Chapter 11 Calculations for Process Heat Conditions

Example11.1 pgno:231

In [17]:
print"\t example 11.1 \t"
print"\t approximate values are mentioned in the book \t"
T1=340.; # inlet hot fluid,F
T2=240.; # outlet hot fluid,F
t1=200.; # inlet cold fluid,F
t2=230.; # outlet cold fluid,F
W=29800; # lb/hr
w=103000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for straw oil \t"
c=0.58; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for straw oil is :  Btu/hr \t",Q
print"\t for naphtha \t"
c=0.56; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for naphtha is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt1=40.;
delt2=T1-t2; # F
delt2=110.
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 0.885 \t" # from fig 18
delt=(0.885*LMTD); # F
print"\t delt is :  F \t",delt
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is :  \t",X
L=16;
Fc=0.405; # from fig.17
Kc=0.23; # crude oil controlling
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
UD1=70; # assume, from table 8a
A1=((Q)/((UD1)*(delt)));
print"\t A1 is :  ft**2 \t",A1
a1=0.1963; # ft**2/lin ft
N1=(A1/(16*a1));
print"\t number of tubes are :\t",N1
N2=124; # assuming two tube passes, from table 9
A2=(N2*L*a1); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(delt)));
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
print"\t hot fluid:shell side,straw oil \t"
ID=15.25; # in
C=0.25; # clearance
B=3.5; # minimum baffle spacing,from eq 11.4,in
PT=1;
As=((ID*C*B)/(144*PT)); # flow area,from eq 7.1,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(W/As); # mass velocity,from eq 7.2,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=3.63; # at 280.5F,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
jH=46; # from fig.28
Z=0.224; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft), at mu3=1.5cp and 35 API
Ho=((jH)*(1/De)*(Z)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ho
phys=1;
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t cold fluid:inner tube side,naphtha \t"
Nt=124;
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 \t",at
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
mu2=1.31; # at 212F,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=102; # from fig.24
Z=0.167; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft), at mu4=0.54cp and 48 API
Hi=((jH)*(1/D)*(Z)); #Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t Hi is :  Btu/(hr)*(ft**2)*(F) \t",Hi
ID=0.62; # 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
phyt=1;
hio=(Hio)*(phyt); # from eq.6.37
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
print"\t pressure drop  for annulus \t"
f=0.00225; # friction factor for reynolds number 7000, using fig.29
s=0.76; # for reynolds number 7000,using fig.6
Ds=15.25/12; # ft
N=(12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t pressure drop  for inner pipe \t"
f=0.0002; # friction factor for reynolds number 31300, using fig.26
s=0.72;
delPt=((f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is :  psi \t",delPt
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is : %.1f Btu/(hr)*(ft**2)*(F) \t",Uc
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t The first trial is disqualified because of failure to meet the required dirt factor \t"
print"\t Proceeding as above and carrying the viscosity correction and pressure drops to completion the new summary is given using a 17.25in. ID shell with 166 tubes on two passes and a 3.5in. baffle space \t"
UD1=60; # assumption for 2 tube passes,3.5 baffle spacing and 17.25in ID
UC1=74.8;
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UC1
UD2=54.2;
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD2
Rd1=0.005; 
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
delPs1=4.7;
print"\t delPs is :  psi \t",delPs1
delPt1=2.1;
print"\t delPt is :  psi \t",delPt1
#end

