initial commit / add all books

13 files changed, 679 insertions, 0 deletions

diff --git a/905/CH2/EX2.1/2_1.sce b/905/CH2/EX2.1/2_1.sce
new file mode 100755
index 000000000..edad20edf
--- /dev/null
+++ b/905/CH2/EX2.1/2_1.sce
@@ -0,0 +1,27 @@
+clear;

+clc;

+

+// Illustration 2.1

+// Page: 94

+

+printf('Illustration 2.1 -  Page: 94\n\n');

+

+// solution

+

+//*****Data*****//

+// a-oxygen   b-stagnant water

+T = 310; // [K]

+// Since the solubility of oxygen in water at 310 K is extremely low, we are dealing with dilute solutions

+k_L = 3.3*10^-5; // [coefficient based on the oxygen concentration difference in the water, m/s]

+row = 993; // [kg/cubic m]

+M_b = 18; // [gram/mole]

+//*****//

+ 

+// Since we are dealing with very dilute solutions

+// Therefore, c = (row/M_avg) = row/M_b

+c = row/M_b; // [kmole/cubic m]

+// Using equation 2.10

+k_x = k_L*c; // [kmole/square m.s]

+printf("The mass-transfer coefficient based on the mole fraction of oxygen in the liquid is %e kmole/square m.s\n\n",k_x);

+

+ 
\ No newline at end of file
diff --git a/905/CH2/EX2.10/2_10.sce b/905/CH2/EX2.10/2_10.sce
new file mode 100755
index 000000000..05eeae308
--- /dev/null
+++ b/905/CH2/EX2.10/2_10.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+

+// Illustration 2.10

+// Page: 127

+

+printf('Illustration 2.10 -  Page: 127\n\n');

+

+// solution

+

+// Example 2.6 using equation 2.73

+// Values of the dimensionless parameters calculated in Example 2.6

+Re = 1223; // [Renoylds Number]

+Sc = 0.905; // [Schmidt Number]

+c = 0.039; // [molar density, kg/cubic m]

+v = 1; // [gas velocity, m/s]

+// Therefore 

+Gm = c*v; // [kmole/square m.s]

+// From equation 2.9 

+// Kg*P = ky

+// Therefore substituting in equation 2.73 we obtain

+ky = 0.281*Gm/(Re^0.4*Sc^0.56); // [kmole/square m.s]

+// Now equation 2.73 were obtained under very dilute concentrations

+// Therefore

+y_bm = 1; 

+// From equation 2.6

+F = ky*y_bm; // [kmole/square m.s]

+printf("Mass transfer coefficient is %e kmole/square m.s\n\n",F);

diff --git a/905/CH2/EX2.11/2_11.sce b/905/CH2/EX2.11/2_11.sce
new file mode 100755
index 000000000..51de881a1
--- /dev/null
+++ b/905/CH2/EX2.11/2_11.sce
@@ -0,0 +1,59 @@
+clear;

+clc;

+

+// Illustration 2.11

+// Page: 129

+

+printf('Illustration 2.11 -  Page: 129\n\n');

+

+// solution

+//*****Data*****//

+// a-water   b-air

+D = 25.4*10^-3; // [diameter of wetted wall tower, m]

+Gy = 10; // [mass velocity, kg/square m.s]

+T1 = 308; // [K]

+P = 101.3; // [kPa]

+p_a1 = 1.95; // [partial pressure of water vapor, kPa]

+R = 8.314; // [cubic m.Pa/mole.K]

+M_a = 18; // [gram/mole]

+Cpa = 1.88; // [kJ/kg.K]

+//*****//

+

+// Properties of dry air at 308 K and 1 atm pressure are

+u = 1.92*10^-5; // [kg/m.s]

+row = 1.14; // [kg/cubic m]

+D_ab = 0.242*10^-4; // [square m/s]

+Sc = 0.696; // [Schmidt number]

+Cp = 1.007; // [kJ/kg.K]

+k = 0.027; // [W/m.K]

+Pr = 0.7; // [Prandtl number]

+

+Re = D*Gy/u; // [Renoylds number]

+// From equation 2,74

+Sh = 0.023*Re^0.83*Sc^0.44; // [Sherwood number]

+// From Table 2.1

+kg = Sh*D_ab/(R*T1*D)*1000; // [mole/square m.s.kPa]

+printf("kg is %e\n",kg);

