next message in archive
no next message in thread
previous message in archive
Index of Subjects
This is a multi-part message in MIME format. ------=_NextPart_000_1DE4_01CFD448.DA0E0740 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Dear All, Sept 18, 2014 I have wondered for years how trees can whip around in the wind and = normally spring back to their original position when the wind eases and = I broached this on Naturens in 2003 (pasted below). Meanwhile another = piece of this puzzle has come to my attention. START OF PASTE\\\\\\\\\\\\\\\\\\\\\\\ Dear All, Feb 14, 2003 <snip> Spruce trees normally=20 have a remarkable ability to bend with the wind and then spring back.=20 <SNIP> Is anyone on naturens familiar with the mechanisms that enable=20 elasticity (molecular or cellular) in living trees ? There will=20 presumably be some reversible water movement, between living cells,=20 walls and intercellular spaces but there must also be some give and take = at the molecular level, e.g. distortion and recovery of normally folded = or coiled molecules in the cell wall or changes in water of hydration;=20 sort of an intramolecular shock absorber. Is lignin involved ? END OF PASTE\\\\\\\\\\\\\\\\\\\\\\\\\\\ NOTICE: The following contains numerous calculations. A copy editor = prize goes to the reader who locates the greatest number of = calculating/transcription errors. Driwater, a commercial product, is 2% Cellulose, 98% water and is = solid at room temperature. I can not directly understand how this is = possible but one must assume that water in the vicinity of Cellulose is = no longer a free liquid and water sufficiently far away is a liquid. So = there is apparently a gradual change in state with distance from = Cellulose, as opposed to the sharp transitions portrayed in = temperature/pressure state diagrams.=20 Also, while browsing through old copies of American Scientist, I = came across an article on Colloidal Macromolecular Phenomena (Battista, = O. A, 1965(20):151-173). Purified mcrofibrils of Cellulose (as in wood = pulp) are schematically represented as very long chains of Cellulose = variously associated with short portions of three to seven other long = chains to form polymer microcrystals which are connected to other = microcrystals by hinges of free (unassociated) lengths of Cellulose = (chain length variable within hinges). With this kind of structure, = Cellulose microfibrils can assume a range of longitudinal or lateral = positions without rupture of chemical bonds.=20 In a vague way this 'explains' how living or steamed wood can bend. = It is possible that the tendency of living trees to return to their = initial shape when wind force abates is driven by elasticity of cell = walls variously distorted into intercellular spaces (when compressed) or = withdrawn from intercellular spaces when extended. Returning to the example of Driwater (2% Cellulose in water being a = solid at room temperature) brings to mind the behavior of water in a = porous medium such as soil.=20 With decrease in effective pore size, water in soil becomes bound in the = soil pores to an increasing degree. At tensions greater than pF 2 (200 = cm of water) water in most soil profiles does not flow downward under = the influence of gravity (i.e. it acts as though it were part of the = solid phase). And as soil water content approaches zero all soils = approach tensions of pF 7 (10^7 cm of water; or expressed another way = 100,000 metres of water tension). Back of envelope approximations: If total dissociation of Cellulose is assumed (no microcrystals) one = can approximate the spacing between polymer chains as follows: Assume the 2% solution implies 2 g Cellulose per 100 mL aqueous = solution=3D=20 20 g Cellulose/dm^3.=20 Glucose (C6H12O6) has a molecular weight of 180.16; call this 180 = and water is 18. So the C6H10O5 units of Cellulose would have a molecular weight of = ~180-18=3D 162 According to=20 http://books.google.