5 files changed, 338 insertions, 337 deletions
diff --git a/839/CH15/EX15.4/Example_15_4.sce b/839/CH15/EX15.4/Example_15_4.sce
index 2a1724600..857829a79 100755
--- a/839/CH15/EX15.4/Example_15_4.sce
+++ b/839/CH15/EX15.4/Example_15_4.sce
@@ -1,54 +1,54 @@
-//clear//
-clear;
-clc;
-
-//Example 15.4
-//Given
-N = 28;
-xF = 0.5/12; // [ft]
-yF = 0.035/12; //[ft]
-km = 26; // [Btu/ft-h-F]
-AT = 2.830; //[ft^2/ft]
-Ab = 0.416; //[ft^2/ft]
-hi = 1500; //[Btu/ft^2-h-F]
-G = 5000; //[lb/h-ft^2]
-Tavg = 130; //[F]
-Tw = 250; //[F]
-mu = 0.046; //[lb/ft-h], from Appendix 8 
-Cp = 0.25; //[Btu/lb-F], from Appendix 15
-k = 0.0162; //[Btu/ft-h-F], from Appendix 12
-ID_shell = 3.068/12; //[ft], from Appendix 5
-OD_pipe = 1.9/12; //[ft], from Appendix 5
-//cross sectional area of shell space
-Ac = %pi/4*(ID_shell^2-OD_pipe^2)-N*xF*yF //[ft^2]
-//The perimeter of air space
-Ap = %pi*ID_shell+AT; //[ft]
-//hydraulic radius
-rh = Ac/Ap; //[ft]
-//equivalent diameter
-De = 4*rh; //[ft]
-//Reynolds Number
-Nre = De*h/mu
-//In computing mu_w the resistance of the wall and the steam film
-//are considered negligible, so
-mu_w = 0.0528; //[lb/ft-h]
-Npr = mu*Cp/k
-//Using Fig. 15.17, the heat transfer factor is
-jh = 0.0031;
-ho = jh*Cp*G*(mu/mu_w)^0.14/Npr^(2/3); //[Btu/ft^2-h-F]
-
-//For rectangular fins, disreagrding the contribution of the ends of the fins to 
-//the perimeter, Lp = 2L and S = Lyf, where yf is the fin thickness and L is the 
-//length of the fin. Then, from Eq.(15.11)
-aFxF = xF*sqrt(2*ho/(km*yF));
-//From Fig. 15.16
-netaF = 0.93;
-Dt = 1.610/12; //[ft], from Appendix 5
-DLbar = (OD_pipe-Dt)/log(OD_pipe/Dt); //[ft]
-Ai = %pi*Dt*1.0; //[ft^2]
-AF = AT-Ab; //[ft^2/ft] 
-xw = (OD_pipe-Dt)/2; //[ft]
-
-//Using Eq.(15.10), the overall coefficient 
-Ut =  1/(Ai/(ho*(netaF*AF+Ab))+(xw*Dt/(km*DLbar))+1/hi);//[Btu/ft^2-h-F]
-disp('Btu/ft^2-h-F',Ut,'The overall heat transfer coefficent is')
+//clear//
+clear;
+clc;
+
+//Example 15.4
+//Given
+N = 28;
+xF = 0.5/12; // [ft]
+yF = 0.035/12; //[ft]
+km = 26; // [Btu/ft-h-F]
+AT = 2.830; //[ft^2/ft]
+Ab = 0.416; //[ft^2/ft]
+hi = 1500; //[Btu/ft^2-h-F]
+G = 5000; //[lb/h-ft^2]
+Tavg = 130; //[F]
+Tw = 250; //[F]
+mu = 0.046; //[lb/ft-h], from Appendix 8 
+Cp = 0.25; //[Btu/lb-F], from Appendix 15
+k = 0.0162; //[Btu/ft-h-F], from Appendix 12
+ID_shell = 3.068/12; //[ft], from Appendix 5
+OD_pipe = 1.9/12; //[ft], from Appendix 5
+//cross sectional area of shell space
+Ac = %pi/4*(ID_shell^2-OD_pipe^2)-N*xF*yF //[ft^2]
+//The perimeter of air space
+Ap = %pi*ID_shell+AT; //[ft]
+//hydraulic radius
+rh = Ac/Ap; //[ft]
+//equivalent diameter
+De = 4*rh; //[ft]
+//Reynolds Number
+Nre = De*G/mu
+//In computing mu_w the resistance of the wall and the steam film
+//are considered negligible, so
+mu_w = 0.0528; //[lb/ft-h]
+Npr = mu*Cp/k
+//Using Fig. 15.17, the heat transfer factor is
+jh = 0.0031;
+ho = jh*Cp*G*(mu/mu_w)^0.14/Npr^(2/3); //[Btu/ft^2-h-F]
+
+//For rectangular fins, disregarding the contribution of the ends of the fins to 
+//the perimeter, Lp = 2L and S = Lyf, where yf is the fin thickness and L is the 
