OK guys, I'm trying to build one of those tiny houses. I want it to be as energy-independent as possible. (For prepper reasons, not global warming BS.) I want to go with solar-heated hot water for as much of the space heating as humanly possible.

So I need to determine 1) Heat loss to conduction through walls (I'm ignoring air infiltration until I tackle this)

Assuming I am making a modest 16x24 house, with R60 walls and R100 ceilings and floor, are these correct?

Energy (in Watts) = DeltaT (internal goal temp - cold night temp in C) * surface area (in sq m) / R-value

Thus:

W= (23- (-1)= 24) * 59.4579 / 60

W = 23.78/hr

W = (23- (-1) = 24) * 71.3495 / 100

W = 17.12/hr

40.9 W/hr, .981 kWh/ day

Assuming these are correct, at what temperature would 300 gallons of heated water need to be to warm the internal 3072 cu ft of air to 70*F? That's where my math failed me. I got 1126 Kelvin.

Any help would be 'preciated.

Don't make it so tight you suffocate, or have humidity issues.

Water at that temperature would require one hell of a pressure vessel.

I intend on forced ventilation with a fan, and a heat recovery ventilator. I agree, I don't want mold issues.

One thing that screwed me up is that R-values are not unitless, and are usually quoted in Imperial, not metric units in the US.  So be careful with your area calculations, or convert your R-values over.

Otherwise, yes, the R value gives the total thermal resistance of the wall.  However, keep in mind that the convection conditions on the interior/exterior also provide some, and perhaps significant thermal resistance.  Thus your calculation could be Watts=area*deltaT/(R_wallcond+R_outsideconv+R_insideconv).

Use U values and not R-values.
U= 1/R and makes a whole lot of the calculations easier.

The units for 'R' are BTU/(h °F ft2)

The necessary temperature of the water depends on the physical shape of the water bath, and how you intend to allow it to conduct/convect heat to your building.  Would you have a fan running inside all the time to move air across the water?

Just use BTUs.
1 BTU is the heat to change 1 pound of water by 1 degree F

You need delta T over the radiator to calculate the delivered heat.
Many are rated for 180 F water in to deliver X BTU/hour.

You can use whatever the heck you want as long as the units are consistent.  Besides, if he's looking at series resistance of a wall including convection resistance, using R values is easier because they can be added.

have you done the heat loss/ gain calculations on the proposed home yet? you way over killed the insulation factor to the point where you will have mold issues. the money spent on framing and  insulating your home to that extreme would require a very long payback too boot not to mention aggravate the humidity level in the home. keep in mind that you may not even see the Sun in January and February so it would be difficult to heat 300 gallons of water to be used to heating

Yeah, when I say R60 I mean TRUE R60. Not R60 cellulose with framing losses, but a cumulative R60 wall.

Yeah, I know it will be ~2 feet thick.

Crap.

Perhaps I was being optimistic, but I'm also assuming the efficiency of the water bath to be 100%, for the reason that once heat gets inside it must heat up the tinyhouse. I could be wrong here.

So just convert BTUs to Watts?

I'm actually using 110*F as a baseline for my heat.
Use too high, and it will never run in Ohio winters. Use too low, and it will be taking in cold water.

Oh believe me, I'm gonna spray foam the f&*( out of it.

Except when they are not just stacked up.
You have studs or framing in that wall.
It 'short circuits' the insulation.

U values make the calculations easier.

If you ONLY have parallel conduction, 1/R is easier.  But when you have both series and parallel conduction, it makes no difference at all which one you use.  Frankly, it makes a piddling difference even in the extreme cases.  It is, however, standard thermal analysis technique to use R.  Do you use 1/R when solving for current in an electrical circuit with parallel loads?

Except when they are in parallel.
Like the studs on each side of the cavity.

That's what I'm trying to do here.

I know the insulation is extreme, that's what I'm shooting for, and why I chose cellulose. It absorbs and redistributes moisture throughout, without allowing any moisture to condense on the framing. I will use forced (negative pressure) ventilation to ensure that no air goes from hot -> cold, which is the main reason for water vapor being trapped in a wall.

Yeah, the payback will likely take decades - but I don't care. This is a prepper house, intended for if fossil fuels or electricity are no longer available. It needs to be as self-sufficient as possible, which is which I'm going to the extremely small size and R-value. The cost is not an issue, as much as the desire for a house with almost no dependence on outside fuel/electricity sources.

I think the humidity would also be kept in check by continuous ventilation, going through a HRV to recover lost heat.

Ohio gets approx 2.5kW/m^2 sunlight in Dec/Jan, for approx 4 hours/day. Assuming a collector efficiency of ~18%, I think ~200ft sq of collector could theoretically provide 100% of space heating needs. I'm just trying to make sure I'm right before I buy spend ~$40k on a 384 square foot cabin and realize I was wrong.

You are going to need a large thermal mass to store the heat.

A couple days in a row of clouds and you better have a decent backup source.

I intend to insulate the water tank to ~R60.

If I do need a backup, I will have a propane/wood burner (have not decided which, yet)

http://www.dickinsonmarine.com/dheaters.php

Just in case your solar dreams don't come true.

If my solar theory doesn't work, I'm lookin' for a woodburning fireplace. Or potentially a propane heater, if I can store several years' worth.

I do like your idea, though. My Father's house runs off of diesel as a backup. He has one of those outdoor woodburners that pipe heated water to your floors.

I guess I was born to be kind of independent?

Just apply the One is None rule and you should be good to go.

What I'm saying is that your R60 is almost certainly an American R-60, which has units of ft^2*degF*hour/BTU, but you are doing the rest of your calculations in SI (meters, Watts, etc).  You need to translate your R-values into SI units.  An SI R60 wall would be like 12 feet thick.