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Posted by Brent Geery on June 27, 2003, 10:55 pm
 
On 26 Jun 2003 23:58:20 -0700, cosolar@yahoo.com (Eric) wrote:


Good point.  I'm going with a home-built tracker, so the cost is less
of a concern than if I was using a retail model.  We are talking about
$.70/watt for the off-spec panels, vs. $.50+ for on-spec ones.  I
have almost 12 Peak-Sun-Hours in the Summer at the building site, and
the PV starts looking like  a reasonable alterative vs the additional
cooling tower construction costs.


Sure, but the alternative cool tower/thermal chimney also raise the
construction costs a good percentage, on this size structure, and will
have times when the humidity won't allow adequate cooling to take
place.  I'm hoping that with the 10-12 PSH during the hottest parts of
the summer, I can offset most/all of the PV costs with the lower
construction costs.


BRENT - The Usenet typo king. :)

Fast Times At Ridgemont High Info http://www.FastTimesAtRidgemontHigh.org
Voted #87 - American Film Institute's Top 100 Funniest American Films

Posted by Nick Pine on June 28, 2003, 12:41 pm
 


You might do the same thing cheaper with 2 1/2" x 11 7/8" I-joists as
studs on 2' centers (Fingerle Lumber sells GP's WIs for $.45/ft)
and sawdust(?) or cellulose fill insulation with fewer voids.


...500ft^2/R84 = 6 Btu/h-F for the attic?


Say 20ft^2/R4 = 5 Btu/h-F? With (2(20+25)8-20)/42 = 17 for the walls...


With excruciating care and lots of well-taped plastic film you might end up
with 15 cfm of natural air leakage, another 15 Btu/h-F or so, making the
total thermal conductance about 6+5+17+15 = 43 Btu/h-F.


The energy savings are probably not worth the expense.


NREL says July is the warmest month, with min, 24h and max average
daily temps of 74.0, 88.9 and 103.9 F and humidity ratio w = 0.0071.
Pa = 29.921/(1+0.62198/w) = 0.338 "Hg, with a 9621/(17.863-ln(Pa))-460
= 48 F dew point. The deep ground temp is 67.7.


NREL's 30 year record high is 116.1... (120-70)43 = 2150 Btu/h, but
evaporative cooling should work well most of the time. You might have
a 4'x25' shallow pond filled with rocks along the north edge of the
house, with water close to the dew point. With an 800 Btu/h-F auto
radiator, you might have something like this:

       1/800  Ti  1/43
48 F ---www---*---www--- 120 F  
            <-- I              

I = (120-48)/0.0245 = 2938 Btu/h, so Ti = 48+2938/800 = 52 F :-)

Nick


Posted by Brent Geery on June 30, 2003, 6:41 pm
 Ok, I still have a of questions, and I'm unclear on a couple of
issues.  To recap, here is my situation:

The structure will be about 450 sq. ft. one story, about
25'x20, using a slab-on-grade foundation, with a
double-wall/staggered-metal studs design, to give me the depth
necessary for the insulation, and act as a thermal break between the
outside and inside.  R-42 insulation in the walls, and R-84 in the
attic.  The roof will be metal with white baked finish "cool
roof/double roof design, with an additional radiant barrier under the
rafters.  Attic ventilation will be full ridge.  Siding will be white
vinyl.  Windows being one each on the north and south side, 3'x3', and
one on the east side, 1'x2', all double glazed low-e with argon fill.
Extra special care will be given to eliminating all breaks in the
vapor barrier envelope.  The only "intentional" breaks will be for the
stove and shower vents.  The main entry door will probably be
fiberglass, with a foam core.

I'm looking to calculate the "worst case" scenario, with a desired
indoor air temp of 65F, with an outdoor air temp of 120F.  I think
this means a dT of 55F.  Humidity varies from 25-33% during the
hottest parts of the year.

The nearest solar data for my construction site is Dagget, CA,  the
links to the comma separated values spreadsheets are:
http://rredc.nrel.gov/solar/old_data/nsrdb/bluebook/data/23161.SBF
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/23161.txt

So you gave me the following calculations and estimations:

Temp of walls, roof, floor:  Estimate 130 for your ceiling, 120 for
south walls, 110 for west walls, and ambient (100?) for east walls.
May have to go even higher with all of these if your ambient get to
115, summer earth temp for your floor, perhaps 70 degrees?

