# Chapter 15 : Vaporizers Evapourators and Reboilers¶

## Example 15.1 pgno:463¶

In [20]:
print"\t example 15.1 \t"
print"\t approximate values are mentioned in the book \t"
ts=250;
T1=400;
T2=300;
w=10000; # lb/hr
W=150000; # lb/hr
l=945.3; # Btu/(lb) , table 7
from math import log10
Q=((w)*(l)); # Btu/hr
print"\t total heat required for steam is :  Btu/hr \t",Q
C=0.63; # Btu/(lb)*(F)
Q=((W)*(C)*(T1-T2)); # Btu/hr
print"\t total heat required for kerosene is :  Btu/hr \t",Q
delt1=T2-ts; #F
delt2=T1-ts; # 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
UD=100;
A=(Q/(UD*LMTD));
print"\t A :  ft**2 \t",A
WC=94500; # Btu/F
vl=0.017; # ft**3/lb, from table 7
vv=13.75; # ft**3/lb, from table 7
print"\t By the law of mixtures \t"
# Assume 80 per cent of the outlet fluid is vapor
v2=(0.8*vv)+(.2*vl);
print"\t v2 : ft**3/lb \t",v2
vav=(WC*(v2-vl)/(UD*A))-((WC*(T2-ts)/(l*w))*(vv-vl))+vl;
print"\t vav :  ft**3/lb \t",vav
print"\t By the approximate method \t"
vav1=(vl+v2)/(2);
print"\t vav :  ft**3/lb \t",vav1
row=62.5;
rowac=(1/vav);
s=(rowac/row);
print"\t actual density :  lb/ft**3 \t",rowac
print"\t s :  \t",s
rowap=(1/vav1);
s=(rowap/row);
print"\t approximate density :  lb/ft**3 \t",rowac
print"\t s :  \t",round(s,4)
# end

	 example 15.1
approximate values are mentioned in the book
total heat required for steam is :  Btu/hr 	9453000.0
total heat required for kerosene is :  Btu/hr 	9450000.0
delt1 is :  F 	50
delt2 is :  F 	150
LMTD is : F 	91.1262293169
A :  ft**2 	1037.02304713
By the law of mixtures
v2 : %.0f ft**3/lb 	11.0034
vav :  ft**3/lb 	3.16417120715
By the approximate method
vav :  ft**3/lb 	5.5102
actual density :  lb/ft**3 	0.316038524635
s :  	0.00505661639416
approximate density :  lb/ft**3 	0.316038524635
s :  	0.0029


## Example 15.2 pgno:464¶

In [21]:
print"\t example 15.2 \t"
print"\t approximate values are mentioned in the book \t"
t1=108; # inlet cold fluid,F
t2=235; # outlet cold fluid,F
Ts=338;
Wp=24700; # lb/hr
Wv=19750; # lb/hr
w=4880; # lb/hr
print"\t 1.for heat balance \t"
Ht1=162; # enthalpy at t1, Btu/lb, fig 9
Ht2=248; # enthalpy at t2, Btu/lb, fig 9
qp=(Wp*(Ht2-Ht1)); # for preheat
print"\t total heat required for preheat of butane is :  Btu/hr \t",qp
Ht3=358; # enthalpy of vapour at t2, Btu/lb, fig 9
qv=Wv*(Ht3-Ht2);
print"\t total heat required for vapourisation of butane is :  Btu/hr \t",qv
Q=qp+qv;
print"\t total heat required for  butane is :  Btu/hr \t",Q
print"\t for steam \t"
l=880.6; # Btu/(lb), table 7
Q=((w)*(l)); # Btu/hr
print"\t total heat required for steam is :  Btu/hr \t",Q
deltp=158.5; # F, from eq 5.14
deltv=103; # F eq 5.14
Wp1=(qp/deltp);
print"\t Wp1 is :  lb/hr \t",Wp1
Wv1=(qv/deltv);
print"\t Wv1 is :  lb/hr \t",Wv1
W=(Wp1+Wv1);
print"\t W is :  lb/hr \t",W
delt=(Q/W);
print"\t weighted delt is : % F \t",delt
Tc=((Ts)+(Ts))/(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,steam \t"
Nt=76;
n=2; # number of passes
L=16; #ft
at1=0.594; # 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
mu1=0.0363; # at 338F, fig 15,lb/(ft)*(hr)
D=0.0725; # ft
Ret=((D)*(Gt)/mu1); # reynolds number
print"\t reynolds number is :  \t",Ret
hio=1500; # condensing steam,Btu/(hr)*(ft**2)*(F)
print"\t hio is :  Btu/(hr)*(ft**2)*(F) \t",hio
print"\t cold fluid:shell side,butane \t"
print"\t preheating \t"
ID=15.25; # in
C=0.25; # clearance
B=5; # baffle spacing,in
PT=1.25;
As=((ID*C*B)/(144*PT)); # flow area,ft**2
print"\t flow area is :  ft**2 \t",As
Gs=(Wp/As); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu2=0.278; # at 172F,lb/(ft)*(hr), from fig.14
De=0.0825; # from fig.28,ft
Res=((De)*(Gs)/mu2); # reynolds number
print"\t reynolds number is :  \t",Res
jH=159; # from fig.28
Z=0.12; # Z=k*((c)*(mu1)/k)**(1/3), fig 16
hop=((jH)*(1/De)*(Z)); #using eq.6.15b,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",hop
Up=((hio)*(hop)/(hio+hop)); # clean overall coefficient,eq 6.38,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient for preheating :  Btu/(hr)*(ft**2)*(F) \t",Up
Ap=(qp/(Up*deltp));
print"\t clean surface required for preheating :  ft**2 \t",Ap
print"\t for vapourisation \t"
mu2=0.242; # at 172F,lb/(ft)*(hr), from fig.14
Res=((De)*(Gs)/mu2); # reynolds number
print"\t reynolds number is :  \t",Res
jH=170; # from fig.28
Z=0.115; # Z=k*((c)*(mu1)/k)**(1/3), fig 16
hov=((jH)*(1/De)*(Z)); #using eq.6.15b,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",hov
Uv=((hio)*(hov)/(hio+hov)); # clean overall coefficient,eq 6.38,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient for vapourisation :  Btu/(hr)*(ft**2)*(F) \t",Uv
Av=(qv/(Uv*deltv));
print"\t clean surface required for vapourisation :  ft**2 \t",Av
Ac=Ap+Av;
print"\t total clean surface :  ft**2 \t",Ac
UC=((Up*Ap)+(Uv*Av))/(Ac);
print"\t weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) \t",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 \t",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
# A total of 170 ft2 are required of which 103 are to be used for vaporization. For the total surface required 318 ft2 will be provided. It can be assumed, then, that the surface provided for vaporization is 193ft**2
# then flux is Q/A=10700, which is with in satisfactory levels.
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"
f=0.000165; # friction factor for reynolds number 62000, using fig.26
s=0.00413;
phyt=1;
delPt=((f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(phyt)))/(2); # using eq.7.45,psi
print"\t delPt is :  psi \t",round(delPt,2)
print"\t allowable delPa is negligible \t"
print"\t pressure drop  for annulus \t"
print"\t preheating \t"
f=0.00145; # friction factor for reynolds number 69200, using fig.29
Lp=(L*Ap/Ac); #ft
print"\t length of preheat zone :  ft \t",Lp
N=(12*Lp/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
s=0.5; # for reynolds number 69200,using fig.6
Ds=1.27; # fig 28
phys=1;
delPsp=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPsp is :  psi \t",delPsp
print"\t vapourisation \t"
f=0.00142;
Lv=9.7; # Lv=L-Lp
Nv=(12*Lv/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",Nv
s=0.28;
delPsv=((f*(Gs**2)*(Ds)*(Nv))/(5.22*(10**10)*(De)*(s)*(1))); # using eq 12.47,psi
print"\t delPsv is :  psi \t",delPsv
delPS=delPsp+delPsv;
print"\t delPS is :  psi \t",round(delPS,2)
print"\t allowable delPa is 5 psi \t"
#end

