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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+//clc();
+
+// Example 6.8
+// Page: 119
+printf("Example-6.8 Page no.-119\n\n");
+
+//***Data***//
+x_sulph = 0.6;
+x_water = 0.4;
+M_i = 18;//[lbm/lbmol]
+Temp = 200;//[F]
+// From Equation 6.11 as given in the book, we have
+// dQ/dm_in = h_i-h_in
+// where h_i is partial molal enthalpy which is taken from the example 6.7 and h_in is the pure species molar enthalpy which is read from the figure 6.8.
+// So at 200F we have
+h_i = 25;//[Btu/lbm]
+h_in = 168;//[Btu/lbm]
+// hence
+dQ_by_dm_in = h_i-h_in;;//[Btu/lbm]
+// Now
+dQ_by_dn_in = M_i*dQ_by_dm_in;//[Btu/lbmol]
+printf("The amount of heat removed to keep the temperature constant is %f Btu/lbm of water added",dQ_by_dm_in);
+// The negative sign shows that this mixing is exothermic; we must remove 143 Btu/lbm of water added. \ No newline at end of file