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re: drum wall thermal storage questions.......................
8 apr 2002
george  wrote:

>thanks for the reply.

you are welcome.

>if i go with 24x36 (706 sq ft after subtracting walls
>and 42" x 12' heat storage area)

i wouldn't subtract them, esp. the heat storage area, since it is
warmer than the house and loses heat energy faster, per square foot.

>with 12" r48 sip walls and 8% window area, to stay 77.5f, the
>house would need 2891 btu/h in january and 10 drums for backup?

yes. with 8% of the floorspace as r4 windows and 15 cfm of air leaks, its
conductance might be 72ft^2/r4 = 18 btu/h-f for windows plus 888/48 = 18.5
for walls plus 864/48 = 18 for the ceiling plus 15 for air, a total of 70
btu/h-f, so it needs (77.5-36.2)70 = 2891 btu/h or 69.4k btu/day to stay
77.5 f in january...

electrical usage of 300 kwh/mo might supply 34.1k btu of that, leaving
35.3k/day from elsewhere, or 5x35.3k = 176.4k for 5 cloudy 36.2 f days,
like 176.4k/(450(125-85)) = 10 drums cooling from 125 to 85 f, assuming
the glazing that heats the drums loses no heat at night, ie the drums are
behind an insulated wall, with a low-thermal-mass sunspace and glazing
to the south of that, with air circulating between the sunspace and
drumspace during the day, and no airflow at night.

on an average day, 32 ft^2 of south windows plus 16/16/8 e/w/n with 50%
solar transmission would provide 0.5(32x1030+16x475+8x230) = 21.2k btu, 
so the house needs 35.3k-21.2k = 14.1k btu from other sources, eg lost
heat from the drums to the house. on an average day, with a ft^2 of r2
air heater glazing with 80% solar transmission and an r48 insulated wall
between the drums and the glazing (the sun does not shine on the drums),
they lose about 6h(125-36.2)a/r2 btu (day) and 18h(125-36.2)a/r50 (night)
and 14.1k btu to the house, so 0.8a1030 = (125-36.2)(6a/2+18a/50)+14.1k,
and a = 26.82 ft^2 min (gathering 22.1k btu/day of sun and losing 14.1k
to the house and 7.1k during the day and 0.9k at night.) you might use
32 ft^2 of air heater glazing, eg a 4' wide x 8' tall panel.  

>if i was to add 3% more window area for total of 11% (using all anderson
>low 8'w (using 8'w for wheelchair can fit) x 6'8" sliding glass door and
>2 sets of windows) how many drums would i need?

more, because the house would lose more heat on a cloudy day...

>would the glazing be okay vertical or should it be angled? 

vertical's fine for winter heating.

>i would like to use the storage as a solar chimney in summer too, 
>pulling air in from earthtubes.

nrel says nashville has an average outdoor temp of 79.3 f in july,
with an average daily min and max of 68.9 and 89.5 and humidity ratio
w = 0.0152, so pa = 29.921/(1+0.62198/w) = 0.714 "hg. at 100% rh, 79.3 f
air has ps = exp(17.863-9621/(460+79.3)) = 1.023 "hg, so the average rh
in july is 100pa/ps = 70%. keeping the house at, say, 75 f, requires 
(79.3-75)70+34.1k/24h = 1722 btu/h or 41.3k btu/day of sensible cooling. 

keeping the rh 50% at 75 f (w = 0.0094) and 15 cfm (900 ft^3/h at 0.075
lb/ft^3) requires about 900(0.075)(0.0152-0.0094)1000btu/lb = 395 btu/h
or 9.5k btu/day of latent cooling. 

>do you know of any builders in middle tn that can build with the sips?

no. again, you might look at http://www.sip.org(?)

>any experience or opinions on sun-liteŽ glazing material?

good stuff, like dynaglas, 51" wide x 12' (or custom length) clear
corrugated polycarbonate greenhouse roofing that overlaps on 4' centers,
with about 6 manufacturers. 

>now, for cooling, would an earthtube array of (4) 12" dia. pvc pipes running
>200' long (open ended outside) leading into (2) 12" dia pvc and into house.
>do you think this would be enough to cool the house on a hot day?

the yearly average temp in nashville is 59.1 f. this corresponds to
the deep earth temp. you might lower that some by putting the tubes
under a crawl space or 4" of stone, and keeping the ground moist.

a foot of 12" pvc pipe has pi ft^2 of surface, so 200x4' has 2513 ft^2.
the soil around them might have 10 btu-in/h-ft^2-f (r0.1) if it's moist,
and the pipe wall is r0.1, so the first inch of cylindrical soil shell
would add r0.1/pift^2 = r0.318, the second would add 0.273, then 0.239,
0.212, and 0.191, a total of r1.23, at least, so au = 2513/1.23 = 2038. 