	 example 11.1 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for straw oil 	
	 total heat required for straw oil is :  Btu/hr 	1728400.0
	 for naphtha 	
	 total heat required for naphtha is :  Btu/hr 	1730400.0
	 delt1 is :  F 	40.0
	 delt2 is :  F 	110.0
	 LMTD is  F 	69.2750233163
	 R is :  	3.33333333333
	 S is :  	0.214285714286
	 FT is 0.885 	
	 delt is :  F 	61.3083956349
	 ratio of two local temperature difference is :  	0.363636363636
	 caloric temperature of hot fluid is :  F 	280.5
	 caloric temperature of cold fluid is :  F 	212.15
	 A1 is :  ft**2 	403.207419539
	 number of tubes are :	128.37729863
	 total surface area is :  ft**2 	389.4592
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	72.4710556785
	 hot fluid:shell side,straw oil 	
	 flow area is :  ft**2 	0.0926649305556
	 mass velocity is :  lb/(hr)*(ft**2) 	321588.758782
	 reynolds number is :  	7013.52894497
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	130.155789474
	 Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	130.155789474
	 cold fluid:inner tube side,naphtha 	
	 flow area is :  ft**2 	0.130027777778
	 mass velocity is :  lb/(hr)*(ft**2) 	792138.431959
	 reynolds number is :  	31262.2572002
	 Hi is :  Btu/(hr)*(ft**2)*(F) 	329.477756286
	 Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	272.36827853
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	272.36827853
	 pressure drop  for annulus 	
	 number of crosses are :  	54.8571428571
	 delPs is :  psi 	5.1651098751
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	2.06675311157
	 clean overall coefficient is : %.1f Btu/(hr)*(ft**2)*(F) 	88.0700339124
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00244401237634
	 The first trial is disqualified because of failure to meet the required dirt factor 	
	 Proceeding as above and carrying the viscosity correction and pressure drops to completion the new summary is given using a 17.25in. ID shell with 166 tubes on two passes and a 3.5in. baffle space 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	74.8
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	54.2
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.005
	 delPs is :  psi 	4.7
	 delPt is :  psi 	2.1

Example 11.2 pgno:235

In [18]:
print"\t example 11.2 \t"
print"\t approximate values are mentioned in the book \t"   
T1=350.; # inlet hot fluid,F
T2=160.; # outlet hot fluid,F
t1=100.; # inlet cold fluid,F
t2=295.; # outlet cold fluid,F
W=84438; # lb/hr
w=86357; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for lean oil \t"
c=0.56; # Btu/(lb)*(F)
Qh=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for lean oil is :  Btu/hr \t",Qh
print"\t for rich oil \t"
c=0.53; # Btu/(lb)*(F)
Qc=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for rich oil is :  Btu/hr \t",Qc
Q=(Qh+Qc)/(2);
print"\t Q is :  V \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 0.875 \t"# for 4-8 exchanger,from fig 21
delt=(0.875*LMTD); # F
print"\t delt is :  F \t",delt
X=((delt1)/(delt2));
print"\t ratio of two local temperature difference is :  \t",X
Fc=0.48; # from fig.17
Kc=0.32; # crude oil controlling
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 :  \t",tc
UD1=50; # assume, from table 8a
A1=((Q)/((UD1)*(delt)));
print"\t A1 is :  ft**2 \t",A1
a1=0.1963; # ft**2/lin ft
N1=(A1/(16*a1*2)); # 2-4 exchanger in series
print"\t number of tubes are :  \t",N1
N2=580; # assuming six tube passes,31in ID, from table 9
A2=(N2*16*a1*2); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(delt)));
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
print"\t hot fluid:inner tube side,lean oil \t"
Nt=580;
n=6; # 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 \t",at
Gt=(W/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
mu2=2.13; # at 212F,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=36.5; # from fig.24
Z=0.185; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft), at mu4=0.88cp and 35 API
Hi=((jH)*(1/D)*(Z)); #Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t Hi is :  Btu/(hr)*(ft**2)*(F) \t",Hi
ID=0.62; # 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
phyt=1;
hio=(Hio)*(phyt); # from eq.6.37
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
print"\t cold fluid:shell side,rich oil \t"
ID=31; # in
C=0.25; # clearance
B=12; # minimum baffle spacing,from eq 11.4,in
PT=1;
As=((ID*C*B)/(144*PT))/(2); # flow area,from eq 7.1,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(w/As); # mass velocity,from eq 7.2,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=3.15; # at 193.5F,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
jH=45; # from fig.28
Z=0.213; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft), at mu3=1.3cp and 35 API
Ho=((jH)*(1/De)*(Z)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ho
phys=1;
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t pressure drop  for inner pipe \t"
f=0.00027; # friction factor for reynolds number 10100, using fig.26
s=0.77;
delPt=((2*f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is :  psi \t",delPt
X1=0.024; # X1=((V**2)/(2*g)), for Gt 1060000,using fig.27
delPr=((4*2*n*X1)/(s)); # using eq.7.46,psi
print"\t delPr is :  psi \t",delPr
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi \t",delPT
print"\t allowable delPT is 10 psi \t"
print"\t pressure drop  for annulus \t"
f=0.0023; # friction factor for reynolds number 6720, using fig.29
s=0.79; # for reynolds number 6720,using fig.6
Ds=31/12; # ft
De=0.0792;
N=(4*12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPa is 10 psi \t"
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
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is : %.4f (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t The initial assumptions have provided an exchanger which very nearly meets all the requirements. Eight-pass units would meet the heat-transfer requirement but would give a tube-side pressure drop of 14 psi. The trial exchanger will be somewhat less suitable when the value of Q, is also taken into account. If the minimum dirt factor of 0.0040 is to be taken literally, it will be necessary to try the next size shell \t"
print"\t Assume a 33 in. ID shell with six1 tube passes and baffies spaced 12-in. apart, since the pressure drop increases with the diameter of the shell for a given mass velocity. \t"
UC1=52.3;
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UC1
UD2=42;
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD2
Rd1=0.0047; 
print"\t calculated Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
Rd2=0.004; 
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd2
delPs1=4.4;
print"\t delPs is :  psi \t",delPs1
delPT1=7.9;
print"\t delPt is :  psi \t",delPT1
#end