+//length of the fin. Then, from Eq.(15.11)
+aFxF = xF*sqrt(2*ho/(km*yF));
+//From Fig. 15.16
+netaF = 0.93;
+Dt = 1.610/12; //[ft], from Appendix 5
+DLbar = (OD_pipe-Dt)/log(OD_pipe/Dt); //[ft]
+Ai = %pi*Dt*1.0; //[ft^2]
+AF = AT-Ab; //[ft^2/ft] 
+xw = (OD_pipe-Dt)/2; //[ft]
+
+//Using Eq.(15.10), the overall coefficient 
+Ut =  1/(Ai/(ho*(netaF*AF+Ab))+(xw*Dt/(km*DLbar))+1/hi);//[Btu/ft^2-h-F]
+disp('Btu/ft^2-h-F',Ut,'The overall heat transfer coefficient is')
+\ No newline at end of file
diff --git a/839/CH22/EX22.5/Example_22_5.sce b/839/CH22/EX22.5/Example_22_5.sce
index 76d8d4dfa..685e0f917 100755
--- a/839/CH22/EX22.5/Example_22_5.sce
+++ b/839/CH22/EX22.5/Example_22_5.sce
@@ -1,99 +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')
+//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')
+\ No newline at end of file
diff --git a/839/CH22/EX22.6/Example_22_6.sce b/839/CH22/EX22.6/Example_22_6.sce
index 326e3f8e9..bea069cab 100755
--- a/839/CH22/EX22.6/Example_22_6.sce
+++ b/839/CH22/EX22.6/Example_22_6.sce
@@ -1,56 +1,57 @@
-//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;
+//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*Gxbar^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*Gxbar^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;
+\ No newline at end of file
diff --git a/839/CH25/EX25.4/Example_25_4.sce b/839/CH25/EX25.4/Example_25_4.sce
index 7369eb582..4a74a8785 100755
--- a/839/CH25/EX25.4/Example_25_4.sce
+++ b/839/CH25/EX25.4/Example_25_4.sce
@@ -1,71 +1,71 @@
-//clear//
-clear;
-clc;
-
-//Example 25.4
-y = 0.0012;
-vdot = 16000; //[ft^3/min]
-P = 760; //[mm Hg]
-rho_b = 30; //[lb/ft^3]
-Lun = 0.5; //[ft]
-
-//Solution
-//(a)
-//Form the hand book
-Pprime = 151; //[mm Hg]
-fs = Pprime; //[mm Hg]
-rho_L = 0.805; //[g/cm^3], at 20C
-Tnb = 79.6; //[C]
-rho_e = 0.75; //[g/cm^3]
-M = 72.1;
-V = M/rho_e; 
-p = y*P; //[mm Hg]
-f = p; //[mm Hg]
-//At 35C
-T = 35+273; //[K]
-A = T/V*log10(fs/f);  
-//Form Fig. 25.4, 
-//the volume adsorbed
-V_ads = 24; //[cm^3/100 g carbon]
-Wsat = V_ads*rho_e; //[g/100 g carbon]
-W0 = 1/3*Wsat; //[g/100 g carbon]
-Working_capacity = Wsat-W0; //[g/100 g carbon]
-//or
-Working_capacity = Working_capacity/100; //[lb/lb carbon]
-disp(Working_capacity,'Working capacity of the bed is')
-
-//(b)
-u0 = 1; //[ft/s]
-A = vdot/u0; //[ft^2]
-D = sqrt(4*A/%pi); //[ft]
-Abed = 10*27; //[ft^2]
-L1 = 4; //[ft]
-c0 = y/359*273/298*72.1; //[lb/ft^3]
-//Form Eq.(25.3)
-tstar = L1*rho_b*(Working_capacity)/(u0*c0*3600); //[h]
-Lu1 = L-Lun; //[ft]
-tb1 = Lu1/L*tstar; //[h]
-
-//if
-L2 = 3; //[ft]
-Lu2 = L2-Lun;
-tb2 = Lu2/L*tstar; //[h]
-//checking for delta_P
-//Using Eq.(7.22)
-phi_s = 0.7; //from Table 28.1
-eps = 0.35; //from Table 7.1
-mu = 1.21*10^-5; //[lb/ft-s]
-rho = 0.074; //[lb/ft^3]
-//For a 4*10-mesh carbon
-Dp = 1.108*10^-2; //[ft]
-deltaPbyL = 150*1*mu*(1-eps)^2/(32.2*phi_s^2*Dp^2*eps^3)+(1.75*rho*1^2*(1-eps)/(32.2*0.7*Dp*eps^3)); //[lbf/ft^2-ft]
-deltaPbyL = deltaPbyL*12/62.4; //[in. H2O/ft]
-//for
-L = 3;
-deltaP = 3*deltaPbyL; //[in. H2O]
-//which satisfactory.
