Re: BTU NEEDS
farmerman wrote:I have a new sun porch that is 36'X8'X8.5'. It is super insulated with a double layer all around of R19 and a foam underlayment of R11. One side of the sun room are six large windows and a glass door.
How do I compute the BTU load so I can hook up a proper length of cast iron baseboard radiator. (I dont have it as a separate zone so if I make the radiator too long, itll turn the room to a sauna) SO, Ive got the approx BTU values per ft of radiator. Now what do I do for the room. PS it faces EAst.
I can just oversize the baseboards but dont want to
First, FM, you need to find out your Outdoor Winter Design Temperature for your city.
http://www.engineeringtoolbox.com/us-outdoor-design-temperature-humidity-d_296.html
Then, Indoor temperature minus OWDT = Delta T [Is there a delta sign on a computer keyboard?]
1. Square foot of insulated wall x U value [the reciprocal of the R factor; R20 = 0.05 U-value] x Delta T = Hourly heat loss for this part of the wall system
EXAMPLE: Columbus, Ohio minus 10; Indoor design temp 70 degrees = Delta T of 80 degrees
Area of wall - 36 x 8.5 = 306sf - [any windows; assume none]
Heat loss = 306sf x .0263 [u-value for R38] x 80 [Delta T] = 643.8 BTUs per hour
2. Repeat for any other wall components, windows, glass door, floor and ceiling.
3. Depending on how air tight you've made it, you'll have to add anywhere from 1/2 an air change per hour down to whatever. You can determine this with an air blower test but to my mind it's not worth it.
Since a heat loss is done for the coldest day one should encounter, the heating system should be fine for all the other days. I'd just go with 1/2 the volume of the room as the hourly air change.
I've forgotten how the air infiltration loss is calculated. Let me rack my brain a bit [or should I waterboard it as this seems to be in vogue] or do some checking.
You should/could add in the solar gain from the east facing windows, though for the OWDT day, it probably won't make a great deal of difference.