The fan BHP shown on the fan rating sheet is the net fan brake horsepower based on the ideal conditions of fan test. The actual operating conditions of cooling tower is quite different from the test conditions of fan maker and the actual fan efficiency will be different from the environmental factor like the inlet and exit air flow conditions, tip clearance, obstructions to air flow, plenum geometry, etc. Therefore, a proper environmental correction factor should be considered to both total pressure and horsepower.

Ventilatoren Sirroco Howden who is supplying the fans had published a paper about the influence on the fan performance as follows;

1) Influence of Fan Inlet Shapes

Refer to example 6-4 how much the resistance is increased for the inlet shape other than R/D = 0.15.

2) Influence of Obstacles present in the air flow of the fan

The influence of fan performance due to the obstacles under the fan depends on the ratio of distance of leading edge of fan blade from the obstacles and the fan stack throat diameter, and on the ratio of area of obstacles and area of fan stack throat. The smaller of the ratio of distance and the larger of the ratio of area, the higher of resistance correction factor. In most cases, the additional pressure drop coefficient due to the obstacles is within 0.1 to 0.15.

3) Influence of Tip Clearance

VSH is describing that the tip clearance less than 1% to the fan diameter does not effect to the fan performance. The author has a different opinion against the publication of VSH and suggests to use the following guideline.

Tip Clearance	Multiplying Factor	Tip Clearance	Multiplying Factor
<= 0.1% to Fan Dia	1.000	<= 0.5% to Fan Dia.	0.950
<= 0.2% to Fan Dia.	0.990	<= 0.6% to Fan Dia.	0.925
<= 0.3% to Fan Dia.	0.975	<= 0.7% to Fan Dia.	0.900
<= 0.4% to Fan Dia.	0.965	<= 0.8% to Fan Dia.	0.875

The additional static pressure increase due to the obstacles could be obtained as adding the pressure drop factor due to the obstacles. The influence of fan performance due to the tip clearance could be achieved as adjusting the power transmission efficiency, which shall be discussed.

Example 8-1. Determine the fan brake horsepower and fan static efficiency for the design conditions dealt above under the assumption that the fan total efficiency is 80.1%.

(Solution)
Fan BHP = Air Volume @Fan in ACFM x Total Pressure in inch Aq. / (Fan Total Efficiency x 6356), or = Air Volume @Fan in ACFM x Static Pressure in inch Aq. / (Fan Static Efficiency x 6356)

Total Static Pressure = PD @Air Inlet + PD @Fill + PD @Drift Eliminator + PD @Fan Inlet = 0.1092 + 0.3011 + 0.0361 + 0.0329 = 0.4793 in Aq.

(Note that the static pressure for rating the fan must be a value of Total Static Pressure - Velocity Recovery unless the venturi height is input to the fan rating program. The suggestion is to use this method instead of inputting the venturi height into the fan rating program, since the efficiency of fan stack used by the fan makers is different each other.)

Total Pressure = Total Static Pressure + Velocity Pressure - Velocity Recovery = 0.4793 + 0.1825 - 0.0178 = 0.6439 inch Aq.
Fan BHP = 1019716.28 x 0.6439 / (0.801 x 6356) = 128.98 BHP
Fan Static Efficiency = Air Volume @ Fan in ACFM x Static Pressure in inch Aq. / (Fan BHP x 6356) = 1019716.28 x (0.4793 - 0.0178) / (128.98 x 6356) = 57.4%

Example 8-2. Determine the motor input power based on the example 8-1.

(Solution)
Actual Fan BHP = Net Fan BHP / System Environmental Correction Factor= 128.98 / 0.95= 135.77 BHP

Motor Shaft BHP = Actual Fan BHP / Efficiency of Power Transmission of Gear Reducer= 135.77 / 0.96= 141.43 BHP

The gear reducer wastes 3 to 5% of motor power, which depends on the number of reduction. The factors influencing the efficiency of gear reducer are:

Frictional loss in bearings
Losses due to pumping or splashing the lubricant oil
Frictional loss in gear tooth action.

All these losses shall be turned to the heat build up of lubricant oil and a proper cooling of lubricant oil is required.

Motor Input Power = Motor Shaft BHP / Motor Efficiency= 141.43 / 0.89 (Motor Efficiency: 89%) = 158.91 BHP

Example 8-3. Determine the rated motor power for above examples.

(Solution)
Minimum Motor Power = Motor Shaft BHP x Motor Minimum Margin x Operation Safety = 141.43 x 1.1 x 1.03 = 160.24 HP

The next available size of motor power is 175 HP. Note that the motor minimum margin depends on the type of cooling tower operation and ambient conditions.