summaryrefslogtreecommitdiff
path: root/1328/CH10
diff options
context:
space:
mode:
Diffstat (limited to '1328/CH10')
-rw-r--r--1328/CH10/EX10.1/10_1.sce73
-rw-r--r--1328/CH10/EX10.2/10_2.sce88
-rw-r--r--1328/CH10/EX10.3/10_3.sce102
-rw-r--r--1328/CH10/EX10.4/10_4.sce40
4 files changed, 303 insertions, 0 deletions
diff --git a/1328/CH10/EX10.1/10_1.sce b/1328/CH10/EX10.1/10_1.sce
new file mode 100644
index 000000000..0fe8afdd0
--- /dev/null
+++ b/1328/CH10/EX10.1/10_1.sce
@@ -0,0 +1,73 @@
+printf("\t example 10.1 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=250; // outlet hot fluid,F
+t1=95; // inlet cold fluid,F
+t2=145; // outlet cold fluid,F
+W=16000; // lb/hr
+w=410; // lb/hr
+printf("\t 1.for heat balance \n");
+printf("\t for crude \n");
+c=0.485; // Btu/(lb)*(F)
+Q=((W)*(c)*(t2-t1)); // Btu/hr
+printf("\t total heat required for crude is : %.2e Btu/hr \n",Q);
+printf("\t for steam \n");
+l=945.5; // 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 :%.0f F \n",LMTD);
+printf("\t On the assumption that the fluids are mixed between passes, each pass must be solved independently. Since only two passes are present in this exchanger, it is simply a matter of assuming the temperature at the end of the first pass. More than half the heat load must be transferred in the first pass; therefore assume ti at the end of the first pass is 125°F \n");
+ti=125; // F
+tc=((t1)+(ti))/2; // caloric temperature of cold fluid,F
+printf("\t caloric temperature of cold fluid is : %.0f F \n",tc);
+printf("\t hot fluid:shell side,steam \n");
+ho=(1500); // condensation of steam Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);
+printf("\t cold fluid:inner tube side,crude \n");
+Nt=86;
+n=2; // number of passes
+L=12; //ft
+at1=0.594; // flow area, in^2,from table 10
+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/(.177)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+mu2=2.95*2.42; // at 145F,lb/(ft)*(hr)
+D=(0.87/12); // ft
+Ret1=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.0f \n",Ret1);
+mu3=4.8*2.42; // at 110F,lb/(ft)*(hr)
+D=(0.87/12); // ft
+Ret2=((D)*(Gt)/mu3); // reynolds number
+printf("\t reynolds number is : %.0f \n",Ret2);
+c=0.485; // Btu/(lb)*(F),at 120F,from fig.2
+k=0.0775; // Btu/(hr)*(ft^2)*(F/ft), from table 4
+Pr=((c)*(mu3)/k); // prandelt number
+printf("\t prandelt number is : %.1f \n",Pr);
+Hi=((1.86)*(k/D)*((Ret2*(D/L)*Pr)^(1/3))); // using eq.6.1,Btu/(hr)*(ft^2)*(F)
+printf("\t Hi is : %.1f Btu/(hr)*(ft^2)*(F) \n",Hi);
+muw=1.2*2.42; // lb/(ft)*(hr),at 249F from fig.14
+phyt=(mu3/muw)^0.14;
+printf("\t phyt is : %.1f \n",phyt); // from fig.24
+hi=(Hi)*(phyt); // from eq.6.37
+printf("\t Correct hi to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);
+tp=(tc)+(((ho)/(hi+ho))*(T1-tc)); // from eq.5.31
+printf("\t tp is : %.0f F \n",tp);
+delt=tp-tc; //F
+printf("\t delt is : %.0f F \n",delt);
+Ai1=0.228 // internal surface per foot of length,ft
+Ai=(Nt*L*Ai1/2); // ft^2
+printf("\t total surface area is : %.1f ft^2 \n",Ai);
+delt3=((hi*Ai*delt)/(W*c)); // delt3=ti-t1, F
