[Greenbuilding] Deep mesh air heater and store
nick pine
nick at early.com
Thu Mar 1 06:31:35 CST 2012
20 S=1.714E-09'Stefan-Boltzmann constant
30 DIM E(5,5)'combined emittance matrix
40 DATA 0,0,70,70,0,1,70,70,0,1,70,70
50 DATA 0.5,0.5,70,70,0.5,0.5,70,70
60 DATA 1,1,70,70
70 'above: solar absorption, IR absorption,
80 ' screen temp, air temp south of screen
90 TAUCUM=1'initialize cumulative layer transmission fraction
100 FOR I=0 TO 5'find each layer's solar absorption fraction
110 READ ASUN(I),AIR(I),TS(I),TE(I)
120 TAUCUML=TAUCUM
130 TAUCUM=TAUCUM*(1-ASUN(I))'cumulative layer transmission
140 ASUN(I)=ASUN(I)*TAUCUML'absorption fraction for layer I
150 NEXT I
160 FOR I=1 TO 4'find IR emittance fraction
170 FOR K=I+1 TO 5'for each pair of layers
180 E=AIR(I)
190 FOR J=I+1 TO K-1
200 E=E*(1-AIR(J))
210 NEXT J
220 E(I,K)=E*AIR(K)
230 E(K,I)=E(I,K)
240 NEXT K
250 NEXT I
260 TA=34'ambient temp (F)
270 SUN=.91^2*1000/6'sun through glazing (Btu/ft^2-h)
280 TR=70'air heater inlet temp (F)
290 CFM=1.2'airflow/ft^2
300 C=10'relaxation damping cap
310 FOR I=1 TO 499'relaxation iterations
320 FOR L=1 TO 5'mesh layer
330 GA=(TE(L-1)-TS(L))/(2/3+1/CFM)'layer air gain
340 GR=0'initialize rad gain
350 FOR K=1 TO 5
360 GR=GR+E(K,L)*S*((TS(K)+460)^4-(TS(L)+460)^4)
370 NEXT K
380 IF L=1 THEN GFLOW=(TA-TS(1))/1 ELSE GFLOW = 0
390 GFLOW=GFLOW+ASUN(L)*SUN+GA+GR'heatflow into layer
400 TS(L)=TS(L)+GFLOW/C'new layer temp
410 TE(L)=TE(L-1)-GA/CFM'air temp leaving layer
420 NEXT L
430 NEXT I
440 PRINT"998'";
450 FOR L=1 TO 4
460 PRINT INT(TS(L)+.5);INT(TE(L)+.5);
470 NEXT L
480 PRINT INT(TS(5)+.5);INT(TE(5)+.5)
490 EFF=100*6*(TE(5)-TR)*CFM*60/55/1000'solar collection efficiency (%)
500 LEFF=100*(1000-6*(TS(1)-TA))/1000'glazing loss-based efficiency
510 PRINT "999'";EFF;LEFF
T1 air T2 air Ts1 air Ts2 air Tb air
73 72 109 93 154 127 156 143 160 152
64.62651 76.43108
This looks good to me now, with 2 layers of IR greenhouse film and
2 layers of 50% black greenhouse shadecloth. A single layer of R1 film
with 70 F air indoors and 34 F air outdoors and no radiation loss would
have a collector efficiency of 100(1000-6h(70-34))/1000 = 78.4%.
I suspect the 2 efficiencies above don't match because I approximated
the heat capacity of a C cfm airstream as C Btu/h-F when it's really
closer to 1.08C, but I'm inclined to believe the 2nd efficiency,
which is nicely close to an ideal 78.4% with no radiation loss.
Here's the solar yard heat store I'm planning now:
12'
................... partial parts list
. . . . . . . . . .
. . . . 15 plastic 55 gallon drums
. . . . 2 10'x16' pieces of EPDM
. . . . 6 tons of 2" clean stone
. . . . 16 3'x4"-D PVC pipes
. . plan view . . 22.9' 2x12'x14' ft^2 wood pallets
. . . . 3x12'x14' ft^2 wire mesh
. . . . 2x12'x16' IR plastic film
. . . . 1x12'x16' 50% greenhouse shadecloth
. . . . 1x12'x16' weed barrier
. . . . . . . . . . 480 ft^2 R30 insulation
................... Grainger 10" fan
12'
................... .
. . / . .
. . 16'. east . <-- 2 12'x16' layers
. . / . elevation . of IR greenhouse film
. south . . . .
. elevation . 13.9' ......collar beam.... \
. . . . . . . . . . . . . . . . . . . . 16'
. . 8'. . . . \
. . . 6'. heat store . .
. . . . . 60.
