1 files changed, 98 insertions, 0 deletions
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