KE-42 3000 5. Cooling Tower

Aalto University School of Science and Technology/ Chemical Engineering 1(19)

23.9.2010

COOLING TOWER

1. Introduction...................................................................................................................................... 2 2. Theory .............................................................................................................................................. 2

2.1 Humid air ................................................................................................................................... 3 2.2 Balance Equations...................................................................................................................... 5 2.3 Operating Line ........................................................................................................................... 7 2.4 Heat and Mass Transfer ............................................................................................................. 8 2.5 Lewis’ Equations........................................................................................................................ 9 2.6 Height of the Tower ................................................................................................................. 10 2.7 Deriving the Equation for Calculating the Height of the Tower.............................................. 10 2.8 NTU ......................................................................................................................................... 11 2.9 Slope of the Tie Line................................................................................................................ 12 2.10 HTU ....................................................................................................................................... 12 2.11 Calculating the Coefficients of Heat and Mass Transfer ....................................................... 12

3. Equipment ...................................................................................................................................... 14 3.1 Cooling Tower ......................................................................................................................... 14 3.2 Measuring Equipment .............................................................................................................. 14

4. Operating the tower........................................................................................................................ 15 4.1 Starting the Work ..................................................................................................................... 15 4.2 Measurements .......................................................................................................................... 15 4.3 Finishing the Work................................................................................................................... 15

5. Report............................................................................................................................................. 16 6. Nomenclature ................................................................................................................................. 17 7. References...................................................................................................................................... 19 8. Appendixes .................................................................................................................................... 19


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1. INTRODUCTION

A cooling tower is a unit where water is cooled with air. Warm water enters the top of the tower, runs down the column and chills because it partially vaporizes to air, which flows upward. Also, the air flow is usually colder than the water flow, so water is chilled because of the heat exchange.

Since both mass and heat transfer phenomena are present, the theory of cooling towers is a bit complicated. Certain simplifications can be made because the liquid phase is pure water, and therefore there is no driving force for mass transfer in that phase.

As many other vapor-liquid mass transfer units, the cooling tower is packed with packing materials, which increases the surface area, in order to enhance the mass and heat transfer.

2. THEORY

In this laboratory work, the height of the cooling tower (the height of the packing) is calculated from the measuring results and the equations derived from mass balances. Obtained result is compared to the real height.

Evaluating the height of the cooling tower is based on theory derived for mass transfer units. The theory will not be presented here entirely but only the parts, which are needed for this work.

Next, let’s consider a stationary cooling tower; a sketch is shown in figure 1. Nomenclature is in chapter 6. Notice, that the dry air flow V is constant and no subscripts are used with it.

′′mA

′′q

b

a

l l

l+dl

Ll, ix,hl

Ll+dl, ix,l+dl

Lb, ix,b, Tx,b

La, ix,a, Tx,a

V, Iy,b, Ty,b, Yb

V, Iy,a, Ty,a, Ya

V, Iy,l+dl

V, Iy,l

Figure 1. Schematic of a cooling tower

Cooling tower is an example of gas–liquid mass transfer equipment. The essential thing in this equipment is to bring the two phases into contact so that the mass and heat transfer between phases is possible. The two phases are usually in countercurrent connection.

Design of a cooling tower is based on equations derived from balances. The most important equation for a mass transfer equipment is the equation of the operating line, which describes how the states of the two phases depend on each other at the same location in the unit (at the same level in the equipment). Here, as usual, the radial (horizontal) mixing is assumed to be perfect or ideal (or mean values are used). Then the properties of a phase are the same on a certain level. The equation of the operating line is always obtained from one (or some) of the balances of the equipment.

In a cooling tower the temperature of the liquid phase Tx and the enthalpy of the gas phase Iy are chosen to describe the states of the phases. Thus the operating line will be represented in Tx,Iy –coordinates, where it is a straight line with certain assumptions. It is possible to represent the temperature of the gas phase Ty in the same coordinates in which case T, Iy –coordinates are used.

Next the so called constitutive equations of the equipment and the mass and heat balances are used to derive the equations needed to calculate a cooling tower. During the derivation many steps and equations are needed because of the simultaneous mass and heat transfer but the achieved result is still quite simple.


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2.1 HUMID AIR

Here only a case where air is the gas and the vapor in it is water is dealt. By definition humidity Y is the mass proportion between water in air and dry air:

B

A

AIR

WATER

mm

mm

Y =≡ , (1)

where air is marked with subscript B and water with A. The mass of the water consists of all the water, which is in the air (solid, liquid, and vapor), but in this work it is assumed that neither ice nor mist is present. The unit of the humidity is

=≡

airdry kg waterkg

mm

YAIR

WATER (2)

Since the humidity is defined as mass proportion rather than mass fraction, the water carried by the air can be calculated as humidity * dry air (not humidity * total air flow). Relative humidity is defined as

fYY

≡ * (3)

Notice that the saturation point Y* and therefore the relative humidity is defined only at temperatures less than the boiling point of water (if the temperature is higher than boiling point, Y* = ∞).

