Installation manual

166753-UIM-C-0706
Unitary Products Group 25
APPLYING FILTER PRESSURE DROP TO
APPLYING FILTER PRESSURE DROP TO APPLYING FILTER PRESSURE DROP TO
APPLYING FILTER PRESSURE DROP TO
DETERMINE SYSTEM AIRFLOW
DETERMINE SYSTEM AIRFLOWDETERMINE SYSTEM AIRFLOW
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 60,000 BTUH (17.58 kW) furnace with a bottom return
opening and operating on high-speed blower, it is found that total sys-
tem static is 0.58” w.c. To determine the system airflow, complete the
following steps:
Obtain the airflow values at 0.50 w.c. (125 Pa) & 0.60 w.c. (150 Pa)
ESP.
Airflow @ 0.50”: 1250 CFM (35.4 m
3
/min)
Airflow @ 0.60”: 1180 CFM (33.4 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.
1180 - 1250 = -70 CFM (-12 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 (-70) = -56
Subtract airflow reduction value to airflow @ 0.50 w.c. (125 Pa) to
obtain actual airflow @ 0.58 in. w.c. (144 Pa) ESP.
1250 - 56 = 1194
TABLE 16: Field Installed Accessories - Non Electrical
MODEL NO. DESCRIPTION USED WITH
1NP0347 PROPANE (LP) CONVERSION KIT ALL MODELS
1PS0301
HIGH ALTITUDE PRESSURE SWITCH KIT
(Does Not Include Orifices)
40, 60 MBH
1PS0302 80 MBH
1PS0311 100, 115, 130 MBH
1SR0302 SIDE RETURN FILTER KIT 1” FILTER ALL MODELS
1SR0200 SIDE RETURN FILTER KIT 1-4” FILTER ALL MODELS
1BR0114 BOTTOM RETURN FILTER KIT 1” FILTER 14-1/2” CABINETS
1BR0214 BOTTOM RETURN FILTER KIT 1-4” FILTER 14-1/2” CABINETS
1BR0117 BOTTOM RETURN FILTER KIT 1” FILTER 17-1/2” CABINETS
1BR0217 BOTTOM RETURN FILTER KIT 1-4” FILTER 17-1/2” CABINETS
1BR0121 BOTTOM RETURN FILTER KIT 1” FILTER 21” CABINETS
1BR0221 BOTTOM RETURN FILTER KIT 1-4” FILTER 21” CABINETS
1BR0124 BOTTOM RETURN FILTER KIT 1” FILTER 24-1/2” CABINETS
1BR0224 BOTTOM RETURN FILTER KIT 1-4” FILTER 24-1/2” CABINETS
1HF0801 INTERNAL FILTER KIT WITH 1” FIBER FILTER ALL MODELS