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diff --git a/542/CH5/EX5.1/Example_5_1.sci b/542/CH5/EX5.1/Example_5_1.sci
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+clear;
+clc;
+printf("\n Example 5.1");
+//Basis 1 kg of solids
+feed_rate_solid = 1.33;  //Mass rate of feed of solids in kg/sec
+U = 1.5;                 //Mass rate of solids in the underflow in kg/sec
+Y = [5.0 ; 4.2; 3.7 ; 3.1; 2.5];
+printf("\n concentration(Y) (kg water/kg solids):\n");
+printf("%.1f\n",Y);
+printf("\n water to overflow (Y-U) (kg water/kg solids):\n");
+O = Y - 1.5;//Amount of water to overflow in kg water/kg solids
+printf("\n %.1f\n",O);
+Uc = [2.00*10^(-4);1.20*10^(-4);0.94*10^(-4);0.70*10^(-4);0.50*10^(-4)];
+printf("\n sedimentation rate uc (m/sec):\n");
+printf("%f  \n",Uc);
+X =[1.75*10^4;2.25*10^4;2.34*10^4;2.29*10^4;2.00*10^4];//X = (Y-U)/Uc
+printf(" \n\n(Y-U)/Uc (s/m):\n");
+printf("\n %d\n",X);
+z = max(X); //prints the maximum value of X
+printf("\nMaximum value of (Y-U)/Uc = %ds/m",z);
+
+//Calculating the require darea of the thickener
+A = z*1.33/1000; //1.33 is the mass feed rate of solids in kg/sec
+//1000 is the density of water in kg/m^3
+printf("\n The required area of the thickener is :%.2fm^2\n",A);