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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
@ 70oF 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
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Square
edge beams and square columns: 1.5
Rounded beams (R = 0.04 x H) and columns (R =
0.04 x W): 1.3
Tapered beams and columns, 30o, H =
0.1 x W: 1.2 |
B) With Louvers
| |
Large,
widely spaced louvers: 2.0 to 3.0
Narrow, small louvers: 2.5 to 3.5 |
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.
Pressure
Drop |
=
= k x (1 / 2) x Air Density x V2 /
115,820 (lb/ft2)
= k x 0.1922 x (1 / 2) x Air Density x V2
/ 115,820 (inch WG = inch Aq. = inch Water)
or,
= k x 0.1922 x (1 / 2) x (Density Ratio x 0.075)
x V2 / 115,820 (inch WG = inch Aq.
= inch Water)
= k x 0.1922 x (1 / 2) x 0.075 / 115,820 x V2
x Density Ratio (inch WG = inch Aq. = inch Water)
= k x V2 x 1 / 16,069,371 x Density
Ratio
= k x V2 x 1 / 4008.72 x Density Ratio
= k x (V / 4008.7)2 x Density Ratio
where,
k: Pressure Drop Coefficient
r: Air Density lb/ft3
V: Air Velocity, ft/min
g: Acceleration Gravity, ft/min2 (1g
= 32.172 ft/sec2 = 115,820 ft/min2)
Density Ratio: Actual Air Density / 0.075
(1 lb/ft2 = 0.1922 inch WG) |
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: 89oF
Inlet (Leaving) Water Temperature: 104oF
Fill Depth: 4 feet
Fill Flute Size: 19 mm
Entering Wet Bulb Temperature: 80oF
Relative Humidity: 80.0%
Site Elevation: 0 feet
Exit Air Temperature: 97oF
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;
| By
Pass Wall Water |
=
[{(Cell Length / Center to Center Distance of
Nozzle) - 2} x 2
+ {(Cell Width / Center to Center Distance of
Nozzle) - 2} x 2] x 10% x GPM / Nozzle
+ 4 Nozzles x 20% x GPM / Nozzle
= [{(42 / 3) - 2} x 2 + {(42 / 3) - 2} x 2] x
10% x 63.776 + 4 x 20% x 63.776
= 357.14 GPM |
3) By-Pass Column Water due
to Spray Nozzles near to Tower Internal Columns
| By-Pass
Column Water |
=
{(Cell Length / Bay Distance) - 1} x {(Cell Width
/ Bay Distance) -1}
x 4 Nozzles x 5% x GPM / Nozzle
= {(42 / 6) - 1} x {(42 / 6) - 1} x 4 x 5% x 63.776
= 459.18 GPM |
|
% By-Pass Water |
=
(By-Pass Wall Water + By-Pass Column Water) /
GPM / 2 x 100(%)
= (357.14 + 459.18) / 12,500 / 2 x 100
= 3.265% |
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 (97oF) = 66.5773 Btu/lb
Air Enthalpy at Inlet (80oF) = 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
To be continued.
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