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DESCRIPTION
DESIGN COOLING TOWER
Citation preview
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Cooling Tower Application, according
1 Data
2 Tower height
3 NTU and HTU
4 Tower area
5 Compensation water
6 Operaqting diagram
7 Cooling tower schematic
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Treibal
Rev. cjc. 30.01.2014
Data for cooling tower application
Main equations and results
Cooling Tower height, NTU and HTU
Free-cross sectional surface of tower
Compensation, elimination, evaporation and entrainment fow rates
Equilibrium curve and operation lines
Schema
Index
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Cooling Tower Application Data
This application will be realized with following numerical data (Note 1).
Data for numerial example
Water flow rate entering the tower L1' = 18.0555556 kg agua/sTemperature of water entering the tower at the top (2) tL2= 45 C
Dry bulb temperature of air entering the tower tdbG1= 30 C
Wet bulb temperature of air entering the tower twbG1 24 C
Local height above sea level H = 0 m
Mximum cooling temperature will be defined with
a differential temperature Dt above air wet bulb temp. Dt = 5 K
Air to water flow rates ratio shall be "r" times its
minimum possible value r = 1.5
The compensation water entering the system wil have
temperature tcomp= 10 C
and will have a hardness da_c= 500 ppm
The in system circulating water sould have a maximum
hardness da_M= 2000 ppm
Mass transfer coefficient in the air ky_kmol= 6.2E-05 kmol / ( m2*s)
Air molecular mass Mair = 28.96 kg/kmol
Tower effective heat or mass transfer surface a = 500 m/m
Liquid unit mass flow rate Lu= 2.7 kg/(s*m
Air unit mass flow rate Gu= 2.0 kg/(s*m
Note 1
Basic data has been taken from [1], pages 278-281.
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Help Variables Cooling Tow
State L1
Water leaving the towertL1= twbG1+ Dt L2' =
tbhG1 24 tL2=
Dt = 5 K
tL1= 29 C
Compensation water
State G1 tacomp= 10 C
Ambient air entering the tower dac= 2000 ppm
tbsG1= 30 C
tbhG1 24 C
H = 0.0 m
h = Psychro_Enthalpy_tdb_twb_H Q
h = N/A kJ/kg
x = Psychro_AbsoluteHumidity_tdb_twb_H
x = N/A kg/lg
Mass transfer coefficient
Mass transfer coefficient in the air
ky_kmol= 6.2E-05 kmol / ( m2*s)
Air molecular mass
Mair = 28.96 kg/kmol
Mas transfer per kilogram daM=ky= ky_kmol* Mair
ky_kmol= 6.2E-05 kmol / ( m2*s)
Mair = 28.96 kg/kmol
ky= 0.0018 kg / ( m2*s)
Product Ky*a
Ky*a = Ky*a
ky= 0.0018 kg / ( m2*s)
a = 500 m/m
Ky*a = 0.90 kg / ( m *s)
Q = 270 W
Bl
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Rev. cjc. 30.01.2014
er Schema
18.0555556 kg / s
45 C Air
Water
tdbG1= 30 C
twbG1 24 C
Water
Air
2000 ppm
Rev. cjc. 30.01.2014
L1G1
Torreenfriadora
G22
owdown water: B
G1
L2
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Cooling Tower height
Tower packing height [2] Number of Tra
The packing height (l) of a tower can be calculated as The number of
is calculated by
[1], eq. (7.53), page 276 Sheet "2.- NTU"example of the
with
Result of NTU
NTU =
NTU =
and [1], eq. (7.54), page 277 Height of Tra
Z : Tower packing height [m]
QS=V : flow rate of dry air (is a constat) [kg/s] HTU =
MB: molar mass of air [kg/kmol] HTU =
ky: mass transfer coefficient in the ai r [kmol/(ms)]
a: effective heat or mass transfer surface [m/m] Tower packin
A : free cross-sectional surface of the tower [m] Z =
Iy : enthalpy in the air phase = enthalpy of humid air [J/kg] HTU =
(in the bulk phase) NTU =
Iy,i: enthalpy in the air phase ( i: at ther boundary, [J/kg] Z =
that is, in saturated condition)
HTU : Height of Transfer Unit
NTU : Number of Transfer Units
Subscripts
B : dry air
y : air phase = humid air
a : top of the tower
b : bottom of the tower
i : corresponds to the boundary (i.e., saturated state)
NTUHTUZ
AakM
QHTU
yB
S
ay,
by,
I
I
y
yiy,
dIII
1NTU
HTU
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Rev. cjc. 30.01.2014
sfer Units
transfer units (NTU)
numerical integtation.
presents a calculation NTU.
xample (sheet NTU)
(hL_a - hL_b) / N * Sf(x)
#VALUE! -
sfer Unit "HTU"
GS / (MB * ky * a * A)
#VALUE! m
height
HTU * NTU
#VALUE! m
#VALUE! -
#VALUE! m
Aak
Q
yB
S
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NTU and HTU calculations
column 1 column 2
Equilibrium curve for saturated air.
