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diff --git a/581/CH3/EX3.6/Example3_6.sci b/581/CH3/EX3.6/Example3_6.sci
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+
+clear;
+clc;
+
+printf("\t Example 3.6\n");
+
+T1=313; //entering temperature of cold water, K
+T2=423; //Entering temperature of hot water, K
+T3=363; //Exit temperature of hot water, K
+Cc=20000; //heat capacity of cold water, W/K
+Ch=10000; //heat capacity of hot water, W/K
+U=500; //overall heat transfer coefficient, w/(m^2*K)
+T4=T1+(Ch/Cc)*(T2-T3);
+e=(T2-T3)/(T2-T1);
+
+NTU=1.15;
+A1=Ch*(NTU)/U;  // since NTU=1.15=U*A/Ch, A can be found by using this formula
+printf("\t area is : %.2f m^2\n",A1);
+
+//another way to calculate the area is by using log mean diameter method
+LMD=(T2-T1-T3+T4)/log(110/20);
+A2=Ch*(T2-T3)/(U*LMD);
+printf("\t area is : %.2f m^2\n",A2);
+printf("\t there is difference of 1 percent in answers which reflects graph reading inaccuracy.");
+// we can see that area calulated is same in above 2 methods.
+//End
+\ No newline at end of file