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diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb new file mode 100755 index 00000000..1d9240ef --- /dev/null +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb @@ -0,0 +1,1051 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Distillation "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.1,Page number:324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "T1 = 303 # [K]\n",
+ "P = 1 # [bar]\n",
+ "D = 0.6 \n",
+ "W = 0.4 \n",
+ "zf = 0.5 \n",
+ " \n",
+ "# Parameters for componenr 'A'\n",
+ "Tc_a = 540.3 # [K]\n",
+ "Pc_a = 27.4 # [bar]\n",
+ "A_a = -7.675 \n",
+ "B_a = 1.371 \n",
+ "C_a =-3.536 \n",
+ "D_a = -3.202 \n",
+ "\n",
+ "# Parameters for component 'B'\n",
+ "Tc_b = 568.8 # [K]\n",
+ "Pc_b = 24.9 # [bar]\n",
+ "A_b = -7.912 \n",
+ "B_b = 1.380 \n",
+ "C_b = -3.804 \n",
+ "D_b = -4.501 \n",
+ "\n",
+ "import math\n",
+ "from numpy import *\n",
+ "from scipy.optimize import fsolve\n",
+ "# Using equation 6.5\n",
+ "# x_a = 1-(T/Tc_a) \n",
+ "# P_a = Pc_a*math.exp((A_a*x_a+B_a*x_a**1.5+C_a*x_a**3+D_a*x_a**6)/(1-x_a)) # [bar]\n",
+ "\n",
+ "# x_b = 1-(T/Tc_b) \n",
+ "# P_b = Pc_b*math.exp((A_b*x_b+B_b*x_b**1.5+C_b*x_b**3+D_b*x_b**6)/(1-x_b)) # [bar]\n",
+ "\n",
+ "# m_a = P_a/P \n",
+ "# m_b = P_b/P \n",
+ "\n",
+ "# Solution of simultaneous equation\n",
+ "def F(e):\n",
+ " f1 = e[1] - (e[2]*Pc_a*math.exp(((A_a*(1-(e[0]/Tc_a))+B_a*(1-(e[0]/Tc_a))**1.5+C_a*(1-(e[0]/Tc_a))**3+D_a*(1-(e[0]/Tc_a))**6))/(1-(1-(e[0]/Tc_a)))))/P \n",
+ " f2 = 1-e[1] - ((1-e[2])*Pc_b*math.exp((A_b*(1-(e[0]/Tc_b))+B_b*(1-(e[0]/Tc_b))**1.5+C_b*(1-(e[0]/Tc_b))**3+D_b*(1-(e[0]/Tc_b))**6)/(1-(1-(e[0]/Tc_b)))))/P \n",
+ " f3 = (-W/D) - ((e[1]-zf)/(e[2]-zf)) \n",
+ " return(f1,f2,f3)\n",
+ "\n",
+ "\n",
+ "# Initial guess\n",
+ "e = [400,0.6,0.4] \n",
+ "y = fsolve(F,e) \n",
+ "T = y[0] # [K] \n",
+ "Yd = y[1] \n",
+ "Xw = y[2] \n",
+ "\n",
+ "print\"The composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage are\",round(Yd,3),round(Xw,3),\"and\",round(T,1),\"K respectively\\n\\n\"\n",
+ "\n",
+ "# For the capculation of the amount of heat to be added per mole of feed\n",
+ "T0 = 298 # [K]\n",
+ "lambdaA = 36.5 # [Latent heats of vaporization at To = 298 K ,kJ/mole]\n",
+ "lambdaB = 41.4 # [Latent heats of vaporization at To = 298 K ,kJ/mole]\n",
+ "CpA = 0.187 # [kJ/mole.K]\n",
+ "CpB = 0.247 # [kJ/mole.K]\n",
+ "CLA1 = 0.218 # [ 298-303 K, kJ/mole.K]\n",
+ "CLB1 = 0.253 # [ 298-303 K, kJ/mole.K]\n",
+ "CLA2 = 0.241 # [ 298-386 K, kJ/mole.K]\n",
+ "CLB2 = 0.268 # [ 298-386 K, kJ/mole.K]\n",
+ "# Bubble point calculated when 'D' approaches 0 and Dew point calculated when 'D' approaches 1\n",
+ "Tbp = 382.2 # [Bubble point of the mixture, K]\n",
+ "Tdp = 387.9 # [Dew point of mixture, K]\n",
+ "\n",
+ "HF = (T1-T0)*(Xw*CLA1+CLB1*(1-Xw)) # [kJ/mole]\n",
+ "HW = (Tbp-T0)*(Xw*CLA2+CLB2*(1-Xw)) # [kJ/mole]\n",
+ "HG = (Tdp-T0)*(Yd*CpA+(1-Yd)*CpB) + Yd*lambdaA +(1-Yd)*lambdaB # [kJ/mole]\n",
+ "\n",
+ "f =1 # [feed]\n",
+ "# Using equation 6.4\n",
+ "def f14(Q):\n",
+ " return(W/D + (HG-(HF+Q/f))/(HW -(HF+Q/f)))\n",
+ "Q = fsolve(f14,40) \n",
+ "print\"The amount of heat to be added per mole of feed is\",round(Q[0],2),\"kJ/mole\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage are 0.576 0.385 and 385.4 K respectively\n",
+ "\n",
+ "\n",
+ "The amount of heat to be added per mole of feed is 42.08 kJ/mole\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.2,Page number:326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a-benzene b-toluene c-orthoxylene\n",
+ "T = 373 # [K]\n",
+ "P = 101.3 # [kPa]\n",
+ "Pa = 182.7 # [kPa]\n",
+ "Pb = 73.3 # [kPa]\n",
+ "Pc= 26.7 # [kPa]\n",
+ "Zfa = 0.5 \n",
