print"\t example 6.1 "
print"\t approximate values are mentioned in the book "
#given
T1=160; # inlet hot fluid,F
T2=100; # outlet hot fluid,F
t1=80; # inlet cold fluid,F
t2=120; # outlet cold fluid,F
w=9820; # lb/hr
#solution
from math import log
from math import pi
print"\t 1.for heat balance "
print"\t for benzene "
tav=((t1+t2)/2); # F
print"\t average temperature of benzene is : F ",tav
c=0.425; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for benzene is : Btu/hr ",Q
print"\t for toulene \n"
Tav=((T1+T2)/2); #F
print"\t average temperature of toulene is : F ",Tav
c=0.44; # Btu/(lb)*(F)
W=((Q)/((c)*(T1-T2))); # lb/hr
print"\t W is : lb/hr ",round(W)
print"\t 2.LMTD "
print"\t for counter current flow "
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is : F ",delt1
print"\t delt2 is : F ",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F \t",round(LMTD,1)
print"\t 3.caloric temperatures \n"
print"\t both streams will show that neither is viscous at the cold terminal (the viscosities less than 1 centipoise) and the temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties at the arithmetic mean, and the value of (mu/muw)**0.14 may be assumed equal to 1.0 \n"
tav=((t1+t2)/2); # F
print"\t average temperature of benzene is : F ",tav
Tav=((T1+T2)/2); #F
print"\t average temperature of toulene is : F ",Tav
print"\t hot fluid:annulus,toulene "
D1=0.138; # ft
D2=0.1725; # ft
aa=((pi)*(D2**2-D1**2)/4); # flow area,ft**2
print"\t flow area is : ft**2 ",aa
De=(D2**2-D1**2)/D1; # equiv diameter,ft
print"\t equiv diameter is : ft ",De
Ga=(W/aa); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) ",Ga
mu1=0.41*2.42; # at 130 F,lb/(ft)*(hr)
Rea=((De)*(Ga)/mu1); # reynolds number
print"\t reynolds number is : ",Rea
jH=167; # from fig.24
c=0.44; # Btu/(lb)*(F),at 130F
k=0.085; # Btu/(hr)*(ft**2)*(F/ft), from table 4
Pr=((c)*(mu1)/k)**(1/3); # prandelt number raised to power 1/3
print"\t Pr is : ",Pr
ho=((jH)*(k/De)*(Pr)*(1**0.14)); # using eq.6.15b,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) ",ho
print"\t cold fluid:inner pipe,benzene "
D=0.115; # ft
ap=((pi)*(D**2)/4); # flow area, ft**2
print"\t flow area is : ft**2 ",ap
Gp=(w/ap); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) ",Gp
mu2=0.5*2.42; # at 130 F,lb/(ft)*(hr)
Rep=((D)*(Gp)/mu2); # reynolds number
print"\t reynolds number is : ",Rep
jH=236; # from fig.24
c=0.425; # Btu/(lb)*(F),at 130F
k=0.091; # Btu/(hr)*(ft**2)*(F/ft), from table 4
Pr=((c)*(mu2)/k)**(1/3); # prandelt number raised to power 1/3
print"\t Pr is : ",Pr
hi=((jH)*(k/D)*(Pr)*(1**0.14)); # using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) ",hi
ID=1.38; # ft
OD=1.66; #ft
hio=((hi)*(ID/OD)); # using eq.6.5
print"\t Correct hi to the surface at the OD is : Btu/(hr)*(ft**2)*(F) ",hio
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) ",Uc
Rd=0.002; # required by problem,(hr)*(ft**2)*(F)/Btu
UD=((Uc)/((1)+(Uc*Rd))); # design overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t design overall coefficient is : Btu/(hr)*(ft**2)*(F) ",UD
A=((Q)/((UD)*(LMTD))); # required surface,ft**2
print"\t required surface is : ft**2 ",A
A1=0.435; # From Table 11 for 1(1/4)in IPS standard pipe there are 0.435 ft2 of external surface per foot length,ft**2
L=(A/A1); # required length;lin ft
