5 files changed, 428 insertions, 0 deletions
diff --git a/1328/CH16/EX16.1/16_1.sce b/1328/CH16/EX16.1/16_1.sce
new file mode 100644
index 000000000..94d8f7383
--- /dev/null
+++ b/1328/CH16/EX16.1/16_1.sce
@@ -0,0 +1,26 @@
+printf("\t example 16.1 \n");
+printf("\t approximate values are mentioned in the book \n");
+Af=(20*0.75*12*2)/(144);
+Ao=((3.14*1.25)-(20*0.035))*(12/144);
+printf("\t fin surface is : %.1f ft^2/lin ft \n",Af);
+printf("\t bare tube surface is : %.3f ft^2/lin ft \n",Ao);
+A=(Af+Ao);
+printf("\t total outside surface : %.2f ft^2/lin ft \n",A);
+Ai=(3.14*1.06*12)/(144);
+printf("\t total inside surface : %.3f ft^2/lin ft \n",Ai);
+printf("\t fin efficiencies \n");
+b=0.0625; // ft
+hf=4; // from table in solution
+m=(5.24*(hf^(1/2))); // m=((hf*P)/(Kax))^(1/2), eq 16.8
+n=(tanh(m*b))/(m*b); // efficiency , eq 16.26
+printf("\n hf      m      n \n "+string(hf)+"      "+string(m)+"      "+string(n)+" \n");
+// similarly efficiencies values are calculated at different hf values
+printf("\t weighted efficiency curve \n");
+hfi=((n*Af)+(Ao))*(hf/Ai); // eq 16.34
+printf("\n hf      hfi \n "+string(hf)+"      "+string(hfi)+" \n");
+// similarly efficiencies values are calculated at different hf values
+hf=[4 16 36 100 400 625 900]; // from 2nd table in the solution
+hfi=[35.4 110.8 193.5 370 935 1295 1700]; // from 2nd table in the solution
+plot2d("oll",hf,hfi);
+xtitle("weighted fin efficiency curve","heat transfer coefficient to fin,Btu/(ft^2)*(hr)","coefficient hf referred to the tube ID");
+//end
diff --git a/1328/CH16/EX16.2/16_2.sce b/1328/CH16/EX16.2/16_2.sce
new file mode 100644
index 000000000..0594ebde2
--- /dev/null
+++ b/1328/CH16/EX16.2/16_2.sce
@@ -0,0 +1,41 @@
+printf("\t example 16.2 \n");
+printf("\t approximate values are mentioned in the book \n");
+Ts=302; // F
+t1=151;
+t2=185;
+w=15200; // lb/hr
+// The dropwise condensation of steam was promoted with oil.
+aa=(3.14*(3.068^2-1.25^2))/(4*144)-((20*0.035*0.75)/(144));
+printf("\t annulus flow area : %.4f ft^2 \n",aa);
+p=(3.14*(1.25/12))-(20*0.035/12)+(20*0.75*2/12);
+printf("\t wetted perimeter : %.2f ft \n",p);
+De=(4*aa/p);
+printf("\t equivalent diameter : %.3f ft \n",De);
+Q=w*0.523*(t2-t1);
+printf("\t heat load : %.2e Btu/hr \n",Q);
+delt1=Ts-t1; //F
+delt2=Ts-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);
+Ai=0.277; // ft^2/ft
+n=20; // number of fins
+Ui=(Q/(Ai*n*LMTD));
+printf("\t Ui : %.0f Btu/(hr)*(ft^2)*(F) \n",Ui);
+hi=3000; // assumed value for dropwise condensation of steam
+hfi=(Ui*hi)/(hi-Ui);
+printf("\t hfi : %.0f Btu/(hr)*(ft^2)*(F) \n",hfi);
+hf=120; // from fig 16.7 for hfi=418
+mu=1.94; // lb/(ft*hr)
+k=0.079;
+Z=2.34; // Z=((c*mu)/k)^(1/3)
+jf=(hf*De/(Z*k)); // eq 16.36
+printf("\t jf : %.0f \n",jf);
+Ga=(w/aa);
+printf("\t Ga : %.2e lb/(hr)*(ft^2) \n",Ga);
+Rea=(De*Ga/mu);
+printf("\t Rea : %.2e \n",Rea);
+// end
+
+
diff --git a/1328/CH16/EX16.3/16_3.sce b/1328/CH16/EX16.3/16_3.sce
new file mode 100644
index 000000000..fb505be8a
--- /dev/null
