# Chapter 16: Extented Surfaces¶

## Example 16.1 pgno: 521¶

In [10]:
print"\t example 16.1 \t"
print"\t approximate values are mentioned in the book \t"
import numpy
import matplotlib
%matplotlib inline
from matplotlib import pyplot
from math import tanh
Af=(20*0.75*12*2)/(144);
Ao=((3.14*1.25)-(20*0.035))*(12/144);
print"\t fin surface is : %.1f ft**2/lin ft \t",Af
print"\t bare tube surface is :  ft**2/lin ft \t",Ao
A=(Af+Ao);
print"\t total outside surface :  ft**2/lin ft \t",A
Ai=(3.14*1.06*12)/(144);
print"\t total inside surface :  ft**2/lin ft \t",Ai
print"\t fin efficiencies \t"
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
print"\t hf      m      n \t "
print hf
print m
print n
# similarly efficiencies values are calculated at different hf values
print"\t weighted efficiency curve \t"
hfi=((n*Af)+(Ao))*(hf/Ai); # eq 16.34
print"\t hf      hfi \t "
print hf
print hfi
# 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
pyplot.plot(hf,hfi)
pyplot.xlabel('heat transfer coefficient to fin,Btu/(ft**2)*(hr)')
pyplot.ylabel('coefficient hf referred to the tube ID')
pyplot.title('weighted fin efficiency curve')
pyplot.show()
#end

	 example 16.1
approximate values are mentioned in the book
fin surface is : %.1f ft**2/lin ft 	2.5
bare tube surface is :  ft**2/lin ft 	0.0
total outside surface :  ft**2/lin ft 	2.5
total inside surface :  ft**2/lin ft 	0.277366666667
fin efficiencies
hf      m      n
4
5.24
0.965717954492
weighted efficiency curve
hf      hfi
4
34.8173760783


## Example 16.2 pgno:525¶

In [11]:
print"\t example 16.2 \t"
print"\t approximate values are mentioned in the book \t"
Ts=302.; # F
t1=151.;
t2=185.;
w=15200.; # lb/hr
from math import log10
# 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));
print"\t annulus flow area : ft**2 \t",aa
p=(3.14*(1.25/12))-(20*0.035/12)+(20*0.75*2/12);
print"\t wetted perimeter : ft \t",p
De=(4*aa/p);
print"\t equivalent diameter : ft \t",De
Q=w*0.523*(t2-t1);
print"\t heat load : Btu/hr \t",Q
delt1=Ts-t1; #F
delt2=Ts-t2; # F
print"\t delt1 is : F \t",delt1
print"\t delt2 is : F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is  F \t",LMTD
Ai=0.277; # ft**2/ft
n=20; # number of fins
Ui=(Q/(Ai*n*LMTD));
print"\t Ui : Btu/(hr)*(ft**2)*(F) \t",Ui
hi=3000; # assumed value for dropwise condensation of steam
hfi=(Ui*hi)/(hi-Ui);
print"\t hfi : Btu/(hr)*(ft**2)*(F) \t",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
print"\t jf : \t",jf
Ga=(w/aa);
print"\t Ga : lb/(hr)*(ft**2) \t",Ga
Rea=(De*Ga/mu);
print"\t Rea : \t",round(Rea)
# end

	 example 16.2
approximate values are mentioned in the book
annulus flow area : ft**2 	0.0391482454167
wetted perimeter : ft 	2.76875
equivalent diameter : ft 	0.0565572845749
delt1 is : F 	151.0
delt2 is : F 	117.0
LMTD is  F 	133.427780969
Ui : Btu/(hr)*(ft**2)*(F) 	365.652178958
hfi : Btu/(hr)*(ft**2)*(F) 	416.40535396
jf : 	36.7135894676
Ga : lb/(hr)*(ft**2) 	388267.720257
Rea : 	11319.0


