Chapter 12 Condensation of Single Vapours

Example 12.1 pgno:274

In [3]:
print"\t example 12.1 \t"
print"\t approximate values are mentioned in the book \t"
T1=244.;# inlet hot fluid,F
T2=244.; # outlet hot fluid,F
t1=85.; # inlet cold fluid,F
t2=120.; # outlet cold fluid,F
W=60000; # lb/hr
w=488000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for propanol \t"
l=285; # Btu/(lb)
Q=((W)*(l)); # Btu/hr
print"\t total heat required for propanol 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
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=101; # assume, from table 8
A1=((Q)/((UD1)*(LMTD)));
print"\t A1 is :  ft**2 \t",A1
a1=0.1963; # ft**2/lin ft
N1=(A1/(8*a1));
print"\t number of tubes are :  \t",N1
N2=766; # assuming 4 tube passes, from table 9
A2=(N2*8*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:shell side,propanol \t"
ID=31; # in
C=0.1875; # clearance
B=31; # baffle spacing,in
PT=0.937;
L=8; # ft
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
G1=(W/(L*N2**(2/3)))# from eq.12.43
print"\t G1 is : %.1f lb/(hr)*(lin ft) \t",G1
print"\t cold fluid:inner tube side,water \t"
Nt=766;
n=4; # number of passes
L=8; #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
V=(Gt/(3600*62.5));
print"\t V is :  fps \t",V
mu2=1.74; # at 102.5F,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=1300; #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)); # using eq.6.5
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio # calculation mistake
ho=200; # assumption
tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \t",tw
tf=(Tc+tw)/(2); # from eq 12.19
print"\t tf is :  F \t",tf
kf=0.094; # Btu/(hr)*(ft**2)*(F/ft), from table 4
sf=0.8; # from table 6
muf=0.62; # cp, from fig 14
ho=172; # Btu/(hr)*(ft**2)*(F), from fig 12.9
print"\t Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t Based on h=172 instead of the assumed 200 a new value of tw,and tf could be obtained to give a more exact value of h based on fluid properties at a value of tf more nearly correct \t"
print"\t pressure drop  for annulus \t"
mu1=0.0242; # lb/(ft)*(hr), fir 15
De=0.0458; # fig 28
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
f=0.00141; # friction factor for reynolds number 84600, using fig.29
s=0.00381; # for reynolds number 84600,using fig.6
Ds=31/12; # ft
phys=1;
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)))/(2); # using eq.12.47,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPa is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00019; # friction factor for reynolds number 36200, 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",round(delPt,1)
X1=0.2; # X1=((V**2)/(2*g)),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",round(delPT,1)
print"\t allowable delPT is 10 psi \t"
Uc=((hio)*(ho)/(hio+ho)); # clean overalcoefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",round(Uc,1)
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",round(Rd,3)
# end

	 example 12.1 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for propanol 	
	 total heat required for propanol is :  Btu/hr 	17100000
	 for water 	
	 total heat required for water is :  Btu/hr 	17080000.0
	 delt1 is :  F 	159.0
	 delt2 is :  F 	124.0
	 LMTD is : F 	140.93382183
	 caloric temperature of hot fluid is :  F 	244.0
	 caloric temperature of cold fluid is :  F 	102.5
	 A1 is :  ft**2 	1199.91715754
	 number of tubes are :  	764.083773266
	 total surface area is :  ft**2 	1202.9264
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	100.74733825
	 hot fluid:shell side,propanol 	
	 flow area is :  ft**2 	1.33543445393
	 mass velocity is :  lb/(hr)*(ft**2) 	44929.1987513
	 G1 is : %.1f lb/(hr)*(lin ft) 	7500
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.401618055556
	 mass velocity is :  lb/(hr)*(ft**2) 	1215084.81317
	 V is :  fps 	5.40037694742
	 reynolds number is :  	36103.3820924
	 hi is :  Btu/(hr)*(ft**2)*(F) 	1300
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	1074.66666667
	 tw is :  F 	124.701882845
	 tf is :  F 	184.350941423
	 Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	172
	 Based on h=172 instead of the assumed 200 a new value of tw,and tf could be obtained to give a more exact value of h based on fluid properties at a value of tf more nearly correct 	
	 pressure drop  for annulus 	
	 reynolds number is :  	85031.2935045
	 number of crosses are :  	3
	 delPs is :  psi 	0.937425488924
	 allowable delPa is 2 psi 	
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	3.32625636683
	 delPr is :  psi 	3.2
	 delPT is :  psi 	6.5
	 allowable delPT is 10 psi 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	148.3
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.003