+// To estimate the heat-transfer coefficient, we use the Dittus-Boelter equation for cooling, equation 2.80

+Nu = 0.023*Re^0.8*Pr^0.3; // [Nusselt number]

+// From Table 2.1

+h = Nu*k/D; // [W/square m.K]

+printf("h is %f\n",h);

+T =373.15; // [K]

+lambda_a = 40.63; // [kJ/mole]

+Tc = 647.1; // [K]

+

+// Solution of simultaneous equation 2.78 and 2.79

+function[f]=F(e) 

+    f(1) = kg*(p_a1 - exp(16.3872-(3885.7/(e(1)-42.98))))-e(2);

+    f(2) = e(2)*M_a*Cpa*(T1-e(1))/(1-exp(-e(2)*M_a*Cpa/h)) + 1000*e(2)*lambda_a*((1-(e(1)/Tc))/(1-(T/Tc)))^0.38;

+    funcprot(0);

+endfunction

+

+// Initial guess

+e = [292 -0.003];

+y = fsolve(e,F);

+Ti = y(1);

+Na = y(2);

+

+printf("The temperature of the liquid water and the rate of water evaporation is %f K and %e mole/square m.s respectively",Ti,Na);
\ No newline at end of file
diff --git a/905/CH2/EX2.12/2_12.sce b/905/CH2/EX2.12/2_12.sce
new file mode 100755
index 000000000..7b6e92a31
--- /dev/null
+++ b/905/CH2/EX2.12/2_12.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+

+// Illustration 2.12

+// Page: 131

+

+printf('Illustration 2.12 -  Page: 131\n\n');

+

+// solution

+//*****Data*****//

+// a-water    b-dry air

+D = 25.4*10^-3; // [Internal diameter of tower, m]

+Z = 1.5; // [length of the wetted section, m]

+Gy = 10; // [mass velocity of air, kg/square m.s]

+Tair = 308; // [K]

+Twater = 295; // [K]

+P = 101.3; // [kPa]

+M_a = 18; // [gram/mole]

+M_b = 29; // [gram/mole]

+R = 8.314; // [cubic m.Pa/mole.K]

+//*****//

+

+// The water vapor partial pressure at the interface remains constant at the vapor pressure of liquid water at 295 K, which is pa1 = Pa = 2.64 kPa

+// The water vapor partial pressure at the bulk of the gas phase increases from pA2 = pAin = 0 for the dry inlet air to pa2= pAout for the air leaving the tower 

+Pa = 2.64; // [kPa]

+

+Gm = Gy/M_b; // [Assuming that gas phase is basically dry air, kmole/square m.s]

+// The properties of dry air at 308 K and 1 atm are (from example 2.9)

+row = 1.14; // [kg/cubic m]

+u = 1.92*10^-5; // [kg/m.s]

+D_ab = 0.242*10^-4; // [square m/s]

+Sc = 0.692; // [Schmidt number]

+

+Re = Gy*D/u; // [Renoylds number]

+

+if(Re<35000 & Re>2000)

+// From equation 2.74

+Sh = 0.023*Re^0.83*Sc^0.44; // [Sherwood number]    

+

+printf("Sherwood number is %f\n\n",Sh);

+else()

+    printf('We cannot use equation 2.74')

+end

+

+c = P/(R*Tair); // [kmole/cubic m]

+// Now using equation 2.89

+Pa_out = Pa*(1-exp((-4*Sh*Z*c*D_ab)/(Gm*D^2))); // [kPa]

+printf("The partial pressure of water in the air leaving the tower is %e kPa\n\n",Pa_out);
\ No newline at end of file
diff --git a/905/CH2/EX2.13/2_13.sce b/905/CH2/EX2.13/2_13.sce
new file mode 100755
index 000000000..86e53f148
--- /dev/null
+++ b/905/CH2/EX2.13/2_13.sce
@@ -0,0 +1,57 @@
+clear;

+clc;

+

+// Illustration 2.13

+// Page: 134

+

+printf('Illustration 2.13 -  Page: 134\n\n');

+

+// solution

+//*****Data*****//

+// a-water    b-dry air

+Gy = 10; // [kg/square m.s]

+dp = 3.5*10^-3; // [diameter of spherical glass beads, m]

+D = 25.4*10^-3; // [Internal diameter of tower, m]

+Tair = 308; // [K]

+Twater = 295; // [K]

+P = 101.3; // [kPa]

+M_a = 18; // [gram/mole]