## Example 16.3 pgno:530¶

In [12]:
print"\t example 16.3 \t"
print"\t approximate values are mentioned in the book \t"
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
from math import log10
print"\t 1.for heat balance \t"
C=0.53; # Btu/(lb)*(F)
Q=((W)*(C)*(T1-T2)); # Btu/hr
print"\t total heat required for gas oil is :  Btu/hr \t",Q
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is :%.0f F \t",LMTD
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is :  \t",X
Fc=0.47; # from fig.17
Kc=0.27;
Tc=((T2)+((Fc)*(T1-T2))); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+((Fc)*(t2-t1))); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,gas oil \t"
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);
print"\t flow area is :  ft**2 \t",aa
p=(3.14*(OD))-(24*0.035)+(24*0.5*2);
print"\t wetted perimeter :  in \t",p
De=(4*aa*12/(p));
print"\t De :  ft \t",De
Ga=(W/aa); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Ga
mu1=2.5*2.42; # at 224F,lb/(ft)*(hr), from fig.14
Rea=((De)*(Ga)/mu1); # reynolds number
print"\t reynolds number is :  \t",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)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Hf
print"\t cold fluid:inner tube side,water \t"
D=0.134; # ft
row=62.5;
at=(3.14*D**2/(4));
print"\t flow area is :  ft**2 \t",at
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
V=(Gt/(3600*row));
print"\t V is :  fps \t",V
mu2=0.72*2.42; # at 99F,lb/(ft)*(hr)
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=(970*0.82); # fig 25
print"\t hi :  Btu/(hr)*(ft**2)*(F) \t",hi
print"\t calculation of tfw \t"
# Tc-tfw=40F assumption from fig 14
tfw=184;
mufw=3.5; # cp, at 184F
phya=(2.5/mufw)**0.14;
print"\t phya is :  \t",phya # from fig.24
hf=(Hf)*(phya); # from eq.6.36
print"\t Correct hf to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hf
Rdo=0.002;
Rf=(1/hf);
print"\t Rf :  \t",Rf
hf1=(1/(Rdo+Rf)); # eq 16.37
print"\t hf1 :  \t",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
print"\t hfi2 :  \t",hfi2
Rmetal=(hfi2-hfi1)/(hfi2*hfi1)# eq 16.39
print"\t Rmetal :  \t",Rmetal
phyt=1; # for cooling water
Rdi=0.003;
Ri=(1/hi);
print"\t Ri :  \t",Ri
hi1=(1/(Rdi+Ri)); # eq 16.40
print"\t hi1 :  \t",hi1
UDi=(hi1*hfi1)/(hi1+hfi1); # eq 16.41
print"\t UDi :  \t",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
print"\t Ai :  ft**2 \t",Ai
L=(Ai/0.422);
print"\t length of pipe required :  lin ft \t",L
# Use two 20-ft hairpins = 80 lin ft
Ai1=(80*0.422); # ft**2
r=(Q/Ai1);
print"\t Q/Ai1 :  Btu/(hr)*(ft**2) \t",r
deltf=(r/hfi2);
deltdo=(r*Rdo/5.76);
print"\t annulus film :  \t",deltf
print"\t annulus dirt :  \t",deltdo
d=deltf+deltdo; # d=Tc-tfw
deltmetal=(r*Rmetal);
deltdi=(r*Rdi);
delti=(r/hi);
print"\t Tc-tfw :  \t",d
print"\t fin and tube metal :  \t",deltmetal
print"\t tube side dirt :  \t",deltdi
print"\t tubeside film :  \t",delti
Td=deltf+deltdo+deltmetal+deltdi+delti;
print"\t total temperature drop :  F \t",Td
print"\t pressure drop  for annulus \t"
De1=0.0359; # ft
Rea1=(De1*Ga/mu1);
print"\t reynolds number :  \t",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
print"\t delPs is :  psi \t",round(delPs,1)
print"\t allowable delPa is 10 psi \t"
print"\t pressure drop  for inner pipe \t"
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
print"\t delPt is :  psi \t",round(delPt,1)
print"\t allowable delPa is 10 psi \t"
#end