+ "Zfb = 0.25 \n",
+ "Zfc = 0.25 \n",
+ "\n",
+ "import math\n",
+ "from numpy import *\n",
+ "from scipy.optimize import fsolve\n",
+ "# Therefore\n",
+ "ma = Pa/P \n",
+ "mb = Pb/P \n",
+ "mc = Pc/P \n",
+ "# Let Feed is 1 kmole\n",
+ "# Therefore D+W = 1\n",
+ "\n",
+ "# Solution of simultaneous equation\n",
+ "def F(e):\n",
+ " f1 = e[0]+e[1]-1 \n",
+ " f2 = e[1]/e[0] + (e[2]-Zfa)/(e[3]-Zfa) \n",
+ " f3 = e[2]-ma*e[3] \n",
+ " f4 = e[4]-mb*e[5] \n",
+ " f5 = 1-e[2]-e[4] -mc*(1-e[3]-e[5]) \n",
+ " f6 = e[1]/e[0] + (e[4]-Zfb)/(e[5]-Zfb) \n",
+ "\n",
+ " return(f1,f2,f3,f4,f5,f6)\n",
+ "\n",
+ "# Initial guess\n",
+ "e = [0.326,0.674,0.719,0.408,0.198,0.272] \n",
+ "y = fsolve(F,e) \n",
+ "D = y[0] \n",
+ "W = y[1] \n",
+ "Yad = y[2] \n",
+ "Xaw = y[3] \n",
+ "Ybd = y[4] \n",
+ "Xbw = y[5] \n",
+ "Ycd = 1-Yad-Ybd \n",
+ "Xcw = 1-Xaw-Xbw \n",
+ "\n",
+ "print\"The amounts of liquid and vapor products are D=\",round(D,3),\"and W=\",round(W,3),\"respectively\"\n",
+ "print\"The vapor compositions of components A(Yad), B(Ybd) and C(Ycd) are\",round(Yad,3),round(Ybd,3),round(Ycd,3),\"respectively\"\n",
+ "print\"The liquid composition of components A(Xaw), B(Xbw) and C(Xcw) are\",round(Xaw,3),round(Xbw,3),round(Xcw,3),\"respectively\\n\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The amounts of liquid and vapor products are D= 0.326 and W= 0.674 respectively\n",
+ "The vapor compositions of components A(Yad), B(Ybd) and C(Ycd) are 0.714 0.199 0.087 respectively\n",
+ "The liquid composition of components A(Xaw), B(Xbw) and C(Xcw) are 0.396 0.275 0.329 respectively\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.3,Page number:328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\t# n-heptane - a n-octane - b\n",
+ "P = 1 \t\t\t\t\t\t# [bar]\n",
+ "\n",
+ "\t# Basis:\n",
+ "F = 100 \t\t\t\t\t# [mole]\n",
+ "\t# Therefore\n",
+ "D = 60 \t\t\t\t\t# [mole]\n",
+ "W = 40 \t\t\t\t\t# [mole]\n",
+ "xf = 0.5 \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "from numpy import *\n",
+ "from pylab import *\n",
+ "y_star = [0.5,0.55,0.60,0.65,0.686,0.70,0.75] \n",
+ "x = [0.317,0.361,0.409,0.460,0.5,0.516,0.577] \n",
+ "#for i in range(1,7):\n",
+ "# f(i-1) = 1/(y_star(i-1)-x(i-1)) \n",
+ "area = matrix([[0.317,5.464],[0.361,5.291],[0.409,5.236],[0.460,5.263],[0.5,5.376],[0.516,5.435],[0.577,7.78]]) \n",
+ "\t# LHS of equation 6.11\n",
+ "a = math.log(F/W) \n",
+ "\n",
+ "\n",
+ "a1=plot(area[:,0],area[:,1],label='$area under curve$') \n",
+ "legend(loc='upper left')\n",
+ "xlabel(\"x\") \n",
+ "ylabel(\"1/(y_satr-x)\") \n",
+ "\n",
+ "# When the area becomes equal to 0.916, integration is stopped this occurs at \n",
+ "xw = 0.33 # [mole fraction of heptane in residue]\n",
+ "yd =( F*xf-W*xw)/D # [mole fraction of heptane]\n",
+ "#Result\n",
+ "\n",
+ "print\"The composition of the composited distillate and the residue are \",round(yd,3),\"and\",round(xw,3),\"respectively\\n\\n\"\n",
+ "show(a1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The composition of the composited distillate and the residue are 0.613 and 0.33 respectively\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x762d0b8>"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.4,Page number:342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "T = 298 \t\t\t\t\t\t\t\t# [K]\n",
+ "Fa = 200 \t\t\t\t\t\t\t\t# [feed, kmole/hr]\n",
+ "zf = 0.6 \n",
+ "yd = 0.95 \n",
+ "xd = yd \n",
+ "xw = 0.05 \n",
+ "q = 0.5 \t\t\t\t\t\t\t\t# [Lf/F]\n",
+ "\n",
+ "\n",
+ "print\"Illustration 6.4(a)\"\n",
+ "\t# Solution (a)\n",
+ "\n",
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "from numpy import *\n",
+ "\n",
+ "\t# Solution of simultaneous equation \n",
+ "def F(e):\n",
