Cooling Tower Kap 65 m3 Per Hr

5/27/2018 Cooling Tower Kap 65 m3 Per Hr

1/42

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


2/42

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


3/42

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.


4/42

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


5/42

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


6/42

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


7/42

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


8/42

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


9/42

25.5 N/A

Top.(2) 45 N/A N/A #VALUE! #VALUE! #VALUE!

42.33 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

''

''


10/42

(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


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


11/42

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


12/42

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


13/42

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 =


according HTU = 2.2 m QS:

ydI

ak

GHTU

y

S

' HTU


14/42

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


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


15/42

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 -


16/42

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


17/42

(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


18/42

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


19/42


20/42


21/42

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


22/42

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)


23/42

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


24/42

Rev. cjc. 30.01.2014

est value e that the

s at least

kg / ( m *s)

m


25/42

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


26/42

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


27/42

ev. c c. . .

Entrainment water

Water

water

Air

Elimination water

L1 G1

Coolingtower

G2

E

W, daC

B, daC


28/42


29/42

Operation Diagram

1 2 3 3a 4 5 6


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!




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


30/42

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


31/42

Rev. cjc. 30.01.2014

2

perature tLC


32/42

Cooling tower schematic [3]


33/42


34/42


35/42

[1] Operaciones de transferencia de masa 2/e

Robert E. Treybal

McGraw Hill,2003

Page 274. Enfriamiento de agua con aire


36/42


37/42


38/42

[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


39/42


40/42

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


41/42

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)


42/42

Documents

Cooling Tower Kap 65 m3 Per Hr