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&dq=3Ddimensions+= of+cellulose+molecule+units&hl=3Den&sa=3DX&ei=3DoEsbVNGCEoOxyASTsYCgBQ&ve= d=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimensions%20of%20cellulose%20molecule%20= units&f=3Dfalse Fig. 3.4 the length of two units, each being C6H10O5), is 1.038 x 10^-9 = M or unit length of 0.519 x 10^-9 M 20/162 (Moles of units) X 6 x 10^23 Avogadro's number (units/mole) = x 0.519 x 10^-9 M (length per unit)=3D 0.384 x 10^14 M or 3.84 x 10^14 = dm (total polymer length). To put this in context, the equatorial radius is 6,378,388 metres = which is equivalent to an equatorial circumference of about 4 x 10^8 dm. = So 20 grams of Cellulose has a total polymer chain length sufficient to = circle the earth at the equator 960,000 times. Holy smoke. [i.e. 38.4 x = 10^13/ 4 x 10^8 =3D 9.6 x 10^5=3D 960,000] So the final step is to space a polymer length of 3.84 x 10^14 dm = equally within a volume of one dm^3. This would be difficult to solve = directly but iteration can quickly reach an approximate solution. =20 And for ease of approximation I will consider one strand to lie = against the side wall of the container so e.g. 100 strands would divide = the distance of 1 dm into 100 equal parts of 1 mm. These equally spaced = strands will run in the X, Y & Z directions i.e. from left to right, = bottom to top and front to back directions. Consequently the number of = strands available for any one direction is 3.84/3 x 10^14 i.e. 1.92 x = 10^14. Considering one direction, say top to bottom: if strands were 1 = micron (10^-5 dm) apart there would be 10^5 strands in each of 10^5 = panels for a total of 10^10 strands; each strand being 1 dm long. This = leaves a residue of 1.92 x 10^4 dm to be distributed so a spacing of one = micron (10^-6 metres) is slightly too large.=20 To add another row and column would require 2 x 10^5 +1 strands so = if the residue were distributed equally to this additional row and = column then the strands would each be (1.92 x 10^4)/(2 x 10^5 +1)=3D = 19200/200001 =3D 0.096 dm long; too insignificant to consider..=20 =20 This requires a leap of faith but if a column of cubes, with edges = defined by hydrophilic polymers, can be considered have to capillary = properties then by calculating the 'capillary rise' the intensity with = which water is bound by this column could be estimated. A column of cubes that has edges of10^-6 metres=3D 10^-4 cm would = have a cross-sectional area 10^-8 cm^2. A capillary tube of this = cross-sectional area would have a radius of 0.00005642 cm.=20 Wetting angle of water on Cellulose is given in the above reference = as ranging from 20-30 degrees. Plugging the above into a general = equation for rise of liquids in capillary tubes: h (cm)=3D (2T Cos = wetting angle)/rdg where T is surface tension of 72.8 (dynes/cm), = wetting angle is 25 degrees, r is radius (cm), d is density g/cm^3 (1), = g is gravitational accelleration (980 cm/sec^2) leads to h=3D 2 x 72.8 x 0.906/(0.00005642 x 980)=3D 2385.8 cm; a very strong = affinity for water. Taking one atmosphere pressure to be 1020 cm water at room = temperature (drawing on memory) then it would take a pressure of about = 2.3 atmospheres to force water from this column into a saturated ceramic = plate that had very fine pores (to prevent air entry). Yt, Dave Webster, Kentville =20 ------=_NextPart_000_1DE4_01CFD448.DA0E0740 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META content=3D"text/html; charset=3Diso-8859-1" = http-equiv=3DContent-Type> <META name=3DGENERATOR content=3D"MSHTML 8.00.6001.23588"> <STYLE></STYLE> </HEAD> <BODY> <DIV>Dear All, =20 = Sept=20 18, 2014</DIV> <DIV> I have wondered for years how trees can whip = around in=20 the wind and normally spring back to their original position when = the wind=20 eases and I broached this on Naturens in 2003 (pasted below).=20 Meanwhile another piece of this puzzle has come to my = attention.