+ " f1 = Fa - e[0]-e[1] \n",
+ " f2 = zf*Fa - yd*e[0] - xw*e[1] \n",
+ " return(f1,f2)\n",
+ "\n",
+ "\n",
+ "# Initial guess\n",
+ "e = [120,70] \n",
+ "y = fsolve(F,e) \n",
+ "D = y[0] \n",
+ "W = y[1] \n",
+ "print\"Quantity of liquid and vapor products are \",round(D,1),\"kmole/h and\",round(W,1),\"kmole/h respectively\"\n",
+ "\n",
+ "\n",
+ "print\"Illustration 6.4(b)\"\n",
+ "# Solution(b)\n",
+ "\t# VLE data is generated in the same manner as generated in Example 6.1 by applying \tRaoult's law\n",
+ "\t# VLE_data = [T,x,y]\n",
+ "VLE_data = matrix([[379.4,0.1,0.21],[375.5,0.2,0.37],[371.7,0.3,0.51],[368.4,0.4,0.64],[365.1,0.5,0.71],[362.6,0.6,0.79],[359.8,0.7,0.86],[357.7,0.8,0.91],[355.3,0.9,0.96]]) \n",
+ "\t# From figure 6.14\n",
+ "\t# The minimum number of equilibrium stages is stepped off between the equilibrium curve and the 45 degree Iine, starting from the top, giving\n",
+ "Nmin = 6.7 \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"The minimum number of theoretical stages is\",Nmin\n",
+ "\n",
+ "print\"Illustration 6.4(c)\"\n",
+ "\t# Solution(c)\n",
+ "\t# Slope of q-line = Lf/F/(1-(Lf/F))\n",
+ "s = q/(1-q) \n",
+ "\t# For minimum reflux ratio\n",
+ "\t# From figure 6.12 y-intercept is\n",
+ "i = 0.457 \n",
+ "# Therefore Rmin is\n",
+ "Rmin = xd/i -1 \n",
+ "\n",
+ "#result\n",
+ "\n",
+ "print\"The minimum reflux ratio is\",round(Rmin,3),\"mole reflux/mole distillate\"\n",
+ "\n",
+ "print\"Illustration 6.4(d)\"\n",
+ "\n",
+ "\t# Solution(d)\n",
+ "R = 1.3*Rmin \n",
+ "\t# The y-intercept of the rectifying-section operating line is\n",
+ "ia = xd/(R+1) \n",
+ "\t# The operating line for the stripping section is drawn to pass through the point x = y = \txw =0.05 on the 45\" line and the point of intersection of the q-line # and the \t\t\trectifying-section operating line.\n",
+ "\t# Therefore from figure 6.15 \n",
+ "Nact = 13 \n",
+ "\t# But it include boiler\n",
+ "Nact1 = Nact-1 \n",
+ "print\"The number of equilibrium stages for the reflux ratio specified is\",Nact1\n",
+ "\t# For the optimal feed-stage location, the transition from one operating line to the \tother occurs \tat the first opportunity\n",
+ "\t# after passing the operating-line intersection \n",
+ "\t# Therefore from figure 6.15 shows that \n",
+ "\n",
+ "\n",
+ "print\"The optimal location of the feed stage for the reflux ratio specified is sixth from the top\"\n",
+ "\n",
+ "print\"Illustration 6.4(e)\"\n",
+ "# Solution(e)\n",
+ "L = R*D \t\t\t\t\t\t\t\t# [kmole/h]\n",
+ "V = L+D \t\t\t\t\t\t\t\t# [kmole/h]\n",
+ "# From equation 6.27\n",
+ "Lst = L+q*Fa \t\t\t\t\t\t\t\t# [kmole/h]\n",
+ "# From equation 6.28\n",
+ "Vst = V+(q-1)*Fa \t\t\t\t\t\t\t# [kmole/h]\n",
+ "\n",
+ "\t# For 50% vaporization of the feed ( zf = 0.60), from calculations similar to those \t\tillustrated in Example 6.1, the separator temperature and the equilibrium # \tcompositions are\n",
+ "Tf = 365.5 \t\t\t\t\t\t\t\t# [K]\n",
+ "yf = 0.707 \n",
+ "xf = 0.493 \n",
+ "\n",
+ "\t# Latent heat vaporisation data at temperature T = 298 K\n",
+ "lambdaA = 33.9 \t\t\t\t\t\t\t# [kJ/mole]\n",
+ "lambdaB = 38 \t\t\t\t\t\t\t\t# [kJ/mole]\n",
+ "\t# Heat capacities of liquids (298-366 K)\n",
+ "Cla = 0.147 \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
+ "Clb = 0.174 \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
+ "\t# Heat capacities of gases, average in the range 298 to 366 K\n",
+ "Cpa = 0.094 \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
+ "Cpb = 0.118 \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
+ "\t# Substituting in equation 6.6 gives\n",
+ "Hf = 0 \n",
+ "Hlf = (Tf-T)*(xf*Cla+(1-xf)*Clb) \t\t\t\t\t# [kJ/mole of liquid \t\t\t\t\t\t\t\t\tfeed]\n",
