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diff --git a/839/CH22/EX22.1/Example_22_1.sce b/839/CH22/EX22.1/Example_22_1.sce
new file mode 100755
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+++ b/839/CH22/EX22.1/Example_22_1.sce
@@ -0,0 +1,39 @@
+//clear//

+clear;

+clc;

+

+//Example 22.1

+//Given

+Dp = 1; //[in.]

+vdot = 25000; //[ft^3/h]

+T = 68; //[F]

+P = 1; //[atm]

+ya = 0.02;

+Mair = 29;

+Mg = 17;

+//Solution

+//The average molecular weiht of the entering gas 

+M = (1-ya)*Mair+ya*Mg;

+rho_y = M*492/(359*(460+68)); //[lb/ft^3]

+

+//(a)

+//Using Fig. 22.5, when Gy =Gx;

+Gy = 0.472; //[lb/ft^2-s]

+Gx = Gy; //[lb/ft^2-h]

+des_value = Gy/2; //[lb/ft^2-h]

+mdot = vdot*rho_y/3600; //[lb/s]

+//Cross-sectional area of the tower

+S = mdot/des_value //[ft^2] 

+//the diameter of the tower is

+Dtower = sqrt(4*S/%pi); //[ft]

+disp('ft',Dtower,'The tower diameter is');

+

+//(b)

+h = 20; //[ft] 

+//Using Fig 22.4, the pressure drop for 

+Gy = 850; //[lb/f^2-h]

+Gx = Gy;

+delta_P = 0.35; //[in.] (H2O/ft)

+//The total pressure drop

+Pt = delta_P*h; //[in.] H2O

+disp('in. H2O' ,Pt,'The pressure drop would be'); 

diff --git a/839/CH22/EX22.2/Example_22_2.sce b/839/CH22/EX22.2/Example_22_2.sce
new file mode 100755
index 000000000..b827fa813
--- /dev/null
+++ b/839/CH22/EX22.2/Example_22_2.sce
@@ -0,0 +1,41 @@
+//clear//

+clear;

+clc;

+

+//Example 22.2

+//Given

+Dp = 1; //[in.]

+vdot = 25000; //[ft^3/h]

+T = 68; //[F]

+P = 1; //[atm]

+ya = 0.02;

+Mair = 29;

+Mg = 17;

+//Solution

+//The average molecular weiht of the entering gas 

+M = (1-ya)*Mair+ya*Mg;

+rho_y = M*492/(359*(460+68)); //[lb/ft^3]

+rho_x = 62.3; //[lb/ft^3]

+//(a)

+//Using Fig.(22.8), from Example 22.1 A = Gx/Gy = 1 and

+//Let 

+A = 1;

+B = A*sqrt(rho_y/rho_x);

+//Form Fig 22.8, the superficial vapor velocity at flooding

+//is uof*sqrt(rho_y/(rho_x-rho_y))=0.11, therefore

+uof = 0.11/sqrt(rho_y/(rho_x-rho_y)); //[m/s]

+//The allowable vapor velocity

+uo = uof*0.5; //[m/s]

+uo = uo*3.28; //[ft/s]

+//the corresponding mass velocity

+Gy = uo*rho_y; //[lb/ft^2-s]

+//The allowable mass velocity in the example was 0.236 lb/ft^2-s.

+//The increase by using structured packing is

+increase = (Gy/0.236)-1;

+disp(increase*100,'The percent increase in mass velocity is');

+

+//(b)

+//The pressure drop

+delta_P = 20*1.22*(0.5/0.9)^1.8; //[in. H2O]

+//This is 1.2 times the pressure drop of 7 in.H2O in the Intolax saddles.

+disp('The pressure drop will be greater than Intolax Saddles')

diff --git a/839/CH22/EX22.3/Example_22_3.sce b/839/CH22/EX22.3/Example_22_3.sce
new file mode 100755
index 000000000..d71021ae7
--- /dev/null
+++ b/839/CH22/EX22.3/Example_22_3.sce
@@ -0,0 +1,90 @@
+//clear//

+clear;

+clc;

+

+//Example 22.3

+//Given

+vdot = 4500; //[SCFM]

+yin = 0.06;

+yout = 0.0002;

+P = 1; //[atm]

+Tiy = 20; //[C]

+Tix = 25; //[C]

+

+//Solution

+//From Perry

+x = [0.0308,0.0406,0.0503,0.0735]';

+y20 = [0.0239,0.0328,0.0417,0.0658]';

+y30 = [0.0389,0.0528,0.0671,0.1049]';

+y40 = [0.0592,0.080,0.1007,0.1579]';

