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 16.2 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 16.3 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 16.4 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 16.5 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