In Civil Engineering, I study the heat lost of building materials. So when I
built my first house I apply this science to my house.
R-Factor is the resistance of a material to heat lost. One R factor is derided
from the heat lost difference of one side of a 1/8 inch of glass to the other
side.
Let's say you build a structure all of 1/8 inch of glass which is a 10 foot
cube. The floor would be 10x10 = 100 sf, also the ceiling at 100 sf and the
walls at 100 sf. This is 6 x 100 sf = 600 sf of all external surfaces between
the outside ambient temperature.
Known: 1/8 of glass = 1 R factor
SF of external areas = 600 square feet
Temperature Difference (TD) in F. between the glass areas. Normally
we
use 100 degree temperature difference in a standard building
structure to
maintain a interior temperature at 70 F. with a exterior
temperature of
- 30 F below.
Conversion: We have to convert the R factor in a U Factor by dividing the R
factor
in to 1 or (1/R)
The Formula: Btur's = SF x U-factor x TD
For our 600 sf glass building example, this becomes
Btur's = 600 sf x (1/1-R factor) x 100 TD
= 600 sf x 1 U factor x 100 TD
= 60,000 Btur's
Using natural gas heat, the Btur's in one cubic foot of gas could be between
between 800 to 1200 Btur's depending on the gas. If the gas company more air
into the gas, the Btur's is less. They try to keep it at 1000 Btur's.
So if our gas btur's rating is 1000 Btur's then the amount of natural gas we
use is 60,000 btur's /1000 cubic feet of gas is 60 cubic of gas per hour. My
natural gas rate would be $0.10 per cubic foot. Our cubic foot of gas is rated
at about 12,000 btur's.
60,000 btur's / 12,000 btur's = 5 cubic foot per hour or about $0.50 per hour.
Lets say the ambient temperature was constant at 30 below zero, then the
monthly cost just for this 10 foot square room would be about $0.50 x 24 hrs x
30 days = $360.00 a month.
Using electric heat which my electricity cost $0.10 per 1000 watt/hr. One
thousand watts or 1kwr is equal to 3412 Btur's. So the amount of kw/hrs we
need for are sample glass room becomes 60,000 btur's /3412 Btur's = 17.58
kw/hrs.
The electricity cost per month becomes $0.10 x 17.58 kwrs x 24 x 30 = $5274.00
Using the above calculations for the amount of insulation you want to use to
reduce these energy cost. You also must have building structure knows to
calculated the total R-Factor of all structure items. Here are some of the
basic R-factors:
1 inch of dry wood = 1 R-Factor. A 2x4 of 3.5 inches is about 3.5 R-Factor.
Steel is about 0.5 R-Factor .
A steel nail driven into the 2x4 to hold on the exterior wall covering reduces
the 2x4 to 3.0 R-Factor. A steel sheet rock screw into the 2x4 reduces the
stud to 2.5 R.
If you use a 1/2 plywood to sheet the outside wall, then this adds another 0.5
R or the studs become 3 R's. Sheet rock on the inside adds another 0.5 R and
you are back up to 3.5 R's.
There are a lot of structures that are now using 5.5 and 7.5 inch studs, but
this is still only 5.5 and 7.5 R-Factor through the wood which conducts heat.
In the northern country, we recommend to break the heat conductance through any
exterior structures. Here is a example of my wall structure:
A 2 by 4 made out of Larch wood kiln dry, ship wrap and further air dry on
spacers for a year. The 2 x 4 space 16 inch on centers and only one plate on
top instead of the double plate. Open door and window box beam headers. The
entire outside sheeting using 1/2 inch Grade A 5 core exterior grade plywood is
industrial glue and a minimum of screws to hold in place while the glue is
drying.
There is a foam base strip rated at 25 lbs per square inch that the 2x4 is bolt
directly onto the concrete foundation wall. The 2x4 wall is set back 1/2 inch
so the 1/2 inch sheeting is flush with the surface of the concrete but is not
touching the concrete. Leave a 1/2 gap and fill with a building caulking to
break the conductance and to prevent outside air or inside air to go out all
these cracks.
Most important thing to do, before putting insulation in the walls, caulk all
the inside joints against all the exterior plywood and the 2x4. I leave a bead
of caulking about 1/8 high so the inside first layer of aluminum cover
insulation foam board does not touch the exterior plywood. 1 inch of this foam
board is 7.5 R-Factor.
The outside of the plywood is cover with 40 lb black water proof paper (not the
building wrap which is normally use for fiberglass insulation). This 40 lb
paper is overlap three times which prevents condensing and allows outside air
to circulated behind the next layers of sheeting.
