initial commit / add all books

6 files changed, 175 insertions, 0 deletions

diff --git a/389/CH2/EX2.1/Example2_1.sce b/389/CH2/EX2.1/Example2_1.sce
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index 000000000..0dcec1356
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+++ b/389/CH2/EX2.1/Example2_1.sce
@@ -0,0 +1,27 @@
+clear;

+clc;

+

+// Illustration 2.1

+// Page: 30

+

+printf('Illustration 2.1 - Page 30\n\n');

+

+// solution

+

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

+// a = O2 & b = CO

+Dab = 1.87*10^(-5);//[square m/s]

+Pt = 10^5;//[N/square m]

+z = 0.002;//[m]

+R = 8314;//[Nm/kmol]

+T = 273;//[K]

+Pa1 = 13*10^(3);//[N/square m]

+Pb1 = 10^(5)-13*10^(3);//[N/square m]

+Pa2 = 6500;//[N/square m]

+Pb2 = 10^(5)-6500;//[N/square m]

+//********//

+

+// Calculation from Eqn. 2.30

+Pbm = (Pb1-Pb2)/log(Pb1/Pb2);//[N/square m]

+Na = Dab*Pt*(Pa1-Pa2)/(R*T*z*Pbm);//[kmol/square m.s]

+printf('Rate of diffusion of oxygen is %e kmol/square m.sec',Na);
\ No newline at end of file
diff --git a/389/CH2/EX2.2/Example2_2.sce b/389/CH2/EX2.2/Example2_2.sce
new file mode 100755
index 000000000..aa3c0e722
--- /dev/null
+++ b/389/CH2/EX2.2/Example2_2.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+

+// Illustration2.2

+// Page: 30

+

+printf('Illustration 2.2 - Page: 30\n\n');

+

+// solution

+

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

+Pt = 10^5;//[N/square m]

+z = 0.002;//[m]

+R = 8314;//[Nm/kmol]

+T = 273;//[K]

+//a = O2 b = CH4 c = H2

+Pa1 = 13*10^(3);//[N/square m]

+Pb1 = 10^(5)-13*10^(3);//[N/square m]

+Pa2 = 6500;//[N/square m]

+Pb2 = 10^(5)-6500;//[N/square m]

+Dac = 6.99*10^(-5);//[N/square m]

+Dab = 1.86*10^(-5);//[N/square m]

+//*******//

+

+// Calculation from Eqn. 2.30

+Pbm = (Pb1-Pb2)/log(Pb1/Pb2);//[N/square m]

+Yb_prime = 2/(2+1);

+Yc_prime = 1-Yb_prime;

+Dam = 1/((Yb_prime/Dab)+(Yc_prime/Dac));//[square m.s]

+Na = Dam*(Pa1-Pa2)*Pt/(R*T*z*Pbm);//[kmol/square m.s]

+printf('Rate of diffusion is %e kmol/square m.sec',Na);
\ No newline at end of file
diff --git a/389/CH2/EX2.3/Example2_3.sce b/389/CH2/EX2.3/Example2_3.sce
new file mode 100755
index 000000000..615b087bc
--- /dev/null
+++ b/389/CH2/EX2.3/Example2_3.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+

+// Illustration2.3

+// Page: 32

+

+printf('Illustration 2.3 - Page: 32\n\n');

+

+// solution

+

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

+// a = C2H5OH b = air

+Pt = 101.3*10^(3);//[N/square m]

+T = 273;//[K]

+//********//

+

+Ma = 46.07;// [kg/kmol]

+Mb = 29;// [kg/kmol]

+//For air from Table 2.2 (Pg 33)

+Eb_by_k = 78.6;// [K]

+rb = 0.3711; // [nm]

+// For C2H5OH using Eqn. 2.38 & 2.39

+// From Table 2.3

+Va = (2*0.0148)+(6*0.0037)+(0.0074);// [cubic m/kmol]

+Tba = 351.4;// [K]

+ra = 1.18*(Va^(1/3));//[nm]

+Ea_by_k = 1.21*Tba;// [K]

+rab = (ra+rb)/2;// [nm]

+Eab_by_k = sqrt(Ea_by_k*Eb_by_k);// [K]

+Collision_value = T/Eab_by_k;

+//From Fig. 2.5 (Page: 32) f(collision value)

+Collision_func = 0.595;