</DIV> <DIV> </DIV> <DIV>START OF PASTE\\\\\\\\\\\\\\\\\\\\\\\</DIV> <DIV>Dear=20 All, &nb= sp; &nbs= p; =20 Feb 14, 2003<BR><snip></DIV> <DIV>Spruce trees normally <BR>have a remarkable ability to bend with = the wind=20 and then spring back. <BR><SNIP></DIV> <DIV>Is anyone on naturens familiar with the mechanisms that enable=20 <BR>elasticity (molecular or cellular) in living trees ? There will=20 <BR>presumably be some reversible water movement, between living cells,=20 <BR>walls and intercellular spaces but there must also be some give and = take=20 <BR>at the molecular level, e.g. distortion and recovery of = normally=20 folded <BR>or coiled molecules in the cell wall or changes in water of=20 hydration; <BR>sort of an intramolecular shock absorber. Is lignin = involved=20 ?</DIV> <DIV>END OF PASTE\\\\\\\\\\\\\\\\\\\\\\\\\\\</DIV> <DIV> NOTICE: The following contains = numerous=20 calculations. A copy editor prize goes to the reader who=20 locates the greatest number=20 of calculating/transcription errors.</DIV> <DIV> </DIV> <DIV> Driwater, a commercial product, is 2% Cellulose, = 98%=20 water and is solid at room temperature. I can = not directly understand=20 how this is possible but one must assume that water in the vicinity of=20 Cellulose is no longer a free liquid and water sufficiently far = away is a=20 liquid. So there is apparently a gradual change in state with distance = from=20 Cellulose, as opposed to the sharp transitions portrayed in = temperature/pressure=20 state diagrams. </DIV> <DIV> </DIV> <DIV> Also, while browsing through old copies of = American=20 Scientist, I came across an article on Colloidal Macromolecular = Phenomena=20 (Battista, O. A, 1965(20):151-173). Purified mcrofibrils of Cellulose = (as in=20 wood pulp) are schematically represented as very long chains of = Cellulose=20 variously associated with short portions of three to seven other long = chains to=20 form polymer microcrystals which are connected to other microcrystals=20 by hinges of free (unassociated) lengths of Cellulose = (chain=20 length variable within hinges). With this kind of structure, Cellulose=20 microfibrils can assume a range of longitudinal or lateral=20 positions without rupture of chemical bonds. </DIV> <DIV> </DIV> <DIV> In a vague way this 'explains' how living or = steamed=20 wood can bend. It is possible that the tendency of living = trees to=20 return to their initial shape when wind force abates is driven by=20 elasticity of cell walls variously distorted into intercellular = spaces=20 (when compressed) or withdrawn from intercellular spaces when = extended.</DIV> <DIV> </DIV> <DIV> </DIV> <DIV> Returning to the example of Driwater (2% = Cellulose in=20 water being a solid at room temperature) brings to mind the behavior of = water in=20 a porous medium such as soil. </DIV> <DIV>With decrease in effective pore size, water in soil = becomes=20 bound in the soil pores to an increasing degree. At = tensions=20 greater than pF 2 (200 cm of water) water in most soil profiles = does not=20 flow downward under the influence of gravity (i.e. it acts as though it = were=20 part of the solid phase). And as soil water content approaches = zero all=20 soils approach tensions of pF 7 (10^7 cm of water; or expressed another = way=20 100,000 metres of water tension).</DIV> <DIV> </DIV> <DIV>Back of envelope approximations:</DIV> <DIV> If total dissociation of Cellulose is assumed = (no=20 microcrystals) one can approximate the spacing between polymer chains as = follows:</DIV> <DIV> Assume the 2% solution implies 2 g Cellulose per = 100 mL=20 aqueous solution=3D </DIV> <DIV>20 g Cellulose/dm^3. </DIV> <DIV> Glucose (C6H12O6) has a molecular weight of = 180.16;=20 call this 180 and water is 18.</DIV> <DIV>So the C6H10O5 units of Cellulose would have a molecular weight of = ~180-18=3D=20 162</DIV> <DIV> According to </DIV> <DIV><A=20 href=3D"http://books.google.