2.1.1 Enthalpy of Humid Air

Next, the following assumptions are made:

O1: Air–water mixture is an ideal mixture of ideal gases. (4)

O2: The heat capacities of air and steam cp,B and cp,A,y are constant. (5) Despite the assumption O2 the heat capacity of humid air is not constant, but depends on humidity as follows:

c c Ycp y p B p A y, , , ,= + (6) The enthalpy of an air-water mixture in temperature T compared to a reference temperature T0 is

[ ] [ ]I I YI

c T T i Y c T T r i

y y B y A

p B B p A y A A

= +

= − + + − + +

, ,

, , , , , ,( ) ( )0 0 0 0 0

(7)

By choosing the enthalpies of the reference temperature T0 to be zero, a following form is obtained:

I c Yc T T Yry p B p A y A= + − +( )( ), , , ,0 0 (8)


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2.1.2 Changes in the Enthalpy of the Humid Air

When the temperature and humidity of humid air change, so according to equation (7) and assumption O2:

[ ] [ ]I I c Y c T T Y r c Y c T T Y ry y p B p A y A p B p A y A, , , , , , , , , ,( )( ) ( )( )2 1 2 2 0 2 0 1 1 0 1 0− = + − + − + − + (9)

This can be written again by using the heat capacity of the humid air cp,y either in state 2 or in state 1. If the state 2 is taken and the term below is both added in and subtracted from the right side of equation (9)

( )( ), , ,c Y c T Tp B p A y+ −2 1 0 , the following is obtained:

[ ] [ ]I I c Y c T T Y r c Y c T T Y r

c Y c T T c Y c T T

c Y c T T c Y c T T

Y Y r

c Y c

y y p B p A y A p B p A y A

p B p A y p B p A y

p B p A y p B p A y

A

p B p A

, , , , , , , , , ,

, , , , , ,

, , , , , ,

,

, , ,

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( )

( )

(

2 1 2 2 0 2 0 1 1 0 1 0

2 2 0 2 1 0

2 1 0 1 1 0

2 1 0

2

− = + − + − + − +

= + − − + −

+ + − − + −

+ −

= +

[ ]y p A y A

p B p A y p A y A

p B p A y A

T T Y Y c T T Y Y r

c Y c T T Y Y c T T r

c Y c T T Y Y I

)( ) ( ) ( ) ( )

( )( ) ( ) ( )

( )( ) ( )

, , ,

, , , , , ,

, , , ,

2 1 2 1 1 0 2 1 0

2 2 1 2 1 1 0 0

2 2 1 2 1 1

− + − − + −

= + − + − − +

= + − + −

Naturally, a suchlike result is obtained when specific heat capacity of air cp,y in the state 1 is used. Altogether:

1,12122,,1,2, )()( Aypyy IYYTTcII −+−=− (10a)

I I c T T Y Y Iy y p y A, , , , ,( ) ( )2 1 1 2 1 2 1 2− = − + − (10b) These equations will be used when deriving the equations for the height of the tower.

2.1.3 Enthalpy of Saturated Humid Air

In order to calculate the height of a mass transfer unit one needs to know how the properties of vapor and liquid phases depend on each other when they are in physical equilibrium. This is described with equilibrium line.

According to the two-film theory the interface between two phases (phase boundary) is in equilibrium state. That is, the temperature curve is continuous and the equilibrium concentrations of the phases prevail. However, the bulk concentrations and temperatures in liquid and vapor phases are different (and not in this equilibrium state).

The equilibrium line will be drawn in the same coordinates than the operating line (Tx,Iy –coordinates). The relationship between temperature and the enthalpy of air must be known when the cooling tower is calculated. Because there’s equilibrium between water and water-air mixture in the system, we need data how the enthalpy of air saturated with water depends on temperature. These values of the equilibrium line are collected for example in the Tables of Keskinen handout (Keskinen 1989). For numerical calculation, an equation that fits the data in the tables is needed.

The fitting of the equilibrium data into an equation with three parameters is shown in appendix 1. The error, which this procedure causes, is approximately 3 %, which is more accurate than the other steps of calculations. So three parameters are adequate enough. As a result, an equation for the relation between the temperature and the enthalpy is obtained:


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I Ty* .. . *= +22 75 0 05126 2 159 (11)

where the reference state is 0 OC and 1 bar.

According to the two-film theory of mass transfer, the states of the phases at their interface are on this equilibrium line (equation (11)), where T is the temperature of the phase boundary and

T Tx i y i, ,= (12)

2.1.4 Humidity of Saturated Humid Air

The humidity of saturated humid air is fitted to a function of three variables in appendix 1. Now an average accuracy of 6 % is obtained.

Y T* .. . * *= + −0 007656 7 86 10 7 2 956 (13)

2.2 BALANCE EQUATIONS

A general balance equation can be described as

accumulation = input - output + generation (14) If the system works in steady state and no reaction occurs, then both the accumulation and the generation terms are zero and

input = output (15)

2.2.1 Whole Tower

The overall water and heat balances for a cooling tower are:

L VY L VYa b b a+ = + (16)

L i VI L i VIa a y b b b y a+ = +, , (17)

2.2.2 Differential Part of the Tower

Next, a part of the process, between arbitrary imaginary levels l and l + dl (which can be thought as actual levels in the equipment), is considered. The change in quantities between the two levels is marked (flow L as an example):

dL L Ll dl h= −+ (18) Water balance in the whole balance area is (flow rate of the dry air is considered to be constant)

L VY L VYl dl l l l dl+ ++ = +( ) ( ) (19) or

dL VdY= (20) Water balance for the water phase is

L L m a Adll dl l A M+ = + ′′ (21) or


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dL m a AdlA M= ′′ (22) Water balance for air phase between the two levels is (with a constant dry air flow rate)

( ) ( )VY m a Adl VYl A M l dl+ ′′ = + (23) or

VdY m a AdlA M= ′′ (24) Combining all the water balances leads to

dL VdY m a AdlA M= = ′′ , (25) which shows that the change in water flow is equal for both phases. So the change of humidity in air related to the height of the tower is equal to the change of flow amount in water phase related to the height of the tower. Because the direction of mass transfer is from water phase into air and water is running downwards, flow tends to decrease in that direction. The change in both phases is equal to the amount of water, which has been transferred.