+deltaH = -8.31*10^3; //[cal/g mol], fro NH3=NH3(aq)

+//Basis:

+gas_in = 100; //[g mol dry]

+air_in = (1-yin)*gas_in; //[mol]

+NH3_in = yin*gas_in; //[mol]

+H2O_in = 2.4; //[mol]

+air_out = air_in; //[mol]

+//The moles of NH3 in the outlet gas,

+NH3_out = air_out*(yout/(1-yout)); //[mol NH3]

+//The amount of NH3 absorbed

+NH3_abs = NH3_in-NH3_out; //[mol]

+//Heat Effects:

+//The heat of absorption 

+Qa = -NH3_abs*deltaH; //[cal]

+//Sensible heat changes in the gas are

+Qair = air_in*7*5; //[cal]

+QH2O =H2O_in*8*5; //[cal]

+Qsy = 3290+96; //[cal]

+//The amount of vaporization of water from the liquid 

+pH2O_20 = 17.5; //[mm Hg], at 20C

+pH2O_25 = 23.7; //[mm Hg], at 25C

+H2O_inlet = gas_in*(pH2O_20/742.5); //[mol]

+H2O_outlet = 94.02*(pH2O_25/736.3); //[mol]

+//The amount of water vaporized

+H2O_vaporized = H2O_outlet-H2O_inlet; //[mol]

+deltaHv = 583; //[cal/g]

+Qv = deltaHv*H2O_vaporized*18.02; //[cal]

+//From Eq.(22.31)

+Qsx = Qa-(Qv+Qsy); //[cal]

+

+Cp = 18; //[cal/g-mol-C]

+xmax = 0.031;

+Tb = 40; //[C]

+Ta = 25; //[C]

+err =1;

+while(err>0.01)

+  Lb = NH3_abs/xmax;

+  Tbnew = Qsx/(Lb*Cp)+Ta;

+  err = Tb-Tbnew;

+  Tb=Tbnew;

+  xmax = xmax+0.002;

+end

+Lmin = Lb-NH3_in; //[mol H2O]

+La = 1.25*Lmin; //[mol]

+Lb = La+NH3_in; //[mol]

+//The temperature rise of the liquuid is

+Tb = Qsx/(Lb*Cp)+Ta; //[C]

+xb = NH3_in/La; //[C]

+ystar = 0.044;

+//Assuming temperature to be linear function of x, so 

+T = 30;

+//x = 0.0137;

+//Using the data given for 30C and interpolating to get the 

+//initial slope for 25 and the final value ystar for 35, the

+//euilibrium line is drawn

+y = [0.06, 0.03,0.01,0.0002]';

+ystar = [0.048,0.017,0.0055,0]';

+delta_y = y-ystar;

+delta_yL = [0.0125, 0.0080,0.00138]';

+delta_NOy = [2.4,2.5,7.1]';

+NOy = sum(delta_NOy);

+disp(NOy,'The value of NOy is');  

+

+

+

+plot(x,y20,x,y30,x,y40);

+xgrid();

+xlabel('x');

+ylabel('y');

+legend('20C','30C','40C');

+title('x vs y of NH3 at different temperatures'); 

diff --git a/839/CH22/EX22.4/Example_22_4.sce b/839/CH22/EX22.4/Example_22_4.sce
new file mode 100755
index 000000000..46c73f521
--- /dev/null
+++ b/839/CH22/EX22.4/Example_22_4.sce
@@ -0,0 +1,60 @@
+//clear//

+clear;

+clc;

+

+//Example 22.4

+//Given

+ieee();

+H = 0.0075; //[TCE]

+T = 20; //[C]

+P = 1; //[atm]

+wa = 6*10^-6; //[g]

+Ca = 6; //[ppm]

+wb = 4.5*10^-9 //[g]

+M = 18;

+

+//Solution

+m = H/P*10^6/M;

+//With this large value of m, the desorption is liquid-phase controlled.

+//At the minimum air rate, the exit gas will be in equilibrium with the

+//incoming solution.

