\t\t#for nitrogen\n",
"Fh = -116.9; \t\t\t#for hydrogen\n",
"Fa = -176.9; \t\t\t#for ammonia\n",
"\n",
"#The function (Ho at 298 K - Ho at 0 K) [J/mol]\n",
"Hn = 8669.; \t\t\t#for nitrogen\n",
"Hh = 8468.; \t\t\t#for hydrogen\n",
"Ha = 9920.; \t\t\t#for methanol\n",
"\n",
"#Free energy of formation at 298 K (J/mol)\n",
"Hf = -46100.;\n",
"import math\n",
"#To calculate equilibrium constant at 500 K\n",
"\n",
"# Calculations\n",
"#Using eq. 9.53 (Page no. 414)\n",
"sum_F = (2*Fa - Fn - 3*Fh) - (2*Ha - Hn - 3*Hh)/T; \t\t\t#(J/mol K)\n",
"#Using eq. 9.57 (Page no.415)\n",
"Go = T*(sum_F + 2*Hf/T)\n",
"K = math.e**(-Go/(R*T))\n",
"\n",
"# Results\n",
"print 'Equilibrium constant is %f'%K\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Equilibrium constant is 0.108493\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"P1 = 1.; \t\t\t#pressure (bar)\n",
"P2 = 2.; \t\t\t#(bar)\n",
"x1 = 0.15; \t\t\t#mol fraction of polymer at 1 bar\n",
"x2 = 0.367; \t\t\t#mol fraction of polymer at 2 bar\n",
"\n",
"# Calculations\n",
"import math\n",
"n = round((math.log(x2/x1)+math.log(2))/(math.log(2)-math.log((1-x1)/(1-x2))))\n",
"\n",
"# Results\n",
"print 'The value of n is %i'%n\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of n is 4\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from scipy.optimize import root\n",
"\n",
"# Variables\n",
"#Reaction: N2 + 3H2 --> 2NH3\n",
"K = 2*10**-4; \t\t\t#equilibrium constant of reaction\n",
"P = 20; \t\t\t#(bar)\n",
"\n",
"# Calculations and Results\n",
"#e(4-2e)/(1-e)**2 = 0.73485\n",
"#e = poly(0,'e')\n",
"def f(e):\n",
" return 2.73845*e**2 - 5.4697*e + 0.73485;\n",
"x = root(f,0)\n",
"\n",
"print '(a) Percentage conversion is %f percent'%(x.x[0]*100)\n",
"\n",
"#(b)\n",
"P = 200; \t\t\t#(bar)\n",
"\n",
"#e(4-2e)/(1-e)**2 = 7.3485\n",
"\n",
"def f2(e):\n",
" return 9.3485*e**2 - 18.697*e + 7.3485;\n",
"x = root(f2,0)\n",
"print ' (b) Percentage conversion is %f percent'%(x.x[0]*100)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Percentage conversion is 14.485445 percent\n",
" (b) Percentage conversion is 53.746561 percent\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.14"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"K = 1; \t\t\t#equilibrium constant for reaction\n",
"\n",
"# Calculations and Results\n",
"e = 1./2;\n",
"print '(a) Fractional dissociation of steam is %i percent'%(e*100)\n",
"\n",
"#On solving we get\n",
"e = 1./2;\n",
"print ' (b) After dilution fractional distillation of steam is %i percent'%(e*100)\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Fractional dissociation of steam is 50 percent\n",
" (b) After dilution fractional distillation of steam is 50 percent\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.15"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from scipy.optimize import root\n",
"\n",
"# Variables\n",
"K = 2.*10**-4; \t\t\t#equilibrium constant of reaction\n",
"P = 20.; \t\t\t#pressure in bar\n",
"\n",
"\n",
"\n",
"def f(e):\n",
" return 1.3674*e**2 - 3.7348*e + 0.3674;\n",
"x = root(f,0)\n",
