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+//Transport Processes and Seperation Process Principles

+//Chapter 8

+//Example 8.4-1

+//Evaporation

+//given data

+Tf=37.8+273.2;//temp in K

+xf=1/100;//wt % of salt in the feed

+xl=1.5/100;//wt % of salt in the extract

+//2 eqns to be solved

+F=9072;//feel flow rate

+xv=0;

+A=[1 1;xl xv];

+B=[F;(F*xf)];

+LV=inv(A)*B;

+L=LV(1,1);//liquid flow rate

+V=LV(2,1);//vapour flow rate

+Cpf=4.14;//heat capacity of the feed

+T1=273.2+100;//bp of water at 101.32 Kpa(as given)

+hf=Cpf*(Tf-T1);//enthalpy of feed

+hl=0;//enthalpy of extract as it is at datum

+hv=2257;//enthalpy of vapour

+lemda=2230;//enthalpy of steam;

+S=(L*hl+V*hv-F*hf)/lemda;//flow rate of steam

+q=S*lemda*(1000/3600);//heat transfered through the heating surface area

+U=1704;//heat transfer coefficient 

+Ts=383.2;

+A=q/(U*(Ts-T1));//heat transfer area

+mprintf("i) extract flow rate= %f kg/h",L)

+mprintf("ii) vapour flow rate= %f kg/h",V)

+mprintf("iii) heat transfer area= %f m2",A)

+//end