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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /389/CH13/EX13.2/Example13_2.sce | |
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diff --git a/389/CH13/EX13.2/Example13_2.sce b/389/CH13/EX13.2/Example13_2.sce new file mode 100755 index 000000000..c2bffa315 --- /dev/null +++ b/389/CH13/EX13.2/Example13_2.sce @@ -0,0 +1,63 @@ +clear;
+clc;
+
+// Illustration 13.2
+// Page: 749
+
+printf('Illustration 13.2 - Page: 749\n\n');
+
+// Solution
+
+//***Data***//
+// Eqb=[x(Wt fraction NaOH in clear solution) N(kg CaCO3/kg soln in settled sludge) y*(wt fraction NaOH in soln of settled sludge)]
+// a=H2O b=CaCO3 c=NaOH
+Eqb = [0.090 0.495 0.0917;0.0700 0.525 0.0762;0.0473 0.568 0.0608;0.0330 0.600 0.0452;0.0208 0.620 0.0295;0.01187 0.650 0.0204;0.00710 0.659 0.01435;0.00450 0.666 0.01015];
+deff('[y]=f80(x)','y=0');
+x = 0:0.01:0.12;
+Mass_c = 0.1;// [kg]
+Mass_b = 0.125;// [kg]
+Mass_a = 0.9;// [kg]
+//**************//
+
+scf(42);
+plot(x,f80,Eqb(:,3),Eqb(:,2));
+xgrid();
+xlabel("x,y Wt. fraction of NaOH in loquid");
+ylabel("N kg CaCO3 / kg solution");
+legend("N Vs x","N Vs Y");
+title("Equilibrium Plot")
+// Basis: 1 kg soln in original mixture.
+// As in Fig. 13.27 (Pg 750)
+// The original mixture corresponds to M1:
+NM1 = 0.125;// [kg CaCO3/kg soln]
+yM1 = 0.1;// [kg NaOH/kg solution]
+// The tie line through M1 is drawn. At point E1 representing the settled sludge:
+N1 = 0.47;// [kg CaCO3/kg soln]
+y1 = 0.100;// [kg NaOH/kg solution]
+E1 = Mass_b/N1;// [kg soln. in sludge]
+Ro = 1-E1;// [kg clear soln drawn]
+
+// Stage 2:
+xo = 0;// [kg NaOH/kg soln]
+// By Eqn. 13.11:
+M2 = E1+Ro;// [kg liquid]
+// By Eqn. 13.12:
+NM2 = Mass_b/(E1+Ro);// [kg CaCO3/kg soln]
+// M2 is located on line RoE1. At this value of N, and the tie line through M2 is drawn. At E2:
+N2 = 0.62;// [kg CaCO3/kg soln]
+y2 = 0.035;// [kg NaOH/kg solution]
+E2 = Mass_b/N2;// [kg soln. in sludge]
+Ro = 1-E2;// [kg clear soln drawn]
+
+// Stage 3:
+xo = 0;// [kg NaOH/kg soln]
+// By Eqn. 13.11:
+M3 = E2+Ro;// [kg liquid]
+// By Eqn. 13.12:
+NM3 = Mass_b/M3;// [kg CaCO3/kg soln]
+// Tie line E3R3 is located through M3.At E3:
+N3 = 0.662;// [kg CaCO3/kg soln]
+y3 = 0.012;// [kg NaOH/kg solution]
+// By Eqn. 13.8:
+E3 = Mass_b/N3;// [kg soln. in sludge]
+printf("The fraction of original NaOH in the slurry: %f \n",E3*y3/Mass_c);
\ No newline at end of file |