Example 12.2 pgno:277

In [6]:
print"\t example 12.2 \t"
print"\t approximate values are mentioned in the book \t"
T1=244.; # inlet hot fluid,F
T2=244.; # outlet hot fluid,F
t1=85.; # inlet cold fluid,F
t2=120.; # outlet cold fluid,F
W=60000; # lb/hr
w=488000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for propanol \t"
l=285; # Btu/(lb)
Q=((W)*(l)); # Btu/hr
print"\t total heat required for propanol 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 F \t",LMTD
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=70; # assume, from table 8
A1=((Q)/((UD1)*(LMTD)));
print"\t A1 is :  ft**2 \t",A1
N2=766; # assuming 4 tube passes, from table 9
a1=0.1963; # ft**2/lin ft
L=(A1/(N2*a1));
print"\t L is :  ft \t",L
A2=(N2*12*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:shell side,propanol \t"
Do=0.0625; # ft
G1=(W/(3.14*N2*Do)); # from eq.12.36
print"\t G1 is :  lb/(hr)*(lin ft) \t",G1
print"\t cold fluid:inner tube side,water \t"
Nt=766;
n=4; # number of passes
L=12; #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
V=(Gt/(3600*62.5));
print"\t V is :  fps \t",V
mu2=1.74; # at 102.5F,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=1300; #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)); # using eq.6.5
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
ho=100; # assumption
tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \t",tw
tf=(Tc+tw)/(2); # from eq 12.19
print"\t tf is :  F \t",tf
kf=0.0945; # Btu/(hr)*(ft**2)*(F/ft), from table 4
sf=0.76; # from table 6
muf=0.65; # cp, from fig 14
ho=102; # Btu/(hr)*(ft**2)*(F), from fig 12.9
print"\t Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t pressure drop  for annulus \t"
ID=31; # in
C=0.1875; # clearance
B=29; # baffle spacing,in
PT=0.937;
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=0.0242; # lb/(ft)*(hr), fig 15
De=0.0458; # fig 28
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
f=0.0014; # friction factor for reynolds number 91000, using fig.29
s=0.00381; # for reynolds number 91000,using fig.6
Ds=31/12; # ft
phys=1;
N=(5); # 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)))/(2); # using eq.12.47,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPa is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00019; # friction factor for reynolds number 36200, 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",round(delPt,1)
X1=0.2; # X1=((V**2)/(2*g)),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",round(delPT,1)
print"\t allowable delPT is 10 psi \t"
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,eq 6.38,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",round(Uc,1)
Rd=((Uc-UD)/((UD)*(Uc))); # eq 6.13,(hr)*(ft**2)*(F)/Btu
print"\t actual Rd is : (hr)*(ft**2)*(F)/Btu \t",round(Rd,3)
# end

	 example 12.2 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for propanol 	
	 total heat required for propanol is :  Btu/hr 	17100000
	 for water 	
	 total heat required for water is :  Btu/hr 	17080000.0
	 delt1 is :  F 	159.0
	 delt2 is :  F 	124.0
	 LMTD is f F 	140.93382183
	 caloric temperature of hot fluid is :  F 	244.0
	 caloric temperature of cold fluid is :  F 	102.5
	 A1 is :  ft**2 	1731.30904159
	 L is :  ft 	11.5139815143
	 total surface area is :  ft**2 	1804.3896
	 correct design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	67.164892167
	 hot fluid:shell side,propanol 	
	 G1 is :  lb/(hr)*(lin ft) 	399.12856929
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.401618055556
	 mass velocity is :  lb/(hr)*(ft**2) 	1215084.81317
	 V is :  fps 	5.40037694742
	 reynolds number is :  	36103.3820924
	 hi is :  Btu/(hr)*(ft**2)*(F) 	1300
	 Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	1074.66666667
	 tw is :  F 	114.545970488
	 tf is :  F 	179.272985244
	 Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	102
	 pressure drop  for annulus 	
	 flow area is :  ft**2 	1.24927739239
	 mass velocity is :  lb/(hr)*(ft**2) 	48027.7641824
	 reynolds number is :  	90895.5206428
	 number of crosses are :  	5
	 delPs is :  psi 	1.7726452141
	 allowable delPa is 2 psi 	
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	5.0
	 delPr is :  psi 	3.2
	 delPT is :  psi 	8.2
	 allowable delPT is 10 psi 	
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	93.2
	 actual Rd is : (hr)*(ft**2)*(F)/Btu 	0.004

