Cheap insulation would be Jablite, perhaps. How to calculate a useful thickness?
It's an old one, with concrete panels slotted into posts. I can get 25mm sheets between the posts, and possibly overlay with more sheets. The floor needs to be about 40mm thick, so I'm thinking 25mm Jablite there, overboarded. Perhaps fibreglass for the ceiling, over ply sheets.
What I'm after is a "how to calculate what power will be needed to keep the place warm when I'm in it" rather than a "Oh use 100mm Celotex throught" answer. Anyone?
Find a U-value calculator and put in all the values, to tell you how much heat loss there is and what it'll require to keep it up to your chosen temperature.
This is the MCS heat loss calculator spreadsheet:
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if you say your 'house' is one room with walls/floor/ceilings as your garage is constructed, it'll work out the heat input required. It knows the U values for a lot of different construction combinations. You'll have to tell it your location, etc so it can work out the climate locally.
Then tweak the construction to see how that affects the heat required.
There are online ready reckonners that you can feed in the dimensions and it will do a guesstimate of losses with and without insulation.
Assuming you only want it warm enough to work in rather than lounging around in a chair then 2" Celotex ought to be enough. My metal garage doors are insulated with 1" expanded polystyrene which makes a big difference overheating in summer. I don't heat my garage though.
I'd be tempted to put a thicker layer on the North facing wall...
ISTR for modest heating above ambient a crude rule of thumb is that +5C roughly doubles your heating costs (eg for greenhouses). AOTBE
Its worth looking for a local supplier of "seconds" grade PIR boards - since they are frequently no more expensive than jablite, and perform much better.
Work out your surface areas, decide on what your worst case acceptable rate of heat loss is, and then divide by your area, you get a target maximum heat loss per square meter.
(note I am over simplifying here by assuming all the surfaces are of uniform heat loss - see the example on the wiki for a spreadsheet based approach)
So if your area is 5.8 x 2.2 x 2 + 4.8 x 2.2 x 2 + 5.8 x 4.8 x 2 = c.
100 m^2, and you want a heat loss below 500W, then you have a maximum limit of 5W / m^2
If you say that your worst case temperature difference is 20 degrees, then you have a target U value of 5 x 20 = 0.25 W/m^2/k
From there you can work back from the known U values of your existing surfaces, and the k value of your chosen insulation materials to work out how much of them you will need.
(sometimes it's easier to take the reciprocal of the U value to get an R value (a measurement of thermal resistance rather than thermal conductance) then you can add the R values for the different layers together and take the reciprocal of the result to get a final U value. ).
I don't know how thick your concrete panels are - but say they are 50mm. Dense concrete has a (very high) k value of 1.4 so divide that by 0.050 to get a scary high U value of 28! So fair to say, they have effectively no insulation properties of their own, so you may as well ignore them!
So if 0.25 is you target U value pick a thickness of insulation that gives you that.
A polyurethane foam board has a k value of 0.025, so divide that by your target U value, and get a thickness of 0.1m
ok so let's gloss over the figure my very rough guestimate above gave then :-)
This will give you chapter and verse:
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So long as you heating power exceeds your worst case heat loss, then it should ultimately get to the temperature you want, however the next question is how quickly.
If you build a spreadsheet as in the example, you could also split down the source data to use computed U values based on the insulation's K value and the thickness. That would make playing with the thickness easy
(in fact you could use the spreadsheet's "goal seek" capability to find the right thickness for you from a specified outcome)
I don't know how prices compare (as I put in mine 15 years ago in my storage shed/railway room), but you can buy chipboard flooring panels with tongue and groove edges and insulation bonded to the back.
I'd have thought heat loss due it standing relatively exposed could be an issue? And drafts, depending on construction. Although quite how you factor that in I'm not sure . . .
Yup it has an effect certainly. It's partially factored in already, but not completely.
Having an exposed wall will tend to mean there is a larger temperature gradient across it and that will be reflected by a higher loss rate (the U value and surface area will be the same, but the Tdelta that those are multiplied by will be more).
Secondly, wind will result in more draughts, and hence more air changes per hour, which are also included in the clacs already. However assessing how much the rate of air change will be very much harder. (I expect you would have to do it empirically by measuring the difference in rate of temperature drop with the heating turned off on a calm day and again on a windy one - from there work out the relative difference in rate)
Lastly extra wind will result in faster convection cooling of the outer surfaces of the building, and possibly some "wind chill" if the outer surfaces are damp and hence can also support evaporative cooling as well.
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