For some people, cold winter weather brings thoughts of hot chocolate by the fireplace, cozy comforters, or maybe skiing.
By contrast, I find myself thinking about insulation and R-values.
So, in the spirit of the holidays, here are two R-value calculations that I’ve been meaning to make.
Heated outdoor faucet cover. Sure, it works in practice,but does it work in theory?
Whenever the weather turns cold, I start getting lots of hits on Post #1412, on making an electrically-heated cover for outdoor faucets. Of late, I’ve been getting more than a hundred hits a day, thanks to this recent cold snap and an offhand reference in an on-line forum for Texas Aggies fans.
One of the interesting findings was how little electricity it takes to keep the inside of the faucet protector warm. For example, a mere 4 watt night-light bulb raised the interior temperature by 28 degrees. That more than meets my needs in any cold snap likely to occur in my area.
But is it really plausible that 4 watts could do that? Or was I (e.g.) mistaking heat leaking out of house for the impact of that small electric light?
Obviously, I could check that empirically by hanging up a standard faucet cover with no added heat, and seeing what the interior temperature was. But, at present, it’s about 15F outside, so I’m ruling that out for now.
Instead, this is a classic cases of “Sure, it works in practice. But does it work in theory?” I’m going to do a theoretical calculation of the temperature rise I should expect, using the R-value (insulating value) of Styrofoam, the dimensions of that faucet cover, and the energy output of a 4-watt bulb.
I’m going to model this as a Styrofoam box with dimensions 4.5″ x 4.5″ x 6″. That effectively covers the open face of the faucet cover with Styrofoam, instead of (in my case) brick. So I’m expecting to see more than 28F temperature increase out of this calculation. The box walls appear to be about 5/8″ thick.
Two final bits of data. The R-value of Styrofoam is listed by most sources as around 5.0 per inch. And 4 watts is equivalent to about 13.5 BTUs per hour (BTUH). (I rounded that down a bit to account for the small amount of energy that escapes from that bulb in the form of light, rather than heat.)
Here’s the calculation, first assuming foam on all sides, and then accounting for one side being brick, with a total R-value (for two inches of brick) of 0.88. (I don’t show the full detail of the brick calculation, only the bottom-line average insulating value of the combined foam/brick container.)
The upshot is that this does, in fact, work in theory. The theoretical temperature rise I get from an all-foam box is 41F, much more than I observed. The theoretical rise I get if I replace one side of the box with brick is 28F, exactly what I observed.
It’s purely a matter of chance that this calculation hits the observed value exactly. The fact that it’s close shows that what worked in practice, does, in fact, work in theory.
3000 gallon insulated tank in the middle of Montana
I’ve been watching Engels Coach Shop on YouTube for some time now. The proprietor is a self-employed wheelwright whose long-standing business builds and fixes all manner of horse-drawn transportation.
This has absolutely no practical relevance to my life, but is purely a pleasure to watch. Not only for the actual work performed, but also because the guy knows how to film, edit, and narrate a video.
Of late, he installed a 3000-gallon above-ground tank for watering his cattle. To which you might reasonably say, so what? Until you realize that he’s in Joliet, Montana. To put it mildly, the combination of an above-ground water tank and a Montana winter constitutes a freeze risk.
On the one hand, it’s heavily insulated (reported R50 on the sides, R120 on the top), and the water itself stores considerable heat energy.
On the other hand, it’s in the middle of Montana.
Source: Western Regional Climate Center
Apparently his YouTube following is deeply divided on whether or not they think this will work. Mr. Engels seemed kind of amused at the folks who thought he was going to end up with a giant ice cube. For my own part, I’m guessing it will work just fine, based solely on the guy who built it. But I don’t quite grasp why he seems amused by the opposite opinion.
So rather than just guess, let me do a couple of crude calculations. From the standpoint of the arithmetic, it’s really no different from my faucet cover. Just bigger.
