Software
Houzz Logo Print
tom_n_6bzone

sunshed redo

17 years ago

A few of you will remember that I made a sunshed, which has its southern roof as polycarbonate twinwall angled down to about 3 feet from the ground. And that that sunshed is well insulated everywhere the polycarbonate is not. And that I use black water barrels and containers to line the northern wall to assist the inside not to freeze. And that I am successful growing greens like leaf lettices and chinese vegs etc past Christmas and I use it to start very early in Spring for harding off veggies.

I've been thinking about removing the T-1 siding on the wall under the southern roof and replacing it with polycarbonate to give me a little more sunlight and I could have a shelf under my main long shelf which is under the roof's polycarbanote. I'd have to remove the insulation on that southern wall. Will the light gain and extra heat gain from the sun on my barrels/containers be worth the refit? There would be more sunlight on my black rubber floor as well. Or will the heat loss taking out the siding/insulation negate the gain?

~tom in Western Md.

Comments (5)

  • 17 years ago
    last modified: 11 years ago

    If we assume that almost all of your heat is being lost through the twinwall, then if you increase your twinwall surface area by 2x, you are going to increase your heat loss by 2x. In a solar-heated building, this means that the difference between inside temperature and outside temperature will decrease by 1/2, or 50%. So if you were heating the inside 20F higher than the outside, then with the mod you will heat the inside only 10F higher than the outside. On the other hand, what if your greenhouse has massive air leaks? Then a lot of your energy may be going into heating up cold air, and the twinwall conversion won't make as big a difference in your inside temperature. Also, if the modification causes 20% more sun energy to enter the building, then your inside temperature will be 10F * (100% + 20%)= 10F * 1.2 = 12F higher than outside. If you want to know the math, please continue.

    This is how I would do it if I were an electrical engineer. (Wait a minute, I AM an electrical engineer! Whoa...) We start with the R-rating of the building. It's hard to understand R, but it's easy to understand 1/R, since it is the heat flow out of the building in BTUs per hour per degree F per square foot of surface area. So now we can write an equation for the BTUs required: BTU = (1/Rbldg) * sq.ft. * delta T, where delta T is the temperature difference from inside to outside. That works for a heated greenhouse, where you keep the temperature constant with a thermostat. What about a solar-heated building, where the BTUs are constant, because they are supplied by the sun, but the delta T is unknown? Now we have to solve for delta T: delta T = Rbldg * BTU/sq.ft. This makes sense. If Rbldg or BTUs go up, the temperature goes up. If square feet go up, the temperature goes down.

    How do you figure out the R-rating of your building? Let's make some simplifying assumptions, such as all of your heat is being lost through the twinwall and the insulation. None is being lost through the floor and through air leaks. If we don't simplify, we're going to be here all night. Now you need to take this on faith: 1/Rbldg = area1/R1 + area2/R2 + ..., where Rbldg is the R-rating of the building, and R1, R2, etc. are the R-ratings of various components, such as twinwall, insulated walls, etc. area1 is the square feet of surface area with an R-rating of R1, and so on.

    So let's say that your twinwall has an R-rating of 2, and an area of 100 sq.ft. Also, your insulated wall has an R-rating of 32, and an area of 100 sq.ft. Then 1/Rbldg = 100/2 + 100/32, which is roughly equal to 100/2, right? So this means that the R-rating of your building is almost completely controlled by the twinwall. The answer: 1/Rbldg = 53 BTUs per hour per degree F.

    Now let's say that you are a cheapskate like me and use R-8 insulation instead of R-32. Now 1/Rbldg = 100/2 + 100/8 = 500/8 = 125/2, which is very close to the previous result of 100/2. So the amount of heat that you saved by going to R-32 instead of R-8 is very small, because the twinwall is the mother of all heat leaks. The answer: 1/Rbldg = 62.5 BTUs per hour per degree F.

    Now let's say that you replace all of the R-32 insulation with twinwall. Now 1/Rbldg = 200/2 + 0/32 = 100. That's roughly double the 53 value from before. That means that Rbldg has gone down by about 50%: cold greenhouse.

    What if ALL of your energy is coming from the sun? Remember the equation for temperature rise: delta T = Rbldg * BTU/sq.ft. So if Rbldg is reduced by 50%, delta T = 0.5 * (original delta T). But what if the increased twinwall area causes 20% more sun energy to come in? That's BTU. So now delta T = 0.5 * 1.2 * (original delta T) = 0.6 * (original delta T). This equation for delta T is really handy for a solar building, since it lets you quickly estimate what's going to happen if you change different variables.

    Conclusion: if you double the twinwall area, then the solar energy had better double, or the inside temperature is going to be a lot colder than before. It is very unlikely that the solar energy is going to double.

    Remember, if you are losing a lot of heat through the floor or air leakage, this analysis gets much more complex, and the twinwall becomes less important in determining temperature rise.

  • 17 years ago
    last modified: 11 years ago

    Nice post, Lilydude. I agree with everything.

    Can you remind us, please, Tom, what is the surface area of the entire structure? What is the R-value of the solid walls? And what is the surface area of the solid wall that you are considering removing?

    Did you have problems with low sunlight last year? I remember your posts on temps and it looked to me like you had a pretty good thing going. Why mess with it?

  • 17 years ago
    last modified: 11 years ago

    Thank you Lilydude, that information is exactly what I can use. Your simple explanation works because I have no air leaks. It is tight where I have to run an exhaust fan during sunlight days for the sake of the plants.

    Hi SB. the shed is 8 x 12. R-value is T-1 siding plus R-19 insulation. The surface area of the south wall is only 3.4 feet high by 12 feet long.

    I have no complaints and yes, you are right, I am very happy with it. The extra shelf addition under my main 12 foot long shelf would give me more growing room. That is the main reason I would want to do this. After calculating with Lilydude's forumula, I think it would be more prudent to add the shelf, leave the wall, and put flourescents under the main shelf. I'll have to ponder it more.
    ~tom

  • 17 years ago
    last modified: 11 years ago

    Tom, I think your solution is the right one. That way, if you have extremely cold weather, you'll still have the higher R-rating, and you'll be able to save your plants by using a heater of some kind.

    Another solution is to have a moveable cover which goes over or under the twinwall, making the R-rating a lot higher at night. From an engineering point of view, this is a really nice solution, since the twinwall is leaking almost all of the heat. Too bad it's such a pain to design and install.

  • 17 years ago
    last modified: 11 years ago

    That's an inspiration lilydude!!! Moveable covers, but perhaps on the inside with insulation attached! aaaah, this is becoming exciting. Runners, perhaps like a sliding glass door on a shower. hmmm, or .... wow... this is good! Thanks so much,
    ~tom