	 example 15.2
approximate values are mentioned in the book
1.for heat balance
total heat required for preheat of butane is :  Btu/hr 	2124200
total heat required for vapourisation of butane is :  Btu/hr 	2172500
total heat required for  butane is :  Btu/hr 	4296700
for steam
total heat required for steam is :  Btu/hr 	4297328.0
Wp1 is :  lb/hr 	13401.8927445
Wv1 is :  lb/hr 	21092
W is :  lb/hr 	34493.8927445
weighted delt is : % F 	124.582285677
caloric temperature of hot fluid is :  F 	338
caloric temperature of cold fluid is :  F 	171
hot fluid:inner tube side,steam
flow area is :  ft**2 	0.15675
mass velocity is :  lb/(hr)*(ft**2) 	31132.3763955
reynolds number is :  	62178.9886688
hio is :  Btu/(hr)*(ft**2)*(F) 	1500
cold fluid:shell side,butane
preheating
flow area is :  ft**2 	0.105902777778
mass velocity is :  lb/(hr)*(ft**2) 	233232.786885
reynolds number is :  	69214.7658922
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	231.272727273
clean overall coefficient for preheating :  Btu/(hr)*(ft**2)*(F) 	200.378071834
clean surface required for preheating :  ft**2 	66.883030772
for vapourisation
reynolds number is :  	79511.1773472
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	236.96969697
clean overall coefficient for vapourisation :  Btu/(hr)*(ft**2)*(F) 	204.640614096
clean surface required for vapourisation :  ft**2 	103.069633088
total clean surface :  ft**2 	169.952663859
weighted clean overall coefficient :  Btu/(hr)*(ft**2)*(F) 	202.96313674
total surface area is :  ft**2 	318.3488
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	108.352513798
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00430213212754
pressure drop  for inner pipe
delPt is :  psi 	0.16
allowable delPa is negligible
pressure drop  for annulus
preheating
length of preheat zone :  ft 	6.29662676683
number of crosses are :  	15.1119042404
delPsp is :  psi 	0.703032923143
vapourisation
number of crosses are :  	23.28
delPsv is :  psi 	1.89396418309
delPS is :  psi 	2.6
allowable delPa is 5 psi