	 example 16.3
approximate values are mentioned in the book
1.for heat balance
total heat required for gas oil is :  Btu/hr 	477000.0
total heat required for water is :  Btu/hr 	478000.0
delt1 is :  F 	120.0
delt2 is :  F 	130.0
LMTD is :%.0f F 	125.073724087
ratio of two local temperature difference is :  	0.923076923077
caloric temperature of hot fluid is :  F 	223.5
caloric temperature of cold fluid is :  F 	98.8
hot fluid:shell side,gas oil
flow area is :  ft**2 	0.0287156933333
wetted perimeter :  in 	29.126
De :  ft 	0.0473238096546
mass velocity is :  lb/(hr)*(ft**2) 	626834.943216
reynolds number is :  	4903.1764525
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	97.2026562015
cold fluid:inner tube side,water
flow area is :  ft**2 	0.01409546
mass velocity is :  lb/(hr)*(ft**2) 	847790.707079
V is :  fps 	3.76795869813
reynolds number is :  	65199.6985471
hi :  Btu/(hr)*(ft**2)*(F) 	795.4
calculation of tfw
phya is :  	0.953986161765
Correct hf to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	92.729988903
Rf :  	0.0107839978396
hf1 :  	78.2227916919
hfi2 :  	450.563280145
Rmetal :  	0.00170212455807
Ri :  	0.00125722906714
hi1 :  	234.894572087
UDi :  	122.267359744
Ai :  ft**2 	31.2572870644
length of pipe required :  lin ft 	74.0694006266
Q/Ai1 :  Btu/(hr)*(ft**2) 	14158.7677725
annulus film :  	31.4245931625
annulus dirt :  	4.9162388099
Tc-tfw :  	36.3408319724
fin and tube metal :  	24.0999863376
tube side dirt :  	42.4763033175
tubeside film :  	17.8008143984
total temperature drop :  F 	120.717936026
pressure drop  for annulus
reynolds number :  	3719.56602669
delPs is :  psi 	7.7
allowable delPa is 10 psi
pressure drop  for inner pipe
delPt is :  psi 	1.6
allowable delPa is 10 psi


## Example 16.4 pgno:535¶

In [13]:
print"\t example 16.4 \t"
print"\t approximate values are mentioned in the book \t"
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
from math import log10
print"\t 1.for heat balance \t"
C=0.225; # Btu/(lb)*(F)
Q=((W)*(C)*(T1-T2)); # Btu/hr
print"\t total heat required for oxygwn is :  Btu/hr \t",Q
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is :  \t",R
S=((t2-t1)/(T1-t1));
print"\t S is :  \t",S
print"\t FT is 0.87 \t"# from fig 18
delt=(0.87*LMTD); # F
print"\t delt is :  F \t",delt
Tc=(T2+T1)/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t2))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,oxygen \t"
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);
print"\t flow area is :  ft**2 \t",As
p=(70*3.14*(OD))-(70*20*0.035)+(70*20*0.5*2);
print"\t wetted perimeter :  in \t",p
De=(4*As*12/(p));
print"\t De :  ft \t",De
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.0545; # at 175F,lb/(ft)*(hr), from fig.15
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",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)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",hf
Rdo=0.003;
hdo=(1/Rdo);
hf1=(hdo*hf)/(hdo+hf); # eq 16.37
print"\t hf1 :  \t",hf1
hfi1=142; # fig 16.9
print"\t cold fluid:inner tube side,water \t"
at1=0.479; # table 10
L=16;
Nt=70;
n=4;
at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48
print"\t flow area is :  ft**2 \t",at
D=0.0652; # ft
row=62.5;
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
V=(Gt/(3600*row));
print"\t V is :  fps \t",V
mu2=1.94; # at 90F,lb/(ft)*(hr)
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=(940*0.96); # fig 25
print"\t hi :  Btu/(hr)*(ft**2)*(F) \t",hi
Rdi=0.003;
hdi=(1/Rdi);
hi1=(hdi*hi)/(hdi+hi);
print"\t hi1 :  Btu/(hr)*(ft**2)*(F) \t",hi1
UDi=((hfi1)*(hi1)/(hi1+hfi1)); # eq 16.41,Btu/(hr)*(ft**2)*(F)
print"\t overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UDi
A2=0.2048; # actual surface supplied for each tube,ft**2,from table 10
A=(Nt*L*A2); # ft**2
print"\t total surface area is :  ft**2 \t",A
UDi1=((Q)/((A)*(delt)));
print"\t design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UDi1
Re=(1/UDi1)-(1/UDi);
print"\t excess fouling factor :  \t",Re
Ro=9.27; #Adding to the outside fouling factor
Rdo1=Rdo+(Re*Ro);
print"\t Rdo :  \t",Rdo1
hf2=(hf/(1+(hf*Rdo1)));
print"\t hf2 :  \t",hf2
hfi2=113;
UDi2=((hfi2)*(hi1)/(hi1+hfi2)); # eq 16.41,Btu/(hr)*(ft**2)*(F)
print"\t overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UDi2
print"\t pressure drop  for annulus \t"
De1=0.0433; # ft
Res1=(De1*Gs/mu1);
print"\t reynolds number :  \t",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
print"\t delPs is :  psi \t",round(delPs,1)
print"\t allowable delPa is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
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
print"\t delPt is :  psi \t",round(delPt)
print"\t allowable delPa is 10 psi \t"
#end

