4 files changed, 168 insertions, 0 deletions
diff --git a/839/CH24/EX24.1/Example_24_1.sce b/839/CH24/EX24.1/Example_24_1.sce
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+++ b/839/CH24/EX24.1/Example_24_1.sce
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+//clear//
+clear;
+clc;
+
+//Example 24.1
+//Given
+Twb = 80; //[F]
+Tdb = 120; //[F]
+v = 3.5; //[ft/s]
+rho = 120; //[lb/ft^3]
+Xe = 0;
+Xc = 0.09;
+lambda = 1049; //[Btu/lb]
+M = 29;
+B = 24; //[in.]
+D = 2; //[in.]
+Dc = 2; //[ft]
+//Solution
+//(a)
+//mass velocity
+G = v*M*492*3600/(359*(460+120)); //[lb/ft^2-h]
+//the coefficent, by Eq.(24.13), in fps units, is
+h = 0.01*G^0.2/2^0.2; //[Btu/ft^2-h-F]
+//Substituting in Eq.(21.15) gives 
+Rc = 1.94*(Tdb-Twb)/(lambda); //[lb/ft^2-h]
+disp('lb/ft^2-h',Rc,'Drying rate during the constant period is')
+
+//(b)
+//Since drying is from both faces, area
+A = Dc*(B/12)^2; //[ft^2]
+//The rate of drying  
+mvdot = Rc*A; //[lb/h] 
+//Volume of the cake
+Vc = (B/12)^2*D/12; //[ft^3]
+//mass of the bone-dry solid is
+mdot_bd = rho*Vc; //[lb]
+//The quantity of moisture to be vaporized is
+X2 = 0.20;
+X1 = 0.10;
+Q = mdot_bd*(X2-X1); //[lb]
+//Drying time
+tT = Q/mvdot; //[h]
+disp('h',tT,'drying time')
diff --git a/839/CH24/EX24.2/Example_24_2.sce b/839/CH24/EX24.2/Example_24_2.sce
new file mode 100755
index 000000000..14fd4eaef
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+++ b/839/CH24/EX24.2/Example_24_2.sce
@@ -0,0 +1,15 @@
+//clear//
+clear;
+clc;
+
+//Example 24.2
+//Given
+X1 = 0.25;
+X = 0.05;
+Dvprime = 8.3*10^-6; //[cm^2/s]
+D = 25.4; //[mm]
+
+//Solution
+s = D/(2*10); //[cm]
+tT = 4*s^2/(%pi^2*Dvprime)*log(8*X1/(%pi^2*X))/3600; //[h] 
+disp('h',tT,'drying time is')
diff --git a/839/CH24/EX24.3/Example_24_3.sce b/839/CH24/EX24.3/Example_24_3.sce
new file mode 100755
index 000000000..2a9dd2297
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+++ b/839/CH24/EX24.3/Example_24_3.sce
@@ -0,0 +1,62 @@
+//clear//
+clear;
+clc;
+
+//Example 24.3
+//Given
+Tw = 80; //[F]
+Tdb = 120; //[F]
+v = 3.5; //[ft/s]
+rho = 120; //[lb/ft^3]
+Xe = 0;
+Xc = 0.09;
+lambda = 1049; //[Btu/lb]
+M = 29;
+B = 24; //[in.]
+D = 2; //[in.]
+Dc = 2; //[ft]
+X2 = 0.20;
+X1 = 0.10;
+Dcyl = 1/4; //[in.]
+L = 4; //[in.]
+Vbar = 3.5; //[ft/s]
+Thb = 120;
+ 
+//Solution
+//Since the Xc is less than 10 percent, all drying takes place
+//in the constant-rate period and the vaporrization temperature,
+//as before, is 80 F. 
