CE427 Fall Semester, 1998

CE427 Fall Semester, 1998 Unit Operations Lab 3 September 8, 1998

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 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. Distillations often involve several, if not many, components, whereas in a cooling tower, it’s just a mixture of water and air. You’d think that would make things simple, but there are sufficient complexities to it that Chapter 23 of McCabe, Smith, and Harriott is devoted to “Humidification Operations.” Cooling towers fall into that class of operations.

Theory from the Textbook

Read Chapter 23 through. 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 so many equations and substitutions in the text treatment that it’s a bit difficult to see “the big picture.” One equation is simply an overall energy balance which becomes analogous to the “operating line” in distillation or absorption. It begins with Equation (23.22). Equation numbers in decimal form are those from MS&H; all other equation numbers pertain to this write-up.

GydHy  d(Gx Hx ) (23.22)

Assuming negligible evaporation, Gxb= Gxa = Gx. The datum plane for enthalpy is arbitrary. For water, it is usually taken as liquid water at a temperature, To . (The table in Perry’s uses 32F). Since only enthalpy differences are involved, this datum plane temperature “washes” out of the analysis.

For liquid water then, Hx = cL(Tx – To) [Equation (23.26)]. Thus, the differential form of Equation (23.22) becomes:

GydHy  GxcLdTx (1)

Integrating from the bottom of the column to any particular point the column, one obtains:

Gy (Hy  Hyb )  GxcL (Tx Txb ) (2)

This is the equation of the Operating Line required for the determination of the number of transfer units.

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 for this. The one in the textbook (Figure 23.2) is 2 not convenient for this because it doesn’t contain enthalpy data. There are several charts in Perry’s that can be used. The spreadsheet referred to in Part (c) calculates the enthalpy directly from the wet- and dry-bulb temperatures.

The value of Hya will be calculated using Equation (2) since GxcL and Gy are known. Humidity measurements at the top using the wet- and dry-bulb temperatures turn out to be erroneous, possibly because of the spray of hot water caused by the air flowing upwards.

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 latent heat of vaporization,o, at the datum plane temperature, and the Lewis Relation, Equation (23.21). The result is:

GydHy = kyMBa(Hi – Hy)dZ (3)

where it has been assumed that the areas for heat transfer (aH) and mass transfer (aM) are the same. Therefore this area is simply called a. In integrated form, this equation is:

H ya zT dH y k y M B a  dz (4)  H  H  G H yb i y 0 y

Now refer to Pages 704 and 705 of the text. It can be seen that the integral on the left has the general form of the integral defining the number of transfer units in terms of the gas film [Equation (22.23), Page 705]:

dH N  y y  (5) H i  H y This equation would be difficult to use for it requires some means for determining the temperature at the air-water interface. [The book, Principles of Unit Operations, by Foust et al., John Wiley & Sons, Inc. (1960), pp. 302-309, gives an example of how this can be done.] However, there is a reasonable way out of this complication through an approximation.

(Something to think about: Read the sentence just below Equation (22.16). Is it obvious that this integral represents the change in vapor concentration divided by the average driving force?)

Refer back to Figures [23.8(a) and 23.8(b)]. There is resistance to heat transfer in both the liquid and gas films, and mass transfer resistance to diffusion in the gas film. As in many such liquid- gas operations, the controlling resistance is in the gas film. The large temperature gradient shown for

Tx in each of the figures is probably exaggerated (for clarity) in comparison with that in the gas film. This means that Tx is not greatly different from Ti and therefore, Hi, the enthalpy at the interface can be replaced by Hx, the enthalpy of air saturated at the bulk temperature of the liquid, Tx. Note, the preceding point was emphasized to the maximum extent possible with word-processing because more people have been confused on this point than on any other. In other words, if you are confused on this point and make an error in your calculations, your results will be in error, big time. 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 Tx. With this assumption, Equation (5 ) takes the form:

H ya dH y N  N   NTU (6) y Oy  H   H H yb x Y Note that from now on, this will be the defining equation for the number of transfer units, NTU. 3

It is particularly interesting to note here that this and the previous few equations indicate or imply that the whole process is driven by enthalpy differences rather than by humidity differences. As pointed out in the Foust reference cited above, 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, think of it as a “pseudo driving force.” See the reference for some discussion of this point if you are interested.