Water temperature at inlet of tower This curve is
tL2 = 45 C hair,sat = f(t)
Water temperature at tower outlet where the function
tL1 = 29 C hair,sat = Psychro_Enthalpy_tdb_HumRel_H
Range: tL2 - tL1 has been used
The curve strats at point
Number of sections 3 (29,96.4)
The range will be divided in a number "N" and ends at point
of sections 4 (45,218.2)
N = 6
Column 1 starts with temperature "tL2"
and ends with temperature "tL1". column 3
Between both temperatures, "N-1" Operation line for r = 1
temperatures are inserted to define where "r" is the ratio between the actual
the N sections. All section are defined mass flow rate and the minimum flow
with the same temperature differential. rate.
The line strts at a point defined by the
Temperature differential inlet air properties (point 1 in operating
DtL = tL2 - tL1 C diagram) also called state "G1"
tL2 = 45 C
tL1 = 29 C State G1 (Point 1)
DtL = 16 K tdbG1 = 29 C
twbG2 = 24 C
Section temperature increment H = 0 mDtL_Sect = DtL / N xG1= soluteHumidity_tdb_twb_H
DtL = 16 K xG1= N/A kg/kg
N = 6 -
DtL_Sect = 2.67 K hG1= N/A kJ/kg
Temperature at point "i+1"
ti+1= ti+ DtL_sect
1 2 3 3a 4 5
Equilibrium Operation Operation
curve for line for line for
saturated air. r = 1 Dh = r = 1.5 Dh =
tL hair,sat hoper_r=1 hair,sat-hop_r=1 hoper_r=1.5 hair,sat-hop_r=1.5
kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg
25 N/A
Ta
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25.5 N/A
Top.(2) 45 N/A N/A #VALUE! #VALUE! #VALUE!
42.33 N/A #VALUE! #VALUE! #VALUE! #VALUE!
39.67 N/A #VALUE! #VALUE! #VALUE! #VALUE!
37.00 N/A #VALUE! #VALUE! #VALUE! #VALUE!
34.33 N/A #VALUE! #VALUE! #VALUE! #VALUE!
31.67 N/A #VALUE! #VALUE! #VALUE! #VALUE!
Bottom (1) 29 N/A N/A #VALUE! N/A #VALUE!
In Operating diagram (sheet 6)
t H
C kJ/kg
Point 1: 29 N/A
Point 2: 45 N/A Line 1-2: Operation line for r = 1
Point 2': 45 #VALUE! Line 1-2': Operation line for r = 1.5
Point 3: 29 N/A
Point 4: 45 N/A Line 3-4: Equilibrium (saturation) line
From column 3a, is possible to see that the differences between the Equilibrium
curve and the operation line for r = 1 are very small (0.53 kJ/kg) at a certain
temperature (39.67 C). Thus, this operation line is near enough a minimum
flow rate line.
In the operating diagram figure (sheet 6), the operating line for r = 1 appears
to be tangent with the Equilibrium curve. This fact visualizes that this line is close
enough to the real minimum flow rate.
The condition of tangency between these two curves, is due to the fact that if
the operating line would cross the equilibrium line, it would not be possible to
have mass transfer in this region.
Height of tower packing Number of Transfer Units "N
From Treibal, Equation (7-5
(7.51a)
(7.51b)
The numerial integration of
performed by means of the t
where H'2 and H'1are the enthalpies of integration method.
the air-water mixture for the actual case, According this method, the i
that is, in this case for r = 1.5. is realized as it is shown in t
6, 7 and 8.
