9 files changed, 485 insertions, 0 deletions
diff --git a/905/CH7/EX7.10/7_10.sce b/905/CH7/EX7.10/7_10.sce
new file mode 100755
index 000000000..815cf58e2
--- /dev/null
+++ b/905/CH7/EX7.10/7_10.sce
@@ -0,0 +1,67 @@
+clear;
+clc;
+
+// Illustration 7.10
+// Page: 461
+
+printf('Illustration 7.10 -  Page: 461\n\n');
+
+// solution
+//*****Data*****//
+Dd = 1.15*10^-9; // [molecular diffusivity of furfural in water, square m/s]
+Dc = 2.15*10^-9; // [molecular diffusivity of furfural in toluene, square m/s]
+m = 10.15; // [equilibrium distribution coefficient, cubic m raffinate/cubic m extract]
+
+printf('Illustration 7.10(a) -  Page: 461\n\n');
+// Solution(a)
+// From example 7.8 and 7.9
+dvs = 3.26*10^-4; // [m]
+Shd = 6.6; // [sherwood number for dispersed phase]
+// From equation 7.52
+kd = Shd*Dd/dvs; // [dispersed phase mass transfer coefficient, m/s]
+printf("The dispersed-phase mass-transfer coefficient is %e m/s.\n\n",kd);
+
+printf('Illustration 7.10(b) -  Page: 461\n\n');
+// Solution(b)
+
+dd = 998;
+dc = 868; // [density of continuous phase, kg/cubic m]
+uc = 0.59*10^-3; // [viscosity of continuous phase, kg/m.s]
+ohm = 182.2; // [rpm]
+g = 9.8; // [square m/s]
+Di = 0.288; // [m]
+sigma = 0.025; // [N/m]
+phiD = 0.385;
+Dt = 0.863; // [m] 
+Scc = uc/(dc*Dc);
+Rec = Di^2*ohm/60*dc/uc;
+Fr = Di*(ohm/60)^2/g;
+Eo = dd*dvs^2*g/sigma;
+
+// From equation 7.53
+Shc = 1.237*10^-5*Rec^(2/3)*Scc^(1/3)*Fr^(5/12)*Eo^(5/4)*phiD^(-1/2)*(Di/dvs)^2*(dvs/Dt)^(1/2);
+// Therefore 
+kc = Shc*Dc/dvs; // [continuous phase mass transfer coefficient, m/s]
+printf("The continuous-phase mass-transfer coefficient is %e m/s.\n\n",kc);
+
+printf('Illustration 7.10(c) -  Page: 462\n\n');
+// Solution(c)
+
+a = 7065; // [square m/cubic m]
+Vt = 0.504; // []
+Qd = 0.097/60; // [cubic m/s]
+Qc = 0.155/60; // [cubic m/s]
+
+// From equation 7.40
+Kod = kd*kc*m/(m*kc+kd); // [m/s]
+// From equation 7.45
+N_tod = Kod*a*Vt/Qd;
+// From equation 7.46
+Emd = N_tod/(1+N_tod); 
+printf("The Murphree dispersed phase efficiency is %f.\n\n",Emd);
+
+printf('Illustration 7.10(d) -  Page: 462\n\n');
+// Solution(d)
+// From equation 7.57
+fext = Emd/(1+Emd*Qd/(m*Qc));
+printf("The fractional extraction of furfural is %f.\n\n",fext);
+\ No newline at end of file
diff --git a/905/CH7/EX7.11/7_11.sce b/905/CH7/EX7.11/7_11.sce
new file mode 100755
index 000000000..15a3b0d4f
--- /dev/null
+++ b/905/CH7/EX7.11/7_11.sce
@@ -0,0 +1,36 @@
+clear;
+clc;
+
+// Illustration 7.11
+// Page: 466
+
+printf('Illustration 7.11 -  Page: 466\n\n');
+
+// solution
+//*****Data*****//
+// Preliminary Design of an RDC
+T = 293; // [K]
+F1 = 12250; // [flow rate for dispersed organic phase, kg/h]
+F2 = 11340; // [flow rate for continuous aqueous phase, kg/h]
+d1 = 858; // [kg/cubic m]
+d2 = 998; // [kg/cubic m]