Calculation of NTU’s

(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:

H’ xa

H Top of Column

Bottom Water Line of H ya Figure 1. Column H’ xb

Therefore, the driving force at any point, H’x – Hy, for this example, would increaseAir Line with Hy: H yb

H H H yb Driving Forcey Sketch ya

H’ - H x y

Figure 2. H H H yb y ya 4

Due to some limitations in using “Drawing” in Word, it was not possible to exactly draw this the way it should have been. Figure 1 would lead to curve exhibiting a minimum about 40% of the way between Hyb and Hya along the abscissa.

The required plot for NTU would take the form shown on the graph below:

1/(H’ -H ) x y

Area = NTU

Figure 3. H H (b) Special case when the H’x lineyb is essentiallyH straight. ya y 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. With both lines straight, the situation is analogous to one that often occurs in heat exchangers, the case in which the logarthmic mean driving force applies. This is shown in Figure 11.5 on Page 316 for the case of a heat exchanger. Under these conditions for a cooling tower, Equation (22.19) on Page 705 applies. 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 of Figure 4 on the following page. 5

H’ x H

T T xb xa

Water Temperature, T x

Figure 4.

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:

2 Hx  Tx  Tx   (7) where  = -2.52683 (8)

 = 0.022464 (9)

 = 100.081 (10)

GxcL Also Hy  Hyb  (Tx Txb ) (11) Gy

This is the operating line equation in terms of temperature. For simplicity, define the slope of this line as s:

GxcL s  (12) Gy Thus Equation (11) can be expressed as:

Hy  sTx  f (13)

where f = Hyb – sTxb. (14)

Hence, the driving force is: 6

2 Hx  Hy  (  f )  (  s)Tx  Tx (15)

It is convenient to set:

a    f (16)

b    s (17)

c   (18) Recall that:

H ya dH y NTU  (6)  H   H H yb x y

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 dTx NTU  (19)  a  bT  cT 2 Txb x x

The integral has the following form:

dx 2 1  2cx  b   tan   (19a)  2   a  bx  cx     where  = 4ac-b2 (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:

2s  1  2cTxa  b  1  2cTxb  b  NTU  tan    tan   (20)        

This whole procedure has been set up on a spreadsheet that will be furnished in Excel format.

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 7 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? Past experience strongly indicates that most people have “cause and effect” very much confused on this issue. They think that everything is all wrapped up in Equation (21); in other words, they think NTU depends on ZT and HTU, and HTU depends on NTU and ZT. The fact of the matter is that the HTU is determined by the fundamental factors of a, and ky. So when you think about what determines the HTU, you should be thinking about what would affect a and ky. Since these two factors can’t be measured individually, at least in our equipment, we resort to looking at their product.

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:

k aM ZT k aM NTU  y B dz  y B Z  T (22) Gy 0 Gy

[Equations (3) and (4) are in terms of Hi before this was changed to H’x.]

In view of Equation (21), then HTU is given by:

Gy HTU  (23) kyaMB

If you had some time on your hands with nothing to do, you would check the units out. HTU definitely has the units of ft, but if you check the units on the right, using the units of the Symbol table on Pages 763 and 764, you’d find that there’s a discrepancy, although not a serious one. Some detective work leads to the conclusion that in the transition of going from Equation (23.25) to the equation just above

Equation (23.36), something got messed up with respect to the driving force units of ky; it is mol fraction in the first equation, and enthalpy in the second. MS&H should have changed the transfer coefficient to something else. For simplicity, the change will be made here, and at the same time, the molecular weight of water, MB, can be included in the new coefficient. The result is:

G K a  y (24) y HTU

3 where Kya has the units of lb dry air/hr,ft .

Kya 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. After all, the theory development here assumed that the evaporation is negligible. This assumption leads to a straight operating line and relatively simple expressions for the NTU. The rate of evaporation, Gxe, is given by the equation:

Gxe  Gy(# a #b ) (25) 8

where the symbols, #a and #b, represent the humidities at the top and bottom of the column, respectively. (The text uses script H’s, but that symbol is not available in the word-processing package used here.) The measured inlet wet- and dry-bulb temperatures are used to get #b. Because the measured wet-bulb temperature at the top of the column is erroneous, the humidity there must be back-calculated from the operating line equation. 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 (25).