[CAUT] MC absorbtion / release rates

[Ric Brekne]

Hi Don

Perhaps I am reading this wrong... but...  it looks preeeettty much to 
me like the time it takes for wood to come to equilibrium with a change 
in climate is dependant on  L² / D... which is explained directly below 
in a snip taken from the web site.   Doesnt look like there is any 
provision in this formula for anything that would allow for wood to give 
off moisture at any different rate then it takes it on. 

If someone can explain otherwise then great.... but it looks like Ron N 
was not only right about not being able to find any support for the 
different rates idea... but that there is direct support for his claim 
that the idea is mistaken.

Cheers
RicB

........................

    "At MC lower than the fiber saturation point (usually 35%), moisture
    change takes place by diffusion within the wood. The standard
    diffusion equation may be written as

    t = L² / D

    where
    D = 1x10-6 cm²/s transverse and radial, 1x10-5 cm²/s lengthwise,
    L is the length along the direction of diffusion.
    t is the time to 1/e of the moisture change, that is to 63% of the
    equilibrium change.

    So, if you have a piece of seasoned wood 2 cm thick that is at the
    15% MC of typical outdoors storage here, and you want to estimate
    how fast it will come to equilibrium in your workshop at 30% RH (7%
    MC) if exposed to air both sides, L = 1 cm and the diffusion
    equation gives t = 1x106 s, 11 days. The equilibrium MC change
    required is 8%, so in 11 days you can expect 63% of 8% = 5% lower
    MC, that is 10% total MC. That leaves 3% to go, and you can expect
    63% of that 3% to take place over the next 11 days, to 8% MC. So, 3
    weeks should be enough time for 2 cm thick wood. If your wood is 4
    cm thick, it will take 3 months, 6 cm thick, 7 months. According to
    the literature, most seasoned temperate woods change moisture at a
    rate within ±20% of this. "