Cooling And Dehumidification - Honeywell AUTOMATIC CONTROL Engineering Manual

COOLING AND DEHUMIDIFICATION

BASIC PROCESS
Cooling and dehumidification can be accomplished in a single
process. The process line moves in a decreasing direction across
both the dry-bulb temperature lines and the constant moisture
lines. This involves both sensible and latent cooling.
Figure 12 illustrates cooling air by removing sensible heat
only. In that illustration, the resulting cooled air was 95 percent
relative humidity, a condition which often calls for reheat (see
DEHUMIDIFICATION AND REHEAT). Figure 25 illustrates
a combination of sensible and latent cooling. Whenever the
surface temperature of the cooling device (Point B), such as a
chilled water coil, is colder than the dew point temperature of
the entering air (Point A), moisture is removed from the air
contacting the cold surface. If the coil is 100 percent efficient,
all entering air contacts the coil and leaving air is the same
temperature as the surface of the coil.
COOLING COIL
85°F DB
63% RH
OA
50°F DB
B
C
50°F DB 60°F DB
93% RH
Fig. 25.
All coils, however, are not 100 percent efficient and all air
does not come in contact with the coil surface or fins. As a
result, the temperature of the air leaving the coil (Point C) is
somewhere between the coolest fin temperature (Point B) and
the entering outdoor air temperature (Point A). To determine
this exact point requires measuring the dry-bulb and wet-bulb
temperatures of the leaving air.
SUPPLY FAN
60°F DB
93% RH
DA
A
0.0166 LB/LB
0.0100 LB/LB
75°F WB
85°F DB 58°F WB
63% RH
C1847
14.1 CF/LB
PSYCHROMETRIC CHART FUNDAMENTALS
To remove moisture, some air must be cooled below its
dew point. By determining the wet-bulb and the dry-bulb
temperatures of the leaving air, the total moisture removed per
pound of dry air can be read on the humidity ratio scale and is
determined as follows:
1. The entering air condition is 85F dry bulb and 63 percent
rh (Point A). The moisture content is 0.0166 pounds of
moisture per pound of dry air.
2. The leaving air condition is 60F dry bulb and 93 percent
rh (Point C). The moisture content is 0.0100 pounds of
moisture per pound of dry air.
3. The moisture removed is:
0.0166 – 0.0100
per pound of dry air
The volume of air per pound at 85F dry bulb and 75F wet bulb
(Point A) is 14.1 cubic feet per pound of dry air. If 5000 cubic
feet of air per minute passes through the coil, the weight of
the air is as follows:
5000 ÷ 14.1 = 355 pounds per minute
The pounds of water removed is as follows:
355 x 0.0066 = 2.34 pounds per minute
or
2.34 x 60 minutes = 140.4 pounds per hour
Since one gallon of water weighs 8.34 pounds, the moisture to
be removed is as follows:
140.4 ÷ 8.34 = 16.8 gallons per hour
AIR WASHERS
Air washers are devices that spray water into the air within a
duct. They are used for cooling and dehumidification or for
humidification only as discussed in the HUMIDIFYING
PROCESS—AIR WASHERS section. Figure 26 illustrates an
air washer system used for cooling and dehumidification. The
chiller maintains the washer water to be sprayed at a constant
50F. This allows the chilled water from the washer to condense
water vapor from the warmer entering air as it falls into the
pan. As a result, more water returns from the washer than has
been delivered because the temperature of the chilled water is
lower than the dew point (saturation temperature) of the air.
The efficiency of the washer is determined by the number and
effectiveness of the spray nozzles used and the speed at which
the air flows through the system. The longer the air is in contact
with the water spray, the more moisture the spray condenses
from the air.
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ENGINEERING MANUAL OF AUTOMATIC CONTROL
= 0.0066 pounds of moisture