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/CH7/EX7.8/7_8.sce | 98 ++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 98 insertions(+) create mode 100644 1328/CH7/EX7.8/7_8.sce (limited to '1328/CH7/EX7.8') diff --git a/1328/CH7/EX7.8/7_8.sce b/1328/CH7/EX7.8/7_8.sce new file mode 100644 index 000000000..d1475bda9 --- /dev/null +++ b/1328/CH7/EX7.8/7_8.sce @@ -0,0 +1,98 @@ +printf("\t example 7.8 \n"); +printf("\t approximate values are mentioned in the book \n"); +T1=228; // inlet hot fluid,F +T2=228; // outlet hot fluid,F +t1=100; // inlet cold fluid,F +t2=122; // outlet cold fluid,F +W=200000; // lb/hr +w=3950; // lb/hr +printf("\t 1.for heat balance \n"); +printf("\t for solution \n"); +c=(0.2*0.30)+(0.8*1); // bcoz of 20 percent solution,Btu/(lb)*(F) +Q1=((W)*(c)*(t2-t1)); // Btu/hr +printf("\t total heat required for solution is : %.2e Btu/hr \n",Q1); +printf("\t for steam \n"); +l=960.1; // latent heat of condensation,Btu/(lb) +Q=((w)*(l)); // Btu/hr +printf("\t total heat required for steam 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 :%.1f F \n",LMTD); +R=((T1-T2)/(t2-t1)); +printf("\t R is : %.0f \n",R); +delt=(LMTD); // when R=0,F +printf("\t delt is : %.1f F \n",delt); +printf("\t The steam coefficient will be very great compared with that for the sugar solution, and the tube wall will be considerably nearer 228°F than the caloric temperature of the fluid. Obtain Fc from U1 and U0 Failure to correct for wall effects, however, will keep the heater calculation on the safe side.\n"); +ta=111; //F +Ta=228; //f +printf("\t hot fluid:tube side,steam \n"); +Nt=76; +n=2; // number of passes +L=16; //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 : %.4f ft^2 \n",at); +Gt=(w/(at)); // mass velocity,lb/(hr)*(ft^2) +printf("\t mass velocity is : %.2elb/(hr)*(ft^2) \n",Gt); +mu2=0.0128*2.42; // at 228F,lb/(ft)*(hr) +D=(0.62/12); // from table 10,ft +Ret=((D)*(Gt)/mu2); // reynolds number +printf("\t reynolds number is : %.2e \n",Ret); +hio=1500; // for condensation of steam +printf("\t Correct hi0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",hio); +printf("\t cold fluid:shell side,sugar solution \n"); +ID=12; // in +d=0.75/12; // diameter of tube,ft +Nt=76; // number of tubes +as=((3.14*(12^2)/4)-(76*3.14*(0.75^2)/4))/144; // flow area,ft^2 +printf("\t flow area is : %.2f ft^2 \n",as); +Gs=(W/as); // mass velocity,lb/(hr)*(ft^2) +printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs); +mu1=1.30*2.42; // at 111F,lb/(ft)*(hr), from fig.14 +De=((4*as)/(Nt*3.14*d)); // from eq.6.3,ft +printf("\t De is : %.3f ft \n",De); +Res=((De)*(Gs)/mu1); // reynolds number +printf("\t reynolds number is : %.2e \n",Res); +jH=61.5; // from fig.24, tube side data +c=0.86; // Btu/(lb)*(F),at 111F,from fig.4 +k=0.333; // Btu/(hr)*(ft^2)*(F/ft) +Pr=((c)*(mu1)/k)^(1/3); // prandelt number raised to power 1/3 +printf("\t Pr is : %.0f \n",Pr); +Ho=((jH)*(k/De)*(Pr)); // H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft^2)*(F) +printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",Ho); +muw=0.51*2.42; // at 210F,lb/(ft)*(hr), from fig.14 +phys=(mu1/muw)^0.14; +printf("\t phys is : %.2f \n",phys); // from fig.24 +ho=(Ho)*(phys); // from eq.6.36 +printf("\t Correct h0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho); +tw=(ta)+(((hio)/(hio+Ho))*(Ta-ta)); // from eq.5.31 +printf("\t tw is : %.0f F \n",tw); +Uc=((hio)*(ho)/(hio+ho)); // clean overall coefficient,Btu/(hr)*(ft^2)*(F) +printf("\t clean overall coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",Uc); +A2=0.1963; // actual surface supplied for each tube,ft^2,from table 10 +A=(Nt*L*A2); // ft^2 +printf("\t total surface area is : %.0f ft^2 \n",A); +UD=((Q)/((A)*(LMTD))); +printf("\t actual design overall coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",UD); +Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu +printf("\t actual Rd is : %.4f (hr)*(ft^2)*(F)/Btu \n",Rd); +printf("\t pressure drop for inner pipe \n"); +f=0.000155; // friction factor for reynolds number 82500, using fig.26 +s=0.0008; +phyt=1; +D=0.0517; +delPt=((f*(Gt^2)*(L)*(2))/(5.22*(10^10)*(D)*(s)*(phyt)))/2; // using eq.7.45,psi +printf("\t delPt is : %.1f psi \n",delPt); +printf("\t pressure drop for annulus \n"); +De1=((4*as)/((Nt*3.14*d)+(3.14*1))); // from eq.6.4,ft +printf("\t De1 is : %.3f ft \n",De1); +Res1=(De1*Gs/mu1); // from eq 7.3 +printf("\t Res1 is : %.2e \n",Res1); +f=0.00025; // friction factor, using fig.26 +s=1.08; // for reynolds number 25300,using fig.6 +delPs=((f*(Gs^2)*(L)*(1))/(5.22*(10^10)*(De1)*(s)*(phys))); // using eq.7.44,psi +printf("\t delPs is : %.2f psi \n",delPs); +//end -- cgit