I have moved in with Nuria Pucci. I am writing this reclining on a foam mattress on the floor of my new bedroom.
I’ve paid the rent until the end of the month. I don’t yet know how to access the internet connection or flush the toilet; in the morning I will see if I can take a shower. I need a Gatorade bottle for emergency urination, since there’s only one bathroom.
I have a minimal amount of stuff here now: a sleeping bag, a sheet, a pillow, Tuga, some food, melatonin, my netbook. I need to move the rest ASAP.
The room is roughly cubical with a side of some four meters, for a total volume of 64m³ or 64kℓ. There are no bookshelves or curtains as such. I suspect I can put bookshelves up.
It’s almost 2 AM, and Nuria is running the microwave and playing the guitar. So I will need earplugs, too.
The room contains about 64kg of air, which has a specific heat of about 1 J/g/K = 1 kJ/kg/K. Raising the temperature of these 64kg of air from its current 10° up to a comfortable 20° thus requires about 640kJ. As I operate at about 100 W, this would take about two hours.
However, the walls are presumably concrete (0.9 J/g/K) or gypsum plaster (1.1 J/g/K), and will probably absorb most of the heat I put into the air — if they don’t simply conduct it outdoors. The walls weigh a great deal more than 64kg, although the speed at which the heat penetrates them will depend on how much of their mass is engaged.
This suggests that hanging fluffy curtains on the walls and perhaps something similar on the ceiling might make the room a great deal more comfortable.
If I were to fill a 20ℓ bucket with hot water (4.2 J/g/K) at 50°, in cooling down to 10°, it would release 160kJ. Thus, four such buckets would be needed to bring the room’s air up to a comfortable temperature.
A eutectic mixture of Glauber’s salt and sodium chloride melts at 18°, with a heat of fusion of 286 kJ/kg, and so it would be adequate to bring the temperature up to there; a slightly smaller amount of sodium chloride would melt instead at 20°, slightly less sharply. 2.2 kilograms of such a mixture in its liquid form would suffice to supply the same heat as all four buckets of water.
If we were to scale up such a system by a factor of 20, we would have a reservoir of 44 kg of phase-change material (30 ℓ at 1.46 g/cc) with 12.6 MJ of phase-change heat capacity; we could also choose a different phase-change temperature. A glazed flat-plate solar heat collector with a water-ethanol mixture pumped through it could provide the heat input; a thermosiphon-driven heat exchanger inside the room, mounted above the reservoir, would silently release the heat into the air when a valve was opened, at which point it would drive convective air circulation through a “chimney” above the heat exchanger.
Running this system for four hours at night would provide 870 watts of heating, which is probably plenty. It could perhaps be charged over a longer period of time during the day; if it only had a single hour to charge in, it would need 3.5 kWt of sunlight; with a 50%-efficient glazed flat-plate collector, that’s about 7 m².
The standard chimney draft calculation is
Q = C A √(2 g H ΔT/T)
where Q is the draft flow rate, A is the cross-sectional area of the chimney, C is a discharge coefficient of about 0.7, g is the gravitational acceleration, H is the height, ΔT is the temperature difference across the chimney wall, and T is the temperature outside.
Let’s say A = 0.01 m², H = 3 m, T = 288 K, ΔT = 5 K; then our flow rate is 7ℓ/s, at which rate sucking all 64kℓ of air through the system would take about 2½ hours, which is a bit too slow. Correcting to 0.05m², we suck all the room’s air through the heater every 30 minutes, which is acceptable; this amounts to a round chimney duct of some 250 mm in diameter.
44 kg of Glauber’s salt at the wholesale price of US$130/tonne would cost US$5.72, currently about AR$80.
The 30 liters would need to have plumbing run through it rather thoroughly in order to be able to freeze all of the salt during the heat-evolution part of the cycle. Wikipedia says that sodium chloride has a thermal conductivity of 6.5 W/m/K at 289 K, but I’m not sure how to apply that information to figure out how closely the pipes need to be spaced.
Well, kind of. Let’s say the pipes are spaced 50 mm apart, so the heat has to travel 25 mm from a pipe to reach the last liquid salt. If you slice 30 ℓ into 50 mm slices, the surface area of the slices is 0.6 m², or 1.2 m² on both sides. The coolant is presumably halfway between the air temperature, which might be 18°, and the salt’s freezing temperature, which might be 22°, so you have a gradient of only 2 K on average over those 25 mm: 80 K/m; multiplying the surface area in, you get 96 K m. That’s 624 W, which is probably acceptable. If the pipe spacing were smaller, you would have proportionally more layers and less distance to the last liquid, so the power goes as the inverse square of the pipe spacing.
Spacing zero-thickness pipes 50 mm apart throughout a 30 ℓ volume in a square pattern would require 12 m of pipes. (I’m not sure how much it affects the surface area of the last or nearly last liquid salt, or whether that matters.) You can do slightly better by using a hexagonal pattern, but then again, you will do slightly worse because the coils have to bend in order to fill the whole space. So this is probably about right.
I’m not sure if it matters to the heat transfer into the salt whether you run the pipes in parallel or purely in series.