-mc = 2*(10*27*3)*30; //[lb]
-
-disp('ft',L2,'Allowing for uncertainties in the calculations, satisfactory bed length will be')
-disp('ft/s',u0,'gas velocity needed')
-disp('lb',mc,'carbon needed')
+//clear//
+clear;
+clc;
+
+//Example 25.4
+y = 0.0012;
+vdot = 16000; //[ft^3/min]
+P = 760; //[mm Hg]
+rho_b = 30; //[lb/ft^3]
+Lun = 0.5; //[ft]
+L = 3;
+
+//Solution
+//(a)
+//Form the hand book
+Pprime = 151; //[mm Hg]
+fs = Pprime; //[mm Hg]
+rho_L = 0.805; //[g/cm^3], at 20C
+Tnb = 79.6; //[C]
+rho_e = 0.75; //[g/cm^3]
+M = 72.1;
+V = M/rho_e; 
+p = y*P; //[mm Hg]
+f = p; //[mm Hg]
+//At 35C
+T = 35+273; //[K]
+A = T/V*log10(fs/f);  
+//Form Fig. 25.4, 
+//the volume adsorbed
+V_ads = 24; //[cm^3/100 g carbon]
+Wsat = V_ads*rho_e; //[g/100 g carbon]
+W0 = 1/3*Wsat; //[g/100 g carbon]
+Working_capacity = Wsat-W0; //[g/100 g carbon]
+//or
+Working_capacity = Working_capacity/100; //[lb/lb carbon]
+disp(Working_capacity,'Working capacity of the bed is')
+
+//(b)
+u0 = 1; //[ft/s]
+A = vdot/u0; //[ft^2]
+D = sqrt(4*A/%pi); //[ft]
+Abed = 10*27; //[ft^2]
+L1 = 4; //[ft]
+c0 = y/359*273/298*72.1; //[lb/ft^3]
+//Form Eq.(25.3)
+tstar = L1*rho_b*(Working_capacity)/(u0*c0*3600); //[h]
+Lu1 = L-Lun; //[ft]
+tb1 = Lu1/L*tstar; //[h]
+
+//if
+L2 = 3; //[ft]
+Lu2 = L2-Lun;
+tb2 = Lu2/L*tstar; //[h]
+//checking for delta_P
+//Using Eq.(7.22)
+phi_s = 0.7; //from Table 28.1
+eps = 0.35; //from Table 7.1
+mu = 1.21*10^-5; //[lb/ft-s]
+rho = 0.074; //[lb/ft^3]
+//For a 4*10-mesh carbon
+Dp = 1.108*10^-2; //[ft]
+deltaPbyL = 150*1*mu*(1-eps)^2/(32.2*phi_s^2*Dp^2*eps^3)+(1.75*rho*1^2*(1-eps)/(32.2*0.7*Dp*eps^3)); //[lbf/ft^2-ft]
+deltaPbyL = deltaPbyL*12/62.4; //[in. H2O/ft]
+//for
+deltaP = 3*deltaPbyL; //[in. H2O]
+//which satisfactory.