## Example 15.3 pgno:475¶

In [22]:
print"\t example 15.3 \t"
print"\t approximate values are mentioned in the book \t"
ts=400;
T1=575;
T2=475;
W=28100; # lb/hr
w=34700; # lb/hr
print"\t 1.for heat balance \t"
HT1=290; # enthalpy at T1, Btu/lb, fig 11
HT2=385; # enthalpy at T2, Btu/lb, fig 11
Q=(W*(HT2-HT1)); # for preheat
print"\t total heat required for gasoline is :  Btu/hr \t",Q
c=0.77; # Btu/(lb), table 7
Q=((w)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for gasoil is :  Btu/hr \t",Q
delt=118; # F eq 5.14
S=75./175.;#((T2-ts)/(T1-ts));
print"\t S is :  \t",S
Kc=0.37; # fig 17
Fc=0.42;
Tc=(T2+(0.42*(T1-T2)));
print"\t Tc is :  F \t",Tc
print"\t hot fluid:inner tube side,gasoil \t"
Nt=68;
n=6; # number of passes
L=12; #ft
at1=0.546; # 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
mu1=0.65; # at 517F, fig 14,lb/(ft)*(hr)
D=0.0694; # ft
Ret=((D)*(Gt)/mu1); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=220; # from fig.24
Z=0.118; # Z=k*((c)*(mu1)/k)**(1/3), fig 16
Hi=((jH)*(1/D)*(Z)); #hi/phyt, Hi=()using eq.6.15d,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Hi
Hio=((Hi)*(0.834/1)); #Hio=(hio/phyp), using eq.6.9
print"\t Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",Hio
# (mu1/muw)**(0.14) is negligible
print"\t cold fluid:shell side,gasoline \t"
ho=300; # assumption
tw=(ts)+(((Hio)/(Hio+ho))*(Tc-ts)); # from eq.5.31
print"\t tw is :  F \t",tw
deltw=(tw-ts);
print"\t deltw :  F \t",deltw
# from fig 15.11, ho>300
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=0.2618; # 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 \t",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
# check for max. flux=Q/A=12500.(satisfactory)
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"
f=0.00015; # friction factor for reynolds number 85700, using fig.26
s=0.71;
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=0.09; # 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",round(delPT,1)
print"\t allowable delPa is 10psi \t"
print"\t delPs is negligible \t"
#end

	 example 15.3
approximate values are mentioned in the book
1.for heat balance
total heat required for gasoline is :  Btu/hr 	2669500
total heat required for gasoil is :  Btu/hr 	2671900.0
S is :  	0.428571428571
Tc is :  F 	517.0
hot fluid:inner tube side,gasoil
flow area is :  ft**2 	0.0429722222222
mass velocity is :  lb/(hr)*(ft**2) 	807498.383969
reynolds number is :  	86215.9813038
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	374.063400576
Correct Hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	311.968876081
cold fluid:shell side,gasoline
tw is :  F 	459.64414193
deltw :  F 	59.6441419296
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	152.933697255
total surface area is :  ft**2 	213.6288
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	105.993294626
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00289577777619
pressure drop  for inner pipe
delPt is :  psi 	2.73790740915
delPr is :  psi 	3.04225352113
delPT is :  psi 	5.8
allowable delPa is 10psi
delPs is negligible


## Example 15.4 pgno:482¶

In [23]:
print"\t example 15.4\t"
print"\t approximate values are mentioned in the book \t"
t1=315.; # inlet cold fluid,F
t2=335.; # outlet cold fluid,F
T1=525.;
T2=400.;
Wv=29000; # lb/hr
Ws=38500; # lb/hr
w=51000; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
Ht1=238; # enthalpy at t1, Btu/lb, fig 9
Ht2=252; # enthalpy at t2, Btu/lb, fig 9
Ht3=378; # enthalpy of vapour at t2
qv=(Wv*(Ht3-Ht2)); # for preheat
print"\t qv is :  Btu/hr \t",qv
qs=Ws*(Ht2-Ht1);
print"\t qs is :  Btu/hr \t",qs
Q=qs+qv;
print"\t total heat required for naphtha is :  Btu/hr \t",Q
c=0.66; # Btu/(lb)(F)
Q=((w)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for gasoil 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 0.97 \t" # from fig 18
delt=(0.97*LMTD); # F
print"\t delt is :  F \t",delt
X=((delt1)/(delt2)); # fig 17
print"\t ratio of two local temperature difference is :  \t",X
Fc=0.41; # from fig.17
Kc=0.42;
Tc=((T2)+((Fc)*(T1-T2))); # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+((Fc)*(t2-t1))); # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:inner tube side,steam \t"
Nt=116;
n=8; # number of passes
L=12; #ft
at1=0.546; # 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
mu1=1.09; # at 451F, fig 14,lb/(ft)*(hr)
D=0.0695; # ft
Ret=((D)*(Gt)/mu1); # reynolds number
print"\t reynolds number is :  \t",Ret
jH=168; # from fig.24
Z=0.142; # Z=k*((c)*(mu1)/k)**(1/3), fig 16
Hi=((jH)*(1/D)*(Z)); #, Hi=(hi/phyt)using eq.6.15d,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Hi
Hio=((Hi)*(0.834/1)); #Hio=(hio/phyp), using eq.6.9
print"\t Correct Hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",Hio
print"\t cold fluid:shell side,naphtha \t"
ho1=200; # assumption
tw=(tc)+(((Hio)/(Hio+ho1))*(Tc-tc)); # from eq.5.31, calculation mistake
print"\t tw is :  F \t",tw
deltw=(tw-tc);
print"\t deltw :  F \t",deltw
# from fig 15.11, hv>300, hs=60
Av=(qv/300);
As=qs/60;
print"\t qv/hv :  \t",Av
print"\t qs/hs :  \t",As
A1=As+Av;
print"\t A :  \t",A1
ho=(Q/A1);
print"\t ho :  \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
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 \t",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
# check for max. flux=Q/A=11500.(satisfactory)
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"
f=0.000168; # friction factor for reynolds number 59200, using fig.26
s=0.73;
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=0.11; # 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",round(delPT,1)
print"\t allowable delPa is negligible \t"
print"\t pressure drop  for annulus \t"
Af=(3.14*(21.25**2-(116))/8);
print"\t flow area :  in**2 \t",Af
As=0.917; # ft**2
p=(3.14*21.25/2)+(3.14*1*116/2)+(21.25)
print"\t wetted perimeter :  in \t",p
De=0.186; # ft
Gs=(Ws/(2*As)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu2=0.435; # at 315F, fig 14,lb/(ft)*(hr)
Res=((De)*(Gs)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
f=0.00028; # using fig.26
row=0.337; # fig 13.14
# soutlet max=0.071,
s=0.35; # using fig.6
phys=1;
delPs=0.25 #((f*(Gs**2)*(L))/(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"
#end

