# Chapter 8 : Flow Arrangements for Increased Heat Recovery¶

## Example 8.1 pgno:181¶

In :
print"\t example 8.1 \n"
print"\t approximate values are mentioned in the book \n"
#given
T1=358.; # inlet hot fluid,F
T2=100.; # outlet hot fluid,F
t1=90.; # inlet cold fluid,F
t2=120.; # outlet cold fluid,F
W=49600.; # lb/hr
w=233000.; # lb/hr
#solution
from math import log
print"\t 1.for heat balance \n"
print"\t for oil \n"
c=0.545; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for oil is : Btu/hr ",Q
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 0.93 " # from fig 19 for 2-4 exchanger
delt=(0.93*LMTD); # F
print"\t delt is :  F ",delt
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is : %.3f \n",X
Fc=0.25; # from fig.17
Kc=0.47; # 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,oil \n"
ID=35; # in
C=0.25; # clearance
B=7; # baffle spacing,in
PT=1.25;
As=((ID*C*B)/(144*PT))/2; # flow area,ft**2,from eq 7.1
print"\t flow area is :  ft**2 ",As
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2),from eq 7.2
print"\t mass velocity is :  lb/(hr)*(ft**2) ",Gs
mu1=1.12*2.42; # at 165F,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=52.5; # from fig.28
Z=0.2; # Z=(k)*(Pr*(1/3)) prandelt number
Ho=((jH)*(1/De)*(Z)); # H0=(h0/phys),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \n",Ho
print"\t cold fluid:inner tube side,water \n"
Nt=454;
n=6; # number of passes
L=12; #ft
at1=0.455; # flow area, in**2
at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48
print"\t flow area is :f 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)); # fps
print"\t V is :  fps ",V
mu2=0.73*2.42; # at 98F,lb/(ft)*(hr),from fig 14
D=(0.76/12); # ft,from table 10
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  ",Ret
hi=1010*0.96; # using fig 25,Btu/(hr)*(ft**2)*(F)
print"\t hi is :  Btu/(hr)*(ft**2)*(F) ",hi
ID=0.76; # ft
OD=1; #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
tw=(tc)+(((Ho)/(hio+Ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \n",tw
muw=1.95*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
Q=6980000; # taking rounded value,Btu/hr
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 \n"
f=0.00215; # friction factor for reynolds number 8900, using fig.29
s=0.82; # for reynolds number 25300,using fig.6
Ds=35/12; # ft
N=(12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  ",N
delPs=((f*(Gs**2)*(Ds)*(2*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 \n"
print"\t pressure drop  for inner pipe \n"
f=0.000195; # friction factor for reynolds number 34900, using fig.26
s=1;
D=0.0633; #ft
phyt=1;
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)
X1=0.13; # 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 8.1

approximate values are mentioned in the book

1.for heat balance

for oil

total heat required for oil is : Btu/hr  6974256.0
for water

total heat required for water is :  Btu/hr  6990000.0
delt1 is :  F  10.0
delt2 is :  F  238.0
LMTD is : F  71.9314248044
R is :  8.6
S is :  0.111940298507
FT is 0.93
delt is :  F  66.8962250681
ratio of two local temperature difference is : %.3f
0.0420168067227
caloric temperature of hot fluid is :  F  164.5
caloric temperature of cold fluid is :  F  97.5
hot fluid:shell side,oil

flow area is :  ft**2  0.170138888889
mass velocity is :  lb/(hr)*(ft**2)  291526.530612
reynolds number is :   8873.57540419
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F)
127.272727273
cold fluid:inner tube side,water

flow area is :f ft**2  0.239085648148
mass velocity is :  lb/(hr)*(ft**2) 974546.158687
V is :  fps  4.33131626083
reynolds number is :   34937.8788163
hi is :  Btu/(hr)*(ft**2)*(F)  969.6
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  736.896
tw is :  F
107.367601613
phys is :   0.925306637672
Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F)  117.76629934
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F)  101.538952854
total surface area is :  ft**2  1426.2864
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F)  73.155514686
actual Rd is :  (hr)*(ft**2)*(F)/Btu  0.00382107226455
pressure drop  for annulus

number of crosses are :   20
delPs is :  psi  4.47363588933
allowable delPs is 10 psi

pressure drop  for inner pipe

delPt is :  psi  4.0
delPr is :  psi  3.1
delPT is :  psi  7.2
allowable delPT is 10 psi