Heat balance (or enthalpy balance) for whole balance area between the two levels is

( ) ( ) ( ) ( )Li VI Li VIl dl l l l dl+ ++ = + (26) or

d Li VdI( ) = (27) Heat balance for water phase between the two levels is

( ) ( ) ,Li Li q a Adl i m a Adll dl l x H A i A M+ = + ′′ + ′′ (28) or

d Li q a Adl i m a Adlx H A i A M( ) ,= ′′ + ′′ (29) Heat balance for air phase is

( ) ( ),VI q a Adl I m a Adl VIl y H A i A M l dl+ ′′ + ′′ = + (30) or

VdI q a Adl I m a Adly H A i A M= ′′ + ′′, (31) Heat balance (27) shows that change of enthalpy in the water phase is equal to the change of enthalpy in air phase. Heat balance (29) shows that total heat flux from water phase to air phase consists of heat flux ′′qx and the enthalpy flux AiA mi ′′, of mass transfer.

Since i. Water is transferred in mass transfer,

ii. The interface of two phases is the balance boundary of the heat balance and iii. Mass transfer in balance (29) is in water phase,

the enthalpy that moves along the mass transfer is same as the enthalpy of water at the phase boundary in water phase. In other words the transferring enthalpy is the enthalpy of liquid water at the interface of two phases iA,i, and iA,i = ix,i (because L phase is pure water).

Heat balance (31) is the corresponding equation in air phase. The total heat flux, which transfers to air, consists of heat flux ′′qy and enthalpy flux I mA i A, ′′ carried by mass transfer. Again, since

i. Water is transferred in mass transfer, ii. The interface of two phases is the balance boundary of heat balance and iii. Mass transfer in balance (31) is in air phase,


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the enthalpy that moves along the mass transfer is same as the enthalpy of water at the phase boundary in air phase. In other words it is the enthalpy of water as steam at the interface of phases IA,i (which is not equal to the enthalpy of V phase at the phase boundary Iy,i).

Notice that since AiAAiA mImi ′′≠′′ ,, (that is i IA i A i, ,≠ ), so equations (29) and (31) substituted in equation (27) gives

′′ ≠ ′′q qy x (32)

2.3 OPERATING LINE

The operating line indicates how the quantities, which describe the states of the phases, depend from each other. The equation of the operating line can always be derived from the balances of equipment.

In the chosen Tx,Iy coordinates, the operating line of the cooling tower can be derived from the heat balance (27) when the following assumptions are made: The flow rate of water is constant (transferring amount is small compared to total flow) and heat capacity cp,x of liquid water is constant:

O3: dL = 0 and m = 0 (33)

O4: i c T Tx p x x REF= −, ( ) (34) Substituting these assumptions to the left side of the heat balance (27) gives (For clarity, L phase is marked with subscript x and V phase with y):

d(Li Ldi i dL Ld c T T i dL Ldc Tx x x p x x REF x p x x) ( ), ,= + = − + = (35) The heat balance (27) becomes to

Lc dT VdIp x x y, = (36) Integrating this over the whole tower gives (V, L, and cp,x are constant):

Lc T T V I Ip x x a x b y a y b, (( ) ), , , ,− = − (37) Integrating between the top of the tower (marked with a) and an arbitrary level l (without subscript) gives

Lc T T V I Ip x x x a y y a, (( ) ), ,− = − (38) or between the bottom of the tower (marked with b) and an arbitrary level l gives

Lc T T V I Ip x x x b y y b, (( ) ), ,− = − (39) Two last equations connect quantities Iy and Tx at an arbitrary height in the tower, so they are the two forms of the operating line. Solving Iy from equations (38) and (39) as a function of Tx gives:

ILc

VT I

Lc

VTy

p xx y b

p xx b= + −,

,,

,( ) (40a)

ILc

VT I

Lc

VTy

p xx y a

p xx a= + −,

,,

,( ) (40b)

Hence, equations (40a) and (40b) are the equations of the operating line.


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2.4 HEAT AND MASS TRANSFER

Coefficients of heat and mass transfer are actually proportionality coefficients, which are defined with the flux equation so that the flux can be calculated from the driving force:

fluxdriving force

resistanceproportionality coefficient * driving force= = (41a)

This equation is commonly used in technology, for example in electrotechnology, heat transfer and mass transfer. The flux equation does not define the mechanism, so all hypothesis about the mechanism are included in the coefficients. So, when publicing a correlation, it must be told, in which circumstances the correlation was derived so that it could be used correctly. (For example, there’s no use for a correlation of laminar flow in case of turbulent circumstances.)

2.4.1 Heat transfer

Heat transfer fluxes in the cooling tower are:

′′ = −q h T Tx x x i( ) (42)

′′ = −q h T Ty y M i y, ( ) (43) Notice that the heat transfer coefficient hy of air phase holds simultaneously for heat and mass transfer. In addition, according to equation (32) ′′ ≠ ′′q qy x , because of the vaporization at the phase boundary. Due to his, the overall coefficient of heat transfer U cannot be defined.