+MTCE = 131.4;

+j = 1.5;

+for i = 1:7

+xa = wa/MTCE*M; 

+ya = m*xa;

+//Per cubic meter of solution fed, the TCE removed is

+VTCE = 10^6*(wa-wb)/MTCE; //[mol]

+//The total amount of gas leaving is

+V = VTCE/ya; //[mol]

+Fmin = V*0.0224; //[std m^3], as 1 gmol = 0.0224 std m^3

+Vmin = Fmin*j;

+//Density at the standard conditions,

+rho = 1.259; //[kg/m^3],

+//so the minimum rate on a mass basis is,

+//Let A = (Gy/Gx)min

+A = Vmin*rho/1000; //[kg air/kg water]

+//If the air rate is 1.5 times the minimum value, then

+ya = ya/j;

+xastar = ya/m;

+Castar = xastar*MTCE/M *10^6; //[ppm]

+delta_Ca = Ca-Castar;

+

+//At bottom

+Cb = 0.0045; //[ppm]

+Cbstar = 0; //[ppm] 

+delta_Cb = Cb-Cbstar; //[ppm]

+delta_CL = (delta_Ca-delta_Cb)/log(delta_Ca/delta_Cb); //[ppm]

+Nox(i) = (Ca-Cb)/delta_CL;

+j = j+0.5;

+end

+

+Hox = 3; //[ft]

+Z = Hox*Nox; //[ft]

+//Going from 1.5 to 2Vmin or from 2 to 3Vmin decreases the tower height

+//considerably, and the reduction in pumping work for water is more than

+//the additional energy needed to force air through the column. Further

+//increase in V does not change Z very much, and the optimum air rate is

+//probably in the range 3 to 5Vmin./

+  

+disp(Nox,'Number of Transfer units with minimum air rates')

diff --git a/839/CH22/EX22.5/Example_22_5.sce b/839/CH22/EX22.5/Example_22_5.sce
new file mode 100755
index 000000000..76d8d4dfa
--- /dev/null
+++ b/839/CH22/EX22.5/Example_22_5.sce
@@ -0,0 +1,99 @@
+//clear//

+clear;

+clc;

+

+//Example 22.5

+//Solution

+//Equlibrium data are shown in Fig.22.22

+//By a heat balance similar to that of Eample 22.3

+//The temperature rise of the liqui was estimated

+//to be

+delta_T = 12.5; //[C]

+//Basis:

+dry_gas_in = 100; //[mol]

+sol_in = 140; //[mol]

+N2_in = 87; //[mol]

+CO2_in = 10; //[mol]

+EO_in = 3; //[mol]

+N2_out = 87; //[mol]

+CO2_out = 10; //[mol]

+EO_out = 3*0.02; //[mol]

+IN = N2_in+CO2_in+EO_in; //[mol]

+OUT = N2_out+CO2_out+EO_out; //[mol]

+//Assuming negligible CO2 absorption and neglect effect of H2O on

+//gas composition.

+//At top:

+xt = 0.004;

+yt = EO_out/OUT;

+//Moles of EO absorbed

+EO_abs = 3*0.98; //[mol]

+//Moles of EO absorbed in water

+EO_H2O = 140*0.0004; //[mol]

+//At bottom:

+xb = (EO_abs+EO_H2O)/(140+EO_abs);

+yb = 0.03;

+//From Fig 22.22

+y = [0.03,0.015,0.005,0.0006]';

+delta_y1 = [0.008,0.0006,0.0024,0.0003]';

+

+for i = 1:length(y)-1

+  delta_y = y(i)-y(i+1);

+  delta_yL = (delta_y1(i)-delta_y1(i+1))/log(delta_y1(i)/delta_y1(i+1));

+  Noy1(i) = delta_y/delta_yL;

+end

+Noy = sum(Noy1);

+

+//Column diameter:

+//Using generalize pressure-drop correlation, Fig.22.6

+//Based on the inlet gas,

+Mbar = 0.87*28+0.1*44+0.03*44;

+//At 40C,

+rho_y = 30.1/359*20*273/313 //[lb/ft^3]

+rho_x = 62.2; //[lb/ft^3]

+//Let A = Gx/Gy*sqrt(rho_y/(rho_x-rho_y))

+A = 1.4*18/(1*30.1)*sqrt(rho_y/(rho_x-rho_y));

+//From Fig. 22.6, for

+delta_P = 0.5; //[in.H2O/ft]

+//Let B = Gy^2*Fp*mux^0.1/(rho_y*(rho_x-rho_y)*gc)

+B = 0.045;

+//From Table 22.1,

+Fp = 40;

+mu = 0.656; //[cP]

+//so 

+Gy = sqrt(B*(rho_y)*(rho_x-rho_y)*32.2/(Fp*mu^0.1)); //[lb/ft^2-h]

+//or 

+Gy = Gy*3600; //[lb/ft^2-s]

+Gx = 1.4*18/(1*Mbar)*Gy; //[lb/f^2-s]

+//For a feed rate 

+F = 10000*Mbar; //[lb/h]

+S = F/Gx; //[ft^2]

+D = sqrt(S*4/%pi); //[ft]

+//Column heigth:

