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diff --git a/1073/CH5/EX5.24/5_24.sce b/1073/CH5/EX5.24/5_24.sce
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+
+clc;
+clear;
+//Example 5.24
+ho=200;     //[W/sq m.K]
+hi=1500;    //[W/sq m.K]
+Cpw=4.2;     //Sp heat of Water in [kJ/(kg.K)]
+Cpo=2.1;     //Sp heat of Oil in [kJ/(kg.K)]
+E=0.8;      //Effectiveness
+k=46;       //[W/m.K]
+m_dot=0.167;    //[kg/s]
+
+mCp_oil=2*m_dot*Cpo*1000      //For oil [W/K]
+//mCp_oil is wrongly calculated as 710.4
+mCp_water=m_dot*Cpw*1000      //For water [W/K]
+//mCp_oil is wrongly calculated as 710.4
+//NOTE:The above two values are wrongly calculated in book as 710.4
+//so we take here:
+mCp_small=710.4     //[W/K]
+//Since both mCp_water and mCp_oil are equal ,therefore:
+C=1;
+
+deff('[x]=f(ntu)','x=E-(ntu/(1+ntu))');
+ntu=fsolve(1,f)
+id=20;      //Internal diameter in [mm]
+od=25;      //External diameter in [mm]
+hio=hi*id/od        //[W/sq m.K]
+Dw=(od-id)/log(od/id)   //[mm]
+Dw=Dw/1000              //[m]
+x=(od-id)/2             //[mm]
+x=x/1000            //[m]
+Do=0.025            //External dia in [m]
+L=2.5;              //Length of tube in [m]
+Uo=1/(1/ho+1/hio+(x/k)*(Dw/Do))     //[W/sq m.K]
+A=ntu*mCp_small/Uo   //Heat transfer area in [sq m]
+n=A/(%pi*Do*L)      //No of tubes
+printf("\nNo. of tubes required = %d",round(n+1));