	 example 15.4
approximate values are mentioned in the book
1.for heat balance
qv is :  Btu/hr 	3654000
qs is :  Btu/hr 	539000
total heat required for naphtha is :  Btu/hr 	4193000
total heat required for gasoil is :  Btu/hr 	4207500.0
delt1 is :  F 	85.0
delt2 is :  F 	190.0
LMTD is : F 	130.683201952
R is :  	6.25
S is :  	0.0952380952381
FT is 0.97
delt is :  F 	126.762705893
ratio of two local temperature difference is :  	0.447368421053
caloric temperature of hot fluid is :  F 	451.25
caloric temperature of cold fluid is :  F 	323.2
hot fluid:inner tube side,steam
flow area is :  ft**2 	0.0549791666667
mass velocity is :  lb/(hr)*(ft**2) 	927624.100038
reynolds number is :  	59146.6742685
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	343.251798561
Correct Hio to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	286.272
cold fluid:shell side,naphtha
tw is :  F 	398.584002369
deltw :  F 	75.384002369
qv/hv :  	12180
qs/hs :  	8983
A :  	21163
ho :  	198.813967774
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	117.329454908
total surface area is :  ft**2 	364.4256
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	91.0801518561
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	0.00245633150105
pressure drop  for inner pipe
delPt is :  psi 	5.24018227225
delPr is :  psi 	4.82191780822
delPT is :  psi 	10.1
allowable delPa is negligible
pressure drop  for annulus
flow area :  in**2 	131.70828125
wetted perimeter :  in 	236.7325
mass velocity is :  lb/(hr)*(ft**2) 	20992.3664122
reynolds number is :  	59146.6742685
delPs is :  psi 	0.25
allowable delPa is 10 psi


## Example 15.5 pgno:488¶

In [24]:
print"\t example 15.5\t"
print"\t approximate values are mentioned in the book \t"
W=40800; # lb/hr
w=4570; # lb/hr
from math import log10
print"\t 1.for heat balance \t"
Ht1=241; # enthalpy of liquid at 228F, Btu/lb, fig 9
Ht2=338; # enthalpy of vapourat 228F, Btu/lb, fig 9
Q=(W*(Ht2-Ht1));
print"\t total heat required for butane is :  Btu/hr \t",Q
l=868; # Btu/(lb), table 7
Q=((w)*(l)); # Btu/hr
print"\t total heat required for steam is : Btu/hr \t",Q
delt=125; # delt=LMTD, isothermal boiling, eq 5.14
# Tc and tc: Both streams are isuthermal
print"\t trail 1 \t"
A1=((Q)/((12000))); # Q/A1 =12000, first trial should always be taken for the maximum allowable flux
print"\t A1 is : ft**2 \t",A1
a1=0.1963; # ft**2/lin ft
L=16;
N1=(A1/(L*a1)); # table 10
print"\t number of tubes are : \t",N1
N2=109; # assuming one tube passes, 13.25-in 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
# Assume 4: 1 recirculation ratio
rowv=(58/(359*(688/492)*(14.7/290))); # eq 15.18
print"\t vapour density : lb/ft**3 \t",rowv
Vv=0.44;
Vl=0.0372; # fig 6
W1=4*W;
print"\t weight flow of recirculated liquid : lb/hr \t",W1
VL=W1*Vl;
VV=W*Vv;
print"\t volume of liquid : ft**3 \t",VL
print"\t volume of vapour : ft**3 \t",VV
V=VL+VV;
print"\t total volume out of reboiler : ft**3 \t",V
vo=(V/(W1+W));
print"\t vo is : ft**3/lb \t",vo
Pl=((2.3*16)/(144*(vo-Vl)))*(log10(vo/Vl));
print"\t pressure leg : psi \t",Pl
print"\t frictional resistance \t"
Nt=109;
n=1; # number of passes
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=((W1+W)/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) \t",Gt
mu1=0.242; # at 228F, fig 14,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu1); # reynolds number
print"\t reynolds number is : \t",Ret
f=0.000127; # using fig.26
s=0.285;
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,2)
P=Pl+delPt;
print"negilgable"
print"\t total resisitance : psi \t",P
F=(16*0.43*62.5/144);
print"\t driving force : psi \t",F
# The resistances are greater than the hydrostatic head can provide; hence the recirculation ratio will be less than 4: 1
print"\t trial 2 \t" # Assume 12'0" tubes and 4:1 recirculation ratio
A1=((Q)/((12000))); # Q/A1 =12000, first trial should always be taken for the maximum allowable flux
print"\t A1 is : ft**2 \t",A1
a1=0.1963; # ft**2/lin ft
L=12;
N1=(A1/(L*a1)); # table 10
print"\t number of tubes are : \t",N1
N2=151; # assuming one tube passes, 15.25-in 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
Pl=((2.3*12)/(144*(vo-Vl)))*(log10(vo/Vl));
print"\t pressure leg : psi \t",Pl
print"\t frictional resistance \t"
Nt=151;
n=1; # number of passes
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=((W1+W)/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) \t",Gt
mu1=0.242; # at 228F, fig 14,lb/(ft)*(hr)
D=0.0517; # ft
Ret=((D)*(Gt)/mu1); # reynolds number
print"\t reynolds number is : \t",Ret
f=0.000135; # using fig.26
s=0.285;
phyt=1;
delPt=((f*(Gt**2)*(12)*(n))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is : psi \t",delPt
P=Pl+delPt;
print"\t total resisitance : psi \t",P
F=(12*0.43*62.5/144);
print"\t driving force : psi \t",F
# Since the driving force is slightly greater than the resistances, a recirculation ratio better than 4:1 is assured.
print"\t hot fluid : shell side,steam \t"
ho=1500; # condensing steam
print"\t cold fluid:inner tube side, butane \t"
jH=330; # from fig.24
Z=0.115; # Z=k*((c)*(mu1)/k)**(1/3), fig 16
Hi=((jH)*(1/D)*(Z)); #, Hi=(hi/phyt)using eq.6.15d,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \t",Hi
Hio=((300)*(0.62/0.75)); #Hio=(hio/phyp), using eq.6.9
print"\t Correct Hio 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
UD=89;
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is : (hr)*(ft**2)*(F)/Btu \t",Rd
# end

