Processing & Handling :: Thermal & Energy Management :: Heat Exchangers, Condensers & Coolers
February 1, 2012
Effective Thermal Design Of Cooling Towers
A stepbystep approach to coolingtower design, with an example calculation to make it clear
Jonny Goyal Air Liquide Engineering and Construction, Lurgi India
Various misconceptions arise when it comes to the thermal design of cooling towers. Sometimes related parameters, such as range, approach, effectiveness, liquidtogas ratio ( L/G), wetbulb temperature, cooling water temperature, relative humidity, number of transfer units (NTU) and other terms create a confusion for the designer in effectively sizing, selecting and evaluating a particular cooling tower. This leads to inadequate design.
The objective of this article is to present a stepwise understanding of how to calculate the NTU for a cooling tower, and thus to understand the basis of thermal design of counterflow cooling towers for optimizing cost and performance.
Definitions
First, let’s look at some of the basic terms and briefly describe their significance and role in cooling tower design and performance.
Drybulb temperature. Drybulb temperature (t_{db}) — usually referred to as the air temperature — is the property of air that is most commonly used. When people refer to the temperature of the air, they are normally referring to its drybulb temperature. The drybulb temperature is an indicator of heat content and is shown along the bottom axis of a psychometric chart. The vertical lines extending upward from this axis are constanttemperature lines.
Wetbulb temperature. Wetbulb temperature (t_{wb}) is the reading when the bulb of a thermometer is covered with a wet cloth, and the instrument is whirled around in a sling. The wetbulb temperature is the lowest temperature that can be reached by evaporation of water only.
Relative humidity (RH). RH is the ratio of the partial pressure of water vapor in air over the saturation vapor pressure at a given temperature. When the relative humidity is 100%, the air is saturated and therefore, water will not evaporate further. Therefore, when the RH is 100% the wetbulb temperature is the same as the drybulb temperature, because the water cannot evaporate any more.
Range. The range is the difference in temperature of inlet hot water (t_{2}) and outlet cold water (t_{1}), t_{2} – t_{1}. A high coolingtower range means that the cooling tower has been able to reduce the water temperature effectively.
Approach. The approach is the difference in temperature of outlet cold water and ambient wetbulb temperature, t_{1} – t_{w}. The lower the approach, the better the cooling tower performance. Although both range and approach should be monitored, the approach is a better indicator of cooling tower performance.
Cooling tower capability. The capability of the cooling tower is a measure of how close the tower can bring the hot water temperature to the wetbulb temperature of the entering air. A larger cooling tower (that is, more air or more fill) will produce a closer approach (colder outlet water) for a given heat load, flowrate and entering air condition. The lower the wetbulb temperature, which indicates either cool air, low humidity or a combination of the two, the lower the cooling tower can cool the water. Capability tests are conducted per the ATC105 Code of the Cooling Tower Institute (CTI; Houston; www.cti.org).
The thermal performance of the cooling tower is thus affected by the entering wetbulb temperature. The entering air drybulb temperature has an insignificant effect on thermal performance.
Effectiveness. A cooling towers effectiveness is quantified by the ratio of the actual range to the ideal range, that is, the difference between cooling water inlet temperature and ambient wetbulb temperature. It is defined in terms of percentage.

(1) 
Liquidtogas ratio (L/G). The L/G ratio of a cooling tower is the ratio of the liquid (water) mass flowrate (L) to gas (air) mass flowrate (G). Cooling towers have certain design values, but seasonal variations require adjustment and tuning of water and air flowrates to get the best cooling tower effectiveness.
Number of transfer units (NTU). Also called the tower coefficient, the NTU is a numerical value that results from theoretical calculations based on a set of performance characteristics. The value of NTU is also representative of the degree of difficulty for the cooling process. The NTU corresponding to a set of hypothetical conditions is called the required coefficient and is an evaluation of the problem. The same calculations applied to a set of test conditions is called the available coefficient of the tower involved. The available coefficient is not a constant but varies with operating conditions. The operating characteristic of a cooling tower is developed from an empirical correlation that shows how the available coefficient varies with operating conditions.
Cooling capacity. The cooling capacity of a tower is the heat rejected [kcal/h or TR (refrigeration tons; 1 TR = 12,000 Btu/h = 3,025.9 kcal/h)], and is determined by the product of mass flowrate of water, times the specific heat times the temperature difference.
Theory — the Merkel equation
In a cooling tower operating in counter current flow, there are two basic principles involved for removing heat by the cooling water:
1.sensible heat transfer due to a difference in temperature levels
2. latent heat equivalent of the mass transfer resulting from the evaporation of a portion of the circulating water

