initial commit / add all books

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diff --git a/839/CH9/EX9.1/Example_9_1.sce b/839/CH9/EX9.1/Example_9_1.sce
new file mode 100755
index 000000000..733b15c17
--- /dev/null
+++ b/839/CH9/EX9.1/Example_9_1.sce
@@ -0,0 +1,20 @@
+//clear//

+clear;

+clc;

+

+//Exapmle 9.1

+//Given

+Dt = 6; //[ft]

+h = 2; //[ft]

+n = 90/60; //[rps]

+mu = 12*6.72*10^-4; //[lb/ft-s]

+g = 32.17; //[ft/s^2]

+rho = 93.5; //[lb/ft^3]

+Da = 2; // [ft]

+

+Nre = Da^2*n*rho/mu; 

+//From curve A of Fig. 9.12

+Np = 5.8

+//Form Eq.(9.20)

+P = Np*rho*n^3*Da^5/g  //[ft-lbf/s]

+P = P/550 //[hp]

diff --git a/839/CH9/EX9.2/Example_9_2.sce b/839/CH9/EX9.2/Example_9_2.sce
new file mode 100755
index 000000000..f14ebb69d
--- /dev/null
+++ b/839/CH9/EX9.2/Example_9_2.sce
@@ -0,0 +1,30 @@
+//clear//

+clear;

+clc;

+

+//Example 9.2

+//Given

+Dt = 6; //[ft]

+h = 2; //[ft]

+n = 90/60; //[rps]

+mu = 12*6.72*10^-4; //[lb/ft-s]

+g = 32.17; //[ft/s^2]

+rho = 93.5; //[lb/ft^3]

+Da = 2; // [ft]

+

+Nre = Da^2*n*rho/mu;

+//Froude number

+Nfr = n^2*Da/g;

+//From Table 9.1 

+a = 1;

+b = 40.0;

+//Using Eq.(9.19)

+m = (a-log(Nre)/2.303)/b;

+//Using Fig. 9.12, curve D,

+Np = 1.07;

+//Corrected valus of Np

+Np =  Np*Nfr^m;

+

+//Form Eq.(9.20)

+P = Np*rho*n^3*Da^5/g //[ft-lbf/s]

+P = P/550 //[hp]

diff --git a/839/CH9/EX9.3/Example_9_3.sce b/839/CH9/EX9.3/Example_9_3.sce
new file mode 100755
index 000000000..19cca12fc
--- /dev/null
+++ b/839/CH9/EX9.3/Example_9_3.sce
@@ -0,0 +1,21 @@
+//clear//

+clear;

+clc;

+

+//Example 9.3

+//Given

+Dt = 6; //[ft]

+h = 2; //[ft]

+n = 90/60; //[rps]

+mu = 1200*6.72*10^-2; //[lb/ft-s]

+g = 32.17; //[ft/s^2]

+rho = 70 //[lb/ft^3]

+Da = 2; // [ft]

+

+Nre = Da^2*n*rho/mu;

+//From Table 9.3

+KL = 65;

+//From Eq.(9.21)

+Np = KL/Nre;

+P = Np*rho*n^3*Da^5/g  //[ft-lbf/s]

+P = P/550 //[hp]

diff --git a/839/CH9/EX9.4/Example_9_4.sce b/839/CH9/EX9.4/Example_9_4.sce
new file mode 100755
index 000000000..5da5391ff
--- /dev/null
+++ b/839/CH9/EX9.4/Example_9_4.sce
@@ -0,0 +1,17 @@
+//clear//

+clear;

+clc;

+

+//Example 9.4

+//Given

+Dt = 6; //[ft]

+Da = 2; //[ft]

+n = 80/60; //[rps]

+T = 70; //[F]

+rho = 62.3; //[lb/ft^3], From Appendix 14

+mu = 6.6*10^-4; // [lb/ft-s], From Appendix 14

+

+Nre = Da^2*n*rho/mu; 

+//From Fig. 9.15

+ntT = 36;

+tT = ntT/1.333 //[s]

diff --git a/839/CH9/EX9.5/Example_9_5.sce b/839/CH9/EX9.5/Example_9_5.sce
new file mode 100755
index 000000000..155063c99
--- /dev/null
+++ b/839/CH9/EX9.5/Example_9_5.sce
@@ -0,0 +1,33 @@
+//clear//

+clear;

+clc;

+

+//Example 9.5

+//Given

+Dt = 6; //[ft]

+H = 8; //[ft]

+T = 70; //[F]

+sp_gr = 3.18;

+w_fr = 0.25;

+Da = 2; //[ft]

+h = 1.5; //[ft]

+gc = 32.17; //[ft-lb/lbf-s^2]

+// (a) 

+//Using data of Buurman et al. in Fig.(9.19)

+//change in nc

+delta_nc = (104/200)^0.2*(2.18/1.59)^0.45*(33.3/11.1)^0.13;

+//change in P

+dalta_P = delta_nc^3;

+

+//Using Fig. 9.19 

+V = %pi/4*Dt^2*H*7.48 ; //[gal]

+P = 3.3*V/1000 //[hp]

+

+//(b)

+//From Table 9.3, for a cour blade turbine,

+KT = 1.27;

+Np = KT;

