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East/West facing roof

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Posted by Simon on May 12, 2004, 9:51 am
 
We want to install solar heating for our hot water system.  However
the pitch of our roof goes the wrong way - the ridge runs north to
south, so if the collector is mounted on the roof tiles, it will face
either East or West.

One alternative would be to have it mounted on the ground facing due
South, but we would then need to have tubes going from the collector
into the house to the tank.  A distance of about 8 metres outdoors.
There is a part of the garden that is pretty much out of the way so it
wouldn' cause too much problem there .

Any suggestions would be welcome.  In particular:

- Would heat loss in the tubes over that 8 metres be a signignificant
consideration even if they were buried into the ground for most of the
distance and well insulated?

- If we stick with having the collector on the roof, is East facing
preferable to West?  Our use throughout the day will be fairly
standard - mainly showers / baths in the morning, but in Winter this
will be before the sun is even up.  Then sometimes a bath in the
evening.

- Is there an approximate percentage difference in efficiency facing
either due East or due West compared with due south facing?  (The
pitch of the roof is approx 1/2)

Thanks in advance for your advice,
Simon.

Posted by Steve Spence on May 12, 2004, 10:36 am
 
depending on south facing wall space, and your latitude, have you thought
about vertical panels on the south wall?

--
Steve Spence
Renewable energy and sustainable living
http://www.green-trust.org
Discuss vegetable oil and biodiesel
powered diesels at
http://www.veggievan.org/discuss/



Posted by Gary on May 12, 2004, 9:23 pm
 Hi,

Simon wrote:

The tables below should help you determine how much of a hit you take
for the E or W orientation, as well as the somewhat low tilt angle.
I have assumed that you are located around 40 deg lat?

The columns shown are:
    Month of year
    Direct Normal solar radiation on 1 ft^2 for a day
        (a panel that tracks the sun would get this)
    Radiation on a panel for the tilt and azimuth shown
        (e.g. for a panel tilted up at 30dg, and facing S)
    Last colum is (Col 2/Col 3) - i.e. the fraction of direct
        normal radiation you are getting

    units are BTU/ft^2-day


Collector panel tilted 40 deg, and facing south
This is the best orientation if your latitude is around 40 deg.
(tilt is equal to local Lat is best)
Month  Typ Idn   Typ Icoltot
    1      2193      1823   0.83
    2      2600      2154   0.83
    3      2917      2389   0.82
    4      3095      2533   0.82
    5      3162      2659   0.84
    6      3181      2709   0.85
    7      3061      2619   0.86
    8      2914      2461   0.84
    9      2713      2289   0.84
   10      2429      2056   0.85
   11      2121      1775   0.84
   12      1974      1634   0.83

--------
This is the orientation you asked about:
Collector Tilted 30deg, and facing East
(get same results for facing west)

Month  Typ Idn   Typ Icoltot
    1      2193       898   0.41
    2      2600      1270   0.49
    3      2917      1692   0.58
    4      3095      2066   0.67
    5      3162      2298   0.73
    6      3181      2374   0.75
    7      3061      2275   0.74
    8      2914      2024   0.69
    9      2713      1644   0.61
   10      2429      1214   0.50
   11      2121       875   0.41
   12      1974       738   0.37

It looks like a well oriented panel gets about 84% of the possible
radiation throughout the year.  A panel facing East or West at 30 deg
tilt gets less than half of the possible radiation in the winter, and
about 70% in the summer.

The numbers come out of a math model, and are probably kind of +- 10%
ish, but should be OK for your purposes.


If you can't do as Steve suggested in the reply above, than this would
probably be preferable to the roof location.  Check to make sure you are
not going to be shaded by building or trees in the on ground location.
A sunchart from http://solardat.uoregon.edu/SunChartProgram.html  might
help with this.

A rough cut at the pipe heat loss:
    assuming you have an average temperature difference between
    the pipe and ground of 28 Kelvin
    and that you are pumping water through the pipes (i.e.
    collecting) for  6hrs a day,
    and assuming 24mm inch pipe insulation on a 18 mm pipe

    Loss rate = 2*Pi*(16m)(0.055w/m-K)(28K)/(ln(21/9)= 180 watts
    
    Qperday = 180w*6hr = 1kw-hr or 3400 BTU per day  of loss

if you have a heater that heats 60 gal of water a day from 50F to 120F
(sorry, I can't think in metric for very long :-), then its producing

    (60gal)(8.3lb/gal)(120F-50F)(1 BTU/F-lb) = 35000 BTU per day

So, your pipe loss would be about 10% of your daily production --
probably not a show stopper??
        


If you have a storage tank, I would think that your local weather might
be a better guide on whether to choose east or west -- e.g. is it
typically more cloudy in the afternoon?  The warmer ambient temperatures
   in the afternoon might reduce your collector losses a bit for a West
orientation.  I think West might be better with a storage tank to keep
the water warm overnight??


See tables above

Hope this helps :-)   Gary

Posted by nicksanspam on May 22, 2004, 2:37 pm
 

One pipe or two? NREL says the yearly average air (= deep ground) temp
in Helena is 44 F. Circulating 130 F water would make the pipe-ground
temp diff 48 K.


