Air temperature and altitude corrections for ventilation systems
Posted by mikeaurelius on January 30, 2009
Altitude and air temperature make a substantial difference in the performance of any ventilation system. The calcuations I have provided here: http://mikeaurelius.wordpress.com/2007/12/25/ventilation-basics-part-two/ are based on sea level and 70 degrees F.
In the calculations, the only variable that needs modifying is VP, Velocity Pressure. The standard calculation is (V/4005) squared.
When applying temperature and altitude corrections, use the following equation: VP = ((V/1096.7) squared) times ‘density of the air in pounds per cubic foot’.
The following table gives standard Density of air in pounds per cubic foot.
| Air Temp |
Altitude in Feet above Sea Level |
||||||||||
|
0 |
1000 |
2000 |
3000 |
4000 |
5000 |
6000 |
7000 |
8000 |
9000 |
10000 |
|
|
-25 |
.092 | .088 | .085 | .082 | .079 | .076 | .073 | .071 | .068 | .066 | .063 |
|
0 |
.086 | .083 | .080 | .077 | .075 | .072 | .069 | .067 | .064 | .062 | .059 |
|
25 |
.082 | .079 | .076 | .074 | .071 | .068 | .066 | .063 | .061 | .059 | .057 |
|
50 |
.078 | .075 | .073 | .070 | .067 | .065 | .062 | .060 | .058 | .056 | .054 |
|
70 |
.075 | .072 | .070 | .067 | .065 | .062 | .060 | .058 | .056 | .054 | .052 |
|
100 |
.071 | .068 | .066 | .064 | .061 | .059 | .057 | .055 | .053 | .051 | .049 |
|
125 |
.068 | .066 | .063 | .061 | .059 | .057 | .055 | .053 | .051 | .049 | .047 |
As an example, let’s go back to example one on the above referenced link and apply both an altitude and temperature correction to the design.
Example:
CFM = 1300
Altitude = 5,000 feet above sea level
Temperature = 100 degrees (summer average)
Duct size = 6″ round (or .196 square feet) Use the area of a circle formula (Pi times radius squared) divided by 144.
Total run = 6 feet.
Number of Bends: 0
Loss Factor: 0.11 (from Table 1)
Solve for Velocity: 1300 / .196 = 6632 Feet per Minute
Solve for VP: (6632/1096.7) squared times .059 (from table above) = 2.15
Solve for SP: 2.15 times 6 times .11 = 1.42 inches of pressure
As you can see from the original calculation in example 1 where the calculated static pressure was 1.80 inches, increasing altitude and/or temperature decreases the static pressure in the system. This is due to the fact that at higher altitudes, air is less dense than at sea level.
However, note the temperatures. If your average temperature is below 70 F, your static pressure is going to increase, because cold air is more dense than warm air. This means that your ventilation system is going to be slightly more inefficient in the winter than it will be in the summer.
Linda said
Thank you very much, Mike. This is most helpful to those of us working at high elevations and colder climates!
Linda
mikeaurelius said
You are very welcome!