2.4.2 Mass transfer

Mass transfer flux from water to air can be calculated only in air phase, because there is no concentration gradient in water phase and thus it has no resistance for mass transfer. In the air phase:

′′ = −Nk

y yAy

yiφ

( ) , (44)

where φ y is the relative speed coefficient, which takes into account the deviations from equimolar mass transfer:

φ yi

i

z y z y

zz yz y

zz y=

− − −−−

= −( ) ( )

ln( ) ln

1 (45)

If the mass transfer is equimolar, then φ y = 1 and if concentrations are dilute, so then φ y ≈ 1. In this case the mass transfer is not equimolar (only water transfers) but the concentrations are assumed to be so small thatφ y ≈ 1 is a proper approximation. So

O5: φ y = 1 (46) This leads to

′′ = −N k y yA y i( ) (47) Multiplying with the molar mass of the transferring component gives

′′ = −m M k y yA A y i( ) (48)


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Instead of mole fractions, the gradient of humidity is used in cooling tower. This can be changed with equation (49):

yn

n

mM

mM

mM

YM

MY

M

AA

A

A

B

B

A

A

A

B A

≡ =+

=+∑ 1

(49)

Equations (48) and (49) gives

′′ =+

−+

=+

−+

m M k

YM

MY

M

YM

MY

M

kY

MY

M

Y

MY

M

A A y

i

A

B

i

A

A

B A

yi

B

i

A B A

1 1

1 1

(50)

According to the assumption O5 (dilute concentrations) Y is small, so the second terms of the denominator are negligible and equation (50) reduces to

′′ = −m M k Y YA B y i( ) (51)

Notify that MB is the molar mass of air, not water!

Since diffusion is not equimolar in the cooling tower (air does not dissolve in water), the assumption O5 of low concentrations has to be used twice.

2.5 LEWIS’ EQUATIONS

It can be shown that

h

M kcy

B yp y= , (52)

holds quite well for air-steam systems. Calculations of cooling tower are based on an assumption that heat and mass transfer can be described with one coefficient. This is possible if Lewis’ equation (52) holds for air phase. In addition the Lewis number has to be one, that is

LeSc

yy

y

=Pr

= 1 (53)

Next, a rough estimate is calculated for steam in air (T = 25 °C) by using physical properties of air. Diffusivity of water in air is about 26*10-6 m2/s (Treyball, 1980). Other values are obtained from ”Kemian laitetekniikan taulukoita ja piirroksia” (Keskinen, 1989):

LeSc

PrDc

. *. * *

. * * . *.

.

..y

y

y p

= = = = =

−

−

−

ηρη

λ

18 43 101186 26 10

18 43 10 1006 100 026

0 600 71

085

6

6

6 3 (54)


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2.6 HEIGHT OF THE TOWER

The height of the tower (packing) can be calculated with either HTU-method or HETP-method.

h N HETP= * (55)

h NTU HTU= * (56) The HTU-method is used in this work, but the HETP-method is tried too. All calculations should be made where the resistance is the greatest; in this work it means the air phase.

2.7 DERIVING THE EQUATION FOR CALCULATING THE HEIGHT OF THE TOWER

Substituting equations (43) and (51) to equation (31) gives:

VdI h T T a Adl I M k Y Y a Adly y M i y H A i B y i M= − + −, ,( ) ( ) (57) Next, the mass transfer is assumed not to affect heat transfer and the surface areas for heat and mass transfer are equal in size. Also the Lewis equation is assumed to hold for the air phase, so:

O6: h hy M y, = (58)

O7: a a aM H= = (59)

O8: h

M kcy

B yp y= , (60)

Substituting assumptions (O6 – O8) into equation (57) gives:

[ ]VdI M k c T T I x x aAdly B y p y i y A i i= − + −, ,( ) ( ) (61)

The expression inside the brackets is the difference of enthalpy of humid air between the phase boundary and bulk as in equation (10). Substituting this into equation (61) results in:

VdI M k I I aAdly B y y i y= −( ), (62) According to equation (63), the change in the enthalpy of air related to the height of tower is proportional to the product of mass transfer coefficient and the difference of enthalpies in the air phase (phase boundary – bulk). So, the driving force for heat transfer is the enthalpy gradient. Equation (62) can be rearranged:

dlV

M k aA I IdI

B y y i yy=

−1

( ),

(63)

This can be integrated over the tower from the bottom (b), where Y is small, to the top (a), where Y is large.

o

l

I

I

B y y,i yydl l

VM k aA (I I )

dIy b

y a

∫ ∫= =

−,

, 1 (64)

If the first quotient of the right side stays constant in the tower, equation (65) is obtained:

lV

M k aA I IdI H N

B y y i yI

I

y y y

y a

y b

=

−=∫

1( ),,

,

, (65)

which is the final equation for calculating the height of the tower.


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In equation (65) H is the height of transfer unit (HTU) and N is the number of transfer units (NTU) in the tower. Subscript y indicates that calculations are made in air phase.

2.8 NTU

The number of transfer units in air phase is

NI I

dIyy i yI

I

y

a

b

=−∫1

( ),

(66)

This integral is the inverse of driving force (enthalpy gradient). This means; the smaller the driving force, the higher the tower.

Water and heat transfers from the water phase to the air phase in a cooling tower, so the cooling tower corresponds to a stripper (figure 2).