+M_b = 29; // [gram/mole]

+R = 8.314; // [cubic m.Pa/mole.K]

+

+// The properties of dry air at 308 K and 1 atm are (from example 2.12)

+row = 1.14; // [kg/cubic m]

+u = 1.92*10^-5; // [kg/m.s]

+D_ab = 0.242*10^-4; // [square m/s]

+Sc = 0.692; // [Schmidt number]

+c = 0.04; // [mole/cubic m]

+Gm = 0.345; // [kmole/square m.s]

+

+Re = Gy*dp/u; // [Renoylds number]

+if(Re<2500 & Re>10)

+    

+// Subsituting in equation 2.90

+jd = 1.17*Re^-0.415;

+printf("Renoylds number is %f\n\n",Re);

+else()

+end

+

+Std = 0.052/(Sc^(2/3));

+// From Table 2.1 

+Sh = Std*Re*Sc; // [Sherwood number]

+// From equation 2.94

+e = 0.406+0.571*(dp/D); // [bed porosity]

+

+printf('Illustration 2.13(a) -  Page: 134\n\n');

+// Solution(a)

+// Now Paout = 0.99*Pa

+// Using equation 2.93 to calculate 'Z'

+deff('[y] = f12(Z)','y = 0.99 - 1 + exp(-6*(1-e)*Sh*c*Z*D_ab/(Gm*dp^2))');

+Z = fsolve(0.06,f12);

+printf("The depth of packing required is %e m\n\n",Z);

+

+printf('Illustration 2.13(b) -  Page: 136\n\n');

+// Solution(b)

+// From equation 2.95

+deltaP = (150*(1-e)/Re + 1.75)*(1-e)*Gy^2*Z/(dp*row*e^3); // [Pa]

+printf("The gas pressure drop through the bed is %f Pa\n\n",deltaP); 
\ No newline at end of file
diff --git a/905/CH2/EX2.14/2_14.sce b/905/CH2/EX2.14/2_14.sce
new file mode 100755
index 000000000..4f42ebfe3
--- /dev/null
+++ b/905/CH2/EX2.14/2_14.sce
@@ -0,0 +1,60 @@
+clear;

+clc;

+

+// Illustration 2.14

+// Page: 138

+

+printf('Illustration 2.14 -  Page: 138\n\n');

+

+// solution

+// a-oxygen     b-water

+// To design the deaerator, We will use commercially available microporous polypropylene hollow fibers in a module

+// Given data:

+m = 40000; // [kg/hr]

+Twater = 298; // [K]

+v = 0.1; // [superficial velocity, m/s]

+P = 101.3; // [kPa]

+V = 40*10^-3; // [Flow rate of nitrogen, cubic m/min]

+d = 2.90*10^-4; // [Outside diameter of fibres, m]

+pf = 0.4; // [Packing factor]

+a = 46.84*100; // [surface area per unit volume, m^-1]

+R = 8.314; // [cubic m.Pa/mole.K]

+// *****//

+

+dw = 1000; // [density of water, kg/cubic m]

+Ql = m/(3600*1000); // [volumetric water flow rate, cubic m/s]

+// Shell diameter

+D = (4*Ql/(%pi*v))^0.5; // [Shell diameter, m]

+

+// the properties of dilute mixtures of oxygen in water at 298 K

+u = 0.9; // [cP]

+// Diffusivity from equation 1.53

+D_ab = 1.93*10^-9; // [square m/s]

+Sc = 467; // [Schmidt number]

+

+Re = d*v*dw/(u*10^-3); // [Renoylds number]

+

+// Substituting in equation (2-97) gives

+Sh = 0.53*(1-1.1*pf)*((1-pf)/pf)^-0.47*(Re^0.53*Sc^0.33);

+

+kl = Sh*D_ab/d; // [mass-transfer coefficient on the shell side, m/s]

+

+// From the specified BFW flow rate

+L = m/(3600*18); // [kmole/s]

+// From ideal gas law

+V1 = V*P/(Twater*R*60); // [kmole/s]

+// From the solubility of oxygen in water at 298 K,

+M = 4.5*10^4;

+A = L/(M*V1); // [Absorption factor]

+printf("Absorption factor is %f\n",A);

+

+// For 99% removal of the dissolved oxygen

+// x_in/x_out = b = 100

+b = 100;

+c = 55.5 // [molar density, kmole/cubic m]