+printf("\t delt3 is : %.1f F \n",delt3);
+ti=t1+delt3; // F
+printf("\t ti is : %.1f F \n",ti);
+printf("\t The oil now enters the second pass at 126.9°F \n");
+// end
+
+
diff --git a/1328/CH10/EX10.2/10_2.sce b/1328/CH10/EX10.2/10_2.sce
new file mode 100644
index 000000000..21a235c4b
--- /dev/null
+++ b/1328/CH10/EX10.2/10_2.sce
@@ -0,0 +1,88 @@
+printf("\t example 10.2 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=250; // outlet hot fluid,F
+t1=95; // inlet cold fluid,F
+t2=145; // outlet cold fluid,F
+W=16000; // lb/hr
+w=423; // lb/hr
+printf("\t 1.for heat balance \n");
+printf("\t for kerosene \n");
+c=0.5; // Btu/(lb)*(F)
+Q=((W)*(c)*(t2-t1)); // Btu/hr
+printf("\t total heat required for kerosene is : %.0f Btu/hr \n",Q);
+printf("\t for steam \n");
+l=945.5; // 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 :%.0f F \n",LMTD);
+tc=((t1)+(t2))/2; // caloric temperature of cold fluid,F
+printf("\t caloric temperature of cold fluid is : %.0f F \n",tc);
+printf("\t hot fluid:shell side,steam \n");
+ho=(1500); // condensation of steam Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);
+printf("\t cold fluid:inner tube side,kerosene \n");
+Nt=86;
+n=2; // number of passes
+L=12; //ft
+at1=0.594; // flow area, in^2,from table 10
+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/(.177)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+mu2=1.36*2.42; // at 145F,lb/(ft)*(hr)
+D=(0.87/12); // ft
+Ret1=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.0f \n",Ret1);
+mu3=1.75*2.42; // at 120F,lb/(ft)*(hr)
+D=(0.87/12); // ft
+Ret2=((D)*(Gt)/mu3); // reynolds number
+printf("\t reynolds number is : %.1e \n",Ret2);
+Z1=331; // Z1=(L*n/D)
+jH=3.1; // from fig 24
+mu4=1.75; // cp and 40 API
+Z2=0.24; // Z2=((k)*(c*mu4/k)^(1/3)), from fig 16
+Hi=((jH)*(1/D)*(Z2)); // using eq.6.15a,Btu/(hr)*(ft^2)*(F)
+printf("\t Hi is : %.2f Btu/(hr)*(ft^2)*(F) \n",Hi);
+ID=0.87; // ft
+OD=1; //ft
+Hio=(Hi*(ID/OD)); //Btu/(hr)*(ft^2)*(F), from eq.6.5
+printf("\t Hio is : %.2f Btu/(hr)*(ft^2)*(F) \n",Hio);
+tw=(tc)+(((ho)/(Hio+ho))*(T1-tc)); // from eq.5.31
+printf("\t tw is : %.0f F \n",tw);
+muw=1.45; // lb/(ft)*(hr),at 249F from fig.14
+phyt=(mu3/muw)^0.14;
+printf("\t phyt is : %.1f \n",phyt); // from fig.24
+hio=(Hio)*(phyt); // from eq.6.37
+printf("\t Correct hio to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio);
+delt=tw-tc; //F
+printf("\t delt is : %.0f F \n",delt);
+printf("\t Since the kerosene has a viscosity of only 1.75 cp at the caloric temperature and delt=129F, free convection should be investigated. \n");
+s=0.8;
+row=50; // lb/ft^3, from fig 6
+s1=0.810; // at 95F
+s2=0.792; // at 145F
+bita=((s1^2-s2^2)/(2*(t2-t1)*s1*s2)); // /F
+printf("\t beta is : %.6f /F \n",bita);
+G=((D^3)*(row^2)*(bita)*(delt)*(4.18*10^8)/(mu3^2));
+printf("\t G is : %.1e \n",G);
+psy=((2.25)*(1+(0.01*G^(1/3)))/(log10(Ret2)));
+printf("\t psy is : %.2f \n",psy);
+hio1=(hio*psy);
+printf("\t corrected hio1 is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio1);