Example 12.3 pgno:285

In [10]:
print"\t example 12.3 \t"
print"\t approximate values are mentioned in the book \t"
T1=200.; # inlet hot fluid,F
T2=130.; # outlet hot fluid,F
T3=125.; # after condensation
t1=65.; # inlet cold fluid,F
t3=100.; # outlet cold fluid,F
W=27958; # lb/hr
w=135500; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for butane \t"
c=0.44; # Btu/(lb)(F)
qd=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for desuperheating of butane is :  Btu/hr \t",qd
HT2=309; # enthalpy at T2, Btu/lb
HT3=170; # enthalpy at T3, Btu/lb
qc=(W*(HT2-HT3)); # for condensation
print"\t total heat required for condensing of butane is :  Btu/hr \t",qc
Q=qd+qc;
print"\t total heat required for  butane is :  Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t3-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
deltw=(qc/w);
print"\t deltw is :  F \t",deltw
t2=t1+deltw;
print"\t t2 is :  F \t",t2
print"\t for desuperheating \t"
delt1=T2-t2; #F
delt2=T1-t3; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTDd=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTDd
w1=(qd/LMTDd);
print"\t w1 is :  lb/hr \t",w1
print"\t for condensing \t"
delt3=T3-t1; #F
delt4=T2-t2; # F
print"\t delt1 is :  F \t",delt3
print"\t delt2 is :  F \t",delt4
LMTDc=((delt4-delt3)/((2.3)*(log10(delt4/delt3))));
print"\t LMTD is : F \t",LMTDc
w2=(qc/LMTDc);
print"\t w1 is :  lb/hr \t",w2
delt=(Q/(w1+w2));
print"\t delt is :  F \t",delt
Tc=((T3)+(T2))/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t3))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,butane \t"
ID=23.25; # in
C=0.25; # clearance
B=12; # baffle spacing,in
PT=1;
As=((ID*C*B)/(144*PT)); # flow area,ft**2
print"\t flow area is :f ft**2 \t",As
print"\t desuperheating \t"
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.0242; # at 165F,lb/(ft)*(hr), from fig.15
De=0.73/12; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=239; # from fig.28
k=0.012; # Btu/(hr)*(ft**2)*(F/ft), from table 5
Z=0.96; # Z=((c)*(mu1)/k)**(1/3)
ho=((jH)*(k/De)*(Z)); # H0=(h0/phya),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,water \t"
Nt=352;
n=4; # number of passes
L=16; #ft
at1=0.302; # flow area,table 10, 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=2.11; # at 82.5F, fig 14,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=800; # fig 25,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)); # using eq.6.5
print"\t Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
Ud=((hio)*(ho)/(hio+ho)); # clean overall coefficient,eq 6.38,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Ud
Ad=(qd/(Ud*LMTDd));
print"\t clean surface required for desuperheating :  ft**2 \t",Ad
print"\t for condensaton \t"
Lc=16*0.6; # condensation occurs 60 of the tube length
G1=(W/(Lc*Nt**(2/3))); # from eq.12.43
print"\t G1 is :  lb/(hr)*(lin ft) \t",G1
ho=200; # assumption
tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \t",tw
tf=(Tc+tw)/(2); # from eq 12.19
print"\t tf is :  F \t",tf
kf=0.075; # Btu/(hr)*(ft**2)*(F/ft)
sf=0.55; # from table 6
muf=0.14; # cp, from fig 14
ho=207; # Btu/(hr)*(ft**2)*(F), from fig 12.9
print"\t Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",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) \t",Uc
Ac=(qc/(Uc*LMTDc));
print"\t clean surface required for desuperheating :  ft**2 \t",Ac
AC=Ad+Ac;
print"\t total clean surface :  ft**2 \t",AC
lc=(Ac/(Ac+Ad));
print"\t assumed condensing length percentage :  \t",lc
UC=((Ud*Ad)+(Uc*Ac))/(AC);
print"\t weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) \t",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 \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",round(Rd,3)
print"\t pressure drop  for annulus \t"
print"\t desuperheating \t"
Ld=6.4; #ft
De=0.0608; # fig 28
f=0.0013; # friction factor for reynolds number 145000, using fig.29
Ds=1.94; # ft
phys=1;
N=(12*Ld/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
row=(58.1/((359)*(625/492)*(14.7/99.7)));
print"\t row is  lb/ft**3 \t",row
s=(row/62.5);
print"\t s is  \t",s
delPsd=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPsd
print"\t condensation \t"
N=(12*Lc/B); # number of crosses,using eq.7.43
print"\t number of crosses are : \t",N
delPsc=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys)))/(2); # using eq 12.47,psi
print"\t delPsc is :  psi \t",delPsc
delPS=delPsd+delPsc;
print"\t delPS is :  psi \t",delPS
print"\t allowable delPa is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00023; # friction factor for reynolds number 17900, 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",round(delPt)
X1=0.075; # X1=((V**2)/(2*g)),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",round(delPT,1)
print"\t allowable delPa is 10 psi \t"
#end