+// Substituting in equation 2.99 yields

+V_T = (L*log(b*(1-A)+A))/(kl*a*c*(1-A)); // [cubic m]

+

+// The module length, Z is

+Z = V_T/(%pi*D^2/4);

+printf("The shell diameter and module length is %f m and %f m respectively\n\n",D,Z);   
\ No newline at end of file
diff --git a/905/CH2/EX2.2/2_2.sce b/905/CH2/EX2.2/2_2.sce
new file mode 100755
index 000000000..4b475119e
--- /dev/null
+++ b/905/CH2/EX2.2/2_2.sce
@@ -0,0 +1,22 @@
+clear;

+clc;

+

+// Illustration 2.2

+// Page: 95

+

+printf('Illustration 2.2 -  Page: 95\n\n');

+

+// solution

+

+//*****Data*****//

+//  a-ammonia   b-air

+T = 300; // [K]

+P = 1; // [atm]

+y_a1 = 0.8; // [ammonia mole fraction in the bulk of the gas phase]

+y_a2 = 0.732; // [ammonia gas-phase mole fraction on interface]

+N_a = 4.3*10^-4; // [ammonia flux,  kmole/square m.s]

+//*****//

+

+// Using equation 2.2

+F_g = N_a/log((1-y_a2)/(1-y_a1)); // [kmole/square m.s]

+printf("The mass-transfer coefficient in the gas phase at that point where flux is 4.3*10^-4 is %e kmole/square m.s\n\n",F_g);
\ No newline at end of file
diff --git a/905/CH2/EX2.3/2_3.sce b/905/CH2/EX2.3/2_3.sce
new file mode 100755
index 000000000..11c530bdb
--- /dev/null
+++ b/905/CH2/EX2.3/2_3.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+

+// Illustration 2.3

+// Page: 96

+

+printf('Illustration 2.3 -  Page: 96\n\n');

+

+// solution

+

+//*****Data*****//

+// a-methanol   b-water

+P = 101.3; // [kPa]

+y_a1 = 0.707; // [mole fraction at interface]

+y_a2 = 0.656; // [mole fraction at bulk of the gas]

+k_g = 1.62*10^-5; // [mass-transfer coefficient at a point in the column, kmole/square m.s.kPa]

+//*****//

+

+// Using equation 2.14

+k_y = k_g*P; // [kmole/square m.s]

+// Using equation 2.12

+N_a = k_y*(y_a1-y_a2); // [kmole/square m.s]

+printf("The methanol flux at the point of given mass transfer coefficient is %e kmole/square m.s\n\n",N_a);

diff --git a/905/CH2/EX2.4/2_4.sce b/905/CH2/EX2.4/2_4.sce
new file mode 100755
index 000000000..301a4dd31
--- /dev/null
+++ b/905/CH2/EX2.4/2_4.sce
@@ -0,0 +1,62 @@
+clear;

+clc;

+

+// Illustration 2.4

+// Page: 99

+

+printf('Illustration 2.4 -  Page: 99\n\n');

+

+// solution

+// Mass Transfer into a Dilute Stream Flowing Under Forced Convection in a Circular Conduit

+

+n = 6; // [number of variables]

+//   Variables                Symbols        Dimensions

+// Tube diameter                 D             L

+// Fluid density                 row           M/L^3 

+// Fluid viscosity               u             M/(Lt)

+// Fluid velocity                v             L/t

+// Mass diffusivity              D_AB          L^2/t

+// Mass-transfer coefficient     kc            L/t

+

+// To determine the number of dimensionless parameters to be formed, we must know the rank, r, of the dimensional matrix.

+// The dimensional matrix is 

+DM = [0,0,1,1,0,0;1,1,-3,-1,2,1;-1,-1,0,0,-1,-1];

+[E,Q,Z ,stair ,rk]=ereduc(DM,1.d-15);

+printf("Rank of matrix is %f\n\n",rk);

+

+//The numbers in the table represent the exponent of M, L, and t in the dimensional expression of each of the six variables involved. For example, the dimensional expression of p is M/Lt; hence the exponents are 1, -1, and -1

+

+// From equation 2.16

+i = n-rk; // [number of dimensional groups]

+// Let the dimensional groups are pi1, pi2 and pi3

+// Therefore pi1 = (D_AB)^a*(row)^b*(D)^c*kc

+//           pi2 = (D_AB)^d*(row)^e*(D)^f*v

+//           pi3 = (D_AB)^g*(row)^h*(D)^i*u

+

+// Solving for pi1

+// M^0*L^0*t^0 = 1 = (L^2/t)^a*(M/L^3)^b*(L)^c*(L/t)

