Chapter 17 Direct Contact Transfer : Cooling Tower¶
Example 71.1 pgno:585¶
In [14]:
print"\t example 17.1 \n"
pw=0.4298; # psia, at 75F, table 7
pt=14.696; # psia
t=75.;
Mw=18.;
Ma=29.;
X=(pw/(pt-pw))*(Mw/Ma);
print"\t humidity is : lb water/lb air ",X
H=(X*t)+(1051.5*X)+(0.24*t); # eq 17.54
print"\t enthalpy at 75F is : Btu/lb dry air ",round(H,1)
# end
Example 17.2 Pgno:602¶
In [15]:
print"\t example 17.2 \t"
print"\t approximate values are mentioned in the book \t"
print"\t by numerical integration \t"
T1=85.;
T2=120.;
A=576.; # ground area, from fig 17.12
L=1500.*(500./576.);
G=1400.;
R=(L/G);
from math import log10
print"\t R is : \t",R
H1=39.1; # fig 17.12
H2=H1+(R*(T2-T1));
print"\t H2 is : Btu \t",H2
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=85F
Hs=50; # fig 17.12
d1=(Hs-H1);
print"\t difference is : \t",d1
#at t=90
Hs=56.7; # fig 17.12
H=43.7; # fig 17.12
d2=Hs-H;
print"\t difference is : \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : \t",d
dT=5; # F
nd1=(dT/d);
print"\t nd1 is : \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.70;
print"\t number of diffusing units : \t",nd
print"\t log mean enthalpy difference \t"
dt=49.9; # diff. of enthalpies at top of the tower, from table in solution
db=10.9; # diff of enthalpies at bottom of the tower,from table in solution
LME=(dt-db)/(2.3*log10(dt/db));
print"\t log mean of enthalpy : Btu/lb \t",LME
nd=(T2-T1)/(LME);
print"\t number of diffusing units are : \t",round(nd,2)
# The error is naturally larger the greater the range
#end
Example 17.3 pgno:604¶
In [16]:
print"\t example 17.3 \t"
print"\t approximate values are mentioned in the book \t"
# Since the loading is based on 1 ft2 of ground area
nd=1.7;
L=1302;
Kxa=115;
Z=(nd*L)/(Kxa);
print"\t Z is : ft \t",Z
HDU=(Z/nd);
print"\t height of diffusion unit : ft \t",round(HDU,1)
# end
Example 17.4 pgno:605¶
In [17]:
print"\t example 17.4 \t"
print"\t approximate values are mentioned in the book \t"
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=79.3F
Hs=43.4; # fig 17.12
H=30.4; # fig 17.12
d1=(Hs-H);
print"\t difference is : \t",d1
#at t=85
Hs=50; # fig 17.12
H=35.7; # fig 17.12
d2=Hs-H;
print"\t difference is : \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : \t",d
dT=(85-79.3); # F
nd1=(dT/d);
print"\t nd1 is : \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.72;
print"\t number of diffusing units : \t",nd
# end
Example 17.5 pgno:609¶
In [18]:
print"\t example 17.5 \t"
print"\t approximate values are mentioned in the book \t"
T1=85;
T2=120;
R=0.93; # R=(L/G), for 1500 gpm
import numpy
import matplotlib
%matplotlib inline
from matplotlib import pyplot
print"\t for 120percent of design \t"
R1=1.2*R;
print"\t R is : %.3f \t",R1
H1=39.1; # at 87.2F
H2=H1+(R1*(T2-T1));
print"\t H2 is : %.1f Btu \t",H2
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=87.2F
Hs=53.1; # from table in the solution
d1=(Hs-H1);
print"\t difference is : %.1f \t",d1
#at t=90
Hs=56.7; # fig 17.12
H=42; # fig 17.12
d2=Hs-H;
print"\t difference is : %.1f \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : %.1f \t",d
dT=(90-87.2); # F
nd1=(dT/d);
print"\t nd1 is : %.3f \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.53;
print"\t number of diffusing units : %.2f \t",nd
print"\t for 80 percent of design \t"
R2=0.8*R;
print"\t R is : %.3f \t",R2
