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.
|
