Altitude/Temperature Correction Factors
Air |
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| Altitude (Ft.) |
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Temp. | 0 | 1000 | 2000 | 3000 | 4000 |
| 5000 |
| 6000 | 7000 | 8000 | 9000 | 10000 |
40 | 1.060 | 1.022 | 0.986 | 0.950 | 0.916 |
| 0.882 |
| 0.849 | 0.818 | 0.788 | 0.758 | 0.729 |
50 | 1.039 | 1.002 | 0.966 | 0.931 | 0.898 |
| 0.864 |
| 0.832 | 0.802 | 0.772 | 0.743 | 0.715 |
60 | 1.019 | 0.982 | 0.948 | 0.913 | 0.880 |
| 0.848 |
| 0.816 | 0.787 | 0.757 | 0.729 | 0.701 |
70 | 1.000 | 0.964 | 0.930 | 0.896 | 0.864 |
| 0.832 |
| 0.801 | 0.772 | 0.743 | 0.715 | 0.688 |
80 | 0.982 | 0.947 | 0.913 | 0.880 | 0.848 |
| 0.817 |
| 0.787 | 0.758 | 0.730 | 0.702 | 0.676 |
90 | 0.964 | 0.929 | 0.897 | 0.864 | 0.833 |
| 0.802 |
| 0.772 | 0.744 | 0.716 | 0.689 | 0.663 |
100 | 0.946 | 0.912 | 0.880 | 0.848 | 0.817 |
| 0.787 |
| 0.758 | 0.730 | 0.703 | 0.676 | 0.651 |
| 1.100 |
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| 1.050 |
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| 1.000 |
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Factor | 0.950 |
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| Sea Level |
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0.900 |
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| 1000 ft | |
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| 2000 ft | ||
Correction |
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0.850 |
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| 3000 ft | |
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| 4000 ft | ||
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0.800 |
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| 5000 ft | |
0.750 |
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| 6000 ft | |
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| 7000 ft | |
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| 0.700 |
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| 8000 ft |
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| 9000 ft | |
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| 0.650 |
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| 10000 ft |
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| 0.600 |
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| 40 | 50 | 60 | 70 | 80 | 90 | 100 |
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| Air Temperature (ºF) |
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Figure 1: 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 6,000 CFM, 1.5 IWC and 4.0 BHP?
Solution: At an elevation of 5,000 ft. the indoor blower will still deliver 6,000 CFM if the rpm is unchanged. However, the Altitude/Temperature Correction Factors table 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. The table shows the correction factor to be 0.832.
Corrected static pressure = 1.5 x 0.832 = 1.248 IWC
Corrected BHP = 4.0 x 0.832 = 3.328
Example 2: A system, located at 5,000 feet of elevation, is to deliver 6,000 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 = 1.5 / .832 = 1.80"
Enter the blower table at 6000 sCFM and static pressure of 1.8". The rpm listed will be the same rpm needed at 5,000 ft.
Suppose that the corresponding BHP listed in the table is 3.2. This value must be corrected for elevation.
BHP at 5,000 ft. = 3.2 x .832 = 2.66
68 | Johnson Controls Unitary Products |