A remote yard boiler

[nick pine]

A water-based version could have a layer of EPDM on the ground with
plastic pallets http://www.uline.com/Product/Detail/H-2094/Pallets/48-x-40-Stackable-Plastic-Pallet
over that and a foot of stone over that and more plastic pallets over
that and 2 24'Lx4'Wx3'H EPDM water troughs over that.

The troughs could be plumbed in series, with 80 F water returning from
a small house tank in the coolest and 140 F solar heated water in the
warmest. To solar heat water, heat the stones with the poly film
pillow, then flood the hot stones with cool water and pump hot water
back into the tank.

When the house tank needs heat, let it flow back to the yard boiler by
gravity, then pump hot water back up into the house tank, via a single
uninsulated bidirectional pipe that's empty most of the time.
http://www.calctool.org/CALC/eng/civil/hazen-williams_g says a
100'x1.5" pipe with a 10' head would have a 35 gpm flow.

Nick

[nick pine]

Carroll Hampleman <trackthesun@...> wrote:

> I use the "HAMP METHOD" of "Going off grid". I use reflecting Sq. Ft. MIRRORS...

We could solar heat the Thomas Paine UU Fellowship in Collegeville PA
with the HAMP METHOD. Make 1000 signal mirrors http://www.dougritter.com/psp_rescueflash.htm
with $1.67 1 ft^2 mirror tiles from Home Depot
http://www.homedepot.com/h_d1/N-5yc1v/R-202300825/h_d2/ProductDisplay?catalogId=10053&langId=-1&keyword=mirror
tiles&storeId=10051 and train 1000 children to stand out in the snow
and aim the mirrors into church windows on Sunday mornings.

A higher-tech system might use heliographs...

Sir Henry Christopher Mance (1840–1926), of British Army Signal Corps,
developed the first apparatus about 1869 while stationed at Karachi,
in the Bombay Presidency in British India. Mance was familiar with
heliotropes by their use for the Great India Survey. The Mance
Heliograph was operated easily by one man, and since it weighed about
seven pounds, the operator could readily carry the device and its
tripod.

Most heliographs were variants of the British army Mance Mark V
version (Fig.1). It used a mirror with a small unsilvered spot in the
centre. The sender aligned the heliograph to the target by looking at
the reflected target in the mirror and moving his head until the
target was hidden by the unsilvered spot. Keeping his head still, he
then adjusted the aiming rod so its cross wires bisected the target.
He then turned up the sighting vane, which covered the cross wires
with a diagram of a cross, and aligned the mirror with the tangent and
elevation screws so the small shadow that was the reflection of the
unsilvered spot hole was on the cross target. This indicated that the
sunbeam was pointing at the target.

http://en.wikipedia.org/wiki/Heliograph

This would foster a great sense of community, no?

The Pottstown UU Fellowship has fewer members and more contrarianism.
A system with 2 800 ft^2 layers of polyethylene greenhouse film over 2
layers of 50% black greenhouse shadecloth and polypropylene weed
barrier fabric over a shallow box with 20 tons of round rocks over 700
ft^3 of water in 3 4'x32'x3'-tall plywood tanks with folded EPDM
rubber liners and a 90 watt 1000 cfm fan and a pump that automatically
floods and drains the rock box back into the tanks at the end of the
day might work better for them.

Yesterday I bought 1000 pounds of 1.25" round river gravel
http://www.hkgroup.com/ProductCatalog/Product.aspx?ProductID=23&LocationID=4
for $20 from Architectual Stone in Stowe, PA.

This looks good, with 19 tons of 3/8" pebbles...

