York W)036 THRU 060 Specifications download pdf (Page 36)

542300-YIM-A-1009

36 Johnson Controls Unitary Products

The examples below will assist in determining the airflow

performance of the product at altitude.

Example 1: What are the corrected CFM, static pressure, and

BHP at an elevation of 5,000 ft. if the blower performance data

is 1,400 CFM, 0.6 IWC and 0.67 BHP?

Solution: At an elevation of 5,000 ft. the indoor blower will still

deliver 1,400 CFM if the rpm is unchanged. However, Table 13

must be used to determine the static pressure and BHP. Since

no temperature data is given, we will assume an air temperature

of 70°F. Table 12 shows the correction factor to be 0.832.

Corrected static pressure = 0.6 x 0.832 = 0.499 IWC

Corrected BHP = 0.67 x 0.832 = 0.56

Example 2: A system, located at 5,000 feet of elevation, is to

deliver 1,400 CFM at a static pressure of 1.5". Use the unit

blower tables to select the blower speed and the BHP

requirement.

Solution: As in the example above, no temperature

information is given so 70°F is assumed.

The 1.5" static pressure given is at an elevation of 5,000 ft. The

first step is to convert this static pressure to equivalent sea level

conditions.

Sea level static pressure = 0.6 / .832 = 0.72"

Enter the blower table at 1,400 sCFM and static pressure of

0.72". The rpm listed will be the same rpm needed at 5,000 ft.

Suppose that the corresponding BHP listed in the table is 0.7.

This value must be corrected for elevation.

BHP at 5,000 ft. = 0.7 x .832 = 0.58

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York W)036 THRU 060 Specifications Page 36