I'da guessed        2*Pi*(16m)(0.033W/m-K)(48K)/(ln(12/9)= 551 watts

It looks like you transposed ln(21..., and 0.055W/mK looks high, compared
to 0.033 for Styrofoam and 0.038 for fiberglass (from the 1998 Schaum's
Outline for Heat Transfer.)

Then again, soil has thermal resistance. The second edition of Kreider
and Rabl's Heating and Cooling of Buildings has an example on page 32:
    
    An uninsulated pipe with an outer surface temp of 100 C and a radius
    r = 15 cm is buried D = 30 cm deep in earth that has a conductivity
    k = 1.7 W/mK (which depends a lot on soil moisture content.) If the
    _surface_ temp of the earth is 20 C, what is the heat loss for a pipe
    with length L = 10 m?

They say use a shape factor S = 2PiL/(cosh^-1(D/r)) = 44.7 m in this case,
when L >> r and D < 3R, which makes the heat loss kSdT = 6474 W. If D > 3R,
S = 2PiL/ln(2D/r). I suppose it wouldn't be very hard to do this calc with
an insulated pipe.

For what it's worth, the local outdoor woodstove installers bury a piece of
4" black plastic flexible corrugated drain pipe 3' underground, then wrap 2
1" black plastic pipes with some 1/2" thick x 12" wide plastic foam, then
tape the foam wrap and slide the wrapped pipes through the 4" outer pipe.
If labor isn't free, this adds a lot to the cost of an outdoor woodstove.
Also, the water jacket inside the woodstove makes it burn inefficiently at
a low temperature, with lots of air pollution. Outdoor woodstoves are
exempt from EPA regs.

I've been thinking about collecting heat from standard PV panels mounted
horizontally, with a reflective wall to the north and above them and a poly
film duct filled with a few inches of water during the day and drained at
night (in wintertime.) When I think about the heat loss and complexity of
underground insulated pipe, it seems to me that a single garden hose might
work better, with the PVs near the ground and an unpressurized tank in
the basement. The hose might only contain water for part of a day, if we
pump water up into the duct in the morning, let it warm to a higher temp,
then pump more cool water up into the duct and allow the warmer water to
overflow back into the basement tank when the pump shuts off.

We might pump 2"/12"x8'x16'x8g/ft^3 = 171 gallons of 110 F water from the
bottom of a basement tank up into a duct that covers an 8'x16' 1 kW array
at 5 gpm over 34 minutes in the morning, let it collect 8 kW (27K Btu/h) of
sun and lose (120-30)128ft^2/R1 = 11.5K Btu/h (a net 15.5K Btu/h) when it's
30 F outdoors until it reaches 130 F 171x8(130-110)/15.5K = 1.8 hours later,
then pump up another 171 gallons of 110 F water, then let the 130 F water
drain down into the basement tank, and so on.

Nick


Posted by Gary on May 22, 2004, 10:43 pm
 Hi Nick,

nicksanspam@ece.villanova.edu wrote:

Opps!  -- I actually meant for the the 21mm to be (24mm insul thickness
+ 9mm pipe radius) = 33mm -- not sure how I managed to get 21 mm

Here is a thought on thick, DIY pipe insulation.
Take a sheet of the 2 inch thick closed cell foam that the hardware
stores sell, rip it into 3.5 inch wide strips, then dado a 3/4inch wide
by 3/8+ inch groove down the middle of each strip.  Put the pipe in the
groove on one strip, apply a bead of "Great Stuff" polyurethane foam
along each of the two flats, and then put another of the 3.5 inch strips
over it with some weights until the PU foam cures (just a few minutes)
(you could actually do this on installation in the trench if need be).
For the joints, use more of the PU foam.  You could cut the ends of each
3.5 inch strip at a 45 to increase the bond area (you could also go
around 90 deg bends this way -- perhaps with a piece of left over foam
board glued in place with more PU foam to make a gusset).
Since the pipe is not bonded to the foam board, they could move around a
bit under thermal expansion without consequence.
With even the worst of table saws you could turn out a mile of this
stuff in an hour.
If I remember correctly, this foam board is less than twenty dollars a
sheet, and you would get about 50 ft per panel, so cost is about 40
cents a foot plus a can of PU foam.

If you had an ingoing an outgoing pipe, you could cut the strips wider,
and make two grooves instead of one.

Some of this foam is used for marina floats, so I am guessing it would
hold up pretty well??
Not sure about the temperature capability of the foam?

The heat loss for the problem above, using the good delta T and k values
becomes:

 >  (was)       2*Pi*(16m)(0.033W/m-K)(48K)/(ln(12/9)= 551 watts
 > (thick foam) 2*Pi*(16m)(0.033W/m-K)(48K)/(ln(50/9)= 93 watts

Quite a difference -- does this work??
The insulation thickness is up by 13X, and heat loss is down by 6X.

I wonder if you put a bit of washed gravel in the trench, if you could
reduce the thermal coupling of the foam insulation with the ground, and
get some further improvement?


Still thinking about the "PV" water heater -- it might improve the
payback on the PV system by an order of magnitude :-)

Gary


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