10 LEW=32'east-west bed length (ft)
20 LM=.3048*LEW'bed length (m)
30 WNS=8'north-south bed width (ft)
40 WM=.3048*WNS'bed width (m)
50 DBED=1'bed depth in direction of flow (ft)
60 L=.3048*DBED'bed depth (m)
70 WBED=4050*LEW*WNS*DBED/27'bed weight (lb)
80 TBED=WBED/2000'bed weight (tons)
90 CBED=.16*WBED'bed heat capacitance (Btu/F)
100 PRINT "900'Tbed (tons):";TBED,"Cbed (Btu/F):";CBED
110 AC=24*32'collector area (ft^2)
120 EC=.75'collector efficiency
130 SUN=1176'sun on collector (Btu/day)
140 HD=6'solar collection hours
150 SHEAT=EC*AC*SUN/HD'heatflow (Btu/h)
160 FI=CBED*(150-80)/SHEAT'flood interval (hours)
170 RHO=1.127'40 C air density (kg/m^3)
180 MU=.000019'air viscosity (Pa-s)
190 FPM=1.2'superficial air velocity
200 V=FPM/196.9'air velocity (m/s)
210 GV=V*RHO'mass velocity (kg/m^2-s)
220 D=.3048*.375/12'pebble diameter (m)
230 DP=L*GV^2/RHO/D*(21+1750*MU/GV/D)'bed pressure drop (Pa)
240 DPH20=INT(1000*DP/249.1+.5)/1000'bed pressure drop ("H20)
250 HV=650*(GV/D)^.7'volumetric conductance (W/m^3K)
260 GBED=1.895*HV*L*LM*WM'bed conductance (Btu/h-F)
270 DT=SHEAT/GBED'air temp diff (F)
280 PRINT "910'DP (inH20):";DPH20,"Gbed (Btu/h-F):";GBED
290 PRINT "920'Sheat (Btu/h):";SHEAT,"DT (F):";DT
300 PRINT "930'Flood interval (hours):";FI

Tbed (tons): 19.2 Cbed (Btu/F): 6144
DP (inH20): .003 Gbed (Btu/h-F): 7102.375
Sheat (Btu/h): 112896 DT (F): 15.89553
Flood interval (hours): 3.809524

Another 6 tons of sand (a 4" layer) would add about 10K Btu/h-F to the
bed conductance and raise the bed pressure drop to about 0.1" and
lower the air temp diff to about 6 F and double the flood interval.

Some passive air heaters on south chuch walls would reduce the
12'Wx32'Lx21'H size of the yard boiler.

Nick

[nick pine]

This seems to work better with Douglassville Quarry's
1/4" anti-skid stone, which PA's DOT spreads on roads
to avoid skids... 77% of it passes a #4 4.75 mm mesh
and 19% passes a #8 2.36 mm mesh, for an average size
of about 3.56 mm (about 1/8".) It has a 2.687 specific
gravity and a 90 lb/ft^3 settled density and it costs
$12/ton... 15 tons could live in a 16'x8'x2.6' deep bed
above the EPDM tanks.

10 LEW=16'east-west bed length (ft)

20 LM=.3048*LEW'bed length (m)
30 WNS=8'north-south bed width (ft)
40 WM=.3048*WNS'bed width (m)

50 DBED=2.6'bed depth in direction of flow (ft)

60 L=.3048*DBED'bed depth (m)

70 WBED=90*LEW*WNS*DBED'bed weight (lb)

80 TBED=WBED/2000'bed weight (tons)
90 CBED=.16*WBED'bed heat capacitance (Btu/F)
100 PRINT "900'Tbed (tons):";TBED,"Cbed (Btu/F):";CBED
110 AC=24*32'collector area (ft^2)
120 EC=.75'collector efficiency
130 SUN=1176'sun on collector (Btu/day)
140 HD=6'solar collection hours
150 SHEAT=EC*AC*SUN/HD'heatflow (Btu/h)
160 FI=CBED*(150-80)/SHEAT'flood interval (hours)
170 RHO=1.127'40 C air density (kg/m^3)
180 MU=.000019'air viscosity (Pa-s)
190 FPM=1.2'superficial air velocity
200 V=FPM/196.9'air velocity (m/s)
210 GV=V*RHO'mass velocity (kg/m^2-s)

220 D=.00356'pebble diameter (m)

230 DP=L*GV^2/RHO/D*(21+1750*MU/GV/D)'bed pressure drop (Pa)
240 DPH20=INT(1000*DP/249.1+.5)/1000'bed pressure drop ("H20)
250 HV=650*(GV/D)^.7'volumetric conductance (W/m^3K)
260 GBED=1.895*HV*L*LM*WM'bed conductance (Btu/h-F)
270 DT=SHEAT/GBED'air temp diff (F)