	 example 12.3 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for butane 	
	 total heat required for desuperheating of butane is :  Btu/hr 	861106.4
	 total heat required for condensing of butane is :  Btu/hr 	3886162
	 total heat required for  butane is :  Btu/hr 	4747268.4
	 for water 	
	 total heat required for water is :  Btu/hr 	4742500.0
	 deltw is :  F 	28
	 t2 is :  F 	93.0
	 for desuperheating 	
	 delt1 is :  F 	37.0
	 delt2 is :  F 	100.0
	 LMTD is : F 	63.4354185149
	 w1 is :  lb/hr 	13574.5364366
	 for condensing 	
	 delt1 is :  F 	60.0
	 delt2 is :  F 	37.0
	 LMTD is : F 	47.6304956503
	 w1 is :  lb/hr 	81589.787109
	 delt is :  F 	49.8348522147
	 caloric temperature of hot fluid is :  F 	127.5
	 caloric temperature of cold fluid is :  F 	82.5
	 hot fluid:shell side,butane 	
	 flow area is :f ft**2 	0.484375
	 desuperheating 	
	 mass velocity is :  lb/(hr)*(ft**2) 	57719.7419355
	 reynolds number is :  	145094.392606
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	45.2593972603
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.184555555556
	 mass velocity is :  lb/(hr)*(ft**2) 	734196.267309
	 V is :  fps 	3.26309452137
	 reynolds number is :  	17989.5483506
	 hi is :  Btu/(hr)*(ft**2)*(F) 	800
	 Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	661.333333333
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	42.3603962493
	 clean surface required for desuperheating :  ft**2 	320.453481046
	 for condensaton 	
	 G1 is :  lb/(hr)*(lin ft) 	2912.29166667
	 tw is :  F 	92.9489164087
	 tf is :  F 	110.224458204
	 Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	207
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	157.653742802
	 clean surface required for desuperheating :  ft**2 	517.525214808
	 total clean surface :  ft**2 	837.978695854
	 assumed condensing length percentage :  	0.617587556066
	 weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) 	113.564132378
	 total surface area is :  ft**2 	1105.5616
	 actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	86.0778119876
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.003
	 pressure drop  for annulus 	
	 desuperheating 	
	 number of crosses are :  	6.4
	 row is  lb/ft**3 	1.09763894416
	 s is  	0.0175622231065
	 delPs is :  psi 	0.964762124732
	 condensation 	
	 number of crosses are : 	9.6
	 delPsc is :  psi 	0.723571593549
	 delPS is :  psi 	1.68833371828
	 allowable delPa is 2 psi 	
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	3.0
	 delPr is :  psi 	1.2
	 delPT is :  psi 	4.1
	 allowable delPa is 10 psi