	 example 11.2 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for lean oil 	
	 total heat required for lean oil is :  Btu/hr 	8984203.2
	 for rich oil 	
	 total heat required for rich oil is :  Btu/hr 	8924995.95
	 Q is :  V 	8954599.575
	 delt1 is :  F 	60.0
	 delt2 is :  F 	55.0
	 LMTD is : F 	57.5283364148
	 R is :  	0.974358974359
	 S is :  	0.78
	 FT is 0.875 	
	 delt is :  F 	50.337294363
	 ratio of two local temperature difference is :  	1.09090909091
	 caloric temperature of hot fluid is :  F 	251.2
	 caloric temperature of cold fluid is :  	193.6
	 A1 is :  ft**2 	3557.83904889
	 number of tubes are :  	566.390577064
	 total surface area is :  ft**2 	3643.328
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	48.8267738849
	 hot fluid:inner tube side,lean oil 	
	 flow area is :  ft**2 	0.202731481481
	 mass velocity is :  lb/(hr)*(ft**2) 	416501.667047
	 reynolds number is :  	10109.4536086
	 Hi is :  Btu/(hr)*(ft**2)*(F) 	130.609284333
	 Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	107.970341715
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	107.970341715
	 cold fluid:shell side,rich oil 	
	 flow area is :  ft**2 	0.322916666667
	 mass velocity is :  lb/(hr)*(ft**2) 	267428.129032
	 reynolds number is :  	6721.07731695
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	121.073684211
	 Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	121.073684211
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	4.32759598913
	 delPr is :  psi 	1.4961038961
	 delPT is :  psi 	5.82369988524
	 allowable delPT is 10 psi 	
	 pressure drop  for annulus 	
	 number of crosses are :  	64
	 delPs is :  psi 	6.44657740525
	 allowable delPa is 10 psi 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	57.073599733
	 actual Rd is : %.4f (hr)*(ft**2)*(F)/Btu 	0.00295933089476
	 The initial assumptions have provided an exchanger which very nearly meets all the requirements. Eight-pass units would meet the heat-transfer requirement but would give a tube-side pressure drop of 14 psi. The trial exchanger will be somewhat less suitable when the value of Q, is also taken into account. If the minimum dirt factor of 0.0040 is to be taken literally, it will be necessary to try the next size shell 	
	 Assume a 33 in. ID shell with six1 tube passes and baffies spaced 12-in. apart, since the pressure drop increases with the diameter of the shell for a given mass velocity. 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	52.3
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	42
	 calculated Rd is :  (hr)*(ft**2)*(F)/Btu 	0.0047
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.004
	 delPs is :  psi 	4.4
	 delPt is :  psi 	7.9

