initial commit / add all books

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diff --git a/542/CH7/EX7.1/Example_7_1.sci b/542/CH7/EX7.1/Example_7_1.sci
new file mode 100755
index 000000000..7ce893d09
--- /dev/null
+++ b/542/CH7/EX7.1/Example_7_1.sci
@@ -0,0 +1,63 @@
+clear;

+clc;

+printf("\n Example 7.1");

+//In the leaf filter filtration is at const pressure from the start

+//V^2 + 2ALV/v = 2(-deltaP)A^2t/(ruv)

+

+//In the filter press,a volume V1 of filtrate is obtained under const rate conditions in time t1,and filtration is then carried out at constant pressure.

+//V1^2 + 2ALV1/v = 2(-deltaP)A^2t1/(ruv) 

+//and (V^2 − V1^2 ) + 2AL/υ(V − V1) = 2(−P)A^2/rμυ(t − t1)

+

+//for the leaf filter

+t2 = 300;          //t2 is in secs

+V2 = 2.5*10^(-4);   //V2 is in m^3

+t3 = 600;           //t3 is in secs

+V3 = 4*10^(-4);     //V3 is in m^3

+A = 0.05;           //A is in m^2

+deltaP = -7.13*10^(4); //it is in N/m^2

+//putting these values in above eq

+

+a = [2*7.13*10^(4)*0.05^(2)*300 -2*0.05*2.5*10^(-4);2*7.13*10^(4)*0.05^(2)*600 -2*0.05*4*10^(-4)];

+b = [(2.5*10^(-4))^2;(4*10^(-4))^2];

+x = inv(a)*b;

+y = [1/x(1);x(2)];

+printf("\n L/υ=%f*10^(-3)   and rμυ = %f*10^(11)",y(2)*10^3,y(1)*10^(-11));

+

+//for the filter press

+V1 = poly([0],'V1');

+s = roots(V1^2 + (2.16*y(2)*V1)-(4*10^(5)*2.16^2)/y(1)*180);

+printf("\n the value of V1 = %fm^3",s(2));

+

+//For a constant pressure period (t - t1)=900secs

+//Calculting the total volume of filtrate

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

+d = roots((V^2-3.33*10^(-4))+(1.512*10^(-2)*(V-1.825*10^(-2))-5.235*10^(-6)*900));

+printf("\n The value of V = %.3f m^3",d(2));

+

+f = (4*10^(5)*(2.16)^2)/(7.13*10^(11)*(6.15*10^(-2) + 2.16*3.5*10^(-3)));

+printf("\n The final rate of filtration is %.2f*10^(-5) m^3/sec",f*10^(5));

+

+// Assuming  viscosity of the filtrate is the same as that of the wash-water

+rw_400 = (0.25)*f;

+printf("\n Rate of washing at 400 kN/m2 = %.1f*10^(-6) m^3/sec",rw_400*10^(6));

+

+rw_275 = rw_400*(275/400);

+printf("\n Rate of washing at 275 kN/m^2 = %.1f*10^(-6) m^3/sec",rw_275*10^6);

+printf("\n Thus the amount of wash-water passing in 600s = %.3f m^3",600*rw_275);

+

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diff --git a/542/CH7/EX7.2/Example_7_2.sci b/542/CH7/EX7.2/Example_7_2.sci
new file mode 100755
index 000000000..318136c9f
--- /dev/null
+++ b/542/CH7/EX7.2/Example_7_2.sci
@@ -0,0 +1,52 @@
+clear;

+clc;

+printf("\n Example 7.2");

+//The slurry contains 100kg whiting/m^3 of water

+printf("\n Volume of 100 kg whiting = %f m^3",100/3000);

+printf("\n Volume of cake = %f m^3",0.0333/0.6);

+printf("\n Volume of liquid in cake = %f m^3",0.05556*0.4);

+printf("\n Volume of filtrate = %.3f m^3",(1-0.0222));

+printf("\n volume of cake/volume of filtrate v = %f",0.0556/0.978);

+A = 10^(-4);                 //area in sq meters

+deltaP = -1.65*10^(5);       //P is in pascals

+l = 0.01;                    //length is in meters

+vol_flow_rate = 2*10^(-8);           //Volume flow rate is in m^3/sec

+u = 10^(-3);                 //vicosity is in Ns/m^2

+

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

+r1 = roots((10^4)*(2*10^(-8)*r)-1.65*10^(5)/(10^(-5)));

+printf("\n r = %.2f*10^(13)/m^2",r1*10^(-13));

+

+function[Lopt]=optimum()

+    Lopt = 1.161*10^(-3)*(900)^(0.5);  //t = 900 secs

+    funcprot(0);