+n = 12; // [Equilibrium stages]
+//*****//
+Qd = F1/d1; // [cubic m/h]
+Qc = F2/d2; // [cubic m/h]
+
+// Assume that based on information in Table 7.5
+// Vd+Vc = V = 22 m/h
+V = 22; // [m/h]
+// Therefore column cross sectional area 
+Ac = (Qd+Qc)/V; // [square m]
+// Column diameter
+Dt = sqrt(4*Ac/%pi); // [m]
+
+// Assume that based on information in Table 7.5
+// 1/HETS = 2.5 to 3.5   m^-1
+// Therefore
+HETS = 1/3; // [m/theoritical stages]
+// Column height
+Z = n*HETS; // [m]
+printf("The height and diameter of an RDC to extract acetone from a dilute toluene-acetone solution is %f m and %f square m respectively\n\n",Z,Dt);  
+\ No newline at end of file
diff --git a/905/CH7/EX7.2/7_2.sce b/905/CH7/EX7.2/7_2.sce
new file mode 100755
index 000000000..851e8f025
--- /dev/null
+++ b/905/CH7/EX7.2/7_2.sce
@@ -0,0 +1,33 @@
+clear;
+clc;
+
+// Illustration 7.2
+// Page: 433
+
+printf('Illustration 7.2 -  Page: 433\n\n');
+
+// solution
+//*****Data*****//
+//  'b'-solvent   'f'-feed   'r'-raffinate   'e'-extract   'c'-one of the   // component in  feed
+F = 50; // [feed rate, kg/h]
+S = 50; // [solvent rate, kg/h]
+xcf = 0.6;
+xbf = 0;
+ycs = 0;
+ybs = 1.0;
+// The equilibrium data for this system can be obtained from Table 7.1 and    // Figure 7.6
+// Plot streams F (xcF = 0.6, xBF = 0.0) and S (yes = 0.0, yBs = 1.0). After  // locating streams F and S, M is on the line FS; its exact location is found // by calculating xcm from
+
+xcm = (F*xcf+S*ycs)/(F+S);
+
+// From figure 7.8
+xcr = 0.189;
+xbr = 0.013;
+yce = 0.334;
+ybe = 0.648;
+M = F+S; // [kg/h]
+// From equation 7.8 
+E = M*(xcm-xcr)/(yce-xcr); // [kg/h]
+R = M-E; // [kg/h]
+printf("The extract and raffinate flow rates are %f kg/h and %f kg/h respectively.\n\n",E,R);
+printf("The compositions when one equilibrium stage is used for the separation is %f and %f in raffinate phase for component b and c respectively and %f and %f in extract phase for component b and c respectively.\n\n",xcr,xbr,yce,ybe);
+\ No newline at end of file
diff --git a/905/CH7/EX7.4/7_4.sce b/905/CH7/EX7.4/7_4.sce
new file mode 100755
index 000000000..5b08fdcde
--- /dev/null
+++ b/905/CH7/EX7.4/7_4.sce
@@ -0,0 +1,57 @@
+clear;
+clc;
+
+// Illustration 7.4
+// Page: 439
+
+printf('Illustration 7.4 -  Page: 439\n\n');
+
+// solution
+//*****Data*****//
+// C-acetic acid    A-water
+// f-feed   r-raffinate   s-solvent
+f = 1000; // [kg/h]
+xCf = 0.35; // [fraction of acid]
+xAf = 1-xCf; // [fraction of water]
+// Solvent is pure
+xAr = 0.02;
+yCs = 0;
+//*****//
+
+printf('Illustration 7.4(a) -  Page: 440\n\n');
+// Solution(a)
+
+// From Figure 7.15
+xCMmin = 0.144;
+// From equation 7.11
+Smin = f*(xCMmin-xCf)/(yCs-xCMmin); // [kg/h]
+printf("The minimum amount of solvent which can be used is %f kg/h.\n\n",Smin);