275 PRINT "910'Sheat (Btu/h):";SHEAT,"Gbed (Btu/h-F):";GBED
280 PRINT "920'DP (inH20):";DPH20,"DT (F):";DT

300 PRINT "930'Flood interval (hours):";FI

Tbed (tons): 14.976 Cbed (Btu/F): 4792.32
Sheat (Btu/h): 112896 Gbed (Btu/h-F): 18388.12
DP (inH20): .052 DT (F): 6.139615
Flood interval (hours): 2.971428

Nick

[nick pine]

>A system with 2 800 ft^2 layers of polyethylene greenhouse film over
2 layers of 50% black greenhouse shadecloth and polypropylene weed

barrier fabric over a shallow box with 10 tons of round rocks over
700 ft^3 of water in plywood tanks might work for UUFP...

Oy, so many rocks! This would horrify Bucky!

A box on the lawn with an R2 cover with a 60 degree slope and 80%
solar transmission could have 3 4'x24'x3'-tall 140 F plywood tanks
with 10'x30' folded EPDM liners with 792 ft^2 of R50 surface that
loses 24h(140-30.4)792ft^2/R50 = 42K Btu/day and stores about
(140-80)864ft^3x62.33 = 3.2 million Btu of useful heat, about
the same as 32 gallons of oil burned at 80% efficiency.

If 1 ft^2 of cover gains 0.8x1122/6h = 150 Btu/h of average daily
sun and a trickle collector warms 80 F water returning from a church
heat exchanger and loses (80-34)1ft^2/R2 = 23 Btu/h for a net gain
of 150-23 = 127 Btu/ft^2 over 6 hours on an average January day,
a 24'x32' 768 ft^2 cover could provide 768x6hx127 = 585K Btu/day
of church heat.

A separate collector that heats 42K Btu of 140 F water would gain
150-(140-34)1ft^2/R2 = 97 Btu/h and require 42K/(97x6h) = 72 ft^2
of glazing, raising the total collector area to 768+72 = 840 ft^2,
eg a 24'x36' tilted south wall.

Alternatively, a dual trickle collector that heats 80 F and 140 F
water simultaneously with an 80 F section of black shadecloth
absorbing sun fraction f on top of a layer of poly film south of
a 140 F opaque deeper trickle collector would look like this:

34 F
----------------------- poly film |
air space |
----------------------- poly film |
air space v
----------------------- poly film 150 Btu/h
shadecloth fraction f 80 F <- 150f|
----------------------- IR poly film |
air space |
----------------------- poly film v
100% shadecloth 140 F <- 150(1-f)
----------------------- poly film

If each layer of poly film has 90% solar transmission and
the 80 F south shadecloth collects Q1 Btu/h of useful heat
with 150f + (140-80)1.5 + (34-80)1ft^2/R2 - Q1 = 0 and
the 140 F north shadecloth collects Q2 Btu/h of useful heat
with 150(1-f) + (80-140)1.5 - Q2 = 0, then Q1+Q2 = 127 Btu/h,
and 6hx127A = 585K+42K makes A = 828 ft^2, with shadecloth
fraction f = (585K/(6x828)-67)/150 = 0.34 and about 1% less
glazing than the 840 ft^2 individual collector total.

With 2 more layers of poly film, a dual collector seems less
economical than 2 single collectors, without accounting for
the lower radiation loss from the dual collector, compared
to 2 single collectors.

Nick

[nick pine]

>If 1 ft^2 of cover gains 0.8x1122/6h = 150 Btu/h of average daily
sun and a trickle collector warms 80 F water returning from a church
heat exchanger and loses (80-34)1ft^2/R2 = 23 Btu/h for a net gain
of 150-23 = 127 Btu/ft^2 over 6 hours on an average January day,
a 24'x32' 768 ft^2 cover could provide 768x6hx127 = 585K Btu/day
of church heat.

>A separate collector that heats 42K Btu of 140 F water would gain
150-(140-34)1ft^2/R2 = 97 Btu/h and require 42K/(97x6h) = 72 ft^2
of glazing, raising the total collector area to 768+72 = 840 ft^2,
eg a 24'x36' tilted south wall.

Alternatively, a single 24'x36' collector could make 42K Btu in
42K/(864ft^2x97) = 0.5 hours, then make 864x127x5.5h = 604K Btu
in the next 5.5 hours of a 6 hour solar collection day.

Nick