8 files changed, 797 insertions, 0 deletions
diff --git a/1328/CH15/EX15.1/15_1.sce b/1328/CH15/EX15.1/15_1.sce
new file mode 100644
index 000000000..f15d6e44b
--- /dev/null
+++ b/1328/CH15/EX15.1/15_1.sce
@@ -0,0 +1,44 @@
+printf("\t example 15.1 \n");
+printf("\t approximate values are mentioned in the book \n");
+ts=250;
+T1=400;
+T2=300;
+w=10000; // lb/hr
+W=150000; // lb/hr
+l=945.3; // Btu/(lb) , table 7
+Q=((w)*(l)); // Btu/hr
+printf("\t total heat required for steam is : %.2e Btu/hr \n",Q);
+C=0.63; // Btu/(lb)*(F)
+Q=((W)*(C)*(T1-T2)); // Btu/hr
+printf("\t total heat required for kerosene is : %.2e Btu/hr \n",Q); 
+delt1=T2-ts; //F
+delt2=T1-ts; // 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);
+UD=100;
+A=(Q/(UD*LMTD));
+printf("\t A : %.2e ft^2 \n",A);
+WC=94500; // Btu/F
+vl=0.017; // ft^3/lb, from table 7
+vv=13.75; // ft^3/lb, from table 7
+printf("\t By the law of mixtures \n");
+// Assume 80 per cent of the outlet fluid is vapor
+v2=(0.8*vv)+(.2*vl);
+printf("\t v2 : %.0f ft^3/lb \n",v2);
+vav=(WC*(v2-vl)/(UD*A))-((WC*(T2-ts)/(l*w))*(vv-vl))+vl;
+printf("\t vav : %.2f ft^3/lb \n",vav);
+printf("\t By the approximate method \n");
+vav1=(vl+v2)/(2);
+printf("\t vav : %.2f ft^3/lb \n",vav1);
+row=62.5;
+rowac=(1/vav);
+s=(rowac/row);
+printf("\t actual density : %.3f lb/ft^3 \n",rowac);
+printf("\t s : %.4f \n",s);
+rowap=(1/vav1);
+s=(rowap/row);
+printf("\t approximate density : %.3f lb/ft^3 \n",rowac);
+printf("\t s : %.4f \n",s);
+// end
diff --git a/1328/CH15/EX15.2/15_2.sce b/1328/CH15/EX15.2/15_2.sce
new file mode 100644
index 000000000..8524e1e2d
--- /dev/null
+++ b/1328/CH15/EX15.2/15_2.sce
@@ -0,0 +1,129 @@
+printf("\t example 15.2 \n");
+printf("\t approximate values are mentioned in the book \n");
+t1=108; // inlet cold fluid,F
+t2=235; // outlet cold fluid,F
+Ts=338;
+Wp=24700; // lb/hr
+Wv=19750; // lb/hr
+w=4880; // lb/hr
+printf("\t 1.for heat balance \n");
+Ht1=162; // enthalpy at t1, Btu/lb, fig 9
+Ht2=248; // enthalpy at t2, Btu/lb, fig 9
+qp=(Wp*(Ht2-Ht1)); // for preheat
+printf("\t total heat required for preheat of butane is : %.2e Btu/hr \n",qp);
+Ht3=358; // enthalpy of vapour at t2, Btu/lb, fig 9
+qv=Wv*(Ht3-Ht2);
+printf("\t total heat required for vapourisation of butane is : %.2e Btu/hr \n",qv);
+Q=qp+qv;
+printf("\t total heat required for  butane is : %.2e Btu/hr \n",Q);
+printf("\t for steam \n");
+l=880.6; // Btu/(lb), table 7
+Q=((w)*(l)); // Btu/hr
+printf("\t total heat required for steam is : %.2e Btu/hr \n",Q);
+deltp=158.5; // F, from eq 5.14
+deltv=103; // F eq 5.14
+Wp1=(qp/deltp);
+printf("\t Wp1 is : %.2e lb/hr \n",Wp1);
+Wv1=(qv/deltv);
+printf("\t Wv1 is : %.2e lb/hr \n",Wv1);
+W=(Wp1+Wv1);
+printf("\t W is : %.2e lb/hr \n",W);
+delt=(Q/W);
+printf("\t weighted delt is : % .1f F \n",delt);
+Tc=((Ts)+(Ts))/(2); // caloric temperature of hot fluid,F
+printf("\t caloric temperature of hot fluid is : %.1f 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:inner tube side,steam \n");
+Nt=76;
+n=2; // number of passes
+L=16; //ft
+at1=0.594; // flow area,table 10, 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);
+mu1=0.0363; // at 338F, fig 15,lb/(ft)*(hr)
+D=0.0725; // ft
+Ret=((D)*(Gt)/mu1); // reynolds number
+printf("\t reynolds number is : %.1e \n",Ret);
+hio=1500; // condensing steam,Btu/(hr)*(ft^2)*(F)
+printf("\t hio is : %.0f Btu/(hr)*(ft^2)*(F) \n",hio);
+printf("\t cold fluid:shell side,butane \n");
+printf("\t preheating \n");
+ID=15.25; // in
+C=0.25; // clearance
