initial commit / add all books

3 files changed, 94 insertions, 0 deletions

diff --git a/542/CH5/EX5.1/Example_5_1.sci b/542/CH5/EX5.1/Example_5_1.sci
new file mode 100755
index 000000000..9f465edd0
--- /dev/null
+++ b/542/CH5/EX5.1/Example_5_1.sci
@@ -0,0 +1,25 @@
+clear;

+clc;

+printf("\n Example 5.1");

+//Basis 1 kg of solids

+feed_rate_solid = 1.33;  //Mass rate of feed of solids in kg/sec

+U = 1.5;                 //Mass rate of solids in the underflow in kg/sec

+Y = [5.0 ; 4.2; 3.7 ; 3.1; 2.5];

+printf("\n concentration(Y) (kg water/kg solids):\n");

+printf("%.1f\n",Y);

+printf("\n water to overflow (Y-U) (kg water/kg solids):\n");

+O = Y - 1.5;//Amount of water to overflow in kg water/kg solids

+printf("\n %.1f\n",O);

+Uc = [2.00*10^(-4);1.20*10^(-4);0.94*10^(-4);0.70*10^(-4);0.50*10^(-4)];

+printf("\n sedimentation rate uc (m/sec):\n");

+printf("%f  \n",Uc);

+X =[1.75*10^4;2.25*10^4;2.34*10^4;2.29*10^4;2.00*10^4];//X = (Y-U)/Uc

+printf(" \n\n(Y-U)/Uc (s/m):\n");

+printf("\n %d\n",X);

+z = max(X); //prints the maximum value of X

+printf("\nMaximum value of (Y-U)/Uc = %ds/m",z);

+

+//Calculating the require darea of the thickener

+A = z*1.33/1000; //1.33 is the mass feed rate of solids in kg/sec

+//1000 is the density of water in kg/m^3

+printf("\n The required area of the thickener is :%.2fm^2\n",A);

diff --git a/542/CH5/EX5.2/Example_5_2.sci b/542/CH5/EX5.2/Example_5_2.sci
new file mode 100755
index 000000000..51b8f5719
--- /dev/null
+++ b/542/CH5/EX5.2/Example_5_2.sci
@@ -0,0 +1,33 @@
+clear;

+clc;

+printf("\n Example 5.2");

+//Area of the tank required to give an underflow concentration of 1200kg/m^3 for a feed rate of 2 m^3/min

+

+//Initial height of slurry in the tank

+H = [900;800;700;600;500;400;300;260;250;220;200;180];

+uc = [13.4;10.76;8.6;6.6;4.9;3.2;1.8;1.21;1.11;0.80;0.60;0.40];

+i=1;

+while i<13

+    c(i)=200*900/H(i);

+    x(i)=1000*(1/c(i)-1/1200);

+    sed(i) = c(i)*uc(i)/(1000*60);

+    y(i)= uc(i)*10^(-3)/((1/c(i)-1/1200)*60);

+    z(i) = 1/y(i);

+    i=i+1;

+end

+printf("\nH(mm)");

+printf("\n%d",H);

+printf("\n c(kg/m^3):\n");

+printf("%d\n",c);

+printf("Sedimentation flux(kg.s/m^2):\n");

+printf("%.4f\n",x);

+printf("uc/(1/c-1/1200)\nkg.sec/m^2:\n");

+printf("%.4f\n",y);

+printf("1000*(1/c-1/cu)\nm^3/kg*10^3\n");

+printf("%.3f \n",x);

+printf("\n\n(1/c-1/1200)/uc\n m^2.kg/sec\n");

+printf("%.1f\n",z);

+m1=max([18.7;20.1;21.3;22.7;23.8;26.0;27.8;30.3;30.0;29.2;27.8;25.0]);

+printf("\n\nthe maximum value of (1/c-1/1200)/uc is %.1f m^2*kg/s",m1);

+A = 2*200*30.3/60;

+printf("\n The area required is A = Qc[(1/c-1/cu)/uc]max = %dm^2",A)

diff --git a/542/CH5/EX5.3/Example_5_3.sci b/542/CH5/EX5.3/Example_5_3.sci
new file mode 100755
index 000000000..c7947ba01
--- /dev/null
+++ b/542/CH5/EX5.3/Example_5_3.sci
@@ -0,0 +1,36 @@
+clear;

+clc;

+printf("\t Example 5.3 ");

+//Assumption: Resistance force F on an isolated sphere is given by Stoke's law:F = 3*%pi(meu)d*u

+

+C = poly([0],'C');

+x=roots(-4.8*C+(1-C));

+printf("\n concentration is:%.3f",x);

+

+//terminal falling velocity u can be calculated by force balance

+//u = d^2*g/(18*meu)*(ps-p)

+function[u]=terminal_velocity()

+    d = 10^(-4);   //diameter is in meters

+    g = 9.81;      //acceleration due to gravity is in m/sec^2

+    meu = 10^(-3); //viscosity is in N.s/m^2

+    ps = 2600;     //density is in kg/m^3

+    p = 1000;      //density is in kg/m^3

+    

+    u = (d^2)*g*(ps-p)/(18*meu);

+    funcprot(0);

+endfunction

+

+

+function[si]=si_max()

+    u=terminal_velocity()

+    printf("\n The terminal falling velocity is %.5f m/sec",u);

+    si=u*x*(1-x)^(4.8);

+    funcprot(0);

+endfunction

+si = si_max();

+printf("\nThe maximum value is %f*10^(-4) m^3/m^2sec",si*10^4)

+

+

+

+

+