Chapter 7 Parallel-Counter flow:Shell and Tube Exchangers¶

Example 7.1 pgno:139¶

In [25]:
print"\t example 7.1 \n"
#given
PT=1; # square pitch,in
do=0.75; # outer diameter,in
#solution
de=((4*(PT**2-(3.14*do**2/4)))/(3.14*do));
print"\t equivalent diameter is :  in ",round(de,2)
De=(de/12); # ft
print"\t De is :  in ",round(De,3)
#end

	 example 7.1

equivalent diameter is :  in  0.95
De is :  in  0.079


Example 7.2 pgno:147¶

In [26]:
print"\t example 7.2 "
print"\t approximate values are mentioned in the book "
print"\t considering 50F approach "
#given
T1=350.; #F
T2=250.; #F
t2=T2-50.; # formula for approach,f
print"\t t2 is :  F ",t2
t2=200.;
print"\t fluids are with equal ranges,so "
t1=t2-(T1-T2);# F
print"\t t1 is :  F ",t1
t1=100.;
#solution

R=((T1-T2)/(t2-t1));
print"\t R is : ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.925 " # from fig 18
print"\t considering 0F approach "
T1=300; #F
T2=200; #F
t2=T2-0; # formula for approach,f
print"\t t2 is :  F ",t2
print"\t fluids are with equal ranges,so "
t1=t2-(T1-T2); # F
print"\t t1 is :  F ",t1
R=((T1-T2)/(t2-t1));
print"\t R is : ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.80 " # from fig 18
print"\t considering 20F cross "
T1=280; #F
T2=180; #F
t2=T2+20; # formula for approach,f
print"\t t2 is :  F ",t2
print"\t fluids are with equal ranges,so \n"
t1=t2-(T1-T2); # F
print"\t t1 is :  F ",t1
R=((T1-T2)/(t2-t1));
print"\t R is : ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.64 " # from fig 18
#end

	 example 7.2
approximate values are mentioned in the book
considering 50F approach
t2 is :  F  200.0
fluids are with equal ranges,so
t1 is :  F  100.0
R is :  1.0
S is :  0.4
FT is 0.925
considering 0F approach
t2 is :  F  200
fluids are with equal ranges,so
t1 is :  F  100
R is :  1
S is :  0
FT is 0.80
considering 20F cross
t2 is :  F  200
fluids are with equal ranges,so

t1 is :  F  100
R is :  1
S is :  0
FT is 0.64


Example 7.3 pgno:151¶

In [27]:
print"\t example 7.3 \n"
print"\t approximate values are mentioned in the book \n"
#given
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
#solution
from math import log
print"\t 1.for heat balance "
print"\t for kerosene "
c=0.605; # Btu/(lb)*(F)
Q1=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for kerosene is :  Btu/hr ",Q1 # calculation mistake in problem
print"\t for crude oil "
c=0.49; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for mid continent crude is :  Btu/hr ",Q # calculation mistake in problem
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F ",delt1
print"\t delt2 is :  F ",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F ",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is : ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.905 " # from fig 18
delt=(0.905*LMTD); # F
print"\t delt is :  F ",delt
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is : ",X
Fc=0.42; # from fig.17
Kc=0.20; # crude oil controlling
Tc=((T2)+((Fc)*(T1-T2))); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F ",Tc
tc=((t1)+((Fc)*(t2-t1))); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F ",tc
print"\t hot fluid:shell side,kerosene "
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
print"\t flow area is :  ft**2 ",As
Gs=(W/As); # mAss velocity,lb/(hr)*(ft**2)
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",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
print"\t reynolds number is : ",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
print"\t Pr is : ",Pr
Ho=((jH)*(k/De)*(Pr)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) ",Ho
print"\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
print"\t flow area is :  ft**2 ",at
Gt=(w/(at)); # mAss velocity,lb/(hr)*(ft**2)
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",Gt
mu2=3.6*2.42; # at 129F,lb/(ft)*(hr)
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is : ",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
print"\t Pr is : ",Pr
Hi=((jH)*(k/D)*(Pr)*(1**0.14)); #Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t Hi is :  Btu/(hr)*(ft**2)*(F) ",Hi
ID=0.81; # ft
OD=1; #ft
Hio=((Hi)*(ID/OD)); #Hio=(hio/phyp), using eq.6.5
print"\t Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",Hio
muw=1.5*2.42; # lb/(ft)*(hr), from fig.14
phyt=(mu2/muw)**0.14;
print"\t phyt is : ",phyt # from fig.24
hio=(Hio)*(phyt); # from eq.6.37
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",hio
tw=(tc)+(((Ho)/(Hio+Ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F ",tw
muw=0.56*2.42; # lb/(ft)*(hr), from fig.14
phys=(mu1/muw)**0.14;
print"\t phys is : ",phys # from fig.24
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",ho
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",Uc
A2=0.2618; # 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 ",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu ",Rd
print"\t pressure drop  for annulus "
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
print"\t number of crosses are :",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi ",delPs
print"\t allowable delPa is 10 psi "
print"\t pressure drop  for inner pipe "
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
print"\t delPt is :  psi ",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
print"\t delPr is :  psi ",delPr
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi ",round(delPT,2)
print"\t allowable delPs is 10 psi "
#end

