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Dewatering of wood in cylinder

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!

RE: Dewatering of wood in cylinder

I thought removing the water would decrease the density of wood, but I suppose the 20ksi is squeezing it out and packing the wood fibers together?

**********************
"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

(OP)
Yep, precisely.  Good point though.

RE: Dewatering of wood in cylinder

Would it be possible a different approach?

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

It makes a difference which wood species is being compressed.  You run the gauntlet from balsa on the extreme light side to ironwood and lignum vitae on the dense side.  This will have a direct impact on the compressive distance if your target of 80#/ft3 needs to be adhered to.  Of course, ironwood and lignum already approach or exceed that figure, but you get the idea.

RE: Dewatering of wood in cylinder

So maybe we would start with Young's modulus and bulk modulus for the wood, but I don't know how to transfer that to the time domain, other than guess it would be near the rate of pressure application.  I'm not sure if you'd necessarily squeeze the water out, or wind up with wet squashed wood, or some combination of both.  Water is pretty good at riding up its pressure to resist, which is what you'd get if you put wood in a pressure chamber.  Same reason that those wierd fish at the bottom of the ocean don't get squeezed in by the pressure and dry out.

**********************
"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

(OP)
I neglected to say that there are self-clearing holes around the periphery for water to flow out of.

RE: Dewatering of wood in cylinder

OK, since this is fluid mechanics, I thought you were applying the pressure using a fluid, like triaxial pressure, but apparently you are applying pressure with a mechanical press to two sides of the wood.

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

(OP)
Moisture content is by weight.

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

Thinking about this logically first.  If wood floats, its density is less than water, so the water in the wet wood is the most dense material.  If you squeeze any of the water out, the incoming material loses density.  So I was correct in my first comment.

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

Right.  You didn't give me anything I could use as a basis for assuming anything else, such as the final density.  Is it that number there, 50 pcf? Or do you want to try to hold it until it doesn't spring back and stays at 80?

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

not final density,  the density of totally compressed wood only, with no water.

**********************
"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

(OP)
The density at exit (AKA final density) is stated to be 80 pcf (see problem statement at beginning of this string).  time to spring back to 50 pcf is irrelevent as it happens after exiting the tube.

RE: Dewatering of wood in cylinder

Ya.  I calculated it anyway.

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

OP said final moisture is 25%.  If this stuff is 65% incoming, and only weighs 12.5 lbs/ft3, this sounds more like loose fill biomass.

Kenc11, what form is the product in?  Is it shavings, fines, or chippings?

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