+B=5; // baffle spacing,in
+PT=1.25;
+as=((ID*C*B)/(144*PT)); // flow area,ft^2
+printf("\t flow area is : %.3f ft^2 \n",as);
+Gs=(Wp/as); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs);
+mu2=0.278; // at 172F,lb/(ft)*(hr), from fig.14
+De=0.0825; // from fig.28,ft
+Res=((De)*(Gs)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Res);
+jH=159; // from fig.28
+Z=0.12; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+hop=((jH)*(1/De)*(Z)); //using eq.6.15b,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",hop);
+Up=((hio)*(hop)/(hio+hop)); // clean overall coefficient,eq 6.38,Btu/(hr)*(ft^2)*(F)
+printf("\t clean overall coefficient for preheating : %.0e Btu/(hr)*(ft^2)*(F) \n",Up);
+Ap=(qp/(Up*deltp));
+printf("\t clean surface required for preheating : %.0f ft^2 \n",Ap);
+printf("\t for vapourisation \n");
+mu2=0.242; // at 172F,lb/(ft)*(hr), from fig.14
+Res=((De)*(Gs)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Res);
+jH=170; // from fig.28
+Z=0.115; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+hov=((jH)*(1/De)*(Z)); //using eq.6.15b,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",hov);
+Uv=((hio)*(hov)/(hio+hov)); // clean overall coefficient,eq 6.38,Btu/(hr)*(ft^2)*(F)
+printf("\t clean overall coefficient for vapourisation : %.0f Btu/(hr)*(ft^2)*(F) \n",Uv);
+Av=(qv/(Uv*deltv));
+printf("\t clean surface required for vapourisation : %.0f ft^2 \n",Av);
+Ac=Ap+Av;
+printf("\t total clean surface : %.1e ft^2 \n",Ac);
+UC=((Up*Ap)+(Uv*Av))/(Ac);
+printf("\t weighted clean overall coefficient : %.0f 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);
+// A total of 170 ft2 are required of which 103 are to be used for vaporization. For the total surface required 318 ft2 will be provided. It can be assumed, then, that the surface provided for vaporization is 193ft^2
+// then flux is Q/A=10700, which is with in satisfactory levels.
+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=0.000165; // friction factor for reynolds number 62000, using fig.26
+s=0.00413;
+phyt=1;
+delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(D)*(s)*(phyt)))/(2); // using eq.7.45,psi
+printf("\t delPt is : %.2f psi \n",delPt);
+printf("\t allowable delPa is negligible \n");
+printf("\t pressure drop  for annulus \n");
+printf("\t preheating \n");
+f=0.00145; // friction factor for reynolds number 69200, using fig.29
+Lp=(L*Ap/Ac); //ft
+printf("\t length of preheat zone : %.1f ft \n",Lp);
+N=(12*Lp/B); // number of crosses,using eq.7.43
+printf("\t number of crosses are : %.0f \n",N);
+s=0.5; // for reynolds number 69200,using fig.6
+Ds=1.27; // fig 28
+phys=1;
+delPsp=((f*(Gs^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys))); // using eq.7.44,psi
+printf("\t delPsp is : %.1f psi \n",delPsp);
+printf("\t vapourisation \n");
+f=0.00142;
+Lv=9.7; // Lv=L-Lp
+Nv=(12*Lv/B); // number of crosses,using eq.7.43
+printf("\t number of crosses are : %.0f \n",Nv);
+s=0.28; 
+delPsv=((f*(Gs^2)*(Ds)*(Nv))/(5.22*(10^10)*(De)*(s)*(1))); // using eq 12.47,psi
+printf("\t delPsv is : %.1f psi \n",delPsv);
+delPS=delPsp+delPsv;
+printf("\t delPS is : %.1f psi \n",delPS);
+printf("\t allowable delPa is 5 psi \n");
+//end
diff --git a/1328/CH15/EX15.3/15_3.sce b/1328/CH15/EX15.3/15_3.sce
new file mode 100644
index 000000000..0e4711306
--- /dev/null
+++ b/1328/CH15/EX15.3/15_3.sce
@@ -0,0 +1,73 @@
+printf("\t example 15.3 \n");
+printf("\t approximate values are mentioned in the book \n");
+ts=400;
+T1=575;
+T2=475;
+W=28100; // lb/hr
+w=34700; // lb/hr