	 example 15.5
approximate values are mentioned in the book
1.for heat balance
total heat required for butane is :  Btu/hr 	3957600
total heat required for steam is : Btu/hr 	3966760
trail 1
A1 is : ft**2 	330
number of tubes are : 	105.068772287
total surface area is : ft**2 	342.3472
correct design overall coefficient is : Btu/(hr)*(ft**2)*(F) 	92.6956025929
vapour density : lb/ft**3 	3.18723589714
weight flow of recirculated liquid : lb/hr 	163200
volume of liquid : ft**3 	6071.04
volume of vapour : ft**3 	17952.0
total volume out of reboiler : ft**3 	24023.04
vo is : ft**3/lb 	0.11776
pressure leg : psi 	1.58756230188
frictional resistance
flow area is : ft**2 	0.228597222222
mass velocity is : lb/(hr)*(ft**2) 	892399.295218
reynolds number is : 	190648.940342
delPt is : psi 	2.1
negilgable
total resisitance : psi 	3.69151627479
driving force : psi 	2.98611111111
trial 2
A1 is : ft**2 	330
number of tubes are : 	140.091696383
total surface area is : ft**2 	355.6956
correct design overall coefficient is : Btu/(hr)*(ft**2)*(F) 	89.21695967
pressure leg : psi 	1.19067172641
frictional resistance
flow area is : ft**2 	0.316680555556
mass velocity is : lb/(hr)*(ft**2) 	644182.272707
reynolds number is : 	137620.75826
delPt is : psi 	0.874030618788
total resisitance : psi 	2.0647023452
driving force : psi 	2.23958333333
hot fluid : shell side,steam
cold fluid:inner tube side, butane
individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) 	734.042553191
Correct Hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) 	248.0
clean overall coefficient is : Btu/(hr)*(ft**2)*(F) 	212.814645309
actual Rd is : (hr)*(ft**2)*(F)/Btu 	0.006537030325


## Example 15.6 pgno492¶

In [25]:
print"\t example 15.6\t"
print"\t approximate values are mentioned in the book \t"
#20000=WD+WB;
#0.99*WD+(0.05*WB)=(20000*.5);
# solving above two eq. we get WD and WB
WD=9570; # lb/hr
WB=10430; # lb/hr
HBl=108; # fig 3 and 12
HDl=85.8; #fig 3 and 12
HDv=253.8; # fig 3 and 12
HFl=92; # fig 3 and 12
l=153; # fig 3 and 12
QR=((2.54+1)*WD*(HDv))-(2.54*WD*HDl)+(WB*HBl)-(20000*HFl);
print"\t total heat duty : Btu/hr \t",round(QR)
Q=QR/153;
print"\t total heat duty : lb/hr \t",round(Q)
# end

	 example 15.6
approximate values are mentioned in the book
total heat duty : Btu/hr 	5799016.0
total heat duty : lb/hr 	37902.0


## Example 15.7 pgno:502¶

In [26]:
print"\t example 15.7 \t"
print"\t approximate values are mentioned in the book \t"

#Basis: One hour
#20000=WD+WB , material balance
#0.99*WD+(0.05*WB)=(20000*0.5) , Benzene balance
# solving above two eq. we get WD and WB
WD=9570; # lb/hr
WB=10430; # lb/hr

#Compositions and Boiling Points
#Feed
l1 = 10000; #Lb/hr , C6H4
l2 = 10000; #Lb/hr , C7H8
lb = l1+l2; #Lb/hr
print"\ttotal Lb/hr is \t",lb
mo1 = 78.1; #Mol. wt., C6H6
mo2 = 93.1; #Mol. wt , C7H8
mh1 = 128.0; #Mol/hr , C6H6
mh2 = 107.5; #Mol/hr , C7H8
mh = mh1 + mh2; # Mol/hr
print"\ttotal Mol/hr is \t",mh
x1 = mh1/mh;
print"\tx1 of C6H6 is \t",x1
x2 = mh2/mh;
print"\tx1 of C7H8 is \t",x2
x = x1+x2;
print"\tTotal x1 is \t",x
Pp1= 1380; # 214Deg F
Pp2=575; # 214Deg F
xp1 = x1*Pp1;
print"\tx1Pp1 of C6H6 is \t",xp1
xp2 = x2*Pp2;
print"\tx1Pp1 of C7H8 is \t",xp2
sxp = xp1 + xp2;
print"\tTotal x1Pp1 is \t",sxp
y1 = xp1/sxp;
print"\ty1 of C6H6 is \t",y1
y2 = xp2/sxp;
print"\ty1 of C7H8 is \t",y2
y = y1+y2;
print"\tTotal y1 is \t",y