+
+printf('Illustration 7.4(b) -  Page: 441\n\n');
+// Solution(b)
+
+S = 1.6*Smin; // [kg/h]
+// From equation 7.11
+xCM = (f*xCf+S*yCs)/(f+S);
+
+// Data for equilibrium line
+// Data_eqml = [xCeq yCeq]
+Data_eqml = [0.0069 0.0018;0.0141 0.0037;0.0289 0.0079;0.0642 0.0193;0.1330 0.0482;0.2530 0.1140;0.3670 0.2160;0.4430 0.3110;0.4640 0.3620];
+
+// Data for operating line
+// Data_opl = [xCop yCop]
+Data_opl = [0.02 0;0.05 0.009;0.1 0.023;0.15 0.037;0.20 0.054;0.25 0.074;0.30 0.096;0.35 0.121];
+
+
+scf(1);
+plot(Data_eqml(:,1),Data_eqml(:,2),Data_opl(:,1),Data_opl(:,2));
+xgrid();
+legend('Equilibrium line,Operating line');
+xlabel("wt fraction of acetic acid in water solutions, xC");
+ylabel("wt fraction of acetic acid in ether solutions, yC");
+
+// Now number of theoritical stages is determined by drawing step by step  // stairs from xC = 0.35 to xC = 0.02
+// From figure 7.16
+// Number of theoritical stages 'N' is
+N = 8; 
+printf("The number of theoretical stages if the solvent rate used is 60 percent above the minimum is %f.\n\n",N);
+\ No newline at end of file
diff --git a/905/CH7/EX7.5/7_5.sce b/905/CH7/EX7.5/7_5.sce
new file mode 100755
index 000000000..ae269489b
--- /dev/null
+++ b/905/CH7/EX7.5/7_5.sce
@@ -0,0 +1,54 @@
+clear;
+clc;
+
+// Illustration 7.5
+// Page: 444
+
+printf('Illustration 7.5 -  Page: 444\n\n');
+
+// solution
+//*****Data*****//
+// C-nicotine    A-water    B-kerosene
+// F-feed    R-raffinate    S-solvent
+F = 1000; // [feed rate, kg/h]
+xAF = 0.99; // [fraction of water in feed]
+// Because the solutions are dilute therefore
+xCF = 0.01; // [fraction of nicotene in feed, kg nicotene/kg water]
+xCR = 0.001; // [fraction of nicotene in raffinate, kg nicotene/kg water ]
+m = 0.926; // [kg water/kg kerosene]
+//*****//
+
+printf('Illustration 7.5(a) -  Page: 444\n\n');
+// Solution(a)
+
+yCS = 0; // [kg nicotene/kg water]
+
+// Because, in this case, both the equilibrium and operating lines are       // straight,if the minimum solvent flow rate Bmin is used, the concentration // of the exiting extract, yCmax, will be in equilibrium with xCF. Therefore
+yCmax = m*xCF; // [kg nicotene/kg kerosene]
+
+A = F*xAF; // [kg water/h]
+// From equation 7.17
+Bmin = A*(xCF-xCR)/(yCmax-yCS); // [kg kerosene/h]
+printf("The minimum amount of solvent which can be used is %f kg kerosene/h.\n\n",Bmin);
+
+printf('Illustration 7.5(b) -  Page: 444\n\n');
+// Solution(b)
+
+B = 1.2*Bmin; // [kg kerosene/h]
+EF = m*B/A;
+Nt = log((xCF-yCS/m)/(xCR-yCS/m)*(1-1/EF)+1/EF)/log(EF);
+
+printf("The number of theoretical stages if the solvent rate used is 20 percent above the minimum is %f .\n\n",Nt);
+
+printf('Illustration 7.5(c) -  Page: 444\n');
+// Solution(c)
+
+Eme = 0.6; // [Murphree stage efficiency]
+// from equation 7.20