+//slurry density

+rho_m = 1/((w_fr/sp_gr)+(1-w_fr))*62; // [lb/ft^3]

+ 

+nc = (P*gc*550/(Np*rho_m*Da^5))^(1/3) // [r/s]

diff --git a/839/CH9/EX9.6/Example_9_6.sce b/839/CH9/EX9.6/Example_9_6.sce
new file mode 100755
index 000000000..ae7b7837f
--- /dev/null
+++ b/839/CH9/EX9.6/Example_9_6.sce
@@ -0,0 +1,53 @@
+//clear//

+clear;

+clc;

+

+//Example 9.6

+//Given

+Dt = 2; //[m]

+Da = 0.667; //[m]

+n = 180/60; //[rps]

+T = 20; //[C]

+qg = 100; //[m^3/h]

+rho = 1000; //[kg/m^3]

+mu = 10^-3; //[kg/m-s] 

+ut = 0.2; //[m/s]

+//(a)

+//The power input is calculated and followed by correction of gas effect

+Nre = n*Da^2*rho/mu;

+//For a flat blade turbine, from Table 9.3

+KT = 5.75;

+//Using Eq.(9.24)

+Po = KT*n^3*Da^5*rho/1000; //[kW] 

+At = %pi/4*Dt^2; //[m^2]

+//Superficial gas velocity

+Vs_bar = At*qg/3600/10 //[m/s]

+//From Fig. 9.20 Pg/Po = 0.60

+Pg = Po*0.6; //[kW] 

+//From Fig.9.7, depth of liquid is equal to diameter of the tank

+//Hence, liquid volume

+V = %pi/4*Dt^2*Dt; //[m^3]

+//The input power per unit volume

+PgbyV = Pg/V ; //[kW/m^3]

+

+//(b)

+sigma = 72.75; //[g/s^2]

+rho_L = 10^-3; //[g/mm]

+PgbyV = PgbyV*10^3 ; //[g/mm-s^2]

+//Using Eq.(9.46)

+//Let x = shi^(0.5)

+//solving the equation as quadratic equation

+a = 1;

+b = -(Vs_bar/ut)^0.5;

+c = -0.216*((PgbyV)^0.4)*(rho_L^0.2)/(sigma^0.6)*(Vs_bar/ut)^(0.5);

+x = (-b+sqrt(b^2-4*a*c))/(2*a);

+shi = x^2;

+

+//(c)

+//To find out mean bubble diameter

+//Using Eq.(9.44)

+Ds_bar = 4.15*sigma^0.6/(PgbyV^0.4*rho_L^0.2)*shi^0.5+0.9 // [mm]

+

+//(d)

+//From Eq.(9.40)

+aprime = 6*shi/Ds_bar //[mm^-1]

diff --git a/839/CH9/EX9.7/Example_9_7.sce b/839/CH9/EX9.7/Example_9_7.sce
new file mode 100755
index 000000000..ab36bb0f1
--- /dev/null
+++ b/839/CH9/EX9.7/Example_9_7.sce
@@ -0,0 +1,24 @@
+//clear//

+clear;

+clc;

+

+//Exapmle 9.7

+//Given

+Dt = 2; //[m]

+Da = 0.667; //[m]

+n = 180/60; //[rps]

+T = 20; //[C]

+qg = 100; //[m^3/h]

+rho = 1000; //[kg/m^3]

+mu = 10^-3; //[kg/m-s] 

+ut = 0.2; //[m/s]

+At = %pi/4*Dt^2; //[m^2]

+//Using values form Example 7.6

+//Assuming Pg/Po decresaes to 0.25

+PgbyV = 0.25*20490/6.28; //[W/m^3]

+//Using Eq.(9.47)

+Vs_barc = 0.114*(PgbyV)*(Dt/1.5)^0.17/1000 //[m/s]

+qg = Vs_barc*At*3600 //[m^3/h]

+//The calculated flooding velocity is beyond the range of the data on which Eq.(9.47)

+//was based, so it may not be relaible. Based on Vs_barc, the highest measured value, qg

+//would be 850 m^3/h.

diff --git a/839/CH9/EX9.8/Example_9_8.sce b/839/CH9/EX9.8/Example_9_8.sce
new file mode 100755
index 000000000..44af9cf3b
--- /dev/null
+++ b/839/CH9/EX9.8/Example_9_8.sce
@@ -0,0 +1,27 @@
+//clear//

+clear;

+clc;

+

+//Example 9.8

+//Given

+D1 = 1; //[ft]

+D6 = 6

+Nre_i = 10^4;

+Da = 4; //[in.]

+t1 = 15; //[s]

+P = 2; //[hp/gal]

+

+//(a)

+//Using Fig. 9.15 

+//the mixing factor ntT is constant and time tT is asumed constant,

+//speed n will be the same in both vessels.

+//Using Eq.(9.24) with consant density

+PbyD_ratio = (D6/D1)^2;

+//The Power input required in the 6-ft vessel is then

+Pin = 2*PbyD_ratio //[hp/1000gal]

+

+//(b)

+//Using Eq.(9.54) with same input power per unit volume in both vessels

+n6byn1 = (D6/D1)^(2/3)

+//blending in the 6-ft vessel would be

+t6 = t1*n6byn1 // [s]