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&= dq=3Ddimensions+of+cellulose+molecule+units&hl=3Den&sa=3DX&ei= =3DoEsbVNGCEoOxyASTsYCgBQ&ved=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimen= sions%20of%20cellulose%20molecule%20units&f=3Dfalse">http://books.goo= gle.ca/books?id=3DNhwSMryGzY0C&pg=3DPA55&dq=3Ddimensions+of+cellu= lose+molecule+units&hl=3Den&sa=3DX&ei=3DoEsbVNGCEoOxyASTsYCgB= Q&ved=3D0CC0Q6AEwAQ#v=3Donepage&q=3Ddimensions%20of%20cellulose%2= 0molecule%20units&f=3Dfalse</A></DIV> <DIV>Fig. 3.4 the length of two units, each being C6H10O5), is = 1.038 x=20 10^-9 M</DIV> <DIV> or unit length of 0.519 x 10^-9 M</DIV> <DIV> 20/162 (Moles of units) X 6 x 10^23 = Avogadro's=20 number (units/mole) x 0.519 x 10^-9 M (length per unit)=3D 0.384 x 10^14 = M or 3.84=20 x 10^14 dm (total polymer length).</DIV> <DIV> </DIV> <DIV> To put this in context, the equatorial radius is = 6,378,388 metres which is equivalent to an equatorial circumference of = about 4 x=20 10^8 dm. So 20 grams of Cellulose has a total polymer chain length = sufficient to=20 circle the earth at the equator 960,000 times. Holy smoke. [i.e. 38.4 x = 10^13/ 4=20 x 10^8 =3D 9.6 x 10^5=3D 960,000]</DIV> <DIV> </DIV> <DIV> So the final step is to space a = polymer length=20 of 3.84 x 10^14 dm equally within a volume of one dm^3. This would be = difficult=20 to solve directly but iteration can quickly reach an approximate = solution.</DIV> <DIV> </DIV> <DIV> And for ease of approximation I will consider = one strand=20 to lie against the side wall of the container so e.g. 100 strands would = divide=20 the distance of 1 dm into 100 equal parts of 1 mm. These equally = spaced=20 strands will run in the <EM>X, Y </EM>& <EM>Z </EM>directions = i.e. from=20 left to right, bottom to top and front to back directions. Consequently = the=20 number of strands available for any one direction is 3.84/3 x 10^14 i.e. = 1.92 x=20 10^14.</DIV> <DIV> </DIV> <DIV> Considering one direction, say top to bottom: if = strands=20 were 1 micron (10^-5 dm) apart there would be 10^5 strands in each of = 10^5=20 panels for a total of 10^10 strands; each strand being 1 dm = long. This=20 leaves a residue of 1.92 x 10^4 dm to be distributed so a spacing of one = micron=20 (10^-6 metres) is slightly too large. </DIV> <DIV> </DIV> <DIV> To add another row and column would require 2 x = 10^5 +1=20 strands so if the residue were distributed equally to this additional = row and=20 column then the strands would each be (1.92 x 10^4)/(2 x 10^5 = +1)=3D=20 19200/200001 =3D 0.096 dm long; too insignificant to = consider.. </DIV> <DIV> </DIV> <DIV><EM> </EM>This requires a leap of faith but if a = column=20 of cubes, with edges defined by hydrophilic polymers, can be considered = have to=20 capillary properties then by calculating the 'capillary rise' the = intensity with=20 which water is bound by this column could be estimated.</DIV> <DIV> </DIV> <DIV> A column of cubes that has edges of10^-6 = metres=3D 10^-4=20 cm would have a cross-sectional area 10^-8 cm^2. A = capillary=20 tube of this cross-sectional area would have a radius of 0.00005642 cm. = </DIV> <DIV> </DIV> <DIV> Wetting angle of water on Cellulose is given in = the=20 above reference as ranging from 20-30 degrees. Plugging the above into a = general=20 equation for rise of liquids in capillary tubes: h (cm)=3D (2T Cos = wetting=20 angle)/rdg where T is surface tension of 72.8 (dynes/cm), wetting angle = is 25=20 degrees, r is radius (cm), d is density g/cm^3 (1), g is gravitational=20 accelleration (980 cm/sec^2) leads to</DIV> <DIV>h=3D 2 x 72.8 x 0.906/(0.00005642 x 980)=3D 2385.8 cm; a very = strong affinity=20 for water.</DIV> <DIV> </DIV> <DIV> Taking one atmosphere pressure to be 1020 cm = water at=20 room temperature (drawing on memory) then it would take a pressure of = about 2.3=20 atmospheres to force water from this column into a saturated ceramic = plate that=20 had very fine pores (to prevent air entry).</DIV> <DIV> </DIV> <DIV>Yt, Dave Webster, Kentville</DIV> <DIV> </DIV> <DIV> </DIV> <DIV> </DIV> <DIV> </DIV></BODY></HTML> ------=_NextPart_000_1DE4_01CFD448.DA0E0740--
next message in archive
no next message in thread
previous message in archive
Index of Subjects