+endfunction

+printf("\n optimum frame thickness = %.1f mm",2*optimum()*1000);

+

+//total cycle time = 1.015L^2 + 900

+//rate of cake production R = L/(1.015L^2 + 900)

+ 

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

+ L1 = roots(1.025*10^(6)*L^2 + 900 - 2.050*10^(6)*L^2);

+ printf("\n Frame thickness = %.2f mm",2*L1(1)*10^3);

+ 

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diff --git a/542/CH7/EX7.3/Example_7_3.sci b/542/CH7/EX7.3/Example_7_3.sci
new file mode 100755
index 000000000..bd0e9e46c
--- /dev/null
+++ b/542/CH7/EX7.3/Example_7_3.sci
@@ -0,0 +1,57 @@
+clear;

+clc;

+printf("\n Example 7.3");

+V = 0.094;         //volume in m^3

+deltaP = -3530;    //P is in kN/m^2

+

+//At t = 1105 secs

+ V1 = 0.166;       //V is in m^3

+ deltaP1 = -5890;   //P is in kN/m^2

+

+a = [2.21*10^(6) -0.094;6.51*10^(6) -0.166];

+b = [0.0088;0.0276];

+x = inv(a)*b;

+y = [x(2);x(1)];

+printf("\n LA/v =%f       A^2/rμv = %f*10^(-7)",y(1),y(2)*10^7);

+printf("\n For the full size plant:");

+printf("\n LA/v = %f       A^2/rμv=%f*10^(-7)",10*y(1),y(2)*10^8);

+

+//Solving LHS of the integral

+LHS = integrate('b+0.154+2.31','b',0,1);

+//Equating LHS = RHS

+t = LHS/(3.46*10^(-3));

+printf("\n t = %d secs",t);

+printf("\n deltaP = %dkN/m^2",(1+0.154)/(4.64*10^(-7)*857));

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diff --git a/542/CH7/EX7.4/Example_7_4.sci b/542/CH7/EX7.4/Example_7_4.sci
new file mode 100755
index 000000000..d24679ebc
--- /dev/null
+++ b/542/CH7/EX7.4/Example_7_4.sci
@@ -0,0 +1,23 @@
+clear;

+clc;

+printf("\n Example 7.4");

+a = [2*84300*0.02^(2)*60 -2*0.02*0.0003;2*84300*0.02^(2)*120 -2*0.02*0.00044];

+b = [0.0003^2;0.00044^2];

+x = inv(a)*b;

+y = [x(2);1/x(1)];

+printf("\n L/v = %f         ruv = %f*10^(10)",y(1),y(2)*10^(-10));

+printf("\n Area of filtering surface = %f m^2",4*(%pi));

+printf("\n Bulk volume of cake deposited =%.3f m^3/revolution",4*(%pi)*0.005);

+

+V = sqrt(1*10^(-6)*143^2);

+printf("\n V = %.3f m^3",V);

+

+t =poly([0],'t');

+t1 = roots(0.141^2 +2*2.19*10^(-3)*0.141-2*84300*(4*(%pi))^(2)*t/(3.48*10^10));

+printf("\n t = %f secs",t1);

+printf("\n time for 1 revolution =%.1f secs",t1/0.4);

+printf("\n speed = %.3fHz",0.4/t1);

+printf("\n rate of filtrate production w = %.2f kg/sec",143/67.3)

+printf("\n mass of slurry S =%.1f kg/sec",1.66*2.11);

+

+

diff --git a/542/CH7/EX7.5/Example_7_5.sci b/542/CH7/EX7.5/Example_7_5.sci
new file mode 100755
index 000000000..7d283ceac
--- /dev/null
+++ b/542/CH7/EX7.5/Example_7_5.sci
@@ -0,0 +1,23 @@
+clear;

+clc;

+printf("\n Example 7.5");

+V1 = 0.00025;               //V is in m^3

+t = 300;                    //t is in secs

+a = [7.14*10^(-6) 2.86*10^(-4);11.42*10^(-6) 2.86*10^(-4)];

+b = [1.2*10^(6);1*10^(6)];

+x = inv(a)*b;

+

+//for the plate and frame filter

+B1 = x(1)/(2*2.2^2*413*10^3);

+B2 = x(2)/(2.2*413*1000);

+

+printf("\nrμv = %d\n",x(1));

+printf("\n rμl = %d",x(2));

+printf("\n B1= %f    B2= %f",B1,B2);

+printf("\n the filtration time for maximum throughput is:");

+t1 = 21.6*10^3;

+t0= t1 +B2*(t1/B1)^(0.5);

