CE427
Unit Operations Lab 3
Fall Semester, 2000
September 24, 2000
COOLING TOWER THEORY
Introduction
Cooling towers are used extensively in the chemical and petroleum industries. We also know
from the thermodynamics course that they are often used for heat removal in power generation
cycles. Undoubtedly more are used in air conditioning cycles than any place else. (Next time you are
in a tall city building, look below and you’ll probably see dozens of them on the rooftops of lower
buildings.) As technology goes, they are pretty old stuff, but they interesting from a technical
standpoint because both heat and mass transfer occur simultaneously, just as in distillation.
Most distillations are carried out in plate columns where one can visualize steps. Cooling
towers usually have grids – a sort of packing, so mass and heat transfer occur essentially
continuously up (and down) the column. Distillation often involves several components, whereas in a
cooling tower, it’s just a mixture of water and air. Chapter 23 of McCabe, Smith, and Harriott (M,
S&H) is devoted to “Humidification Operations” including Cooling towers.
Theory from the Textbook
Read Chapter 23. You may be somewhat acquainted with the first half or so of this chapter
because the material is covered in CE212. The equation development of particular importance to us
is on Pages 757 through 760. It is built around Figure 23-8(a). Thus, there is a temperature gradient
in the liquid which supplies sensible heat to the air-water interface. This heat is then carried away
into the gas stream by two mechanisms, one being sensible heat and the other latent heat from the
vaporization of a small amount of water at the surface.
There are a large number of equations in the text. We present an abridged version of these
equations here. Equation numbers in decimal form are those from MS&H; all other equation
numbers pertain to this write-up. First, Equation (23.22) below is the overall energy balance
equation, which is analogous to the “operating line” in distillation or absorption. Here H denotes the
enthalpy (Btu/lb or J/g), while G denotes the mass flow rate (lb/ft2/hr or kg/m2/hr). The subscript x is
associated with the liquid phase, while y with the gas phase.
Gy dH y  d (Gx H x )
(23.22)
Assuming that only a small fraction of the liquid evaporates in this process compared to the total
amount of liquid fed into the tank, we can write Gxb Gxa Gx : In these equations the subscript a
depicts the top and b the bottom of the Cooling tower. The datum plane (or reference temperature, T0
value) for all enthalpy calculations is arbitrary and it eventually cancels out on both sides of the
equations. For water, it is usually taken at a temperature of To=32F (according to Perry’s
Handbook). For liquid water then, Hx = cL(Tx – To) [Equation (23.26)]. Thus, the differential form of
Equation (23.22) becomes:
Gy dH y  Gx cLdTx
(1)
Integrating from the bottom of the column (i.e. end b) to any particular point the column, one obtains:
Gy (H y  H yb )  Gx cL (Tx  Txb )
(2)
This is the equation of the “Operating Line” required for the determination of the number of transfer
units (NTU) as described later.
2
The value of Hyb can be determined from the measured wet- and dry-bulb temperatures of
the entering air. One can use a psychrometric chart from Perry’s (Chapter 12-5) for this. The one in
the textbook (Figure 23.2) is not convenient because it doesn’t contain enthalpy data. The
spreadsheet referred to in Part (c) also calculates the enthalpy directly from the wet- and dry-bulb
temperatures.
The value of Hya is be calculated using Equation (2) since GxcL and Gy are known. These
estimates will also be verified based on the measured dry- and wet-bulb temperatures of the exit airstream.
As mentioned previously, heat is transferred into the gas stream by two mechanisms. The
first is the sensible heat transfer and is given by Equation (23.24). The second is the latent heat
transfer into the gas stream. This involves the mass transfer rate [Equation (23.25)]. The equation
23.25 is multiplied by the latent heat of vaporization,o, at the datum plane temperature, and the
Lewis Relation [Equation (23.21)] is employed together with it to obtain [this is Eq. above (23.36)]:
GydHy = kyMBa(Hi – Hy)dZ
(3)
Here MB is the molecular weight of the liquid; a is the interfacial area between the two phases which
is the same for both heat and mass transfer; ky is the mass transfer coefficient in lb-mol/ft2/hr. or
equivalent SI units; and Z is the height of the element being considered measured from the base of
the Cooling tower. Hi and Hy are the gas-phase enthalpies measured at either the interface (i) or the
bulk (y) of the gas phase at the particular height Z in the Cooling tower. In integrated form, this
equation is:
H ya

