Herrmidifier Load Calculator - Manual (Page 12)

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
12 w w w. h e r r m i d i f i e r- h v a c . c o m
Heating Load
To determine the maximum demand load due to heating ob-
serve the following steps.
Step 1. Determine the amount of moisture which must be
added per cubic foot of air to be humidified. As you know,
heating air lowers its R.H., but does not obliterate it, so the
moisture already in the cool air before heating must be de-
termined and subtracted since this moisture is already avail-
able in the heated air. Using Table 2-A, first determine the
maximum design indoor temperature and record the grains
of moisture listed to the right of that temperature, then multi-
ply the grains (at 100% R.H.) times the desired level of R.H.
(a percentage) to arrive at the grains of moisture per cubic
foot of air necessary to maintain the desired R.H.
Now take the minimum design outdoor temperature (you are
essentially heating infiltrated outdoor air) and its grains of
moisture (at 100% R.H.) times the minimum expected R.H.
to arrive at the moisture content already available before
heating.
Next simply subtract the outdoor design moisture from the
indoor design moisture and arrive at the grains of moisture
per cubic foot of air needed to be added to the heated air to
reach design conditions of R. H.
Example: (From Table 2-A)
Indoor Design Moisture @ 70°F 65% R.H.
8.10 gr. x .65(%) = 5.2 gr./cu. ft.
Outdoor Design Moisture @ 0°F 50% R.H.
.48gr.x.50(%) = .24 gr./cu. Ft.
Indoor Design Moisture = 5.26 gr./cu. ft.
-Outdoor Design Moisture = .24 gr./cu. ft.
Moisture to be added = 5.02 gr./cu. ft
Step 2. Determine the maximum volume of air per hour that
will need to have moisture added to it. Natural ventilation,
exhaust and makeup air must be considered. You should use
the largest of these, but not all three.
Example: A large woodworking plant.
1. The area of the plant is 100’ x 200’ x 15’H = 300,000
cu. ft. - If it is tight construction, you would use one air
change per hour. If it is loose or old construction - 1-1/2
air changes per hour. Three or four exposed walls and
shipping doors - 2 air changes per hour. Never use less
than one air change per hour unless sealed off with no
traffic in or out. In this example, the building has four ex-
posed walls and two shipping doors. Volume of air to be
humidified is 300,000 cu. ft. x 2 = 600,000 CFH.
2. The plant contains exhaust equipment exhausting 20,000
CFM to the outside. This volume would be 20,000 CFM
x 60 min./hr. = 1,200,000 CFH. (Note that this volume is
greater than the volume calculated by plant area.)
3. The plant has a makeup air system that supplies 15,000
CFM of outside air to the plant. This volume is 15,000
CFM x 60 min./hr. = 900,000 CFH.
The greatest volume of air is due to the exhaust - 1,200,000
CFH. This is the maximum volume of air that must be hu-
midified in any given hour. You will note that we are exhaust-
ing 1,200,000 CFH but making up only 900,000 CFH. The
balance of 300,000 CFH is brought in by forced infiltration
through cracks around doors, windows, etc. It is not this ex-
haust that we wish to humidify but rather, the air replacing it.
Step 3. THE MAXIMUM HEATING DEMAND LOAD is now
calculated using the following formula:
(step 1) (step 2)
gr. Moisture needed/cu. ft. x CFH air =
7,000 gr./lbs.
lbs. of water per hour required.
or, from the example:
5.02 gr./cu. ft. x 1,200,000 CFH = 860 lbs./hr.water
7,000 gr./lbs.
ALWAYS CONSULT AN
EXPERT AT THE START