+ "\t# From equation 6.7\n",
+ "Hvf = (Tf-T)*(yf*Cpa+(1-yf)*Cpb) + yf*lambdaA + (1-yf)*lambdaB \t# [kJ/mole of vapor feed]\n",
+ "\n",
+ "Lf = Fa*q \t\t\t\t\t# [kmole/h]\n",
+ "Vf = Fa*(1-q) \t\t\t\t\t# [kmole/h]\n",
+ "\t# From equation 6.3\n",
+ "Qf = (Hvf*Vf +Hlf*Lf-Fa*Hf)*1000.0/3600.0 # [kW]\n",
+ "\n",
+ "\n",
+ "Tlo = 354.3 \t\t\t\t\t# [Bubble point temperature, K]\n",
+ "T1 = 355.8 \t\t\t\t\t# [Dew point temperature, K]\n",
+ "y1 = 0.95 \t\t\t\t\t# [composition of saturated vapor at dew point]\n",
+ "x0 = 0.95 \t\t\t\t\t# [composition of saturated liquid at bubble \t\t\t\t\t\tpoint]\n",
+ "Hv1 = (T1-T)*(y1*Cpa+(1-y1)*Cpb) + y1*lambdaA + (1-y1)*lambdaB # [kJ/mole of vapor feed]\n",
+ "Hlo = (Tlo-T)*(x0*Cla+(1-x0)*Clb) \t\t# [kJ/mole of liquid feed]\n",
+ "\n",
+ "\t# An energy balance around condenser\n",
+ "Qc = V*(Hv1-Hlo)*1000/3600 \t\t\t# [kW]\n",
+ "\n",
+ "\t# A flash-vaporization calculation is done in which the fraction vaporized is known \t(53.8/75.4 = \t0.714) and the concentration\n",
+ "\t# of the liquid residue is fixed at xw = 0.05\n",
+ "\t# The calculations yield\n",
+ "Tr = 381.6 \t\t\t\t\t# [K]\n",
+ "x12 = 0.093 \n",
+ "y13 = 0.111 \n",
+ "T12 = 379.7 \t\t\t\t\t# [Bubble point of the liquid entering in the \t\t\t\t\t\treboiler, K]\n",
+ "\n",
+ "Hl12 = (T12-T)*(x12*Cla+(1-x12)*Clb) \t\t# [kJ/mole of liquid feed]\n",
+ "Hv13 = (Tr-T)*(y13*Cpa+(1-y13)*Cpb) + y13*lambdaA + (1-y13)*lambdaB # [kJ/mole of vapor feed]\n",
+ "\n",
+ "Hlw = (Tr-T)*(xw*Cla+(1-xw)*Clb) # [kJ/mole of liquid feed]\n",
+ "\n",
+ "\t# An energy balance around the reboiler \n",
+ "Qr = (Vst*Hv13+W*Hlw-Lst*Hl12)*1000.0/3600.0 \t# [kW]\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"The thermal load of the condenser, reboiler, and feed preheater are\",round(Qc),\"kW\",round(Qr),\"kW\",\"and\", round(Qf),\"kW respectively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 6.4(a)\n",
+ "Quantity of liquid and vapor products are 122.2 kmole/h and 77.8 kmole/h respectively\n",
+ "Illustration 6.4(b)\n",
+ "The minimum number of theoretical stages is 6.7\n",
+ "Illustration 6.4(c)\n",
+ "The minimum reflux ratio is 1.079 mole reflux/mole distillate\n",
+ "Illustration 6.4(d)\n",
+ "The number of equilibrium stages for the reflux ratio specified is 12\n",
+ "The optimal location of the feed stage for the reflux ratio specified is sixth from the top\n",
+ "Illustration 6.4(e)\n",
+ "The thermal load of the condenser, reboiler, and feed preheater are 2549.0 kW 1794.0 kW and 1466.0 kW respectively\n"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stderr",
+ "text": [
+ "C:\\Anaconda\\lib\\site-packages\\scipy\\optimize\\minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
+ " improvement from the last ten iterations.\n",
+ " warnings.warn(msg, RuntimeWarning)\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.7,Page number:357"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\t# a-benzene b-toluene\n",
+ "xa = 0.46 \n",
+ "xb = 0.54 \n",
+ "Tb = 395 \t\t\t\t# [bottom temp., K]\n",
+ "Tt = 360 \t\t\t\t# [top temp., K]\n",
+ "alphab = 2.26 \n",
+ "alphat = 2.52 \n",
+ "D = 1.53 \t\t\t\t# [diameter of column, m]\n",
+ "f = 0.81 \t\t\t\t# [flooding]\n",
+ "deltaP = 700 \t\t\t\t# [average gas-pressure drop, Pa/tray]\n",
+ "import math\n",
+ "\n",
+ "Tavg = (Tb+Tt)/2 \t\t\t# [K]\n",
+ "alpha_avg = (alphab+alphat)/2 \n",
+ "\n",
+ "print \"Solution6.7(a)\" \n",
+ "\t# Solution(a)\n",
+ "\t\n",
+ "\t# Constants for components 'a' and 'b'\n",
+ "Aa = 4.612 \n",
+ "Ba = 148.9 \n",
+ "Ca = -0.0254 \n",
+ "Da = 2.222*10**-5 \n",
+ "ua =math.exp(Aa+Ba/Tavg+Ca*Tavg+Da*Tavg**2) \t\t\t# [cP]\n",
+ "\n",
+ "Ab = -5.878 \n",
+ "Bb = 1287 \n",
+ "Cb = 0.00458 \n",