Example 12.4 pgno:290

In [1]:
print"\t example 12.4 \t"
print"\t approximate values are mentioned in the book \t"
T1=130.; # inlet hot fluid,F
T2=125.; # outlet hot fluid,F
T3=100.; # after sucooling
t1=80.; # inlet cold fluid,F
t3=100.; # outlet cold fluid,F
W=21000; # lb/hr
w=167000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for pentane \t"
HT1=315; # enthalpy at T1, Btu/lb
HT2=170; # enthalpy at T2, Btu/lb
qc=(W*(HT1-HT2)); # for condensation
print"\t total heat required for condensing of pentane is :  Btu/hr \t",qc
c=0.57; # Btu/(lb)(F)
qs=((W)*(c)*(T2-T3)); # Btu/hr
print"\t total heat required for subcooling of pentane is :  Btu/hr \t",qs
Q=qs+qc;
print"\t total heat required for  pentane is :  Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t3-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
deltw=(qc/w);
print"\t deltw is : F \t",deltw
t2=t3-deltw;
print"\t t2 is : F \t",t2
print"\t for condensing \t"
delt1=T2-t2; #F
delt2=T1-t3; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTDc=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTDc
w1=(qc/LMTDc);
print"\t w1 is : lb/hr \t",w1
print"\t subcooling \t"
delt3=T3-t1; #F
delt4=T2-t2;
print"\t delt1 is : F \t",delt3
print"\t delt2 is :  F \t",delt4
LMTDs=((delt4-delt3)/((2.3)*(log10(delt4/delt3))));
print"\t LMTD is f F \t",LMTDs
w2=(qs/LMTDs);
print"\t w1 is :  lb/hr \t",w2
delt=(Q/(w1+w2));
print"\t delt is :  F \t",delt
Tc=((T1)+(T2))/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t3))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,pentane \t"
print"\t for condensaton \t"
Do=0.0625; # ft
Nt=370; # number of tubes
G1=(W/(3.14*Nt*Do)); # from eq.12.42
print"\t G1 is :  lb/(hr)*(lin ft) \t",G1
print"\t cold fluid:inner tube side,water \t"
n=4; # 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
V=(Gt/(3600*62.5));
print"\t V is :  fps \t",V
mu2=1.98; # at 90F,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is : %.2e \t",Ret
hi=940; #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)); # using eq.6.5
print"\t Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
ho=125; # assumption
tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \t",tw
tf=(Tc+tw)/(2); # from eq 12.19
print"\t tf is :  F \t",tf
kf=0.077; # Btu/(hr)*(ft**2)*(F/ft), table 4
sf=0.6; # from table 6
muf=0.19; # cp, from fig 14
ho=120; # Btu/(hr)*(ft**2)*(F), from fig 12.9
print"\t Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",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) \t",Uc
Ac=(3040000/(104*36.4));
print"\t clean surface required for dcondensation :  ft**2 \t",Ac
print"\t subcooling \t"
ID=25; # in
C=0.25; # clearance
B=12; # 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=0.46; # at 112.5F,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=46.5; # from fig.28
k=0.077; # Btu/(hr)*(ft**2)*(F/ft), from table 4
Z=1.51; # Z=((c)*(mu1)/k)**(1/3)
ho=((jH)*(k/De)*(Z)); # using eq.6.15b,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",ho
Us=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \t",Us
As=(qs/(Us*LMTDs));
print"\t clean surface required for desuperheating :  ft**2 \t",As
AC=As+Ac;
print"\t total clean surface :  ft**2 \t",AC
UC=((Us*As)+(Uc*Ac))/(AC);
print"\t weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) \t",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 \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"
print"\t condensation \t"
Lc=13.4; #ft
De=0.0792; # fig 28
f=0.0012; # friction factor for reynolds number 193000, using fig.29
mu3=0.0165; # at 127.5F
Ds=2.08; # ft
phys=1;
Res1=(De*Gs/mu3);
print"\t reynolds number is  \t",Res1
rowvap=(72.2/((359)*(590/492)*(14.7/25)));
print"\t rowvapour is  ld/ft**3 \t",rowvap
s=(rowvap/62.5);
print"\t s is  \t",s
N=(12*Lc/B)+(1); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPsc=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys)))/(2); # using eq.12.47,psi
print"\t delPsc is :  psi \t",delPsc
print"\t delPss is negligible \t"
print"\t allowable delPa is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.00022; # friction factor for reynolds number 22500, 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",round(delPt,1)
X1=0.1; # X1=((V**2)/(2*g)),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",round(delPT,1)
print"\t allowable delPT is 10 psi \t"
#end

	 example 12.4 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for pentane 	
	 total heat required for condensing of pentane is :  Btu/hr 	3045000
	 total heat required for subcooling of pentane is :  Btu/hr 	299250.0
	 total heat required for  pentane is :  Btu/hr 	3344250.0
	 for water 	
	 total heat required for water is :  Btu/hr 	3340000.0
	 deltw is : F 	18
	 t2 is : F 	82.0
	 for condensing 	
	 delt1 is :  F 	43.0
	 delt2 is :  F 	30.0
	 LMTD is : F 	36.1514237668
	 w1 is : lb/hr 	84229.0477863
	 subcooling 	
	 delt1 is : F 	20.0
	 delt2 is :  F 	43.0
	 LMTD is f F 	30.0807554052
	 w1 is :  lb/hr 	9948.22091297
	 delt is :  F 	35.465033613
	 caloric temperature of hot fluid is :  F 	127.5
	 caloric temperature of cold fluid is :  F 	90.0
	 hot fluid:shell side,pentane 	
	 for condensaton 	
	 G1 is :  lb/(hr)*(lin ft) 	289.206403856
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.193993055556
	 mass velocity is :  lb/(hr)*(ft**2) 	860855.557544
	 V is :  fps 	3.8260247002
	 reynolds number is : %.2e 	22477.8951137
	 hi is :  Btu/(hr)*(ft**2)*(F) 	940
	 Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	777.066666667
	 tw is :  F 	95.1964008573
	 tf is :  F 	111.348200429
	 Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	120
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	103.947681332
	 clean surface required for dcondensation :  ft**2 	803.043110735
	 subcooling 	
	 flow area is :  ft**2 	0.520833333333
	 mass velocity is :  lb/(hr)*(ft**2) 	40320.0
	 reynolds number is :  	6939.13043478
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	68.2933263158
	 clean overall coefficient is : Btu/(hr)*(ft**2)*(F) 	62.7761756841
	 clean surface required for desuperheating :  ft**2 	158.471279981
	 total clean surface :  ft**2 	961.514390716
	 weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) 	97.1620302154
	 total surface area is :  ft**2 	1162.096
	 actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	81.0408681377
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00204736690467
	 pressure drop  for annulus 	
	 condensation 	
	 reynolds number is  	193536.0
	 rowvapour is  ld/ft**3 	0.342030962803
	 s is  	0.00547249540485
	 number of crosses are :  	14.4
	 delPsc is :  psi 	1.29132938612
	 delPss is negligible 	
	 allowable delPa is 2 psi 	
	 pressure drop  for inner pipe 	
	 delPt is :  psi 	3.9
	 delPr is :  psi 	1.6
	 delPT is :  psi 	5.5
	 allowable delPT is 10 psi