Example 11.3 pgno:238

In [19]:
print"\t example 11.3 \t"
print"\t approximate values are mentioned in the book \t"
T1=190.; # inlet hot fluid,F
T2=120.; # outlet hot fluid,F
t1=80.; # inlet cold fluid,F
t2=120.; # outlet cold fluid,F
W=100000; # lb/hr
w=154000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for caustic \t"
c=0.88; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for caustic is :  Btu/hr \t",Q
print"\t for water \t"
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 0.815 \t" # for 4-8 exchanger,from fig 21
delt=(0.815*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 :  \t",tc
UD1=250; # assume, from table 8
A1=((Q)/((UD1)*(delt)));
print"\t A1 is :  ft**2 \t",A1
a1=0.2618; # ft**2/lin ft
L=16;
N1=(A1/(16*a1));
print"\t number of tubes are :  \t",N1
N2=140; # assuming four tube passes,19.25in ID, from table 9
A2=(N2*L*a1); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(delt)));
print"\t correct design overall coefficient is : Btu/(hr)*(ft**2)*(F) \t",UD
print"\t hot fluid:shell side,caustic \t"
ID=19.25; # in
C=0.25; # clearance
B=7; # minimum baffle spacing,from eq 11.4,in
PT=1.25;
As=((ID*C*B)/(144*PT)); # flow area,from eq 7.1,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(W/As); # mass velocity,from eq 7.2,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=1.84; # at 155F,lb/(ft)*(hr), from fig.14
De=0.72/12; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=75; # from fig.28
Z=0.575; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft)
Ho=((jH)*(1/De)*(Z)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ho
phys=1; # low viscosity
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t cold fluid:inner tube side,water \t"
Nt=140;
n=4; # number of passes
L=16; #ft
at1=0.546; # flow area, in**2
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
V=Gt/(3600*62.5);
print"\t V is  fps \t",V
mu2=1.74; # at 100F,lb/(ft)*(hr)
D=0.0695; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=1240*0.94; # from fig 25
print"\t Hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=0.834; # 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) \t",hio
print"\t pressure drop  for annulus \t"
f=0.0019; # friction factor for reynolds number 17400, using fig.29
s=1.115; # for reynolds number 17400,using fig.6
Ds=19.25/12; # ft
De=0.06;
N=(12*L/B)+1; # number of crosses,using eq.7.43
print"\t number of crosses are : %.0f \t",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPa is 10 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00018; # friction factor for reynolds number 46300, using fig.26
s=1;
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 \t",delPt
X1=0.18; # 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 \t",delPr
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi \t",delPT
print"\t allowable delPa is 10 psi \t"
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
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t Adjustment of the baffie space to use the full 10 psi will still not permit the exchanger to make the 0.002 dirt factor. The value of UD has been assumed too high \t"
print"\t Try a 21.25 in ID shell with four tube passes and a 6 in baffie space This corresponds to 170 tubes \t"
UC1=39
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UC1
UD2=200;
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD2
Rd1=0.0024; 
print"\t calculated Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
Rd2=0.002; 
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd2
delPs1=9.8;
print"\t delPs is :  psi \t",delPs1
delPT1=4.9;
print"\t delPt is :  psi \t",delPT1
#end

	 example 11.3 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for caustic 	
	 total heat required for caustic is :  Btu/hr 	6160000.0
	 for water 	
	 total heat required for water is :  Btu/hr 	6160000.0
	 delt1 is :  F 	40.0
	 delt2 is :  F 	70.0
	 LMTD is : F 	53.6684619193
	 R is :  	1.75
	 S is :  	0.363636363636
	 FT is 0.815 	
	 delt is :  F 	43.7397964643
	 caloric temperature of hot fluid is :  F 	155.0
	 caloric temperature of cold fluid is :  	100.0
	 A1 is :  ft**2 	563.331382215
	 number of tubes are :  	134.485146633
	 total surface area is :  ft**2 	586.432
	 correct design overall coefficient is : Btu/(hr)*(ft**2)*(F) 	240.152047558
	 hot fluid:shell side,caustic 	
	 flow area is :  ft**2 	0.187152777778
	 mass velocity is :  lb/(hr)*(ft**2) 	534322.820037
	 reynolds number is :  	17423.5702186
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	718.75
	 Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	718.75
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.132708333333
	 mass velocity is :  lb/(hr)*(ft**2) 	1160439.56044
	 V is  fps 	5.15750915751
	 reynolds number is :  	46350.8904888
	 Hi is :  Btu/(hr)*(ft**2)*(F) 	1165.6
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	972.1104
	 pressure drop  for annulus 	
	 number of crosses are : %.0f 	28
	 delPs is :  psi 	6.97705175383
	 allowable delPa is 10 psi 	
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	4.27604456957
	 delPr is :  psi 	2.88
	 delPT is :  psi 	7.15604456957
	 allowable delPa is 10 psi 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	413.224149078
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00174403453004
	 Adjustment of the baffie space to use the full 10 psi will still not permit the exchanger to make the 0.002 dirt factor. The value of UD has been assumed too high 	
	 Try a 21.25 in ID shell with four tube passes and a 6 in baffie space This corresponds to 170 tubes 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	39
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	200
	 calculated Rd is :  (hr)*(ft**2)*(F)/Btu 	0.0024
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.002
	 delPs is :  psi 	9.8
	 delPt is :  psi 	4.9