+Eo = log(1+Eme*(EF-1))/log(EF); // [overall efficiency]
+Nr = Nt/Eo; // [number of real stages]
+disp(Nr);
+// The nearest integer to number of real stages is 11
+// Therefore
+Nr = 11;
+printf("The number of real stages required is %f.\n\n",Nr);
+\ No newline at end of file
diff --git a/905/CH7/EX7.6/7_6.sce b/905/CH7/EX7.6/7_6.sce
new file mode 100755
index 000000000..e63bdb005
--- /dev/null
+++ b/905/CH7/EX7.6/7_6.sce
@@ -0,0 +1,96 @@
+clear;
+clc;
+
+// Illustration 7.6
+// Page: 449
+
+printf('Illustration 7.6 -  Page: 449\n\n');
+
+// solution
+//*****Data*****//
+// C-styrene    A-ethylbenzene    B-diethylene glycol
+F = 1000; // [kg/h]
+XF = 0.6; // [wt fraction of styrene]
+XPE = 0.9;
+XN = 0.1;
+// All above fractions are on solvent basis
+// Equilibrium Data for Ethylbenzene (A)-Diethylene Glycol (B)-Styrene (C) at 298 K
+// Data_eqm = [X Y];
+// X - kg C/kg (A+C) in raffinate solution
+// Y - kg C/kg (A+C) in extract solution
+Data_eqm = [0 0;0.087 0.1429;0.1883 0.273;0.288 0.386;0.384 0.48;0.458 0.557;0.464 0.565;0.561 0.655;0.573 0.674;0.781 0.863;0.9 0.95;1 1];
+//*****//
+
+printf('Illustration 7.6(a) -  Page: 449\n\n');
+// Solution(a)
+
+// Minimum theoretical stages are determined on the XY equilibrium distribution diagram, stepping them off from the diagonal line to the equilibrium curve, beginning at XPE = 0.9 and ending at XN = 0.1
+
+Data_opl = [0 0;0.09 0.09;0.18 0.18;0.27 0.27;0.36 0.36;0.45 0.45;0.54 0.54;0.63 0.63;0.72 0.72;0.81 0.81;0.90 0.90;1 1;];
+
+scf(1);
+plot(Data_eqm(:,1),Data_eqm(:,2),Data_opl(:,1),Data_opl(:,2));
+xgrid();
+legend('Equilibrium line','Operating line');
+xlabel("X,kg C/kg (A+C) in raffinate solution");
+ylabel("Y,kg C/kg (A+C) in extract solution");
+
+// Figure 7.20
+Nmin = 9; // [number of ideal stages]
+
+printf("The minimum number of theoretical stages are %f.\n\n",Nmin);
+
+printf('Illustration 7.6(b) -  Page: 450\n\n');
+// Solution(b)
+
+// Since the equilibrium-distribution curve is everywhere concave downward// ,the tie line which when extended passes through F provides the minimum
+// reflux ratio
+// From figure 7.19
+NdeltaEm = 11.04;
+NE1 = 3.1;
+// From equation 7.30
+// Y = R_O/P_E, external reflux ratio
+Ymin = (NdeltaEm-NE1)/NE1; // [kg reflux/kg extract product]
+
+printf("The minimum extract reflux ratio is %f kg reflux/kg extract product.\n\n",Ymin);
+
+printf('Illustration 7.6(c) -  Page: 450\n\n');
+// Solution(c)
+
+Y = 1.5*Ymin; // [kg reflux/kg extract product]
+// From equation 7.30
+NdeltaE = Y*NE1+NE1;
+// From figure 7.19
+NdeltaR = -24.90;
+// From figure 7.21
+N = 17.5; // [number of equilibrium stages]
+
+// From figure 7.19
+// For XN = 0.1 NRN = 0.0083
+NRN = 0.0083;
+// Basis: 1 hour
+
+// e = [P_E R_N] 
+// Solution of simultaneous equation
+function[f]=G(e)