+printf("\t 1.for heat balance \n");
+HT1=290; // enthalpy at T1, Btu/lb, fig 11
+HT2=385; // enthalpy at T2, Btu/lb, fig 11
+Q=(W*(HT2-HT1)); // for preheat
+printf("\t total heat required for gasoline is : %.2e Btu/hr \n",Q);
+c=0.77; // Btu/(lb), table 7
+Q=((w)*(c)*(T1-T2)); // Btu/hr
+printf("\t total heat required for gasoil is : %.2e Btu/hr \n",Q);
+delt=118; // F eq 5.14
+S=((T2-ts)/(T1-ts));
+printf("\t S is : %.3f \n",S);
+Kc=0.37; // fig 17
+Fc=0.42;
+Tc=(T2+(0.42*(T1-T2)));
+printf("\t Tc is : %.0f F \n",Tc);
+printf("\t hot fluid:inner tube side,gasoil \n");
+Nt=68;
+n=6; // number of passes
+L=12; //ft
+at1=0.546; // flow area,table 10, 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);
+mu1=0.65; // at 517F, fig 14,lb/(ft)*(hr)
+D=0.0694; // ft
+Ret=((D)*(Gt)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+jH=220; // from fig.24
+Z=0.118; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+Hi=((jH)*(1/D)*(Z)); //hi/phyt, Hi=()using eq.6.15d,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hi);
+Hio=((Hi)*(0.834/1)); //Hio=(hio/phyp), using eq.6.9
+printf("\t Correct Hi0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hio);
+// (mu1/muw)^(0.14) is negligible
+printf("\t cold fluid:shell side,gasoline \n");
+ho=300; // assumption
+tw=(ts)+(((Hio)/(Hio+ho))*(Tc-ts)); // from eq.5.31
+printf("\t tw is : %.0f F \n",tw);
+deltw=(tw-ts);
+printf("\t deltw : %.0f F \n",deltw);
+// from fig 15.11, ho>300
+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=0.2618; // 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);
+// check for max. flux=Q/A=12500.(satisfactory)
+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=0.00015; // friction factor for reynolds number 85700, using fig.26
+s=0.71;
+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=0.09; // 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 delPa is 10psi \n");
+printf("\t delPs is negligible \n");
+//end
diff --git a/1328/CH15/EX15.4/15_4.sce b/1328/CH15/EX15.4/15_4.sce
new file mode 100644
index 000000000..9b4bde2be
--- /dev/null
+++ b/1328/CH15/EX15.4/15_4.sce
@@ -0,0 +1,120 @@
+printf("\t example 15.4\n");
+printf("\t approximate values are mentioned in the book \n");
+t1=315; // inlet cold fluid,F
+t2=335; // outlet cold fluid,F
+T1=525;
+T2=400;
+Wv=29000; // lb/hr
+Ws=38500; // lb/hr
+w=51000; // lb/hr
+printf("\t 1.for heat balance \n");
+Ht1=238; // enthalpy at t1, Btu/lb, fig 9
+Ht2=252; // enthalpy at t2, Btu/lb, fig 9
+Ht3=378; // enthalpy of vapour at t2 
+qv=(Wv*(Ht3-Ht2)); // for preheat
+printf("\t qv is : %.2e Btu/hr \n",qv);
+qs=Ws*(Ht2-Ht1);
+printf("\t qs is : %.2e Btu/hr \n",qs);
+Q=qs+qv;
+printf("\t total heat required for naphtha is : %.2e Btu/hr \n",Q);
+c=0.66; // Btu/(lb)(F)
+Q=((w)*(c)*(T1-T2)); // Btu/hr
+printf("\t total heat required for gasoil 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);
+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 0.97 \n"); // from fig 18
+delt=(0.97*LMTD); // F
+printf("\t delt is : %.0f F \n",delt);
+X=((delt1)/(delt2)); // fig 17
+printf("\t ratio of two local temperature difference is : %.3f \n",X);
+Fc=0.41; // from fig.17
+Kc=0.42;
+Tc=((T2)+((Fc)*(T1-T2))); // caloric temperature of hot fluid,F
+printf("\t caloric temperature of hot fluid is : %.0f F \n",Tc);
+tc=((t1)+((Fc)*(t2-t1))); // caloric temperature of cold fluid,F
+printf("\t caloric temperature of cold fluid is : %.0f F \n",tc);
+printf("\t hot fluid:inner tube side,steam \n");
+Nt=116;
+n=8; // number of passes
