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.