+Dab = (10^(-4)*(1.084-(0.249*sqrt((1/Ma)+(1/Mb))))*T^(3/2)*sqrt((1/Ma)+(1/Mb)))/(Pt*(rab^2)*Collision_func);//[square m/s]

+printf('The diffusivity of ethanol through air at 1 atm. & 0C is %e square m/s\n',Dab);

+printf('The observed value (Table 2.1) is 1.02*10^(-5) square m/s')
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diff --git a/389/CH2/EX2.4/Example2_4.sce b/389/CH2/EX2.4/Example2_4.sce
new file mode 100755
index 000000000..aa6a029e9
--- /dev/null
+++ b/389/CH2/EX2.4/Example2_4.sce
@@ -0,0 +1,34 @@
+clear;

+clc;

+

+// Illustration 2.4

+// Page: 34

+

+printf('Illustration 2.4 - Page: 34\n\n');

+

+// solution

+

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

+// a = acetic acid b = H2O

+z = 0.001;// [m]

+Dab = 0.95*10^(-9);//[square m/s]

+//************//

+

+Ma = 60.03;// [kg/kmol]

+Mb = 18.02;// [kg/kmol]

+//At 17 C & 9% solution

+density1 = 1012; //[kg/cubic m]

+Xa1 = (0.09/Ma)/((0.09/Ma)+(0.91/Mb));

+Xb1 = 1-Xa1;

+M1 = 1/((0.09/Ma)+(0.91/Mb));// [kg/kmol]

+//At 17 C & 3% solution

+density2 = 1003.2; //[kg/cubic m]

+Xa2 = (0.03/Ma)/((0.03/Ma)+(0.97/Mb));

+Xb2 = 1-Xa2;

+M2 = 1/((0.03/Ma)+(0.97/Mb));// [kg/kmol]

+avg_density_by_M = ((density1/M1)+(density2/M2))/2;//[kmol/cubic m]

+// From Eqn. 2.42

+Xbm = (Xb2-Xb1)/log(Xb2/Xb1);

+// From Eqn. 2.41

+Na = Dab*(avg_density_by_M)*(Xa1-Xa2)/(Xbm*z); //[square m/s]

+printf('The rate of diffusion is %e square m/s',Na);
\ No newline at end of file
diff --git a/389/CH2/EX2.5/Example2_5.sce b/389/CH2/EX2.5/Example2_5.sce
new file mode 100755
index 000000000..ae51e5e1a
--- /dev/null
+++ b/389/CH2/EX2.5/Example2_5.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+

+// Illustration 2.5

+// Page: 37

+

+printf('Illustration 2.5 - Page: 37\n\n');

+

+// solution

+

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

+// a = mannitol b = H2O

+T = 293; // [K]

+//*****//

+

+Mb = 18.02;// [kg/kmol]

+// From Table 2.3 (Pg 33)

+Va = (0.0148*6)+(0.0037*14)+(0.0074*6); // [cubic m/kmol]

+viscosity = 0.001005; // [kg/m.s]

+association_factor = 2.26; // [water as a solvent]

+Dab = (117.3*10^(-18))*((association_factor*Mb)^0.5)*T/(viscosity*Va^0.6); // [square m/s]

+printf('Diffusivity of mannitol is %e square m/s\n',Dab);

+printf('Observed value is 0.56*10^(-9) square m/s');
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diff --git a/389/CH2/EX2.6/Example2_6.sce b/389/CH2/EX2.6/Example2_6.sce
new file mode 100755
index 000000000..f0b3123b4
--- /dev/null
+++ b/389/CH2/EX2.6/Example2_6.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+

+// Illustration 2.6

+// Page: 37

+

+printf('Illustration 2.6 - Page 37\n\n');

+

+// solution

+

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

+T2 = 70+273;// [K]

+//**********//

+

+// a = mannitol b = H2O

+// From Illustration 2.5 at 20 C

+viscosity1 = 1.005*10^(-3); // [kg/m.s]

+Dab1 = 0.56*10^(-9); //[m^2/s]

+T1 = 273+20;// [K]

+// At 70 C

+viscosity2 = 0.4061*10^(-3); // kg/m.s

+// Eqn. 2.44 indicates Dab*viscocity/T  =  constnt

+Dab2 = Dab1*(T2)*(viscosity1)/(T1*viscosity2);// [square m/s]

+printf('Diffusivity of mannitol at 70 OC is %e square/s\n',Dab2);

+printf('Observed value at 70 OC is 1.56*10^(-9) square m/s');
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