+L=12; //ft
+at1=0.546; // flow area,table 10, 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);
+mu1=1.09; // at 451F, fig 14,lb/(ft)*(hr)
+D=0.0695; // ft
+Ret=((D)*(Gt)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+jH=168; // from fig.24
+Z=0.142; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+Hi=((jH)*(1/D)*(Z)); //, Hi=(hi/phyt)using eq.6.15d,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hi);
+Hio=((Hi)*(0.834/1)); //Hio=(hio/phyp), using eq.6.9
+printf("\t Correct Hio to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hio);
+printf("\t cold fluid:shell side,naphtha \n");
+ho1=200; // assumption
+tw=(tc)+(((Hio)/(Hio+ho1))*(Tc-tc)); // from eq.5.31, calculation mistake
+printf("\t tw is : %.0f F \n",tw);
+deltw=(tw-tc);
+printf("\t deltw : %.0f F \n",deltw);
+// from fig 15.11, hv>300, hs=60
+Av=(qv/300);
+As=qs/60;
+printf("\t qv/hv : %.3e \n",Av);
+printf("\t qs/hs : %.0e \n",As);
+A1=As+Av;
+printf("\t A : %.3e \n",A1);
+ho=(Q/A1);
+printf("\t ho : %.0f \n",ho);
+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=0.2618; // 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);
+// check for max. flux=Q/A=11500.(satisfactory)
+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=0.000168; // friction factor for reynolds number 59200, using fig.26
+s=0.73;
+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=0.11; // 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 delPa is negligible \n");
+printf("\t pressure drop  for annulus \n");
+Af=(3.14*(21.25^2-(116))/8);
+printf("\t flow area : %.0f in^2 \n",Af);
+as=0.917; // ft^2
+p=(3.14*21.25/2)+(3.14*1*116/2)+(21.25);
+printf("\t wetted perimeter : %.1f in \n",p);
+De=0.186; // ft
+Gs=(Ws/(2*as)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.1e lb/(hr)*(ft^2) \n",Gs);
+mu2=0.435; // at 315F, fig 14,lb/(ft)*(hr)
+Res=((De)*(Gs)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+f=0.00028; // using fig.26
+row=0.337; // fig 13.14
+// soutlet max=0.071,
+s=0.35; // using fig.6
+phys=1;
+delPs=((f*(Gs^2)*(L))/(5.22*(10^10)*(De)*(s)*(phys))); // using eq.7.44,psi
+printf("\t delPs is : %.4f psi \n",delPs);
+printf("\t allowable delPa is .25 psi \n");
+//end
diff --git a/1328/CH15/EX15.5/15_5.sce b/1328/CH15/EX15.5/15_5.sce
new file mode 100644
index 000000000..9607778b2
--- /dev/null
+++ b/1328/CH15/EX15.5/15_5.sce
@@ -0,0 +1,116 @@
+printf("\t example 15.5\n");
+printf("\t approximate values are mentioned in the book \n");
+W=40800; // lb/hr
+w=4570; // lb/hr
+printf("\t 1.for heat balance \n");
+Ht1=241; // enthalpy of liquid at 228F, Btu/lb, fig 9
+Ht2=338; // enthalpy of vapourat 228F, Btu/lb, fig 9
+Q=(W*(Ht2-Ht1));
+printf("\t total heat required for butane is : %.2e Btu/hr \n",Q);
+l=868; // Btu/(lb), table 7
+Q=((w)*(l)); // Btu/hr
+printf("\t total heat required for steam is : %.2e Btu/hr \n",Q);
+delt=125; // delt=LMTD, isothermal boiling, eq 5.14
+// Tc and tc: Both streams are isuthermal
+printf("\t trail 1 \n");
+A1=((Q)/((12000))); // Q/A1 =12000, first trial should always be taken for the maximum allowable flux
+printf("\t A1 is : %.1e ft^2 \n",A1);
+a1=0.1963; // ft^2/lin ft
+L=16;
+N1=(A1/(L*a1)); // table 10
+printf("\t number of tubes are : %.0f \n",N1);
+N2=109; // assuming one tube passes, 13.25-in ID, from table 9
+A2=(N2*L*a1); // ft^2
+printf("\t total surface area is : %.0f ft^2 \n",A2);
+UD=((Q)/((A2)*(delt)));
+printf("\t correct design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD);
+// Assume 4: 1 recirculation ratio
