From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 1328/CH20/EX20.3/20_3.sce | 64 +++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 64 insertions(+) create mode 100644 1328/CH20/EX20.3/20_3.sce (limited to '1328/CH20/EX20.3/20_3.sce') diff --git a/1328/CH20/EX20.3/20_3.sce b/1328/CH20/EX20.3/20_3.sce new file mode 100644 index 000000000..36aac87e9 --- /dev/null +++ b/1328/CH20/EX20.3/20_3.sce @@ -0,0 +1,64 @@ +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 -- cgit