Can you predict the airflow through lumber stacks, if you know the kiln size, fan outputs, etc? There must be a formula for this. How does static pressure enter into the calculations?
From contributor N:
The first step is to determine the fan performance. Since we are pushing air through a lumber stack, we encounter some resistance, which increases as the airflow increases. This is called static pressure. For my calculation, I normally use 1/2", but I know most people use a value between 1/4"-3/4". You can obtain these fan performance graphs from the manufacturer.
For the purpose of this example, we will use a kiln that is 27' wide and a stacking height of 12'. As a rule of thumb, half the "window" is lumber and half is free air on 4/4" lumber. We will be trying to get about 400 feet per minute airflow.
Window of open area: 27'x12'/2= 162 ft2
Air volume needed for 400 ft/min.: 400x162 = 64800 ft3
Our 42" fans will provide you with 20000 ft3/minute at 1/2" static pressure. Three fans should be sufficient for an airflow of nearly 400 feet per minute.
Total air volume from three 42" fans: 60000 ft3
Estimated airflow with three 42" fans: (60000/162)= 370 ft/min.
In reality, you must allow for some losses, where air goes around or under the load. Even in a well-baffled kiln, there will be some losses around the ends of the boards. And if you have a pile of random length wood, or bundles or differing lengths, these can be even higher.
When we calculate estimated air velocity, we might assume as high as 30% losses, depending on the load configuration the customer is expecting. In a tiny kiln, your numbers will be better. In a large commercial kiln, these losses become a more important part of the equation.
If one is to consider 4/4" lumber as 1 1/8" with 3/4" stickers, the actual opening is 40% of the face. (075 / (1.125+.75) = 40%.
This means that there is included 10% extra allowance for additional opening. Or in other words, 25% losses (25% of 40% is 10%). So I believe contributor B and I almost agree on the calculation.
I can only agree that the air velocity required will depend on the species, thickness, and depth of the kiln charge.
Although this may be controversial, I like to install a little extra fan capacity and then either install frequency drive (generally only affordable on larger kilns) or operate the fans in intervals. We have many customers who believe the interval-operated fans reduce energy consumption and improve drying quality.
Next step is to figure out the area of the sticker openings that you have. Multiply the height of the sticker (inches) times the length of the lumber (feet). If you have several stacks, end to end, like both an 8' and 12' stack to give 20' overall, use 20'. Then divide the answer by 12 to get it into square feet. Then multiply this answer times the number of sticker openings. This gives the sticker area.
Next multiply the total area times the velocity to obtain cubic feet per minute GOING THROUGH THE LOAD.
As mentioned, there will also be some losses through the 4x4s, around the ends of the load, and so on. Therefore, you need to account for this loss as well. In a well-loaded kiln, you would have to add another 50% to the original cfm value.
Gene Wengert, forum technical advisor
We address part of this problem by asking for 500 to 600 fpm through the load, but you also have to consider the load width. In fact, for white woods, we suggest loading the kiln half full for the first day or two in order to achieve the correct conditions throughout the load and also flash off the surface moisture. Then, add the remaining half of the lumber, blowing air initially through the dry part of the load first. More info on this is in DRYING HARDWOOD LUMBER.
Gene Wengert, forum technical advisor