Herrmidifier Load Calculator - Manual (Page 14)

L o a d C a l c u l a t i o n G u i d e
L o a d C a l c u l a t i o n G u i d e
14 w w w. h e r r m i d i f i e r- h v a c . c o m
Notice that the MAXIMUM HEATING DEMAND LOAD in this
example is 393 lbs./hr. which occurs at 55°F, not at design
outdoor temperature as with traditional heating systems.
Therefore, any humidification system installed to handle this
load will have to have a maximum capacity of 393 lbs./hr.
water output.
Cooling Load
As we have seen, moisture is removed by condensation dur-
ing the cooling process as the air is cooled below its dew
point and then reheats to room temperature. This condensed
moisture must be replaced when its loss causes the R.H. to
fall too low. To determine the maximum demand load due to
cooling observe the following steps:
Step 1. Determine the amount of moisture per cubic foot of
air that will be removed by condensation during the cooling
process as this will have to be replaced. This is done by us-
ing Table 2-A to determine the grains of moisture per cubic
foot of air at indoor design temperature and R.H., then deter-
mining the grains of moisture per cubic foot of air after it has
been cooled and subtracting this figure from that obtained for
the design temperature and R.H.
In determining the amount of moisture in the air after it has
been cooled, it is necessary to know the air temperature
drop across the cooling coils. This is usually a 15°F drop
with standard air conditioning equipment, but can vary from
manufacturer to manufacturer. It is best to check. Further,
the R.H. of the air after cooling will be about 85% rather than
100% because, only that portion of the air in contact with
the cooling coils will be cooled to coil temperature with con-
densation resulting. The coil temperature will be less than
the exit air temperature. The following example will illustrate
Step 1.
Example:
60 tons air conditioning (standard) maintaining 70°F at 65%
R.H.
Moisture content of air at design
70°F, 65% R.H. = 8.10 gr./CF x .65(%) = 5.26 gr./CF
-Moisture-content after cooling
55°F, 85% R.H. = 4.89 gr./CF x .85(%) = 4.15 gr./CF
Moisture removed during cooling= 1.11 gr./CF
Therefore we must replace 1.11 grains of moisture per cubic
foot of air to return the air to the desired condition of 70°F,
65% R.H. after cooling in this example.
Step 2. Determine the volume of air passing over the cool-
ing coils per hour. This is usually about 400 CFM per ton of
rated capacity for standard air conditioning equipment, but
this may vary with the make of equipment. It is always best
to check. Computer room air conditioning equipment is often
designed for 500-800 CFM per ton of rated capacity.
Example:
60 ton air conditioning
60 x 400 CFM = 24,000 x 60 min./hr. = 1,440,000 CFH
Step 3. Since almost all standard air conditioning equipment
is usually slightly overrated to allow it to shut down periodi-
cally and defrost, it is necessary to determine the maximum
“ON” time during any given hour, since this is when mois-
ture is being removed from the air. This is usually 80%, but
should be checked, as special equipment may be designed
to cool rapidly and use less “ON” time.
Step 4. The MAXIMUM COOLING DEMAND LOAD for hu-
midification can now be calculated using the following for-
mula:
gr./CF needed x CFH x “ON” time
7,000 gr/lb. = lbs./water req.
Using the preceding examples in Steps 1-3:
1.11 gr./CF x 1,440,000 CFH x.80 “ON” = 183 lbs.
7,000 gr./lb. water/hr.
In this example, the MAXIMUM COOLING DEMAND LOAD
is 183 lbs. water/hr. Therefore, any humidification system
installed to handle this load will have to have a maximum
capacity of 183 lbs./hr. water output.
Cooling Load- Low Temperature (32°F-47°F)
In special applications, such as cold storage, the cooling
equipment (refrigeration) is usually designed to have an air
temperature drop across the coils of less than 15°F, to move
a greater volume of air across the coils than the 400 CFM per
ton experienced with standard air conditioning equipment
and to have a definite “OFF” cycle for defrosting.