The air pressures are always dropped in the area where the direction of air flow is changed or the velocity of air flow is decreased suddenly. Representative areas where the pressure losses of air are occurring in the induced draft counter flow cooling tower are as follows;

• Air Inlet (Entrance Loss)
• Fill
• Water Distribution Piping
• Drift Eliminator
• Fan Inlet (Sometimes called plenum losses)

Most of air pressure drops at all the areas excepting fill section can be easily calculated as per the well known formula of K x (Air Velocity / 4008.7)² x Density Ratio. K is a pressure drop coefficient and depends on the shape of obstruction laid in the air stream. Density ratio is an actual air density divided by 0.075 lb/ft³ @ 70^oF dry air conditions. In cooling tower, these pressure losses are called "Static Pressure Loss", just "Static Pressure", or "System Resistance. The performance of cooling tower fans depends on the calculation degree of static pressures at the cooling tower.

The minimum value of pressure drop coefficient at the air inlet is including the two turns of air stream directions and is 1.0 for a hypothetical perfect bell inlet. As a guide line, K values at the air inlet are as below;

A) Without Louvers

B) With Louvers

In most cases, the pressure drops at the water distribution piping zone are included into the pressure drops at drift eliminators because the drift eliminators are installed onto the water distribution pipes or within 2 feet from pipes. In this case, K values is in the range of 1.6 to 3.0. Of course, it must be based on the data provided by manufacturer. The pressure drop coefficient at the fan inlet will be discussed in the examples related to the fans later again, but it is in the range of 0.1 to 0.3.

Therefore, a constant of 4008.7 is obtained from above in order to convert the unit of pressure drop to inch Aq. using the ft/min unit of air velocity and lb/ft³ unit of air density.

It is important to predict the obstructions in the air stream. The obstructions which must be considered in designing the cooling tower are as follows;

• Obstruction at the air inlet area
• Obstruction at the fill

The obstructions at the air inlet area are the area of preventing the air flow and are a summation of area projected to the air inlet with the columns, beams, or bracing, etc. There is no need to consider the air flow obstruction due to the inlet louvers. The obstruction at the fill is a plain area which is not filled due to the columns or bracing, etc.

The types of air inlet for the counter flow induced draft cooling tower as below are being used.

• One Side Open: This arrangement is useful for the area where the obstruction to be able to disturb the air flow or to increase the inlet wet bulb temperature due to the adjacent building or the heat sources to be able to affect the entering wet bulb temperature are located to the one side of cooling tower. When to design the cooling tower with this arrangement of air inlet, a special attention is required for the even air distribution into the fill section.
• Two Side Open & Ends Closed: This arrangement is most general for the industrial cooling towers.
• All Around Cell Group
• Back To Back & Open All Round: This is useful for a case where the area is limited.

Example 6-1. Determine the pressure drop at the air inlet for the below given conditions.

Given,
Cell Length: 42.0 feet
Cell Width: 42.0 feet
Air Inlet Height: 15.0 feet
Number of Spray Nozzle: 196 each (Center to Center Distance of Nozzles: 3 feet)
Water Flow Rate: 12500 GPM
Exit (Entering) Water Temperature: 89^oF
Inlet (Leaving) Water Temperature: 104^oF
Fill Depth: 4 feet
Fill Flute Size: 19 mm
Entering Wet Bulb Temperature: 80^oF
Relative Humidity: 80.0%
Site Elevation: 0 feet
Exit Air Temperature: 97^oF
Arrangement of Air Inlet: Two Sides Open & Ends Closed
Material of Tower Framework: Wood
Type of Air Inlet Louver: Large, Widely Spaced

(Solution)
In order to obtain the air mass flow, the following calculation must be first accomplished. The actual cooling range through the tower must be calculated because there is a by-pass wall water in the tower.

New Tower Range = Design Range / (1 - % by pass wall water / 100)
(Note: This was already discussed in example 5-2.)
% By-Pass Water Calculation is as follows:

1) Water Flow Rate per Nozzle = Design Water Flow Rate / Total Number of Nozzles= 12,500 GPM / 196 = 63.78 GPM/Nozzle

2) By-Pass Wall Water from Spray Nozzles;

3) By-Pass Column Water due to Spray Nozzles near to Tower Internal Columns

Therefore, the actual range through tower is obtained from relation of Design Range / (1 - % By-Pass Water / 100)