"\n",
"# Results\n",
"print 'Percentage coversion in presence of argon is %f percent'%(x.x[0]*100)\n",
"print ' while in absence of argon is 14.48 percent' \t\t\t#From example 9.13\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage coversion in presence of argon is 10.219587 percent\n",
" while in absence of argon is 14.48 percent\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"P = 1.; \t\t\t#pressure in bar\n",
"K = 1.; \t\t\t#equilibrium constant of reaction\n",
"\n",
"e)(2-e)] = K = 1% so\n",
"\n",
"e = 2./3;\n",
"print '(a). The conversion of steam is %f percent'%(e*100)\n",
"\n",
"#(b). CO supplied is only 50% of the theoretical requirement\n",
"#Mole fraction of components\n",
"#CO: (0.5-e)/1.5\n",
"#H20: (1-e)/1.5\n",
"#CO2: e/1.5\n",
"#H2: e/1.5\n",
"\n",
"#e**2/[(0.5-e)(1-e)] = K = 1\n",
"#1.5e-0.5 = 1\n",
"e = 0.5/1.5;\n",
"print ' (b). Percentage conversion of steam is %f percent'%(e*100)\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a). The conversion of steam is 66.666667 percent\n",
" (b). Percentage conversion of steam is 33.333333 percent\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"K = 1.; \t\t\t#equilibrium constant of reaction\n",
"\n",
"# Calculations\n",
"e = 1./3;\n",
"\n",
"# Results\n",
"print 'Percentage conversion of steam is %f percent'%(e*100)\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage conversion of steam is 33.333333 percent\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"#Reaction: CO(g) + 2H2(g) --> CH3OH(g)\n",
"Kf = 4.9*10**-5;\n",
"Kfi = 0.35;\n",
"P = 300.; \t\t\t#pressure in bar\n",
"n_CO = 25.;\n",
"n_H2 = 55.;\n",
"n_inert = 20.;\n",
"v = -1-2+1; \t\t\t#change in number of moles in reaction\n",
"\n",
"# Calculations\n",
"#Mole fractions in the equilibrium mixture\n",
"#CO = (25-e)/(100-2e)\n",
"#H2 = (55-2e)/(100-2e)\n",
"#CH3OH = e/(100-2e)\n",
"\n",
"Ky = (Kf/Kfi)*P**(-v)\n",
"\t\t\t#[e/(100-2e)]/[(25-e)/(100-2e)][(55-2e)/(100-2e)]**2 = Ky% so\n",
"\n",
"def f(e):\n",
" return (4+4*Ky)*e**3 - (400+320*Ky)*e**2 + (10000+8525*Ky)*e - 75625*Ky\n",
"\n",
"x = root(f,0)\n",
"conv = x.x[0]/n_CO; \t\t\t#first two roots are complex\n",
"\n",
"# Results\n",
"print 'Percentage conversion of CO is %f percent'%(conv*100)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage conversion of CO is 61.015734 percent\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"#Reaction: 1/2N2 + 3/2H2 --> NH3\n",
"Kp = 1.25*10**-2 ;\t\t\t#equilibrium constant\n",
"P = 50; \t\t\t#pressure in bar\n",
"v = 1-(3./2)-(1./2) \t\t\t#change in number of moles in reaction\n",
"\n",
"#Initial composition of gas mixture\n",
"n_h = 60.;\n",
"n_n = 20.;\n",
"n_inert = 100-n_h-n_n;\n",
"\n",
"# Calculations\n",
"\n",
"Ky = Kp*(P**-v)\n",
"#e/(100-e)/[(20-(e/2)]**1/2[{60-(3e/2)}/(100-e)]**3/2 = Ky\n",
"#e = poly(0%'e'\n",
"\n",
"def f(e):\n",
" return (1.6875*Ky**2-1)*e**4 - (270*Ky**2+200)*e**3 + (16200*Ky**2-10000)*e**2 - (334800*Ky**2)*e + 4320000*Ky**2;\n",
"x = root(f,0)\n",
"e = x.x[0]\n",
"\n",
"#x(4) being the only positive root is the percentage conversion\n",