+Uc=((hio1)*(ho)/(hio1+ho)); // clean overall coefficient,Btu/(hr)*(ft^2)*(F)
+printf("\t clean overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc);
+A2=0.2618; // 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)*(delt)));
+printf("\t actual design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD);
+Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu
+printf("\t actual Rd is : %.2f (hr)*(ft^2)*(F)/Btu \n",Rd);
+// end
+
diff --git a/1328/CH10/EX10.3/10_3.sce b/1328/CH10/EX10.3/10_3.sce
new file mode 100644
index 000000000..389fa7413
--- /dev/null
+++ b/1328/CH10/EX10.3/10_3.sce
@@ -0,0 +1,102 @@
+printf("\t example 10.3 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=250; // outlet hot fluid,F
+t1=105; // inlet cold fluid,F
+t2=130; // outlet cold fluid,F
+w=50000; // lb/hr
+W=622; // lb/hr
+printf("\t 1.for heat balance \n");
+printf("\t for gas oil \n");
+c=0.47; // Btu/(lb)*(F)
+Q=((w)*(c)*(t2-t1)); // Btu/hr
+printf("\t total heat required for gas oil is : %.2e Btu/hr \n",Q);
+printf("\t for steam \n");
+l=945.5; // 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);
+tc=((t1)+(t2))/2; // caloric temperature of cold fluid,F
+printf("\t caloric temperature of cold fluid is : %.1f F \n",tc);
+printf("\t hot fluid:shell side,steam \n");
+ID=15.25; // in
+C=0.25; // clearance
+B=15; // baffle spacing,in
+PT=1.25;
+as=((ID*C*B)/(144*PT)); // flow area,ft^2, eq 7.1
+printf("\t flow area is : %.3f ft^2 \n",as);
+Gs=(6220/as); // mass velocity,lb/(hr)*(ft^2), calculation mistake
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs);
+mu1=0.0314; // at 250F,lb/(ft)*(hr), from fig.15
+De=0.060; // from fig.29,ft
+Res=((De)*(Gs)/mu1); // reynolds number, calculation mistake
+printf("\t reynolds number is : %.2e \n",Res);
+ho=1500; //Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);
+printf("\t cold fluid:inner tube side,crude oil \n");
+d1=0.5; // in
+d2=0.87; // in
+at1=((3.14*(d2^2-d1^2))/4);
+printf("\t at1 is : %.1f in^2 \n",at1);
+Nt=86;
+n=2; // number of passes
+L=12; //ft
+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 : %.1e lb/(hr)*(ft^2) \n",Gt);
+De=(d2^2-d1^2)/(12*d2);
+printf("\t De is : %.4f ft \n",De);
+mu2=16.7; // at 117F,lb/(ft)*(hr)
+Ret=((De)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+jH=3.1; // from fig.24
+Z=0.35; // Z=(K*(c*mu3/k)^(1/3)),Btu/(hr)(ft^2)(F/ft), at mu3=6.9cp and 28 API
+Hi=((jH)*(1/De)*(Z)); //Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft^2)*(F)
+printf("\t Hi is : %.1f Btu/(hr)*(ft^2)*(F) \n",Hi);
+ID=0.87; // ft
+OD=1; //ft
+Hio=((Hi)*(ID/OD)); //Hio=(hio/phyp), using eq.6.5
+printf("\t Correct Hi0 to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",Hio);
+muw=4.84; // lb/(ft)*(hr), from fig.14
+phyt=(mu2/muw)^0.14;
+printf("\t phyt is : %.2f \n",phyt); // from fig.24
+hio=(Hio)*(phyt); // from eq.6.37
+printf("\t Correct hi0 to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio);
+tw=(tc)+(((ho)/(hio+ho))*(T1-tc)); // 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 : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc);