The final evaluation is done
iy
I
III
NTU
ay
by
,
1,
,
'2
'
1
''
'H
H i HH
dHNTU
GS'
mHH
dH
ak
GZ
H
H iy
S '
2
'
1
''
''
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(7.51c) following equation
(7.51d)
NTU = (hL_a- hL_b) /(2
Also where
hL_in_r=1.5= #VALUE!
hL_out=r=1.5= N/A
N = 6
Sf(x) = #VALUE!
Numerical results shown are from next NTU = #VALUE!
calculation sheets.
Treybal [2] result is
NTU = 3.25
The difference comes from t
the psychrometric properties
takes its values from graphi
the example are taken formfunctios. Also, the numerical
method is not indicated.
Comparison between the example calculation table and the tabvle from Treibal
1 2 3 4 5 6
Curva de Lnea de Lnea de
equilibrio para operacin operacin
aire saturado para r = 1 para r = 1.5 Dh = 1/Dht hair,sat hoper_r=1 hoper_r=1.5 hair,sat-hop_r=1.5
kJ/kg kJ/kg kJ/kg kJ/kg 1/(kJ/kg)
25 N/A
25.5 N/A
29 N/A 72 72 #VALUE! #VALUE!
31.67 N/A 96 87.7 #VALUE! #VALUE!
34.33 N/A 119 103.4 #VALUE! #VALUE!
37.00 N/A 143 119.1 #VALUE! #VALUE!
39.67 N/A 166 134.9 #VALUE! #VALUE!
42.33 N/A 190 150.6 #VALUE! #VALUE!
45.00 N/A 213 166.27 #VALUE! #VALUE!
1 2 3 4 5 6
Curva de Lnea de Lnea de
equilibrio para operacin operacin
aire saturado para r = 1 para r = 1.5 Dh = 1/Dh
t hair,sat hoper_r=1 hoper_r=1.5 hair,sat-hop_r=1.5
kJ/kg kJ/kg kJ/kg kJ/kg 1/(kJ/kg)
25 N/A
Calculation table, using psychrometric functions.
Table from Treybal [1], page 280
2
__
N
hhNTU
bLaL
mNTUHTUZ
maky
AakM
GHTU
yB
S
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25.5 N/A
29 100.0 72 72 28.0 0.0357
31.67 114.0 96 92.0 22.0 0.0455
34.33 129.8 119 106.5 23.3 0.0429
37.00 147.0 143 121.0 26.0 0.0385
39.67 166.8 166 135.5 31.3 0.0319
42.33 191.0 190 149.5 41.5 0.0241
45.00 216.0 213 163.50 52.5 0.0190
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The line ends in a point defined by the Slope of line with r = 1.5 Air flow rate
properties of the leaving satutared air S = m = L * Cpw / Gs (7.54)
Point 2' L = 18.0555556 kg/s From heat bal
Estado G2' Cpw = 4.1868 kJ/(kg*K)
tbsG2 = 45 C Gs = #VALUE! kg as/s
f = 100 % S = #VALUE! (kJ/kg)/K
H = 0 m Exit enthalpy
hG2= Psychro_Enthalpy_tdb_HumRel_H S = (hG2- hG1) / (tbsG2- tbsG1)
hG2'= N/A kJ/kg hG2= hG1+ S * (tbsG2- tbsG1)
The operation linefor r = 1, is the
straight line 1con: r = G / Gmin hG1= N/A J/kg
Slope of operation line witrh r = 1 S = #VALUE! (kJ/kg)/K
Sr=1= (hG2'- hG1) / (tL2- tL1) tbsG2= 45 C
hG2'= N/A kJ/kg tbsG1= 29 C [1], Eq, (7.54)
hG1= N/A kJ/kg hG2= #VALUE! J/kg
tL2= 45 C
tL1= 29 C column 5 GS: gas flow
Sr=1= #VALUE! (kJ/kg)/K Driving enthalpy difference at a point "i" hG2: exit air e
GS= Gr=1= 1/m * L * cpw (7.54) Dhi= hair,sat_i-hop_r=1.5_i hG1: inlet air e
m = Sr=1 L: liquid flow r
Sr=1= #VALUE! (kJ/kg)/K column 6 cpw: liquid sp
m = #VALUE! Reciproc of driving enthalpy difference tL2: inlet wate
L = 18.0555556 kg agua/s tL1: exir water
Cpw = 4.1868 kJ/(kg*K) column 7
Gr=1= #VALUE! kg as/s Coefficients for numerical integration
Ci= 1 at both ends
column 4 2 in the other elements
Operation line for r = 1.5
The line starts from the same point 1 column 8
the properties at the inlet defined as the Numerical integration elements
state G1. f(xi) = Ci* (1/Dhi)
Gs = r * Gr=1
r = 1.5
Gr=1= #VALUE! kg as/s
Gs = #VALUE! kg as/s
6 7 8
Numerical Air conditions in the tower,for r = 1
integration Conditions at the bottom of the tower (poin
1/Dh coefficient Point "1"
Ci f(x) tbs1= 29.0 C
1/(kJ/kg) hG1= N/A kJ/kg
Conditions at the top of the tower (point 2'
le 1. Tower packing height calculation
m
cLG
tt
hhm
cLG
hhG
bottom
top
HH
pw
S
LL
GG
pwS
GGS
Liair
DD
12
12
2
:1
:2
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Point 2'
#VALUE! 1 #VALUE! tbs2'= 45 C
#VALUE! 2 #VALUE! h2'= N/A kJ/kg
#VALUE! 2 #VALUE!