+    f(1) = F - e(1) - e(2);
+    f(2) = F*XF-e(1)*XPE-e(2)*XN;
+    funcprot(0);
+endfunction
+// Initial guess:
+e = [600 300];
+y = fsolve(e,G);
+P_E = y(1); // [kg/h]
+R_N = y(2); // [kg/h]
+
+R_O = Y*P_E; // [kg/h]
+E_1 = R_O+P_E; // [kg/h]
+
+B_E = E_1*NE1; // [kg/h]
+E1 = B_E+E_1; // [kg/h]
+RN = R_N*(1+NRN); // [kg/h]
+S = B_E+R_N*NRN; // [kg/h]
+
+printf("The number of theoretical stages are %f.\n",N);
+printf('The important flow quantities at an extract reflux ratio of 1.5 times the minimum value are\n\n');
+printf(" PE = %f kg/h\n RN = %f kg/h\n RO = %f kg/h\n E1 = %f kg/h\n BE = %f kg/h\n E1 = %f kg/h\n RN = %f kg/h\n S = %f kg/h\n",P_E,R_N,R_O,E_1,B_E,E1,RN,S);
+\ No newline at end of file
diff --git a/905/CH7/EX7.7/7_7.sce b/905/CH7/EX7.7/7_7.sce
new file mode 100755
index 000000000..d270799f0
--- /dev/null
+++ b/905/CH7/EX7.7/7_7.sce
@@ -0,0 +1,49 @@
+clear;
+clc;
+
+// Illustration 7.7
+// Page: 454
+
+printf('Illustration 7.7 -  Page: 454\n\n');
+
+// solution
+//*****Data*****//
+Ff = 1.89; // [cubic m/min]
+Fs = 2.84; // [cubic m/min]
+t = 2; // [min]
+//*****//
+
+printf('Illustration 7.7(a) -  Page: 454\n\n');
+// Solution(a)
+
+Q = Ff+Fs; // [total flow rate, cubic m/min]
+Vt = Q*t; // [cubic m]
+// For a cylindrical vessel H = Dt
+Dt = (4*Vt/%pi)^(1/3); // [m]
+H = Dt; // [m]
+printf("The diameter and height of each mixing vessel is %f m and %f m respectively.\n\n",Dt,H);
+
+printf('Illustration 7.7(b) -  Page: 454\n\n');
+// Solution(b)
+// Based on a recommendation of Flynn and Treybal (1955),
+P = 0.788*Vt; // [mixer power, kW]
+printf("The agitator power for each mixer is %f kW.\n\n",P);
+
+printf('Illustration 7.7(c) -  Page: 454\n\n');
+// Solution(c)
+
+// Based on the recommendation by Ryan et al. (1959), the disengaging area  // in the settler is
+// Dt1*L1 = Q/a = Y
+a = 0.2; // [cubic m/min-square m]
+Y = Q/a; // [square m]
+// For L/Dt = 4
+Dt1 = (Y/4)^0.5; // [m]
+L1 = 4*Dt1; // [m]
+printf("The diameter and length of a settling vessel is %f m and %f m respectively.\n\n",Dt1,L1);
+
+printf('Illustration 7.7(d) -  Page: 454\n\n');
+// Solution(d)
+// Total volume of settler
+Vt1 = %pi*Dt1^2*L1/4; // [cubic m]
+tres1 = Vt1/Q; // [min]
+printf("The residence time in the settling vessel is %f min.\n\n",tres1); 
+\ No newline at end of file
diff --git a/905/CH7/EX7.8/7_8.sce b/905/CH7/EX7.8/7_8.sce
new file mode 100755
index 000000000..559006465
--- /dev/null
+++ b/905/CH7/EX7.8/7_8.sce
@@ -0,0 +1,67 @@
+clear;
+clc;
+
+// Illustration 7.8
+// Page: 456
+
+printf('Illustration 7.8 -  Page: 456\n\n');
+
+// solution
+//*****Data*****//
+Ff = 1.61; // [flow rate of feed, kg/s]
+Fs = 2.24; // [flow rate of solvent, kg/s]
+t = 2*60; // [residence time in each mixer, s]
+df = 998; // [density of feed, kg/cubic m]