Figure 1. The Merkel equation is derived by considering
a falling water droplet surrounded by saturated air

Merkel developed the basic equation based on the above principles. The Merkel model is universally accepted for designing and rating of counterflow cooling towers. The model is based on a drop of water falling through an upstream flow of unsaturated air at a wetbulb temperature of
t_{wb} with enthalpy
h_{A} (
Figure 1), in a counterflow cooling tower. The drop of water is assumed to be surrounded by a film of saturated air at the water temperature
WT with saturation enthalpy
h_{W}. As the drop travels downward, heat and mass transfer takes place from the interface air film to the upstream air, thereby cooling the water from hot temperature to a cold temperature.
The main assumptions of Merkel theory are the following:
1. The saturated air film is at the temperature of the bulk water.
2. The saturated air film offers no resistance to heat transfer.
3. The vapor content of the air is proportional to the partial pressure of the water vapor.
4. The heat transferred from the air to the film by convection is proportional to the heat transferred from the film to the ambient air by evaporation.
5. The specific heat of the airwater vapor mixture and the heat of vaporization are constant.
6. The loss of water by evaporation is neglected.
7. The force driving heat transfer is the differential enthalpy between the saturated and bulk air.

Figure 2. This plot, known as the driving force diagram, shows the enthalpy versus
temperature for water and air


Figure 3. Solving the Merkel equation (Equation 2), is usually done graphically,
where the integral is equal to the area under the curve

This cooling process can best be explained on a psychometric chart, which plots enthalpy versus temperature. The process is illustrated in the socalled drivingforce diagram shown in
Figure 2. The air film is represented by the water operating line on the saturation curve. The main air is represented by the air operating line, the slope of which is the ratio of liquid (water) to air (
L/G). The cooling characteristic, a degree of difficulty to cooling is represented by the Merkel equation:

(2) 
Where:
K = overall enthalpy transfer coefficient, lb/hft^{2}
a = Surface area per unit tower volume, ft^{2}/ft^{3}
V = Effective tower volume, ft^{3}
L = Water mass flowrate, lb/h
Equation 2 basically says that at any point in the tower, heat and water vapor are transferred into the air due (approximately) to the difference in the enthalpy of the air at the surface of the water and the main stream of the air. Thus, the driving force at any point is the vertical distance between the two operating lines. And therefore, the performance demanded from the cooling tower is the inverse of this difference. The solution of the Merkel equation can be represented by the performance demand diagram shown in Figure 3. The KaV/L value is equal to the area under the curve, and represents the sum of NTUs defined for a cooling tower range.
An increase in the entering t_{wb} moves the air operating line towards the right and upward to establish equilibrium. Both the cold water temperature (CWT) and hot water temperature (HWT) increases, while the approach decreases. The curvature of the saturation line is such that the approach decreases at a progressively slower rate as the t_{wb} increases. An increase in the heat load increases the cooling ranges and increases the length of the air operating line. To maintain equilibrium, the line shifts to the right increasing the HWT, CWT, and approach. The increase causes the hot water temperature to increase considerably faster than does the cold water temperature. In both these cases, the KaV/L should remain constant. However, a change in L/G will change the KaV/L value.
Cooling tower design
On the basis of the above discussion, it is clear that there are five parameters that, in combination, dictate and define the performance of a cooling tower, namely:
1. Hot water temperature, HWT
2. Cold water temperature, CWT
3. Wet bulb temperature, t _{wb}
4. Water mass flowrate, L
5. Air mass flowrate, G
The first four parameters are determined by the user of the cooling tower. It is the fifth quantity, G, that is selected by the designer of the cooling tower. Once these five quantities are available, the tower characteristic ( KaV/L), can be calculated through the Merkel equation.
The first step in designing a cooling tower is the generation of a demand curve. In this curve, the KaV/L values are plotted against varying L/G ratios. The next step is to superimpose fillcharacteristic curves and demand curves. The Cooling Technology Institute has tested a variety of fill configurations and generated fill characteristic curves for each type; CTI’s Technical Paper TP88_05 can be referred to in this regard.
Cooling tower design is basically an iterative process. The factors that effect the selection of design L/G and consequently the fill height are: cell dimensions, water loading, air velocities across various cooling tower sections and pressure drops, and fan selection.
The classical method of thermal rating of cooling towers is to estimate the ratio of liquid to gas first and then find the proper tower volume by the means of trial and error using the tower performance curve. The L/G is the most important factor in designing the cooling tower and related to the construction and operating cost of cooling tower.
Finally we can summarize the importance of the L/G ratio with the following points.
A high L/G ratio means:
• More water to less air
• Air is more saturated — driving force is reduced
• More residence time of water needed
• Less cooling in given time
• Increase in required fan power
• Decrease in height of tower
• Low evaporation loss (under same water flowrate)
An example makes it clear
As an example, let us design a cooling tower with the following data:
Capacity ( F): 3,000 m^{3}/h
Wet bulb temperature (t_{wb}): 29°C Relative humidity () 92%
Cooling water inlet (t_{2}): 43°C
Cooling water outlet (t_{1}): 33°C Altitude ( Z): 10 m
Step I. This step involves heat load calculations as follows:
1. Range = (t_{2 }– t_{1}) = 43 – 33 = 10°C
2. Approach = (t_{1} – t_{wb}) = 33 – 29 = 4°C
3. Heat load, Q = mC_{p}(t_{2} – t_{1})
= 998.13 x F x Range
= 998.13 x 3,000 x 10
= 29,943,900 kcal/h
Step II.This step involves total psychometric calculations as follows:
1. Barometric pressure (p) at the given altitude (Z) is calculated by using the following equation:

(3) 
For an altitute of 10 m, this becomes
p = 101.2 kPa
2. Assume a dry bulb temperature of say, t_{db} = 32°C
3. Calculate water vapor saturation pressure (p_{ws}) at the assumed t_{db} for the temperature range of 0 to 200°C using the equation:

(4) 
Where:
C _{1} = –5.8002206 x 10^{3}
C _{2} = 1.3914993 x 10^{0}
C _{3} = – 4.8640239 x 10^{–2}
C _{4} = 4.1764768 x 10^{–5}
C _{5} =–1.4452093 x 10^{–8}
C _{6} =6.5459673 x 10^{0}
and T represents the dry bulb temperature in Kelvin. This results in the value:
p_{ws} = 4.7585 kPa
4. The partial pressure of water (p_{w}) at given relative humidity is found using the following equation:

(5) 
p_{w} = 4.3779 kPa
5. The partial pressure (p_{ws}) is again calculated using Equation 4. This time T represents the wet bulb temperature in Kelvin, which calculates to:
p_{ws} = 4.0083 kPa
6. Using p_{ws} calculated in Step 5 we recalculate t_{wb} using the Carrier equation:

(6) 
which gives the result:
t _{wb} = 37.7°C
7. This step is an iterative process, whereby the assumed value of t_{db} in Step 2 is varied in such a way that the calculated t_{wb} in Step 6 equals the actual (real) t_{wb}.
8. After a number of iterations, the calculated t _{db} value converges to 30.12°C.
Step III.This step involves the calculation of the inlet air enthalpy (h_{1}) as follows:
1. The humidity ratio (W) for dry air is calculated using the following equation:

(7) 
W = 0.02515 kg water/kg dry air
2. The specific volume (v) for dry air is calculated using the following equation:

(8) 
v = 0.89511 m^{3}/kg, dry air
3. Calculate the enthalpy of inlet air (h_{1}) using the following equation:

(9) 
h _{1} = 94.750 kJ/kg
4. Calculate the humidity ratio at saturation (W_{s}) for wet air using same Equation 7. Here we now use p_{ws}:
W_{s} = 0.02743 kg water/kg moist air
5. Calculate the specific volume (v) for wet air using Equation 8 with W_{s}.
v = 0.89827 m^{3}/kg moist air
Step IV.This step involves the calculation of the exit air properties, as follows:
1. Assume some value of the L/G ratio, say 1.575, and calculate h_{2} for exit air using the following equation:

(10) 
h _{2} = 160.50 kJ/kg
2. Assume that the exit air has a relative humidity of 97–99% (design RH at the outlet), and also assume some value of exit air drybulb temperature.
3. Use the same partial pressure and humidity equations as discussed in Step II and Step III, to calculate the enthalpy of exit air at these assumed values in Point 2 above. At an assumed RH of 98.5% and exit air t _{db} of 37°C, we recalculate
h_{2} = 141.18 kJ/kg
4. This is again an iterative process. Next, assume the value of exit air t_{db} in Point 2 (at same relative humidity) in such a way that the calculated h_{2} in Point 3 equals the h_{2} calculated in Point 1.
5. After a number of iterations, the calculated exit air t_{db} value converge to 39.55°C.
6. Now that the drybulb temperature and RH are known values, recalculate the wetbulb temperature using Equation 6.
t_{wb} = 39.31°C
7. Calculate the dry and wet specific volume of exit air using Equation 8. Also calculate the density of dry air and wet air (for inlet and exit).
v = 0.9540 m^{3}/kg, dry air
v = 0.9551 m^{3}/kg, moist air
Average density dry = 1.0827 kg/m^{3}
Average density wet = 1.0801 kg/m^{3}
Step V. This step is to help draw the driving force diagram as follows:
1. Take different temperature ranges (covering cooling water inlet and outlet temperature) and calculate the enthalpy of air using Equation 9 and psychometric calculations discussed above. Plot the air saturation curve (enthalpy versus temperature) as shown in Figure 2.
2. Take cooling water outlet temperature and calculate the enthalpy of fin (h’) using Equation 9. Here the partial pressure, saturation pressure and humidity ratio are calculated for the corresponding temperature taken (Table 1).
h’ at 33°C = 116.569 kJ/kg
This enthalpy specifies Point B on the graph of Figure 2, and is the starting point of the water operating line.
3. Similarly calculate enthalpy of air h_{a}, at wetbulb temperature (Table 1).
h_{a} = 94.667 kJ/kg
This enthalpy specifies Point A on the graph of Figure 2 and is the starting point of the air operating line.
4. Take incremental change in temperature (say 0.5 or 1.0) up to the hot water temperature and calculate the h' and h_{a}. The ending points are shown as C and D in Figure 2 on the water operating and air operating line respectively.
5. The difference between h’ and h_{a} will give you the enthalpy driving force for incremental change in temperature.
6. Take the inverse of enthalpy difference in each incremental step (Table 1).
7. Calculate NTU = 4.18 x ∆t x (Average of incremental increase in inverse of enthalpy difference).
Or for 0.5°C increment in temperature, calculated NTU = 0.096
8. Similarly, calculate the NTU for each step and add to get the total NTU for the particular assumed L/G ratio (Table 1). Or, for an assumed L/G of 1.575, Total NTU = 1.7533 — this is KaV/L.
9. Now to plot the tower characteristic curve, first we vary the L/G ratio and repeat all calculations discussed above to generate the data for various NTU to plot. The curve represents “Design NTU” on the graph, shown in Figure 4.
10. Take the design L/G ratio and plot the tower characteristic curve by assuming the slope of the line (m), which usually varies between –0.5 to –0.8. One can also consult with vendors for this value as it also depends on the type of fins used.
11. Calculate the value of the constant C, related to cooling tower design using equation:
NTU = C x ( L/G) ^{m}
C = 2.522

Figure 4. The intersection of the tower characteristic curve and the design NTU curve gives the design L/G ratio

Table 1. 
Water Temperature, t 
Water Vapor Saturation Pressure, pws 
Partial Press. of H_{2}O Vapor, pw 
Humidity Ratio, W 
Enthalpy of Film, h' 
Enthalpy of Air, ha 
Enthalpy Difference, (h'ha) 
1/(h'ha) 
∆t 
NTU 
∑NTU 
Cumulative Cooling Range 
°C 
K 
kPa 
kPa 
kg water/kg dry air 
kJ/kg 
kJ/kg 
kJ/kg 
kg/kJ 




