Your calculator is not valid. It assumes raising the temp of a volume of air 30 degrees and calculates the energy to do that. The result is in BTU’s. You interpreted it to be BTU/hr. Did you look at the calculations you just did?
The author of the formula then tried to convert BTU’s to KW. This does not compute. One is energy, and one is power. Fail.
You need a formula that takes into account surface area of the shed, a temp difference and a R value. I can heat a 26 ft travel trailer and maintain it at 70 F with an outside temp of 30 F using two 1500 watt electrical heaters. That’s a 40 degree difference. Your 1.5 in your calculator assumes an insulated value yet comes up with 8 kw. Not accurate and actually the wrong units. Also, if you are having trouble heating your large bay, understandable.
Also, the OP only needs a 12 degree rise, not 30 as you used. I’ll stick with my claim he will be fine with two 1500 watt heaters. If we can agree on the thickness of the wood siding on the shed, I’ll give you a better formula.
No, I get it. There are unit factors that come into play that were fudged a bit to provide assumptions related to continuous losses (a shed will be a lossy building), and raising the temperature to a thaw point in a practical worst case scenario (0F overnight is very viable, if cold soaked - think cold spell with loss of power), how does op keep things with water in them from freezing, which was part of one of the original posts.
So one can ask how many Joules, or BTU are required to heat a mass of air from x to y. It isn’t instantaneous, so there needs to be some rate that it is supplied, Then you consider that in a leaky outbuilding, there are losses as you push power in, and once you achieve “steady state”. I took the calculation for BTU and arbitrarily added the /hr because it is an uninsukated structure with significant air turnover, so it’s as good as needing to fully replenish the volume of air at that heat rise level continuously. Heating the shed is barely better than heating the outdoors….
So thus we consider btu/hr, or j/s… thus watts. Long term losses get us the sustained type values, Wh or the like.
The issue is how do you get an easy to use calculation that gives some insights for what it takes to get a building to a temperature from a worst case, then keeps it there. A travel trailer, which notionally has sound and thermal insulation, if not multiple layers of walls and structure, is going to have a vastly different loss factor than an open building with high ceilings (major issue #1 compared to a travel trailer), losses everywhere, air leaks, high air turnover rates, etc. a shed with a high roof will have double or more the air volume compared to a travel trailer of similar dimensions, to begin with.
Inagree the units aren’t right, there is a time factor that needs to be considered. But practically speaking, lossy structures are lossy structures and need a LOT of energy to keep temperatures up, especially when cold soaked. My lossy, (similar to op) buildings and practical experiences, including using 1500w and 5kw resistive heaters validate that.