#VALUE! 2 #VALUE!
#VALUE! 2 #VALUE!
#VALUE! 2 #VALUE!
#VALUE! 1 #VALUE!
Sf(x) = #VALUE!
TU" Height of Transfer Unit "HTU" Height of Tra
1)
HTU =
withGs =
TU is MB=
rapezoidal HTU = G'S/( ky * a) ky_kmol=
G'S: 2.0 kg/(m*s) a =
ntegration ky*a : 0.9 kg / ( m *s) A =
he columns HTU = 2.2 m HTU =
Treybal [2] result is
according HTU = 2.2 m QS:
ydI
ak
GHTU
y
S
' HTU
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MB:
Height of packing tower ky:
a:
A :
Z = HTU * NTU
* N) * Sf(x) HTU = 2.2 m
NTU = #VALUE!
kJ/kg Z = #VALUE! m
kJ/kg
Treybal [2] result is
Z = 7.22 m
-
-
he values of
. Treybal
s, and in
psychrometric l integration
Trapezoidal numerical integration rule
7 8
Numerical
integration
coefficientCi f(x)
NTU=(hL2- hL1)/2*N* (f(x1) + 2*f(x2) + 2*f(x3) + .+ 2*f(xN-1) + f(xN) )
1 #VALUE! NTU = (hL_a- hL_b) /(2* N) * Sf(x)
2 #VALUE! where
2 #VALUE! hL_in_r=1.5= 166.3 kJ/kg
2 #VALUE! hL_out=r=1.5= 72.0 kJ/kg
2 #VALUE! N = 6
2 #VALUE! Sf(x) = #VALUE!
1 #VALUE!Sf(x) = #VALUE! NTU = #VALUE! -
7 8
Numerical
integration
coefficient
Ci f(x) Treybal table differs from the calculation
table in the values of the psichrometric
properties.
)(xf
)1(...22
11
2)(
1
Niparag
Nyiparag
fgN
abdxxf
i
i
k
N
k
i
b
a
NTUHTUZ
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Additionaly, Treibal uses a different
1 0.03575 numerical integration method, where
1 0.04545 the numerical integration coefficienst are
1 0.04292 not required (or Ci= 1)
1 0.03846 The numerical integration used is not
1 0.03195 indicated and Treybal gives as a
1 0.02410 final result a NTU value
1 0.01905
Sf(x) = 0.23767 NTU = 3.25 -
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Rev. cjc. 30.01.2014
Page 1 of 4
ance
, page 277
ate [kg as/ s]
nthalpy (top) [kJ/kg]
nthalpya (bottom) [kJ/kg]
ate [kg/ s]. (assumed constant)
cific heat [kJ/(kg*K)]
r temperature (top) [C]
temperature (bottom) [C]
Rev. cjc. 30.01.2014
Page 2 of 4
t "1" in diagram)
(Column 1)
(Column 3)
in diagram)
hh
tt
ttcL
GG
LL
LLpw
iquid
1
12
12
121
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(Column 1)
(Column 3)
Page 3 of 4
sfer Unit "HTU"
GS / (MB * ky * a * A)
#VALUE! kg as/s
28.96 kg/kmol
6.2E-05 kmol / ( m2*s)
500 m/m
#VALUE! m
#VALUE! m
flow rate of dry air (is a constat) [kg/s]
AakM
G
yB
S
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molar mass of air [kg/kmol]
mass transfer coefficient in the air [kmol/(ms)]
effective heat or mass transfer surface [m/m]
free cross-sectional surface of the tower [m]
Page 4 of 4
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HTU Treibal
Equation (7-51)
HTU = V / (MB * ky * a * A)
withV = 56.84 kg as/s
MB = 28.96 kg/kmol
ky_kmol = 6.2E-05 kmol / ( m2*s)
a = 500 m/m
A = 5.33 m
HTU = 11.9 m
Tower height
l = NTU * HTU
AakM
VHTU
yB
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NTU = 4.25 -
HTU = 11.9 m
l = 50.4 m
V = G * A
G = 10.66 kg as/(s*m)
5.33 m
56.84 kg as/s
G'S /( ky * a)
2 kg/(m*s)
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Free-cross sectional surface of tower
Area of cross sectional surface Seleted area
From borh results, the smallL = Lu* A should be selected, to ensur
L: liquid flow rate [kg/s] value of the product "ky*a" h
Lu: unit flow rate (for unit of cross the indicated value of
sectional surface): [kg/ ( m*s)]
A: area of cross section ky*a = 0.90
So
A = L / Lu A = #VALUE!