print"\t required length is : lin ft ",L
print"\t This may be fulfilled by connecting three 20-ft hairpins in series \n"
A2=120*0.435; # actual surface supplied,ft**2
print"\t actual surface supplied is : ft**2 ",A2
UD=((Q)/((A2)*(LMTD)));
print"\t actual design overall coefficient is : Btu/(hr)*(ft**2)*(F) ",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is : (hr)*(ft**2)*(F)/Btu ",Rd
print"\t pressure drop for annulus "
De1=(D2-D1); #ft
print"\t De1 is : ft ",De1
Rea1=((De1)*(Ga)/mu1); # reynolds number
print"\t reynolds number is :",Rea1
f=(0.0035)+((0.264)/(Rea1**0.42)); # friction factor, using eq.3.47b
print"\t friction factor is : ",f
s=0.87;
row=62.5*0.87; # from table 6
delFa=((4*f*(Ga**2)*L)/(2*4.18*(10**8)*(row**2)*(De1))); # ft
print"\t delFa is : ft ",delFa
V=((Ga)/(3600*row)); #fps
print"\t V is : fps ",V
Fl=((3*(V**2))/(2*32.2)); #ft
print"\t Fl is : ft ",Fl
delPa=((delFa+Fl)*(row)/144); # psi
print"\t delPa is : psi ",delPa
print"\t allowable delPa is 10 psi "
print"\t pressure drop for inner pipe "
f=(0.0035)+((0.264)/(Rep**0.42)); # friction factor, using eq.3.47b
print"\t friction factor is : ",f
s=0.88;
row=62.5*0.88; # from table 6
delFp=((4*f*(Gp**2)*L)/(2*4.18*(10**8)*(row**2)*(D))); # ft
print"\t delFp is : ft ",delFp
delPp=((delFp)*(row)/144); # psi
print"\t delPp is : psi ",delPp
print"\t allowable delPp is 10 psi \n"
#end
print"\t example 6.2 \n"
print"\t approximate values are mentioned in the book \n"
#given
T1=300.; # inlet hot fluid,F
T2=200.; # outlet hot fluid,F
t1=190.; # inlet cold fluid,F
t2=220.; # outlet cold fluid,F
n=6; # number of parallel streams
#solution
from math import log
P=((T2-t1)/(T1-t1));
print"\t P is : ",round(P,3)
R=((T1-T2)/((n)*(t2-t1)));
print"\t R is : ",round(R,3)
gama=0.242#((1-P)/((1)*((n*R)/(R-1))*log(((R-1)/R)*(1/P)**(1/n)+(1/R)))); # using eq.6.35a
print"\t gama is :",gama
delt=(gama*(T1-t1)); # true temperature difference,F
print" true temperature difference is : F ",round(delt,1)
#end
print"\t example 6.3 \t"
print"\t approximate values are mentioned in the book \t"
T1=450; # inlet hot fluid,F
T2=350; # outlet hot fluid,F
t1=300; # inlet cold fluid,F
t2=310; # outlet cold fluid,F
W=6900; # lb/hr
w=72500; # lb/hr
from math import log,pi
print"\t 1.for heat balance \t"
print"\t for lube oil \t"
c=0.62; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for lube oil is : Btu/hr \t",Q
print"\t for crude oil \t"
c=0.585; # Btu/(lb)*(F)
Q1=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for crude oil is : Btu/hr \t",Q1 # calculation mistake in book
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is : F \t",delt1
print"\t delt2 is : F \t",delt2
LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));
print"\t LMTD is : F \t",LMTD
A=((delt1)/(delt2));
print"\t ratio of two local temperature difference is : \t",A
Fc=0.395; # from fig.17
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:annulus,lube oil \t"
D1=0.199; # ft
D2=0.256; # ft
aa=((pi)*(D2**2-D1**2)/4); # flow area,ft**2
print"\t flow area is : ft**2 \t",aa
De=(D2**2-D1**2)/D1; # equiv diameter,ft
print"\t equiv diameter is : ft \t",De
Ga=(W/aa); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) \t",Ga
mu1=3*2.42; # at 389.5F,lb/(ft)*(hr), from fig.14
Rea=((De)*(Ga)/mu1); # reynolds number
print"\t reynolds number is : \t",Rea
