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diff --git a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch11.ipynb b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch11.ipynb index fff7b6e8..24212b64 100644 --- a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch11.ipynb +++ b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch11.ipynb @@ -27,31 +27,22 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The total number of the drops of the benznene required to saturate the water\n", "\n", "import math \n", "\n", - "# Variables\n", "V_water = 1. #[L] volume of the water\n", "Temp = 25. #[C]\n", "d_benzene = 0.88 #[g/ml] density of the benzene \n", "M_benzene = 78. #[g/mol] molecular weight of the benzene\n", "M_water = 18. #[g/mol]\n", - "# Typically a buret will deliver about 20 drops of lliquid per millimeter, so moles of benzene in one drop is\n", "\n", - "# Calculations\n", "n_1drop = 1./20*(d_benzene/M_benzene) #[mol/drop] \n", "\n", - "# No of moles in 1 litre of the water is \n", "n_water = 1000./M_water #[mol]\n", - "# Because 1 litre = 1000 g\n", - "# Now from the table 11.1 (page 273), at the saturated condition at the temperature 25C, solubility of benzene in the water is\n", "s_benzene = 0.000405\n", "n_benzene_saturate = s_benzene*n_water #[mol]\n", - "# Thus no of the drops of the benzene is\n", "N_benzene = n_benzene_saturate/n_1drop #[drops]\n", "\n", - "# Results\n", "print \"The total number of the drops of the benznene required to saturate the water is %0.0f drops\"%(N_benzene)\n" ], "language": "python", @@ -79,29 +70,17 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The amount of the ground water \n", "\n", "import math \n", "\n", - "# Variables\n", "m_benzene = 1000. #[lbm]\n", "M_benzene = 78. #[lbm/lbmol]\n", - "# The total moles benzene are \n", "n_benzene = m_benzene/M_benzene #[lbmol]\n", - "# To saturate the water with benzene \n", - "# the mole fraction of the benzene in the water will be \n", "x_benzene = 0.000405\n", - "# and \n", - "# n_benzene = x_benzene*n_T\n", - "# in this case n_benzene << n_water, so n_T = n_water\n", - "# Thus\n", - "#n_benzene = x_benzene*n_water\n", "\n", - "# Calculations\n", "n_water = n_benzene/x_benzene #[lbmol]\n", "m_water = n_water*18 #[lbm] \n", "\n", - "# Results\n", "print \"The amount of the ground water that will make a saturated solution will be %e lbm\"%(m_water)\n" ], "language": "python", @@ -137,32 +116,23 @@ "'''\n", "import math \n", "\n", - "# Variables\n", "Temp = 25. #[C]\n", "n_water = 3.75 #[mol]\n", "n_ethanol = 2.5 #[mol]\n", "n_benzene = 3.75 #[mol]\n", "\n", - "# By the simple stoichiomtetry the overall mole fractions are \n", "x_water = 0.375\n", "x_ethanol = 0.250\n", "x_benzene = 0.375\n", "\n", - "# Calculations\n", - "# We locate the point corresponding to this concentration on the diagram 11.1 (page 277), by drawing any two of the three straight lines corresponding to those mole fractions.\n", - "# We find that the point falls almost exactly on the fifth tie line from the top.\n", - "#For this the end-point values are read from the table 11.4 ( page 276 ), which is fifth row from the bottom \n", - "# Thus in water reach phase we have the composition as\n", "x_water_w = 64.9 #[%]\n", "x_ethanol_w = 31.75 #[%]\n", "x_benzene_w = 3.37 #[%]\n", "\n", - "# and in the benzene reach phase composition is \n", "x_water_b = 6.43 #[%]\n", "x_ethanol_b = 18.94 #[%]\n", "x_benzene_b = 74.62 #[%]\n", "\n", - "# Results\n", "print \"The composition of the two equilibrium phases i.e. water-reach phase and benzene reach phase is as \"\n", "print \"\\t\\t\\t\\tWater-reach