+ "Db = -0.450*10**-5 \n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "ub = math.exp(Ab+Bb/Tavg+Cb*Tavg+Db*Tavg**2) \t\t\t# [cP]\n",
+ "\n",
+ "\t# At the average column temperature \n",
+ "ul = ua**xa*ub**xb \t\t\t\t\t\t# [cP]\n",
+ "K = alpha_avg*ul \n",
+ "\t# From the O\u2019Connell correlation\n",
+ "Eo = 0.52782-0.27511*math.log10(K) + 0.044923*(math.log10(K))**2 \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"The overall tray efficiency using the O\u2019Connell correlation is \",round(Eo,1) \n",
+ "\n",
+ "print \"Example 6.7(b)\"\n",
+ "\t# Solution(b)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "Nideal = 20 \t\t\t\t\t\t\t# [number of ideal stages]\n",
+ "Nreal = Nideal/(Eo) \t\t\t\t\t\t# [nnumber of real stages]\n",
+ "print Nreal \n",
+ "\t# Since real stages cannot be fractional, therefore\n",
+ "Nreal = 34 \n",
+ "\t# From Table 4.3 tray spacing \n",
+ "t = 0.6 \t\t\t\t\t\t\t# [m]\n",
+ "\t# Adding 1 m over the top tray as an entrainment separator and 3 m beneath # the bottom \ttray for bottoms surge capacity, the total column height is\n",
+ "Z = 4+Nreal*t \t\t\t\t\t\t\t# [m]\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"The number of real trays and the total tower height are\",Nreal,\"and\",Z,\" m respectively\"\n",
+ "\n",
+ "print \"Solution 6.7(c)\"\n",
+ "\t# Solution(c)\n",
+ "\n",
+ "\t# Total gas pressure drop\n",
+ "deltaPc = deltaP*Nreal/1000 \t\t\t\t\t# [kPa]\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"The total gas-pressure drop through the column is\",deltaPc,\"kPa\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Solution6.7(a)\n",
+ "The overall tray efficiency using the O\u2019Connell correlation is 0.6\n",
+ "Example 6.7(b)\n",
+ "32.1801990089\n",
+ "The number of real trays and the total tower height are 34 and 24.4 m respectively\n",
+ "Solution 6.7(c)\n",
+ "The total gas-pressure drop through the column is 23 kPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.10,Page number:371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable declaration\n",
+ "\t# A-toluene B-1,2,3-trimethyl benzene C-benzene\n",
+ "\t# Solution of above three are ideal \n",
+ "\t# Feed\n",
+ "za = 0.40 \n",
+ "zb = 0.30 \n",
+ "zc = 0.30 \n",
+ "\t# Bottom \n",
+ "FRAd = 0.95 \t\t\t\t# [recovery of toluene in distillate]\n",
+ "FRBw = 0.95 \t\t\t\t# [recovery of 1,2,3-trimethyl benzene in the bottom]\n",
+ "P = 1 \t\t\t\t\t# [atm]\n",
+ "\n",
+ "\t# First estimate of distillate composition \n",
+ "xc = 40/70 \n",
+ "xa = 30/70 \n",
+ "xb = 0 \n",
+ "\t# The bubble point temperature for this solution is \n",
+ "Tb = 390 \t\t\t\t# [K]\n",
+ "\t# The corresponding parameters for benzene, toluene and 1,2,3-trimethyl benzene\n",
+ "\t# For toluene\n",
+ "Tc_a = 568.8 \t\t\t\t\t# [K]\n",
+ "Pc_a = 24.9 \t\t\t\t\t# [bar]\n",
+ "A_a = -7.912 \n",
+ "B_a = 1.380 \n",
+ "C_a =-3.804 \n",
+ "D_a = -4.501 \n",
+ "\t# For 1,2,3-trimethyl benzene\n",
+ "Tc_b = 664.5 \t\t\t\t\t# [K]\n",
+ "Pc_b = 34.5 \t\t\t\t\t# [bar]\n",
+ "A_b = -8.442 \n",
+ "B_b = 2.922 \n",
+ "C_b =-5.667 \n",
+ "D_b = -2.281 \n",
+ "\t# For benzene\n",
+ "Tc_c = 540.3 \t\t\t\t\t# [K]\n",
+ "Pc_c = 27.4 \t\t\t\t\t# [bar]\n",
+ "A_c = -7.675 \n",
+ "B_c = 1.371 \n",
+ "C_c =-3.536 \n",
+ "D_c = -3.202 \n",
+ "\n",
+ "\n",
+ "\t# At the estimated reboiler temperature of 449.3 K\n",
+ "Tr = 449.3 \t\t\t\t\t# [K]\n",
+ "\t# P = [Toluene 1,2,3-trimethyl benzene Benzene]\n",
+ "\n",
+ "\t# P = [Tc Pc A B C D]\n",
+ "P1=zeros((3,6))\n",
+ "P1= matrix([[568.8,24.9,-7.912,1.380,-3.804,-4.501],[664.5,34.5,-8.442,2.922,-5.667,2.281],[540.3,27.4,-7.675,1.371,-3.536,-3.202]]) \n",
+ "\n",
+ "for i in range(0,3):\n",