+

+// Solution of simultaneous equation

+function[f]=F(e)

+    f(1) = 2*e(1)-3*e(2)+e(3)+1;

+    f(2) = -e(1)-1;

+    f(3) = -e(2);

+    funcprot(0);

+endfunction

+

+// Initial guess:

+e = [0.1 0.8 0.5];

+y = fsolve(e,F);

+a = y(1);

+b = y(2);

+c = y(3);

+printf("The coefficients of pi1 are %f %f %f\n\n",a,b,c);

+// Similarly the coefficients of pi2 and pi3 are calculated

+// Therefore we get pi1 = kc*D/D_AB = Sh i.e. Sherwood Number

+//                  pi2 = v*D/D_AB = P_ed i.e. Peclet Number

+//                  pi3 = u/(row*D_AB) = Sc i.e. Schmidt Number

+

+// Dividing pi2 by pi3 gives

+//       pi2/pi3 = D*v*row/u = Re i.e. Renoylds number

+

+printf('The result of the dimensional analysis of forced-convection mass transfer in a circular conduit indicates that a correlating relation could be of the form\n Sh = function(Re,Sc)\n which is analogous to the heat transfer correlation \n Nu = function(Re,Pr)');
\ No newline at end of file
diff --git a/905/CH2/EX2.6/2_6.sce b/905/CH2/EX2.6/2_6.sce
new file mode 100755
index 000000000..7ae43e74f
--- /dev/null
+++ b/905/CH2/EX2.6/2_6.sce
@@ -0,0 +1,79 @@
+clear;

+clc;

+

+// Illustration 2.6

+// Page: 111

+

+printf('Illustration 2.6 -  Page: 111\n\n');

+

+// solution

+//*****Data*****//

+// a-UF6    b-air

+M_a = 352; // [molecular weight of UF6, gram/mole]

+M_b = 29; // [gram/mole]

+d = 0.01; // [diameter, m]

+x = 0.1; // [length exposed to air stream, m]

+v = 1; // [m/s]

+Ts = 303; // [surface temperature of solid, K]

+P_a = 27; // [vapor pressure of UF6, kPa]

+Tb = 325; // [bulk temperature of solid ,K]

+P = 101.3; // [kPa]

+R = 8.314; // [cubic m.Pa/mole.K]

+//*****//

+

+y_a1 = P_a/P; // [mole fraction at point 1]

+y_a2 = 0; // [mole fraction at point 2]

+

+// Along the mass-transfer path-cylinder surface (point 1) to bulk air (point 2)

+Tavg = (Ts+Tb)/2; // [K]

+

+// At point 1, the gas is saturated with UF6 vapor, while at point 2 the gas is virtually free of UF6

+// Therefore

+Pavg = (P_a+0)/2; // [average partial pressure, kPa]

+y_a = Pavg/P; // [mole fraction of UF6]

+

+Mavg = M_a*y_a+M_b*(1-y_a); // [gram/mole]

+row_avg = P*Mavg/(R*Tavg); // [kg/cubic m]

+

+// Parameter for c-O2, d-N2 and a-UF6

+yi_c = 0.182;   yi_d = 0.685;  yi_a = 0.133; 

+Tc_c = 154.6;   Tc_d = 126.2;  Tc_a = 505.8; // [K]

+Pc_c = 50.4;    Pc_d = 33.9;   Pc_a = 46.6;  // [bar]

+M_c = 32;       M_d  = 28;     M_a  = 352;  // [gram/mole]

+V_c = 73.4;     V_d  = 89.8;  V_a  = 250; // [cubic cm/mole]

+Z_c = 0.288;    Z_d  = 0.290;  Z_a  = 0.277;

+

+// From equation 2.52 and 2.53

+Tcm = yi_c*Tc_c+yi_d*Tc_d+yi_a*Tc_a; // [K]

+Pcm = 10^6*R*Tcm*(yi_c*Z_c+yi_d*Z_d+yi_a*Z_a)/((yi_c*V_c+yi_d*V_d+yi_a*V_a)*100000); // [bar]

+M_avg = yi_c*M_c+yi_d*M_d+yi_a*M_a; // [gram/mole]

