What kind of R-value for ferm chamber?
 

I got a lead on some discounted 4x8 sheets of 3" polystyrene insulation. It's got an R-value of 5 per inch. So it'd be R-15 at 3". Is that plenty? If I double layer it to R-30 will I see a big difference? It's possible I might want to keep my ferm chamber outside, so I'm wondering if the beefier level of insulation would make more sense? I'm planning to cool it with a mini-fridge. Will that work outisde? Any insight is appreciated. Thanks!

The higher the R-value the better. Depending on how hot it gets around you, you may want the beefier walls. It also depends on how big of a chamber you're making. I'm not an expert on ferm chambers or insulation for that matter, but thicker couldn't hurt, especially if its discount prices.

Hm, the email that I got sent had an equation in it. I liked the equation! :) I'm currently concerned about the possible need to have this outside. As far as size goes, I had planed 4' x 3' x 30" tall. Right now I'm thinking about building it on top of a pallet.

I had an equation, but changed it because I figured not everyone would understand it. (If you look at my reason for edit, it says "too complicated" haha)

The equation is: Q = (T_outside - T_chamber) * Surface Area/R-value

The only problem now is figuring out units. Temps are in Fahrenheit, Area in sq. ft., and I'm guessing the R-value is in °F*ft^2/BTU. So assuming outside temp of 75, chamber temp of 65 (?), surface area of 4*3*2.5=30 sq. ft., and an R-value of 15, your fridge would do at least 20 BTUs of cooling (energy usage would be higher due to efficiency of fridge). 20=(75-65)*30/15

The only thing I can't figure out is if the R-Value is already in those units... sorry I can't help more.

According to wikipedia, US R-value units are ft^2 * degrees F * per Hour all over Btus.

oh yeah, per hour. duh. from the pictures I've seen of other people's fermentation chambers, 3 inches will probably be fine.

Found a site, http://www.beananimal.com/articles/dorm-fridge-aquarium-chiller.aspx, that says "The average cooling capacity of a dorm fridge is somewhere around 150 BTU/hr"

Sooo....

Say I get up to 100 in the summer, its outside, and I want to ferment at 60. So temp dif is 40 degrees f.

Say I go 4' x 3' x 3' for round numbers. That's 75 ft^2.

Therefore at R-15 I have Q = 40 degrees F * 75 ft^2 / R-15, or Q = 200. Which would be BTUs per hour needed to keep temp, I think... So a mini fridge cooling an R15 insulated chamber at that size in the hot weather outside might not be able to keep up.

But, if I use R-30, then Q = 100 BTUs (?) needed per hour to keep temp.

I love math.

R-20 would put me right at 150 BTUs an hour..

All of that sounds right to me. It may be hard to get R-20, unless you can somehow cut the insulation in half and you'd get about R-22.5. That may be difficult tho.

Couldn't follow the 4'x3'x3' = 75 ft^2
I got 36 ft^3 but i think i know hat you meant

anyways use exterior surface area:
resulting in interior dimensions of 2'x2.5'x3.5'
assuming a rectangular box without a sloped top:
2(2.5'x3')+2(2.5'x4')+2(4'x3')= 59 sqft

Q= (1/R-value) x A x ΔT

where Q is the maximum possible flow of heat in BTU/hr for the enclosure
(1/R-value#) is conductance expressed in BTU/Hour*sqft*°F
A is the surface area of the enclosure in square feet
and ΔT is the change in temperature across the wall in °F

Q=(1/15)x59x40
Q=157.3 BTU/hr


using interior dimensions of 2.5'x3'x4
assuming a rectangular box without a sloped top:
2(3.5'x3')+2(3.5'x4.5')+2(4.5'x3')= 79.5 sqft
Q=(1/15)x79x40
Q=210.7 BTU/hr


but if you assume a lower ΔT on the base of the unit (say only 10°F)
Q=[(1/15)x63.25x40]+[(1/15)x15.75x10]
Q=179.2 BTU/hr