Example 11.4 pgno:241

In [20]:
print"\t example 11.4 \t"
print"\t approximate values are mentioned in the book \t"
T1=225.; # inlet hot fluid,F
T2=225.; # outlet hot fluid,F
t1=80.; # inlet cold fluid,F
t2=200.; # outlet cold fluid,F
W=10350; # lb/hr
w=115000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for steam \t"
l=962; # Btu/(lb)
Qh=((W)*(l)); # Btu/hr
print"\t total heat required for steam is :  Btu/hr \t",Qh
print"\t for alcohol \t"
c=0.72; # Btu/(lb)*(F)
Qc=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for alcohol is :  Btu/hr \t",Qc
Q=(Qh+Qc)/(2);
print"\t Q is :  V \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
Tc=((T2)+(T1)); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t2)); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  \t",tc
L=12;
UD1=200; # assume, from table 8
A1=((Q)/((UD1)*(LMTD)));
print"\t A1 is :f ft**2 \t",A1
a1=0.2618; # ft**2/lin ft
N1=(A1/(12*a1));
print"\t number of tubes are :  \t",N1
N2=232; # assuming two tube passes,23.25in ID, from table 9
A2=(N2*L*a1); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(LMTD)));
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
print"\t hot fluid:inner tube side,steam \t"
Nt=232;
n=2; # number of passes
L=12; #ft
at1=0.546; # flow area, in**2
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=0.0314; # at 225F,lb/(ft)*(hr)
D=0.0695; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hio=1500; # condensation of steam
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
print"\t cold fluid:shell side,alcohol \t"
ID=23.25; # in
C=0.25; # clearance
B=7; # minimum baffle spacing,from eq 11.4,in
PT=1.25;
As=((ID*C*B)/(144*PT)); # flow area,from eq 7.1,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(w/As); # mass velocity,from eq 7.2,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=1.45; # at 193.5F,lb/(ft)*(hr), from fig.14
De=0.72/12; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=83; # from fig.28
Z=0.195; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft)
Ho=((jH)*(1/De)*(Z)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ho
phys=1;
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t pressure drop  for inner pipe \t"
f=0.000175; # friction factor for reynolds number 52000, using fig.26
s=0.00076;
delPt=((f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(1)))/(2); # using eq.7.45,psi
print"\t delPt is :  psi \t",delPt
print"\t delPr is negligible \t"
print"\t allowable delPa is negligible \t"
print"\t pressure drop  for annulus \t"
f=0.0018; # friction factor for reynolds number 21000, using fig.29
s=0.78; # for reynolds number 21000,using fig.6
Ds=1.94; # ft
De=0.06;
N=(12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPa is 10 psi \t"
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
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t This is clearly an instance in which UD was assumed too high.It is now a question of how much too high. With the aid of the summary it is apparent thatin a larger shell a clean overall coefficient of about 200 may be expected \t"
print"\t Assume a 27in. ID shell with 2 tube passes,334 tubes and baffies spaced 7in. apart, since the pressure drop increases with the diameter of the shell for a given mass velocity. \t"
UC1=214;
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UC1
UD2=138.5;
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD2
Rd1=0.0025; 
print"\t calculated Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
Rd2=0.002; 
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd2
delPs1=0.23;
print"\t delPs is :  psi \t",delPs1
delPT1=7.1;
print"\t delPt is :  psi \t",delPT1
#end