with a heat capacity rate (airflow in cfm) ch and an entering air temp
thi = 75 f and a tho exit temp, 2117 btu/h of cooling requires ch(75-tho)
> 2117, approximately. an exchange effectiveness e = (thi-tho)/(thi-tco)
makes tho = 75-15.9e, so ch(75-(75-15.9e)) > 2117, ie e > 133/ch, but
e = 1-exp(-2038/ch) here, and 133/ch < 1-exp(-2038/ch) works with a mere
134 cfm: 133/134 = 0.9925 < 1-exp(-2038/134) = 0.99999975...

it looks like this heat exchanger is oversized, and one requiring more
airflow would be more economical. if ch = 500 cfm, 1-exp(-au/500)
> 133/500, so a = 1.23x155 = 190 ft^2. you might use 60' of 12" pipe.

but the ground warms up, and bugs and mold and varmints may get into the
pipe, so maybe it's better to forget the drums and just use a small earth
tube to dehumidify fresh air and use a 24' diameter x 4' deep swimming pool
with some foamboard around the edge and on top for heating and cooling.
that way, you could take advantage of better than average days and store
coolth for a month or so... 

you could also use this for heating, with a semicylindrical reflector
that extends 12' above the south wall of the pool and r10 sidewall and
floating foamboard. on an average jan day, the pool surface might collect
0.8x452ft^2x730 = 264k btu of overhead sun plus 0.9^3x12x24x1030 = 216k
of south wall sun, a total of 480k btu.

if it supplies 14.1k btu to the house on an average day and loses
6h(t-36.2)452ft^2/r1 (day) and 18h((t-36.2)452/r11 (night) from the top,
and 24h(t-36.2)301/r10 from the side and 24h(t-59.1)452/r10 from the bottom,
480k = 14.1k+4177(t-36.2)+1085(t-59.1), so 5262t = 681.2k, and t = 129 f.

with r10 foamboard on top and sides, the pool has a thermal conductance
of about 120 btu/h-f on a cloudy day. if 115.8k pounds of water supplies
176.4k btu/day to the house and cools from 129 f to tf over 5 cloudy
36.2 f days, (129-tf)115.8k = 176.4k + 5x24((129-tf)/2-36.2)120, roughly,
so t = 117 f after 5 days. with a 2-drum greywater heat exchanger, the
pool might also provide all the hot water needed for showers.

for a water-air heat exchanger, you might screw a 16'x24' piece of plastic
birdmesh or greenhouse shadecloth to the ceiling with 5x24' of 1x3 sleepers
and slip a 100' roll of 30" diameter greenhouse poly film duct ($62 from
griffin greenhouse supplies at (717) 656-0809) with 8 right angle folds
around diagonal strings between the shadecloth and the ceiling to make
328 ft^2 of surface with a slow-moving air plus radiation conductance of
about 2.5 btu/h-f-ft^2, ie 960 btu/h-f. with a ceiling fan, this should
provide 1471 btu/h of heat on a cloudy day with a 1471/960 = 1.5 f temp
difference :-) on an average july day, the pool and heat exchanger might
keep the room 72.6 f. 

>would i be better off having the array closed loop and using a fan to push
>air thru the loop to cool it instead of using solar chimney and having air
>being pulled into the top of the thermal storage area?

yes. 

nick

10 ddbt=79.3'daytime dry bulb temp (f)
20 nmin=68.9'average night min (f)
30 ndbt=(ddbt+nmin)/2'average night temp (f)
40 w=.0152'humidity ratio
50 pa=29.921/(1+.62198/w)'vapor pressure ("hg)
60 tdp=9621/(17.863-log(pa))-460'dew point (f)
70 twbe=(ndbt+tdp)/2'initial wet bulb estimate (f)
80 twb=9621/(22.47-log(ndbt+100*pa-twbe))-460'wet bulb calc
90 if abs(twbe-twb)>.001*twb then twbe=twb: goto 80'iterate to 0.1%
100 print tdp,nmin,twb,ndbt
110 v=6.8'windspeed (mph)
120 t=70.6'pond temp (f)
130 a=.002056*tdp+.7378'term in rad loss
140 qr=1.63e-09*((t+460)^4-a*(ndbt+460)^4)'rad loss (btu/h-ft^2)
150 qc=(.74+.3*v)*(t-ndbt)'convection loss (btu/h-ft^2)
160 b=3.01*(.74+.3*v)*((t+twb)/65-1)'term in evap loss
170 qe=b*(t-twb)-qc'evaporation loss (btu/h-ft^2)
180 q=qr+qc+qe'total loss (btu/h-ft^2)
190 a=452'pond area (ft^2)
200 h=6'flooding time (hours)
210 qn=a*h*q'night loss (btu)
220 print qr,qc,qe,qn
230 tr=72.6'room temp (f)
240 qnh=24*(79.3-tr)*70+34100!'daily cooling requirement (btu)
250 th=tr-qnh/24/960'water temp required for cooling (f)
260 print tr,th,qnh,qn

68.62067      68.9          70.1294       74.10001
12.62215     -9.730021      14.31797      46673.78
72.6          70.63143      45356.01      46673.78




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