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Home design and AC load calculation

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Posted by Brent Geery on June 26, 2003, 10:48 pm
 
Could someone point out how to go about predicting the cooling
capacity I'll need for a super insulated "cabin" in the So. Cal.
desert?

The structure will be about 450 sq. ft. one story structure, about
25'x20, using a slab-on-grade foundation, with a
double-wall/staggered-studs metal design, to give me the depth
necessary for the insulation, and act as a thermal break between the
outside and inside.  I will be using metal studs throughout, on 24"
centers (between the inner and outer wall, give effectively 12"
spacing.)  R-42 insulation in the walls, and R-84 in the attic.  The
roof will be metal with white baked finish metal "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' double
glazed low-e with argon fill, and one on the east side, 1'x2' double
glazed low-e with argon fill.  Extra special care will be given to
eliminating all breaks in the vapor barrier envelope.  The main entry
door will probably be fiberglass, with a foam core.  I am thinking of
having a double-entry door, to decrease the heat transfer through the
entryway.  Have I missed anything?

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

Originally, was going to go with a cool tower/thermal chimney combo
for cooling, but with the low cost of off-spec PV on a two-axis
tracker, I'm thinking of going with a conventional AC, and offsetting
the extra PV costs with lower construction costs (would cut about
1/3-1/2 the materials cost!) and much simpler construction 1 story
construction (good bonus, as it's being 100% owner-built).  I'm hoping
a 5000 BTU AC will be more than enough capacity, and only have to run
a fraction of the time, even on a record-setting 115-120 degree day.

Please help me understand how to calculate what I need, based on my
assumptions above, and using the data available at the above links.

--
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 Newby on June 27, 2003, 4:11 am
 
How many cubic feet of space do you need to cool?  How many people will
occupy the area?



Posted by Mark or Sue on June 27, 2003, 5:37 am
 
You need to calculate the gains from conduction, air infiltration, solar
radiation, and internal heat sources. dT is the difference between the
temperature you want inside, and that outside. Note that roof temperatures
can be way above air temp, and southern & western walls will be warmer than
eastern walls. So perhaps use a temp of 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 ambients get to 115. Have to guess the
summer earth temp for your floor, perhaps 70 degrees?

Floor conduction: perimeter in feet * 0.8 * dT (this may be a cooling
benefit, a negative dT, as the ground is cooling the floor). Insulate your
floor perimeter to help keep the earth below at a more stable and
comfortable temperature.
Wall conduction: (Area - window and door area) * dT / RValue  (do this for
each wall and add together)
Window, door, and roof conduction is the same, just use appropriate areas
and R Values and add to above

Air Infiltration: This one is complicated depending on how much air change
you want, the outside humidity, and whether you have a heat exchanger or
not. I'd WAG   0.4 * cu ft / 60 * 4.5 * 25

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.

Internal heat: Add all electrical power in the house used while you are
cooling -- the cooling fan, lights, appliances, computers, etc. Also count
people at 0.1 KW each. So add all of the kilowatts, and multiply by 3412 to
get BTUs generated.

All of the equations produce BTU, so add them all up to find a rough heat
gain. You need an A/C that is this big to keep you cool. There are HVAC book
much more detailed than this, but this should get you within about 25% if
you know the numbers. However, you'll have to guess with some of these
numbers, but you'll quickly see what matters and what does not. May give you
some ideas on where you need to improve or realocate money.

--
Mark
Kent, WA




Posted by Tim on June 27, 2003, 10:59 am
 

Brent Geery wrote:


An evaporative cooler will probably be fine in a desert location, depending
on the maximum wet bulb temperature for your region. I would estimate the
total cooling load for you situation, allowing for no shading, to be around
5400 btu, slightly less than 1/2 ton. I would go with the additional 600 btu
and get a 6000 unit since humidity probably won't be a major concern. This
will allow for slightly shorter run times. Don't forget to allow some way for
outside air infiltration, or humidity, mold and mildew, and allergen
concentrations will be a problem, as well as outgassing from building
materials. This has the potential for a nasty indoor environment in a hame as
tight as what you describe.


Posted by Brent Geery on June 27, 2003, 10:55 pm
 wrote:


How did you deduce the heat load of 5400 BTU?  At what indoor/outdoor
temp difference?  I'm looking at a worst-case scenario of say 70
indoors and 120 outdoors - a 50F difference.  I'd want the AC sized to
run about full-time under these worst-case conditions.  In more normal
summer situations, it will be more like 25-30F temp difference.

Regarding air exchange, I believe the standard approach is to use an
outside/inside air heat exchanger, or just use the vent setting on a
standard wall/window air conditioner during the cooler evening/night
hours.

--
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

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