12 m of 5mm-diameter pipe holds 236mℓ of coolant, almost exactly one US cup, which is a comfortingly small number for the most hazardous part of the system. If it had the specific heat of water, with that same 2° ΔT, it would transfer about 2.0 kJ of heat each circuit; this is too small, because at 870 W, it would have to make a circuit of the entire reservoir every 2.3 seconds, 5.3 meters per second. That’s far too fast, but it might be bearable if you can run a lot of pipes in parallel instead of in series. Just in case, though, it’s probably better to think in terms of 10mm-diameter pipe, 940mℓ of coolant, 7.9 kJ, and 9.1 seconds of reservoir transit time.
We probably need turbulent flow for maximum heat transfer power, so we want the Reynolds number in the pipes to be over 4000; but not too much over 4000, because we lose efficiency. If Re = vD/ν, solving for v, we get v = ν Re / D. If D = 10mm, ν = 1 cSt = 1 mm²/s, and Re = 4000, then v = 0.4 m/s; at 9.1 seconds of reservoir transit time, this suggests we’d need to split the reservoir coolant into branches of about 3.6 m each.
400 mm/s seems like a lot to expect out of a thermosiphon. How can we fix that? If the pipes were thinner, we would need an even higher velocity to hit Re = 4000, which seems counterintuitive; for example, at 5mm, we need 0.8 m/s. But that really is the way it works: thicker pipes, where viscosity matters less, are more prone to turbulence. So apparently the cure is to use thicker pipes still in order to get turbulent flow at a lower speed.
Let’s say we go to 20mm diameter, at which point we can afford 36 seconds of reservoir transit time, because the coolant holds 32 kJ, because there are 3.8 liters of it. Now we get turbulent flow at 0.2 m/s, at which speed we would transit all 12 meters of pipe serially in 60 seconds, so we still do better if we split the pipe into two parallel runs.
How much pressure do we need to get to 0.2 m/s in 12 meters of 20mm pipe, though?
The Darcy-Weisbach equation says
Δp = f L ρ v² / (2 D)
where f is the Darcy friction factor, L is the length of the pipe, ρ is the density of the liquid, v is the velocity as before, and D is the diameter as before.
Unfortunately, I don’t have a Moody diagram handy to figure out what f should be. The Haaland equation, a simple approximation to the recurrent Colebrook equation, says:
1/√f = -1.8 log₁₀ ((ε/D/3.7)¹·¹¹ + 6.9/Re)
where ε is the surface roughness and D is the diameter, as before. The relative roughness ε/D here should be about 0.01 to 0.001, so that part of the sum could work out to be as high as .0014, or much lower; 6.9/4000 is about .0017. So our logarithm here is about -2.5 to -2.8, so 1/√f ≈ 4.5 to 5, so f should be in the neighborhood of 0.04 or 0.05. [Later I checked this against a Moody chart and, yes, .045 to .055.]
The original Colebrook equation says
1/√f = -2 log₁₀ (ε/D/3.7 + 2.51/(Re√f))
Just checking here, Re√f ≈ 800, ε/D/3.7 ≈ .0027, 2.51/800 ≈ .0031, the logarithm is about -2.2, and yes, that means 1/√f ≈ 4.5.
Back to the Darcy-Weisbach equation, we have f = 0.04, L = 12 m, ρ = 1 g/cc, v = 0.2 m/s, D = 20 mm, which works out to 480 Pa of head loss. Or only 240 Pa if we split it in half. That’s 0.03 psi, which kind of sounds reasonable, but it’s 24.5 mm of water head, which seems like it might be hard to get out of a thermosiphon.
This all depends on the thermal coefficient of expansion of water, which notoriously falls to zero at 4°, then goes negative. Mixing ethanol into the water might help with this, both because it serves as antifreeze and because its own TCE is very large, much larger than water’s. Wikipedia [[Thermal expansion]] claims that water’s volumetric coefficient is 207 ppm/K at 20°; if we have a 2 K difference between the hot side and the cold side of the thermosiphon, then we have 414 ppm. That would mean we needed 59 meters of height to get our thermosiphon to siphon at 240 Pa with only 2 K difference.
So, no, you won’t get turbulent flow through the heat exchanger from the thermosiphon. That’s too much to ask from a poor little bedroom thermosiphon.
What will happen in reality if you have, say, 1 m of thermosiphon height? (We’re running out of height here in the bedroom...)
414 ppm of a meter is 414 microns, which works out to about 4 Pa. Let’s solve Darcy-Weisbach for v:
Δp = f L ρ v² / (2 D)
Δp 2 D / f L ρ = v²
v = √(Δp 2 D / f L ρ)
So in this case we have Δp = 4 Pa, D = 20 mm, f we don’t know but for laminar flow it’s 64/Re, L = 6 m, ρ = 1 g/cc. But wait, actually L depends on how many pipes we put in parallel through the reservoir; it might be 500mm if we put 24 parallel pipes there instead of two. Or 250mm if we put 48 pipes in. We were only going with two in order to try to get turbulent flow.
I think that shows that we can get almost arbitrarily high heat transfer powers out of almost arbitrarily low coolant volumes by putting the coolant through lots of pipes in parallel. But I’m not going to finish the calculations tonight.