	 example 16.4
approximate values are mentioned in the book
1.for heat balance
total heat required for oxygwn is :  Btu/hr 	1012500.0
total heat required for water is :  Btu/hr 	1010000
delt1 is :  F 	20
delt2 is :  F 	150
LMTD is : F 	66.8818722257
R is :  	7
S is :  	0
FT is 0.87
delt is :  F 	58.1872288364
caloric temperature of hot fluid is :  F 	175
caloric temperature of cold fluid is :  F 	90
hot fluid:shell side,oxygen
flow area is :  ft**2 	1.46834418403
wetted perimeter :  in 	1570.8
De :  ft 	0.0448691882056
mass velocity is :  lb/(hr)*(ft**2) 	20431.1770539
reynolds number is :  	16820.7399724
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	20.6536497998
hf1 :  	19.4485963081
cold fluid:inner tube side,water
flow area is :  ft**2 	0.0582118055556
mass velocity is :  lb/(hr)*(ft**2) 	867521.622428
V is :  fps 	3.85565165523
reynolds number is :  	29155.8813311
hi :  Btu/(hr)*(ft**2)*(F) 	902.4
hi1 :  Btu/(hr)*(ft**2)*(F) 	243.418213207
overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	89.6828045249
total surface area is :  ft**2 	229.376
design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	75.673831631
excess fouling factor :  	0.00206419817483
Rdo :  	0.0221351170807
hf2 :  	14.1738000666
overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	77.1741091592
pressure drop  for annulus
reynolds number :  	16232.4764484
delPs is :  psi 	0.6
allowable delPa is 2 psi
pressure drop  for inner pipe
delPt is :  psi 	3.0
allowable delPa is 10 psi


