re: cordwood masonry SOLAR.KnowledgePublications.com

re: cordwood masonry
12 may 1997
a local sawmill sells 3 ft^3 bags of sawdust for 50 cents each, about half
the price of fiberglass insulation, cheaper by the truckload, no doubt...
a 12" sawdust or strawbale wall with 6" cement/sawdust walls inside and
outside might work better with only occasional cordwood ties. maybe the
inner and outer walls can be thinner than that, eg 3" thick, with wood ties
every 2', with some radial nails or strings through holes in the ends of 
the wood ties or some clay wattle and daub, with a "slenderness ratio" of
about 8, so the masonry does not buckle.

leslie gratton says the us r-value of 9 parts sand, 1 part wet sawdust, 1 part
lime and 1 part portland cement (was that the recipe?) is something like 0.16
ft^2-f-hr/btu per inch of thickness. two 6" cement/sawdust walls, ie a total
of 12 inches of this mortar would have an r-value of 12 x 0.16 = 1.92. not
a very large thermal resistance, about like a double pane window... 

if sawdust insulation has us r3.5 per inch of thickness, 12 inches of it has
r3.5 x 12 = 42. where do we get the sawdust? "wood chips" with some lime to
add insect and fire resistance? coarser wood chips may have a lower r-value
than finer sawdust, since air can move more freely though them. 

so far, the sawdust/mortar and the sawdust insulation are in series, like 2
electrical resistors, so the resistances add--about r44 together. 

dry softwood logs might have r1.3 per inch of thickness, lengthwise, so 24"
logs might have 1.3 x 24 = r31.2, depending on dryness and density. the ashrae
hof lists "hem-fir" as having a us r-value across the grain of 1.1-1.33 per
inch of thickness, for a square foot section, at 12% moisture content. they say
thermal conductivity increases linearly with density, and give this formula:

k (= 1/r) = 0.1791 + (0.001874 + 0.0005753m)rho/(1+0.01m),
            where rho is the density of moist wood in lb/ft^3
	    and m is the moisture content in %. 

it seems to me that logs should have more thermal conductivity and less
thermal resistance lengthwise, as with strawbales, where airflow and heatflow
are greater parallel to the straw... 

these wood thermal resistors are in parallel with the sawdust, and they are
better heat conductors, so they lower the thermal resistance of the wall.
since wood is a better heat conductor than sawdust, the more sawdust and less
wood we use in the wall, the higher the wall's r-value will be, no? (unless
it falls down :-) so a cordwood wall packed with wood will have a lower 
r-value than one built with, say, one log in every 2'x2' section.

suppose the wall is 60% wood and 40% cement/sawdust, ie looking from outdoors
to indoors, we see 60% 24" logs with an r-value of 31.2 and 40% cement and
sawdust with an r-value of 43.92. we have something like a simple equivalent
dc electrical circuit like this:

                 rwood = 31.2/0.6
         -------------www---------------
outdoors|                               |indoors
--------                                 ------
        |      rsawdust = 43.92/0.4     |
         -------------www---------------

the equivalent parallel resistance of two resistors r1 and r1 in parallel is 
rp = r1r2/(r1+r2) or 1/rp = 1/r1 + 1/r2... ie 1/rp = 0.6/31.2 + 0.4/43.92,
so rp = 1/0.0283 = 35.3.

from a different point of view: currents add, for resistors in parallel. the
heat that passes from indoors to outdoors is the sum of the heat that passes
through each resistor. a square foot section of wall, looking from outdoors to
indoors, will contain 60% wood and 40% mortar/sawdust. the wood has a thermal
conductance (1/r) of 0.6 ft^2/r31.2 = 0.0192 btu/hr-f, and the cement+sawdust
has a conductance of 0.4 ft^2/r43.92 = 0.0091. the sum is 0.0283, eg when it's
69 f outdoors and 70 f indoors, 0.0283 btu per hour of heat will flow out of
the wall. that makes the overall r-value of the wall 1/0.0283 = r35.3.

looks pretty good, but 6" of the end of each log is buried in mortar with a
low thermal resistance, which tends to "short out" the higher resistance of
that part of the log. the remaining 12' log has a resistance of only 1.3x12
= 15.6, so the equivalent circuit is really closer to this:

                rwood = 15.6/0.6
         -------------www---------------
outdoors|                               |indoors
--------                                 ------
        |      rsawdust = 43.92/0.4     |
         -------------www---------------
                   
and the sum of reciprocals is 0.6/15.6 + 0.4/43.92 = 0.0476, so the parallel
combination has an r-value of 1/0.0476 = 21. not bad, but with one 4" log in
every 2'x2' section, ie about 2% logs, which might also serve as nailers for
portable plywood forms, 2' high, the wall is 1/(0.02/15.6 + 0.98/43.92) =
r42.4, and we need 30 times less wood...

a 30" thick wall including 6" masonry walls with 60% wood might give

                rwood = 18x1.3/0.6
         -------------www---------------
outdoors|                               |indoors
--------                                 ------
        |      rsawdust = 30x3.5/0.4    |
         -------------www---------------

and the sum of reciprocals is 0.6/10.4 + 0.4/29.3 = 0.0295, so the parallel
combination has an r-value of 40. with one 4" log in every 2'x2' section, ie
about 2% logs, this might be an r98 wall... sand stores about 18 btu/f-ft^3,
so the 6" inside wall would store about 9 btu/f-ft^3, making the rc time
constant of the wall about 880 hours or 37 days, which might make for a good
solar house. 

a 30x40' one-story house with 1120 ft^2/r98 = 11 btu/hr-f of walls and, say,
48 ft^2/r3 = 16 btu/hr-f of windows, and 1400 ft^2/r100 = 14 btu/hr-f of
ceiling and a little air infiltration might have a total thermal conductance
of about 50 btu/hr-f, requiring about 24(70-30)50 = 48k btu/day to stay warm 
in the winter, which might come from 64 ft^2 of vertical solar glazing on a
low-thermal-mass, thermally-isolated sunspace, eg a plastic film sunspace 8'
tall x 12' deep x 16' long. the inside walls of the house would have about
1120 x 9 = 10k btu/f of thermal capacitance, for a natural rc time constant of
about 10k/50 = 200 hours. if the house were 75 f inside on a 30 f day, after 
5 days without sun the indoor temp would be about 30 + (75-30)exp(-5x24/200)
= 55 f. not bad. a concrete floor might raise this temperature.

a solar closet could also help, as well as provide hot water. keeping this
house at exactly 68 f for 5 days without sun requires about 240k btu of heat,
which might come from 10 55 gallon drums or 100 5 gallon plastic buckets of 
130 f water cooling to 80 f. 

how to make a vapor barrier inside the walls, and a weather barrier outside?
and how to make a foundation? dig a shallow trench, line it with 8' wide poly
film, build the wall on top, pull up the extra poly film to attach to the
outside wall, 2' above the ground, and berm some earth around it?

nick