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+
+
+//example 10.3
+clc; funcprot(0);
+//exapple 10.3 
+// Initialization of Variable
+rho1=2600;//density lighter
+rho2=5100;//density heavier
+pd1=0.000015:0.000010:0.000095;//particle diameter lighter
+pd2=0.000025:0.00001:0.000095;//particle diameter heavier
+wp1=[0 22 35 47 59 68 75 81 100];//weight distribution lighter
+wp2=[0 21 33.5 48 57.5 67 75 100];//weight distribution heavier
+rho=998.6;//density water
+mu=1.03/1000;//viscosity water
+g=9.81;
+u=0.004;//velocity of water
+d=95/1000000;//paeticle diameter maximum
+//calculation
+//part 1
+Re=d*u*rho/mu;
+d1=sqrt(18*mu*u/g/(rho1-rho));
+d2=sqrt(18*mu*u/g/(rho2-rho));
+function[a]=inter(d,f,g,b);//interpolation linear
+    for i=1:b
+        if d<=f(i+1)& d>f(i)  then
+            break
+        else 
+            continue
+        end
+        break
+    end
+    a=(d-f(i))/(f(i+1)-f(i))*(g(i+1)-g(i))+g(i);
+endfunction
+[a]=inter(d1,pd1,wp1,9);
+[b]=inter(d2,pd2,wp2,8);
+v2=1/(1+5)*100-b/100*1/(1+5)*100;
+v1=5/(1+5)*100-a/100*5/(1+5)*100;
+pl2=(v2)/(v2+v1);
+disp(pl2, "The fraction of heavy ore remained in bottom");
+ //part 2
+ rho=1500;
+ mu=6.25/10000;
+ a=log10(2*d^3*rho*g*(rho1-rho)*3*mu^2);//log10(Re^2(R/rho/mu^2))
+  //using value from chart(graph)
+Re=10^0.2136;
+u=Re*mu/rho/d;
+d2=sqrt(18*mu*u/g/(rho1-rho));
+[b]=inter(d2,pd2,wp2,8);
+disp(100-b+3.5,"The percentage of heavy ore left in this case");
+//part 3
+a=0.75//% of heavy ore in overhead product
+s=100*5/6/(100*5/6+0.75*100/6);
+disp(s,"the fraction of light ore in overhead product:");
+//part 4
+da=pd2(1);
+db=pd1(9);
+rho=(da^2*rho2-db^2*rho1)/(-db^2+da^2);
+  disp(rho,"The minimum density required to seperate 2 ores in kg/m^3:")