+

+// From equation 2.50

+Em =  0.176*(Tcm/(M_avg^3*Pcm^4))^(1/6); // [uP]^-1

+

+// From equation 2.51

+Trm = Tavg/Tcm;

+f_Trm = (0.807*Trm^0.618)-(0.357*exp(-0.449*Trm))+(0.340*exp(-4.058*Trm))+0.018; 

+// From equation 2.49 

+u = f_Trm/Em; // [uP]

+u = u*10^-7; // [viscosity, kg/m.s]

+

+Re = d*v*row_avg/u; // [Renoylds number]

+

+// Diffusivity of UF6 vapors in air at 314 K and 1 atm from equation 1.49

+D_ab = 0.0904; // [square cm/s]

+

+Sc = u/(row_avg*D_ab*10^-4); // [Schmidt number]

+

+Sh_avg = 0.43 + 0.532*Re^0.5*Sc^0.31; // [Sherwood number]

+// From equation 1.7

+c = P/(R*Tavg); // [kmole/cubic m]

+// From Table 2.1 

+F_av = Sh_avg*D_ab*c*10^-4/d; // [kmole/square m.s]

+

+// From equation 2.2

+N_avg = F_av*log((1-y_a2)/(1-y_a1)); // [kmole/square m.s]

+S = 2*%pi*d^2/4 +%pi*d*x; // [total surface area of the cylinder, square m]

+

+w_a = N_avg*S*M_a; // [rate of sublimation of the solid, kg/s] 

+printf("Rate of sublimation of a cylinder of UF6 is %e kg/s\n\n",w_a);
\ No newline at end of file
diff --git a/905/CH2/EX2.7/2_7.sce b/905/CH2/EX2.7/2_7.sce
new file mode 100755
index 000000000..e4f6629df
--- /dev/null
+++ b/905/CH2/EX2.7/2_7.sce
@@ -0,0 +1,78 @@
+clear;

+clc;

+

+// Illustration 2.7

+// Page: 116

+

+printf('Illustration 2.7 -  Page: 116\n\n');

+

+// solution

+//*****Data*****//

+// a-benzene      b-nitrogen

+T = 300; // [K]

+P = 101.3; // [kPa]

+v =10; // [m/s]

+R = 8.314; // [cubic m.Pa/mole.K]

+//*****//

+

+// Combining the given correlation with the definitions of j-H, and St_H, from Table 2.1 yields

+//            j_H = h*Pr^(2/3)/(Cp*row*v) = h*Pr^(2/3)/(Cp*Gy) = f(Re)

+// Therefore 

+//            h = Cp*Gy*f(Re)/(Pr)^(2/3) = 20*(Gy)^0.5    for carbon dioxide

+

+// Since Re = row*v*l/u = Gy*l/u, where 'l' is a characteristic length, the function f(Re) must be compatible with 20*Gy^0.5 .Therefore, let f(Re) = bRe^n, where 'b'   and 'n' are constants to be evaluated. Then,

+

+//            h =  (Cp*Gy*b/Pr^(2/3))*(l*Gy/u)^n = 20*Gy^0.5

+// Comparing both sides of equation, we get

+//            n+1 =0.5

+// Therefore

+n = -0.5;

+// Data on the properties of C02 at 300 K and 1 bar

+u = 1.5*10^-5; // [viscosity, P]

+Pr = 0.77; // [Prandtl number]

+Cp = 853; // [J/kg.K]

+// Therefore

+//           b = 5.086*l^0.5

+//           j_D = j_H = f(Re) = 5.086*(l^0.5)*Re^-0.5

+// From Table 2.1

+//           F = j_D*c*v/Sc^(2/3) = 5.086*(l^0.5)*c*v/(Re^0.5*Sc^(2/3)) = 5.086*(row*v*u)^0.5/(Mavg*Sc^(2/3))

+

+// Vapor pressure of benzene

+P_a = exp(15.9008-(2788.51/(T-52.36))); // [mm of Hg]

+P_a = P_a*101.3/760; // [kPa]

+

+// Parameter for a-benzene, b-nitrogen 

+yi_a = 0.07;     yi_b = 0.93;   

+Tc_a = 562.2;    Tc_b = 126.2;  // [K]

+Pc_a = 48.9;     Pc_b = 33.9;  // [bar]

+M_a = 78.1;      M_b  = 28;   // [gram/mole]

+V_a = 259;       V_b  = 89.8; // [cubic cm/mole]