33 
306.15 
5.0343 
5.0343 
0.0326 
116.5686 
94.6668 
21.9018 
0.0457 




33.5 
306.65 
5.1774 
5.1774 
0.0335 
119.5982 
97.9585 
21.6397 
0.0462 
0.5 
0.096 
0.096 
0.5 
34 
307.15 
5.3239 
5.3239 
0.0345 
122.6986 
101.2503 
21.4483 
0.0466 
0.5 
0.097 
0.193 
1 
34.5 
307.65 
5.4740 
5.4740 
0.0356 
125.8718 
104.5420 
21.3298 
0.0469 
0.5 
0.098 
0.291 
1.5 
35 
308.15 
5.6278 
5.6278 
0.0366 
129.1197 
107.8338 
21.2859 
0.0470 
0.5 
0.098 
0.389 
2 
35.5 
308.65 
5.7853 
5.7853 
0.0377 
132.4444 
111.1255 
21.3189 
0.0469 
0.5 
0.098 
0.487 
2.5 
36 
309.15 
5.9466 
5.9466 
0.0388 
135.8480 
114.4173 
21.4307 
0.0467 
0.5 
0.098 
0.585 
3 
37 
310.15 
6.2810 
6.2810 
0.0412 
142.9006 
121.0008 
21.8999 
0.0457 
1.0 
0.193 
0.778 
4 
38 
311.15 
6.6315 
6.6315 
0.0436 
150.2958 
127.5843 
22.7116 
0.0440 
1.0 
0.187 
0.965 
5 
39 
312.15 
6.9987 
6.9987 
0.0462 
158.0530 
134.1678 
23.8852 
0.0419 
1.0 
0.180 
1.145 
6 
40 
313.15 
7.3835 
7.3835 
0.0489 
166.1928 
140.7513 
25.4415 
0.0393 
1.0 
0.170 
1.314 
7 
41 
314.15 
7.7863 
7.7863 
0.0518 
174.7371 
147.3348 
27.4023 
0.0365 
1.0 
0.158 
1.473 
8 
42 
315.15 
8.2080 
8.2080 
0.0549 
183.7094 
153.9183 
29.7911 
0.0336 
1.0 
0.146 
1.619 
9 
43 
316.15 
8.6492 
8.6492 
0.0581 
193.1348 
160.5018 
32.6330 
0.0306 
1.0 
0.134 
1.753 
10 

∑NTU 
1.75334674 

Table 2. TYPICAL THERMAL CALCULATIONS OF COUNTERFLOW COOLING TOWER 
Flowrate 
3,000 
m^{3}/h 

Hot water temperature 
43 
°C 
Wetbulb temperature 
29 
°C 
Cold water temperature 
33 
°C 
Approach 
4 
°C 
Range 
10 
°C 
Assumed drybulb temperature 
30.12 
°C 
Heat load 
29943888.32 
kcal/h 
Assumed L/G 
1.575 

Barometric pressure (p) 
101.2 
kPa 
No. of cells 
3 




Cell length 
14 
m 
Water vapor saturation pressure (p_{ws}), at t_{db} 
4.2754 
kPa 
Cell width 
14 
m 
Partial pressure of water vapor (p_{w}) 
3.9333 
kPa 
Air inlet height 
5.5 
m 
Water vapor saturation pressure (p_{ws}), at t_{wb} 
4.0083 
kPa 
Design RH 
92% 

Recalculating t_{wb} 
29.0 
°C 
Density of water 
1,000 
kg/m^{3} 
Difference 
0.0 
°C 
Altitude 
10 
m 
Dew point temperature (t_{d}) of moist air (for temperature between 0 and 93°C) 
28.66 
°C 


Inlet Air Properties 
Exit Air Properties 
Inlet t_{wb} 
29 
°C 
Enthalpy 
160.50 
kJ/kg 
RH 
92% 

Exit air temperature (t_{db}) 
39.55 
°C 
Inlet t_{db} at above RH 
30.12 
°C 
Water vapor saturation pressure (p_{ws}), at t_{db} 
7.2097 
kPa 
Humidity ratio (W) 
0.0252 
kg water/kg dry air 
Put some value of exit RH (Assuming it is in between 97 and 99%) 
98.5 
% 
Specific volume (v) 
0.8951 
m^{3}/kg, dry air 
Partial pressure of water vapor (p_{w}) 
7.1016 
kPa 
Density 
1.1172 
kg/m^{3}, dry air 
Humidity ratio (W) at above t_{db} 
0.04692 
kg water/kg dry air 
Humidity ratio at saturation (W_{s}) 
0.0274 
kg water/kg moist air 
Enthalpy of exit air 
160.50 
kJ/kg 
Specific volume, (v) at saturation 
0.8983 
m^{3}/kg, moist air 
Difference 
0.00 