	 example 7.3

approximate values are mentioned in the book

1.for heat balance
for kerosene
total heat required for kerosene is :  Btu/hr  5034810.0
for crude oil
total heat required for mid continent crude is :  Btu/hr  5110700.0
delt1 is :  F  100.0
delt2 is :  F  220.0
LMTD is : F  152.195928445
R is :  2.71428571429
S is :  0.241379310345
FT is 0.905
delt is :  F  137.737315243
ratio of two local temperature difference is :  0.454545454545
caloric temperature of hot fluid is :  F  279.8
caloric temperature of cold fluid is :  F  129.4
hot fluid:shell side,kerosene
flow area is :  ft**2  0.147569444444
mAss velocity is :  lb/(hr)*(ft**2)  296809.411765
reynolds number is :  25296.2566845
Pr is :  1.0
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F)  86.2363636364
cold fluid:inner tube side,crude oil

flow area is :  ft**2  0.141267361111
mAss velocity is :  lb/(hr)*(ft**2)  1054737.61829
reynolds number is :  8172.03733177
Pr is :  1.0
Hi is :  Btu/(hr)*(ft**2)*(F)  35.362962963
Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  28.644
phyt is :  1.1303932977
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  32.3789856193
tw is :  F  242.299617309
phys is :  0.953986161765
Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  82.26829755
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F)  23.2344278003
total surface area is :  ft**2  661.8304
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F)  56.0637396161
actual Rd is :  (hr)*(ft**2)*(F)/Btu  -0.0252027385623
pressure drop  for annulus
number of crosses are : 38
delPs is :  psi  3.45911725247
allowable delPa is 10 psi
pressure drop  for inner pipe
delPt is :  psi  6.13806896405
delPr is :  psi  2.89156626506
delPT is :  psi  9.03
allowable delPs is 10 psi


Example 7.4 pgno:155¶

In [28]:
print"\t example 7.4 "
print"\t approximate values are mentioned in the book "
#given
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
#solution
from math import log
print"\t 1.for heat balance "
print"\t for distilled water "
c=1; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for distilled water is : Btu/hr ",Q
print"\t for raw water "
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for raw water is :  Btu/hr ",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F ",delt1
print"\t delt2 is :  F ",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F ",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is :  ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.945 " # from fig 18
delt=(0.945*LMTD); # F
print"\t delt is :  F ",delt
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is : ",X
Fc=0.42; # from fig.17
Kc=0.20; # crude oil controlling
Tc=((T2)+(T1))/2; # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F ",Tc
tc=((t1)+(t2))/2; # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F ",tc
print"\t hot fluid:shell side,distilled water "
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
print"\t flow area is :  ft**2 ",As
Gs=(W/As); # mAss velocity,lb/(hr)*(ft**2),using eq.7.2
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",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
print"\t reynolds number is : ",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
print"\t Pr is : ",Pr
ho=((jH)*(k/De)*(Pr)); # using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) ",ho
print"\t cold fluid:inner tube side,raw water "
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
print"\t flow area is :  ft**2 ",at
Gt=(w/(at)); # mAss velocity,lb/(hr)*(ft**2)
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",Gt
V=(Gt/(3600*62.5));
print"\t V is  fps ",V
mu2=0.92*2.42; # at 77.5F,lb/(ft)*(hr)
D=0.65/12; #ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is : ",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
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",hio
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",Uc
print"\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 "
A2=0.1963; # 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 ",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu ",Rd
print"\t pressure drop  for annulus "
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
print"\t number of crosses are : ",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi ",delPs
print"\t allowable delPs is 10 psi "
print"\t pressure drop  for inner pipe "
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
print"\t delPt is :  psi ",round(delPt,1)
X1=0.33; # X1=((V**2)/(2*g)), for Gt 1060000,using fig.27
delPr=((4*n*X1)/(s)); # using eq.7.46,psi
print"\t delPr is :  psi ",round(delPr,1)
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi ",round(delPT,1)
print"\t allowable delPT is 10 psi "
#end

