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+clear ;
+clc;
+// Example 28.3
+printf('Example 28.3\n\n');
+//page no. 875
+// Solution fig. 28.3
+
+// Given 
+//Input analysis 
+soln1 = 600 ; // Mass flow rate of entering solution 1 -[lb/hr]
+c1_NaOH = 10/100 ;// Fraction of NaOH in entering solution 1
+T1 = 200 ;// Temperature at entry 
+soln2 = 400 ;// Mass flow rate of another solution 2 entering -[lb/hr]
+c2_NaOH = 50/100 ;// Fraction of NaOH in another entering solution 2
+
+// Additional data is obtained from steam table and NaOH-H2O enthalpy-concentration chart in Appendix I at given reference temperature (del_H = 0 , 32 degree F for pure water)
+F = soln1 + soln2; // Mass flow rate of final solution - [lb/hr]
+
+ // Material balance to get composition of final solution 
+F_NaOH =  c1_NaOH * soln1 + c2_NaOH * soln2 ;// Mass of NaOH in final solution-[lb]
+F_H2O = F - F_NaOH ;// Mass of H2O in final solution-[lb]
+
+// Enthalpy data from H-x chart , according to book it is as follows
+H_soln1 = 152 ;// Specific enthalpy change for solution 1-[Btu/lb]
+H_soln2 = 290 ;// Specific enthalpy change for solution 2-[Btu/lb]
+
+// Energy balance
+H_F = (soln1*H_soln1 + soln2*H_soln2)/F ;// Specific enthalpy change for final solution -[Btu/lb]
+
+//(a)
+printf(' (a) The final temperature of the exit solution from figure E28.3 using the obtained condition of final solution is 232 degree F \n');
+
+//(b)
+cF = F_NaOH*100/F; // Concentration of final solution -[wt % NaOH ]
+printf('  (b) The concentration of final solution is %.0f wt.%% NaOH . \n',cF);
+
+//(c)
+// For fraction of H2O vapour . By interpolation , draw the tie line through the point x = .26 .H = 270 (make it parallel to 220 and 250 degree F line ). The final temperature of the exit solution from figure E28.3 using the obtained condition of final solution is 232 degree ; the enthalpy of the liquid at the bubble point at this temperature is about 175 Btu/lb . The enthalpy of saturated water vapour fro the steam table at 232 degree F is 1158 Btu/lb . Let x be the water vapour evaporated , therefore
+x = (F*H_F - F*175)/(1158 - 175) ;// H2O evaporated per hour -[lb]
+
+printf('  (c) H2O evaporated per hour is %.1f lb . \n',x);
+\ No newline at end of file