Tx,a, Iy,a

Tx,b, Iy,b

equilibrium

Iy,i

Tx,

Iy Iy,i-Iy

Tx,i-Tx

Tx

α = −h

M kx

B y

Operating line

Figure 2. Determining the number of transfer units.

The integral has to be calculated numerically, which gives

NI I

dII I

Iyy i yI

I

yy i k y kk

K

y k

y b

y a

=−

=−∫ ∑

=

1 1

1( ) ( ), , , ,,

,

,

∆ (67)

In order to calculate the value of the integral, a relationship between the enthalpy of air at the phase boundary Iy,i and in the bulk phase Iy is needed on every level of the tower. This refers to the slope of tie line, which connects these two states (figure 2).

The integral is calculated numerically as follows:

1. The interval being integrated (the length of the operating line) is portioned out to ∆Iy 2. From each subinterval a value for Iy is chosen (Iy,k). 3. The corresponding Tx is calculated.

• Now the operating line point (Tx,k, Iy,k) is known. 4. The phase boundary point (Tx,i,k, Iy,i,k) corresponding to the operating line point (Tx,k, Iy,k) is

calculated. • The phase boundary point cannot be explicitly calculated from the operating line point

but it is determined by the operating line point and the tie line. • The phase boundary point is to be determined by iterating either graphically or

numerically. 5. When the operating line point and (Tx,k, Iy,k) and the corresponding phase boundary point

(Tx,i,k, Iy,i,k) are known, the value of the integral in a subinterval k is calculated. 6. Finally, all calculated values are summed.

There is an example of calculating a cooling tower with MS.EXCEL program in appendix 1. The equilibrium line (or phase boundary) points are solved with SOLVER (the difference of a-a’ or α-α‘


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is being minimized). Notice that cooling tower corresponds theoretically to a stripper, so the operating line lies below the equilibrium line but the integral is the difference of enthalpies in the equilibrium line and in the operating line, so it is positive.

2.9 SLOPE OF THE TIE LINE

According to equation (27) d Li VdI( ) = . Substituting this into the left side of equation (62) gives

d Li M k I I aAdlx B y y i y( ) ( ),= − (68) Substituting the equation of heat transfer in the L phase (42) into the heat balance of the L phase (29) gives

d Li h T T a Adl i m a Adlx x i H A i A M( ) ( ) ,= − + ′′ (69) Since the amount of water, which transfers from the water phase to the air phase, is negligible ( ′′ =mA 0 ), the enthalpy carried by the transferred water is negligible, too:

d Li h T T a AAdlx x x i H( ) ( )= − (70) Equations (68) and (70) and assumption O7 give

M k I I aA l h T T aA lB y y i y x x i( ) ( ), − = −d d (71) Now the slope of tie lines is obtained:

mI I

T Th

M ky i y

i x

x

B y

=−

−= −

( )

( ), (72)

If the coefficient of heat transfer is small in the water phase, the slope is small too, and the tie line approaches a horizontal line. If the coefficient of heat transfer is much greater than the coefficient of heat transfer in the air phase, the tie line approaches a vertical line.

2.10 HTU

The height of transfer unit in the air phase is obtained from equation (65):

HV

M k aAyB y

=

(73)

Since the conditions vary in the tower, gives equation (73) different values for the height of transfer unit in the top and in the bottom of the tower.

2.11 CALCULATING THE COEFFICIENTS OF HEAT AND MASS TRANSFER

The required mass and heat transfer coefficients can be calculated from empirical correlations based on experimental data. These correlations are usually in the form of:

Sh bRe Scn m= Nu bRe Prn m= (74)

St bRe ScMn 1 m 1= − − St bRe PrH

n 1 m 1= − − (75) Due to the analogy of heat and mass transfer, the parameters b, n, and m are same in equations (74) and (75). Defining the coefficients for mass transfer is such a difficult task that these correlations typically exists only for cases where transfer is between a fluid and a solid surface, and the analogy of heat and mass transfer can be used (heat transfer between solid surface and fluid is a typical heat transfer problem in process industry).


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Now the liquid can be thought to percolate along the surface of the packing material and air flows upwards. The following correlation, which describes mass transfer in turbulent flow, would then be correct enough:

Sh Re Sc= 0 023 0 8 1 3. . / Nu Re Pr= 0 023 0 8 1 3. . / (76) Although the Reynolds number in the layer will not be higher than 2100, this correlation is used. Situation for the water phase is more complicated. The velocity of the water is very low, so its Reynolds number is small and furthermore the calculated Nu-number and heat transfer coefficient may become very small. In order to avoid this, the following correlation is used in the water phase:

Nu Re Pr= +2 0 0 5 1 2 1 3. . / / (77) This will ensure that the Nu number will not be smaller than 2,0. Different correlations are used for air and water phase, which is a very common procedure.

The definitions of dimensionless numbers are:

ScDAb

≡η

ρ Sh

kd

cDp

Ab

≡ Pr ≡η

λ

cp λ

phdNu ≡ (78)

StSh

SckcuM ≡ =

Re

pH uc

hSc

NuSt

ρ=≡

Re (79)

It should be defined in every correlation how the values in the correlations are calculated. Especially, when calculating height of packing, the definitions and calculations of flow velocities ought to be well-defined.