+Z_a = 0.271;     Z_b  = 0.290;

+sigma_a = 5.349; sigma_b = 3.798; // [Angstrom]

+ek_a = 412.3;    ek_b = 71.4; // [E/k, K]

+

+

+// From equation 2.52 and 2.53

+Tcm = yi_b*Tc_b+yi_a*Tc_a; // [K]

+Pcm = 10^6*R*Tcm*(yi_b*Z_b+yi_a*Z_a)/((yi_b*V_b+yi_a*V_a)*100000); // [bar]

+M_avg = yi_b*M_b+yi_a*M_a; // [kg/kmole]

+printf("Average molecular weight is %f kg/kmole\n\n",M_avg);

+row = P*M_avg/(R*T); // [kg/cubic m]

+printf("Density of mixture is %f kg/cubic m\n\n",row);

+// From equation 2.50

+Em =  0.176*(Tcm/(M_avg^3*Pcm^4))^(1/6); // [uP]^-1

+

+// From equation 2.51

+Trm = T/Tcm;

+f_Trm = (0.807*Trm^0.618)-(0.357*exp(-0.449*Trm))+(0.340*exp(-4.058*Trm))+0.018; 

+// From equation 2.49 

+u = f_Trm/Em; // [uP]

+u = u*10^-7; // [viscosity, kg/m.s]

+printf("Average viscosity of mixture is %e kg/m.s\n\n",u);

+

+// Calculating diffusivity of benzene using equation 1.49

+D_ab = 0.0986; // [square cm/s]

+Sc = u/(row*D_ab*10^-4); // [Schmidt number]

+

+F = 5.086*(row*v*u)^0.5/(M_avg*Sc^(2/3)); // [kmole/square m.s]

+printf("The required mass transfer coefficient is %e kmole/square m.s\n\n",F);
\ No newline at end of file
diff --git a/905/CH2/EX2.8/2_8.sce b/905/CH2/EX2.8/2_8.sce
new file mode 100755
index 000000000..f92feb271
--- /dev/null
+++ b/905/CH2/EX2.8/2_8.sce
@@ -0,0 +1,52 @@
+clear;

+clc;

+

+// Illustration 2.8

+// Page: 120

+

+printf('Illustration 2.8 -  Page: 120\n\n');

+

+// solution

+//*****Data*****//

+//   a-liquid benzene     b-nitrogen

+T = 300; // [K]

+l = 3; // [length of vertical plate, m]

+b = 1.5; // [width of vertical plate, m]

+P = 101.3; // [kPa]

+v = 5; // [velocity across the width of plate, m/s]

+row_a = 0.88; // [gram/cubic cm]

+//*****//

+

+y_a1 = 0.139; // [mole fraction of benzene at inner edge]

+y_a2 = 0; 

+

+// The film conditions, and average properties, are identical to those in Example 2.7, only the geometry is different

+// Therefore

+M_avg = 31.4; // [kg/kmole]

+row = 1.2; // [kg/cubic m]

+u = 161*10^-7; // [kg/m.s]

+D_ab = 0.0986; // [square cm/s]

+Sc = 1.3; // [Schmidt Number]

+Re = row*v*b/u; // [Renoylds Number]

+

+if(Re>4000)

+    printf('The flow across the plate is turbulent\n\n');

+    else(Re<2000)

+    printf('The flow across the plate is laminar\n\n');

+    end

+    

+// Using equation 2.57

+Sh_l = 0.036*Re^0.8*Sc^(1/3); 

+

+// Nitrogen (component B) does not react with benzene (component A), neither dissolves in the liquid; therefore, NB = 0 and siA = 1. The F-form of the mass-transfer coefficient should be used    

+F = Sh_l*1.26*D_ab*10^-4/(M_avg*b); // [kmole/square m.s]

+N_a = F*log((1-y_a2)/(1-y_a1)); // [kmole/square m.s]

+

+// The total mass rate of evaporation over the surface of the plate

+S = 1.5*3; // [square m]

+M_a = 78.1; // [gram/mole]

+wa = N_a*S*M_a*60*1000; // [gram/min]

+

+V = wa/row_a; // [volumetric flow rate, ml/min]