phase\\t\\tbenzene-reach phase\"\n", "print \"Mol water\\t\\t\\t%f\\t\\t\\t\\t%f\"%(x_water_w,x_water_b)\n", @@ -205,34 +175,20 @@ "'''\n", "import math \n", "\n", - "# Variables\n", "Temp = 25. #[C]\n", - "# Here we assume benzene to be component 1 and water to be componenet 2\n", - "# From table 11.1 given in the book(page 273) \n", - "# The mole fraction of benzene in water is\n", "x_1in2 = 405. #[ppm]\n", - "# and the mole fraction of water in benzene is \n", "x_2in1 = 3000. #[ppm]\n", "\n", - "# Calculations\n", - "# Thus mole fraction of water in water rich phase is\n", "x_water_w = (10**(6)-405.)/(10**(6))\n", "x_benzene_w = 1-x_water_w\n", "\n", - "# and mole fraction of the benzene in benzene rich phase is\n", "x_benzene_b =(10.**(6.)-3000.)/(10**(6))\n", "x_water_b = 1-x_benzene_b\n", "\n", - "# Here both x_water and x_benzene are nearly equal to 1\n", - "# Thus assumption used for derivation of the equation 11.4(page 282) are suitable here and the equation is \n", - "# x_i_1 = y_i_1 , where y_i_1 is activity coefficient\n", "\n", - "# So activity coefficient of benzene in water is\n", "y_benzene = 1/(x_benzene_w)\n", - "# and activty coefficient of the water in benzene is\n", "y_water = 1/(x_water_b)\n", "\n", - "# Results\n", "print \" Activity coefficient of benzene in water is %f\"%(y_benzene)\n", "print \" Activity coefficient of water in benzene is %f\"%(y_water)\n" ], @@ -262,7 +218,6 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Draw Gibb's free energy vs mole fractions\n", "\n", "import math \n", "from numpy import * \n", @@ -270,13 +225,11 @@ "\n", "%pylab inline\n", "\n", - "# Variables\n", "R = 8.314 #[J/(mol*K)] Universal gas consmath.tant\n", "T = 298.15 #[K] Temperature\n", "g_a_0 = 2. #[kj/mol] Gibb's free energy of the pure species 'a'\n", "g_b_0 = 1. #[kj/mol] Gibb's free energy of the pure species 'b'\n", "\n", - "# Calculations\n", "for a in range(4):\n", " def f(x): \n", " return x*g_a_0 + (1-x)*g_b_0 + (R*T)/1000*(x*log(x) + (1-x)*log(1-x) + x*a*(1-x)**(2) + (1-x)*a*(x)**(2))\n", @@ -294,7 +247,6 @@ "suptitle(\"Calculated values of the molar Gibbs energy of a binary mixture at 25 \u0014 C, assuming \\\n", "the symmetrical activity coefficient equation, with various values of A.\")\n", "\n", - "# Results\n", "show()\n", "print \" The plot is shown in the graphic window.\"\n" ], @@ -346,30 +298,20 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Draw two liquid phase\n", "\n", "import math \n", "from matplotlib.pyplot import *\n", "from numpy import *\n", "\n", "\n", - "# Variables\n", "T = 92 + 273.15 #[K] Temperature of the system\n", "R = 8.314 #[m**(3)*Pa/(mol*K)] universal gas consmath.tant\n", "\n", - "# Van Laar equation coefficients are given\n", "A = 1.2739\n", "B = 3.9771\n", "\n", - "# From van laar equation, for water, we have\n", - "# lnY_a = B**(2)*A*(1-x_a)**(2)/(A*x_a + B*(1-x_a))**(2)\n", - "# Similarily for n-bumath.tanol\n", - "# lnY_b = A**(2)*B*x_a**(2)/(A*x_a + B*(1-x_a))**(2)\n", "\n", - "# Let us say, g = g_mix - sum(x_i*g_i_0)\n", - "# So, plotting g vs x_i we have\n", "\n", - "# Calculations\n", "def f(x_a): \n", "\t return (R*T/1000)*( x_a*log(x_a) + (1-x_a)*log(1-x_a) + \\\n", " x_a*(B**(2)*A*(1-x_a)**(2)/(A*x_a + B*(1-x_a))**(2)) + (1-x_a)* \\\n", @@ -378,23 +320,15 @@ "x_a = linspace(0.000001,0.99999,100)\n", "a = f(x_a)\n", "plot(x_a,a)\n", - "#xlabel(\" Mol fraction of species a, x_a\")\n", - "#ylabel(\" g_mix - sum(x_i*g_i0)\")\n", "\n", - "# Now drawing math.tangent \n", - "#x = linspace(0.000001,0.99999,100)\n", "x = linspace(0.0,0.99999,100)\n", "a = 1.2090312*x-1.251495764\n", - "#plot(x,a)\n", "plot(x,1.2090312*x-1.251495764)\n", "xlabel(\" Mol fraction of species a, x_a\");\n", "ylabel(\" g_mix - sum(x_i*g_i0)\");\n", "\n", "show()\n", "\n", - "# Results\n", - "# Figure shows the results of the calculation of the whole range of x_a \n", - "# Drawing the math.tangent to the curve, the line touches the curve at two points x_a = 0.47 and 0.97\n", "print \"Thus based on the several assumptions that the Van Laar equation is an accurate representation of LLE\"\n", "print \"we would conclude that at 92 deg C water-n-butanol does form two liquid phases.