initial commit / add all books

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diff --git a/1328/CH20/EX20.3/20_3.sce b/1328/CH20/EX20.3/20_3.sce
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+printf("\t example 20.3 \n");

+printf("\t approximate values are mentioned in the book \n");

+T1=675; // inlet hot fluid,F

+T2=200; // outlet hot fluid,F

+t1=120; // inlet cold fluid,F

+t2=140; // outlet cold fluid,F

+W=33100; // lb/hr

+w=510000; // lb/hr

+printf("\t 1.for heat balance \n");

+printf("\t for oil \n");

+c=0.64; // Btu/(lb)*(F)

+Q=((W)*(c)*(T1-T2)); // Btu/hr

+printf("\t total heat required for oil is : %.2e Btu/hr \n",Q);

+printf("\t for water \n");

+c=1; // Btu/(lb)*(F)

+Q=((w)*(c)*(t2-t1)); // Btu/hr

+printf("\t total heat required for water is : %.2e Btu/hr \n",Q);

+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=230;

+printf("\t LMTD is :%.0f F \n",LMTD);

+Tc=((T2)+(T1))/(2); // caloric temperature of hot fluid,F

+printf("\t caloric temperature of hot fluid is : %.1f F \n",Tc);

+tc=((t1)+(t2))/(2); // caloric temperature of cold fluid,F

+printf("\t caloric temperature of cold fluid is : %.0f \n",tc);

+printf("\t hot fluid:inner tube side, oil \n");

+at=0.0458; // flow area, ft^2, table 11

+printf("\t flow area is : %.4f ft^2 \n",at);

+Gt=(W/(at)); // mass velocity,lb/(hr)*(ft^2)

+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);

+mu2=5.56; // at 400F,lb/(ft)*(hr)

+D=0.242; // ft, table 11

+Ret=((D)*(Gt)/mu2); // reynolds number

+printf("\t reynolds number is : %.2e \n",Ret);

+jH=100; // from fig.24

+Z=0.245; // Z=(k(c*mu/k)^(1/3)), Btu/(hr)*(ft)*(F/ft), fig 16

+hi=((jH)*(Z/D)); //Hi=(hi/phyp),using eq.6.15,Btu/(hr)*(ft^2)*(F)

+printf("\t hi is : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);

+ID=2.9; // ft

+OD=3.5; // ft

+hio=((hi)*(ID/OD)); // using eq.6.5

+printf("\t Correct hio to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio);

+ho=150; // Btu/(hr)*(ft^2)

+tw=(tc)+(((hio)/(hio+ho))*(Tc-tc)); // from eq.5.31

+printf("\t tw is : %.0f F \n",tw);

+tf=(tw+tc)/2;

+printf("\t tf is : %.0f F \n",tf);

+delt=110; // F

+d0=3.5; // in, fig 10.4

+Uc=((ho*hio)/(ho+hio)); // from eq 6.38

+printf("\t Uc is : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc);

+Rd=0.01;

+hd=(1/Rd);

+printf("\t hd is : %.0f \n",hd);

+UD=((Uc*hd)/(Uc+hd));

+printf("\t UD is : %.0f Btu/(hr)*(ft^2)*(F) \n",UD);

+A=(Q/(UD*(LMTD)));

+printf("\t Area is : %.0f ft^2 \n",A);

+a=0.917; // ft^2/ft, table 11

+L=(A/(a*24));

+printf("\t pipe length : %.0f \n",L);

+// end