	 example 7.4
approximate values are mentioned in the book
1.for heat balance
for distilled water
total heat required for distilled water is : Btu/hr  1400000.0
for raw water
total heat required for raw water is :  Btu/hr  1400000.0
delt1 is :  F  10.0
delt2 is :  F  13.0
LMTD is : F  11.4344840601
R is :   1.6
S is :  0.277777777778
FT is 0.945
delt is :  F  10.8055874368
ratio of two local temperature difference is :  0.769230769231
caloric temperature of hot fluid is :  F  89.0
caloric temperature of cold fluid is :  F  77.5
hot fluid:shell side,distilled water
flow area is :  ft**2  0.254166666667
mAss velocity is :  lb/(hr)*(ft**2)  688524.590164
reynolds number is :  16099.0598149
Pr is :  1.0
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F)  573.381818182
cold fluid:inner tube side,raw water
flow area is :  ft**2  0.185555555556
mAss velocity is :  lb/(hr)*(ft**2)  1508982.03593
V is  fps  6.70658682635
reynolds number is :  36712.4177803
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  1158.3
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F)  383.527824238
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
total surface area is :  ft**2  502.528
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F)  257.821653196
actual Rd is :  (hr)*(ft**2)*(F)/Btu  0.00127127741812
pressure drop  for annulus
number of crosses are :  16
delPs is :  psi  7.65506963359
allowable delPs is 10 psi
pressure drop  for inner pipe
delPt is :  psi  4.9
delPr is :  psi  2.6
delPT is :  psi  7.6
allowable delPT is 10 psi


Exxample 7.5 pgno:159¶

In [29]:
print"\t example 7.5 "
print"\t approximate values are mentioned in the book "
#given
T1=175.; # inlet hot fluid,F
T2=150.; # outlet hot fluid,F
t1=85.; # inlet cold fluid,F
#solution
delt1=T2-t1; #F
print"\t delt1 is :  F ",delt1
U=15; # assumption,Btu/(hr)*(ft^2)*(F)
theta=8000; # operating hours,hr
CW=(0.01/8300); # water cost,\$/lb
print"\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));
print"\t X is : ",X
Y=((T1-T2)/delt1);
print"\t Y is : ",Y
A=0.96; # A=(delt2/delt1), from fig 7.24
delt2=0.96*delt1;
print"\t delt2 is :  F ",delt2
t2=T1-delt2; # F
print"\t t2 is :  F ",t2
#end

	 example 7.5
approximate values are mentioned in the book
delt1 is :  F  65.0
For annual charges assume 20 per cent repair and maintenanc.e and 10 per cent depreciation

X is :  0.120481927711
Y is :  0.384615384615
delt2 is :  F  62.4
t2 is :  F  112.6