Example 12.5 pgno:295

In [5]:
print"\t example 12.5 \t"
print"\t approximate values are mentioned in the book \t"
T1=130; # inlet hot fluid,F
T2=125; # outlet hot fluid,F
T3=100; # after subcooling
t1=80; # inlet cold fluid,F
t3=100; # outlet cold fluid,F
W=21000; # lb/hr
w=167000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for pentane \t"
c=0.57; # Btu/(lb)(
qs=((W)*(c)*(T2-T3)); # Btu/hr
print"\t total heat required for subcooling of pentane is :  Btu/hr \t",qs
HT1=315; # enthalpy at T1, Btu/lb
HT2=170; # enthalpy at T2, Btu/lb
qc=(W*(HT1-HT2)); # for condensation
print"\t total heat required for condensing of pentane is :  Btu/hr \t",qc
Q=qs+qc;
print"\t total heat required for  pentane is :  Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)-++++++++-
Q=((w)*(c)*(t3-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
deltw=18.2;
print"\t deltw is :  F \t",deltw
t2=t3-deltw;
print"\t t2 is :  F \t",t2
print"\t for condensing \t"
delt1=T2-t2; #F
delt2=T1-t3; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTDc=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));
print"\t LMTD is : F \t",LMTDc
w1=(qc/LMTDc);
print"\t w1 is :  lb/hr \t",w1
print"\t subcooling \t"
delt3=T3-t1; #F
delt4=T2-t2; # F
print"\t delt1 is :  F \t",delt3
print"\t delt2 is :  F \t",delt4
LMTDs=((delt4-delt3)/((2.3)*(log10(delt4/delt3))));
print"\t LMTD is : F \t",LMTDs
w2=(qs/LMTDs);
print"\t w1 is :  lb/hr \t",w2
delt=(Q/(w1+w2));
print"\t delt is : % F \t",delt
Tc=((T1)+(T2))/(2); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t3))/(2); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,pentane \t"
C1=0.198; # for 0.3Ds
Ds=25; # in
L=16; # ft
N=370
a=(C1*Ds**2);
print"\t a is :  in**2 \t",a
N1=((N*a*4)/(3.14*Ds**2));
print"\t number of submerged tubes are :  \t",N1
Nt=N-N1;
print"\t number of tubes for condensation are :  \t",Nt
Af=(N1/N);
print"\t flooded surface :  \t",Af
print"\t for condensaton \t"
G1=(W/(L*Nt**(2/3))); # from eq.12.43
print"\t G1 is :  lb/(hr)*(lin ft) \t",G1
print"\t cold fluid:inner tube side,water \t"
n=4; # 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
V=(Gt/(3600*62.5));
print"\t V is :  fps \t",V
mu2=1.98; # lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=940; #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)); # using eq.6.5
print"\t Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
ho=251; # Btu/(hr)*(ft**2)*(F), from fig 12.9
print"\t Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",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) \t",Uc
Ac=(qc/(Uc*LMTDc));
print"\t clean surface required for dcondensation :  ft**2 \t",Ac
print"\t subcooling \t"
ho=50; # Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",ho
Us=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Us
As=(qs/(Us*LMTDs));
print"\t clean surface required for desuperheating :  ft**2 \t",As
AC=As+Ac;
print"\t total clean surface :  ft**2 \t",AC
UC=((Us*As)+(Uc*Ac))/(AC);
print"\t weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) \t",UC
A=1160; # 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",round(Rd,4)
print"\t pressure drop  for annulus \t"
print"\t condensation \t"
print"\t It will be necessary to spread the batHes to a spacing of 18in.to compensate for the reduction in crossfiow area due to the flooded subcooling zone. The tube-side pressure drop will be the same as before. Assume bundle flooded to 0.3Ds.\t"
As=0.547; # ft**2
Gs=(W/(As)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
De=0.0792; # fig 28
Res=((De)*(Gs)/0.0165); # reynolds number
print"\t reynolds number is :  \t",Res
f=0.00121; # friction factor for reynolds number 193000, using fig.29
s=0.00454; # for reynolds number 193000,using fig.6
Ds=2.08; # ft
B=18
phys=1;
N=(12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
delPsc=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys)))/(2); # using eq.12.47,psi
print"\t delPsc is :  psi \t",round(delPsc)
print"\t delPss is negligible \t"
print"\t allowable delPT is 10 psi \t"
#e 