The velocity of a flow in a packing layer can be defined as superficial velocity or as interstitial velocity. In correlations (76) the velocities are the interstitial velocities in a pipe, so the they should be used when calculating the height of the tower. The ratio of the superficial velocity to the interstitial velocity is the void fraction of bed ε. The Reynolds number in a bed is then

Re xp x x

x

d u=

ρ

η Re y

p y y

y

d u=

ρ

η (80)

where the characteristic dimension of packing layer is the particle diameter dp. So, when using correlations (76), Reynolds number is calculated with interstitial velocities, which are:

uLAx

x

=ρ ε

uVAy

y

=ρ ε

(81)

in which it is assumed that humidity is negligible, and the mass flow of dry air can be used. Notice that the velocity of liquid flow will be very small since it is calculated per the whole cross-sectional area (of which the gas flow takes the most, which is ignored).

Since flows L and V change as a function of height, the calculations will give local L and V, a local Re, local Sh and Nu, and finally local k and h. An example of calculations is shown in appendix 1.


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3. EQUIPMENT

3.1 COOLING TOWER

In the laboratory work a cooling tower made of plexiglass, a schema of which is shown in figure 3, is used. The tower is in the machinery room E334 of chemical engineering laboratory.

Exiting air

Air flow

Hot water

Water tank Fan

Figure 3. Cooling tower.

Plastic Pall rings of 2 in are placed as packing material inside the tower. The height of the packing is 2000 mm and the inner diameter of the tower is 580 mm. The rotameter of water flow gives the flow rate as percentage terms of the maximum flow rate, which is 6018 l/h (20 °C).

3.2 MEASURING EQUIPMENT

In the laboratory equipment there are readily attached: § Rotameter of incoming water § Thermometer for incoming and exiting water and air flows and for water tank.

Also a psychrometer is needed for the work.


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4. OPERATING THE TOWER

4.1 STARTING THE WORK

The work is started according to the assistant’s instructions. Measurements are done with two different incoming water flow rates, which are given by the assistant. Temperatures and the humidity are measured every four (4) minutes until the temperature of the water tank stays constant.

4.2 MEASUREMENTS

Measurements are done with two different incoming water flow rates. In both measurements the following values are documented after the system has stabilized:

• Incoming water flow rate from (rotameter) • Humidity of incoming air from the inlet pipe (near the fan) • Temperature of incoming water • Temperature of incoming air • Temperature of exiting water • Temperature of exiting air • Temperature of water tank • Also measuring accuracy of every value is estimated and documented.

When all measurements are done, the results are shown to assistant.

4.3 FINISHING THE WORK

After the assistant has approved the results, the laboratory equipment is shut down, the surroundings is cleaned up, and all other equipment is returned where they belong.


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5. REPORT

1. Operating line § Assume that the exiting air is saturated. § Assume that water flow in the tower is constant and equals to La. § Calculate the flow rate of dry air from the heat balance of the whole tower. § Define the operating line equation and its end points in (Tx, Iy) -coordinates.

2. Calculate the amount of vaporized water in the tower. Compare it to the water flow. 3. Calculate hx, hy and ky at the top and bottom of bed § Use the physical properties of air in its temperature. § Diffusivity of water in air is 26*10-6 m2/s. § Velocities are interstitial velocities in the bed (use the flow rate of dry air). The void fraction

of bed can be obtained for example from Perry (1973). § Calculate the concentration in Sh number from ideal gas law. Molar mass of air is M = MB

=0.029 kg/mol

4. Calculate the Le number from equation (53) at the top and bottom of the bed. Calculate the left side of the equation (52). Compare it to the heat capacity of dry air.

5. N and HETP § Draw the operating line and the equilibrium line graphically or with computer. § Define number of needed ideal stages N by stepping off the tower manually. § Assume that HETP is 3 ft. § Calculate the height of the tower

6. NTU and HTU § Calculate the height of transfer unit HTU. § Calculate the number of transfer units NTU (for example, as in appendix 1). § Calculate the height of the tower. § (If hx >> ky, the temperature of phase boundary is the same as the temperature of the water

phase.)

7. Incorrect estimate as in general instructions. § A numerical incorrect estimate is not needed. § Pay special attention to gross and systematic errors and to general estimate of the results.


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6. NOMENCLATURE

A free cross-sectional surface of the tower, m2 a surface for mass or heat transfer, m2/m3 aH effective heat transfer surface, m2/m3 aM effective mass transfer surface, m2/m3 b parameter, dimensionless c molar concentration, mol/m3 cp,A,y heat capacity of steam, J/kg cp,B heat capacity of dry air, J/kg cp,x heat capacity of the water phase = heat capacity of liquid water, J/kg cp,y heat capacity of the air phase = heat capacity of humid air, J/kg dp characteristic dimension of the packing, m DAB diffusivity, m2/s f relative humidity, dimensionless l height of the packing, m hx heat transfer coefficient in the water phase, W/m2/K hy heat transfer coefficient in the air phase, W/m2/K Hy height of transfer unit in the air phase, m ix enthalpy in the water phase = enthalpy of liquid water, J/kg IA enthalpy of steam, J/kg Iy enthalpy in the air phase = enthalpy of humid air, J/kg ky mass transfer coefficient in the air, kmol/s/m2 L flow rate of water, kg/s m parameter, dimensionless mi mass of component i, kg

′′mA mass transfer flux of water to the air phase, kg/m2/s M molar mass, kg/mol n parameter, dimensionless

′′N A mass transfer flux of water, mol A/m2/s Ny number of transfer units in the air phase, dimensionless p total pressure