L = 18.0555556 kg/s
Lu= 2.7 kg/(m*s)
A = 6.69 m
Using the gas flow rate
A = GS/ GSu
G: gas rate [kg/s]
Gu: unit flow rate (for unit of cross
sectional surface): [kg/ ( m*s)]
A: area of cross section
Gs = #VALUE! kg as/s
GSu= 2 kg/(m*s)
A = #VALUE! m
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Rev. cjc. 30.01.2014
est value e that the
s at least
kg / ( m *s)
m
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Compensation water
Cosidering a compensation and a Entrainment loss rate "W", water that is Application
continuous elimination, the mass being transported with the exit air,
balance is leaving from de top of the tower as a Evaporation
(a) loss of water.1.- Absolute h
Evaporation rate "E", water that is Assuming tha
A water hardness balance is evapotated in the air flow producing saturated at P
de cooling of the water flow fO
(b) From equation (d) The enthalpy
calculation Ta
and therefore hO=
Assuming initi
tO=
(.c) the correspon
assumed tem
tO=
Eliminating M from (a) and (.c) fO
H =
h =
(d) Now, using S
temperature t
With calculate
M : compensation rate [kg/h] the correspon
B : elimination rate [kg/h] can be be cal
E : evaporation rate [kg/h] t =
W : entrainment loss rate [kg/h] fO
daC : hardness weight fraction of H =
circulating water [kg/kg] or [ppm] xa2=
daM : hardness weight fraction of
compensation water [kg/kg] or [ppm] 2.- Absolute h
From sheet 2
Elimination rate "B", required to replace xa1=
water with a maximum allowable salts
content with fresh water with the in this 3.- Humidity c
water existing salt content. This is Dx2-1=
called the compensation rate. xa2=
(e) xa1=Dx2-1=
WEBM
CM daWBdaM
M
C
da
daWBM
M
C
da
daWBWEB
Wdada
daEB
MC
M
Wdada
daEB
da
dada
EWB
da
dada
EWB
da
da
EWB
Eda
daW
da
daB
Eda
daW
da
daB
Eda
daW
da
daW
da
daB
da
da
WEda
da
Wda
da
B
WEda
daW
da
daBB
da
daW
da
daBWEB
MC
M
M
MC
M
CM
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
M
C
1
11
11
1
1
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E = GS* Dx2-1
Dry air flow rate (sheet 2)
ate "E" Gs = #VALUE! kg as/s
Dx2-1= #VALUE! kg/kg umidity of exit air E = #VALUE! kg/s
the leaving air is basically
oint "O" Entrainment loss "W"
100 % To estimate the entrainment losses,
t this point, from the one assumes that these losses are
ble 1, is a percentage &W of the water flow rate
#VALUE! kJ/kg &L = 0.2 %
ally a temperature value The water flow rate is
30 C L = 18.0555556 kg/s
ding enthalpy for this W = L * &L Compensatio
erature is with L = 18.0555556 kg/s
40.0 C &L = 0.002 -
100 % W = 0.036 kg/s
0 m.a.s.l.