+rowv=(58/(359*(688/492)*(14.7/290))); // eq 15.18
+printf("\t vapour density : %.2f lb/ft^3 \n",rowv);
+Vv=0.44; 
+Vl=0.0372; // fig 6
+W1=4*W;
+printf("\t weight flow of recirculated liquid : %.3e lb/hr \n",W1);
+VL=W1*Vl;
+VV=W*Vv;
+printf("\t volume of liquid : %.2e ft^3 \n",VL);
+printf("\t volume of vapour : %.3e ft^3 \n",VV);
+V=VL+VV;
+printf("\t total volume out of reboiler : %.3e ft^3 \n",V);
+vo=(V/(W1+W));
+printf("\t vo is : %.4f ft^3/lb \n",vo);
+Pl=((2.3*16)/(144*(vo-Vl)))*(log10(vo/Vl));
+printf("\t pressure leg : %.1f psi \n",Pl);
+printf("\t frictional resistance \n");
+Nt=109;
+n=1; // number of passes
+at1=0.302; // flow area,table 10, 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=((W1+W)/(at)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+mu1=0.242; // at 228F, fig 14,lb/(ft)*(hr)
+D=0.0517; // ft
+Ret=((D)*(Gt)/mu1); // reynolds number
+printf("\t reynolds number is : %.1e \n",Ret);
+f=0.000127; // using fig.26
+s=0.285;
+phyt=1;
+delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(D)*(s)*(phyt))); // using eq.7.45,psi
+printf("\t delPt is : %.2f psi \n",delPt);
+P=Pl+delPt;
+printf("\t total resisitance : %.2f psi \n",P);
+F=(16*0.43*62.5/144);
+printf("\t driving force : %.2f psi \n",F);
+// The resistances are greater than the hydrostatic head can provide; hence the recirculation ratio will be less than 4: 1
+printf("\t trial 2 \n"); // Assume 12'0" tubes and 4:1 recirculation ratio
+A1=((Q)/((12000))); // Q/A1 =12000, first trial should always be taken for the maximum allowable flux
+printf("\t A1 is : %.1e ft^2 \n",A1);
+a1=0.1963; // ft^2/lin ft
+L=12;
+N1=(A1/(L*a1)); // table 10
+printf("\t number of tubes are : %.0f \n",N1);
+N2=151; // assuming one tube passes, 15.25-in ID, from table 9
+A2=(N2*L*a1); // ft^2
+printf("\t total surface area is : %.0f ft^2 \n",A2);
+UD=((Q)/((A2)*(delt)));
+printf("\t correct design overall coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",UD);
+Pl=((2.3*12)/(144*(vo-Vl)))*(log10(vo/Vl));
+printf("\t pressure leg : %.1f psi \n",Pl);
+printf("\t frictional resistance \n");
+Nt=151;
+n=1; // number of passes
+at1=0.302; // flow area,table 10, 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=((W1+W)/(at)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+mu1=0.242; // at 228F, fig 14,lb/(ft)*(hr)
+D=0.0517; // ft
+Ret=((D)*(Gt)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+f=0.000135; // using fig.26
+s=0.285;
+phyt=1;
+delPt=((f*(Gt^2)*(12)*(n))/(5.22*(10^10)*(D)*(s)*(phyt))); // using eq.7.45,psi
+printf("\t delPt is : %.2f psi \n",delPt);
+P=Pl+delPt;
+printf("\t total resisitance : %.2f psi \n",P);
+F=(12*0.43*62.5/144);
+printf("\t driving force : %.2f psi \n",F);
+// Since the driving force is slightly greater than the resistances, a recirculation ratio better than 4:1 is assured.
+printf("\t hot fluid : shell side,steam \n");
+ho=1500; // condensing steam
+printf("\t cold fluid:inner tube side, butane \n");
+jH=330; // from fig.24
+Z=0.115; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+Hi=((jH)*(1/D)*(Z)); //, Hi=(hi/phyt)using eq.6.15d,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hi);
+Hio=((300)*(0.62/0.75)); //Hio=(hio/phyp), using eq.6.9
+printf("\t Correct Hio 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);
+UD=89;
+Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu
+printf("\t actual Rd is : %.4f (hr)*(ft^2)*(F)/Btu \n",Rd);
+// end
diff --git a/1328/CH15/EX15.6/15_6.sce b/1328/CH15/EX15.6/15_6.sce
new file mode 100644