+ " P1[i]= P1[i,1]*math.exp((P1[i,2]*(1-Tr/P1[i,0])+P1[i,3]*(1-Tr/P1[i,0])**1.5+P1[i,4]*(1-Tr/P1[i,0])**3+P1[i,5]*(1-Tr/P1[i,0])**6)/(1-(1-Tr/P1[i,0]))) \n",
+ "\n",
+ "PA1 = P1.item(0) \t\t\t\t\t# [bar]\n",
+ "PB1 = P1.item(6) \t\t\t\t\t# [bar]\n",
+ "PC1 = P1.item(12)\t\t\t\t# [bar]\n",
+ "alphaAB1 = PA1/PB1 \n",
+ "alphaCB1 = PC1/PB1 \n",
+ "\n",
+ "\t\t# At the estimated distillate temperature of 390 K\n",
+ "Td = 390 \t\t\t\t\t# [K]\n",
+ "\t# P = [Toluene 1,2,3-trimethyl benzene Benzene]\n",
+ "\t# P = [Tc,Pc,A,B,C,D]\n",
+ "P2 = zeros((3,6)) \n",
+ "P2= matrix([[568.8,24.9,-7.912,1.380,-3.804,-4.501],[664.5,34.5,-8.442,2.922,-5.667,2.281],[540.3,27.4,-7.675,1.371,-3.536,-3.202]]) \n",
+ "for i in range(0,3):\n",
+ " P2[i] = P2[i,1]*math.exp((P2[i,2]*(1-Td/P2[i,0])+P2[i,3]*(1-Td/P2[i,0])**1.5+P2[i,4]*(1-Td/P2[i,0])**3+P2[i,5]*(1-Td/P2[i,0])**6)/(1-(1-Td/P2[i,0])))\n",
+ " \n",
+ "PA2 = P2.item(0) # [bar]\n",
+ "PB2 = P2.item(6) # [bar]\n",
+ "PC2 = P2.item(12) # [bar]\n",
+ "alphaAB2 = PA2/PB2 \n",
+ "alphaCB2 = PC2/PB2 \n",
+ "# The geometric-average relative volatilities are\n",
+ "alphaAB_avg = math.sqrt(alphaAB1*alphaAB2) \n",
+ "alphaCB_avg = math.sqrt(alphaCB1*alphaCB2) \n",
+ "\n",
+ "# From equation 6.66\n",
+ "Nmin = math.log(FRAd*FRBw/((1-FRAd)*(1-FRBw)))/log(alphaAB_avg) \n",
+ "\n",
+ "# From equation 6.67\n",
+ "FRCd = alphaCB_avg**Nmin/((FRBw/(1-FRBw))+alphaCB_avg**Nmin) # [fractional recovery of benzene in the distillate]\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "\n",
+ "print\"The number of equilibrium stages required at total reflux is\",round(Nmin,2)\n",
+ "print\"The recovery fraction of benzene in the distillate is \",round(FRCd,3)\n",
+ "print\"\\n\\nThus, the assumption that virtually all of the LNK will be recovered in the distillate is justified\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of equilibrium stages required at total reflux is 4.32\n",
+ "The recovery fraction of benzene in the distillate is 0.997\n",
+ "\n",
+ "\n",
+ "Thus, the assumption that virtually all of the LNK will be recovered in the distillate is justified\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.11,Page number:376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\t# 1-toluene 2-1,2,3--trimethylbenzene 3-benzene\n",
+ "\t# Basis: 100 kmol/h of feed\n",
+ "F = 100 \t\t\t\t\t\t# [kmole/h]\n",
+ "\t# Since feed is saturated, therefore\n",
+ "q = 0 \n",
+ "\t# From example 6.10\n",
+ "x1d = 0.3 \n",
+ "x2d = 0.3 \n",
+ "x3d = 0.4 \n",
+ "a12 = 3.91 \n",
+ "a32 = 7.77 \n",
+ "a22 = 1 \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "\t# Equ 6.78 gives\n",
+ "def f14(Q):\n",
+ " return(1- a12*x1d/(a12-Q)-a22*x2d/(a22-Q)-a32*x3d/(a32-Q)) \n",
+ "Q = fsolve(f14,2) \n",
+ "\n",
+ "\t# From the problem statement\n",
+ "\t# d1 = D*x1d d2 = D*x2d\n",
+ "d1 = F*x1d*0.95 \t\t\t\t\t# [kmol/h]\n",
+ "d2 = F*x2d*0.05 \t\t\t\t\t# [kmol/h]\n",
+ "d3 = F*x3d*0.997 \t\t\t\t\t# [kmol/h]\n",
+ "\n",
+ "\t\t# Summing the three distillate, d1,d2 and d3\n",
+ "D = d1+d2+d3 \t\t\t\t\t\t# [kmole/h]\n",
+ "\n",
+ "Vmin = a12*d1/(a12-Q)+a22*d2/(a22-Q)+a32*d3/(a32-Q) \n",
+ "\n",
+ "\t\t# From the mass balance \n",
+ "Lmin = Vmin-D \t\t\t\t\t\t# [kmol/h]\n",
+ "\t# Minimum reflux ratio\n",
+ "Rmin = Lmin/D \n",
+ "\n",
+ "#Results\n",
+ "print\"The minimum reflux ratio is \",round(Rmin[0],3) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum reflux ratio is 0.717\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.12,Page number:377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# x-mole fraction a-relative volatility\n",
+ "xA = 0.25 \n",
+ "aA = 4.08 \n",
+ "xB = 0.11 \n",
+ "aB = 1.00 \n",
+ "xC = 0.26 \n",
+ "aC = 39.47 \n",
+ "xD = 0.09 \n",
+ "aD = 10.00 \n",