	 example 11.4 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for steam 	
	 total heat required for steam is :  Btu/hr 	9956700
	 for alcohol 	
	 total heat required for alcohol is :  Btu/hr 	9936000.0
	 Q is :  V 	9946350.0
	 delt1 is :  F 	145.0
	 delt2 is :  F 	25.0
	 LMTD is : F 	68.3416294443
	 caloric temperature of hot fluid is :  F 	450.0
	 caloric temperature of cold fluid is :  	280.0
	 A1 is :f ft**2 	727.693360612
	 number of tubes are :  	231.631449138
	 total surface area is :  ft**2 	728.8512
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	199.68228374
	 hot fluid:inner tube side,steam 	
	 flow area is :  ft**2 	0.439833333333
	 mass velocity is :  lb/(hr)*(ft**2) 	23531.640773
	 reynolds number is :  	52084.3641314
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	1500
	 cold fluid:shell side,alcohol 	
	 flow area is :  ft**2 	0.226041666667
	 mass velocity is :  lb/(hr)*(ft**2) 	508755.760369
	 reynolds number is :  	21051.962498
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	269.75
	 Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	269.75
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	0.421749731499
	 delPr is negligible 	
	 allowable delPa is negligible 	
	 pressure drop  for annulus 	
	 number of crosses are :  	20
	 delPs is :  psi 	7.39957120534
	 allowable delPa is 10 psi 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	228.633987851
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.000634152640616
	 This is clearly an instance in which UD was assumed too high.It is now a question of how much too high. With the aid of the summary it is apparent thatin a larger shell a clean overall coefficient of about 200 may be expected 	
	 Assume a 27in. ID shell with 2 tube passes,334 tubes and baffies spaced 7in. apart, since the pressure drop increases with the diameter of the shell for a given mass velocity. 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	214
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	138.5
	 calculated Rd is :  (hr)*(ft**2)*(F)/Btu 	0.0025
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.002
	 delPs is :  psi 	0.23
	 delPt is :  psi 	7.1

Example 11.5 pgno:246

In [21]:
print"\t example 11.5 \t"
print"\t approximate values are mentioned in the book \t"
T1=250.; # inlet hot fluid,F
T2=125.; # outlet hot fluid,F
t1=80.; # inlet cold fluid,F
t2=100.; # outlet cold fluid,F
W=41300.; # lb/hr
w=64500.; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for gas \t"
c=0.25; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for gas is :  Btu/hr \t",Q
print"\t for water \t"
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 0.935 \t" # from fig 18
delt=(0.935*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
UD1=14.8; # assume, from table 8
A1=((Q)/((UD1)*(delt)));
print"\t A1 is :  ft**2 \t",A1
a1=0.2618; # ft**2/lin ft
N1=(A1/(12*a1));
print"\t number of tubes are :  \t",N1
N2=358; # assuming 12 tube passes, from table 9
L=12;
A2=(N2*L*a1); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(delt)));
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
print"\t When solved in a manner identical with the preceding examples and using the smallest integral number of bundle crosses (five) corresponding to a 28.8 in spacing \t"
UC1=22.7;
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UC1
UD2=14;
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD2
Rd1=0.027; 
print"\t calculated Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
Rd1=0.005; 
print"\t required Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd1
delPs1=5.2;
print"\t delPs is :  psi \t",delPs1
delPt1=1.0;
print"\t delPt is :  psi \t",delPt1
print"\t The first trial is disqualified because of failure to meet the required dirt factor and the the pressure drop is five times greater than the allowable \t"
print"\t This would be unsatisfactory, since gases require large inlet connections and the flow distribution on the first and third bundle crosses would be poor and the conditions of allowable pressure drop would still not be met \t"
UD1=15; # assume, from table 8
A1=((Q)/((UD1)*(delt)));
print"\t A1 is :  ft**2 \t",A1
a1=0.2618; # ft**2/lin ft
N1=(A1/(12*a1));
print"\t number of tubes are : %. \t",N1
N2=340; # assuming eight tube passes, from table 9
A2=(N2*L*a1); # ft**2
print"\t total surface area is :  ft**2 \t",A2
UD=((Q)/((A2)*(delt)));
print"\t correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
print"\t hot fluid:shell side,gas \t"
ID=31; # in
C=0.25; # clearance
B=24; # baffle spacing,in
PT=1.25;
As=((ID*C*B)/(144*PT)); # flow area,from eq 7.1,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(W/As)/(2); # mass velocity,from eq 7.2,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.050; # at 187.5F,lb/(ft)*(hr), from fig.15
De=0.99/12; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=105; # from fig.28
k=0.015; # Btu/(hr)(ft**2)( degree F/ft)
Z=0.94; # Z=((c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft)
Ho=((jH)*(k/De)*(Z)); # H0=(h0/phya),using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ho
phys=1;
ho=(Ho)*(phys); # from eq.6.36
print"\t Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t cold fluid:inner tube side,crude oil \t"
Nt=340;
n=12; # number of passes
L=12; #ft
at1=0.546; # flow area, in**2
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
V=(Gt/(3600*62.5));
print"\t V is :  fps \t",V
mu2=1.96; # at 90F,lb/(ft)*(hr)
D=0.0695; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=667; #Btu/(hr)*(ft**2)*(F)
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=0.83; # 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) \t",hio # calculation mistake
phyt=1;
print"\t pressure drop  for annulus \t"
f=0.0017; # friction factor for reynolds number 33000, using fig.29
s=0.0012; # for reynolds number 33000,using fig.6
Ds=31/12; # ft
N=(3); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t pressure drop  for inner pipe \t"
f=0.00022; # friction factor for reynolds number 21300, using fig.26
s=1;
delPt=((f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is :  psi \t",delPt
X1=0.052; # 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 \t",delPr
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi \t",delPT
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",round(Uc,1)
Rd=-1*(((Uc-UD)/((UD)*(Uc)))-0.02); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",round(Rd,4)
# end