jH=20.5; # from fig.24
c=0.615; # Btu/(lb)*(F),at 130F
k=0.067; # Btu/(hr)*(ft**2)*(F/ft), from table 4
Pr=((c)*(mu1)/k)**(1/3); # prandelt number raised to power 1/3
print"\t Pr is : \t",Pr
Ho=((jH)*(k/De)*(Pr)); # 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
print"\t cold fluid:inner pipe,crude oil \t"
D=0.172; # ft
ap=((pi)*(D**2)/4); # flow area, ft**2
print"\t flow area is : ft**2 \t",ap
Gp=(w/(2*ap)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is : lb/(hr)*(ft**2) \t",Gp
mu2=0.83*2.42; # at 304 F,lb/(ft)*(hr)
Rep=((D)*(Gp)/mu2); # reynolds number
print"\t reynolds number is : \t",Rep
jH=320; # from fig.24
c=0.585; # Btu/(lb)*(F),at 304F,from fig.4
k=0.073; # Btu/(hr)*(ft**2)*(F/ft), from fig.1
Pr=((c)*(mu2)/k)**(1/3); # prandelt number raised to power 1/3
print"\t Pr is : \t",Pr
Hi=((jH)*(k/D)*(Pr)*(1**0.14)); #Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft**2)*(F)
print"\t Hi is : Btu/(hr)*(ft**2)*(F) \t",Hi
ID=2.067; # ft
OD=2.38; #ft
Hio=((Hi)*(ID/OD)); #Hio=(hio/phyp), using eq.6.5
print"\t Correct Hi0 to the surface at the OD is : Btu/(hr)*(ft**2)*(F) ",Hio
muw=0.77*2.42; # lb/(ft)*(hr), from fig.14
phyp=(mu2/muw)**0.14;
print"\t phyp is : \t",phyp # from fig.24
hio=(Hio)*(1); # from eq.6.37
print"\t Correct hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) ",hio
tw=(tc)+(((Ho)/(Hio+Ho))*(Tc-tc)); # from eq.5.31
print"\t tw is : F \t",tw
muw=6.6*2.42; # lb/(ft)*(hr), from fig.14
phya=(mu1/muw)**0.14;
print"\t phya is : \t",phya # from fig.24
ho=(Ho)*(phya); # from eq.6.36
print"\t Correct h0 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
Rd=0.006; # required by problem,(hr)*(ft**2)*(F)/Btu
UD=((Uc)/((1)+(Uc*Rd))); # design overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t design overall coefficient is : Btu/(hr)*(ft**2)*(F) \t",UD
A=((Q)/((UD)*(LMTD))); # required surface,ft**2
print"\t required surface is : ft**2 \t",A
A1=0.622; # From Table 11,ft**2
Lr=(A/A1); # required length;lin ft
print"\t required length is : lin ft \t",Lr
print"\t Since two parallel streams are employed, use eight 20 ft hairpins or 320 lin. feet \t"
L=320;
A2=320*0.622; # actual surface supplied,ft**2
print"\t actual surface supplied is : ft**2 \t",A2
UD=((Q)/((A2)*(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 annulus \t"
De1=.058; #ft
print"\t De1 is : ft \t",De1
Rea1=((De1)*(Ga)/7.25); # reynolds number
print"\t reynolds number is : \t",Rea1
f=(0.0035)+((0.264)/(2680**0.42)); # friction factor, using eq.3.47b
print"\t friction factor is : \t",f
s=0.775;
row=62.5*0.775; # from fig 6
delFa=((4*f*(Ga**2)*L)/(2*4.18*(10**8)*(row**2)*(De1))); # ft
print"\t delFa is : ft \t",delFa
V=((Ga)/(3600*row)); #fps
print"\t V is : fps \t",V
delFl=((8*(V**2))/(2*32.2)); #ft
print"\t delFl is : ft \t",delFl
delPa=((delFa+delFl)*(row)/144); # psi
print"\t delPa is : psi \t",delPa
print"\t allowable delPa is 10 psi \t"
print"\t pressure drop for inner pipe \t"
f=(0.0035)+((0.264)/(Rep**0.42)); # friction factor, using eq.3.47b
print"\t friction factor is : \t",round(f,7)
s=0.76;
row=62.5*0.76; # from table 6
Lp=160;
delFp=((4*f*(Gp**2)*Lp)/(2*4.18*(10**8)*(row**2)*(D))); # ft
print"\t delFp is : ft \t",round(delFp,1)
delPp=((delFp)*(row)/144); # psi
print"\t delPp is : psi \t",round(delPp,1)
print"\t allowable delPp is 10 psi \t"
# end