+ "xE = 0.17 \n",
+ "aE = 2.11 \n",
+ "xF = 0.12 \n",
+ "aF = 0.50 \n",
+ "\n",
+ "FRlkd = 0.98 \n",
+ "FRhkd = 0.01 \n",
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "from numpy import *\n",
+ "\t# For methane\n",
+ "D_CR = (aC-1)/(aA-1)*FRlkd + (aA-aC)/(aA-1)*FRhkd \n",
+ "\t# For ethane\n",
+ "D_DR = (aD-1)/(aA-1)*FRlkd + (aA-aD)/(aA-1)*FRhkd \n",
+ "\t# For butane\n",
+ "D_ER = (aE-1)/(aA-1)*FRlkd + (aA-aE)/(aA-1)*FRhkd \n",
+ "\t# For hexane\n",
+ "D_FR = (aF-1)/(aA-1)*FRlkd + (aA-aF)/(aA-1)*FRhkd \n",
+ "\t# Since the feed is 66% vaporized\n",
+ "q = 1-0.66 \n",
+ "\n",
+ "\t# Now equation 6.82 is solved for two values of Q\n",
+ "def f14(Q1):\n",
+ " return(0.66 - aA*xA/(aA-Q1)-aB*xB/(aB-Q1)-aC*xC/(aC-Q1)-aD*xD/(aD-Q1)-aE*xE/(aE-Q1)-aF*xF/(aF-Q1)) \n",
+ "Q1 = fsolve(f14,1.2) \n",
+ "\n",
+ "def f15(Q2):\n",
+ " return(0.66 - aA*xA/(aA-Q2)-aB*xB/(aB-Q2)-aC*xC/(aC-Q2)-aD*xD/(aD-Q2)-aE*xE/(aE-Q2)-aF*xF/(aF-Q2)) \n",
+ "Q2 = fsolve(f15,2.5) \n",
+ "\n",
+ "\t# Basis: 100 mole of feed\n",
+ "F = 100 \t\t\t\t\t\t\t# [mole]\n",
+ "\t# Let d1 = Dxad, d2 = Dxbd, d3 = Dxcd, and so on\n",
+ "d1 = F*xA*FRlkd \t\t\t\t\t\t# [moles of propane]\n",
+ "d2 = F*xB*FRhkd \t\t\t\t\t\t# [moles of pentane]\n",
+ "d3 = F*xC \t\t\t\t\t\t\t# [moles of methane]\n",
+ "d4 = F*xD \t\t\t\t\t\t\t# [moles of ethane]\n",
+ "d6 = F*xF*0 \t\t\t\t\t\t\t# [moles of hexane]\n",
+ "\t# And d5 is unknown\n",
+ "\t# Applying equation 6,78 for each value of Q\n",
+ "\n",
+ "\t# Solution of simultaneous equation \n",
+ "#Vmin=aA*d1/(aA-Q1)+aB*d2/(aB-Q1)+aC*d3/(aC-Q1)+aD*d4/(aD-Q1)+aE*d5/(aE-Q1)+aF*d6/(aF-Q1)\n",
+ "#Vmin=aA*d1/(aA-Q2)+aB*d2/(aB-Q2)+aC*d3/(aC-Q2)+aD*d4/(aD-Q2)+aE*d5/(aE-Q2)+aF*d6/(aF-Q2)\n",
+ "# we get\n",
+ "d5=-(72.243-121.614)/(2.494+2.863)\n",
+ "Vmin=72.243 + 2.494*D5\n",
+ "\n",
+ "\t# From equ 6.84\n",
+ "D = d1+d2+d3+d4+d5+d6 \t\t\t\t\t\t# [mole]\n",
+ "# From mass balance \n",
+ "Lmin = Vmin-D \t\t\t\t\t\t\t# [mole]\n",
+ "# For minimum reflux ratio\n",
+ "Rmin = Lmin/D \n",
+ "print\"The minimum reflux ratio is\",round(Rmin,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum reflux ratio is 0.384\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.13,Page number:380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable declaration\n",
+ "\t# A-benzene B-toluene C-1,2,3-trimethylbenzene\n",
+ "\t# From example 6.10\n",
+ "Nmin = 4.32 # [stages]\n",
+ "\t# From example 6.11\n",
+ "Rmin = 0.717 \t\t\t\t\t\t# [minimum reflux ratio]\n",
+ "\t# For R = 1\n",
+ "R = 1 \n",
+ "X = (R-Rmin)/(R+1) \n",
+ "\t# From equ 6.88\n",
+ "Y = 1-math.exp((1+54.4*X)/(11+117.2*X)*(X-1)/math.sqrt(X)) \n",
+ "\t# Fro equ 6.86\n",
+ "N = (Y+Nmin)/(1-Y) \n",
+ "\t# From example 6.10 99.7% of the LNK (benzene) is recovered in the distillate# , 95% of \t\tthe light key is in the distillate, and 95% of the heavy key is in# the bottoms\n",
+ "\n",
+ "\t# For a basis of 100 mol of feed, the material balances for three components # are\n",
+ "\t# For distillate\n",
+ "nAd = 39.88 \t\t\t\t\t\t# [LNK, moles of benzene]\n",
+ "nBd = 28.5 \t\t\t\t\t\t# [LK, moles of toluene]\n",
+ "nCd = 1.50 \t\t\t\t\t\t# [HK, moles of 1,2,3-trimethylbenzene]\n",
+ "nTd = nAd+nBd+nCd \t\t\t\t\t# [total number of moles]\n",
+ "xAd = nAd/nTd \n",
+ "xBd = nBd/(nTd) \n",
+ "xCd = nCd/(nTd) \n",
+ "\n",
+ "\t# For bottoms\n",
+ "nAb = 0.12 \n",
+ "nBb = 1.50 \n",
+ "nCb = 28.50 \n",
+ "nTb = nAb+nBb+nCb \n",
+ "xAb = nAb/nTb \n",
+ "xBb = nBb/nTb \n",
+ "xCb = nCb/nTb \n",
+ "\n",
+ "D = nTd \n",
+ "W = nTb \n",
+ "\t# From problem statement\n",
+ "Zlk = 0.3 \n",
+ "Zhk = Zlk \n",
+ "\t# Substituting in equation 6.89\n",
+ "\t# T = Nr/Ns\n",
+ "T = (Zhk/Zlk*W/D*(xBb/xCd)**2)**0.206 \n",
