Uline H-3030 Manual Del Usuario

H-3030, H-3784
EVAPORATIVE COOLER
WHAT IS EVAPORATIVE COOLING?
When trying to understand evaporative cooling, it may
be best to think of air as being like a sponge, meaning
that air has an ability to absorb moisture that it comes
in contact with. The amount of moisture that the air will
absorb depends on the state of the air, or specifically,
how much moisture the air already contains and the
temperature of the air. If the air is warm and contains
only a small amount of moisture, it will more readily
absorb moisture. As air cools, its volume decreases, and
with it, its ability to absorb moisture decreases.
The term "relative humidity" describes the quantity of
water in the air in relation to its total capacity.
Any volume of air at any given temperature has an
ability to hold a certain quantity of moisture. If the air
contains 20% of its total capacity to hold moisture, the
relative humidity is said to be 20%. Whereas, a humidity
of 100% indicates that the air at this temperature and
pressure is holding all the moisture it can. If the air has
less that 100% relative humidity when entering the unit,
then it has the ability to hold more moisture, and will
thus evaporate more water and cool more effectively.
When describing the amount of moisture in the air, the
term relative humidity is used because the absorption
capacity of air changes relative to air temperature. The
warmer the air, the more absorbent it becomes, and
can consequently hold more water. That is to say that
air that has a 100% relative humidity can hold no more
water vapor. However, if the air is heated, it expands,
and, as a result, the relative humidity decreases even
though the total amount of water vapor in the air has
not changed. As a result, we must describe the level of
humidity relative to its maximum capacity. Is it a 50˚F
sponge or an 80˚F sponge? An 80˚F sponge will hold
more water at 50% humidity than a 50˚F sponge.
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INTRODUCTION
How is cooling produced? In order to evaporate water,
heat (energy) is required. In fact, the evaporation of one
gallon of water requires almost 8,700 BTU's. Where does
this heat come from? The heat comes from whatever the
water is in contact with as it evaporates. This could be a
hot sidewalk, your body, a tree, or from the air itself.
As the heat is removed from an object, the temperature
of that object is decreased. In the case of the unit, heat
is removed from the air, reducing the temperature
of the air.
It is important to realize that the temperature of the
water does not have a great effect upon the cooling
produced by the evaporation. If you were to place
a gallon of 50˚F water on a warm sidewalk, it would
consume 9,000 BTU's during its evaporation, thus
making the sidewalk 9,000 BTU's cooler. A gallon of
90˚F water would produce 8,700 BTU's of cooling, only
a 3% difference in the total result. This translates into a
difference of less than 1˚F in the performance of the unit.
The following table demonstrates the BTU's removed from
the air based on a given amount of water evaporated
in an hour by the unit.
U.S. GALLON/HOUR
10 (3.8 liters or 8.3 Imperial Gallons)
12 (45.4 liters or 10.0 Imperial Gallons)
14 (53.0 liters or 11.7 Imperial Gallons)
In simple terms, evaporative cooling is nature's way of
cooling. The unit utilizes the same phenomenon, but in
an extremely efficient manner.
Para Español, vea páginas 8-14
Pour le français, consulter les pages 15-21
TOTAL BTU'S REMOVED
87,000
104,400
121,800
0621 IH-3030