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clear;
clc;
printf("\t Example 7.13\n");
C=800; //crystal formed in kg/hr
t2=49; //temp. of the entering fed
t1=27; //temp. of the product
t3=21; //temp. of the leaving cooling water
t4=15; //temp. of the enetring cooling water
U=175; //overall heat transfer coefficient
F=140*151.85/277.85; //feed concentration
xf=F/240; //concentration in feed solution
P=74*151.85/277.85; //product concentration
xm=P/174; //concentration of FeSO4 in product solution
xc=151.85/277.85; //
//mass balance F = M+C ----eqn 1st
//sloute balance F*xf = M*xm + C*xc ----eqn 2nd
//solving these we get
F=800*.3141/0.0866; //feed conc.
M=F-C; //product concentration
//making energy balance
//heat to be removed by cooling water =heat to be removed from solution + heat of crystallization
cp=.7; //specific heat capacity
dt=(t2-t1); //change in temp.
dh=15.8; //heat of crystallization
Q=F*cp*dt+dh*C; //heat to be removed by cooling water
cp=1; //specific heat capacity of water
dt=(t3-t4); //change in temp.
mw=Q/(cp*dt); //cooling water needed
printf("\n cooling water requiement is :%f kg/hr",mw);
//Q=U*A*(dtlm)
dtlm=((t2-t3)-(t1-t4))/(log((t2-t3)/(t1-t4)));//log mean temp. difference
A=Q/(U*dtlm); //area of the crystallizer section
l=A/1.3;
printf("\n length of crystallliser sections needed is :%f m",l);
//end
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