H yb
dH y
Hi  H y

zT

0
kyM Ba
G y
dz
(4)
It can be seen that the integral on the left has the general form of the integral defining the
number of transfer units (NTU) in terms of the gas film [Equation (22.23), Page 705] i.e.:
Ny  
dH y
Hi  H y
(5)
There are two issues to consider with respect to this equation:
A) In this equation the enthalpy appears to be the driving force for the mass transfer rather than by
humidity differences. Here the numerator depicts the change in vapor concentration, while the
denominator is the average driving force. While it appears from the equation that it the enthalpy that
drives the mass-transfer this is in fact not the case: It is the simplifying assumptions made during the
Cooling tower derivation that give this appearance. As pointed out in the Foust etr al. [Principles of
Unit Operations, John Wiley & Sons, Inc. (1960), pp. 302-309], enthalpy is an extensive
thermodynamic property and as such, cannot be a driving force for any transfer operation. If it makes
you feel any better, you can think of this as a “pseudo driving force.” See the reference for some
discussion of this point if you are interested.
B) The equation (5) is difficult to use since it requires some means for determining the temperature at
the air-water interface in order to calculate Hi. However, there is a reasonable way out of this
complication if we assume that the resistance to mass transfer is predominantly in the gas-phase.
This is a reasonable assumption for this system since air (in comparison to water) is a much poorer
medium for heat transfer. Refer back to Figures 23.8(a) then, it appears that the temperature Tx in the
liquid bulk is not very different from Ti. Then the enthalpy at the interface Hi, can be replaced by Hx,
the enthalpy of air saturated at the bulk temperature of the liquid, T x. Note that the symbol, Hx ,
was earlier defined as the enthalpy of the liquid, but here the prime sign emphasizes the fact that this
is the saturation enthalpy of the vapor phase at temperature Tx.
3
With this assumption, Equation (5 ) takes the form:
H ya
N y  N Oy 
dH y

H x  H y
H yb
 NTU
(6)
This Eq (6) will be the defining equation for the number of transfer units, NTU in the project
report.
Calculation of NTU’s
Three methods denoted below as (a), (b) and (c) are used to estimate the NTU.
(a) Graphical Integration
In graphical integration, one plots 1/(Hx-Hy) versus Hy between the limits of Hyb and Hya.
To begin with, a plot of Hy and Hx might take the form (Figure 1):
H’xa
Top of Column
H
Water Line
Bottom
of
Column
Hya
H’xb
Air Line
Hyb
Hyb
Hya
Hy
Therefore, the driving force at any point, H’x – Hy, for this example, would increase with Hy (Figure 2):
Driving Force Sketch
H’x - Hy
Hyb
Hy
Hya
4
The required plot for NTU would take the form shown on the graph below (Figure 3):
1/(H’x-Hy)
Area = NTU
Hyb
Hya
Hy
(b) Special case when the H’x line is essentially straight.
In Figure 1, the curvature of the upper line (“water line”) comes about mainly from the
nonlinearity of the curve of water vapor pressure versus temperature. Sometimes, this curvature is
not very great. The operating line – that is, the“air line” – will be straight if the liquid water and dry air
rates are constant. Even though there is some evaporation (perhaps three or four per cent) this can
be ignored. Under these conditions for a cooling tower, Equation (22.19) on Page 705 applies (i.e.
simple logarithmic mean equations can be used to calculate NTU just as your learnt in the heat
exchanger problems last semester). Hence, no graphical integration is required.
(c) Analytical Integration for NTU
The procedure for calculating the NTU’s by graphical integration actually begins with a plot of
H’x and Hy versus Tx rather than versus Hy. The cooling water inlet and exit temperatures, namely Txa
and Txb, are the primary physical parameters of interest and are measured. This plot takes the form
given below (Figure 4):
H’
H
x
Hy
Txb
Bottom
Water Temperature, Tx
Txa
Top
5
This will be the first plot that you will prepare for the graphical integration procedure. It is also the
beginning plot for the development of the analytical method. Because the curvature of the H’x line is
not very great, it can be very well represented by a second order equation:
Hx  Tx  Tx2  
where
(all parameters in fps units) (7)
 = -2.52683
(8)
 = 0.022464
(9)
 = 100.081
Also
(10)
H y  H yb 
Gx c L
(Tx  Txb )
Gy
(assuming GX is constant with length)
(11)
This is the operating line equation in terms of temperature. For simplicity, define the slope of this line
as s:
s
Gx cL
Gy
(12)
Thus Equation (11) can be expressed as:
H y  sTx  f
(13)
where f = Hyb – sTxb.
(14)
Hence, the driving force is:
Hx  Hy  (  f )  (  s)Tx  Tx2
(15)
It is convenient to set:
a   f
(16)
b  s
(17)
c
(18)
Recall that:
H ya
NTU 

H yb
dH y
H x  H y
(6)
Note that H’x - Hy in Equation ( 6 ) is now in terms of Tx, not Hy. The dummy variable must now be
expressed in terms of temperature. Since Hy = sTx + d, dHy = sdTx. Hence, Equation (6) becomes:
Txa
NTU 
s dTx
2
x  cTx
 a  bT
Txb
(19)
6
The integral has the following form:
dx
 a  bx  cx
2
 2cx  b 
tan 1 


 

2

where  = 4ac-b2
(19a)
(19b)
There are several alternate forms of the integral in integral tables, but this is the form we will use.
Thus, the NTU’s are given by the equation:
NTU 
 2cTxb  b 
2s  1  2cTxa  b 
  tan 1 

tan 
 
 
 