+//From Exapmle 24.1, mass of water to be evaporated
+mdot = 8*(X2-X1); //[lb]
+//The quantity of heat to be transferred 
+QT = mdot*lambda; //[Btu]
+//mass of the dry soild in one cylinder is
+mp = %pi/4*(Dcyl/12)^2*(L/12)*rho; //[lb]
+//surface area of one cylinder is
+Ap = %pi*(Dcyl/12)*(L/12); //[ft^2]
+//Total area exposed by 8 lb solids
+A = 8/mp*Ap; //[ft^2]
+//The heat transfer coefficient is found from the 
+//equivalent form of Eq.(21.62)
+//hDbyk = 1.17*Nre^0.585*Npr^(1/3)
+//For air at 1 atm and 120F, the properties are
+rho_a = M/359*492/580; //[lb/ft^3]
+mu_a = 0.019; //[cP], from Appendix 8
+k_a = 0.0162; //[Btu/ft-h-F], from Appendix 12
+Cp_a = 0.25; //[Btu/lb-F], from Appendix 15
+Nre = 1/48*Vbar*rho_a/(mu_a*6.72*10^-4);
+Npr = mu_a*2.42*Cp_a/k_a;
+//Form Eq.(21.62)
+h = (k_a*1.17*Nre^0.585*Npr^(1/3))/(1/48); //[Btu/ft^2-h-F]
+mdot_g = v*3600*rho_a; //[lb]
+//From Fig. 23.2
+cs = 0.25;
+delta_Thb = Thb-Tw; //[F]
+delta_Tha = 8.24; //[F]
+//The heat transferred form the gas to a thin section of the bed
+delta_TL = (delta_Thb-delta_Tha)/log(delta_Thb/delta_Tha); //[F]
+//rate of heat transfer
+qT = h*A*delta_TL; //[Btu/h]
+//drying time
+tT = QT/qT; //[h]
+disp('h',tT,'Required drying time is')
diff --git a/839/CH24/EX24.4/Example_24_4.sce b/839/CH24/EX24.4/Example_24_4.sce
new file mode 100755
index 000000000..3f365ead6
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+++ b/839/CH24/EX24.4/Example_24_4.sce
@@ -0,0 +1,48 @@
+//clear//
+clear;
+clc;
+
+//Example 24.4
+//Given
+msdot = 2800; //[lb/h]
+Xa = 0.15;
+Xb = 0.005;
+Ti = 80; //[F]
+To = 125; //[F]
+Thb = 260; //[F]
+Hb = 0.01; //[lb water/lb dry air]
+G = 700; //[lb/ft^2-h]
+Cps = 0.52; //[Btu/lb-F] 
+
+//Solution
+//Counter current operation will be used.
+//Assuming
+Nt = 1.5; //NTU
+//From Fig. 23.2
+Twb = 102; //[F]
+//From Eq. (2.48)
+Tha = (Thb-Twb)/exp(Nt)+Twb; //[F]
+Tsb = To; //[F]
+lambda = 1036; //[Btu/lb], at 102 F, from Appendix 7
+Cpv = 0.45; //[Btu/lb-F], from Appendix 15
+Cpl = 1.0; //[Btu/lb-F]
+//From Eq.(24.9)
+mvdot = msdot*(Xa-Xb); //[lb/h]
+//The heat duty is found form substitution in Eq.(24.1)
+qTdot = Cps*(To-Ti)+Xa*Cpl*(Twb-Ti)+(Xa-Xb)*lambda+Xb*Cpl*(To-Twb)+(Xa-Xb)*Cpv*(Tha-Twb); //[Btu/lb]
+qT = qTdot*msdot; //[Btu/h]
+//The flow rate of the entering air is found from a heat balance and the humid heat csb.
+//From Fig. 23.2
+csb = 0.245; //[Btu/lb-F],
+mgdot = qT/(csb*(Thb-Tha)*(1+Hb)); //[lb/h of dry air]
+//From Eq.(24.10), The outlet humidity
+Ha = Hb+mvdot/mgdot; //[lb/lb]
+
+//For a given flow rate, the cross-sectional area of the dryer must be
+Ac = qT/(csb*(Thb-Tha))/G; //[ft^2]
+//The dryer diameter is
+D = (4*Ac/%pi)^0.5; //[ft]
+delta_TL = ((Thb-Twb)-(Tha-Twb))/log((Thb-Twb)/(Tha-Twb)); //[F]
+//Using Eq.(24.29), the dryer length
+L = qT/(0.125*%pi*D*G^0.67*delta_TL); //[ft]
+disp('respectively','ft',L,'ft',D,'Required diameter and length of the dryer is')