A 2-inch thick Dow Corning Blue poly high dense 25 lbs per square inch rating
is screw on to the 5 core plywood, NOT THE 2X4 WALL STUDS with 3 inch building
screws through a 2 inch plastic foam washer which is sink below the surface of
the foam. These screw washers are than cover with a foam using a commercial
foam gun and trowel flat. This foam sheets is not butted tight together. Leave
a 1/4 inch space and insert foam into all these joints for a air tight surface.
This foam sheet adds 10 R-factor to the 7.5 R-factor to the first later of foil
cover internal sheeting. Note - Do not use this type of aluminum foil foam
sheet for the outside sheeting because then ends of the sheets are not cover.
I use a real wood siding 8 inches wide which has be dip into a sealant and
primer coat and into a color latex stain coating. This siding is also held on
with 4-1/2 building screws which only go into the plywood not the building
stud. No screws are exposed when installing this type of covering.
This wood siding and felt covering adds about 1-R factor which is now 18 R
Factor.
On the inside between the spaces of the 2 x 4 studs, we add two more 1 inch
sheets of the 7.5 R-Factor foil cover sheets. First caulk these foam sheets
that touches the 2 by 4 studs about a 1/8 high bead so the next layer of foam
is place against it. The R factor is now 18 + 7.5 + 75 = 35 R Factor.
Note: Every 3/4 inch of dead air space is = to 3.5 R Factor.
These foil cover foam sheets are now flush with the the surface of the wall.
The expose 2x4 studs are then cover with aluminum tape.
Before we continue to add more insulation to these walls, the ceiling is first
insulated. The ceiling rafters have a drop down ceiling joist. The exterior
walls are 10 feet high where the top board of the rafter is setting on the wall
which extends out another 2 feet where it is about 9 feet above the inside
floor level. The ceiling joist drops down into the room another foot for a 9
foot ceiling height.
This rafter type allows to have about 18 inches of insulation right at the
exterior walls. Most houses may only have 4 inches of insulation at this spot.
Foam air sheets are install for 10 feet in from the exterior walls to allow
air to circulated
from the soffit vents to the roof vents.
The first layer of drop down ceiling of 12 inches is a combination of
fiberglass and form blow in. Then there is two layers of 16 inch thick fiber
glass roll on into two overlapping direction. This is about 44 inches thick in
the middle of the ceiling but tapers down to 16 inches next to the wall. The
calculation R-Factor becomes about 110 R factor average.
When designing a rafter like these, I specified a catwalk to be place at about
3 foot above the wall height which is about 4 feet above the ceiling level, so
you do not have to walk on the ceiling rafters.
Now we are not done yet with the inside foam sheeting. The inside form
sheeting will now be 2 inch thick Dow Corning Blue foam which is design for
plaster. I did not use the plaster type which is now a combination of foam and
plaster where is spray on up to a 1.5 inch thick by contractors.
I first install a 1/2 inch plywood on the ceiling glue and screw to the ceiling
joist. Then install electrical conduit and 4 inch square junction boxes 2-1/8
inch deep with a 3/4 inch dry wall ring onto the plywood. Next install another
layer of 2 inch form on the ceiling only screwing this foam to the plywood, not
to the joist.
Now I can now continue to add two more 2 inch form sheets to the exterior
walls. The first sheet is screw to the wood studs with the foam washer screws.
Then we install the electrical conduit and boxes glue to this foam layer. And
finish up with the next foam layer.
A window and door openings also was foam cover and a 3/4 plywood layer cover
these sides opening. The plywood and door and window openings frames also had
a wood break where the window and door trim cover these joints.
So far we had 35 R-Factor and adding two more sheets of 2 inch poly form
becomes 55 R-factor in the none stud areas. In the stud areas this subtracts
3.5 factor in these areas.
The last covering which is 5/8 sheet rock which is fully glue using a 1/4
groove trowel adds about another 1/8 inch of air space which is estimate to 2.5
R Factor.
In designing the Btur's lost, you have to calculated each type of material
using the above formula and add all the results together.
To test out the Btur's lost for a building. Turn on the heating system and
note the amount of nature gas is use in one hour or electricity in one hour.
Then use the above formula noting TD temperature difference and the raise of
the interior temperature in one hour.
Roland
----- Original Message -----
From: tomw<mailto:[email protected]>
To: [email protected]<mailto:[email protected]>
Sent: Saturday, October 12, 2013 9:37 AM
Subject: Re: [EVDL] Another method of heating a EV
Hi Roland,
What insulation (type, thickness, installation method) are you using to
achieve these R values?
/"Before the EV would leave the garage, which the walls were insulated to 55
R-factor, the ceiling to 110 R-Factor... "/
I'm looking for the highest R/$ way to insulate my metal garage, and I'm
sure you looked into the best installation method to obtain an overall R
value close to the rating of the insulation.
--
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