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+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-1
+//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(6)-V(5));//slope of st line in s/m6
+Kp=Kpby2*2;
+mu=8.937e-4;
+A=0.0439;
+delP=338*1000;
+//Now Kp=mu*alpha*cs/(A*A*delP)
+Cs=23.47;
+alpha=Kp*(A*A*delP)/(mu*Cs);
+//also B=(mu*Rm)/(A*A*(-delP))
+Rm=B*(A*(delP))/mu
+mprintf("alpha= %f m/kg",alpha)
+mprintf(" Rm= %f m-1",Rm)
+//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