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diff --git a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch9.ipynb b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch9.ipynb index 1e6bece8..cadd834d 100644 --- a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch9.ipynb +++ b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch9.ipynb @@ -35,22 +35,14 @@ "\n", "import math \n", "\n", - "# Variables\n", "x_isopropanol = 0.4720\n", "x_water = 0.5280\n", - "# From the table A.7 (page 427) reported in the book the Van Laar coefficients for isopropanol-water system at 1atm are given by\n", "A = 1.0728\n", "B = 0.4750\n", "\n", - "# Calculations\n", - "# Van Laar equations are given \n", - "# math.log10(Y_a) = A*x_b**(2)/(A/B*x_a+x_b)**(2)\n", - "# math.log10(Y_b) = B*x_a**(2)/(B/A*x_b+x_a)**(2)\n", - "# We calculate Y_isopropanol and Y_water as\n", "Y_isopropanol = 10**(A*x_water**(2)/(A/B*x_isopropanol+x_water)**(2))\n", "Y_water = 10**(B*x_isopropanol**(2)/(B/A*x_water+x_isopropanol)**(2))\n", "\n", - "# Results\n", "print \" Value of the liquid-phase activity coefficient for isopropanol is %f\"%(Y_isopropanol)\n", "print \" And value of the liquid-phase activity coefficient for water is %f\"%(Y_water)\n" ], @@ -87,23 +79,14 @@ "'''\n", "import math \n", "\n", - "# Variables\n", - "# Recieving the VLE data from the example 8.2, we have\n", "x_acetone = 0.05\n", "x_water = 0.95\n", - "# And the activity coefficient is given by\n", "y_acetone = 7.04\n", "y_water = 1.01\n", - "# we hve the relation g_E/RT = summation(x_i*math.log(y_i))\n", - "# let C = g_E/RT , so\n", "\n", - "# Calculations\n", "C = (x_acetone*math.log(y_acetone)+x_water*math.log(y_water))\n", - "# Now let M = (g_E/RT )/(x_acetone*x_water)\n", - "# So\n", "M = C/(x_acetone*x_water)\n", "\n", - "# Results\n", "print \"The value of g_E/RT for acetone-water solution at 1 atm pressure is %f\"%(C)\n", "print \"The value of g_E/RT)/x_a*x_b) for acetone-water solution at 1 atm pressure is %f\"%(M)\n" ], @@ -133,22 +116,16 @@ "cell_type": "code", "collapsed": false, "input": [ - "# calculate Values of the constants in Morgules equation for acetone-water\n", "\n", "import math \n", "\n", - "# Variables\n", "y_acetone_infinity = 10.\n", "y_water_infinty = 5.\n", "Pressure = 1. #[atm]\n", "\n", - "# Calculations\n", - "# From equation 9.L and 9.M (page 224) as reported in the book, we have \n", - "# Constants in morgules equation b and c as\n", "b = math.log(y_acetone_infinity)\n", "c = math.log(y_water_infinty)\n", "\n", - "# Results\n", "print \"Values of the consmath.tants in Morgules equation for acetone-water at 1 atm are b = %f\"%(b)\n", "print \" and c = %f\"%(c)\n" ], @@ -178,52 +155,30 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The activity coefficient of ethanol\n", "\n", "import math \n", "\n", - "# Variables\n", "P_1 = 10. #[atm]\n", "x_a_1 = 0.1238 # mole fraction