+++ b/1328/CH16/EX16.3/16_3.sce
@@ -0,0 +1,128 @@
+printf("\t example 16.3 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=200; // outlet hot fluid,F
+t1=80; // inlet cold fluid,F
+t2=120; // outlet cold fluid,F
+W=18000; // lb/hr
+w=11950; // lb/hr
+printf("\t 1.for heat balance \n")
+C=0.53; // Btu/(lb)*(F)
+Q=((W)*(C)*(T1-T2)); // Btu/hr
+printf("\t total heat required for gas oil is : %.2e Btu/hr \n",Q);
+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 :%.0f F \n",LMTD);
+X=((delt1)/(delt2));
+printf("\t ratio of two local temperature difference is : %.2f \n",X);
+Fc=0.47; // from fig.17
+Kc=0.27; 
+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,gas oil \n");
+ID=3.068; // in, table 11
+OD=1.9; // in, table 11
+af=0.0175; // fin cross section,table 10
+aa=((3.14*ID^2/(4))-(3.14*OD^2/(4))-(24*af))/(144);
+printf("\t flow area is : %.4f ft^2 \n",aa);
+p=(3.14*(OD))-(24*0.035)+(24*0.5*2);
+printf("\t wetted perimeter : %.2f in \n",p);
+De=(4*aa*12/(p));
+printf("\t De : %.4f ft \n",De);
+Ga=(W/aa); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Ga);
+mu1=2.5*2.42; // at 224F,lb/(ft)*(hr), from fig.14
+Rea=((De)*(Ga)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Rea);
+jf=18.4; // from fig.16.10
+Z=0.25; // Z=k*((c)*(mu1)/k)^(1/3), fig 16
+Hf=((jf)*(1/De)*(Z)); // Hf=(hf/phya),using eq.6.15,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",Hf);
+printf("\t cold fluid:inner tube side,water \n");
+D=0.134; // ft
+row=62.5;
+at=(3.14*D^2/(4));
+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*row));
+printf("\t V is : %.2f fps \n",V);
+mu2=0.72*2.42; // at 99F,lb/(ft)*(hr)
+Ret=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.1e \n",Ret);
+hi=(970*0.82); // fig 25
+printf("\t hi : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);
+printf("\t calculation of tfw \n");
+// Tc-tfw=40F assumption from fig 14
+tfw=184;
+mufw=3.5; // cp, at 184F
+phya=(2.5/mufw)^0.14;
+printf("\t phya is : %.2f \n",phya); // from fig.24
+hf=(Hf)*(phya); // from eq.6.36
+printf("\t Correct hf to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",hf);
+Rdo=0.002;
+Rf=(1/hf);
+printf("\t Rf : %.4f \n",Rf);
+hf1=(1/(Rdo+Rf)); // eq 16.37
+printf("\t hf1 : %.1f \n",hf1);
+hfi1=255; // fig 16.9
+hfi2=(hf1*5.76); // eq 16.38 and fig 16.9,((Af+Ao)/(Ai))=5.76 from previous prblm
+printf("\t hfi2 : %.0f \n",hfi2);
+Rmetal=(hfi2-hfi1)/(hfi2*hfi1); // eq 16.39
+printf("\t Rmetal : %.5f \n",Rmetal);
+phyt=1; // for cooling water
+Rdi=0.003;
+Ri=(1/hi);
+printf("\t Ri : %.5f \n",Ri);
+hi1=(1/(Rdi+Ri)); // eq 16.40
+printf("\t hi1 : %.1f \n",hi1);
+UDi=(hi1*hfi1)/(hi1+hfi1); // eq 16.41
+printf("\t UDi : %.0f \n",UDi);
+// To obtain the true flux the heat load must be divided by the actual heat-transfer surface.For a 1}2-in. IPS pipe there are 0.422 ft2/lin foot, from table 11
+// trial
+Ai=(Q/(UDi*LMTD)); // LMTD=delt
+printf("\t Ai : %.1f ft^2 \n",Ai);
+L=(Ai/0.422);
+printf("\t length of pipe required : %.1f lin ft \n",L);
+// Use two 20-ft hairpins = 80 lin ft