Example 7.6 pgno:161¶

In [30]:
print"\t example 7.6 \n"
print"\t approximate values are mentioned in the book \n"
#given
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
#solution
from math import log
print"\t 1.for heat balance \n"
print"\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
print"\t total heat required for solution is :  Btu/hr ",Q1
print"\t for water \n"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr ",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F ",delt1
print"\t delt2 is :  F ",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F ",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is : ",R
S=((t2-t1)/(T1-t1));
print"\t S is : ",S
print"\t FT is 0.81 " # from fig 18
delt=(0.81*LMTD); # F
print"\t delt is :  F ",delt
Tc=((T2)+(T1))/2; # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is : F",Tc
tc=((t1)+(t2))/2; # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F ",tc
print"\t hot fluid:shell side,phosphate solution "
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
print"\t flow area is :  ft**2 ",As
Gs=(W/As); # mAss velocity,lb/(hr)*(ft**2),using eq.7.2
print"\t mAss velocity is : lb/(hr)*(ft**2) ",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
print"\t reynolds number is : ",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
print"\t Pr is : ",Pr
ho=((jH)*(k/De)*(Pr)); # using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) ",ho
print"\t cold fluid:inner tube side,raw water "
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
print"\t flow area is :  ft**2 ",at
Gt=(w/(at)); # mAss velocity,lb/(hr)*(ft**2)
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",Gt
V=(Gt/(3600*62.5));
print"\t V is  fps ",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
print"\t reynolds number is : ",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
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",hio
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",Uc
A2=0.1963; # 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 ",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu ",Rd
print"\t pressure drop  for annulus "
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
print"\t number of crosses are : ",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi ",delPs
print"\t allowable delPs is 10 psi "
print"\t pressure drop  for inner pipe "
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
print"\t delPt is :  psi ",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
print"\t delPr is :  psi ",round(delPr,1)
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi ",round(delPT,1)
print"\t allowable delPT is 10 psi \n"
#end

	 example 7.6

approximate values are mentioned in the book

1.for heat balance

for solution

total heat required for solution is :  Btu/hr  915667.2
for water

total heat required for water is :  Btu/hr  915200.0
delt1 is :  F  22.0
delt2 is :  F  60.0
LMTD is : F  37.8749328485
R is :  2.72727272727
S is :  0.268292682927
FT is 0.81
delt is :  F  30.6786956073
caloric temperature of hot fluid is : F 120.0
caloric temperature of cold fluid is :  F  79.0
hot fluid:shell side,phosphate solution
flow area is :  ft**2  0.0347916666667
mAss velocity is : lb/(hr)*(ft**2)  579449.101796
reynolds number is :  15796.5061612
Pr is :  1.0
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F)  295.957894737
cold fluid:inner tube side,raw water
flow area is :  ft**2  0.0545277777778
mAss velocity is :  lb/(hr)*(ft**2)  762913.907285
V is  fps  3.39072847682
reynolds number is :  17899.0185011
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  661.333333333
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F)  204.459014471
total surface area is :  ft**2  163.3216
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F)  182.656652912
actual Rd is :  (hr)*(ft**2)*(F)/Btu  0.000583796892885
pressure drop  for annulus
number of crosses are :  96
delPs is :  psi  9.51887629626
allowable delPs is 10 psi
pressure drop  for inner pipe
delPt is :  psi  1.58733222032
delPr is :  psi  0.6
delPT is :  psi  2.2
allowable delPT is 10 psi



Example 7.7 pgno:166¶

In [31]:
print"\t example 7.7 \n"
print"\t approximate values are mentioned in the book \n"
#given
U=50; # Btu/(hr)*(ft**2)*(F)
TP=328; # F
TE=228; # F
#solution
import numpy

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
print"\t X is : ",X
a=(1); # coefficient of t**2
b=(-556); # coefficient of t
c=(74784-X); # constant
print"\t coefficient of t**2 is : ",a
print"\t coefficient of t is : ",b
print"\t constant term is : ",c
P=numpy.array([c, b, a])
t=numpy.roots(P)
print"\t t is :",t
print"\t t cannot be greater than 328F t is 218F "
#end

	 example 7.7

approximate values are mentioned in the book

X is :  1050.97060273
coefficient of t**2 is :  1
coefficient of t is :  -556
constant term is :  73733.0293973
t is : [ 0.00457855  0.00296217]
t cannot be greater than 328F t is 218F


