Static pressure in ducts and vessel

[GordonC]

If air is venting thru a 10" duct and there is a static pressure drop of
2” water column for every 100’ of length at 2000 cfm and the duct opens to a 1000 cubic feet airtight vessel and exits through another 100’ of 10” duct how can you find the pressure drop for the tank alone?

 

[Q_Goest]

Hi Gordon. You can add up the pressure losses for each individual portion of the system to come up with a total pressure loss. The loss for the tank, assuming it's large relative to the duct diameter, is only the entrance and exit losses. There's some discussion of that here:
http://utwired.engr.utexas.edu/che354/mom/pfit.ppt
and here:
https://www.physicsforums.com/showthread.php?t=105066&page=2

 

[ravenprp]

To find the pressure drop for the tank alone, we would need to calculate the total static pressure drop in the system and then subtract the pressure drop in the ducts.

First, we can calculate the total static pressure drop in the system by multiplying the static pressure drop per 100 feet (2" water column) by the total length of the ducts (200 feet). This gives us a total static pressure drop of 4" water column.

Next, we need to determine the pressure drop in the ducts. Since the air is venting through two 10" ducts, the total cross-sectional area is 200 square inches (2 x 10 x 10). We can use this information to calculate the velocity of the air, which is 2000 cubic feet per minute divided by 200 square inches, giving us a velocity of 10 feet per minute.

Using the Bernoulli's equation, we can then calculate the pressure drop in the ducts, which is equal to the dynamic pressure (0.5 x density of air x velocity squared) divided by the acceleration due to gravity. Assuming standard conditions, the density of air is approximately 0.075 pounds per cubic foot, and the acceleration due to gravity is 32.2 feet per second squared. Plugging in these values, we get a dynamic pressure of 0.002 pounds per square inch, which is equivalent to 0.048 inches of water column.

Therefore, the total pressure drop in the ducts is 0.048 inches of water column.

Finally, we can find the pressure drop for the tank alone by subtracting the pressure drop in the ducts (0.048 inches of water column) from the total static pressure drop in the system (4 inches of water column). This gives us a pressure drop of 3.952 inches of water column for the tank alone.

In summary, to find the pressure drop for the tank alone, we need to calculate the total static pressure drop in the system and then subtract the pressure drop in the ducts. This can be done by using the Bernoulli's equation and taking into account the velocity and cross-sectional area of the ducts.