+A=270; // ft^2
+printf("\t total surface area is : %.0f ft^2 \n",A);
+UD=((Q)/((A)*(LMTD)));
+printf("\t actual design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD);
+Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu
+printf("\t actual Rd is : %.3f (hr)*(ft^2)*(F)/Btu \n",Rd);
+printf("\t pressure drop for annulus \n");
+f=0.0016; // friction factor for reynolds number 25300, using fig.29
+s=0.00116; // for reynolds number 25300,using fig.6
+Ds=15.25/12; // ft
+phys=1;
+N=(12*L/B); // number of crosses,using eq.7.43
+printf("\t number of crosses are : %.0f \n",N);
+delPs=((f*(19600^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys)))/(2); // using eq.7.44,psi
+printf("\t delPs is : %.1f psi \n",delPs);
+printf("\t pressure drop for inner pipe \n");
+dt=(d2-d1)/(12); // ft
+printf("\t dt is : %.4f ft \n",dt);
+Ret2=(dt*Gt/mu2);
+printf("\t Ret2 is : %.0f \n",Ret2);
+f=0.00066; // friction factor for reynolds number 8220, using fig.26
+phyt=1.35; // fig 6
+printf("\t phyt is : %.2f \n",phyt);
+s=0.85;
+delPt=((f*(420000^2)*(L)*(n))/(5.22*(10^10)*(0.0309)*(s)*(phyt))); // using eq.7.45,psi
+printf("\t delPt is : %.1f psi \n",delPt);
+printf("\t delPr is negligible \n");
+//end
diff --git a/1328/CH10/EX10.4/10_4.sce b/1328/CH10/EX10.4/10_4.sce
new file mode 100644
index 000000000..84daa1b65
--- /dev/null
+++ b/1328/CH10/EX10.4/10_4.sce
@@ -0,0 +1,40 @@
+printf("\t example 10.4 \n");
+printf("\t approximate values are mentioned in the book \n");
+t1=100; // F
+t2=0; // F
+T1abs=100+460; // R
+T2abs=460; //R
+delt=t1-t2;
+printf("\t delt is : %.f F \n",delt);
+hc=0.3*(delt^0.25); // convection loss, Btu/(hr)*(ft^2)*(°F)
+printf("\t convection loss is : %.2f Btu/(hr)(ft^2)(F) \n",hc);
+e=0.8; // emissivity
+hr=((0.173*e*((T1abs/100)^4-(T2abs/100)^4))/(T1abs-T2abs)); // radiation rate, from 4.32, Btu/(hr)(ft^2)(F)
+printf("\t radiation loss is : %.2f Btu/(hr)(ft^2)(F) \n",hr);
+hl=hc+hr; // combined loss, Btu/(hr)(ft^2)(F)
+printf("\t combined loss is : %.1f Btu/(hr)(ft^2)(F) \n",hl);
+D=5; // ft
+L=12; // ft
+A1=((2*3.14*D^2)/(4))+(3.14*D*L); // total tank area
+printf("\t total tank area is : %.1f ft^2 \n",A1);
+Q=(hl*A1*delt); // total heat loss
+printf("\t total heat loss : %.2e Btu/hr \n",Q);
+printf("\t This heat must be supplied by the pipe bundle,Assuming exhaust steam to be at 212°F \n");
+d0=1.32;
+X=(delt/d0);
+tf=((t1+212)/2); // F
+printf("\t X is : %.0f \n",X);
+printf("\t tf is : %.0f F \n",tf);
+hio=48; // from fig 10.4, Btu/(hr)(ft^2)(F)
+ho=1500; // condensation of steam,Btu/(hr)(ft^2)(F)
+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=0.02; // dirt factor, (hr)(ft^2)(F)/Btu
+UD=((Uc)/((1)+(Uc*Rd))); // design overall coefficient,Btu/(hr)*(ft^2)*(F)
+printf("\t design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD);
+A2=((Q)/((UD)*(212-100))); // total surface,ft^2
+printf("\t total surface is : %.1f ft^2 \n",A2);
+A3=2.06; // area/pipe
+N=(A2/A3);
+printf("\t number of pipes are : %.0f \n",N);
+//end
generated by cgit (git 2.34.1) at 2024-12-19 12:59:42 +0000