+Ai1=(80*0.422); // ft^2
+r=(Q/Ai1);
+printf("\t Q/Ai1 : %.2e Btu/(hr)*(ft^2) \n",r);
+deltf=(r/hfi2);
+deltdo=(r*Rdo/5.76);
+printf("\t annulus film : %.1f \n",deltf);
+printf("\t annulus dirt : %.1f \n",deltdo);
+d=deltf+deltdo; // d=Tc-tfw
+deltmetal=(r*Rmetal);
+deltdi=(r*Rdi);
+delti=(r/hi);
+printf("\t Tc-tfw : %.1f \n",d);
+printf("\t fin and tube metal : %.1f \n",deltmetal);
+printf("\t tube side dirt : %.1f \n",deltdi);
+printf("\t tubeside film : %.1f \n",delti);
+Td=deltf+deltdo+deltmetal+deltdi+delti;
+printf("\t total temperature drop : %.1f F \n",Td);
+printf("\t pressure drop  for annulus \n");
+De1=0.0359; // ft
+Rea1=(De1*Ga/mu1);
+printf("\t reynolds number : %.2e \n",Rea1);
+f=0.00036; // fig 16.10
+s=0.82; //using fig.6
+delPs=((f*(Ga^2)*(80))/(5.22*(10^10)*(De1)*(s)*(phya))); // 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=0.000192; // friction factor for reynolds number 65000, using fig.26
+s=1;
+delPt=((f*(Gt^2)*(80))/(5.22*(10^10)*(0.134)*(s)*(1))); // using eq.7.45,psi
+printf("\t delPt is : %.1f psi \n",delPt);
+printf("\t allowable delPa is 10 psi \n");
+//end
diff --git a/1328/CH16/EX16.4/16_4.sce b/1328/CH16/EX16.4/16_4.sce
new file mode 100644
index 000000000..d40926e9a
--- /dev/null
+++ b/1328/CH16/EX16.4/16_4.sce
@@ -0,0 +1,111 @@
+printf("\t example 16.4 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=100; // outlet hot fluid,F
+t1=80; // inlet cold fluid,F
+t2=100; // outlet cold fluid,F
+W=30000; // lb/hr
+w=50500; // lb/hr
+printf("\t 1.for heat balance \n")
+C=0.225; // Btu/(lb)*(F)
+Q=((W)*(C)*(T1-T2)); // Btu/hr
+printf("\t total heat required for oxygwn is : %.2e Btu/hr \n",Q);
+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 : %.1f \n",R);
+S=((t2-t1)/(T1-t1));
+printf("\t S is : %.4f \n",S);
+printf("\t FT is 0.87 \n"); // from fig 18
+delt=(0.87*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 : %.0f F \n",tc);
+printf("\t hot fluid:shell side,oxygen \n");
+ID=19.25; // in, table 11
+OD=1; // in, table 11
+as=((3.14*ID^2/(4))-(70*3.14*OD^2/(4))-(70*20*0.035*0.5))/(144);
+printf("\t flow area is : %.2f ft^2 \n",as);
+p=(70*3.14*(OD))-(70*20*0.035)+(70*20*0.5*2);
+printf("\t wetted perimeter : %.2e in \n",p);
+De=(4*as*12/(p));
+printf("\t De : %.3f ft \n",De);
+Gs=(W/as); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs);
+mu1=0.0545; // at 175F,lb/(ft)*(hr), from fig.15
+Res=((De)*(Gs)/mu1); // reynolds number
+printf("\t reynolds number is : %.3e \n",Res);
+jH=59.5; // from fig.16.10a
+k=0.0175;
+Z=0.89; // Z=((c)*(mu1)/k)^(1/3), fig
+hf=((jH)*(k/De)*(Z)); //using eq.6.15,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",hf);
+Rdo=0.003;
+hdo=(1/Rdo);
+hf1=(hdo*hf)/(hdo+hf); // eq 16.37
+printf("\t hf1 : %.1f \n",hf1);
+hfi1=142; // fig 16.9
+printf("\t cold fluid:inner tube side,water \n");
+at1=0.479; // table 10
+L=16;
+Nt=70;
+n=4;
+at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48
+printf("\t flow area is : %.4f ft^2 \n",at);
+D=0.0652; // ft
+row=62.5;