+mc = 2*(10*27*3)*30; //[lb]
+
+disp('ft',L2,'Allowing for uncertainties in the calculations, satisfactory bed length will be')
+disp('ft/s',u0,'gas velocity needed')
+disp('lb',mc,'carbon needed')
+\ No newline at end of file
diff --git a/839/CH6/EX6.2/Example_6_2.sce b/839/CH6/EX6.2/Example_6_2.sce
index 4abb1d09a..ed2213d65 100755
--- a/839/CH6/EX6.2/Example_6_2.sce
+++ b/839/CH6/EX6.2/Example_6_2.sce
@@ -1,57 +1,57 @@
-//clear//
-clear;
-clc;
-
-//Example 6.2
-//Given
-Tr = 1000; //[R] 
-pr = 20; //[atm]
-Ma_a = 0.05;
-gama = 1.4;
-gc = 32.174; //[ft-lb/lbf-s^2]
-M = 29;
-R = 1545;
-//(a)
-//Using Eq.(6.45)
-A = 2*(1+((gama-1)/2)*Ma_a^2)/((gama+1)*Ma_a^2);
-fLmax_rh = (1/Ma_a^2-1-(gama+1)*log(A)/2)/gama  
-
-//(b)
-//Using Eq.(6.28), the pressure at the end of the isentropic nozzle pa
-A = (1+(gama-1)*(Ma_a^2)/2); 
-pa = pr/(A^(gama/(gama-1)))  // [atm]
-//From Example 6.1, the density of air at 20atm and 1000R is 0.795 lb/ft^3
-//Using Eq.(6.17), the acoustic velocity 
-Aa  = sqrt(gc*gama*Tr*R/M) //[m/s]
-//The velocity at the entrance of the pipe 
-ua  = Ma_a*Aa //[m/s]
-//When L_b = L_max, the gas leaves the pipe at the asterisk conditions, where 
-Ma_b = 1;
-// Using Eq.(6.43)
-A = (gama-1)/2;
-Tstar = Tr *(1+A*Ma_a^2)/(1+A*Ma_b^2) // [K]
-// Using Eq.(6.44)
-rho_star = 0.795*Ma_a/sqrt(2*(1+(gama-1)*Ma_a^2/2)/(2.4)) //[lb/ft^3]
-//Using Eq.(6.39)
-pstar = p0*Ma_a/sqrt(1.2) // [atm]
-//Mass velocity through the entire pipe
-G = 0.795*ua //[lb/ft^2-s]
-ustar = G/rho_star //[ft/s]
-
-//(c)
-//Using Eq.(6.45) with f_Lmax_rh = 400
-
-err = 1;
-eps = 10^-3;
-Ma_ac = rand(1,1);
-i =1;
-while((err>eps))  
-  A = 2*(1+((gama-1)/2)*Ma_ac^2)/((gama+1)*Ma_ac^2);
-  B = gama*400+1+(gama+1)*log(A)/2;
-  Ma_anew  = sqrt(1/B);
-  err = Ma_ac-Ma_anew;
-  Ma_ac = Ma_anew;
-end
-Ma_ac;
-uac  = Ma_ac*ua/Ma_a //[ft/s]
-Gc = uac*0.795 //[lb/ft^2-s]
+//clear//
+clear;
+clc;
+
+//Example 6.2
+//Given
+Tr = 1000; //[R] 
+pr = 20; //[atm]
+Ma_a = 0.05;
+gama = 1.4;
+gc = 32.174; //[ft-lb/lbf-s^2]
+M = 29;
+R = 1545;
+//(a)
+//Using Eq.(6.45)
+A = 2*(1+((gama-1)/2)*Ma_a^2)/((gama+1)*Ma_a^2);
+fLmax_rh = (1/Ma_a^2-1-(gama+1)*log(A)/2)/gama  
+
+//(b)
+//Using Eq.(6.28), the pressure at the end of the isentropic nozzle pa
+A = (1+(gama-1)*(Ma_a^2)/2); 
+pa = pr/(A^(gama/(gama-1)))  // [atm]
+//From Example 6.1, the density of air at 20atm and 1000R is 0.795 lb/ft^3
+//Using Eq.(6.17), the acoustic velocity 
+Aa  = sqrt(gc*gama*Tr*R/M) //[m/s]
+//The velocity at the entrance of the pipe 
+ua  = Ma_a*Aa //[m/s]
+//When L_b = L_max, the gas leaves the pipe at the asterisk conditions, where 
+Ma_b = 1;
+// Using Eq.(6.43)
+A = (gama-1)/2;
+Tstar = Tr *(1+A*Ma_a^2)/(1+A*Ma_b^2) // [K]
+// Using Eq.(6.44)
+rho_star = 0.795*Ma_a/sqrt(2*(1+(gama-1)*Ma_a^2/2)/(2.4)) //[lb/ft^3]
+//Using Eq.(6.39)
+pstar = pa*Ma_a/sqrt(1.2) // [atm]
+//Mass velocity through the entire pipe
+G = 0.795*ua //[lb/ft^2-s]
+ustar = G/rho_star //[ft/s]
+
+//(c)
+//Using Eq.(6.45) with f_Lmax_rh = 400
+
+err = 1;
+eps = 10^-3;
+Ma_ac = rand(1,1);
+i =1;
+while((err > eps))  
+  A = 2*(1+((gama-1)/2)*Ma_ac^2)/((gama+1)*Ma_ac^2);
+  B = gama*400+1+(gama+1)*log(A)/2;
+  Ma_anew  = sqrt(1/B);
+  err = Ma_ac-Ma_anew;
+  Ma_ac = Ma_anew;
+end
+Ma_ac;
+uac  = Ma_ac*ua/Ma_a //[ft/s]
+Gc = uac*0.795 //[lb/ft^2-s]
+\ No newline at end of file