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diff --git a/1328/CH12/EX12.1/12_1.sce b/1328/CH12/EX12.1/12_1.sce
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+printf("\t example 12.1 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=244; // inlet hot fluid,F
+T2=244; // outlet hot fluid,F
+t1=85; // inlet cold fluid,F
+t2=120; // outlet cold fluid,F
+W=60000; // lb/hr
+w=488000; // lb/hr
+printf("\t 1.for heat balance \n");
+printf("\t for propanol \n");
+l=285; // Btu/(lb)
+Q=((W)*(l)); // Btu/hr
+printf("\t total heat required for propanol 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=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
+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 : %.0f F \n",Tc);
+tc=((t1)+(t2))/(2); // caloric temperature of cold fluid,F
+printf("\t caloric temperature of cold fluid is : %.1f F \n",tc);
+UD1=100; // assume, from table 8
+A1=((Q)/((UD1)*(LMTD)));
+printf("\t A1 is : %.0f ft^2 \n",A1);
+a1=0.1963; // ft^2/lin ft
+N1=(A1/(8*a1));
+printf("\t number of tubes are : %.0f \n",N1);
+N2=766; // assuming 4 tube passes, from table 9
+A2=(N2*8*a1); // ft^2
+printf("\t total surface area is : %.0f ft^2 \n",A2);
+UD=((Q)/((A2)*(LMTD)));
+printf("\t correct design overall coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",UD);
+printf("\t hot fluid:shell side,propanol \n");
+ID=31; // in
+C=0.1875; // clearance
+B=31; // baffle spacing,in
+PT=0.937;
+L=8; // ft
+as=((ID*C*B)/(144*PT)); // flow area,from eq 7.1,ft^2
+printf("\t flow area is : %.2f ft^2 \n",as);
+Gs=(W/as); // mass velocity,from eq 7.2,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs);
+G1=(W/(L*N2^(2/3))); // from eq.12.43
+printf("\t G1 is : %.1f lb/(hr)*(lin ft) \n",G1);
+printf("\t cold fluid:inner tube side,water \n");
+Nt=766;
+n=4; // number of passes
+L=8; //ft
+at1=0.302; // flow area, in^2
+at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48
+printf("\t flow area is : %.3f 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);
+V=(Gt/(3600*62.5));
+printf("\t V is : %.2f fps \n",V);
+mu2=1.74; // at 102.5F,lb/(ft)*(hr)
+D=0.0517; // ft
+Ret=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+hi=1300; //Btu/(hr)*(ft^2)*(F)
+printf("\t hi is : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);
+ID=0.62; // ft
+OD=0.75; //ft
+hio=((hi)*(ID/OD)); // using eq.6.5
+printf("\t Correct hi0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",hio); // calculation mistake
+ho=200; // assumption
+tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); // from eq.5.31
+printf("\t tw is : %.0f F \n",tw);
+tf=(Tc+tw)/(2); // from eq 12.19
+printf("\t tf is : %.1f F \n",tf);
+kf=0.094; // Btu/(hr)*(ft^2)*(F/ft), from table 4
+sf=0.8; // from table 6
+muf=0.62; // cp, from fig 14
+ho=172; // Btu/(hr)*(ft^2)*(F), from fig 12.9
+printf("\t Correct ho to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);
+printf("\t Based on h=172 instead of the assumed 200 a new value of tw,and tf could be obtained to give a more exact value of h based on fluid properties at a value of tf more nearly correct \n");
+printf("\t pressure drop  for annulus \n");
+mu1=0.0242; // lb/(ft)*(hr), fir 15
+De=0.0458; // fig 28
+Res=((De)*(Gs)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Res);
+f=0.00141; // friction factor for reynolds number 84600, using fig.29
+s=0.00381; // for reynolds number 84600,using fig.6
+Ds=31/12; // ft
+phys=1;
+N=(3); // number of crosses,using eq.7.43
+printf("\t number of crosses are : %.0f \n",N);
+delPs=((f*(Gs^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys)))/(2); // using eq.12.47,psi
+printf("\t delPs is : %.1f psi \n",delPs);
+printf("\t allowable delPa is 2 psi \n");
+printf("\t pressure drop  for inner pipe \n");
+f=0.00019; // friction factor for reynolds number 36200, using fig.26
+s=1;
+phyt=1;
+delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(D)*(s)*(phyt))); // using eq.7.45,psi
+printf("\t delPt is : %.1f psi \n",delPt);
+X1=0.2; // X1=((V^2)/(2*g)),using fig.27
+delPr=((4*n*X1)/(s)); // using eq.7.46,psi
+printf("\t delPr is : %.1f psi \n",delPr);
+delPT=delPt+delPr; // using eq.7.47,psi
+printf("\t delPT is : %.1f psi \n",delPT);
+printf("\t allowable delPT is 10 psi \n");
+Uc=((hio)*(ho)/(hio+ho)); // clean overall coefficient,Btu/(hr)*(ft^2)*(F)
+printf("\t clean overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc);
+Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu
+printf("\t actual Rd is : %.4f (hr)*(ft^2)*(F)/Btu \n",Rd);
+// end