index 000000000..90f25d19c
--- /dev/null
+++ b/1328/CH15/EX15.6/15_6.sce
@@ -0,0 +1,17 @@
+printf("\t example 15.6\n");
+printf("\t approximate values are mentioned in the book \n");
+//20000=WD+WB;
+//0.99*WD+(0.05*WB)=(20000*.5);
+// solving above two eq. we get WD and WB
+WD=9570; // lb/hr
+WB=10430; // lb/hr
+HBl=108; // fig 3 and 12
+HDl=85.8; //fig 3 and 12
+HDv=253.8; // fig 3 and 12
+HFl=92; // fig 3 and 12
+l=153; // fig 3 and 12
+QR=((2.54+1)*WD*(HDv))-(2.54*WD*HDl)+(WB*HBl)-(20000*HFl);
+printf("\t total heat duty : %.1e Btu/hr \n",QR);
+Q=QR/153;
+printf("\t total heat duty : %.2e lb/hr \n",Q);
+// end
diff --git a/1328/CH15/EX15.7/15_7.sce b/1328/CH15/EX15.7/15_7.sce
new file mode 100644
index 000000000..b0a41c96d
--- /dev/null
+++ b/1328/CH15/EX15.7/15_7.sce
@@ -0,0 +1,100 @@
+printf("\t example 15.7 \n");
+printf("\t approximate values are mentioned in the book \n");
+
+//Basis: One hour
+//20000=WD+WB , material balance
+//0.99*WD+(0.05*WB)=(20000*0.5) , Benzene balance
+// solving above two eq. we get WD and WB
+WD=9570; // lb/hr
+WB=10430; // lb/hr
+
+//Compositions and Boiling Points
+//Feed
+l1 = 10000; //Lb/hr , C6H4
+l2 = 10000; //Lb/hr , C7H8
+lb = l1+l2; //Lb/hr
+printf("\ttotal Lb/hr is %.0f\n",lb);
+mo1 = 78.1; //Mol. wt., C6H6
+mo2 = 93.1; //Mol. wt , C7H8
+mh1 = 128.0; //Mol/hr , C6H6
+mh2 = 107.5; //Mol/hr , C7H8
+mh = mh1 + mh2; // Mol/hr
+printf("\ttotal Mol/hr is %.1f\n",mh);
+x1 = mh1/mh;
+printf("\tx1 of C6H6 is %.3f\n",x1);
+x2 = mh2/mh;
+printf("\tx1 of C7H8 is %.3f\n",x2);
+x = x1+x2;
+printf("\tTotal x1 is %.3f\n",x);
+Pp1= 1380; // 214°F
+Pp2=575; // 214°F
+xp1 = x1*Pp1;
+printf("\tx1Pp1 of C6H6 is %.0f\n",xp1);
+xp2 = x2*Pp2;
+printf("\tx1Pp1 of C7H8 is %.0f\n",xp2);
+sxp = xp1 + xp2;
+printf("\tTotal x1Pp1 is %.0f\n",sxp);
+y1 = xp1/sxp;
+printf("\ty1 of C6H6 is %.3f\n",y1);
+y2 = xp2/sxp;
+printf("\ty1 of C7H8 is %.3f\n",y2);
+y = y1+y2;
+printf("\tTotal y1 is %.3f\n",y);
+
+
+w1 = 0.558; //from eq 15.42
+printf("\t(WR`/V =((xD - yF)/.(xD - xF))) = %.3fmol/mol\n",w1);
+wD=1;
+xD = 0.992;
+//V = WR' + WD
+// WR'/V = 0.558
+//Solving, WR' = (WR' * 0.558) + (0.558 * WD)
+Wr = 1.27; // mol reflux/mol distillate
+printf("\tWR` = %.2f (mol reflux)/(mol distillate)\n",Wr);
+Wr1 = Wr * 2; // mol/ mol distillate
+printf("\tAssumed 200 percent of the theoretical minimum reflux as economic\n\tWR = %.2f(mol)/(mil distillate)\n",Wr1);
+in = (wD * xD)/(Wr1 + 1); //intercept for the upper operating line
+printf("\tThe intercept for the upper operating line = %.3f\n",in);
+p = 13; // From fig. 15.23, connecting the corresponding lines
+printf("\tConnecting the corresponding line in Fig. 15.23, plates required: %.0f\n",p);
+fp = 7; // From fig. 15.23, connecting the corresponding lines
+printf("\tFeed plate is %.0fth(from top)\n",fp);
+d=122.5;
+tf = Wr1 * d;
+printf("\tTotal reflux is %.1f\n",tf);
+printf("\t\t\t\t\tHeat balances");
+
+//Heat Balances
+l1 = 33900;
+l2 = 9570;
+l3 = 24330;
+b1 = 253.8;
+b2 = 85.8;
+b3 = 85.8;
+bt1 = b1*l1;
+bt2 = b2*l2;
+bt3 = b3*l3;
+bt4 = 5688000;
+printf("\n\t\t\t\tMol/hr\tMol.wt.\tLb/hr\tTemp,°F\tBtu/lb\tBtu/hr\n\t________________________________________________________________________\n\tHeat balance \n\taround condenser:\n");
+printf("\t  Heat in:\n\t  Top plate vapor.......433\t87.3\t%.0f\t195\t%.1f\t%.0f\n",l1,b1,bt1);
+printf("\t  Heat out:\n\t  Distillate............");
+printf("122.5\t78.3\t%.0f\t195\t%.1f\t%.0f\n",l2,b2,bt2);
+printf("\t  Reflux................");
