035-19670-003 Rev. A (1004)
Unitary Products Group
21
* Washable Fibers are the type supplied with furnace (if supplied).
APPLYING FILTER PRESSURE DROP TO
DETERMINE SYSTEM AIRFLOW
To determine the approximate airflow of the unit with a filter in place, fol-
low the steps below:
1.
Select the filter type.
2.
Select the number of return air openings or calculate the return
opening size in square inches to determine the proper filter pres-
sure drop.
3.
Determine the External System Static Pressure (ESP) without the
filter.
4.
Select a filter pressure drop from the table based upon the number
of return air openings or return air opening size and add to the
ESP from Step 3 to determine the total system static.
5.
If total system static matches a ESP value in the airflow table (i.e.
0.20 w.c. (50 Pa), 0.60 w.c. (150 Pa), etc,) the system airflow cor-
responds to the intersection of the ESP column and Model/Blower
Speed row.
6.
If the total system static falls between ESP values in the table (i.e.
0.58 w.c. (144 Pa), 0.75 w.c. (187 Pa), etc.), the static pressure
may be rounded to the nearest value in the table determining the
airflow using Step 5 or calculate the airflow by using the following
example.
Example: For a 130,000 BTUH (38.06 kW) furnace with 2 return open-
ings and operating on high-speed blower, it is found that total system
static is 0.58” w.c. To determine the system airflow, complete the follow-
ing steps:
Obtain the airflow values at 0.50 w.c. (125 Pa) & 0.60 w.c. (150 Pa)
ESP.
Airflow @ 0.50”: 2125 CFM (60.17 m
3
/min)
Airflow @ 0.60”: 2035 CFM (57.62 m
3
/min)
Subtract the airflow @ 0.50 w.c. (125 Pa) from the airflow @ 0.60 w.c.
(150 Pa) to obtain airflow difference.
2035 - 2125 = -90 CFM (2.55 m
3
/min)
Subtract the total system static from 0.50 w.c. (125 Pa) and divide this
difference by the difference in ESP values in the table, 0.60 w.c.
(150 Pa) - 0.50 w.c. (125 Pa), to obtain a percentage.
(0.58 - 0.50) / (0.60 - 0.50) = 0.8
Multiply percentage by airflow difference to obtain airflow reduction.
(0.8) X (-90) = -72
Subtract airflow reduction value to airflow @ 0.50 w.c. (125 Pa) to
obtain actual airflow @ 0.58 inwc (144 Pa) ESP.
2125 - 72 = 2053
TABLE 16:
Filter Performance - Pressure Drop Inches W.C. and (kPa)
Airflow Range
Minimum Opening Size
Filter Type
Disposable WASHABLE
FIBERS* Pleated
1 Opening
2 Openings
1 Opening
2 Openings
1 Opening
2 Openings
1 Opening
2 Openings
CFM
Cm/m
In³
m³
In³
m³
inwc
kPa
inwc
kPa
inwc
kPa
inwc
kPa
inwc
kPa
inwc
kPa
0 - 750
0 - 21.24
230 0.0038
0.01 0.0025
0.01 0.0025
0.15 0.0374
751 - 1000
21.27 - 28.32
330 0.0054
0.05 0.0125
0.05 0.0125
0.2
0.0498
1001 - 1250
28.35 - 35.40
330 0.0054
0.1
0.0249
0.1
0.0249
0.2
0.0498
1251 - 1500
35.42 - 42.47
330 0.0054
0.1
0.0249
0.1
0.0249
0.25 0.0623
1501 - 1750
42.50 - 49.55
380 0.0062 658 0.0108 0.15 0.0374 0.09 0.0224 0.14 0.0349 0.08 0.0199
0.3
0.0747 0.17 0.0423
1751 - 2000
49.58 - 56.63
380 0.0062 658 0.0108 0.19 0.0473 0.11 0.0274 0.18 0.0448
0.1
0.0249
0.3
0.0747 0.17 0.0423
2001 & Above 56.66 & Above 463 0.0076 658 0.0108 0.19 0.0473 0.11 0.0274 0.18 0.0448
0.1
0.0249
0.3
0.0747 0.17 0.0423