w1 = 0.558; #from eq 15.42
print"\t(WR/V =((xD - yF)/.(xD - xF))) = mol/mol\t",w1
wD=1;
xD = 0.992;
#V = WR' + WD
# WR'/V = 0.558
#Solving, WR' = (WR' * 0.558) + (0.558 * WD)
Wr = 1.27; # mol reflux/mol distillate
print"\tWR = %.2f (mol reflux)/(mol distillate)\t",Wr
Wr1 = Wr * 2; # mol/ mol distillate
print"\tAssumed 200 percent of the theoretical minimum reflux as economic\t\tWR = (mol)/(mil distillate)\t",Wr1
In = (wD * xD)/(Wr1 + 1); #intercept for the upper operating line
print"\tThe intercept for the upper operating line = \t",In
p = 13; # From fig. 15.23, connecting the corresponding lines
print"\tConnecting the corresponding line in Fig. 15.23, plates required: \t",p
fp = 7; # From fig. 15.23, connecting the corresponding lines
print"\tFeed plate is th(from top)\t",fp
d=122.5;
tf = Wr1 * d;
print"\tTotal reflux is \t",tf
print"\t\t\t\t\tHeat balances"

#Heat Balances
l1 = 33900;
l2 = 9570;
l3 = 24330;
b1 = 253.8;
b2 = 85.8;
b3 = 85.8;
bt1 = b1*l1;
bt2 = b2*l2;
bt3 = b3*l3;
bt4 = 5688000;
print"\t\t\t\t\tMol/hr\tMol.wt.\tLb/hr\tTemp,DegF\tBtu/lb\tBtu/hr\t\t________________________________________________________________________\t\tHeat balance \t\taround condenser:\t"
print"\t  Heat in:\t\t  Top plate vapor.......433\t87.3\t\t195\t%.1f\t\t",l1,b1,bt1
print"\t  Heat out:\t\t  Distillate............"
print"122.5\t78.3\t\t195\t\t\t",l2,b2,bt2
print"\t  Reflux................"
print"310.5\t78.3\t\t195\t\t\t",l3,b3,bt3
print"\t  Condenser duty, by\t\t  difference........... .....   ....    ......  .."
print".     .....   5688000\t"
print"\t\t\t\t\t\t\t\t\t_______\t\t\t\t\t\t\t\t\t\t8600000\t\t"
tho=7640000;

lam = 153; # At 246 DegF, Btu/hr
rv = 5800000/153; #Lb/hr
print"\tReboiler vapor is  lb/hr\t",rv
to = rv + 10430; #Lb/hr
print"\tTrapout is  lb/hr\t",to

print"\t\t\t\t\tMol/hr\tMol.wt.\tLb/hr\tTemp,DegF\tBtu/lb\tBtu/hr\t\t________________________________________________________________________\t"
print"\tHeat in:\t\t  Trapout...............522\t92.8\t\t246\t108.0\t5230000\t",to
print"\t  Reboiler duty, \t\t  by difference.......  ....    ....    .....   ...     .....   5800000\t"
print"\t\t\t\t\t\t\t\t\t_______\t\t\t\t\t\t\t\t\t\t11030000\t\t"
print"\t\tReboiler requirements are\t"
print"\t\tTotal liquid to reboiler\t48330 lb/hr\t\t\tVaporization\t\t\t37900 lb/hr\t\t\tTemperature(nearly isothermal)\t246DegF\t\t\tPressure\t\t\t5 psig\t\t\tHeat load\t\t\t5800000 Btu/hr\t"
print"total heat out is btu/hr",tho
#end

	 example 15.7
approximate values are mentioned in the book
total Lb/hr is 	20000
total Mol/hr is 	235.5
x1 of C6H6 is 	0.543524416136
x1 of C7H8 is 	0.456475583864
Total x1 is 	1.0
x1Pp1 of C6H6 is 	750.063694268
x1Pp1 of C7H8 is 	262.473460722
Total x1Pp1 is 	1012.53715499
y1 of C6H6 is 	0.740776464914
y1 of C7H8 is 	0.259223535086
Total y1 is 	1.0
(WR/V =((xD - yF)/.(xD - xF))) = mol/mol	0.558
WR = %.2f (mol reflux)/(mol distillate)	1.27
Assumed 200 percent of the theoretical minimum reflux as economic		WR = (mol)/(mil distillate)	2.54
The intercept for the upper operating line = 	0.280225988701
Connecting the corresponding line in Fig. 15.23, plates required: 	13
Feed plate is th(from top)	7
Total reflux is 	311.15
Heat balances
Mol/hr	Mol.wt.	Lb/hr	Temp,DegF	Btu/lb	Btu/hr		________________________________________________________________________		Heat balance 		around condenser:
Heat in:		  Top plate vapor.......433	87.3		195	%.1f		33900 253.8 8603820.0
Heat out:		  Distillate............
122.5	78.3		195			9570 85.8 821106.0
Reflux................
310.5	78.3		195			24330 85.8 2087514.0
Condenser duty, by		  difference........... .....   ....    ......  ..
.     .....   5688000
_______										8600000
Reboiler vapor is  lb/hr	37908
Trapout is  lb/hr	48338
Mol/hr	Mol.wt.	Lb/hr	Temp,DegF	Btu/lb	Btu/hr		________________________________________________________________________
Heat in:		  Trapout...............522	92.8		246	108.0	5230000	48338
Reboiler duty, 		  by difference.......  ....    ....    .....   ...     .....   5800000
_______										11030000
Reboiler requirements are
Total liquid to reboiler	48330 lb/hr			Vaporization			37900 lb/hr			Temperature(nearly isothermal)	246DegF			Pressure			5 psig			Heat load			5800000 Btu/hr
total heat out is btu/hr 7640000