+

+printf("Liquid benzene should be supplied at the top of the plate at the rate %f ml/min so that evaporation will just prevent it from reaching the bottom of the plate.\n\n",V);
\ No newline at end of file
diff --git a/905/CH2/EX2.9/2_9.sce b/905/CH2/EX2.9/2_9.sce
new file mode 100755
index 000000000..427383d23
--- /dev/null
+++ b/905/CH2/EX2.9/2_9.sce
@@ -0,0 +1,84 @@
+clear;

+clc;

+

+// Illustration 2.9

+// Page: 123

+

+printf('Illustration 2.9 -  Page: 123\n\n');

+

+// solution

+//*****Data*****//

+// a-water    b-air

+dp1 = 10^-3; // [diameter of spherical drop of water, m]

+Tair = 323; // [K]

+P = 101.3; // [kPa]

+Twater = 293; // [K]

+R = 8.314; // [cubic m.Pa/mole.K]

+M_a = 18; // [gram/mole]

+M_b = 29; // [gram/mole]

+//*****//

+

+dp2 = (1/2)^(1/3)*dp1; // [m]

+dp = (dp1+dp2)/2; // [m]

+

+row_p = 995; // [density of water, kg/cubic m]

+row1b = 1.094; // [density of air, kg/cubic m]

+u = 1.95*10^-5; // [kg/m.s]

+row_pr = row_p-row1b; // [kg/cubic m]

+g = 9.8; // [accleration due to gravity, square m/s]

+// Combining equation 2.68 and 2.69

+Ga = 4*dp^3*row1b*row_pr*g/(3*u^2); // [Galileo Number]

+

+// Relationship between Re and Cd

+// Re/Cd = Re^3/Ga = 3*row^2*vt^3/(4*g*u*row_pr)

+

+// The following correlation is used to relate Re/Cd, to Ga

+// ln(Re/Cd)^(1/3) = -3.194 + 2.153*ln(Ga)^(1/3) - 0.238*(ln(Ga)^(1/3))^2 + 0.01068*(ln(Ga)^(1/3))^3

+// Therefore let A = (Re/Cd)

+A = exp(-3.194 + 2.153*log(Ga^(1/3)) - 0.238*(log(Ga^(1/3)))^2 + 0.01068*(log(Ga^(1/3)))^3);

+

+// Therefore 'vt' will be

+vt = A*(4*g*row_pr*u/(3*row1b^2))^(1/3); // [Terminal velocity of particle, m/s]

+printf("Terminal velocity of particle is %f m/s\n\n",vt); 

+

+P_w = 2.34; // [vapor pressure of water, kPa]

+y_w = P_w/P; // [mole fraction of water at the inner edge of the gas film]

+M_avg = 18*y_w+29*(1-y_w); // [gram/mole]

+

+row2b = P*M_avg/(R*Twater); // [kg/cubic.m]

+delta_row = row2b - row1b; // [kg/cubic.m]

+

+Tavg = (Tair+Twater)/2; // [K]

+// At Temperature equal to Tavg density and viscosity are

+row3 = 1.14; // [kg/cubic.m]

+u1 = 1.92*10^-5; // [kg/m.s]

+

+Grd = g*row3*delta_row*(dp^3)/(u1^2);

+

+// Diffusivity of water at Tavg and 1 atm is

+D_ab = 0.242*10^-4; // [square m/s]

+Sc = u1/(row3*D_ab); // [Schmidt Number]

+Re = dp*row3*vt/u1; // [Renoylds Number]

+

+// From equation 2.65 Re is greater than  0.4*Grd^0.5*Sc^(-1/6)

+// Therfore equation 2.64 can be used to calculate mass transfer coefficient

+

+Sh = 2+0.552*(Re^0.5)*Sc^(1/3); // [Sherwood Number]

+// From Table 2.1

+// Sh = kc*P_bm*dp/(P*D_ab), since P_bm is almost equal to P

+// Therefore 

+// Sh = kc*dp/D_ab;

+kc = Sh*D_ab/dp; // [m/s]

+

+ca2 = 0; // [dry air concentration]

+ca1 = P_w/(R*Twater); // [interface concentration, kmole/cubic.m]

+// Average rate of evaporation 

+wa = %pi*dp^2*M_a*kc*(ca1-ca2)*1000; // [g/s]

+

+// Amount of water evaporated

+m = row_p*%pi*dp1^3/12*1000; // [g]

+// Time necessary to reduce the volume by 50%

+t = m/wa; // [s]

+

+D = t*vt; // [distance of fall, m]

+printf("The distance of fall is %f m\n\n",D); 
\ No newline at end of file