"#Mole fractions in equilibrium mixture\n",
"x_n = (20-(e/2))/(100-e)\n",
"x_h = (60-3*(e/2))/(100-e)\n",
"x_a = e/(100-e)\n",
"x_inert = 1 - x_n - x_h - x_a;\n",
"\n",
"# Results\n",
"print 'Composition of gas leaving the reactor is'\n",
"print ' Nitrogen : %f percent'%(x_n*100)\n",
"print ' Hydrogen : %f percent'%(x_h*100)\n",
"print ' Ammonia : %f percent'%(x_a*100)\n",
"print ' Inert gas : %f percent'%(x_inert*100)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Composition of gas leaving the reactor is\n",
" Nitrogen : 17.048802 percent\n",
" Hydrogen : 51.146406 percent\n",
" Ammonia : 9.837327 percent\n",
" Inert gas : 21.967465 percent\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"P = 85.; \t\t\t#pressure in bar\n",
"n_e = 0.015; \t\t\t#mol percent of ethanol\n",
"n_w = 0.95; \t\t\t#mole percent of water\n",
"n_a = 0.48; \t\t\t#mol percent of ethylene in vapour phase\n",
"M = 18.; \t\t\t#molecular mass of water\n",
"fc = 0.9; \t\t\t#fugacity coeffecient for ethylene\n",
"\n",
"#To evaluate the equilibrium constant\n",
"#K = a_c/(a_a*a_b)\n",
"# Calculations\n",
"m_e = n_e/(n_w*M*10**-3) \t\t\t#mol/kg water\n",
"a_c = m_e;\n",
"fa = fc*n_a*P; \t\t\t#bar\n",
"a_a = fa;\n",
"\n",
"#Since mol fraction of water is close to unity% so fugacity coeffecient of water is assumed to be 1\n",
"a_b = n_w;\n",
"K = a_c/(a_a*a_b)\n",
"\n",
"# Results\n",
"print 'The equilibrium constant is %5.4e (mol C2H4)/(kg water bar)'%K\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The equilibrium constant is 2.5146e-02 (mol C2H4)/(kg water bar)\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.21"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"T = 1000.; \t\t\t#temperature of reaction in K\n",
"P = 1.; \t\t\t#pressure in bar\n",
"R = 8.314; \t\t\t#ideal gas constant\n",
"import math\n",
"\n",
"#Function for standard free energy of the reaction\n",
"def G(T):\n",
" y = 1.8856*10**5 - 243.42*T + 11.8478*T*math.log(T) - 3.1045*10**-3*T**2 + 1.7271*10**-6*T**3 - (4.1784*10**5)/T\n",
" return y\n",
"\n",
"# Calculations and Results\n",
"K = math.e**(-Go/(R*T))\n",
"#Using eq. 9.75 (Page no. 432)\n",
"p_CO2 = K; \t\t\t#decomposition pressure\n",
"print 'Decomposition pressure of limestone at 1000 K s %f bar'%p_CO2\n",
"\n",
"#At pressure = 1 bar\n",
"K = 1.;\n",
"Go = 0.; \t\t\t#since K = 1\n",
"\n",
"T = 1160.; \t\t\t#assumed temperature (K)\n",
"flag = 1.;\n",
"while(flag==1):\n",
" res = round(G(T))\n",
" if(res<=0):\n",
" flag = 0;\n",
" else:\n",
" T = T+1;\n",
"\n",
"print 'Decomposition temperature at 1 bar is %i K'%T\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Decomposition pressure of limestone at 1000 K s 0.048344 bar\n",
"Decomposition temperature at 1 bar is 1169 K\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"K = 0.403; \t\t\t#equilibrium constant of reaction\n",
"T = 1200.; \t\t\t#temperature of reaction (K)\n",
"To = 273.; \t\t\t#standard temperature (K)\n",
"Vo = 22.4*10**-3; \t\t\t#molar volume at STP \n",
"M = 55.8; \t\t\t#molecular mass of iron\n",