Density 
1.1133 
kg/m^{3}, moist air 
Exit t_{wb} 
39.31 
°C 
Enthalpy of moist air 
94.5702 
kJ/kg 

Water vapor saturation pressure (p_{ws}) at exit t_{wb} 
7.1170 
kPa 
Humidity Ratio at t_{wb} 
0.0256 
kg water/kg moist air 

Recalculating t_{wb} 
39.31 
°C 
Enthalpy at t_{wb} 
94.6668 
kJ/kg 

Difference 
0.00 



Dew point temperature (t_{d}) of moist air (for temperature between 0 and 93°C) 
39.24 
°C 
Air Flow 

Specific volume (v) 
0.9540 
m^{3}/kg, dry air 
Average density dry 
1.0827 
kg/m^{3} 

Density 
1.0482 
kg/m^{3}, dry air 
Average density wet 
1.0801 
kg/m3 

Humidity ratio (W_{s}) at saturation 
0.0477 
kg water/kg moist air 
Air flowrate at fin 
162.90 
m^{3}/s, per cell 

Specific volume (v) at saturation 
0.9551 
m^{3}/kg, moist air 
Air flow at inlet, at rain zone 
157.87 
m^{3}/s, per cell 

Density 
1.0470 
kg/m^{3}, moist air 
Air flowrate at fan 
168.26 
m^{3}/s, per cell 

KaV/L 
1.7533 



Constant related to tower design (C) 
2.522 

The significance of these calculations is that now we can directly calculate the cooling tower characteristic by using our equations and can compare with the vendor data if the provided height of the cell is adequate to meet the calculated NTU. Obviously the height also depends on the type of packing, but along with vendor input we can create a complete economical design of our cooling tower. Further, we can develop a calculation sheet in Microsoft Excel, which gives all the results of the psychometric chart as well as the cooling tower design. Typical thermal calculations for a counterflow cooling tower can be seen in Table 2.
Edited by Gerald Ondrey
Reference:
1) Cooling Tower Thermal Design Manual,” Daeil Aqua Co., Ltd., available as download at: che.sharif.ir/~heatlab/Lab/Benefit%20Book%20&%20Journal/Benefit%20book/Cooling%20Tower%20Thermal%20Design%20Manual.pdf
2) Rosaler, Robert C., The Standard Handbook of Plant Engineering, 2nd Edition, McGrawHill, New York, 1995.
3) Green, Don W., others, Perrys Chemical Engineers Handbook, 6th Edition, McGrawHill, New York, 1984.
4) Psychometric Data from ASAE D271.2 December 1994.
5) ASHRAE Handbook — Fundamentals, Chapter 1, American Soc. of Heating, Refrigerating and AirConditioning Engineers, Atlanta, Ga.
Author
Jonny Goyal is an assistant lead engineer at Air Liquide Engineering and Construction, c/o Lurgi India Co. Pvt. Ltd. (A22/10, Mohan Cooperative Industrial Estate, Mathura Road, New Delhi 110044 India; Phone: +9142595400; Fax: +9142595051: Email: jonny.goyal@airliquide.com), where he is involved in process designing activities related to basic and detailed engineering projects. On the technology and business development front, he is consistently involved in working with and developing Lurgi FixedBed, DryBottom gasification technology and downstream acidgas removal. Prior to joing Lurgi in 2007, Goyal had two years work experience in Mojj Engineering Pune in the design and commissioning of bioethanol plants from molasses or grains as feedstock, and also has experience in plant operation of distillation columns in the solvent recovery plant of Nectar Lifesciences Ltd. Goyal has a B.Tech.Ch.E. from SBSCET, Ferozepur, which is affilated with PTU Jalandhar, Punjab.
In the original, print article (February 2012, pp. 24–29), there were mistakes in Equations (4) and (6). The correct equations are as follows:

(4)




(6)

These typos do not affect any of the numbers in the Tables or the example problem because the correct equations (not those printed) were used for the calculations. Theonline version of the article has been co
Regards,
Jonny
thanks for your nice detail but while design any cooling tower the Air flow rate is differ by
1.volume of the tank
2.Inlet temp of tower
3.atmospheric pressure
4.Inlet water flow rate
so how we consider the air flow rate required for cooling tower.is there is any thumb rule for L/G ratio or air flow requirement.
please let me know..
thank in advance..