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(5) Fan Starting Requirements: Fan equipment inertia must be overcome in order to bring the fan up to speed. The time taken to accelerate the fan to full speed must be within the motor winding limitations or the life of the motor will be reduced. It is important to know the inertia moment for fan, gear reducer and coupling shaft, the speed torque curve for fan including the efficiency of fan system and gear reducer, the motor capability (acceleration torque = pull up torque), and allowable starting time for the motor and the starter.

The inertia of an driver indicates the energy required to keep the system running. However, starting or accelerating the system requires extra energy. The torque required to accelerate a body is equal to the WR² of the body, times the change in rpm, divided by 308 times the interval (in second) in which this acceleration takes places:

Accelerating Torque, T_acc = WR² x N / (308 x t) (lb-ft)

Where,
N: Change in rpm
W: Weight in lbs
R: Radius of Gyration
t: Time of Acceleration in seconds
WR²: Inertia Moment
308: This constant is derived by transferring linear motion to angular motion, and considering acceleration due to gravity.

The fan is reached to the full speed of fan in 3-5 seconds. This means that it takes 3-5 seconds motor to reach the full speed of motor. The time that it takes to accelerate an induction motor from zero to full speed may be found from the following equation.

Motor Starting Time
= (WR_m² + WR_s² + WR_g² + WR_f²) x N / (308 x T_acc) (second)

Where,
WR²: Moment of Inertia
WR_m²: Inertia of Motor (lb-ft²)
WR_s²: Inertia of Shaft (lb-ft²)
WR_g²: Inertia of Gearbox (lb-ft²) at the speed of motor
WR_f²: Inertia of Fan (lb-ft²) at the speed of motor
T_acc (Average Accelerating Torque) = 0.8 x {(T_s + T_m)} / 2 - TL
= Pull Up Torque
T_s: Starting Torque
T_m: Maximum Torque (= Breakdown Torque)
TL: Full Load Torque
N: Motor Speed

Since the moment of inertia of fan is in reference to the fan shaft speed, it should be converted to the motor speed by the following relationship:

WR_f² at motor speed = WR_f² at fan speed x (1/Gear Ratio)²

In order to find the total time required to accelerate the motor and mechanical equipment of cooling tower, the area between the motor speed-torque curve and the fan speed-torque curve which is including the efficiencies of gear reducer and cooling tower system, is divided into strips, the ends of which approximate straight lines. Each strip corresponds to a speed increment which takes place within a definite time interval.

Let's calculate the acceleration torque and horsepower about below sample.

• ●  WR² of Motor: 64.3 lb-ft² (1,770 RPM, 3 seconds of acceleration time)
• ●  WR² of Shaft: 2.4 lb-ft²
• ●  WR² of Gear Reducer: 4.350 lb-ft²at high speed shaft (or 852.60 lb-ft² at low speed shaft)
• ●  WR² of Fan: 34,260 lb-ft²
• ●  Gear Reduction Ratio: 14.0:1

Then, Equivalent WR²= 64.3 + 2.4 + 852.6 (1/14.0)² + 34,260 (1/14.0) ² = 245.85 lb-ft²

Now, accelerating torque can be determined:

T_acc = (Equivalent WR² x N) / (308 x t)
= (245.85 x 1,770) / (308 x 3)
= 470.94 lb-ft
(Note that the engineers always have to check the acceleration torque of motor for maintaining the starting time required and the motor life service.)

Now, horsepower to accelerate is:

HP_acc = (T_acc x N) / 5250
= 470.94 x 1,770 / 5250
= 158.77 hp

From this, the size of the motor which will accelerate and run the load can be determined. Losses must be considered here. Throughout these example, we have assumed no losses due to friction, heat, etc. Obviously, there are friction losses, and if, say 10% of power is lost, it at be made up.

HP_Losses = (10%) x (HP_acc)
= 10% x 158.77 = 15.88

Therefore, total horsepower is:

HP_Total = HP_acc + HP_Losses
= 158.77 + 15.88 = 174.65 or approximately 175 HP

Thus, it takes 175 horsepower to accelerate and run this system.

4) Motor Input Power Rating

The horse power shown on the fan rating sheet is the net shaft break horsepower that the fan requires. Thus, the necessary input motor power must be calculated referring factors below.

• ●  Motor efficiency
• ●  Gear reducer drive efficiency
• ●  System efficiency

Motor efficiencies are obtained from motor manufacturers and usually range from 0.85 to 0.96. Gear reducer drive efficiencies range typically 0.92 to 0.97. System efficiencies are more difficult to assess. As you see the footnote of fan rating sheet, Hudson fans are tested in nearly perfect conditions of very close tip clearance, good inlet conditions and minimal obstructions from beams. It is likely the actual tower system will not be as ideal as Hudson's performance curve conditions. Unless you have compared your system to Hudson's one by means of wind tunnel or other method, we suggest using a system efficiency of 0.95.

Since actual operating conditions of commercial equipment will vary from wind tunnel test conditions, proper environmental correction factors must be applied to both total pressure and horsepower. These environmental factors include inlet and exit air flow conditions, tip clearance, obstructions to air flow, plenum geometry, etc. Therefore actual fan efficiencies will be different from those indicated unless the actual environmental system is equivalent to the test conditions.

Input Horsepower = BHP / (Eff_m x Eff_g x Eff_s)

Where,
Eff_m: Motor efficiency
Eff_g: Gear reducer efficiency
Eff_s: System efficiency

Air horsepower which is the actual output of the fan is calculated by:

HP_Air = Total Pressure x CFM / (6356 x Efficiency)

5) Motor Output Power Rating

• ●  Power required = Fan BHP / (Efficiencies of Fan System and Gear Reducer)
• ●  Minimum Power to be installed = Power required x 1.10 of Design Margin

In case of previous sample job,

Power required = 139.8 hp / (0.95 of fan system efficiency x 0.95 of gear reducer efficiency)
= 154.90 hp

Minimum Power to be installed = 154.90 x 1.1 of design margin
= 170.39 hp

Therefore, the next available size of motor power is 175 hp.

6) Consequent Pole Speed Controls

This is limited to the case of using the polypole change motor. Pole changing is handled by switching circuits that resemble some of those controls. The choice of which motor type to use depends on the speed versus torque requirements of the driven load. In the application of the cooling tower axial fan, the type of load is a variable torque, since the fan power required at the low speed is reduced to a cubic speed ratio (= low speed/high speed). The actual circuit used will require a three-terminal contactor for one speed and a five-terminal contactor for the other. There are three classes of consequent pole motors depending on their intended service. The NEMA has standardized the terms as follows:

• ●  Constant Torque Motor: This control is to obtain the same rated torque at both high and low speed. These motors are connected in parallel wye for high speed and series delta for low speed.
• ●  Constant Horsepower Motor: This control is to obtain the same rated power at both high and low speed. These motors are connected in-series delta for high speed and parallel wye for low speed.
• ●  Variable Torque Motor: This is to obtain the rated torque directly proportional to speed at either high or low speed. These motors are connected in parallel wye for high speed and series wye for low speed.