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+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-2
+//Mechanical-Physical Seperation Processes
+//given data
+t=[4.4 9.5 16.3 24.6 34.7 46.1 59 73.6 89.4 107.3];
+V=(1/1000)*[0.498 1 1.501 2 2.498 3.002 3.506 4.004 4.502 5.009]
+tbyVd=[];
+for(i=1:10)
+    tbyVd(i)=(t(1,i)/V(1,i))
+end
+tbyV=(1/1000)*tbyVd
+plot2d((V*1000),tbyV);
+xtitle("Determination of the constants","Vx1000","(t/V)x(10^-3)")
+//as seen from the graph,
+B=6400;//intercept in s/m3
+//as it is seen the given graph resembles a straight line we can find Kp/2
+Kpby2=[tbyV(6)-tbyV(5)]*1000/(V(1,6)-V(1,5));//slope of st line in s/m6
+Kp=Kpby2*2;
+A=20*0.837;//total area= (no. of plates)*(area of one plate)
+Aold=0.0439;
+Kpn=Kp*((Aold/A)^2);
+Bn=B*((Aold/A));
+V=3.37;//amt6 of filtrate to be recovered inm3
+t=((Kpn/2)*V*V)+(Bn*V);
+mprintf("the time taken to recover the filtrate= %f s",t)
+//As the slope has been calculated in numerical value deviations have been found.Kpby2=6*10^6 according to the example but the value calculated here is 5.815*10^6