of ethanol at 10 atm pressure\n", "Temp = 273.15+85.3 # [K]\n", "R = 0.08206 #[(L*atm)/(mol*K)]\n", "P_0 = 1. #[atm]\n", - "# so\n", "delta_P = (P_1-P_0) #[atm]\n", - "# Molecular weight of ethanol and water are respectively\n", "M_ethanol = 46. #[g/mol]\n", "M_water = 18. #[g/mol]\n", "\n", - "# Calculations\n", - "# Now changing the mol fraction of ethanol in the wt fraction \n", "m_a_1 = x_a_1*M_ethanol/(x_a_1*M_ethanol+(1-x_a_1)*M_water)\n", - "# From example 8.9(page 188) we know that at this T and 1 atm and x_a_0, activity coefficient for ethanol \n", "y_ethanol_0 = 2.9235\n", - "# Now from figure 6.15(page 129), we read that at 20C and m_a_1 mass fraction ethanol ,\n", "v_ethanol_1 = 1.16 #[cm**(3)/g]\n", - "# Similarily for mass fraction corresponding to mole fraction x_a_1 \n", "v_ethanol_0 = 1.27 #[cm**(3)/]\n", - "# Difference of thes etwo values is \n", "v = v_ethanol_1-v_ethanol_0 #[cm**(3)/g]\n", "v = v*46. #[L/g]\n", - "# If we assume that this value is more or less independent of temperature, we can use it as the corresponding value at 85.3C, and compute \n", - "# From equation 7.31(page 225)\n", - "# d(math.log(y_i))/dP = (v_1-v_0)/(R*T) at consmath.tant temperature and mole fraction \n", - "# Let d(math.log(y_i))/dP = C, then\n", "C = (v_ethanol_1-v_ethanol_0)/(R*Temp)\n", "\n", - "# Also we can have \n", - "# delta_math.log(y_i) = (d(math.log(y_i))/dP)*delta_P\n", - "# or \n", - "# delta_math.log(y_i) = C*delta_P\n", - "# and delta_math.log(y_i) = math.log(y_ehmath.tanol_1)-math.log(y_ethanol_0)\n", - "# So\n", "y_ethanol_1 = math.exp(math.log(y_ethanol_0)+C*delta_P)\n", "\n", - "# Results\n", "print \"The activity coefficient of ethanol in the solution at 10 atm pressure is %f\"%(y_ethanol_1)\n", "\n", - "# Note : Answer is different because of rouding error. Please calculate manually.\n" ], "language": "python", "metadata": {}, @@ -250,39 +205,26 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The activity coefficient of ethanol\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", - "# Variables\n", "x_ethanol = 0.1238\n", "Temp_1 = 273.15+85.3 #[K]\n", "P = 1. #[atm]\n", "Temp_2 = 273.15+70 #[K]\n", "R = 8.314 #[j/(mol*K)]\n", - "# From example 8.9, at temperature 85.3C the activity coefficient is \n", "y_ethanol_1 = 2.9235\n", - "# From figure 9.5[4] (page 227) as reported in the book, we read the value of (h_i_average-h_i_0) at temperatures 90C and 70C for ethanol.\n", - "# which are respectively\n", "delta_h_2 = 0.2 #[kJ/mol]\n", "delta_h_1 = 1.0 #[kJ/mol]\n", "\n", - "# Calculations\n", - "# Taking the average of these two values we have \n", "delta_h_average = (delta_h_1+delta_h_1)/2*1000. #[J/mol]\n", - "# From the equation 7.32 (page 225) reported in the book \n", - "# d(math.log(y_i))/dT = (h_i_average-h_i_0)/(R*T**(2)) at consmath.tant pressure and mole fraction\n", - "# So\n", - "# it can be taken approximately as \n", - "# So\n", "\n", "def f7(T): \n", "\t return (1/T**(2))\n", "\n", "y_ethanol_2 = y_ethanol_1*math.exp((delta_h_average/R)* (quad(f7,Temp_1,Temp_2))[0])\n", "\n", - "# Results\n", "print \"The activity coefficient for ethanol in the solution at 70 deg C and 1 atm is %f\"%(y_ethanol_2)\n", "\n" ], @@ -311,54 +253,32 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find activity coefficient of acetone\n", "\n", "import math \n", "\n", - "# Variables\n", - "# In this solution we will give the identity to the three species as\n", - "# a- Acetone \n", - "# b- Methanol\n", - "# c- Water\n", - "# Given\n", "x_a = 0.1200\n", "x_b = 0.1280\n", "x_c = 0.7520\n", "Temp = 66.70 #[C]\n", "P = 1. #[atm] pressure\n", - "# As reported in the book that from [5] we get the following values \n", - "# acetone-methanol(a-b)\n", "A_ab = 0.2634\n", "A_ba = 0.2798\n", - "# acetone-water(a-c)\n", "A_ac = 0.9709\n", "A_ca = 0.5579\n", - "# methanol-water(b-c)\n", "A_bc = 0.3794\n", "A_cb = 0.2211\n", "\n", - "# Calculations\n", - "# Now consider the equation 9.10 (page 228) \n", - "# The first term on the right of the equation is\n", "T_1 = round(x_b**(2)*(A_ab+2*x_a*(A_ba-A_ab)),5)\n", - "# similarily the second and third terms are given respectively as \n", "T_2 = round(x_c**(2)*(A_ac+2*x_a*(A_ca-A_ac)),3)\n", "T_3 = 0.0550 #x_b*x_c*(0.5*(A_ba+A_ab+A_ac-A_bc-A_cb)+x_a*(A_bc-A_ab+A_ca-A_ac)+(x_b-x_c)*(A_bc-A_cb)-(1-2*x_a)*0.00)\n", - "# thus whole term on the right hand side is\n", "T = T_1+T_2+T_3\n", - "# So \n", "y_a = 10**(T)\n", - "# for this temperature vapour pressure of the acetone is calculated as \n", "p_acetone = 1.417 #[atm]\n", - "# So that we estimate\n", "y_acetone = x_a*y_a*p_acetone\n", "\n", - "# Results\n", "print \" yacetone : %.3f\"%y_acetone\n", "print \"The activity coefficient of acetone in the given mixture is %f\"%(y_a)\n", - "# The experimental value is y_acetone = 0.698\n", "\n", - "# Answer is different because of rounding error." ], "language": "python", "metadata": {}, @@ -389,7 +309,6 @@ "\n", "import math \n", "\n", - "# Variables\n", "T = 85.3+273.15 #[K] Temperature\n", "P = 1. #[atm] Pressure of the mixture\n", "R = 8.314 #[(Pa*m(3)/(K*mol))]\n", @@ -397,101 +316,74 @@ "y_i = 0.1238 # mole fraction of the ethanol in the vapor phase\n", "y_j = (1-y_i) # mole fraction of the water vapor in the vapor phase\n", "\n", - "# From the table A.1( table 417), the properties of water and ethanol are given as \n", - "# Critical temperatures are \n", "T_c_ii = 513.9 #[K] Critical temperature of the ethanol\n", "T_c_jj = 647.1 #[K] Criatical temperature of water\n", "\n", - "# Critical pressure are \n", "P_c_ii = 61.48 #[bar] Critical pressure of ethanol \n", "P_c_jj = 220.55 #[bar] Critical pressure of water\n", "\n", - "# A# Resultsentric factor\n", "w_ii = 0.645 # accentric factor of the ethanol \n", "w_jj = 0.345 # accentric factor of the