initial commit / add all books

9 files changed, 556 insertions, 0 deletions

diff --git a/3516/CH7/EX7.1/Ex7_1.sce b/3516/CH7/EX7.1/Ex7_1.sce
new file mode 100644
index 000000000..54b1aebd6
--- /dev/null
+++ b/3516/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,8 @@
+printf("\t example 7.1 \n");

+PT=1; // square pitch,in

+do=0.75; // outer diameter,in

+de=((4*(PT^2-(3.14*do^2/4)))/(3.14*do));

+printf("\t equivalent diameter is : %.2f in \n",de);

+De=(de/12); // ft

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

+//end

diff --git a/3516/CH7/EX7.2/Ex7_2.sce b/3516/CH7/EX7.2/Ex7_2.sce
new file mode 100644
index 000000000..a9fe79bb9
--- /dev/null
+++ b/3516/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,42 @@
+printf("\t example 7.2 \n");

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

+printf("\t considering 50F approach \n");

+T1=350; //F

+T2=250; //F

+t2=T2-50; // formula for approach,f

+printf("\t t2 is : %.0f F \n",t2);

+printf("\t fluids are with equal ranges,so \n");

+t1=t2-(T1-T2); // F

+printf("\t t1 is : %.0f F \n",t1);

+R=((T1-T2)/(t2-t1));

+printf("\t R is : %.0f \n",R);

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

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

+printf("\t FT is 0.925 \n"); // from fig 18

+printf("\t considering 0F approach \n");

+T1=300; //F

+T2=200; //F

+t2=T2-0; // formula for approach,f

+printf("\t t2 is : %.0f F \n",t2);

+printf("\t fluids are with equal ranges,so \n");

+t1=t2-(T1-T2); // F

+printf("\t t1 is : %.0f F \n",t1);

+R=((T1-T2)/(t2-t1));

+printf("\t R is : %.0f \n",R);

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

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

+printf("\t FT is 0.80 \n"); // from fig 18

+printf("\t considering 20F cross \n");

+T1=280; //F

+T2=180; //F

+t2=T2+20; // formula for approach,f

+printf("\t t2 is : %.0f F \n",t2);

+printf("\t fluids are with equal ranges,so \n");

+t1=t2-(T1-T2); // F

+printf("\t t1 is : %.0f F \n",t1);

+R=((T1-T2)/(t2-t1));

+printf("\t R is : %.0f \n",R);

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

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

+printf("\t FT is 0.64 \n"); // from fig 18

+//end

diff --git a/3516/CH7/EX7.3/Ex7_3.sce b/3516/CH7/EX7.3/Ex7_3.sce
new file mode 100644
index 000000000..0c5e7dc86
--- /dev/null
+++ b/3516/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,124 @@
+printf("\t example 7.3 \n");

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

+T1=390; // inlet hot fluid,F

+T2=200; // outlet hot fluid,F

+t1=100; // inlet cold fluid,F

+t2=170; // outlet cold fluid,F

+W=43800; // lb/hr

+w=149000; // lb/hr

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

+printf("\t for kerosene \n");

+c=0.605; // Btu/(lb)*(F)

+Q1=((W)*(c)*(T1-T2)); // Btu/hr

+printf("\t total heat required for kerosene is : %.1e Btu/hr \n",Q1); // calculation mistake in problem

+printf("\t for crude oil \n");

+c=0.49; // Btu/(lb)*(F)

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

+printf("\t total heat required for mid continent crude is : %.1e Btu/hr \n",Q); // calculation mistake in problem

+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 0.905 \n"); // from fig 18

+delt=(0.905*LMTD); // F

+printf("\t delt is : %.0f 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)+((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:shell side,kerosene \n");

+ID=21.25; // 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 : %.4f 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=0.40*2.42; // at 280F,lb/(ft)*(hr), from fig.14

+De=0.99/12; // from fig.28,ft

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

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

+jH=93; // from fig.28

+c=0.59; // Btu/(lb)*(F),at 280F,from fig.4

+k=0.0765; // Btu/(hr)*(ft^2)*(F/ft), from fig.1

+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)); // 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);

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

+D=0.0675; // ft

+Nt=158;