+uf = 0.89*10^-3; // [viscosity of feed, kg/m.s]
+ds = 868; // [density of solvent, kg/cubic m]
+us = 0.59*10^-3; // [viscosity of solvent, kg/m.s]
+sigma = 0.025; // [interfacial tension, N/m]
+g = 9.8; // [square m/s]
+//*****//
+
+Qf = Ff/df; // [volumetric flow rate of feed, cubic m/s]
+Qs = Fs/ds; // [volumetric flow rate of solvent, cubic m/s]
+// Volume fractions in the combined feed and solvent entering the mixer 
+phiE = Qs/(Qs+Qf); 
+phiR = 1-phiE;
+
+printf('Illustration 7.8(a) -  Page: 457\n\n');
+// Solution(a)
+
+Q = Qf+Qs; // [total flow rate, cubic m/s]
+Vt = Q*t; // [vessel volume, cubic m]
+// For a cylindrical vessel,  H = Dt
+// Therefore,    Vt = %pi*Dt^3/4
+Dt = (4*Vt/%pi)^(1/3); // [ diameter, m]
+H = Dt; // [height, m]
+Di = Dt/3; // [m]
+printf("The height and diameter of the mixing vessel are %f m and %f m respectively.\n",Dt,H);
+printf("The diameter of the flat-blade impeller is %f m.\n\n",Di);
+
+printf('Illustration 7.8(b) -  Page: 457\n\n');
+// Solution(b)
+
+// For the raffinate phase dispersed:
+phiD = phiR;
+phiC = phiE;
+deltad = df-ds; // [kg/cubic m]
+rowM = phiD*df+phiC*ds; // [kg/cubic m]
+uM = us/phiC*(1 + 1.5*uf*phiD/(us+uf)); // [kg/m.s]
+// Substituting in equation 7.34
+ohm_min = sqrt(1.03*phiD^0.106*g*deltad*(Dt/Di)^2.76*(uM^2*sigma/(Di^5*rowM*g^2*deltad^2))^0.084/(Di*rowM))*60; // [rpm]
+printf("The minimum rate of rotation of the impeller for complete and uniform dispersion.is %f rpm.\n\n",ohm_min);
+
+printf('Illustration 7.8(c) -  Page: 457\n\n');
+// Solution(c)
+
+ohm = 1.2*ohm_min; // [rpm]
+
+// From equation 7.37
+Re = ohm/60*Di^2*rowM/uM; // [Renoylds number]
+// Then according to Laity and Treybal (1957), the power number, Po = 5.7
+Po = 5.7
+// From equation 7.37
+P = Po*(ohm/60)^3*Di^5*rowM/1000; // [kW]
+// Power density
+Pd = P/Vt; // [kW/cubic m]
+printf("The power requirement of the agitator at 1.20 times the minimum rotation rate is %f kW.\n\n",P);
+\ No newline at end of file
diff --git a/905/CH7/EX7.9/7_9.sce b/905/CH7/EX7.9/7_9.sce
new file mode 100755
index 000000000..21d7601f8
--- /dev/null
+++ b/905/CH7/EX7.9/7_9.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+
+// Illustration 7.9
+// Page: 460
+
+printf('Illustration 7.9 -  Page: 460\n\n');
+
+// solution
+//*****Data*****//
+// From example 7.8
+Di = 0.288; // [m]
+sigma = 0.025; // [N/m]
+ohm = 152*1.2/60; // [rps]
+ds = 868; // [kg/cubic m]
+phiD = 0.385;
+
+// Therefore from equation 7.49
+We = Di^3*ohm^2*ds/sigma; // [Weber number]
+
+// From equation 7.50
+dvs = Di*0.052*(We)^-0.6*exp(4*phiD); // [m]
+disp(dvs);
+// Substituting in equation 7.48
+a = 6*phiD/dvs; // [square m/cubic m]
+printf("The Sauter mean drop diameter and the interfacial area is %e m and %f square m/cubic m respectively.\n\n",dvs,a); 
+\ No newline at end of file