+Gt=(w/(at)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+V=(Gt/(3600*row));
+printf("\t V is : %.2f fps \n",V);
+mu2=1.94; // at 90F,lb/(ft)*(hr)
+Ret=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+hi=(940*0.96); // fig 25
+printf("\t hi : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);
+Rdi=0.003;
+hdi=(1/Rdi);
+hi1=(hdi*hi)/(hdi+hi);
+printf("\t hi1 : %.0f Btu/(hr)*(ft^2)*(F) \n",hi1);
+UDi=((hfi1)*(hi1)/(hi1+hfi1)); // eq 16.41,Btu/(hr)*(ft^2)*(F)
+printf("\t overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UDi);
+A2=0.2048; // 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);
+UDi1=((Q)/((A)*(delt)));
+printf("\t design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UDi1);
+Re=(1/UDi1)-(1/UDi);
+printf("\t excess fouling factor : %.5f \n",Re);
+Ro=9.27; //Adding to the outside fouling factor
+Rdo1=Rdo+(Re*Ro);
+printf("\t Rdo : %.4f \n",Rdo1);
+hf2=(hf/(1+(hf*Rdo1)));
+printf("\t hf2 : %.1f \n",hf2);
+hfi2=113;
+UDi2=((hfi2)*(hi1)/(hi1+hfi2)); // eq 16.41,Btu/(hr)*(ft^2)*(F)
+printf("\t overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UDi2);
+printf("\t pressure drop  for annulus \n");
+De1=0.0433; // ft
+Res1=(De1*Gs/mu1);
+printf("\t reynolds number : %.2e \n",Res1);
+f=0.00025; // fig 16.10
+s=0.00133;
+delPs=((f*(Gs^2)*(L))/(5.22*(10^10)*(De1)*(s)*(1))); // using eq.7.44,psi
+printf("\t delPs is : %.1f psi \n",delPs);
+printf("\t allowable delPa is 2 psi \n");
+printf("\t pressure drop  for inner pipe \n");
+f=0.00021; // friction factor for reynolds number 29100, using fig.26
+s=1;
+delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(0.0625)*(s)*(1))); // using eq.7.45,psi
+printf("\t delPt is : %.0f psi \n",delPt);
+printf("\t allowable delPa is 10 psi \n");
+//end
diff --git a/1328/CH16/EX16.5/16_5.sce b/1328/CH16/EX16.5/16_5.sce
new file mode 100644
index 000000000..f563ea2f3
--- /dev/null
+++ b/1328/CH16/EX16.5/16_5.sce
@@ -0,0 +1,122 @@
+printf("\t example 16.5 \n");
+printf("\t approximate values are mentioned in the book \n");
+T1=250; // inlet hot fluid,F
+T2=200; // outlet hot fluid,F
+t1=150; // inlet cold fluid,F
+t2=190; // outlet cold fluid,F
+W=100000; // lb/hr
+w=31200; // lb/hr
+printf("\t 1.for heat balance \n")
+C=0.25; // Btu/(lb)*(F)
+Q=((W)*(C)*(T1-T2)); // Btu/hr
+printf("\t total heat required for air is : %.2e Btu/hr \n",Q);
+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 : %.1f \n",R);
+S=((t2-t1)/(T1-t1));
+printf("\t S is : %.4f \n",S);
+printf("\t FT is 0.985 \n"); // from fig 18
+delt=(0.985*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 : %.0f F \n",tc);
+Af=(3.14*2*8*12*(1.75^2-1^2))/(4);
+Ao=((3.14*1*12)-(3.14*1*8*0.035*12));
+printf("\t fin surface is : %.0f in^2/lin ft \n",Af);
+printf("\t bare tube surface is : %.1f in^2/lin ft \n",Ao);
+A=(Af+Ao);
+printf("\t total outside surface : %.1f ft^2/lin ft \n",A);
+p=(2*3*2*8*12/8)+(((12)-(8*0.035*12))*(2));
+printf("\t projected perimeter : %.1f in/ft \n",p);
+De=(2*A/(3.14*p*12)); // eq 16.104
+printf("\t De : %.3f ft \n",De);