\"\n", "\n" @@ -440,26 +374,14 @@ "'''\n", "import math \n", "\n", - "# Variables\n", "R = 8.314 #[J/(ml*K)]\n", - "# We find that the water in benzene least squares fit line has the equation \n", - "# math.log(x_water(benzene-reach phase)) = 4.175-2967.7/T\n", "\n", - "# equation 11.7 in the book (page 286) is \n", - "# math.log(x_i_1) = (h_i_0-h_i_average)/(RT) + consmath.tant of integration\n", "\n", - "# Comparing the two equations term by term, we have\n", "\n", - "# Calculations\n", - "# (h_i_0-h_i_average)/(RT) = -2967.7/T\n", - "# let us say (h_i_0-h_i_average) = h_mix\n", - "# So that \n", "h_mix = -2967.7*R/1000 #[kJ/mol] Heat of mixing of water-in-benzene\n", "\n", - "# Now, for benzene-in-water the consmath.tant in the above equation is -522.9K, so\n", "h_mix_1 = 522.9*R/1000 #[kJ/mol] Heat of mixing of benzene-in-water\n", "\n", - "# Results\n", "print \" Heat of mixing of water-in-benzene is given as %f kJ/mol\"%(h_mix)\n", "print \" Heat of mixing of benzene-in-water is given as %f kJ/mol\"%(h_mix_1)\n" ], @@ -489,30 +411,22 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The amount of water that will come out of the solution in the gasoline\n", "import math \n", "\n", - "# Variables\n", "T_i = 50. #[F] Initial temperature of the system \n", "T_f = 20. #[F] Final temperature of the system\n", "M_gas = 115. #[g/mol] Molecular weight of gasoline at room temperature\n", "M_water = 18. #[g/mol] Molecular weight of water at the room temperaature\n", "d = 720. #[g/L] density of gasoline at the room temperature\n", "\n", - "# From Figure 11.10 ( page 288 ), solubility of the water in gasoline ( similar to solubility of water in cyclohexane ) at 50 deg F is given as \n", "s_50 = 0.00026 #[mol fraction]\n", "\n", - "# And linearly extraploting the cyclohexane curve in figure 11.10 to 20 deg F, we get the solubility of water at 20deg F as \n", "s_20 = 0.0001 #[mol fraction]\n", "\n", - "# Calculations\n", - "# So, rejected water is\n", "s_rej = s_50 - s_20 # mol of water per mole of gasoline \n", "\n", - "# In terms of weight, rejected water will be\n", "w = (s_rej*d*M_water)/M_gas #[g water/L gasoline]\n", "\n", - "# Results\n", "print \" The amount of water that will come out of the solution in the gasoline will be %f g water/L gasoline\"%(w)\n", "print \" At 20 deg F we would expect this water to become solid ice, forming a piece large enough to plug the fuel line of a parked auto.\"\n" ], @@ -552,47 +466,30 @@ "from matplotlib.pyplot import *\n", "import math \n", "\n", - "# Variables\n", "Temp = 25. #[C]\n", "x_water = 5. #[mo]\n", "x_benzene = 0.1 #[mol]\n", "\n", - "# The fugacity of the ethanol must be same in both phases so that we have distribution coefficient \n", "\n", - "# K = ( distribution coefficient ) = x_ethanol(water_rich phase)/x_ethanol(benzene-rich phase) = y_ethanol(benzene-rich phase)/y_ethanol(water-rich phase)\n", - "# Here distribution coefficient is defined as rhe ratio of mole fractions of the distributed solute between the two phases.