+printf("310.5\t78.3\t%.0f\t195\t%.1f\t%.0f\n",l3,b3,bt3);
+printf("\t  Condenser duty, by\n\t  difference........... .....   ....    ......  ..");
+printf(".     .....   5688000\n");
+printf("\t\t\t\t\t\t\t\t\t_______\n\t\t\t\t\t\t\t\t\t8600000\n\n");
+
+lam = 153; // At 246 °F, Btu/hr
+rv = 5800000/153; //Lb/hr
+printf("\tReboiler vapor is %.2e lb/hr\n",rv);
+to = rv + 10430; //Lb/hr
+printf("\tTrapout is %.3e lb/hr\n",to);
+
+printf("\n\t\t\t\tMol/hr\tMol.wt.\tLb/hr\tTemp,°F\tBtu/lb\tBtu/hr\n\t________________________________________________________________________\n");
+printf("\tHeat in:\n\t  Trapout...............522\t92.8\t%.0f\t246\t108.0\t5230000\n",to);
+printf("\t  Reboiler duty, \n\t  by difference.......  ....    ....    .....   ...     .....   5800000\n");
+printf("\t\t\t\t\t\t\t\t\t_______\n\t\t\t\t\t\t\t\t\t11030000\n\n");
+printf("\n\tReboiler requirements are\n");
+printf("\t\tTotal liquid to reboiler\t48330 lb/hr\n\t\tVaporization\t\t\t37900 lb/hr\n\t\tTemperature(nearly isothermal)\t246°F\n\t\tPressure\t\t\t5 psig\n\t\tHeat load\t\t\t5800000 Btu/hr\n")
+//end
diff --git a/1328/CH15/EX15.8/15_8.sce b/1328/CH15/EX15.8/15_8.sce
new file mode 100644
index 000000000..00ece1613
--- /dev/null
+++ b/1328/CH15/EX15.8/15_8.sce
@@ -0,0 +1,198 @@
+printf("\t example 15.8 \n");
+printf("\t approximate values are mentioned in the book \n");
+//Dew point of Overhead
+vc(1) = 6.4; // Mol/hr
+vc(2) = 219.7; //Mol/hr
+vc(3) = 2.3; //Mol/hr
+
+K(1) = 2.8; //at 148°F and 40 psia
+K(2) = 1.01; //at 148°F and 40 psia
+K(3) = 0.34; //at 148°F and 40 psia
+
+i=1;
+while(i<4)
+    v(i)=vc(i)/K(i);
+    i=i+1;
+end
+
+printf("\n\t\tDEW POINT OF OVERHEAD");
+printf("\n\t\tMol/hr\t\tK(148°F,40 psia)\tV/K\n");
+printf("\t\t--------------------------------------------\n");
+i=1;
+while(i<4)
+    printf("\tC"+string(i+3) + "\t%.1f\t\t%.1f\t\t\t%.1f\n",vc(i),K(i),v(i));
+    i = i+1
+end
+
+
+bc(1)=4.1; //Mol/hr
+bc(2)=49.3; //Mol/hr
+bc(3)=71.9; //Mol/hr
+bc(4)=52.5; //Mol/hr
+bc(5)=54.7; //Mol/hr
+bc(6)=82.5; //Mol/hr
+bc(7)=76.6; //Mol/hr
+bc(8)=22.4; //Mol/hr
+tbc = 0;
+i=1;
+while(i<9)
+    tbc = tbc+bc(i);
+    i=i+1;
+end
+
+bK(1)=5.8; //at 330°F, 40 psia
+bK(2)=3.0; //at 330°F, 40 psia
+bK(3)=1.68; //at 330°F, 40 psia
+bK(4)=0.98; //at 330°F, 40 psia
+bK(5)=0.57; //at 330°F, 40 psia
+bK(6)=0.35; //at 330°F, 40 psia
+bK(7)=0.21; //at 330°F, 40 psia
+bK(8)=0.13; //at 330°F, 40 psia
+
+KL(1)=23.8;
+KL(2)=148.0;
+KL(3)=120.8;
+KL(4)=51.4;
+KL(5)=31.2;
+KL(6)=28.9;
+KL(7)=16.1;
+KL(8)=2.9;
+tk =0;
+i=1;
+while(i<9)
+    tk = tk + KL(i);
+    i=i+1;
+end
+
+l(1)=1700; //Lb/hr
+l(2)=13900; //Lb/hr
+l(3)=13030; //Lb/hr
+l(4)=6260; //Lb/hr
+l(5)=4240; //Lb/hr
+l(6)=4330; //Lb/hr
+l(7)=2640; //Lb/hr
+l(8)=520; //Lb/hr
+
+tl=0;
+i=1;
+while(i<9)
+    tl = tl+l(i);
+    i=i+1;
+end
+
+printf("\n\t\tBUBBLE POINTS OF BOTTOMS\n");
+printf("\t\tMol/hr\t\tK(330°F,40psia)\t\tKL\t\tLb/hr\n");
+printf("\t\t--------------------------------------------------------------\n");
+i=1;
+while(i<9)
+    printf("\tC"+string(i+4)+"\t%.1f\t\t%.2f\t\t\t%.1f\t\t%.0f\n",bc(i),bK(i),KL(i),l(i));
+    i=i+1;
+end
+printf("\t\t____\t\t\t\t\t____\t\t____\n");
+printf("\t\t%.1f\t\t\t\t\t%.1f\t\t%.0f\n",tbc,tk,tl);
+av = tl/tk;
+printf("\tAverage mol. wt. %.1f\n",av);
+
+lh(1)=48894;//Lb/hr
+lh(2)=16298;//Lb/hr
+lh(3)=32596;//Lb/hr
+bl(1)=286;//Btu/hr
+bl(2)=129;//Btu/hr
+bl(3)=129;//Btu/hr
+i=1;
+while(i<4)
+    bh(i)=lh(i)*bl(i); //Btu/hr
+    i=i+1;
+end
+
+//Heat Balances
+printf("\n\n\t\t\t\t\t\tHEAT BALANCES:");