+//From Fig. 22.20 at Gy = 500 and Gx = 1500

+Hy_NH3 = 1.4; //[ft]

+mu_40 =0.0181*10^-2; //[P], Appendix 8

+Dv = 7.01*10^-3; //[cm^2/s], from Eq.(21.25)

+rho = 2.34*10^-2; //[lb/ft^3]

+Nsc = mu_40/(rho*Dv);

+//Form Table 22.1,

+fp = 1.36;

+Hy_EO = 1.4*(1.1/0.66)^0.5*1/1.36*(Gy/500)^0.3*(1500/Gx)^0.4; //[ft]

+//Form Fig. 22.19,

+Hx_O2 = 0.9; //[ft]

+Gx1 = 1500;

+mu1 = 0.00656; //[P]

+rho1 = 1; //[lb/ft^3]

+//Using Eq.(21.28)

+Dv1 = 2.15*10^-5; //[cm^2/s]

+Nsc1 = mu1/(rho1*Dv1);

+//Using Eq.(22.35), with the correction factor fp and Nsc = 381, 

+//for O2 in water at 25 C

+Hx_EO = Hx_O2*(Gx/(mu1*100)/(Gx1/0.894))^0.3*(Nsc1/381)^0.5/1.36; //[ft]

+//From Fig 22.22, the average value of m

+m = 1.0;

+//From Eq.(22.30)

+HOy = 1.71+(1*0.96)/1.4; //[ft] 

+

+disp(NOy,'number of transfer units required')

+disp('ft',D,'diameter of the column')

+disp('ft',HOy,'packing height')

diff --git a/839/CH22/EX22.6/Example_22_6.sce b/839/CH22/EX22.6/Example_22_6.sce
new file mode 100755
index 000000000..326e3f8e9
--- /dev/null
+++ b/839/CH22/EX22.6/Example_22_6.sce
@@ -0,0 +1,56 @@
+//clear//

+clear;

+clc;

+

+//Example 22.6

+//Solution

+rho_m = 62.2/18; //[mol/ft^3]

+//kya = 0.025*Gy^0.7*Gx^0.25

+H2ObySO2 = 2*0.98964/0.01036; 

+//and

+xb = 1/(H2ObySO2+1);

+//The molal mass velocity of the feed gas Gm is

+Gm_in = 200/29*(1/0.8); //[mol/ft^2-h]

+SO2_in = Gm_in*0.2; //[mol/ft^2-h]

+Air_in = Gm_in*0.8; //[mol/ft^2-h]

+Air_out = Air_in; //[mol/ft^2-h]

+SO2_out = Air_out*(0.005/(1-0.005)); //[mol/ft^2-h]

+SO2_abs = SO2_in-SO2_out; //[mol/ft^2-h]

+H2O_in = H2ObySO2*SO2_abs; //[mol/ft^2-h]

+//Operating line

+x = 0:6;

+x = x/10^3;

+A = x./(1-x);

+B = H2O_in/Air_in*A+(0.005/0.995);

+y = B./(B+1);

+plot(x,y)

+xgrid();

+xlabel('x');

+ylabel('y');

+//legend('20C','30C','40C');

+title('x vs y');

+Gxbar = H2O_in*18.02+SO2_abs*64.1/2; //[lb/ft^2-h]

+kxa = 0.131*Gxbar^0.82; //[mol/ft^3-h]

+//The gas film coefficients are calculated for the bottom

+//and the top of the tower:

+//At bottom:

+Gy_B = (Air_in*29)+(SO2_in*64.1); //[lb/ft^2-h]

+kya_B = 0.025*Gy_B^0.7*Gx^0.25; //[mol/ft^3-h]

+//At top:

+Gy_T = (Air_out*29)+(SO2_out*64.1); //[lb/ft^2-h]

+kya_T = 0.025*Gy_T^0.7*Gx^0.25; //[mol/ft^3-h]  

+//Assuming 

+yLbar = 0.82

+C = kxa*yLbar/kya_B;

+//a line from (yb,xb) with a slope of -C, gives

+yi = 0.164;

+yLbar = 0.818;

+m = 20.1

+Kya_prime = 1/(yLbar/kya_B+m/kxa); //[mol/ft^3-h]

+//The fraction of the total resistance that is in the liquid is

+Rf = m/kxa/(1/Kya_prime);

+//For different values of y1

+y1 =[0.2,0.15,0.1,0.05,0.02,0.005]';

+delta_y1 = [0.103,0.084,0.062,0.034,0.015,0.005]';

+y1i = [0.164,0.118,0.074,0.034,0.012,0.002]';

+delta_yi = y1-y1i;