First, I wanted to check out the water tower in Joliet, MT. Just to be sure that a big enough tank, with enough throughput, would not freeze in that climate. But when I tried a trick that always works for finding water towers on the East Coast — use Google Earth, set the perspective flat, and look for a water tower to stick up above the houses, because they are all 120 feet tall, more-or-less — that didn’t work. This, despite the fact that there is a municipal water system with a 160,000 gallon tank.
That’s because the Joliet water tower is mostly underground. Like so. I have no idea whether that was driven by economics, or by threat of freezing.
Source: Laurel Outlook
So, is a well-insulated tank, above ground, a problem or not?
The first hint that it’s not a problem is that the total heat loss of this tank is maybe 16 times the heat loss of my faucet cover. This tank is enormously larger. But it’s also enormously better insulated. The combination of having about 300 times the surface area, and maybe 20 times the average insulation, is that, by calculation (below, highlighted in yellow), this tank only loses a bit over five BTUs per hour per degree F. That’s just 16 times the heat loss in my Styrofoam faucet cover.
Here, I’ve assumed a tank shaped like a cube, with an average R-value of 60 on all surfaces. Should be close enough for a rough cut like this:
Well, given that a four-watt bulb would heat my faucet cover, it should be no surprise that even a modest heat input would (eventually) result in a large temperature differential between the inside and outside of that tank. Where four watts was enough to create a 41F difference in my all-foam faucet cover, here, a typical stock tank heater (150 W) would (eventually) generate a massive 94F difference between interior and exterior of the tank.
That’s a big enough difference that (arguably) this simple linear R-value calculation does not exactly hold. I don’t think that much matters. If for no other reason that, given the tiny heat input (about the same as you would use to heat a cup of water to boiling for tea), it would take years to reach equilibrium.
(Well, might as well calculate that roughly. This is about 25,000 pound of water. To raise that by 94F, with zero losses, using a 150W heater, would take just over half a year. With losses, yeah, a couple of years. If then.)
I’m going to go out on a limb and say that, if the tank is well-mixed, running a 150W stock tank heater inside it would, in fact, guarantee that it would not freeze under almost any conceivable circumstances in that climate.
But there’s no electricity at that site. Instead, the tank has to “coast” all winter, using just the energy embodied in the water in the tank itself.
So, how much energy is there in that water? How much heat would you have to remove to take water, at a typical late-summer temperature for that area, and bring it down to 32F?
By definition, a BTU is the amount of energy required to raise one pound of water by 1 degree F. So if (say) the water starts out around 62F (late summer/early fall), it would have to lose over three-quarters of a million BTUs in order to reach 32F. As shown below, bottom line.
Now I’m going to do a little hypothetical calculation. Let me plop that tank down in January, in Joliet, MT, and see how much it cools off over the month. That is, let me start with that tank at 62F, and let it sit for 31 days with an average external temperature of 24F — the actual average temperature for that month and location. This should be a worst-case scenario for temperature loss, because it’s the largest temperature differential you could hope to see. Water temperature from late summer, against dead-of-winter air temperatures.
Here’s the simulation. I just calculate the daily heat loss, and then drop the temperature each day, using that heat loss (in BTUSs) as a fraction of the total heat embodied in the 62F vs 32F water. (That is, I pro-rate the BTUs of daily heat loss over the total 750K BTUs that would take the water from 62F to 32F).
OK, I finally get the joke. Worst case, this tank ought to lose just over 5F per month, in the coldest month of the year. And note that the cooler the tank gets, the slower the additional temperature loss gets. For all practical purposes, the likelihood that the tank will freeze is zero.
(Note that the calculation is linear in temperature, so that it doesn’t really matter if the temperature does up and down in January. The average heat loss is going to match the average temperature. There are more refined physics calculations that will add some slight non-linearity to this, but not enough to matter).
Unsurprisingly, this tank isn’t just built for that climate. It’s over-built. Some of my assumptions might be a bit off. The tank is a cylinder, not a cube. Likely I could have calculated the average insulation value better. I don’t really know the insulation value for the bottom of the tank. And so on. But even with that, this seems to have been built with a huge margin of safety.
I should have expected no less.