H1=39.1; # at 87.2F
H2=H1+(R2*(T2-T1));
print"\t H2 is : Btu \t",H2
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=82.5F
Hs=47.2; # from table in the solution
d1=(Hs-H1);
print"\t difference is : \t",d1
#at t=85
Hs=50; # fig 17.12
H=40.8; # fig 17.12
d2=Hs-H;
print"\t difference is : \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : \t",d
dT=(85-82.5); # F
nd1=(dT/d);
print"\t nd1 is : \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.92;
print"\t number of diffusing units : \t",nd
X=[1.115, 0.93, 0.74];
Y=[1.53, 1.70, 1.92];
pyplot.plot(X,Y)
pyplot.xlabel('L/G')
pyplot.ylabel('nd')
pyplot.title('KxaV/L vs L/G')
pyplot.show()
print"\t trial 1 \t"
R3=1.1;
print"\t R is : \t",R3
H1=34.5; # at 87.2F
H2=H1+(R3*(T2-T1));
print"\t H2 is : Btu \t",H2
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=85F
Hs=50; # from table in the solution
d1=(Hs-H1);
print"\t difference is : \t",d1
#at t=90
Hs=56.7; # fig 17.12
H=40; # fig 17.12
d2=Hs-H;
print"\t difference is : \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : \t",d
dT=(90-85); # F
nd1=(dT/d);
print"\t nd1 is : \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.48;
print"\t number of diffusing units : \t",nd
R3=1.19; # from fig 17.14
print"\t L/G is : \t",R3
print"\t trial 2 \t"
R4=1.2;
print"\t R4 is : \t",R4
H1=34.5; # at 87.2F
H2=H1+(R4*(T2-T1));
print"\t H2 is : Btu \t",H2
# The area between the saturation line and the operating line represents the potential for heat transfer
# at T=85F
Hs=50; # from table in the solution
d1=(Hs-H1);
print"\t difference is : \t",d1
#at t=90
Hs=56.7; # fig 17.12
H=40.5; # fig 17.12
d2=Hs-H;
print"\t difference is : \t",d2
d=(d1+d2)/(2);
print"\t average of difference is : \t",d
dT=(90-85); # F
nd1=(dT/d);
print"\t nd1 is : \t",nd1
# similarly calculating nd at each temperature and adding them will give you total nd value
nd=1.56;
print"\t number of diffusing units : \t",nd
R3=1.08; # from fig 17.14
print"\t L/G is : \t",R3
# end
Example 17.6 pgno:615¶
In [19]:
print"\t example 17.6 \t"
print"\t approximate values are mentioned in the book \t"
# basis 1ft^2 ground area
#Assumption: 20 per cent of the initial vapor content of the gas enters the water body
X1=(1.69/(14.7-1.69))*(18./29.);
print"\t X1 : lb/lb \t",X1
G=1500.;
w1=G*X1;
print"\t total water in inlet gas : lb/hr \t",w1
# The inlet gas is at 300F and a 120F dew point. Use 0.25 Btu/(lb)(DegF) for the specific heat of nitrogen
H1=(0.0807*120)+(0.0807*1025.8)+(0.45*0.0807*(300-120))+(0.25*300); # eq 17.55
print"\t H1 : Btu/lb dry air \t",H1
X2=(w1*(1-.2)/G);
print"\t outlet gas humidity : lb/lb \t",X2
pw=(X2*29*14.7/18)/(1+(X2*29/18));
print"\t pw : psia \t",pw
Tw=112.9; # F, from table 7 for above pw
# The outlet gas has a temperature of 200DegF and a 112.9DegF dew point
H2=(X2*Tw)+(X2*1029.8)+(X2*0.45*(200-Tw))+(0.25*200); # eq 17.55
print"\t H2 : Btu/lb dry air \t",H2
q=G*(H1-H2);
print"\t total heat load : Btu/hr \t",q
w2=q/(120-85);
print"\t water loading : lb/hr \t",w2
print"\t interval 1 \t"
# (Kxa*delV/L)= 0 t0 0.05
nd=0.05; # nd=Kxa*V/L
Le=0.93; # fig 17.4 at 300F
C=(0.25)+(0.45*X1);
print"\t C : Btu/(lb)*(F) \t",C
haV=(nd*w2*Le*C);
print"\t haV : Btu/(hr)*(F) \t",haV
qc=(haV*(300-120));
print"\t qc : Btu/hr \t",qc
delT=(qc/(C*G));
print"\t delT : F \t",delT
T1=(300-delT);
