initial commit / add all books

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diff --git a/1328/CH7/EX7.4/7_4.sce b/1328/CH7/EX7.4/7_4.sce
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+printf("\t example 7.4 \n");

+printf("\t approximate values are mentioned in the book \n");

+T1=93; // inlet hot fluid,F

+T2=85; // outlet hot fluid,F

+t1=75; // inlet cold fluid,F

+t2=80; // outlet cold fluid,F

+W=175000; // lb/hr

+w=280000; // lb/hr

+printf("\t 1.for heat balance \n");

+printf("\t for distilled water \n");

+c=1; // Btu/(lb)*(F)

+Q=((W)*(c)*(T1-T2)); // Btu/hr

+printf("\t total heat required for distilled water is : %.1e Btu/hr \n",Q);

+printf("\t for raw water \n");

+c=1; // Btu/(lb)*(F)

+Q=((w)*(c)*(t2-t1)); // Btu/hr

+printf("\t total heat required for raw water is : %.1e 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 : %.2f \n",R);

+S=((t2-t1)/(T1-t1));

+printf("\t S is : %.3f \n",S);

+printf("\t FT is 0.945 \n"); // from fig 18

+delt=(0.945*LMTD); // F

+printf("\t delt is : %.2f F \n",delt);

+X=((delt1)/(delt2));

+printf("\t ratio of two local temperature difference is : %.3f \n",X);

+Fc=0.42; // from fig.17

+Kc=0.20; // crude oil controlling

+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);

+printf("\t hot fluid:shell side,distilled water \n");

+ID=15.25; // in

+C=0.1875; // clearance

+B=12; // baffle spacing,in

+PT=0.9375;

+as=((ID*C*B)/(144*PT)); // flow area,ft^2,using eq.7.1

+printf("\t flow area is : %.3f ft^2 \n",as);

+Gs=(W/as); // mass velocity,lb/(hr)*(ft^2),using eq.7.2

+printf("\t mass velocity is : %.1e lb/(hr)*(ft^2) \n",Gs);

+mu1=0.81*2.42; // at 89F,lb/(ft)*(hr), from fig.14

+De=0.55/12; // from fig.28,ft

+Res=((De)*(Gs)/mu1); // reynolds number

+printf("\t reynolds number is : %.2e \n",Res);

+jH=73; // from fig.28

+c=1; // Btu/(lb)*(F),at 89F,from fig.table 4

+k=0.36; // Btu/(hr)*(ft^2)*(F/ft), from table 4

+Pr=((c)*(mu1)/k)^(1/3); // prandelt number raised to power 1/3

+printf("\t Pr is : %.3f \n",Pr);

+ho=((jH)*(k/De)*(Pr)); // using eq.6.15,Btu/(hr)*(ft^2)*(F)

+printf("\t individual heat transfer coefficient is : %.2e Btu/(hr)*(ft^2)*(F) \n",ho);

+printf("\t cold fluid:inner tube side,raw water \n");

+Nt=160;

+n=2; // number of passes

+L=16; //ft

+at1=0.334; // 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/(at)); // mass velocity,lb/(hr)*(ft^2)

+printf("\t mass velocity is : %.3e lb/(hr)*(ft^2) \n",Gt);

+V=(Gt/(3600*62.5));

+printf("\t V is %.1f fps \n",V);

+mu2=0.92*2.42; // at 77.5F,lb/(ft)*(hr)

+D=0.65/12; //ft

+Ret=((D)*(Gt)/mu2); // reynolds number

+printf("\t reynolds number is : %.2e \n",Ret);

+hi=1350*0.99; //using fig.25,Btu/(hr)*(ft^2)*(F)

+ID=0.65; // 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);

+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);

+printf("\t ·when both. film coefficients are high the thermal resistance of the tube metal is not necessarily insignificant as assumed in the derivation of Eq. (6.38). For a steel 1.8 BWG tube Rm= 0.00017 and for copper Rm= 0.000017 \n");

+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)*(delt)));

+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 annulus \n");

+f=0.0019; // friction factor for reynolds number 16200, using fig.29

+s=1; // 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*(Gs^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys))); // using eq.7.44,psi

+printf("\t delPs is : %.1f psi \n",delPs);

+printf("\t allowable delPs is 10 psi \n");

+printf("\t pressure drop  for inner pipe \n");

+f=0.00019; // friction factor for reynolds number 36400, using fig.26

+s=1;

+phyt=1;

+D=0.054; // ft

+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=0.33; // X1=((V^2)/(2*g)), for Gt 1060000,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");

+//end