+printf("\n\t\t\t\tMol/hr\t\tMol.wt.\t\tLb/hr\t\tTemp,°F\t\tBtu/lb\t\tBtu/hr\n\t");
+printf("\t\t\t----------------------------------------------------------------------------------------");
+printf("\n\tHeat Balance onCondeser\n\t  Heat in:\n\t   Top plate vapor......");
+printf("685.2\t\t71.3\t\t" + string(lh(1)) + "\t\t148\t\t" +string(bl(1)) + "\t\t" + string(bh(1)) + "\n");
+printf("\t  Heat out:\n\t   Distillate...........");
+printf("228.4\t\t71.3\t\t" + string(lh(2)) + "\t\t124\t\t" +string(bl(2)) + "\t\t" + string(bh(2)) + "\n");
+printf("\t   Reflux, (2-1)........");
+printf("456.8\t\t71.3\t\t" + string(lh(3)) + "\t\t129\t\t" +string(bl(3)) + "\t\t" + string(bh(3)) + "\n");
+printf("\t   Condenser duty, by\n\t    difference......... ");
+printf(".....\t\t.....\t\t.....\t\t.....\t\t......\t\t7680000\n");
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t________\n");
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t"+string(bh(1))+"\n");
+//Heat Balances on reboiler
+//Assume 30° difference between reboiler and bottom plate giving bottom-plate temperature of 300°F
+//Mol/hr from Eq. 15.47
+rl(1)=78177;//Lb/hr
+rl(2)=22700;//Lb/hr
+rl(3)=55477;//Lb/hr
+rb(1)=234;//Btu/lb
+rb(2)=369;//Btu/lb
+rb(3)=256;//Btu/lb
+
+i=1;
+while(i<4)
+    rr(i)=rl(i)*rb(i);//Btu/hr
+    i=i+1;
+end
+tt = rr(1)+4280000;// Btu/hr
+printf("\t\t\t\t\t\tHEAT BALANCES on reboiler:");
+printf("\n\tHeat in:\n\t  Trapout...............");
+printf("619.7\t\t126.6\t\t"+string(rl(1))+"\t\t300\t\t"+string(rb(1))+"\t\t%.2e\n",rr(1));
+printf("\t  Reboiler duty........");
+printf(" .....\t\t.....\t\t......\t\t...\t\t...\t\t4280000\n");
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t________\n");
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t%.3e",tt);
+printf("\n\tHeat out:\n\t  Reboiler vapor........");
+printf("205.7\t\t110.3\t\t"+string(rl(2))+"\t\t330\t\t"+string(rb(2))+"\t\t%.2e\n",rr(2));
+printf("\t  Reboiler vapor........");
+printf("414.0\t\t134.0\t\t"+string(rl(3))+"\t\t330\t\t"+string(rb(3))+"\t\t%.2e\n",rr(3));
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t________\n");
+printf("\t\t\t\t\t\t\t\t\t\t\t\t\t\t%.3e",rr(2)+rr(3));
+
+//y*
+pc(1)=0.056;
+pc(2)=0.350;
+pc(3)=0.285;
+pc(4)=0.122;
+pc(5)=0.074;
+pc(6)=0.068;
+pc(7)=0.038;
+pc(8)=0.007;
+
+//K(300°F,40psia)
+pK(1)=4.5;
+pK(2)=2.25;
+pK(3)=1.20;
+pK(4)=0.66;
+pK(5)=0.38;
+pK(6)=0.22;
+pK(7)=0.13;
+pK(8)=0.07;
+
+printf("\n\n\t\tCALCULATION OF BOTTOM PLATE TEMPERATURE\n");
+printf("\t\ty*\t\t\tReboiler vapor\t\t\t\tK(300°F,40psia)\tMol*K\n\t\t\t\tV = y*205.7 +\tBottoms\t=\tTrapout\n");
+printf("\t\t----------------------------------------------------------------------------------------\n");
+
+i=1;
+pcs=0;
+pc2=0;
+bcs=0;
+tcs=0;
+gg=0;
+while(i<9)
+    temp = pc(i)*205.7;
+    temp2 = temp + bc(i);
+    printf("\tC"+ string(i+4)+ "\t" +string(pc(i))+ "\t\t%.1f\t\t" + string(bc(i))+"\t\t%.1f\t\t"+string(pK(i))+"\t\t%.2f\n",temp,temp2,temp2*pK(i));
+    
+    pcs=pcs+pc(i);
+    pc2=pc2+temp;
+    bcs=bcs+bc(i);
+    tcs=tcs+temp2;
+    gg=gg+(temp2*pK(i));
+    i=i+1;
+end
+printf("\t\t----------------------------------------------------------------------------------------\n");
+printf("\t\t%.3f\t\t%.1f\t\t%.1f\t\t%.1f\t\t\t\t%.1f\n",pcs,pc2,bcs,tcs,gg);
+printf("\n\tReboiler requirements are\n");
+printf("\t\tVaporization\t\t\t22700 lb/hr\n\t\tTotal liquor to reboiler\t78177 lb/hr\n\t\tHeat load\t\t\t4280000 Btu/hr\n\t\tTemperature range\t\t300-330°F\n\t\tOperating pressure\t\t40psia")
+//end