## Example 16.5 pgno:556¶

In [14]:
print"\t example 16.5 \t"
print"\t approximate values are mentioned in the book \t"
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\
from math import log10
print"\t 1.for heat balance \t"
C=0.25; # Btu/(lb)*(F)
Q=((W)*(C)*(T1-T2)); # Btu/hr
print"\t total heat required for air is :  Btu/hr \t",Q
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is :  \t",R
S=((t2-t1)/(T1-t1));
print"\t S is :  \t",S
print"\t FT is 0.985 \t" # from fig 18
delt=(0.985*LMTD); # F
print"\t delt is :  F \t",delt
Tc=(T2+T1)/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t2))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",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));
print"\t fin surface is :  in**2/lin ft \t",Af
print"\t bare tube surface is :  in**2/lin ft \t",Ao
A=(Af+Ao);
print"\t total outside surface :  ft**2/lin ft \t",A
p=(2*3*2*8*12/8)+(((12)-(8*0.035*12))*(2));
print"\t projected perimeter :  in/ft \t",p
De=(2*A/(3.14*p*12)); # eq 16.104
print"\t De :  ft \t",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
print"\t flow area :  ft**2 \t",As
print"\t hot fluid:shell side,oxygen \t"
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.052; # at 225F,lb/(ft)*(hr), from fig.15
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",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)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",hf
Rdo=0.003;
hdo=(1/Rdo);
hf1=(hdo*hf)/(hdo+hf); # eq 16.37
print"\t hf1 :  \t",hf1
hfi1=142; # fig 16.9
print"\t cold fluid:inner tube side,water \t"
at1=0.546; # table 10
L=4;
Nt=21;
n=1;
at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48
print"\t flow area is :  ft**2 \t",at
D=0.0695; # ft
row=62.5;
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
V=(Gt/(3600*row));
print"\t V is :  fps \t",V
mu2=0.895; # at 170F,lb/(ft)*(hr)
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=(710*0.94); # fig 25
print"\t hi :  Btu/(hr)*(ft**2)*(F) \t",hi
Rdi=0.003;
hdi=(1/Rdi);
hi1=(hdi*hi)/(hdi+hi); # 16.40
print"\t hi1 :  Btu/(hr)*(ft**2)*(F) \t",hi1
k1=60; # table 3 , for brass
# yb=0.00146 ft
X=((0.875-0.5)/12)*(21.5/(60*0.00146))**(1/2);
print"\t X : \t",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
print"\t hfi2 :  \t",hfi2
UDi=((hfi2)*(hi1)/(hi1+hfi2)); # eq 16.41,Btu/(hr)*(ft**2)*(F)
print"\t overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UDi
A=(21*4*Ai); # ft**2
print"\t inside surface per bank is :  ft**2 \t",A
Ai1=(Q/(UDi*delt));
print"\t Ai1 :  ft**2 \t",Ai1
Nb=(Ai1/A);
print"\t number of banks :  \t",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
print"\t net free volume :  ft**3 \t",Vn
Af1=(41*2.34*4/2);
print"\t frictional surface :  ft**2 \t",Af1
print"\t pressure drop  for annulus \t"
De1=(4*Vn/Af1); # ft
print"\t De1 :  ft \t",De1
Res1=(De1*Gs/mu1);
print"\t reynolds number :  \t",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)));
print"\t delPs is :  psi \t",delPs
print"\t pressure drop  for inner pipe \t"
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
print"\t delPt is :  psi \t",round(delPt,2)
#end

	 example 16.5
approximate values are mentioned in the book
1.for heat balance
total heat required for air is :  Btu/hr 	1250000.0
total heat required for water is :  Btu/hr 	1248000.0
delt1 is :  F 	50.0
delt2 is :  F 	60.0
LMTD is : F 	54.9097962457
R is :  	1.25
S is :  	0.4
FT is 0.985
delt is :  F 	54.086149302
caloric temperature of hot fluid is :  F 	225.0
caloric temperature of cold fluid is :  F 	170.0
fin surface is :  in**2/lin ft 	310.86
bare tube surface is :  in**2/lin ft 	27.1296
total outside surface :  ft**2/lin ft 	337.9896
projected perimeter :  in/ft 	161.28
De :  ft 	0.111235119048
flow area :  ft**2 	7.53
hot fluid:shell side,oxygen
mass velocity is :  lb/(hr)*(ft**2) 	13280.2124834
reynolds number is :  	28408.1926263
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	22.9878748896
hf1 :  	21.5048242607
cold fluid:inner tube side,water
flow area is :  ft**2 	0.079625
mass velocity is :  lb/(hr)*(ft**2) 	391836.734694
V is :  fps 	1.74149659864
reynolds number is :  	30427.5453198
hi :  Btu/(hr)*(ft**2)*(F) 	667.4
hi1 :  Btu/(hr)*(ft**2)*(F) 	222.303643994
X : 	0.03125
hfi2 :  	212.371237248
overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	108.611980947
inside surface per bank is :  ft**2 	18.312
Ai1 :  ft**2 	212.447090593
number of banks :  	11.6015230774
net free volume :  ft**3 	1.89483559028
frictional surface :  ft**2 	191.88
pressure drop  for annulus
De1 :  ft 	0.0395004292324
reynolds number :  	10087.9633345
delPs is :  psi 	0.232069401484
pressure drop  for inner pipe
delPt is :  psi 	0.39