## Example 15.8 pgno:506¶

In [27]:
print"\t example 15.8 \n"
print"\t approximate values are mentioned in the book \n"
import numpy
vc=numpy.array([6.4, 219.7, 2.3])
#vc(1) = 6.4; # Mol/hr
#vc(2) = 219.7; #Mol/hr
#vc(3) = 2.3; #Mol/hr
K=numpy.array([2.8, 1.01, 0.34])
#K(1) = 2.8; #at 148DegF and 40 psia
#K(2) = 1.01; #at 148DegF and 40 psia
#K(3) = 0.34; #at 148DegF and 40 psia
v=numpy.array([0, 0, 0])
i=0;
while (i<3):
v[0]=vc[0]*K[0];
v[1]=vc[1]*K[1];
v[2]=vc[1]*K[2];
i=i+1;

print"\n\t\tMol/hr\t\tK(148DegF,40 psia)\tV/K\n"
print"\t\t--------------------------------------------\n"

print vc,K,v

bc=numpy.array([4.1, 49.3, 71.9, 52.5, 54.7, 82.5, 76.6, 22.4])
#bc(1)=4.1; #Mol/hr
#bc(2)=49.3; #Mol/hr
#c(3)=71.9; #Mol/hr
#bc(4)=52.5; #Mol/hr
#bc(5)=54.7; #Mol/hr
#bc(6)=82.5; #Mol/hr
#bc(7)=76.6; #Mol/hr
#bc(8)=22.4; #Mol/hr
tbc=numpy.array([0, 0, 0, 0, 0, 0, 0, 0])
tbc[0]=tbc[0]+bc[0];
i=1
while (i<8):
tbc[i]=tbc[i-1]+bc[i];
i=i+1;

bK=numpy.array([5.8, 3.0, 1.68, 0.98, 0.57, 0.35, 0.21, 0.13])
#bK(1)=5.8; #at 330DegF, 40 psia
#bK(2)=3.0; #at 330DegF, 40 psia
#bK(3)=1.68; #at 330DegF, 40 psia
#bK(4)=0.98; #at 330DegF, 40 psia
#bK(5)=0.57; #at 330DegF, 40 psia
#bK(6)=0.35; #at 330DegF, 40 psia
#bK(7)=0.21; #at 330DegF, 40 psia
#bK(8)=0.13; #at 330DegF, 40 psia

KL=numpy.array([23.8, 148.0, 120.0, 51.4, 31.2, 28.9, 16.1, 2.9])

#KL(1)=23.8;
#KL(2)=148.0;
#KL(3)=120.8;
#KL(4)=51.4;
#KL(5)=31.2;
#KL(6)=28.9;
#KL(7)=16.1;
#KL(8)=2.9;

tk=([0, 0, 0, 0, 0, 0, 0, 0])
i=1;
op=40;
tk[0]=KL[0]
while (i<8):
tk[i]=tk[i-1]+KL[i];
i=i+1;

l=numpy.array([1700, 13900, 13030, 6260, 4240, 4330, 2640, 520])
#l(1)=1700; #Lb/hr
#l(2)=13900; #Lb/hr
#l(3)=13030; #Lb/hr
#l(4)=6260; #Lb/hr
#l(5)=4240; #Lb/hr
#l(6)=4330; #Lb/hr
#l(7)=2640; #Lb/hr
#l(8)=520; #Lb/hr

tl=numpy.array([0, 0, 0, 0, 0, 0, 0, 0,])

i=0;
while (i<8):
tl[i]=tl[i]+l[i];
i=i+1;

print"\n\t\tBUBBLE POINTS OF BOTTOMS\n"
print"\t\tMol/hr\t\tK(330DegF,40psia)\t\tKL\t\tLb/hr\n"
print"\t\t--------------------------------------------------------------\n"
tlr=78177;
i=0;
while (i<8):

print bc[i],bK[i],KL[i],l[i]
i=i+1;

print"\t\t____\t\t\t\t\t____\t\t____\n"

print"\t\t",tbc
print"\t\t\t\t\t",tk
print"\t\t",tl
av = numpy.array([0, 0, 0, 0, 0, 0, 0, 0])
av=tl/tk
hl=4280000;
print"\tAverage mol. wt. \n",av

lh=numpy.array([48894, 16298, 32596])
bl=numpy.array([286, 129, 129])
#lh(1)=48894;#Lb/hr
#lh(2)=16298;#Lb/hr
#lh(3)=32596;#Lb/hr
#bl(1)=286;#Btu/hr
#bl(2)=129;#Btu/hr
#bl(3)=129;#Btu/hr

bh=numpy.array([0, 0, 0])
vap=22700;
i=0;

while (i<3):
bh[0]=lh[0]*bl[0];
i=i+1;

#Heat Balances
print"\n\n\t\t\t\t\t\tHEAT BALANCES:"
print"\n\t\t\t\tMol/hr\t\tMol.wt.\t\tLb/hr\t\tTemp,DegF\t\tBtu/lb\t\tBtu/hr\n\t"
print"\t\t\t----------------------------------------------------------------------------------------"
print"\n\tHeat Balance onCondeser\n\t  Heat in:\n\t   Top plate vapor......"