	 example 12.5 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for pentane 	
	 total heat required for subcooling of pentane is :  Btu/hr 	299250.0
	 total heat required for condensing of pentane is :  Btu/hr 	3045000
	 total heat required for  pentane is :  Btu/hr 	3344250.0
	 for water 	
	 total heat required for water is :  Btu/hr 	3340000
	 deltw is :  F 	18.2
	 t2 is :  F 	81.8
	 for condensing 	
	 delt1 is :  F 	43.2
	 delt2 is :  F 	30
	 LMTD is : F 	36.2404655222
	 w1 is :  lb/hr 	84022.0994992
	 subcooling 	
	 delt1 is :  F 	20
	 delt2 is :  F 	43.2
	 LMTD is : F 	30.1594958556
	 w1 is :  lb/hr 	9922.24808506
	 delt is : % F 	35.5529639184
	 caloric temperature of hot fluid is :  F 	127
	 caloric temperature of cold fluid is :  F 	90
	 hot fluid:shell side,pentane 	
	 a is :  in**2 	123.75
	 number of submerged tubes are :  	93.3248407643
	 number of tubes for condensation are :  	276.675159236
	 flooded surface :  	0.252229299363
	 for condensaton 	
	 G1 is :  lb/(hr)*(lin ft) 	1312.5
	 cold fluid:inner tube side,water 	
	 flow area is :  ft**2 	0.145062323071
	 mass velocity is :  lb/(hr)*(ft**2) 	1151229.32312
	 V is :  fps 	5.11657476943
	 reynolds number is :  	30059.8767704
	 hi is :  Btu/(hr)*(ft**2)*(F) 	940
	 Correct hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	777.066666667
	 Correct ho to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	251
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	189.718954672
	 clean surface required for dcondensation :  ft**2 	442.876673258
	 subcooling 	
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	50
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	46.9772690634
	 clean surface required for desuperheating :  ft**2 	211.213812188
	 total clean surface :  ft**2 	654.090485446
	 weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) 	143.625919769
	 total surface area is :  ft**2 	1160
	 actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	80.9865065381
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.0054
	 pressure drop  for annulus 	
	 condensation 	
	 It will be necessary to spread the batHes to a spacing of 18in.to compensate for the reduction in crossfiow area due to the flooded subcooling zone. The tube-side pressure drop will be the same as before. Assume bundle flooded to 0.3Ds.	
	 mass velocity is :  lb/(hr)*(ft**2) 	38391.2248629
	 reynolds number is :  	184277.879342
	 number of crosses are :  	10
	 delPsc is :  psi 	1.0
	 delPss is negligible 	
	 allowable delPa is 2 psi

Example 12.6 pgno:299

In [8]:
print"\t example 12.6 \t"
print"\t approximate values are mentioned in the book \t"
T1=176.; # inlet hot fluid,F
T2=176.; # outlet hot fluid,F
t1=85.; # inlet cold fluid,F
t2=120.; # outlet cold fluid,F
W=30000; # lb/hr
w=120000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
print"\t for carbon disulfide \t"
l=140; # Btu/(lb)
Q=((W)*l); # Btu/hr
print"\t total heat required for carbon disulfide 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
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:inner tube side,carbon disulfide \t"
hio=300; # Btu/(hr)*(ft**2)*(F)
print"\t cold fluid:shell side,water \t"
ID=17.25; # in
C=0.25; # clearance
B=6; # 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=1.7; # at 280F,lb/(ft)*(hr), from fig.14
De=0.0792; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=103; # from fig.28
k=0.36; # Btu/(hr)*(ft**2)*(F/ft), from fig.1
Z=1.68; # Z=((c)*(mu1)/k)**(1/3); # prandelt number
ho=((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",ho
tw=(tc)+(((hio)/(hio+ho))*(Tc-tc)); # from eq.5.31
print"\t tw is :  F \t",tw
tf=(Tc+tw)/(2); # from eq 12.19
print"\t tf is :  F \t",tf
print"\t hot fluid:inner tube side,carbon disulfide \t"
kf=0.09; # Btu/(hr)*(ft**2)*(F/ft), from fig 14
sf=1.26; # from table 6
rowf=78.8; # lb/ft**3
muf=0.68; # cp, from fig 24
Nt=177;
D=0.0517; # ft
G1=(W/(3.14*Nt*D));
print"\t G1 is : lb/(hr)*(lin ft) \t",G1
Ret=((4)*(G1)/muf); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=(0.251*(((kf**3)*(rowf**2)*(4.17*10**8))/(muf**2))**(1/3)); # hi*(((kf**3)*(rowf**2)*(4.17*10**8))/(muf**2))**(-1)=0.251, from fig 12.12
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=0.62; # ft
OD=.75; #ft
hio1=((hi)*(ID/OD)); #Hio=(hio/phyp), using eq.6.5
print"\t Correct hio1 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio1
Uc=((hio1)*(ho)/(hio1+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
L=16;
A=(Nt*L*A2); # ft**2
print"\t total surface area is :  ft**2 \t",A
UD=((Q)/((A)*(LMTD)));
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 inner pipe \t"
n=1; # number of passes
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=(30000/(0.372)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
mu2=0.029; # at inlet,lb/(ft)*(hr)
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
row=(76.1/((359)*(636/492)*(14.7/39.7)));
print"\t row is  ld/ft**3 \t",row
s=(row/62.5);
print"\t s is  \t",s
f=0.000138; # friction factor for reynolds number 143000, using fig.26
delPt=((f*(Gt**2)*(16)*(1))/(5.22*(10**10)*(0.0517)*(s)))/(2); # using eq.7.45,psi
print"\t delPt is :  psi \t",round(delPt+0.1,1)
print"\t allowable delPa is negligible psi \t"
print"\t pressure drop  for annulus \t"
f=0.0017; # friction factor for reynolds number 31000, using fig.29
s=1; # for reynolds number 31000,using fig.6
Ds=17.25/12.; # ft
B=6.;
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))); # using eq.7.44,psi
print"\t delPs is :  psi \t",round(delPs,1)
print"\t allowable delPT is 2 psi \t"
#end