print"\t T(0.05) : F \t",T1
delt=(qc/w2);
print"\t delt : F \t",delt
t1=(120-delt);
print"\t t(0.05) : F \t",t1
print"\t interval 2 \t"
# (Kxa*delV/L)= 0.05 to 0.15
nd1=0.1;
haV1=(nd1*w2*Le*C);
print"\t haV1 : Btu/(hr)*(F) \t",haV1
qc1=(haV1*(T1-t1));
print"\t qc1 : Btu/hr \t",qc1
delT1=(qc1/(C*G));
print"\t delT1 : F \t",delT1
T2=(T1-delT1);
print"\t T(0.15) : F \t",T2
X3=0.0748; # at 117.6F
w3=(nd1*w2*(0.0807-X3));
print"\t water diffused during interval : lb/hr \t",w3
w4=(w1-w3);
print"\t water remaining : lb/hr \t",w4
l1=1027; # Btu/lb, l1= lamda at 117.6F
qd=(w3*l1);
print"\t qd : Btu/hr \t",qd
q1=(qd+qc1);
print"\t q1 : Btu/hr \t",q1
delt1=(q1/w2);
print"\t delt1 : F \t",delt1
t2=(t1-delt1);
print"\t t(0.15) : F \t",t2
X4=0.0640; # at 112.5
X5=(w4/G);
print"\t X(112.5F) : lb/lb \t",X5
print"\t interval 3 \t"
# (Kxa*delV/L)= 0.15 to 0.25
nd1=0.1;
haV1=(nd1*w2*Le*C);
print"\t haV1 : Btu/(hr)*(F) \t",haV1
qc2=(haV1*(T2-t2));
print"\t qc2 : Btu/hr \t",qc2
delT2=(qc2/(C*G));
print"\t delT2 : F \t",delT2
T3=(T2-delT2);
print"\t T(0.25) : F \t",T3
w5=(nd1*w2*(X5-X4));
print"\t water diffused during interval : lb/hr \t",w5
w6=(w4-w5);
print"\t water remaining : lb/hr \t",w6
l2=1030; # Btu/lb, l1= lamda at 112.5F
qd1=(w5*l2);
print"\t qd1 : Btu/hr \t",qd1
q2=(qd1+qc2);
print"\t q2 : Btu/hr \t",q2
delt2=(q2/w2);
print"\t delt2 : F \t",delt2
t3=(t2-delt2);
print"\t t(0.25) : F \t",t3
X6=0.0533; # at 106.5
X7=(w6/G);
print"\t X(106.5F) : lb/lb \t",X7
# The calculations of the remaining intervals until a. gas temperature of 200DegF is reached are shown in Fig. 17.17
w7=21.92; # total water diffused from table in solution
d=(w7/w1)*100;
print"\t calculated diffusion : \t",d
print"\t Using some standard low-pressure-drop data \t"
# For G = 1500, extrapolate to L = 2040 on logarithmic coordinates. Kxa = 510.
ndt=.54; # from 1st table in solution
Kxa=510; # from 2nd table in solution
Z=(ndt*w2/Kxa);
print"\t tower height : ft \t",Z
A=(50000/G);
print"\t cross section : ft^2 \t",round(A,1)
# end
Example 17.7 pgno:620¶
In [20]:
print"\t example 17.7 \t"
print"\t approximate values are mentioned in the book \t"
C=0.28; # assumption
w=50000.; # lb/hr
G=1500.;
Qs=(w*C*(500-200));
Qd=(w/G)*(22685); # qd=22685, from previous prblm
from math import log
print"\t sensible heat : Btu/hr \t",Qs
print"\t approximate diffusion : Btu/hr \t",Qd
Q=(Qs+Qd);
print"\t total heat : Btu/hr \t",Q
# an allowance as high as 30 per cent of the sensible load can be made and the excess water compensated for by throttling when the tower is in operation
w1=(Q/(120-85));
print"\t total water quantity : lb/hr \t",w1
# If the maximum liquid loading is taken as 2040 lb/(hr)(ft'!), the required tower cross section
A=(w1/2040);
print"\t tower cross section : ft**2 \t",A
w3=(w/A);
print"\t new gas rate : lb/(hr)(ft**2) \t",w3
# The two terminal temperature differences are (200 - 85) and (500 - 120).
LMTD=((500-120)-(200-85))/(log((500-120)/(200-85)));
print"\t LMTD : \t",LMTD
dt=35;
N=(dt/LMTD); # eq 17.88
print"\t haV/L : \t",N
Le=0.93;
nd=(N/(C*Le));
print"\t number diffusion units : \t",nd
# By extrapolation for G = 718 and L = 2040,Kxa=215
L=2040;
Kxa=215;
Z=(nd*L/Kxa); # calculation mistake
print"\t height of tower : ft \t",Z
di=(A)**(1/2);
print" ground dimensions : ft \t",di
# ground dimensions are 5.8*8.3*8.3 ft
# end