\n", "\n", - "# We observe that for the first experimental data set in table 11.4 in the book(page 276)\n", "x_ethanol_water_rich = 3.817 #[%]\n", "x_ethanol_benzene_rich = 1.010 #[%]\n", "\n", - "# So \n", "K = x_ethanol_water_rich/x_ethanol_benzene_rich\n", "\n", - "# Calculations\n", - "# Now for all the data set in the table 11.4 in the book(page 276), we wiil draw a plot \n", "X = array([3.817,7.968,12.977,18.134,23.540,24.069,27.892,31.725,35.510,39.382,41.062,41.771])\n", "Y = array([1.010,3.323,5.860,9.121,12.939,13.340,16.090,18.943,22.444,26.216,29.341,33.093])\n", "Z = X/Y\n", "\n", - "# Plotting the graph between 'Z' and 'Y' \n", "plot(Y,Z)\n", "xlabel(\"Mol% ethanol in benzene-rich phase \");\n", "ylabel(\"Distribution coefficient of ethanol, K_ethanol\");\n", - "#xgrid()\n", - "#xlabel(\"Mol% ethanol in benzene-rich phase \")\n", - "#ylabel(\"Distribution coefficient of ethanol, K_ethanol\")\n", "show()\n", - "# We see from the plot that at the low mole percent of ethanol , the distribution coefficient is approximately \n", "K_1 = 4\n", "\n", - "# Thus ratio of the amount of the ethanol distributed in the two phases will be 4\n", - "#the amount in mol % is \n", - "# for water rich phase \n", "m_water_rich = 100*K_1/(K_1+1)\n", "m_benzene_rich = 100/(K_1+1)\n", "\n", - "# Results\n", "print \" Ethanol''s 0.1 mol distributed in the water rich phase will be %f mol%% of the total mol\"%(m_water_rich)\n", "print \" Ethanol''s 0.1 mol distributed in the benzene rich phase will be %f mol%% of the total mol\"%(m_benzene_rich)\n" ], @@ -638,25 +535,17 @@ "\n", "import math \n", "\n", - "# Variables\n", "Temp = 20. #[C]\n", - "# At this temperature solubility of NaCl is \n", "s = 36.0 #[g per 100 g of water]\n", "M_NaCl = 58.5 #[g/mol] molecular weight of NaCl \n", "M_water = 18. #[g/mol] molecular weight of water\n", "\n", - "# Calculations\n", - "# weight fraction of NaCl\n", "w = s/(s+100)\n", - "# In weight percentage \n", "w_percent = w*100 #[wt %]\n", "\n", - "# Mol fraction of the NaCl\n", "x = (s/M_NaCl)/((s/M_NaCl)+(100/M_water))\n", - "# In mol percentage\n", "x_percent = x*100 #[mol %]\n", "\n", - "# Results\n", "print \" Weight fraction of the NaCl in the saturated solution is %0.1f wt %%\"%(w_percent)\n", "print \" Mol fraction of the NaCl in the saturated solution is %0.0f mol %%\"%(x_percent)\n", "\n", @@ -696,26 +585,19 @@ "\n", "import math \n", "\n", - "# Variables\n", "T_inlet = 68. #[F]\n", "T_outlet = 110. #[F]\n", "\n", - "# from the figure 11.13 we read that at 68F the solubility of CaCO3 and CaSO4.2H2O are \n", "s_inlet_carbonate = 60. #[ppm]\n", "s_inlet_sulphate = 2020. #[ppm]\n", "\n", - "# At 110F the solubility of the CaCO3 is \n", "s_outlet_carbonate = 30. #[ppm]\n", - "# at 110F the least soluble form of the CaSO4 is anhydride with the solubility \n", "s_outlet_sulphate = 2000. #[ppm]\n", " # This is close enough to the solubility of the gypsum at 68F \n", " # so we conclude that we would not expect either form of CaSO4 to prdcipitate\n", " \n", - "# Calculations\n", - "# Thus total amount of the calcium carbonate which will cime out of the solution and will remain in the heater will be \n", "w = s_inlet_carbonate - s_outlet_carbonate #[ppm]\n", "\n", - "# Results\n", "print \" Total amount of the solid left behind in the heater will be %0.1f ppm\"%(w)\n", "\n", " # Now if a typical houshold water heater heats 100 gallons/per day , we would expect to deposite \n", @@ -756,25 +638,14 @@ "\n", "import math \n", "\n", - "# Variables\n", "x_2 = 0.1\n", - "#y_i_1 = (x_i__ideal /x_i_1) , at consmath.tant temperature\n", - "# From figures 11.15 and 11.16 given in the book (page 298-299) ( or the equations of the lines on those figures, presented in [14] ) we can compute the values in Table 11.6\n", - "# We see that at x_solute = 10% \n", - "# T_m/T for the solution in benzene at which math.log(x_experimental) = -1 is equal to 1.332\n", - "# and that for the solution in CCl4 is equal to 1.288\n", - "#Now at the that value of the T_m/T \n", "x_ideal_benzene = 0.114\n", "x_ideal_CCl4 = 0.152\n", "\n", - "# Calculations\n", - "# In benzene the average these compounds is\n", "y_i_1 = x_ideal_benzene/x_2# corresponding to practically ideal solution\n", "\n", - "# and in benzene the average of these compounds is \n", "y_i_2 = x_ideal_CCl4/x_2# corresponding to mild type II behavior \n", "\n", - "# Results\n", "print \" Activity coefficient in benzene corresponding to practically ideal solution is %0.2f\"%(y_i_1)\n", "print \" Activity coefficient in CCl4 corresponding to mild type II behavior is %0.2f\"%(y_i_2)\n" ], @@ -804,30 +675,20 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find The solubility of the NaCl in water\n", "import math \n", "\n", - "# Variables\n", "T = 273.15+20 #[K]\n", "\n", - "# The equation 11.15 (page 297) is given by \n", - "# log(1/(x_i_1*y_i_1)) = math.log(p_i_solid_phase/p_i_subcooled_liquid) = delta_h_solid_to_liquid/(R*T_melting_point)*(T_melting_point/T-1)\n", - "# Ignoring the moment the wild extraplation involved, we simply insert the appropriate values\n", "T_m = 273.15+800 #[K]\n", "delta_h_fusion = 30219. #[J/g]\n", "R = 8.314 #[J/(mol*K)]\n", "\n", - "# Calculations\n", - "# Let log(1/(x_i_1*y_i_1)) = a\n", "a = delta_h_fusion/(R*T)*(T_m/T-1)\n", "\n", - "# Now \n", "x_NaCl_into_y_i_1 = 1/math.exp(a)\n", "\n", - "# If we make the plausible assumption that y_i_1 = 1.00, then\n", "x_NaCl = 1/math.exp(a)*1\n", "\n", - "# Results\n", "print \" The solubility of the NaCl in water at 20 deg C is %e \"%( x_NaCl)\n", "print \" But the experimental value is 0.1 so Similar to the results in book our results are very far wrong\"\n" ], @@ -857,24 +718,15 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find Equilibrium concentration of water vapour\n", "import math \n", "\n", - "# Variables\n", "P = 1*14.7 #[psia]\n", "T = 30. #[F]\n", - "#******#\n", - "#The vapour pressure of ice at 30F is 0.0808 psia i.e.\n", "p_ice = 0.0808 #[psia]\n", - "# We may assume that the solubility of nitrogen and oxygen in solid ice is negligible\n", - "#Thus\n", "x_water_in_ice = 1.00\n", - "#and thus use Raoult's law,finding\n", "\n", - "# Calculations\n", "y_water_vapour = x_water_in_ice*p_ice/P\n", "\n", - "# Results\n", "print \" Equilibrium concentration of water vapour in the air is %0.4f\"%(y_water_vapour)\n" ], "language": "python", @@ -902,35 +754,24 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find Estimated solubility of naphthalene in CO2\n", "\n", "import math \n", "\n", - "# Variables\n", "T = 273.15+35 #[K]\n", "p_d = 100. #[atm]\n", "R = 82.06 #[(cm**(3)*atm)/(mol*K)]\n", "\n", - "#The calculated vapour pressure of naphthalene at 35C is\n", "p_naphthalene = 0.00065 #[atm]\n", "\n", - "#The solid is practically pure naphthalene\n", "x_naphthalene = 1.00\n", - "#Total pressure \n", "P = p_d\n", "\n", - "# Calculations\n", - "#By Raoult's law\n", "y_naphthalene = x_naphthalene*p_naphthalene/P\n", - "#At this high a pressure the volume of solid naphthalene is\n", "v = 132. #[cm**(3)/mol]\n", - "# We have equation math.log(f_d/f_c) = v/(R*T)*(p_d-p_c)\n", "p_c = 1. #[atm]\n", "f_d_by_f_c = math.exp(v/(R*T)*(p_d-p_c))\n", - "#and the estimated\n", "y_naphthalene = f_d_by_f_c*y_naphthalene\n", "\n", - "# Results\n", "print \"Estimated solubility of naphthalene in CO2 is %e\"%(y_naphthalene)\n", "\n" ], |