Example 7.8 pgno:167¶

In [32]:
print"\t example 7.8 \n"
print"\t approximate values are mentioned in the book \n"
#given
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
#solution
import sys
print sys.getdefaultencoding()
from math import log
print"\t 1.for heat balance \n"
print"\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
print"\t total heat required for solution is :  Btu/hr ",Q1
print"\t for steam \n"
l=960.1; # latent heat of condensation,Btu/(lb)
Q=((w)*(l)); # Btu/hr
print"\t total heat required for steam is :  Btu/hr ",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F ",delt1
print"\t delt2 is :  F ",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F ",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is : ",R
delt=(LMTD); # when R=0,F
print"\t delt is :  F ",delt

ta=111; #F
Ta=228; #f
print"\t hot fluid:tube side,steam "
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
print"\t flow area is :  ft**2 ",at
Gt=(w/(at)); # mAss velocity,lb/(hr)*(ft**2)b
print"\t mAss velocity is : lb/(hr)*(ft**2) ",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
print"\t reynolds number is : ",Ret
hio=1500; # for condensation of steam
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",hio
print"\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
print"\t flow area is :  ft**2 ",As
Gs=(W/As); # mAss velocity,lb/(hr)*(ft**2)
print"\t mAss velocity is :  lb/(hr)*(ft**2) ",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
print"\t De is :  ft ",De
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is : ",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
print"\t Pr is :",Pr
Ho=((jH)*(k/De)*(Pr)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) ",Ho
muw=0.51*2.42; #  at 210F,lb/(ft)*(hr), from fig.14
phys=(mu1/muw)**0.14;
print"\t phys is : ",phys # from fig.24
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) ",ho
tw=(ta)+(((hio)/(hio+Ho))*(Ta-ta)); # from eq.5.31
print"\t tw is :  F ",tw
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",Uc
A2=0.1963; # 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 ",A
UD=((Q)/((A)*(LMTD)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) ",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu ",Rd
print"\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
print"\t delPt is :  psi ",round(delPt,1)
print"\t pressure drop  for annulus \n"
De1=((4*As)/((Nt*3.14*d)+(3.14*1))); # from eq.6.4,ft
print"\t De1 is :  ft",round(De1,3)
Res1=(De1*Gs/mu1); # from eq 7.3
print"\t Res1 is : ",round(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
print"\t delPs is :  psi ",round(delPs,2)
#end

	 example 7.8

approximate values are mentioned in the book

ascii
1.for heat balance

for solution

total heat required for solution is :  Btu/hr  3784000.0
for steam

total heat required for steam is :  Btu/hr  3792395.0
delt1 is :  F  128.0
delt2 is :  F  106.0
LMTD is : F  116.654454302
R is :  0.0
delt is :  F  116.654454302
hot fluid:tube side,steam
flow area is :  ft**2  0.0796944444444
mAss velocity is : lb/(hr)*(ft**2)  49564.3081213
reynolds number is :  82671.1836992
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  1500
cold fluid:shell side,sugar solution

flow area is :  ft**2  0.551953125
mAss velocity is :  lb/(hr)*(ft**2)  362349.610757
De is :  ft  0.148026315789
reynolds number is :  17049.3572499
Pr is : 1.0
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F)  138.3504
phys is :  1.13996690865
Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  157.714877798
tw is :  F  218.119942108
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F)  142.70989533
total surface area is :  ft**2  238.7008
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F)  136.194122026
actual Rd is :  (hr)*(ft**2)*(F)/Btu  0.000335238210724
pressure drop  for inner pipe

delPt is :  psi  2.8
pressure drop  for annulus

De1 is :  ft 0.122
Res1 is :  14084.0
delPs is :  psi  0.07


Example 7.9 pgno:170¶

In [33]:
print"\t example 7.9 \n"
print"\t approximate values are mentioned in the book \n"
#given
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)
#solution
X=((U*A)/(w*c));
print"\t X is : ",X
R=((w*c)/(W*C));
print"\t R is : ",R
S=0.265; # from fig 7.25, by comparing X an R
t2=(t1)+((0.265)*(T1-t1)); # S=((t2-t1)/(T1-t1))
print"\t t2 is :  F ",t2
T2=((T1)-((R)*(t2-t1))); # R=((T1-T2)/(t2-t1))
print"\t T2 is :  F ",round(T2)
# end

	 example 7.9

approximate values are mentioned in the book

X is :  0.628360498562
R is :  2.77815829528
t2 is :  F  176.85
T2 is :  F  176.0