+printf("\n t = %f secs",t0);

+V = (t1/B1)^(0.5);

+printf("\n V= %f m^3",V);

+printf("\nMean rate of filtration is: %.2f *10^-6 m^3/s",(V/(t1+t0))/10^-6);
\ No newline at end of file
diff --git a/542/CH7/EX7.6/Example_7_6.sci b/542/CH7/EX7.6/Example_7_6.sci
new file mode 100755
index 000000000..40ee58c6e
--- /dev/null
+++ b/542/CH7/EX7.6/Example_7_6.sci
@@ -0,0 +1,64 @@
+clear;

+clc;

+printf("\n Example 7.6");

+A=0.6*0.6*%pi;       //in m^2

+rate=1.25*10^-4;     // in m^3/s

+

+v_w=0.2/(3*10^3);

+v_f=10^-3-v_w;

+

+v=v_w/v_f;

+v_rate=rate*v;

+w=360*0.2;

+

+t=v_rate*w/A;

+printf("\nThickness of cake produced is : %.1f mm",t/10^-4);

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

+K1 = roots((1.25*10^(-4)*360)^2-K*(6.5*10^(4)*(0.36*(%pi))^(2)*72));

+printf("\n The value of K is %.2f*10^(-10)",K1*10^(10));

+

+//Filter press

+//Using a filter press with n frames of thickness b m the total time, for one complete cycle of the press =(tf+120n+240),where tf is the time during which filtration is occurring

+//overall rate of filtration = Vf/(tf + 120n + 240)

+

+// Vf = 0.3^(2)*n*b/0.143

+//tf = 2.064*10^5 b^2

+

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

+b1 = roots(b^2 - 0.0458*b - 0.001162);

+printf("\n The thickness is %.4f m",b1(1));

+

+function[n]=number_of_plates()

+    

+    n = (0.030 + 25.8*b1(1)^2)/(0.629*b1(1)-0.015);

+    funcprot(0);

+endfunction

+n = number_of_plates();

+printf("\n The minimum number of plates required is %d",ceil(n));

+

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

+d1 = roots(ceil(n)*(0.629*d-0.015)-0.030-25.8*d^2);

+printf("\n The sizes of frames which will give exactly the required rate of filtration when six are used are %f mm",d1*10^3);

+printf("\n\n\n  Thus any frame thickness between 47 and 99 mm will be satisfactory. In practice,50 mm (2 in) frames would probably be used.")

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diff --git a/542/CH7/EX7.7/Example_7_7.sci b/542/CH7/EX7.7/Example_7_7.sci
new file mode 100755
index 000000000..dba6cd4be
--- /dev/null
+++ b/542/CH7/EX7.7/Example_7_7.sci
@@ -0,0 +1,48 @@
+clear;

+clc;

+printf("\n Example 7.7");

+//Case 1

+

+                 //dV/dt = A^2(-deltaP)/vru(V + AL/v) = a/V+b

+                 //For constant rate filtration:

+                 //Vo/to = a/Vo + b

+                 //Vo^2 + bVo = ato

+                 //For constant pressure filtration

+                 //0.5(V^2 - Vo^2)+b(V-Vo)=a(t-to)

+                 //to=600s,t-to=3600s,Vo=V/4

+                 //V^2/16 +bV/4 = 600a

+                 //o.5(V^2 - V^2/16)+b(V-V/4)=3600a

+                 //3600a = (15/32)V^2 +3/4(bV) = 3/8(V^2) + 3/2(bV)

+                 //b = V/8

+                // a = (V^2/16 + V^2/32)/600 = (3/19200)V^2 

+                //Total cycle time = 900 + 4200 = 5100secs

+                //Filtration rate = V/5100 = 0.000196V

+

+//Case 2

+                //V1/t1 = a/ (V1 +b/4)=Vo/to=a/(Vo+b)

+                //0.5*(49/64V^2 - V1^2)+b/4(7/8V-V1)=a(t-t1)

+                //V/2400 = (3/19200)V^2/(V1+V/32)

+                //t1 = (to/Vo)V1

+  t1 = 600/(1/4)*(11/32);

+  printf("\n t1 = %dsecs",t1);

+  //Substituting gives 

+  deltaT = (19200/3)*(784-121+34)/2048;

+  printf("\n t -t1 = %d secs",deltaT); 

+  Cycle_time = 180+900+t1+deltaT;

+  printf("\n cycle time = %d secs",Cycle_time);

+  Increase = (0.000214 - 0.000196)/(0.000196)*100;

+  printf("\n Increase in filtration rate is %.1f per cent",Increase);

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\ No newline at end of file