+// 21 tubes may be fit in one :vertical bank (Fig. 16.19b) ,20 tubes in alternating banks for triangular pitch
+as=((4^2*12^2)-(21*1*48)-((21)*(2*0.035*3*8*48/8)))/(144); // fig 16.19
+printf("\t flow area : %.1f ft^2 \n",as);
+printf("\t hot fluid:shell side,oxygen \n");
+Gs=(W/as); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs);
+mu1=0.052; // at 225F,lb/(ft)*(hr), from fig.15
+Res=((De)*(Gs)/mu1); // reynolds number
+printf("\t reynolds number is : %.2e \n",Res);
+jf=157; // from fig.16.18a
+k=0.0183;
+Z=0.89; // Z=((c)*(mu1)/k)^(1/3), fig
+phys=1;
+hf=((jf)*(k/De)*(Z)); //using eq.6.15,Btu/(hr)*(ft^2)*(F)
+printf("\t individual heat transfer coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",hf);
+Rdo=0.003;
+hdo=(1/Rdo);
+hf1=(hdo*hf)/(hdo+hf); // eq 16.37
+printf("\t hf1 : %.1f \n",hf1);
+hfi1=142; // fig 16.9
+printf("\t cold fluid:inner tube side,water \n");
+at1=0.546; // table 10
+L=4;
+Nt=21;
+n=1;
+at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48
+printf("\t flow area is : %.4f ft^2 \n",at);
+D=0.0695; // ft
+row=62.5;
+Gt=(w/(at)); // mass velocity,lb/(hr)*(ft^2)
+printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt);
+V=(Gt/(3600*row));
+printf("\t V is : %.2f fps \n",V);
+mu2=0.895; // at 170F,lb/(ft)*(hr)
+Ret=((D)*(Gt)/mu2); // reynolds number
+printf("\t reynolds number is : %.2e \n",Ret);
+hi=(710*0.94); // fig 25
+printf("\t hi : %.0f Btu/(hr)*(ft^2)*(F) \n",hi);
+Rdi=0.003;
+hdi=(1/Rdi);
+hi1=(hdi*hi)/(hdi+hi); // 16.40
+printf("\t hi1 : %.0f Btu/(hr)*(ft^2)*(F) \n",hi1);
+k1=60; // table 3 , for brass
+// yb=0.00146 ft
+X=((0.875-0.5)/12)*(21.5/(60*0.00146))^(1/2);
+printf("\t X :%.2f \n",X);
+nf=0.91; // from fig 16.13a , by comparing X value
+Ai=0.218; // ft^2/ft
+hfi2=((nf*Af/144)+(Ao/144))*(hf1/Ai); // eq 16.34
+printf("\t hfi2 : %.0f \n",hfi2);
+UDi=((hfi2)*(hi1)/(hi1+hfi2)); // eq 16.41,Btu/(hr)*(ft^2)*(F)
+printf("\t overall coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",UDi);
+A=(21*4*Ai); // ft^2
+printf("\t inside surface per bank is : %.1f ft^2 \n",A);
+Ai1=(Q/(UDi*delt));
+printf("\t Ai1 : %.0f ft^2 \n",Ai1);
+Nb=(Ai1/A);
+printf("\t number of banks : %.0f \n",Nb);
+Vn=(4*4*1.95/12)-(41*3.14*1*4/(2*4*144))-((41*3.14*0.035*8*4/(144*2*4))*(1.75^2-1^2)); // fig 16.19b
+printf("\t net free volume : %.2f ft^3 \n",Vn);
+Af1=(41*2.34*4/2);
+printf("\t frictional surface : %.0f ft^2 \n",Af1);
+printf("\t pressure drop  for annulus \n");
+De1=(4*Vn/Af1); // ft
+printf("\t De1 : %.2f ft \n",De1);
+Res1=(De1*Gs/mu1);
+printf("\t reynolds number : %.2e \n",Res1);
+f=0.0024; // fig 16.18b
+s=0.000928;
+Lp=1.95;
+R1=0.538; // R1=(De1/ST)^(0.4)
+R2=1; // R2=(SL/ST)^0.6
+delPs=((f*(Gs^2)*(Lp)*(R1)*(R2))/(5.22*(10^10)*(De1)*(s)*(1)));
+printf("\t delPs is : %.2f psi \n",delPs);
+printf("\t pressure drop  for inner pipe \n");
+f=0.0002; // friction factor for reynolds number 30400, using fig.26
+s=1;
+delPt=((f*(Gt^2)*(L)*(Nb))/(5.22*(10^10)*(0.0695)*(s)*(1))); // using eq.7.45,psi
+printf("\t delPt is : %.2f psi \n",delPt);
+//end