So, With my worst case of 120F ambient, I'll then assume 155 roof, 145
west wall, 135 south wall, and 120 for east and north walls?

Floor conduction: perimeter in feet * 0.8 * dT

So that gives 90 (feet) * 0.8 * 5 (70 ground/65 inside) = 360 (BTU/h)

Wall/roof/door conduction:
Area - window and door area * dT / RValue

So that gives 500 (ft^2 roof) * 90F / 84 = 536 (BTU/h) +
160 (ft^2 west wall) * 80F / 42 = 305 (BTU/h) +
157 (ft^2 east wall - window) * 55F / 42 = 206 (BTU/h) +
191 (ft^2 south wall - window) * 70F / 42 = 318 (BTU/h) +
170 (ft^2 north wall - window) * 55F / 42 = 223 (BTU/h) +
21 (ft^2 entry door) * 55F / 7 = 165 (BTU/h) = 1753 (BTU/h) all total

Air Infiltration:
0.4 * cu ft / 60 * 4.5 * 25 (but may be as low as 12 or even less)

So that gives 3648 (ft^3) * 0.4 / 60 * 4.5 * 18(?) = 1970 (BTU/h)

Solar radiation:
For each south and west window, calculate area * shading
coefficient * 125. The shading coefficient depends on shade sources --
trees, blinds, and LowE or not. Perhaps use 0.2 if you're using slat
type blinds, 0.4 if not.

I don't have a clue how to use this calculation.  The north window
will be shaded almost all day by the roof overhang (is that the
shading you refer to?)  And what do I do about calculating the north
window?

Internal heat load:
Add all electrical power usage.  Also count people at 0.1 KW each.
kilowatts * 3412 to get BTUs generated.

So, assuming I'm consuming 1 KW/h during the daytime, that would
create a cooling load of 3412 BTU/h!?  ;(  That doesn't seem right to
me, for some reason.  That means almost 70% of the capacity of a
5000BTU AC?  Is that really correct?

So far, under my worst-case conditions, I'm at about 4100 BTUs/h plus
the windows and the electrical usage heat load, which I'm still a
little unclear on how to calculate.  Thanks again for the help.

--
BRENT - The Usenet typo king. :)

Fast Times At Ridgemont High Info http://www.FastTimesAtRidgemontHigh.org
Voted #87 - American Film Institute's Top 100 Funniest American Films

Posted by Mark or Sue on July 1, 2003, 5:22 am
 

Where are your window conduction losses? Do the same thing -- window area *
dT / RVal, but their R values are usually around 4 (look for a U value on
the windows you're thinking about and divide that number into 1 to get R
value).



Shading is anything that blocks light from coming directly into your
window -- overhangs, trees, blinds, anything. If it a total light blocking
shade, then radiation is 0. North windows don't get much solar radiation
except for early summer mornings and late summer days. So the time duration
is less, and the angle is oblique, but if you have no shade or blind on that
window you get solar heating through it. You also need to worry about this
if you have a water "reflecting pool" in front of your house (south side).
Basically, any direct light counts against you, reflected or direct. The
more it is filtered/blocked, the lower the shading coefficient. The
multiplier of 125 can get you quickly, so you want to watch the direct
radiation. Maybe 0.05 for the north window, 0.1 for the east, and 0 for the
south....I'm just guessing here.


Yep. All electrical power you use gets dissipated as heat. Get any major
power users outside if you can, or in unconditioned space (like a garage or
utility room you can close off). The compressor in an airconditioner should
be cooled by the outside air. But a fan in the house will give off heat as
it operates. Think low power here -- fluorescent lights, low power fans, no
large TV's, etc. If you have a window air conditioner, hopefully it is
designed to vent all of its heat outside! But a swamp cooler is probably
going to heat your inside air (but their fans are rather wimpy).


Sure. I'm no expert at this either, but something backed by math makes me
feel better than people's guesses. One other thing you could do, if you have
access to the property now, and that is to lay out some objects in the sun
and measure their temperature on a hot day. Use the same materials -- white
roof section, wall section, etc.

One other thing to consider is the thermal mass of your house. We didn't
take that into consideration, and if you have enough you may be able absorb
the days heat and get rid of it at night (but I don't know how cool you get
at night).


--
Mark
Kent, WA




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