	 example 11.5 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for gas 	
	 total heat required for gas is :  Btu/hr 	1290625.0
	 for water 	
	 total heat required for water is :  Btu/hr 	1290000.0
	 delt1 is :  F 	45.0
	 delt2 is :  F 	150.0
	 LMTD is : F 	87.3092936462
	 R is :  	6.25
	 S is :  	0.117647058824
	 FT is 0.935 	
	 delt is :  F 	81.6341895592
	 caloric temperature of hot fluid is :  F 	187.5
	 caloric temperature of cold fluid is :  F 	90.0
	 A1 is :  ft**2 	1067.71638982
	 number of tubes are :  	339.863887771
	 total surface area is :  ft**2 	1124.6928
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	14.0502389358
	 When solved in a manner identical with the preceding examples and using the smallest integral number of bundle crosses (five) corresponding to a 28.8 in spacing 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	22.7
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	14
	 calculated Rd is :  (hr)*(ft**2)*(F)/Btu 	0.027
	 required Rd is :  (hr)*(ft**2)*(F)/Btu 	0.005
	 delPs is :  psi 	5.2
	 delPt is :  psi 	1.0
	 The first trial is disqualified because of failure to meet the required dirt factor and the the pressure drop is five times greater than the allowable 	
	 This would be unsatisfactory, since gases require large inlet connections and the flow distribution on the first and third bundle crosses would be poor and the conditions of allowable pressure drop would still not be met 	
	 A1 is :  ft**2 	1053.48017129
	 number of tubes are : %. 	335.332369267
	 total surface area is :  ft**2 	1068.144
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	14.7940751147
	 hot fluid:shell side,gas 	
	 flow area is :  ft**2 	1.03333333333
	 mass velocity is :  lb/(hr)*(ft**2) 	19983.8709677
	 reynolds number is :  	32973.3870968
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	17.9454545455
	 Correct h0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	17.9454545455
	 cold fluid:inner tube side,crude oil 	
	 flow area is :  ft**2 	0.107430555556
	 mass velocity is : %lb/(hr)*(ft**2) 	600387.847447
	 V is :  fps 	2.6683904331
	 reynolds number is :  	21289.2629579
	 hi is :  Btu/(hr)*(ft**2)*(F) 	667
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	553.61
	 pressure drop  for annulus 	
	 number of crosses are :  	3
	 delPs is :  psi 	0.788231357313
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	3.14770229996
	 delPr is :  psi 	2.496
	 delPT is :  psi 	5.64370229996
	 clean overall coefficient is : Btu/(hr)*(ft**2)*(F) 	17.4
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.0099