Home design and AC load calculation

[Brent Geery]

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.

 

[Mark or Sue]

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.

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.

 

[Brent Geery]

not sure I would use off-spec modules with a tracker, due to the cost
of trackers, the space is at a premium. just something to look at.

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
$2.70/watt for the off-spec panels, vs. $4.50+ 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.

also, swamp coolers use a lot less electricity than ac.

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 Geery]

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.

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.

 

[Nick Pine]

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.

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 $1.45/ft)
and sawdust(?) or cellulose fill insulation with fewer voids.

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.

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

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.

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

Extra special care will be given to eliminating all breaks in the vapor
barrier envelope.

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.

...I am thinking of having a double-entry door, to decrease the heat
transfer through the entryway.

The energy savings are probably not worth the expense.

The nearest Solar data for my construction site is Dagget, CA...

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.

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... 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.

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

 

[Brent Geery]

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.

 

[Mark or Sue]

Ok, I still have a of questions, and I'm unclear on a couple of
issues. To recap, here is my situation:

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

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).

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?

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.

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?

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).

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.

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).