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diff --git a/1328/CH20/EX20.5/20_5.sce b/1328/CH20/EX20.5/20_5.sce
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+printf("\t example 20.5 \n");
+printf("\t approximate values are mentioned in the book \n");
+Nt=25; // number of tubes
+A=50; // total projected area
+Tav=100; // F
+s=28; // assumption spray, lb/(min)*(ft^2)
+Do=0.0833; // ft
+PH=0.1562;
+Y=0.874;
+Z=0.466;
+E=(0.171*(Do*Y*Z)^0.1); // (E/(Do*Y*Z)^0.1)=0.171, from fig 20.10
+printf("\t evaporation percentage is : %.2f \n",E);
+Q=(295*500*(143-130));
+printf("\t heat load is : %.2e Btu/hr \n",Q);
+Q1=(Q*(1-0.12));
+printf("\t sensible heat is : %.2e Btu/hr \n",Q1);
+t2=(90)+(Q1/(28*60*50));
+printf("\t final spray temperature is : %.0f F \n",t2);
+w=(s*60*50);
+printf("\t total spray : %.1e lb/hr \n",w);
+m=(w/(2*4*12));
+printf("\t m is : %.0f lb/(hr)*(ft^2) \n",m);
+mu=1.84; // lb/(ft)*(hr)
+Z=((m^0.3)*Do*Y*Z/(mu*0.125));
+printf("\t Z is : %.2f \n",Z);
+N=3; // assume 3 horizontal rows
+ho=300*(N^0.05); // (ho/(N^0.05))=300, from fig 20.11
+printf("\t ho is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);
+printf("\t tube side coefficient \n");
+printf("\t assuming even number of passes and tube side velocity about 8fps \n");
+at=0.0775; // ft^2
+Gt=(295*500/(at)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.1e lb/(hr)*(ft^2) \n",Gt);
+V=(Gt/(3600*62.5));
+printf("\t velocity is : %.2f fps \n",V);
+hi=2140; // Btu/(hr)*(ft^2)*(F), fig 25
+ID=0.87; // ft
+OD=1; // ft
+hio=((hi)*(ID/OD)); // using eq.6.5
+printf("\t Correct hio to the surface at the OD is : %.2e Btu/(hr)*(ft^2)*(F) \n",hio);
+Uc=((ho*hio)/(ho+hio)); // from eq 6.38
+printf("\t Uc is : %.0f Btu/(hr)*(ft^2)*(F) \n",Uc);
+a=0.2618; // ft^2, table 11
+A1=(2*3*25*12*a);
+printf("\t total surface is : %.0f ft^2 \n",A1);
+T1=143; // inlet hot fluid,F
+T2=130; // outlet hot fluid,F
+t1=90; // inlet cold fluid,F
+t2=110; // outlet cold fluid,F
+delt1=T2-t1; //F
+delt2=T1-t2; // F
+printf("\t delt1 is : %.0f F \n",delt1);
+printf("\t delt2 is : %.0f F \n",delt2);
+LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
+printf("\t LMTD is :%.1f F \n",LMTD); // calculation mistake in book
+R=0.65;
+S=0.377;
+FT=0.97; // fig 18
+delt=(FT*LMTD);
+printf("\t delt is : %.1f F \n",delt);
+UD=(Q/(A1*(delt)));
+printf("\t UD is : %.0f Btu/(hr)*(ft^2)*(F) \n",UD);
+Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu
+printf("\t actual Rd is : %.4f (hr)*(ft^2)*(F)/Btu \n",Rd);
+printf("\t The assumption of three horizontal rows is satisfactory, since a dirt factor of 0.004 was required \n");
+// end
+
+
+