(20)
This whole procedure has been set up in the analysis Excel spreadsheet.
Height of a Transfer Unit (HTU)
Since the height of the cooling tower is Z, the height of a transfer unit, HTU, is simply Z
divided by the number of transfer units, NTU:
HTU  ZT / NTU
(21)
In designing a cooling tower, the rate of incoming hot water is known as well as the condition of the
air that will be used to cool it – that is, the wet- and dry-bulb air temperatures. Some optimization of
the air rate to be used is required. If the air rate is high, then the power requirements will be high, and
a relatively short tower can be used. On the other hand, if the air rate is low, then the power
requirement is low, but the tower will be relatively high. The air rate affects the slope of the operating
line and therefore the separation of the “water” and “air” lines in Figure 1. At high air rates, the
separation of these lines is large, and the number of transfer units decreases, and vice versa. The
idea that you should get from this is that the NTU’s required are a measure of the difficulty of the
cooling job to be done. The larger the NTU’s, the more difficult the job is.
Once the NTU requirement is known, one can calculate the height of tower required by
multiplying the NTU’s by the HTU. But what determines the height of a transfer unit? Answer: HTU
is determined by the fundamental features of the system: the interfacial area for heat/mass transfer,
a, and the mass-transfer coefficient, ky. Since these two factors cannot be measured individually, at
least in our equipment, we prefer to look at their product kya. It is important to note that not though
Equation (21) relates NTU and HTU to the height of the Cooling tower (Z); the parameters (i.e. NTU
and HTU) are in fact independent of each other: while NTU is directly dependent on the driving force
for mass-transfer, HTU depends on the physical features of the mass transfer via a and ky.
Refer back to Equation (4). We assume that the product, kyMBa/G’y, is constant. Combining
the forms and ideas of Equations (3), (4), and (6), one finds:
NTU 
k y aM B
G y
ZT
 dz 
0
k y aM B
G y
ZT
(22)
In view of Equation (21), then HTU is given by:
HTU 
Gy
k y aM B
(23)
7
Here, MBkya (or also called Kya in data sheet and elsewhere) is of more fundamental interest than
HTU and will be calculated from the results of the experiment.
Evaporation Rate
There is one more item of interest, namely, the evaporation rate of water. The theory
development here assumes that the evaporation is negligible (at least in the liquid phase). This
assumption leads to a straight operating line and relatively simple expressions for the NTU. Let us
examine of this is indeed the case. The rate of evaporation, Gxe, is given by the equation:
Gxe  Gy (# a  # b )
(24)
where the symbols, #a and #b, represent the humidities at the top and bottom of the column,
respectively. (Same as cursive H in MS&H) The measured inlet wet- and dry-bulb temperatures are
used to get #b. This equation is used to get Hya. This value, in combination with Equation (23.8)
leads to a value of #a. The evaporation rate can then be calculated using Equation (24).
Another way to get #a is to use the exit dry and wet bulb temperature. Unlike previous years,
where there was a tremendous spray from the exit port of this column, we have made “engineering”
efforts to reduce the spray. The dry and wet-bulb temperatures measured should be reasonable and
they should be used in conjunction with Eq. (24) above, to get a second estimate of the evaporation
rate.
Lab Instructions:
1) Zero the air manometer. At the start of the experiments, under no-flow conditions, the
manometer should read zero, else adjust it using the black knob at the right-top corner of the
manometer.
2) Fill up the inlet water tank (made of copper: does not have a heating coil) with hot water.
To do this, drain the inlet water for a few minutes/seconds until you are sure that you are getting
"hot" water.
Then fill the copper tank with the water. You will need to do this every 1-2 runs since the exit
water that comes out of the Cooling tower does not recirculate in the system, it drains out of the
system.
3) Ensure that the wicks of the wet-bulb temp thermocouples are completely wet at all times.
Climb up to the top of the tower if necessary to confirm this before each run.
4) The power panel for the experiment has three switches: Tower (to power up the system)
followed by Air (for the air blower) and Pump (for water flow). The air flow-rate may be adjusted
by changing the damper setting from 1 through 5. Pick any four damper settings for your
experiment. Simultaneously, the water flow rate can be adjusted using the valve below the
manometer. Use 4, 8 and 12 liters/min for you experiment. Therefore, you will perform
experiments under 12 conditions (4 air flow rates X 3 water flow rates).
5) Ensure that the system has reached steady state before starting acquisition of temperature 16 data using the Labview program as we did in the previous experiment. Note that the data in the
first column of your output file is Time (irrelevant in this case since we are only interested in
steady-state results). The second to seventh column have temperature values (in degree C) at
steady state. The eighth column is packed with zero. Jot down what column corresponds to what
temperature.
Caution: Ensure that the experiment is clean and the place is dry. This way you will get good data.
Good Luck!