′′qx heat flux leaving from water, W/m2 ′′q y heat flux entering air, W/m2

Ti temperature at the phase boundary, °C Tx bulk temperature in the water phase, °C Ty bulk temperature in the air phase, °C V flow rate of dry air, kg/s ux velocity of water, m/s uy velocity of dry air, m/s Y bulk humidity of air, kg H2O/kg dry air Yi humidity of air in the phase boundary, kg H2O/kg dry air y mole fraction of water, dimensionless yi mole fraction of ´water at the phase boundary, dimensionless

Greek letters α slope of the tie line, J/kgK ε empty space, void fraction, dimensionless λ heat conductivity, W/mK ρx density of water, kg/m3 ρy density of humid air, kg/m3 φ relative speed coefficient in mass transfer, dimensionless


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Subscripts a top of the tower b bottom of the tower A water B dry air i phase boundary H heat transfer M mass transfer y air phase = humid air x water phase = water

Superscripts * equilibrium value

mean value

Dimensionless numbers Le Lewis number Nu Nusselt number Pr Prandtl number Re Reynolds number Sc Schmidt number Sh Sherwood number StH Stanton number for heat transfer StM Stanton number for mass transfer


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7. REFERENCES

Keskinen, K.I., Kemian laitetekniikan taulukoita ja piirroksia, Otakustantamo, 1989 McCabe, W.L., Smith, J.C. and Harriot, P., Unit Operations of Chemical Engineering, McGraw-Hill, 5th ed., 1993 Perry, R.H. and Chilton, C.H., eds., Chemical Engineers’ Handbook, 5th ed., McGraw-Hill, 1973. Treyball, R.E., Mass-Transfer Operations, McGraw-Hill, 3rd. ed., 1980

8. APPENDIXES

1. Sample calculations by Martti Pekkanen (values for packing materials from Perry (1973)).

Appendix 1 23.9.2010 1

23.9.2010

Cooling towerSaturated air-water system T I* Y*I*=aa+bb*T^cc Y*=aaa+bbb*T^ccc 10 30.1 0.0084

aa= 22.725754 aaa= 0.007656 20 55.8 0.0132bb= 0.051216 bbb= 7.860E-07 30 102.0 0.0259cc= 2.159413 ccc= 2.956000 40 173.2 0.0504

Tref=0 50 264.6 0.0904

Physical properties of air Physical properties of water

cp,B= 1.00 kJ/kg/K cp,x= 4.18 kJ/kg/K IA(Tref)= 2500.00 kJ/kgMB= 0.0290 kg/mol iA(Tref)= 0.00 kJ/kg

cp,y= 1.88 kJ/kg/K rA(Tref)= 2500.00 kJ/kg

Operating linea b, saturated b

V*I= 63.85 34.90 22.05 =V*I kWI= 102.00 55.75 35.23 =I kJ/kg

V*Y= 0.0162 0.0082 0.0038 =V*Y kg water/sY= 0.0259 0.0132 0.0060 =Y kg water/kg dry air

Ty= 30.00 20.00 <------ 20.00 =Ty CV= 0.63 0.63 0.63 =V kg dry air/s

tower initialL= 0.50 0.50 0.50 =L kg/s

Tx= 45.00 31.15 25.00 =Tx Cix= 188.10 130.20 104.50 =ix kJ/kg

L*ix= 94.05 65.10 52.25 =L*ix kW

b, saturated is the state of saturated air in the feed temperature T=Tyb of the air phase

Water feed flow rate L= 0.50 kg/sAmount of vaporized water V*(Ya-Yb)= 0.01 kg/sProportion of vaporization Vaporization / water flow rate= 2.50 %

Heat power of cooling tower L*(ixa-ixb)= 41.80 kWPower of vaporization V*(Ia-Ib)*ra= 31.19 kWProportion of vaporization Vapoization / heat= 74.61 %

HeightHy= 0.16 m From page 2Ny= 1.00 From page 3

h= 0.16 m

Appendix 1 23.9.2010 2

23.9.2010

Tower a= 500 m2/m3e= 0.8 A= 0.36 m2

dp= 0.003 m

Transfer coefficientswater phase air phase

a b a bT= 45.00 25.00 30 20 C temperature measured

eta= 0.000599 0.000894 0.000019 0.000018 kg/m/s viscosity from literaturecp= 4178.70 4178.80 1006.60 1006.30 J/kg/K heat capacity from literature

lam= 0.637100 0.606300 0.026400 0.026000 J/m/K/s heat conductivity from literaturePr= 3.93 6.16 0.71 0.71 Pr number =eta*cp/lam

eta= 0.000019 0.000018 kg/m/s viscosity from literatureroo= 1.166900 1.205000 kg/m3 density from literature

D= 0.000026 0.000026 m2/s diffusivity assumptionSc= 0.62 0.58 Sc number =eta/roo/D

m= 0.50 0.50 0.63 0.63 kg/s mass flow rate measuredroo= 1000.00 1000.00 1.166900 1.205000 kg/m3 density from literature

V= 0.000500 0.000500 0.536476 0.519513 m3/s volume flow rate =m/roou= 0.001736 0.001736 1.862764 1.803866 m/s velocity =V/A/e

Re= 8.70 5.83 349.28 353.82 Re number =roo*u*dp/lam

Nu= 4.33 4.21 2.22 2.25 Nu numberSh= 2.12 2.10 Sh number

c= 40.09 41.46 mol/m3 molar concentration =p/RT

h= 918.77 851.36 19.57 19.49 J/m2/K/s heat transfer coefficient =Nu*lam/dp

k= 0.7360 0.7542 mol/m2/s mass transfer coefficient =Sh*c*D/dpk'= 0.018358 0.018190 m/s mass transfer coefficient =Sh*D/dp