"n = 100.; \t\t\t#moles of gas entering\n",
"n_C = 20.; \t\t\t#moles of carbon mono oxide\n",
"n_N = 80.; \t\t\t#moles of nitrogen\n",
"\n",
"\n",
"# Calculations\n",
"#Let e be the extent of reaction\n",
"#Mole fractions in equilibrium mixture\n",
"#CO = (20-e)/100\n",
"#CO2 = e/100\n",
"#e/(20-e) = K\n",
"e = (20*K)/(1+K)\n",
"n_CO2 = e; \t\t\t#moles of CO2 at equilibrium\n",
"n_Fe = n_CO2; \t\t\t#by stoichiometry\n",
"V = (n*Vo*T)/To; \t\t\t#volume of 100 mol of gas at 1200 K and 1 bar\n",
"\n",
"#Let m be iron produced per 100 m**3 gas\n",
"m = (n_Fe*100*M)/V;\n",
"\n",
"# Results\n",
"print 'Iron produced per 100 cubic m of gas is %f kg'%(m/1000)\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Iron produced per 100 cubic m of gas is 3.255704 kg\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from scipy.optimize import fsolve\n",
"\n",
"# Variables\n",
"P = 1.; \t\t\t#pressure in bar\n",
"K1 = 0.574; \t\t\t#equilibrium constant for eq. 9.88 (Page no. 437)\n",
"K2 = 2.21; \t\t\t#equilibrium constant for eq. 9.89 (Page no. 437)\n",
"\n",
"v1 = 1+3-1-1;\n",
"v2 = 1+1-1-1;\n",
"Ky1 = K1*P**-v1;\n",
"Ky2 = K2*P**-v2;\n",
"\n",
"# Calculations\n",
"#mole fractions in equilibrium mixture are:\n",
"#CH4: (1-e1)/(6+2e1)\n",
"#H2O: (5-e1-e2)/(6+2e1)\n",
"#CO: (e1-e2)/(6+2e1)\n",
"#H2: (3e1+e2)/(6+2e1)\n",
"#CO2: e2/(6+2e1)\n",
"\n",
"#For 1st reaction:\n",
"#Ky1 = [(e1-e2)(3e1+e2)**3]/[(1-e1)(5-e1-e2)(6+2e1)**2]\n",
"#For 2nd reaction:\n",
"#Ky2 = [e2(3e1+e2)]/[(e1-e2)(5-e1-e2)]\n",
"#on solving% we get:\n",
"\n",
"def f2(e):\n",
" f_1 = ((e[0]-e[1])*(3*e[0]+e[1])**3)/((1-e[0])*(5-e[0]-e[1])*(6+2*e[0])**2)-Ky1\n",
" f_2 = (e[1]*(3*e[0]+e[1]))/((e[0]-e[1])*(5-e[0]-e[1]))-Ky2\n",
" y = [f_1,f_2]\n",
" return y\n",
"eo = [0.9,0.6]; \t\t\t#initial guesses\n",
"e = fsolve(f2,eo)\n",
"\n",
"\t\t\t#Mole fraction of components:\n",
"n_m = (1-e[0])/(6+2*e[0])\n",
"n_w = (5-e[0]-e[1])/(6+2*e[0])\n",
"n_CO = (e[0]-e[1])/(6+2*e[0])\n",
"n_h = (3*e[0]+e[1])/(6+2*e[0])\n",
"n_c = e[1]/(6+2*e[0])\n",
"\n",
"# Results\n",
"print 'Mole fraction of the components are:'\n",
"print ' Methane = %f'%n_m\n",
"print ' Water = %f'%n_w\n",
"print ' Carbon monoxide = %f'% n_CO\n",
"print ' Hydrogen = %f'%n_h\n",
"print ' Carbon dioxide = %f'%n_c\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mole fraction of the components are:\n",
" Methane = 0.011240\n",
" Water = 0.441597\n",
" Carbon monoxide = 0.035687\n",
" Hydrogen = 0.430593\n",
" Carbon dioxide = 0.080883\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"#A system consisting of CO% CO2% H2% H2O% CH4\n",
"\n",
"\n",
"r = 2.;\n",
"C = 5.; \t\t\t#no. of components\n",
"pi = 1.; \t\t\t#no. of phases\n",
"\n",
"# Calculations\n",
"#From eq. 9.90 (Page no. 438)\n",
"F = C-pi-r+2;\n",
"\n",
"# Results\n",
"print 'The number of degrees of freedom are %i'%F\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of degrees of freedom are 4\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n"
],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}