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Anleitung Zusammenfassung
(kPa) Input/ Airflow/cabinet 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 CFM CFM CFM CFM CFM CFM CFM CFM CFM CFM PT9C20N100UP11 HIGH 2465 2380 2295 2195 2095 1995 1875 1760 1620 1470 M-HI 2085 2035 1960 1880 1800 1705 1605 1485 1360 1235 M-LO 1725 1625 1595 1540 1485 1405 1315 1235 1125 995 LOW 1420 1360 1310 1255 1180 1120 1070 970 950 845 PT9D20N120UP11 HIGH 2615 2535 2450 2385 2285 2175 2075 1945 1825 1670 M-HI 2055 2045 2015 1985 1932 1855 1785 1730 1605 1470 M-LO 1690 1650 1620 1600 1570 1525 1470 1395 1300 1200 LOW 1345 1335 1335 1285 1250 1230 1180 1115 1010 850 NOTES: 1. Airflow expressed in standard cubic feet per minute (CFM) and in cubic meters per minute (m3/min). 2. Return air is through side opposite motor (left side). 3. Airflows above 1800 CFM (50.97 m3/min) require either return from two sides or one side plus bottom. 4. Motor voltage at 115 V. 5. NR = Operation at this static pressure is not recommended. BLK WHT GRN BLK (HOT) WHT (NEUTRAL) GRN NOMINAL 120 VOLT ROOM THERMOSTAT R W G Y C FURNACE CONTROL CONDENSING UNIT TO AIR CONDITIONER CONTROLS R G C COMMON T’STAT CONNECTION W1/W Y/Y2 LINE WIRING CONNECTIONS SINGLE STAGE HEATING THERMOSTAT CONNECTIONS ROOM THERMOSTAT R W1 W2 G Y2 Y1 C FURNACE CONTROL CONDENSING UNIT TO AIR CONDITIONER CONTROLS R G COMMON T’STAT CONNECTION Y1 W1/W C W2 Y/Y2 ROOM THERMOSTAT R W G Y1 C Y2 FURNACE CONTROL CONDENSING UNIT TO AIR CONDITIONER CONTROLS R G C COMMON T’STAT CONNECTION Y1 W1/W Y/Y2 TWO-STAGE HEATING AND TWO-STAGE COOLING SINGLE STAGE HEATING AND TWO-STAGE COOLING THERMOSTAT CONNECTIONS THERMOSTAT CONNECTIONS Unitary Products Group 036-21581-001 Rev. B (0904) FILTER PERFORMANCE NOTE: The filter pressure drop values in the “Filter Perfor- The airflow capacity data published in the “Blower Perfor-mance” table shown below are typical values for the type of mance” table listed above represents blower performance filter listed and should only be used as a guideline. Actual WITHOUT filters. To determine the approximate blower per-pressure drop ratings for each filter type vary between filter formance of the system, apply the filter drop value for the fil-manufacturer. ter being used or select an appropriate value from the “Filter Performance” table shown below. FILTER PERFORMANCE - PRESSURE DROP INCHES W.C. AND (KPA) Airflow Range Minimum Opening Size Filter Type Disposable WASHABLE FIBER* Pleated 1 Opening 2 Openings 1 Opening 2 Opening 1 Opening 2 Opening 1 Opening 2 Opening Sq. in. Sq. in. In w.c. Pa In w.c. Pa In w.c. Pa In w.c. Pa In w.c. Pa In w.c. Pa 0 - 750 230 0.01 2.5 0.01 2.5 0.15 37 751 - 1000 330 0.04 10 0.03 7.5 0.20 50 1001 - 1250 330 0.08 20 0.07 17 0.20 50 1251 - 1500 330 0.08 20 0.07 17 0.25 62 1501 - 1750 380 658 0.14 35 0.08 20 0.13 32 0.06 15 0.30 75 0.17 42 1751 - 2000 380 658 0.17 42 0.09 22 0.15 37 0.07 17 0.30 75 0.17 42 2001 & Above 463 658 0.17 42 0.09 22 0.15 37 0.07 17 0.30 75 0.17 42 APPLYING FILTER PRESSURE DROP TO DETERMINE SYSTEM AIRFLOW To determine the approximate airflow of the unit with a filter in place, follow 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 pressure 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, 0.60, etc,) the system airflow corresponds 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, 0.75, 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 120,000 Btuh furnace with 2 return openings 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 following steps: 1. Obtain the airflow values at 0.50" & 0.60" ESP. Airflow @ 0.50": 2285 CFM Airflow @ 0.60": 2175 CFM 2. Subtract the airflow @ 0.50" from the airflow @ 0.60" to obtain airflow difference. 2175 - 2285 = -110 CFM 3. Subtract the total system static from 0.50" and divide this difference by the difference in ESP values in the table, 0.60" - 0.50", to obtain a percentage. (0.58 - 0.50) / (0.60 - 0.50) = 0.8 4. Multiply percentage by airflow difference to obtain airflow reduction. (0.8) x (-110) = -88 5. Subract airflow reduction value to airflow @ 0.50" to obtain actual airflow @ 0.58" ESP. 2288 - 88 = 2197 Unitary Products Group 036-21581-001 Rev. B (0904) ACCESSORIES PROPANE (LP) CONVERSION KIT 1NP0580 - All units This accessory conversion kit may be used to convert natural gas (N) units for propane (LP) operation. Conversions must b...