Actual Range = (104 - 89) / (1 - 3.265 / 100) = 15.5063
A value of L/G is obtained from the equation of ha₂ = ha₁ + L/G x New Tower Range.
L/G = (ha₂ - ha₁) / New Tower Range
Air Enthalpy at Exit (97^oF) = 66.5773 Btu/lb
Air Enthalpy at Inlet (80^oF) = 43.6907 Btu/lb
Therefore, L/G = (66.5773 - 43.6907) / 15.5063 = 1.4760

The air mass is calculated from the relation of G = L / (L/G). Here the value of L is a net water flow rate through the cooling tower. That is, L = Design Water Flow Rate x (500 / 60) x (1 - % By-Pass Water / 100) = 12,500 x (500 / 60) x (1 - 3.265 / 100). (Note: (500 / 60) is a constant to covert water flow rate in GPM to lb/min unit.) Then, the value of air mass flow = 12,500 x (500 / 60) x (1 - 3.265 / 100) / 1.4760 = 68,271.5 lb/min

Second, let's calculate the area of obstruction in the air inlet. In case of wood structure, one bay (between center of columns) is based on 6 feet and the traversal member is based on 6 feet in the height. Therefore, the number of bay for the 42 feet of cell length is 7 and the width of column is 4 inch. In the traversal member, two beams are required for this air inlet height.

Area of Obstruction due to Columns = No. of Bay x Width of Column x Air Inlet Height x No. of Air Inlet = 7 x (4 / 12) x 15 x 2= 70 ft²

Area of Obstruction due to Traversal Members = No. of Members x Height of Members x Cell Length x No. of Air Inlet = 2 x (4 / 12) x 42 x 2= 56 ft²

Total Area of Obstructions = 70 + 56 = 126 ft²

Overall Area of Air Inlet = Cell Length x Air Inlet Height x No. of Air Inlet = 42 x 15 x 2= 1,260 ft²

% Obstruction @ Air Inlet = Total Area of Obstructions / Overall Area of Air Inlet x 100 (%)= 126 / 1,260 x 100(%)= 10.0%
Net Area of Air Inlet = 1,260 - 126 = 1,134 ft²

Air Density and Specific Volume @ Air Inlet must be based on the dry bulb temperature at a relative humidity, not on wet bulb temperature. Let's find a dry bulb temperature from Psychrometric chart or from the following computer calculation method.

The dry bulb temperature corresponding 80% RH at 80^oF WBT is 85.24^oF. (Note: Some engineers are using the air density and specific volume at the air inlet using the web bulb temperature. This is totally wrong and is quite different from the value at the dry bulb temperature & relative humidity.)

Specific Volume @ 85.24 DBT & 80% RH = 14.2230 ft³/lb
Airflow Volume @ Air Inlet = Air Mass Flow x Specific Volume @Air Inlet= 68,271.5 x 14.2230= 971,028 ft³/min
(For reference, the specific volume at the given wet bulb temperature is 14.1126 ft³/lb and airflow volume becomes 963,485 ft³/min. Compare this value with above airflow volume.)

Air Velocity @ Air Inlet = Airflow Volume @ Air Inlet / Net Area of Air Inlet= 971,028 / 1,134 = 856.29 ft/min (FPM)
Air Density @ 85.24 DBT & 80% RH = 0.0718 lb/ft³
Pressure Drop Coefficient for this arrangement = 2.5

Then, pressure drop is obtained from below:

Pressure Drop = K (V / 4008.7)² x Density Ratio= 2.5 x (856.29 / 4008.7)² x (0.0718 / 0.0750)= 0.1092 inch Aq.
(For reference, the air density at the given wet bulb temperature is 0.0724 lb/ft^3. Compare this with the previous value of air density.)

Download the example file (exe6_1.zip)

Example 6-2. Determine the pressure drop at the fill for the same example 6-1.

(Solution)
First, it is to calculate the average air velocity through the fill. The reasons why the average air velocity must be calculated are based the assumptions below;

1) The heat exchange in the rain zone is negligible and there is no change in the air between the entering air into the tower inlet and into the bottom of fill.
2) The heat is completely exchanged at the fill section & water distribution zone.
3) The exit air from the fill is 100% saturated and the heat of exit air transferred from the water is considered as an adiabatic process.

To calculate the average air velocity, the average air volume and specific volume through the fill must be calculated.

Average Specific Volume = 2 / (1 / Specific Volume @ Tower Inlet Temp. + 1 / Specific Volume @ Tower Exit Air Temp.)

Specific Volume @ 85.24 DBT & 80% RH = 14.2230 ft³/lb
Specific Volume @ 97.0 DBT & 100% RH = 14.9362 ft³/lb (The exit temp. was guessed.)