+n=4; // number of passes

+L=16; //ft

+at1=0.515; // flow area, 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);

+mu2=3.6*2.42; // at 129F,lb/(ft)*(hr)

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

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

+jH=31; // from fig.24

+c=0.49; // Btu/(lb)*(F),at 304F,from fig.4

+k=0.077; // Btu/(hr)*(ft^2)*(F/ft), from fig.1

+Pr=((c)*(mu2)/k)^(1/3); // prandelt number raised to power 1/3

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

+Hi=((jH)*(k/D)*(Pr)*(1^0.14)); //Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft^2)*(F)

+printf("\t Hi is : %.0f Btu/(hr)*(ft^2)*(F) \n",Hi);

+ID=0.81; // 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 : %.0f Btu/(hr)*(ft^2)*(F) \n",Hio);

+muw=1.5*2.42; // 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 : %.0f Btu/(hr)*(ft^2)*(F) \n",hio);

+tw=(tc)+(((Ho)/(Hio+Ho))*(Tc-tc)); // from eq.5.31

+printf("\t tw is : %.0f F \n",tw);

+muw=0.56*2.42; // 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);

+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);

+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 : %.5f (hr)*(ft^2)*(F)/Btu \n",Rd);

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

+f=0.00175; // friction factor for reynolds number 25300, using fig.29

+s=0.73; // for reynolds number 25300,using fig.6

+Ds=21.25/12; // ft

+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 delPa is 10 psi \n");

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

+f=0.000285; // friction factor for reynolds number 8220, using fig.26

+s=0.83;

+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=0.15; // 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 delPs is 10 psi \n");

+//end

diff --git a/3516/CH7/EX7.4/Ex7_4.sce b/3516/CH7/EX7.4/Ex7_4.sce
new file mode 100644
index 000000000..046cd5410
--- /dev/null
+++ b/3516/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,112 @@
+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

diff --git a/3516/CH7/EX7.5/Ex7_5.sce b/3516/CH7/EX7.5/Ex7_5.sce
new file mode 100644
index 000000000..f5d3b4c92
--- /dev/null
+++ b/3516/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,23 @@
+printf("\t example 7.5 \n");

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

+T1=175; // inlet hot fluid,F

+T2=150; // outlet hot fluid,F

+t1=85; // inlet cold fluid,F

+delt1=T2-t1; //F

+printf("\t delt1 is : %.0f F \n",delt1);

+U=15; // assumption,Btu/(hr)*(ft^2)*(F)

+theta=8000; // operating hours,hr

+CW=(0.01/8300); // water cost,$/lb

+printf("\t For annual charges assume 20 per cent repair and maintenanc.e and 10 per cent depreciation \n");

+CF=(0.3*4); // annual fixed charges/ft^2

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

+X=((U)*(theta)*(CW)/(CF*c));

+printf("\t X is : %.4f \n",X);

+Y=((T1-T2)/delt1);

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

+A=0.96; // A=(delt2/delt1), from fig 7.24

+delt2=0.96*delt1;

+printf("\t delt2 is : %.1f F \n",delt2);

+t2=T1-delt2; // F

+printf("\t t2 is : %.1f F \n",t2);

+//end

diff --git a/3516/CH7/EX7.6/Ex7_6.sce b/3516/CH7/EX7.6/Ex7_6.sce
new file mode 100644
index 000000000..0b79c9742
--- /dev/null
+++ b/3516/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,107 @@
+printf("\t example 7.6 \n");

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

+T1=150; // inlet hot fluid,F

+T2=90; // outlet hot fluid,F

+t1=68; // inlet cold fluid,F

+t2=90; // outlet cold fluid,F

+W=20160; // lb/hr

+w=41600; // lb/hr

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

+printf("\t for solution \n");

+c=(0.3*0.19)+(0.7*1); // Btu/(lb)*(F), bcoz of 30 percent of solution

+Q1=((W)*(c)*(T1-T2)); // Btu/hr

+printf("\t total heat required for solution is : %.2e Btu/hr \n",Q1);

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

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

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

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

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

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

+printf("\t FT is 0.81 \n"); // from fig 18

+delt=(0.81*LMTD); // F

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

+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,phosphate solution \n");

