initial commit / add all books

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diff --git a/542/CH8/EX8.2/Example_8_2.sci b/542/CH8/EX8.2/Example_8_2.sci
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index 000000000..c4d187e29
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+clear; 

+clc;

+printf("\n Example 8.2");

+

+//From the gel polarisation model:

+         //J = (1/A)(dV/dt) = hD ln(Cg/Cf)

+         //Cf = Co(Vo/V)

+  //where Co and Vo are the initial concentration and volume,respectively and Cf and V are the values at subsequent times

+  //Combining these eq gives

+        //dV/dt = A(hDln(Cg/Co)-hDln(Vo/V))

+V = [10 5 3 2 1];

+y = [9.90 13.64 18.92 27.30 112.40];

+plot(V,y)

+xtitle("Area under the curve is 184.4","Volume(m^3)","(J - hDln(Vo/V))^(-1)")

+

+

+//(b)

+Jo = 0.04*log(250/20);

+printf("\n Jo = %.3f m/h",Jo);

+Jf = 0.04*log(250/200);

+printf("\n Jf = %f m/h",Jf);

+Jav = Jf + 0.27*(0.101-0.008);

+printf("\n Jav = %f m/h",Jav);

+//For the removal of 9m^3 filtrate in 4 hours

+Area = (9/4)/Jav;

+printf("\n Area = %fm^2",Area);

+

diff --git a/542/CH8/EX8.3/Example_8_3.sci b/542/CH8/EX8.3/Example_8_3.sci
new file mode 100755
index 000000000..b771af109
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+++ b/542/CH8/EX8.3/Example_8_3.sci
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+clear;

+clc;

+printf("\n Example 8.3");

+//It is assumed that Q0 is the volumetric flowrate of feed

+// Q2 the volumetric flowrate of concentrate

+//C0 the solute concentration in the feed

+// C2 the solute concentration in the concentrate

+// F the volumetric flowrate of membrane permeate

+// A the required membrane area.

+// It is also assumed that there is no loss of solute through the membrane.

+Cl = 3;

+while 1

+    Clnew = Cl -(0.04-0.02*log(30/Cl))/(Cl^(-1)/50);

+    if Clnew == Cl then

+        break;

+    end

+    Cl = Clnew;

+end

+    printf("\n Cl = %d kg/m^3",Cl);

+printf("\n below this concentration the membrane flux is 0.04 m/h");

+

+//This does not pose a constraint for the single stage as the concentration of solute C2 will be that of the final concentrate, 20 kg/m3.

+//Conservation of solute gives:QoCo = Q2C2

+//A fluid balance gives : Qo = F + Q2

+//Combining these eq and substituting Known values:

+A = (2.438/0.02)/log(30/20);

+printf("\n A = %d m^2",A);

+//The tubular membranes to be used are available as 30 m^2modules.

+printf("\n the no of required modules are %d ",A/30);

+

+//Part(b)

+

+//Conservation of solute gives = QoCo = Q1C1 = Q2C2

+//A fluid balance on stage 2 gives Q1 = Q2 + F2

+//A fluid balance on stage 2 gives Q1 = Q2 +F2

+//Substituting given values in above eqns

+//2.5 = 1.25/C1 + 0.02A1ln(30/C1)

+function[A1]=a(C1)

+    A1 = (2.5-1.25/C1)/(0.02*log(30/C1));

+    funcprot(0);

+endfunction

+function[A2]=b(C1)

+    A2 = (1.25/C1 - 0.0625)/0.00811

+    funcprot(0);

+endfunction

+

+printf("\n The procedure is to use trial and error to estimate the value of C1 that gives the optimum values of A1 and A2");

+printf("\n If C1 = 5kg/m^3 then A1 = %d m^2 and A2 = %d m^2",a(5),b(5));

+printf("\n an arrangement of 3 modules −1 module is required.");

+printf("\n\n\n  If C1 = 4 kg/m^3 then A1 = %dm^2 and A2 = %dm^2",a(4),b(4));

+printf("\n an arrangement of 2 modules −1 module is almost sufficient.");

+printf("\n\n\n   If C1 = 4.5 kg/m^3 then A1 = %dm^2 and A2 = %d m^2",a(4.5),b(4.5));

+printf("\n an arrangement of 2 modules −1 module which meets the requirement");

+printf("\n\n This arrangement requires the minimum number of modules.");

+

+

+

+

+

+

+

+

+

+