Therefore, the average specific volume at the fill = 14.5709 ft³/lb
Then, the average air volume at the fill is obtained from Average Specific Volume x Air Mass Flow. That is, the average air volume at the fill = 994,776.8 ft³/min
Average Air Velocity = Average Air Volume / Net Fill Area
Net Fill Area = (Cell Length x Cell Width) x (1 - % Fill Obstruction / 100)
% Fill Obstruction = (Sectional Area of Column x Number of Columns) / (Cell Length x Cell Width) x Margin x 100(%)= (4 x 4 / 144 x 7 x 7) / (42 x 42) x 3.6 x 100 (Note: Safety margin for wood tower is about 3.6)= 1.11%
Therefore, the net fill area = (42 x 42) x (1 - 1.11 / 100) = 1,730.7 ft²
Average Air Velocity @Fill = Average Air Volume @Fill / Net Fill Area= 574.78 ft/min

Second, the water loading calculation is required as follows;

Water Loading = Tower Water Flow Rate / Net Fill Area
= Design Water Flow Rate x (1 - % By-Pass Water / 100) / Net Fill Area = 12,500 x (1 - 3.27 / 100) / 1,730.7 = 6.99 GPM/ft²

Air Density @ 85.24 DBT & 80% RH = 0.0718lb/ft³
Air Density @ 97.0 DBT & 100% RH = 0.0696 lb/ft³
Then, average air density at fill = 0.0707 lb/ft³

Now, all the parameters are ready to compute the pressure drop at the fill. The calculation of pressure drop at the fill is very complicated and it is impossible to predict the pressure drop if
the formula for the pressure drop is not available. The formula of calculating the pressure drop at the fill is a proprietary data of fill maker.

Pressure Drop @Fill = 0.3011 inch WG.

Example 6-3. Determine the pressure drop at the drift eliminator per the given conditions in example 6-1.

(Solution)
In general, the obstruction area in the drift eliminator is considered same as the fill obstruction area. Therefore, the net drift eliminator area = (42 x 42) x (1 - 1.11 / 100) = 1,730.7 ft². There is no change in the air mass flow through out the cooling tower. Therefore, the value of air mass flow is same as the above obtained value of 68,271.47 lb/min. The air density and specific volume at 97^oF 100% RH are 0.0696 lb/ft³, 14.9362 ft³/lb respectively.

Then, the air volume at the drift eliminator is obtained from Specific Volume x Air Mass Flow.
That is, the air volume at the drift eliminator = 1,019,716.3 ft³/min
Air Velocity @ Drift Eliminator = Airflow Volume @ Drift Eliminator / Net Area of Drift Eliminator
Air Velocity @ Drift Eliminator = 589.19 ft/min
Pressure Drop Coefficient for a general module type of drift eliminator = 1.6 to 2.0

Then, pressure drop is obtained from below:

Pressure Drop = K (V / 4008.7)² x Density Ratio= 1.8 x (589.19 / 4008.7)² x (0.0696 / 0.0750)= 0.0361 inch Aq.

Example 6-4. Determine the pressure drop at the fan inlet of fan stack per the given conditions in example 6-1. Let's assume that the 28 feet of fan in the diameter with the 88 inch of air seal disk was used and the fan inlet shape is rounded. (R/D = 0.10)

(Solution)
The pressure drop is occurring at the fan inlet of fan stack unless the shape of fan inlet is elliptical bell and no obstruction under the fan in case of induced draft fan arrangement. The following table could be applied to the cooling tower fan stack as a guide line in choosing the pressure drop coefficient.

In practice, it is quite essential to add some extra to the above K value since there are a lot of obstructions under the fan. It is considered that there is no change in the heat from the drift eliminator to the fan. Accordingly, the specific volume at the fan is same as the value at the drift eliminator. Let's calculate the net fan area.

Fan Net Area = 3.1416 / 4 x (Fan Dia² -Air Seal Disk²)= 573.52 ft²
Air Velocity @ Fan = Airflow Volume @ Fan / Net Fan Area= 1019716.3 / 573.52 = 1778.00
Air Velocity @ Fan = 1,778.00 ft/min

(Note: The air volume at fan is same as the air volume at the drift eliminator.)

Then, pressure drop is obtained from below:
Pressure Drop = K (V / 4008.7)² x Density Ratio= 0.18 x (1778.0 / 4008.7)² x (0.0696 / 0.0750)= 0.0329 inch Aq.