Dewatering of wood in cylinder
Dewatering of wood in cylinder
(OP)
Hi. Just wondering if anyone has any time on their hands to help with modeling a process wherein wood chips are packed into a cylinder and subjected to compression (up to about 20,000 PSI). the objective is to determine the length of the tube that provides 15 seconds of residence time at an incoming flow rate of 1 lb/sec. Relevent parameters: initail moisture content (MC) = 65%; desired moisture content = 25% MC. Incoming density = 12.5 lb/cu ft. density at exit = 80 lb/cu.ft. This seems like a differential equation with both the density and moisture content varying along the length of the cylinder. the formula should describe cylinder geometry (assume uniform cylinder diameter, the varying density, and moisture content over time. Thanks if you can help. Been a long time since college so I'm struggling with this one.
Thanks for taking a look!
Thanks for taking a look!





RE: Dewatering of wood in cylinder
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
RE: Dewatering of wood in cylinder
Heat up the wood cylinder until you reach the desired moisture content and then apply pressure, not to squeeze out moisture, but just to compact wood fibers.
RE: Dewatering of wood in cylinder
RE: Dewatering of wood in cylinder
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
RE: Dewatering of wood in cylinder
I was going to say, model flow through the capiliaries, but with applied pressure, not capiliary, but you'll probably be creating a lot of other escape routes when the wood splits up.
I think fluid mechanics has little to do with this. I think that you should assume that any water will be forced out immediately when you get to 20,000 psi, or maybe before.
So, are those moisture contents by weight, or by volume?
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
I think it's a differential equation with moisture content (MC) changing over time (d(MC)/dt) and density changing over time (d(rho)/dt) as a result of an applied force (assumed constant). Since there's no acceleration ocurring, it's a first order equation. Perhaps 2 equations that are solved simultaneously? Probably very simple to solve once it is modeled correctly.
Comments?
RE: Dewatering of wood in cylinder
Say the incoming material is composed of two things, wood fiber and water. The density of that is 12.5, as you've said. If you happen to have 12.5 lbs of the stuff, then its 65% water by weight, that's 8.125 lbs of water and 4.375 lbs of wood fiber and the total volume is 1 ft3. Water has a density of 62.4 lbs/ft3, so the water volume is 0.1302 ft3 and the wood volume is 0.8698 ft3, total is 1 ft3
That tells me the density of the wood fiber is 4.375 lbs/0.8698 ft3 = 5.030 lbs/ft3
Now if you squeeeeeeeeeezzzzzzzzzze all the water out of the incoming material, you'll have only wood left, so its density will be 5.030 pcf when you squeeze it bone dry. You'd have to add rocks to it to get it to 80 pcf.
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
RE: Dewatering of wood in cylinder
How long does it take to compress it to 80 pcf and by using what force? Also, how long does it take to spring back to 50?
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
RE: Dewatering of wood in cylinder
IN
vol wt density
total 1 12.5 12.5
H2O 0.1302 8.125 62.4
woodfib 0.0437 4.375 100.2
FreeH20 0 0 62.4
OUT
wood vol wt density
total 0.09375 7.500 80
H2O 0.0501 3.125 62.4
woodfib 0.0437 4.375 100.2
OUT
FreeH20 0.0801 5.000 62.4
Now what to do with it?
I don't see any differential equations there.
So, we want to calculate the length of tube to hold for 15 seconds, given inlet flow of 1 lb/sec.
If inlet density is 12.5 lbs/ft3, then 1 lb is a volume of 0.08 ft3, so the volumetric flowrate is 0.08ft3/sec
Length of tube?
Q ft3/s = AV
A ft2 = π D^2/4 , D in feet
V = 0.08/(π D^2/4), V in fps
L = 15 seconds * 0.08/(π D^2/4)
Plug in the diameter in feet and get the length of pipe in feet.
**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/
RE: Dewatering of wood in cylinder
Kenc11, what form is the product in? Is it shavings, fines, or chippings?