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hot air heat
9 feb 1999
roger "g43"  writes of:

>...a 20 x 24 garage that i use for my woodworking hobby.  the roof faces
>south and i would like to build a hot air collector that will aid in

you might insulate the attic floor and reroof with transparent material,
but i guess that means using a fan to bring the heat down to the garage. 
why not add a sunspace on the south side instead, or cover the south wall
with something transparent and an air gap to make it a solar air heater? 
steve baer suggests making the air gap at least 1/15th of the height for
good natural convection. dampers might prevent reverse air flow at night.

and "barrie hopkinson"  writes:

>does anyone have any information/ideas about the practicality of using air
>as a media for solar heating panels, with a view to heating a holiday home
>whilst place is vacant.

again, an inexpensive lean-to sunspace on the south side with some usable 
floorspace seems practical. how warm can we make the holiday home if it's
a 3m metric r9 strawbale cube with no air leaks and no internal heat gain?

where i live in december the average outdoor temp is about 0 c and about
3.1 kwh/m^2-day of solar energy falls on a south wall. if the transparent
cover transmits 90% of that, a 3mx3m wall gains about 25 kwh/day. if the 
solar energy that flows into the air heater equals the heat energy that
flows out of the whole structure on an average day, and the cube has lots
of thermal mass inside with a constant internal temperature t (c), and the
air heater works for 6 hours a day with outlet temperature t + dt, and the
air heater glazing and passive plastic film one-way dampers have a thermal 
conductance of 6 w/m^2-c, and the dampers each have area a m^2, we have

    25 kwh = 6h(t+dt-0c)9m^2x6w/m^2-c    air heater day loss
          + 18h(t-0c)(9m^2-2a)/r9        south wall night loss
          + 18h(t-0c)2ax6w/m^2-c         damper night loss
          + 24h(t-0c)45m^2/r9            other 5 surfaces

so t = (25k-324dt)/(462+212a)            equation 1.

using one empirical chimney formula, the airflow volume is about
0.2asqrt(hxdt) m^3/s and the heatflow into the cube is about
250asqrt(h)dt^(3/2) watts when the air heater is working, so if h = 3m,

  dt = ((138t+212a)t/(433a))^(2/3)       equation 2.

what is the optimum damper area? large dampers mean lots of heatflow
and a low air heater temperature, so the loss to the outdoors is small,
but the nightime heat loss is large. small dampers make small heatflow
and a hot air heater and a large heat loss to the outdoors and a small
heat loss at night. so there must be a good size inbetween...

let's try finding the optimum area by solving equations 1 and 2. suppose
a = 0.25 m^2 and dt = 0 for starters. then we have t = 25k/504 = 49.56 c
and dt = (2.083(49.56))^(2/3) = 22 c. plugging 22 c back into (1) gives
t = 35.42, which makes dt = 17.59. after 4 more iterations, these settle
down to t = 37.76 and dt = 18.36 c, within 0.03 c. doing this for
a = 0.25, 0.3 and then 0.275 gives an optimal area a = 0.275 m^2, where
t = 38.19 c (101 f) and dt = 15.82 c, making the sunspace outlet air
temperature 54 c (129 f.) making the damper area about 3% of the wall
area works well in this case, but the performance is not very sensitive
to the choice of area.

for this to work well, the temperature difference between the air in
the cube and the thermal mass needs to be small when the air heater is
working, so the thermal mass needs to have lots of surface, on the order
of 10x the air heater glazing area. a dark mesh screen in the air heater
could improve the solar collection efficiency by keeping cooler cube air
(vs solar warmed air) near the cold glazing before it flows through the
mesh to be heated and returns to the cube. the mesh could also reduce
reradiation loss and increase absorptance and the absorber heat transfer
area, which lowers the absorber temperature and its radiation loss. 

the cube's thermal conductance is about 3mx3mx6/r9 = 6 w/c on a cloudy
day when the air heater is not working, so adding say 8 200 liter water
drums makes the time constant rc = 8x200kgx1.16wh/kg-c/6w/c = 309 hours, 
and it theoretically cools to 20 c when 20 = 0 + (38-0)exp(-t/309), or
t = -309ln(20/38) = 198 hours or 8.3 cloudy days in a row. wrapping 4
of the drums with a layer of 1.3 cm (1/2") hardware cloth around 5 cm
(2") of 2 cm (3/4") stone could increase the thermal mass surface...


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