water\n", "\n", - "# Compressibility factor are\n", "z_c_ii = 0.24 # compressibility factor of ethanol\n", "z_c_jj = 0.229 # compressibility factor of the water\n", "\n", - "# Calculations\n", - "# Critical volume are given by \n", "V_c_ii = z_c_ii*R*T_c_ii/(P_c_ii*100000)*10**(6) # critical volume the ethanol\n", "V_c_jj = z_c_jj*R*T_c_jj/(P_c_jj*100000)*10**(6) # critical volume the ethanol\n", "\n", - "# Now\n", - "# for k_ij = 0.0\n", "T_c_ij_0 = (T_c_ii*T_c_jj)**(1./2) #[K]\n", "w_ij = (w_ii + w_jj)/2.\n", "z_c_ij = (z_c_ii + z_c_jj)/2.\n", "V_c_ij = ( (V_c_ii**(1./3) + V_c_jj**(1./3))/2.)**(3)\n", "P_c_ij_0 = (z_c_ij*R*T_c_ij_0)/(V_c_ij/10.**(6.))/10.**(5) #[bar]\n", "\n", - "# again for k_ij = 0.01\n", "T_c_ij_1 = (T_c_ii*T_c_jj)**(1./2.)*(1-0.01) #[K]\n", "P_c_ij_1 = (z_c_ij*R*T_c_ij_1)/(V_c_ij/10.**(6))/10**(5) #[bar]\n", "\n", - "# Now \n", "T_r_ii = T/T_c_ii\n", "T_r_jj = T/T_c_jj\n", "T_r_ij_0 = T/T_c_ij_0\n", "T_r_ij_1 = T/T_c_ij_1\n", "\n", - "# and\n", "P_r_ii = P/P_c_ii\n", "P_r_jj = P/P_c_jj\n", "P_r_ij_0 = P/P_c_ij_0\n", "P_r_ij_1 = P/P_c_ij_1\n", "\n", - "# Now we will calculate f(T_r) for each component and mixture\n", "f_Tr_ii = ( 0.083 - 0.422/T_r_ii**(1.6) ) + w_ii*( 0.139 - 0.172/T_r_ii**(4.2))\n", "f_Tr_jj = ( 0.083 - 0.422/T_r_jj**(1.6) ) + w_jj*( 0.139 - 0.172/T_r_jj**(4.2))\n", "f_Tr_ij0 = ( 0.083 - 0.422/T_r_ij_0**(1.6) ) + w_ij*( 0.139 - 0.172/T_r_ij_0**(4.2))\n", "f_Tr_ij1 = ( 0.083 - 0.422/T_r_ij_1**(1.6) ) + w_ij*( 0.139 - 0.172/T_r_ij_1**(4.2))\n", "\n", - "# Let us define A = (P_r*f(T_r)/T_r) , so\n", "A_ii = P_r_ii*f_Tr_ii/T_r_ii\n", "A_jj = P_r_jj*f_Tr_jj/T_r_jj\n", "\n", - "# We are given\n", "v_ii = 0.975\n", "v_jj = 0.986\n", "\n", - "# Now,\n", "B_ii = ( f_Tr_ii*R*T_c_ii/P_c_ii)*(10.**(3)/10**(5)) #[L/mol]\n", "B_jj = ( f_Tr_jj*R*T_c_jj/P_c_jj)*(10.**(3)/10**(5)) #[L/mol]\n", "B_ij0 = ( f_Tr_ij0*R*T_c_ij_0/P_c_ij_0)*(10.**(3)/10**(5)) #[L/mol]\n", "B_ij1 = ( f_Tr_ij1*R*T_c_ij_1/P_c_ij_1)*(10.**(3)/10**(5)) #[L/mol]\n", "\n", - "# now we will calculate 'delta'\n", "delta_ij0 = 2*B_ij0 - B_ii - B_jj #[L/mol]\n", "delta_ij1 = 2*B_ij1 - B_ii - B_jj #[L/mol]\n", "\n", - "# We have \n", - "# b_a = B_aa + y_b**(2)*delta and b_b = B_bb + y_a**(2)*delta\n", - "# so,\n", "b_ethanol0 = B_ii + y_j**(2)*delta_ij0 #[L/mol]\n", "b_water0 = B_jj + y_i**(2)*delta_ij0 #[L/mol]\n", "b_ethanol1 = B_ii + y_j**(2)*delta_ij1 #[L/mol]\n", "b_water1 = B_jj + y_i**(2)*delta_ij1 #[L/mol]\n", "\n", - "# Now \n", - "# phi_i = exp(b_i*P/(R*T))\n", - "# So,\n", "phi_ethanol0 = math.exp((b_ethanol0*P)/(R_1*T))\n", "phi_water0 = math.exp((b_water0*P)/(R_1*T))\n", "phi_ethanol1 = math.exp((b_ethanol1*P)/(R_1*T))\n", "phi_water1 = math.exp((b_water1*P)/(R_1*T))\n", "\n", - "# and\n", - "# Y_i = phi_i/v_i\n", - "# So,\n", "Y_ethanol0 = phi_ethanol0/v_ii\n", "Y_water0 = phi_water0/v_jj\n", "Y_ethanol1 = phi_ethanol1/v_ii\n", "Y_water1 = phi_water1/v_jj\n", "\n", - "# Results\n", "print \" The results are summarize in the following table\"\n", "print \" Property \\t\\t\\t Mix ij Assuming k_ij = 0.0 \\t\\t\\t Mix ij Assuming k_ij = 0.01\"\n", "print \" phi_ethanol \\t\\t\\t\\t %f \\t\\t\\t\\t\\t %f \"%(phi_ethanol0,phi_ethanol1)\n", @@ -541,25 +433,16 @@ "\n", "import math \n", "\n", - "# Variables\n", "T = 65+273.15 #[K] Temperature\n", "R = 8.314 #[(m**(3)*Pa)/(mol*K)] Universal gas consmath.tant \n", - "# From the table 9.C ( page 239 ) given in the book the molar volumes and solubility of n-hexane and diethylketone at 25 deg C are given as \n", "v_hex = 131.6 #[ml/mol] Molar volume of n-Hexane\n", "v_dketone = 106.4 #[ml/mol] Molar volume of diethylketone\n", "s_hex = 14.9 #[MPa**(0.5)] Solubility of n-Hexane\n", "s_dketone = 18.1 #[MPa**(0.5)] Solubility of diethylketone\n", "\n", - "# Calculations\n", - "# Here we will use these values with the assumption that Y_i,65C = Y_i,25C\n", - "# At infinite dilution, the volume fraction of the other species is 1.00, so, \n", - "# math.logY_a = v_a*phi_b**(2)*(delta_a - delta_b)**(2)/(R*T)\n", - "# so, for n-Hexane\n", "Y_hex = math.exp(v_hex*1**(2)*(s_hex - s_dketone)**(2)/(R*T))\n", - "# And that for diethylketone\n", "Y_dketone = math.exp(v_dketone*1**(2)*( s_dketone - s_hex )**(2)/(R*T))\n", "\n", - "# Results\n", "print \" The infinite dilution activity coefficient of n-Hexane is %f\"%(Y_hex)\n", "print \" The infinite dilution activity coefficient of diethlyketone is %f\"%(Y_dketone)\n" ], @@ -596,26 +479,18 @@ "'''\n", "import math \n", "\n", - "# Variables\n", "P = 1. #[atm]\n", "T = 25. #[C]\n", "y_i = 1.00 # amount of the oxygen in the vapour \n", - "# Umath.sing the consmath.tants for O2 in table A.2 \n", "A = 6.69147\n", "B = 319.0117\n", "C = 266.7\n", "\n", - "# Calculations\n", - "# By Antoine equation \n", - "# log10(P_i) = A-B/(T+C)\n", "P_i = 10**(A-B/(T+C)) #[mmHg]\n", "P_i = P_i/760. #[atm]\n", - "# This is extrapolated vapour pressure of O2 at 25C\n", - "# We will take this value as equal to the Henry's law consmath.tant\n", "H_i = P_i\n", "x_i = y_i*P/H_i\n", "\n", - "# Results\n", "print \" Henry's law constant for O2 is %f atm\"%(P_i)\n", "print \" solubility of O2 is %e\"%(x_i)\n" ], @@ -645,22 +520,16 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The activity coefficient of the oxygen\n", "\n", "import math \n", "\n", - "# Variables\n", "y_a = 1.00\n", "P = 1.00 #[atm]\n", "x_a = 0.231*10**(-4)\n", - "# Using the constants for O2 in table A.2 in the Antoine equation , we find the vapour pressure of the oxygen at 25C viz.\n", "p_a = 521.15 #[atm]\n", - "# Thus activity coefficient is calculated by rewriting the equation 8.6 and umath.sing the above values \n", "\n", - "# Calculations\n", "Y_O2 = (y_a*P)/(x_a*p_a)\n", "\n", - "# Results\n", "print \"The activity coefficient of the oxygen in the water is %f\"%(Y_O2)\n" ], "language": "python", |