1
$\begingroup$

I am trying to calculate the effect of insulation and weather conditions on a cottage. The cottage is made from a timber frame while the panels are of 4mm plywood. The weather conditions vary throughout the year between 0 and 30 Celsius. Cottage is around 25 m2 with 2.2 m height, consider as single room.

  • Question 1: What is the indoor temperature range if the insulation material used has an R-2.0?

  • Question 2: If we are to install a timber furnace, how much timber do we need to maintain the room at optimal conditions during the winter ? How long will it take to reach the optimal temperature ? (assume optimal is 25 Celsius)

Improve this question
$\endgroup$
  • 1
    $\begingroup$ Your 'Question 1' depends on the heat source in the cottage. If there is no source of heat, then the temperature is roughly the average of the outdoor temperature over the last ~10 hours. $\endgroup$ – Chris Mueller Mar 19 '15 at 12:47
  • $\begingroup$ To give the most accurate answer, we would need to know the specific location of the cottage, location and size of trees or other shade structures, type and pitch of roof, whether there is an attic or not, orientation of the cottage, number of windows on each wall, wall thickness, leakage rate, specific details about the windows (size, configuration, coating, number of panes, interstitial gas, frame material)... $\endgroup$ – Air Mar 19 '15 at 18:06
3
$\begingroup$

The indoor temperature in the unheated case will be just the low pass filtered outdoor temperature. In steady state, therefore, the indoor temperature will be the same as the outdoor - at first glance. However, sunlight heating one or two sides or coming in through a window can make a big difference. Basically, there are too many variables to give a reasonable prediction in the real-life case.

However, we can do the math on the parts you have told us. I think those values are highly suspect, since the overall effective insulation will be quite different from just what the walls are rated as, but I'll use your numbers anyway. This is less of a useful answer to your particular problem than to show how to analyze something like this if you did actually have all the pertinent information.

The common "R value" rating is how many °F rise needs to be across one square foot to transfer 1 BTU per hour. You say to approximate the room as a square box that has 25 m² floor and ceiling area, and 2.2 m wall height. The box therefore is 5 x 5 x 2.2 m in dimension. It's not clear whether the floor should count towards the heat loss. Let's ignore it for this analysis to simplify things. Each wall is (5 m)(2.2 m) = 11 m², and you have 4 of them, so 44 m² for the walls. That plus the 25 m² roof is a total area of 69 m² = 743 ft².

Now we have to find the temperature difference that is to be maintained across this surface. You say you want to maintain 25°C = 77°F (wow, that's wastefully warm!) inside, but that's only half the spec. We also need to know how cold it is on the outside of that wall, which you say is 0°C = 32°F worst case. That means the temperature difference is 45°F.

So now we've defined the problem as how much heat power does it take to maintain 45°F across 743 ft² of R-2 insulation. From the description of R-value above, you can see that each square foot will transfer 22.5 BTU/h.

(22.5 BTU / h ft²)(743 ft²) = 16.7 kBTU/hour

So, you want a heater that can put out 15,000-20,000 BTU/hour. This is a spec that should be directly given for any commercially made heater.

Improve this answer
$\endgroup$
  • $\begingroup$ The question says the low range of outdoor temperature is 0 °C, quite a bit warmer than your assumption. Also, if you're modeling the max heating load, you'll want to assume a clear night sky and account for radiative cooling. $\endgroup$ – Air Mar 19 '15 at 18:10
  • $\begingroup$ @Air: I somehow didn't see the 0C spec for coldest outside. I have updated the answer accordingly. Yes, there are lots of other factors that make this calculation pretty much useless, with radiative cooling only one of them. This "cabin" is probably not that air tight, so convective cooling is probably significant, especially with wind. Then things like windows, doors, and other openings will make a significant difference too, the roof is unlikely flat and the same R value, etc. I also ignored heat loss thru the floor, as I said. $\endgroup$ – Olin Lathrop Mar 19 '15 at 18:47
  • $\begingroup$ True; I specifically mentioned sky cooling because it seems the easiest to account for given the sparse information we have to work with and the one room box model. Way too many assumptions to make to properly address insolation or drafts, IMO, though I have never seen loss through the floor taken into account with such simple model. $\endgroup$ – Air Mar 19 '15 at 18:55
  • $\begingroup$ You and Chris appear to assume that the cottage cannot make a net solar gain wrt mean ambient. It should be easy to provide differential energy transfer with solar energy entering at shorter wavelengths and heatin the interior which reradiates at much longer wavelength. Add a window with wavelength bandpass characteristics (such as glass) and you have an energy trap. This is of course a standard "solar heater" arrangement. To what extent anything like this happens serendipitously in a given case is TBD. $\endgroup$ – Russell McMahon Mar 20 '15 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.