#Heat Balances on reboiler
#Assume 30Deg difference between reboiler and bottom plate giving bottom-plate temperature of 300DegF
#Mol/hr from Eq. 15.47

pc=numpy.array([0.056, 0.35, 0.285, 0.122, 0.074, 0.068, 0.038, 0.007])
#pc(1)=0.056;
#pc(2)=0.350;
##pc(3)=0.285;
#pc(4)=0.122;
#pc(5)=0.074;
#pc(6)=0.068;
#pc(7)=0.038;
#pc(8)=0.007;

pk=numpy.array([4.5, 2.25, 1.2, 0.66, 0.38, 0.22, 0.13, 0.07])
#pK(1)=4.5;
#pK(2)=2.25;
#pK(3)=1.20;
#pK(4)=0.66;
#pK(5)=0.38;
#pK(6)=0.22;
#pK(7)=0.13;
#pK(8)=0.07;

print"\n\n\t\tCALCULATION OF BOTTOM PLATE TEMPERATURE\n"
print"\t\ty*\t\t\tReboiler vapor\t\t\t\tK(300DegF,40psia)\tMol*K\n\t\t\t\tV = y*205.7 +\tBottoms\t=\tTrapout\n"
print"\t\t----------------------------------------------------------------------------------------\n"

print "CALCULATION OF BOTTOM PLATE TEMPERATURE"
print"      y*          Reboiler vapor              K(300DegF,40psia)   Mol*K"
print"              V = y*205.7 +   Bottoms =   Trapout"
print"      ----------------------------------------------------------------------------------------"
print   "C5 0.056       11.5        4.1     15.6        4.5     70.29"
print   "C6 0.35        72.0        49.3        121.3       2.25        272.91"
print   "C7 0.285       58.6        71.9        130.5       1.2     156.63"
print   "C8 0.122       25.1        52.5        77.6        0.66        51.21"
print   "C9 0.074       15.2        54.7        69.9        0.38        26.57"
print   "C10    0.068       14.0        82.5        96.5        0.22        21.23"
print   "C11    0.038       7.8     76.6        84.4        0.13        10.97"
print   "C12    0.007       1.4     22.4        23.8        0.07        1.67"
print       "----------------------------------------------------------------------------------------"
print       "1.000      205.7       414.0       619.7               611.5"

print"Reboiler requirements are"
print"      Vaporization lb/hr",vap
print"      Total liquor to reboiler  lb/hr",tlr
print"      Temperature range 300-330 def F"
print"      Operating pressure  psi",op

	 example 15.8

approximate values are mentioned in the book

Mol/hr		K(148DegF,40 psia)	V/K

--------------------------------------------

[   6.4  219.7    2.3] [ 2.8   1.01  0.34] [ 17 221  74]

BUBBLE POINTS OF BOTTOMS

Mol/hr		K(330DegF,40psia)		KL		Lb/hr

--------------------------------------------------------------

4.1 5.8 23.8 1700
49.3 3.0 148.0 13900
71.9 1.68 120.0 13030
52.5 0.98 51.4 6260
54.7 0.57 31.2 4240
82.5 0.35 28.9 4330
76.6 0.21 16.1 2640
22.4 0.13 2.9 520
____					____		____

[  4  53 124 176 230 312 388 410]
[23.800000000000001, 171.80000000000001, 291.80000000000001, 343.19999999999999, 374.39999999999998, 403.29999999999995, 419.39999999999998, 422.29999999999995]
[ 1700 13900 13030  6260  4240  4330  2640   520]
Average mol. wt.
[ 71.42857143  80.9080326   44.65387252  18.24009324  11.32478632
10.7364245    6.29470672   1.23135212]

HEAT BALANCES:

Mol/hr		Mol.wt.		Lb/hr		Temp,DegF		Btu/lb		Btu/hr

----------------------------------------------------------------------------------------

Heat Balance onCondeser
Heat in:
Top plate vapor......

CALCULATION OF BOTTOM PLATE TEMPERATURE

y*			Reboiler vapor				K(300DegF,40psia)	Mol*K
V = y*205.7 +	Bottoms	=	Trapout

----------------------------------------------------------------------------------------

CALCULATION OF BOTTOM PLATE TEMPERATURE
y*          Reboiler vapor              K(300DegF,40psia)   Mol*K
V = y*205.7 +   Bottoms =   Trapout
----------------------------------------------------------------------------------------
C5 0.056       11.5        4.1     15.6        4.5     70.29
C6 0.35        72.0        49.3        121.3       2.25        272.91
C7 0.285       58.6        71.9        130.5       1.2     156.63
C8 0.122       25.1        52.5        77.6        0.66        51.21
C9 0.074       15.2        54.7        69.9        0.38        26.57
C10    0.068       14.0        82.5        96.5        0.22        21.23
C11    0.038       7.8     76.6        84.4        0.13        10.97
C12    0.007       1.4     22.4        23.8        0.07        1.67
----------------------------------------------------------------------------------------
1.000      205.7       414.0       619.7               611.5
Reboiler requirements are
Vaporization lb/hr 22700
Total liquor to reboiler  lb/hr 78177