## Example 8.2 pgno:184¶

In :
print"\t example 8.2 \t"
print"\t approximate values are mentioned in the book \t"
#given
T1=250; # inlet hot fluid,F
T2=100; # outlet hot fluid,F
t1=90; # inlet cold fluid,F
t2=150; # outlet cold fluid,F
W=60000; # lb/hr
w=168000; # lb/hr
#solution
from math import log
print"\t 1.for heat balance \t"
print"\t for acetone \t"
c=0.57; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for acetone is :  Btu/hr \t",Q # calculation mistake in problem
print"\t for acetic acid \t"
c=0.51; # Btu/(lb)*(F)
Q1=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for acetic acid is :  Btu/hr \t",Q1 # calculation mistake in problem
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)/((1)*(log(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 0.88 \t" # from fig 20,for 3-6 exchanger
delt=(0.88*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,acetone \t"
ID=21.25; # in
C=0.25; # clearance
B=5; # baffle spacing,in
PT=1;
As=((ID*C*B)/(144*PT)); # flow area,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.20*2.42; # at 175F,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 :  \t",Res
phys=1;
jH=137; # from fig.28
c=0.63; # Btu/(lb)*(F),at 175F,from fig.2
k=0.095; # 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 :  \t",Pr
ho=((jH)*(k/De)*(Pr)*1); # using eq.6.15b,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t cold fluid:inner tube side,acetic acid \t"
Nt=270;
n=2; # number of passes
L=16; #ft
at1=0.268; # 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 \t",at
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
mu2=0.85*2.42; # at 120F,lb/(ft)*(hr)
D=(0.584/12); # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=56; # from fig.24
c=0.51; # Btu/(lb)*(F),at 120F,from fig.2
k=0.098; # Btu/(hr)*(ft**2)*(F/ft), from table 4
phyt=1;
Pr=((c)*(mu2)/k)**(1/3); # prandelt number raised to power 1/3
print"\t Pr is :  \t",Pr
hi=((jH)*(k/D)*(Pr)*(1**0.14)); # using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=0.584; # ft
OD=0.75; #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) \t",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) \t",Uc
A2=0.1963; # actual surface supplied for each tube,ft**2,from table 10
A=3*(Nt*L*A2); # ft**2
print"\t total surface area is :  ft**2 \t",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t pressure drop  for annulus \t"
f=0.00155; # friction factor for reynolds number 52200, using fig.29
s=0.79; # for reynolds number 25300,using table.6
Ds=21.25/12; # ft
N=(12*L/B)+1; # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPs=((f*(Gs**2)*(Ds)*(3*N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPs is 10 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00024; # friction factor for reynolds number 158000, using fig.26
s=1.07;
D=0.0487; # ft
delPt=((f*(Gt**2)*(L)*(n*3))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is :  psi \t",round(delPt)
X1=0.063; # X1=((V**2)/(2*g)), for Gt 1060000,using fig.27
delPr=(3)*((4*n*X1)/(s)); # using eq.7.46,psi
print"\t delPr is :  psi \t",round(delPr,1)
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi \t",round(delPT,1)
print"\t allowable delPT is 10 psi \t"
#end

	 example 8.2
approximate values are mentioned in the book
1.for heat balance
for acetone
total heat required for acetone is :  Btu/hr 	5130000.0
for acetic acid
total heat required for acetic acid is :  Btu/hr 	5140800.0
delt1 is :  F 	10
delt2 is :  F 	100
LMTD is : F 	39.0865033713
R is :  	2
S is :  	0
FT is 0.88
delt is :  F 	34.3961229667
caloric temperature of hot fluid is :  F 	175
caloric temperature of cold fluid is :  F 	120
hot fluid:shell side,acetone
flow area is :  ft**2 	0.184461805556
mass velocity is :  lb/(hr)*(ft**2) 	325270.588235
reynolds number is :  	53203.694701
Pr is :  	1.0
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	164.4
cold fluid:inner tube side,acetic acid
flow area is :  ft**2 	0.25125
mass velocity is :  lb/(hr)*(ft**2) 	668656.716418
reynolds number is :  	15819.782952
Pr is :  	1.0
hi is :  Btu/(hr)*(ft**2)*(F) 	112.767123288
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	87.808
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	57.2370234092
total surface area is :  ft**2 	2544.048
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	58.6249615773
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	-0.000413628549958
pressure drop  for annulus
number of crosses are :  	39
delPs is :  psi 	10.407463223
allowable delPs is 10 psi
pressure drop  for inner pipe
delPt is :  psi 	4.0
delPr is :  psi 	1.4
delPT is :  psi 	5.2
allowable delPT is 10 psi