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+
+clear;
+clc;
+printf("\t Example 7.12\n");
+
+                //FeSO4.7H2O
+C=1000;                      //crystal formed in kg
+hf=26.002;                   //enthalpy of the feed at 80 degree in cal/g
+hl=-1.33;                   //enthalpy of the saturated sol at 30 degree in cal/g
+hc=-50.56;                 //enthalpy of crystal
+xf=40/(100+40);
+xm=30/(100+30);
+xc=151.84/277.85;        //151.84 is the weight of FeSO4
+        //component balance
+//    F*xf = M*xm + C*xc    ------eqn 1st
+//    F = M + 10000 + V     ------eqn 2nd
+//    F*Hf = V*Hv + M*Hm +C*Hc-----eqn 3rd
+Hf=26.002;                //enthalpy of the feed at 80 degree in cal/g
+Hv=612;                  //
+Hm=-1.33;               //enthalpy of the saturated sol at 30 degree in cal/g
+Hc=-50.56;             //enthalpy of crystal leaving the crystalliser
+
+//solving these we gt 
+a=[1 -1 -1;.286 -.231 0;26.002 1.33 -612]
+b=[10000;5470;-505600]
+x=inv(a)*b;                    //solving out the values using matrices 
+t1=x(1);                        //3 solution of the eqn
+t2=x(2);
+t3=x(3);
+printf("\n the feed rate F= : %f kg/hr \n value of M= : %f kg/hr\n value of V=: %f kg/hr",t1,t2,t3);
+//end
+\ No newline at end of file