From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 3516/CH12/EX12.1/Ex12_1.sce | 113 ++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 113 insertions(+) create mode 100644 3516/CH12/EX12.1/Ex12_1.sce (limited to '3516/CH12/EX12.1') diff --git a/3516/CH12/EX12.1/Ex12_1.sce b/3516/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..6a52aecc6 --- /dev/null +++ b/3516/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,113 @@ +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 -- cgit