Hy= 0.16 0.16 m height of transfer unit =m/MB/k/a/A

Lewisall values are from tha air phase air phase

a bLe= 0.86 0.81 Le number =Sc/Pr

left side of the lewis equation = h/MB/k= 916.74 890.92 J/kg/KRight side of the lewis equation = cp= 1006.60 1006.30 J/kg/K heat capacity from literature

abs. rel. difference % = 8.93 11.47

Appendix 1 23.9.2010 3

23.9.2010

Tie lineshx= 900.0000 a= -44335ky= 0.7000

Ny DIy= 5

operating line phase boundary NyIy Tx Iy,i Tx,i a' Iy,i-Iy DNy Ny (a'-a)^435.23 25.00 0.0040.23 26.50 83.35 26.50 -43800 43.13 0.12 0.12 8.20E+1045.23 28.00 90.99 27.99 -43681 45.77 0.11 0.23 1.83E+1150.23 29.49 99.13 29.49 -43789 48.90 0.10 0.33 8.87E+1055.23 30.99 107.75 30.99 -43716 52.53 0.10 0.42 1.47E+1160.23 32.49 116.87 32.49 -43778 56.65 0.09 0.51 9.62E+1065.23 33.99 126.50 33.98 -44005 61.27 0.08 0.59 1.19E+1070.23 35.48 136.62 35.48 -43798 66.40 0.08 0.67 8.30E+1075.23 36.98 147.26 36.98 -43916 72.03 0.07 0.74 3.07E+1080.23 38.48 158.40 38.48 -43853 78.18 0.06 0.80 5.38E+1085.23 39.98 170.06 39.97 -43870 84.84 0.06 0.86 4.68E+1090.23 41.47 182.24 41.47 -43954 92.02 0.05 0.91 2.11E+1095.23 42.97 194.94 42.97 -43962 99.71 0.05 0.96 1.93E+10

100.23 44.47 208.16 44.47 -44204 107.94 0.05 1.01 2.90E+08105.23 45.97 221.91 45.96 -43901 116.68 0.04 1.05 3.53E+10

9.00E+11

Tie lineshx >> ky => Tx,i = Tx Tie lines are vertical

Ny DIy= 5

operating line phase boundary NyIy Tx Iy,i Tx,i Iy,i-Iy DNy Ny35.23 25.00 0.0040.23 26.50 83.36 26.50 43.13 0.12 0.1245.23 28.00 91.00 28.00 45.77 0.11 0.2350.23 29.49 99.13 29.49 48.91 0.10 0.3355.23 30.99 107.76 30.99 52.53 0.10 0.4260.23 32.49 116.88 32.49 56.65 0.09 0.5165.23 33.99 126.50 33.99 61.28 0.08 0.5970.23 35.48 136.63 35.48 66.41 0.08 0.6775.23 36.98 147.27 36.98 72.04 0.07 0.7480.23 38.48 158.42 38.48 78.19 0.06 0.8085.23 39.98 170.08 39.98 84.85 0.06 0.8690.23 41.47 182.26 41.47 92.03 0.05 0.9195.23 42.97 194.96 42.97 99.73 0.05 0.96

100.23 44.47 208.18 44.47 107.96 0.05 1.01105.23 45.97 221.94 45.97 116.71 0.04 1.05

Appendix 1 23.9.2010 4

23.9.2010

Saturated air-water mixture Enthalpy

I1=aa+bb*T^cc aa= 22.725754bb= 0.051216cc= 2.159413dd=ee=

Values from literature ABS, %T I I1 DI DI / I DI / I (DI / I)^2C kJ/kg k.i.

10 29.29 30.12 -0.83 -0.028 2.83 0.00080015 42.03 40.47 1.56 0.037 3.71 0.00137720 57.43 55.75 1.68 0.029 2.92 0.00085325 76.36 76.20 0.16 0.002 0.21 0.00000430 99.77 102.00 -2.23 -0.022 2.23 0.00049835 129.1 133.31 -4.21 -0.033 3.26 0.00106240 166.1 170.27 -4.17 -0.025 2.51 0.00062945 213.4 213.00 0.40 0.002 0.19 0.00000450 274 261.61 12.39 0.045 4.52 0.002046

2.49 0.007274

Saturated air-water mixture Humidity

Y2=aaa+bbb*T^ccc aaa= 0.007656bbb= 7.86E-07ccc= 2.956543ddd=eee=

Values from literature ABS, %T Y Y2 DY DY / Y DY / Y (DY / Y)^2C kg /kg k.i.

10 0.007636 0.008367 -0.000731 -0.096 9.57 0.00916815 0.010660 0.010015 0.000645 0.061 6.05 0.00366520 0.014710 0.013178 0.001532 0.104 10.42 0.01085125 0.020110 0.018337 0.001773 0.088 8.82 0.00777630 0.027230 0.025967 0.001263 0.046 4.64 0.00215335 0.036610 0.036539 0.000071 0.002 0.20 0.00000440 0.048920 0.050520 -0.001600 -0.033 3.27 0.00107045 0.065080 0.068375 -0.003295 -0.051 5.06 0.00256450 0.086320 0.090567 -0.004247 -0.049 4.92 0.002421

5.88 0.039671

KE-42 3000 5. Cooling Tower

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