5188236-BIM-A-0316
Johnson Controls Unitary Products
43
Figure 22: Altitude/Temperature Correction Factors
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 15
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 14 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
0.600
0.650
0.700
0.750
0.800
0.850
0.900
0.950
1.000
1.050
1.100
40
50
60
70
80
90
100
Air Temperature (ºF)
Correction Factor
Sea Level
1000 ft
2000 ft
3000 ft
4000 ft
6000 ft
7000 ft
8000 ft
9000 ft
10000 ft
5000 ft
