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')

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