This fall, it looks like PGW may finally shut off the gas for 40,000
non-paying customers in Philadelphia. What would they do for space heat
and showers and cooking?
An equilateral A-frame on a flat Phila roof with a $200 12'x16' Dynaglas
polycarbonate south wall might collect 0.9x16(6x620+12cos(30)1000)
= 203.2K Btu of sun and lose 6h(Tg-34)12x16/R1 = 1152Tg - 39.2K on
an average January day, where Tg (F) is the A-frame air temp. If it
collects Q Btu/day of useful heat, Tg = 210.4-Q/1152.
If a north wall reflects 0.9x16(6x620+0.9x12cos(30)1000) = 188.4K Btu
into a polyethylene film Tw (F) water duct on the floor and it loses
6h(Tw-Tg)1.5x12x16 = 1728(Tw-Tg), Q = 188.3K-1728(Tw-(210.4-Q/1152))
= 691(319-Tw). Tw = 100 F makes Q = 152K Btu (45 kWh for PV fans
per day, with Tg = 79 F. How can we distribute this heat?
If ridge R of the A-frame reflects direct sun rays S from the horizon
vertically down onto edge Q of a duct that extends 6' to the south from
base B of the north wall, the ridge needs a 45 degree north wall slope.
Two north slopes might join at J, like this, viewed in a fixed font:
R ---
U .
. 45
(xo,yo) J . . . . <-- S south -->
. |
y . . .
. beta 10.4' = 12'cos(30)
^ . . .
| L. |
| . . .
| .
| 45 . beta alpha
----O----------B--------------Q--------------N--> x
| 4.4' | 6' | (not to scale)
| 10.4' |
With a beta degree tilt at bottom B, reflected ray angle BJQ = beta,
so alpha = 180-2beta. If x=y along line OR, tan(beta) = xo/(xo-4.4),
so xo = 4.4tan(beta)/(tan(beta)-1) = 10.4tan(beta)/(tan(beta)+1), ie
beta = 67.96 degrees, which makes the lower slope L = 7.97' and the
upper slope U = 4.25'. We might make L = 8' and U = 4', with 4'x8'
double-foil polyisocyanurate boards and a low vertical wall at B.
This would reduce the heat losing surface and raise the solar collection
efficiency in times of direct sun, with 2 suns from B to Q, so we might
produce 2 kW of electrical power with 1 kW of standard PV panels under
that 6'x16' of duct. Polyethylene and water and glass have similar
refractive indices, so the ducts might transmit most of the sun to
the PVs. When I put a water-filled duct on a PV panel, the output
current decreased 6%, in a crude experiment. Can we make 3 suns?
A draindown EPDM bladder with an upper constraint (eg welded-wire fence)
from Q to N might collect diffuse sun, with warm water naturally
thermosyphoning up through the BQ duct when available.
Nick
non-paying customers in Philadelphia. What would they do for space heat
and showers and cooking?
An equilateral A-frame on a flat Phila roof with a $200 12'x16' Dynaglas
polycarbonate south wall might collect 0.9x16(6x620+12cos(30)1000)
= 203.2K Btu of sun and lose 6h(Tg-34)12x16/R1 = 1152Tg - 39.2K on
an average January day, where Tg (F) is the A-frame air temp. If it
collects Q Btu/day of useful heat, Tg = 210.4-Q/1152.
If a north wall reflects 0.9x16(6x620+0.9x12cos(30)1000) = 188.4K Btu
into a polyethylene film Tw (F) water duct on the floor and it loses
6h(Tw-Tg)1.5x12x16 = 1728(Tw-Tg), Q = 188.3K-1728(Tw-(210.4-Q/1152))
= 691(319-Tw). Tw = 100 F makes Q = 152K Btu (45 kWh for PV fans
per day, with Tg = 79 F. How can we distribute this heat?
If ridge R of the A-frame reflects direct sun rays S from the horizon
vertically down onto edge Q of a duct that extends 6' to the south from
base B of the north wall, the ridge needs a 45 degree north wall slope.
Two north slopes might join at J, like this, viewed in a fixed font:
R ---
U .
. 45
(xo,yo) J . . . . <-- S south -->
. |
y . . .
. beta 10.4' = 12'cos(30)
^ . . .
| L. |
| . . .
| .
| 45 . beta alpha
----O----------B--------------Q--------------N--> x
| 4.4' | 6' | (not to scale)
| 10.4' |
With a beta degree tilt at bottom B, reflected ray angle BJQ = beta,
so alpha = 180-2beta. If x=y along line OR, tan(beta) = xo/(xo-4.4),
so xo = 4.4tan(beta)/(tan(beta)-1) = 10.4tan(beta)/(tan(beta)+1), ie
beta = 67.96 degrees, which makes the lower slope L = 7.97' and the
upper slope U = 4.25'. We might make L = 8' and U = 4', with 4'x8'
double-foil polyisocyanurate boards and a low vertical wall at B.
This would reduce the heat losing surface and raise the solar collection
efficiency in times of direct sun, with 2 suns from B to Q, so we might
produce 2 kW of electrical power with 1 kW of standard PV panels under
that 6'x16' of duct. Polyethylene and water and glass have similar
refractive indices, so the ducts might transmit most of the sun to
the PVs. When I put a water-filled duct on a PV panel, the output
current decreased 6%, in a crude experiment. Can we make 3 suns?
A draindown EPDM bladder with an upper constraint (eg welded-wire fence)
from Q to N might collect diffuse sun, with warm water naturally
thermosyphoning up through the BQ duct when available.
Nick