.......................................................
|4.3'| 14' | 4.6'|
... 70 F air would inflate the space between the 2 layers of IR
greenhouse film and flow through 1/2" holes on a 14” grid in
the inner film into 3 N-S rows of 5 plastic 55 gallon drums in
a 12'x14'x6'H box which support 2 3'Wx4'Hx10'L water troughs
surrounded by 6 tons of rocks with about 500 ft^3 (4000 gallons)
of water, like this:
south south
^ ^
| 12' |
.................................
i insulation i
. . . . . . . . .
n n
. . D D D D D . .
s p p p p s
. . 4'x10'x3'H . . plan view
u u
. . EPDM tank . .
l p p p p l
. . D D D D D . . 14'
a p p p p a
. . 4'x10'x3'H . . east -->
t t
. . EPDM tank . .
i p p p p i p are 3'x4" pipes to
. . D D D D D . . connect the airspaces
o o above and below the drums.
. . . . . . . . .
n insulation n
.................................
south south
^ ^
| 6' |
.........................................
i insulation 8"
. . . . . . . . . . . --> down
n w | 4" w | 4" w | w f
. . D R U M . . 24"
s i a o i p o i p i o
. . 4'x10'x3'H . .
u r i f r a f r a r a 48"
. . EPDM tank . .
l e r s e l s e l e m
. . D R U M . . 24" 14'
a m g t m l t m l m b
. . 4'x10'x3'H . .
t e a o e e o e e e o 48"
. . EPDM tank . .
i s p n s t n s t s r
. . D R U M . . 24"
o h | e h | e h | h d
. . . . . . . . . . . --> down
n insulation 8"
.........................................
8" 4" 4" 5" 36" 4" 5" 2"
This would have about C = 3'x12'x14'x62.33 = 31.4K Btu/F of thermal
capacitance. Cooling it from 140 to 70 F would release (140-70)C
= 2.2 million Btu, equivalent to 2.2M/130K = 17 gallons of oil.
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/data/13739.SBF
says 620 Btu/ft^2 of sun falls on the ground and 1000 falls on south
walls on an average 30.4 F January day with a 37.9 high and a 30.4
average daytime temp near Philadelphia. On a clear day, 890 falls
on the ground and 1880 falls on south walls.
On an average day, a south wall with a 60 degree slope would receive
620cos60+1000sin60 = 1176 Btu/ft^2 of sun. This store could collect
0.8x1176x12'x16' = 180.6K Btu/day and lose 6h(70-34)12'x16'/R1 =
41.5K from the glazing, for a net gain of 139.1K Btu.
Storing 139K Btu over a 6 hour January solar collection day at a rate
of 139.1K/6h = 23.2K Btu/h with air cooling from 140 to 70 F requires
about 23.2K/(140-70) = 331 cfm of airflow, which could come from
a 26 watt Grainger 10" fan. If the entire stone bed warms from 70 to
140 F, it needs a 139.1K/(140-70) = 1987 Btu/F heat capacitance, eg
1987/0.16 = 12.4K lb of stone, ie 12.4K/3500 = 3.5 yd^3 or 100 ft^3
in 2 12x100ft^3/(12'x14')/2 = 4" layers above and below the drums.
A 12'x14'x6' tall 140 F box with 648 ft^2 of R30 surface and a 648/30
= 22 Btu/h-F thermal conductance would lose 24h(140-30)22 = 57K Btu
on an average 30 F January day near Phila, for a net gain of about
139.1K-57K = 82.1K, which should be plenty for hot water for showers
and trickle charging for cloudy days, since 5 cloudy days in a row
only happen 3% of the time, if cloudy days are like coin flips.
A yard furnace that heats a house on average days needs more glazing.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where
L = 0.2032m (8/12') is the length of the bed in the flow direction.
If Rhoair = 1.059 kg/m^3 and V = 331cfm/12'/14' = 2 fpm (0.01 m/s)
and G = 1.059V = 0.011 kg/m^2-s and Mu = 1.99E-5 Pa-s at 60 C (140 F)
with equivalent pebble diameter D = 23.5 mm (0.92"), then dP =
0.2032G^2/(1.059x0.0235)(21+1750x1.99E-5/(0.0235G)) = 0.16 Pascals,
ie 0.000063 "H20 :-)
SETP equation 3.16.16 says the volumetric heat transfer coefficient
Hv = 650(G/D)^0.7 = 357 W/m^3K, ie 357x8/12x12x14x0.3048^3 = 1134 W/K
for the whole bed, ie 2148 Btu/h-F, like 2.1 car radiators. The air-
stone temp diff would be 15K/2148 = 6 F. The bottom layer could use
more stone.
A $70 Lasko 2155A 3-speed reversible 16" window fan can move up to
2470 cfm with 90 watts. On a cloudy day, it could push lots of house
air into the bottom. At 0.1" H20, 25 Pa = 0.2032G^2/(1.059x0.0235)
(21+1750x1.99E-5/(0.0235G)) makes G = 0.35 = 1.059V, ie V = 0.33 m/s,
ie 65 fpm, ie 65x12'x14' = 10.8K cfm. At 2000 cfm, 2000/12'/14' = 12
fpm, ie V = 0.06 m/s, so G = 0.064 and Hv = 1311 W/K, ie 4159 W/K
for the whole bed, ie 7.9K Btu/h-F, like 8 car radiators.
Nick
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