+ID=10.02; // in

+C=0.25; // clearance

+B=2; // baffle spacing,in

+PT=1;

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

+printf("\t flow area is : %.4f ft^2 \n",as);

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

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

+mu1=1.20*2.42; // at 120F,lb/(ft)*(hr), from fig.14

+De=0.95/12; // from fig.28,ft

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

+printf("\t reynolds number is : %.3e \n",Res);

+jH=71; // from fig.28

+c=1; // Btu/(lb)*(F),at 120F,from fig.table 4

+k=0.33; // Btu/(hr)*(ft^2)*(F/ft), from table 4

+Pr=((0.757)*(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 : %.0f Btu/(hr)*(ft^2)*(F) \n",ho);

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

+Nt=52;

+n=2; // number of passes

+L=16; //ft

+at1=0.302; // 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 : %.4f 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);

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

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

+mu2=0.91*2.42; // at 79F,lb/(ft)*(hr),from table 14

+D=(0.62/12); // from table 10

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

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

+hi=800*1; //using fig.25,Btu/(hr)*(ft^2)*(F)

+ID=0.62; // 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);

+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 : %.5f (hr)*(ft^2)*(F)/Btu \n",Rd);

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

+f=0.0019; // friction factor for reynolds number 15750, using fig.29

+s=1.3; // for reynolds number 25300,using fig.6

+Ds=10.02/12; // 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=0.00023; // friction factor for reynolds number 17900, using fig.26

+s=1;

+phyt=1;

+D=0.0517; // 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=0.08; // 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

diff --git a/3516/CH7/EX7.7/Ex7_7.sce b/3516/CH7/EX7.7/Ex7_7.sce
new file mode 100644
index 000000000..f56c20bcd
--- /dev/null
+++ b/3516/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,22 @@
+printf("\t example 7.7 \n");

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

+U=50; // Btu/(hr)*(ft^2)*(F)

+TP=328; // F

+TE=228; // F

+CP=(0.30/(888.8*1000));

+CE=(0.05/(960*1000));

+CF=1.20;

+theta=8000; // annual hours

+X=((CF*(TP-TE))/((CP-CE)*U*theta)); // from eq 7.53

+printf("\t X is : %.9f \n",X);

+a=(1); // coefficient of t^2

+b=(-556); // coefficient of t

+c=(74784-X); // constant

+printf("\t coefficient of t^2 is : %.2f \n",a);

+printf("\t coefficient of t is : %.2f \n",b);

+printf("\t constant term is : %.9f \n",c);

+P=poly([c b a], 't','c');

+t=roots(P);

+printf("\t t is :%.0f \n",t);

+printf("\t t cannot be greater than 328F \n \t t is 218F \n");

+//end

diff --git a/3516/CH7/EX7.8/Ex7_8.sce b/3516/CH7/EX7.8/Ex7_8.sce
new file mode 100644
index 000000000..d1475bda9
--- /dev/null
+++ b/3516/CH7/EX7.8/Ex7_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

diff --git a/3516/CH7/EX7.9/Ex7_9.sce b/3516/CH7/EX7.9/Ex7_9.sce
new file mode 100644
index 000000000..915e8f1ca
--- /dev/null
+++ b/3516/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,20 @@
+printf("\t example 7.9 \n");

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

+T1=390; // F

+t1=100; // F

+U=69.3; // Btu/(hr)*(ft^2)*(F)

+A=662; // ft^2

+W=43800; // lb/hr

+w=149000; // lb/hr

+C=0.60; // Btu/(lb)*(F)

+c=0.49; // Btu/(lb)*(F)

+X=((U*A)/(w*c));

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

+R=((w*c)/(W*C));

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

+S=0.265; // from fig 7.25, by comparing X an R

+t2=(t1)+((0.265)*(T1-t1)); // S=((t2-t1)/(T1-t1))

+printf("\t t2 is : %.0f F \n",t2);

+T2=((T1)-((R)*(t2-t1))); // R=((T1-T2)/(t2-t1))

+printf("\t T2 is : %.0f F \n",T2);

+// end