N/A kJ/kg Elimination rate "B"
lver, find a value of the B = E * ( daM/ (daC- daM) ) - W
to obtain that h = hO E = #VALUE! kg/s
d exit air temperature W = 0.036 kg/s
ding absolute humidity da_M= 500 ppm
ulated da_c= 2000 ppm
40.0 C B = #VALUE! kg/s
100 %
0 m.a.s.l. Compensation rate "M"
N/A kg/kg M = (B + W) * daC/ daM
B = #VALUE! kg/s
umidity of inlet air W 0.036 kg/s
da_c= 2000 ppm
N/A kg/kg da_M= 500 ppm
M = #VALUE! kg/s
hange
xa2- xa1 kg/kg
N/A kg/kg
N/A kg/kg#VALUE! kg/kg
M, daM
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ev. c c. . .
Entrainment water
Water
water
Air
Elimination water
L1 G1
Coolingtower
G2
E
W, daC
B, daC
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Operation Diagram
1 2 3 3a 4 5 6
Curva de Lnea de Lnea de
equilibrio para operacin operacinaire saturado para r = 1 Dh = para r = 1.5 Dh = 1/Dh
tL hair,sat hoper_r=1 hair,sat-hop_r=1 hoper_r=1.5 hair,sat-hop_r=1.5
kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg 1/(kJ/kg)
25.0 N/A
25.5 N/A
29.0 N/A N/A #VALUE! N/A #VALUE! #VALUE!
31.7 N/A #VALUE! #VALUE! #VALUE! #VALUE! #VALUE!
34.3 N/A #VALUE! #VALUE! #VALUE! #VALUE! #VALUE!
37.0 N/A #VALUE! #VALUE! #VALUE! #VALUE! #VALUE!
39.7 N/A #VALUE! #VALUE! #VALUE! #VALUE! #VALUE!
42.3 N/A #VALUE! #VALUE! #VALUE! #VALUE! #VALUE!45.0 N/A N/A #VALUE! #VALUE! #VALUE! #VALUE!
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
25.0 30.0 35.0 40.0 45.0 50
Enthalpy
air-vapor
[kJ/kg
da]
Liquid temperature [C]
Operation Diagram of Cooling Tower
Equilibrium curve, for saturated air
Operating line with r = 1
Operating line with r = 1.5
2
1
2'
3
4
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7 8
Numerical H'
integration (gas-vapor mixture)coefficient kJ/kg as
Ci f(x)
H'*2 4
1 #VALUE! H'2
2 #VALUE!
2 #VALUE!
2 #VALUE!
2 #VALUE!
2 #VALUE!1 #VALUE! H'*1 3
Sf(x) = #VALUE!
H'1
1
Liquid tem
tL1 tL2
.0
Equilibrium curve
Op. L, r = 1
Op. L. r = 1.5
R
ST
U
',HtL
*,HtL
', ii Ht
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Rev. cjc. 30.01.2014
2
perature tLC
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Cooling tower schematic [3]
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[1] Operaciones de transferencia de masa 2/e
Robert E. Treybal
McGraw Hill,2003
Page 274. Enfriamiento de agua con aire
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[1] Operaciones de transferencia de masa 2/e
Robert E. Treybal
McGraw Hill, 2003
[2]
[3]
http://library.kfupm.edu.sa/ISI/2006/5-2006.pdf
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Packing height and free-cross sectional surface of a tower [2]
1.- Packing height
The packing height of a tower can be calculated
according [2], equation (65)
Z = I : Tower packinQS=V : flow rate of dr
MB: molar mass o
Naming the first term as "Height of Transfer ky: mass transfer
Unit (HTU) " a: effective heat
A : free cross-se
Iy : enthalpy in th
(in the bulk ph
Iy,i: enthalpy in th
and the second term as the Numbert of Transfer that is, in satu
Units (NTU)
HTU : Height of Tra
NTU : Number of Tr
Subscripts
B : dry air
the packinh height becomes y : air phase = h
a : top of the tow
b : bottom of the
i : corresponds t
yyiy
I
IyB
S dIIIAakM
QZ
ay
by
,
1,
,
AakM
VHTU
yB
yyiy
I
I
dIII
NTU
ay
by
,1
,
,
NTUHTUZ
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height [m] air (is a constat) [kg/s]
air [kg/kmol]
coefficient in the air [kmol/(ms)]
or mass transfer surface [m/m]
tional surface of the tower [m]
air phase = enthalpy of humid air [J/kg]
ase)
air phase (i: at ther boundary,
rated condition)
sfer Unit
nsfer Units
mid air
r
tower
o the boundary (i.e., saturated state)
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