+ "\n",
+ "\t# Solution of simultaneous equation \n",
+ "def H(e):\n",
+ " f1 = e[0]-e[1]*T \n",
+ " f2 = e[0]+e[1]-N \n",
+ " return(f1,f2) \n",
+ "\n",
+ "\n",
+ "\t# Initial guess\n",
+ "e = [5,4] \n",
+ "y = fsolve(H,e) \n",
+ "Nr = y[0] \t\t\t\t\t# [number of stages in rectifying section]\n",
+ "Ns = y[1] \t\t\t\t\t# [number of stages in stripping section]\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Nr=\",round(Nr,2),\"Ns=\",round(Ns,2),\"\\n\"\n",
+ "print\"Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nr= 5.39 Ns= 4.53 \n",
+ "\n",
+ "Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.14,Page number:387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\t# a-acetone b-methanol c-water\n",
+ "yna = 0.2971 \n",
+ "yn1a = 0.17 \n",
+ "ynIa = 0.3521 \n",
+ "mnIa = 2.759 \n",
+ "xna = 0.1459 \n",
+ "ynb = 0.4631 \n",
+ "yn1b = 0.429 \n",
+ "ynIb = 0.4677 \n",
+ "mnIb = 1.225 \n",
+ "xnb = 0.3865 \n",
+ "ync = 0.2398 \n",
+ "yn1c = 0.4010 \n",
+ "ynIc = 0.1802 \n",
+ "mnIc = 0.3673 \n",
+ "xnc = 0.4676 \n",
+ "\n",
+ "Fabv = 4.927 \t\t\t\t\t\t# [mol/square m.s]\n",
+ "Facv = 6.066 \t\t\t\t\t\t# [mol/square m.s]\n",
+ "Fbcv = 7.048 \t\t\t\t\t\t# [mol/square m.s]\n",
+ "aI = 50 \t\t\t\t\t\t# [square m]\n",
+ "Vn1 = 188 \t\t\t\t\t\t# [mol/s]\n",
+ "Vn = 194.8 \t\t\t\t\t\t# [mol/s]\n",
+ "\n",
+ "print \"Solution6.14(a)\" \n",
+ "from numpy import *\n",
+ "import math\n",
+ "\t# Solution(a)\n",
+ "\n",
+ "ya = (yna+ynIa)/2 \n",
+ "yb = (ynb+ynIb)/2 \n",
+ "yc = (ync+ynIc)/2 \n",
+ "\n",
+ "Rav = ya/Facv+yb/Fabv+yc/Facv \n",
+ "Rbv = yb/Fbcv+ya/Fabv+yc/Fbcv \n",
+ "\n",
+ "Rabv = -ya*(1/Fabv-1/Facv) \n",
+ "Rbav = -yb*(1/Fabv-1/Fbcv) \n",
+ "\t# Thus in matrix form\n",
+ "Rv =matrix([[Rav,Rabv],[Rbav,Rbv]]) \n",
+ "kv = Rv.I \t\t\t# [inverse of Rv]\n",
+ "\t# From equ 6.99\n",
+ "b =matrix([[yna-ynIa],[ynb-ynIb]]) \n",
+ "J = kv*b \n",
+ "\n",
+ "\t# From equ 6.98\n",
+ "Jc = -sum(J) \t\t\t\t# [mol/square m.s]\n",
+ "\n",
+ "print\"The molar diffusional rates of acetone, methanol and water are\",round(J[0][0],4),\"mol/square m.s\",round(J[1][0],4),\"mol/square m.s and\",round(Jc,3) ,\"mol/square m.s respectively\"\n",
+ "print \"Solution 6.14(b)\\n\" \n",
+ "\t# Solution(b)\n",
+ "Ntv = Vn1-Vn \t\t\t\t\t\t# [mol/s]\n",
+ "\n",
+ "\t# From equation 6.94\n",
+ "Nta = aI*J[0][0]+ya*Ntv \n",
+ "Ntb = aI*J[1][0]+yb*Ntv \n",
+ "Ntc = aI*Jc+yc*Ntv \n",
+ "print\"The mass transfer rates of acetone, methanol and water are\",round(Nta,1),\"mol/s\" ,round(Ntb,1),\" mol/s and\", round(Ntc),\" mol/s respectively\"\n",
+ "\n",
+ "print \"Example6.14(c)\\n\"\n",
+ "\t# Solution(c)\n",
+ "\n",
+ "\t# Approximate values of Murphree vapor tray efficiency are obtained from # equation \t6.105\n",
+ "\n",
+ "EMG_a = (yna-yn1a)/(mnIa*xna-yn1a) \n",
+ "EMG_b = (ynb-yn1b)/(mnIb*xnb-yn1b) \n",
+ "EMG_c = (ync-yn1c)/(mnIc*xnc-yn1c) \n",
+ "\n",
+ "print\"The Murphree vapor tray efficiencies for acetone, methanol and water are\",round(EMG_a,3), round(EMG_b,3),\"and\",round(EMG_c,3), \"respectively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Solution6.14(a)\n",
+ "The molar diffusional rates of acetone, methanol and water are"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " -0.3068 mol/square m.s -0.0824 mol/square m.s and 0.389 mol/square m.s respectively\n",
+ "Solution 6.14(b)\n",
+ "\n",
+ "The mass transfer rates of acetone, methanol and water are -17.5 mol/s -7.3 mol/s and 18.0 mol/s respectively\n",
+ "Example6.14(c)\n",
+ "\n",
+ "The Murphree vapor tray efficiencies for acetone, methanol and water are 0.547 0.767 and 0.703 respectively\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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