	 example 12.6 	
	 approximate values are mentioned in the book 	
	 1.for heat balance 	
	 for carbon disulfide 	
	 total heat required for carbon disulfide is :  Btu/hr 	4200000
	 for water 	
	 total heat required for water is :  Btu/hr 	4200000.0
	 delt1 is :  F 	91.0
	 delt2 is :  F 	56.0
	 LMTD is : F 	72.1704928998
	 caloric temperature of hot fluid is :  F 	176.0
	 caloric temperature of cold fluid is :  F 	102.5
	 hot fluid:inner tube side,carbon disulfide 	
	 cold fluid:shell side,water 	
	 flow area is :  ft**2 	0.1796875
	 mass velocity is :  lb/(hr)*(ft**2) 	667826.086957
	 reynolds number is :  	31112.8388747
	 individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	786.545454545
	 tw is :  F 	122.793674699
	 tf is :  F 	149.396837349
	 hot fluid:inner tube side,carbon disulfide 	
	 G1 is : lb/(hr)*(lin ft) 	1044.06562495
	 reynolds number is :  	6141.56249973
	 hi is :  Btu/(hr)*(ft**2)*(F) 	0.251
	 Correct hio1 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	0.207493333333
	 clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	0.207438610331
	 total surface area is :  ft**2 	555.9216
	 actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	104.682978368
	 actual Rd is :  (hr)*(ft**2)*(F)/Btu 	-4.81115068056
	 pressure drop  for inner pipe 	
	 flow area is :  ft**2 	0.371208333333
	 mass velocity is :  lb/(hr)*(ft**2) 	80645.1612903
	 reynolds number is :  	143770.856507
	 row is  ld/ft**3 	0.572484035397
	 s is  	0.00915974456635
	 delPt is :  psi 	0.4
	 allowable delPa is negligible psi 	
	 pressure drop  for annulus 	
	 number of crosses are :  	32.0
	 delPs is :  psi 	8.4
	 allowable delPT is 2 psi

Example 12.7 pgno:308

In [14]:
print"\t example 12.7 \t"
print"\t approximate values are mentioned in the book \t"
V=7.5; # fps
W=250000.;
CCl=0.85;
CT=1.;
CL=1.;
Ct=263.;
UD=(CCl*CT*CL*Ct*(V**(1/2)));
print"\t design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
A=(W/8);
print"\t area is :  ft**2 \t",A
a1=0.229; # ft**2/ft, table 10
at=0.475; # in**2, table 10
t1=70;
Ts=91.72; #F
n=2;
L=26;
t2=((Ts)-((Ts-t1)/((10)**(0.000279*UD*L*n*a1/(V*at))))+8); 
print"\t t2 is :  F \t",round(t2) # calculation mistake in book
Go=(W*950)/((t2-t1)*500);
print"\t circulation rate is :  gpm \t",round(Go)
# end

	 example 12.7 	
	 approximate values are mentioned in the book 	
	 design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	223.55
	 area is :  ft**2 	31250.0
	 t2 is :  F 	86.0
	 circulation rate is :  gpm 	29176.0