Background.
Fair warning: This is a technical post. It wanders off into a discussion of heat content, conduction, radiation, and so on. Calculations will be openly displayed, without apology.
If that’s not your thing, then skip it. Sometimes the juice is not worth the squeeze.
Let me get to the bottom line. In the situation I’m studying — the details of which are spelled out below — on a cold spring night, a raised garden bed loses more heat by radiating it into space as infrared than it does by conducting heat into the cold air. In the right circumstances, much more. That’s why covering a raised bed with a radiant barrier such as a space blanket is a cheap and convenient way to keep a raised bed warm at night. And the heat content of soil is low enough that you probably get some material benefit from adding some gallons of sun-warmed water under that radiant barrier. The result is that a workable setup for keeping a raised bed warm on cool spring nights is to place a handful of solar-heated water jugs on the bed, then cover it all with a space blanket (or equivalent).
Now that you know the answer, this post goes step-by-step to show why it’s not such a goofy idea, after all, to put a space blanket on a raised garden bed.
Let me start by saying that I didn’t think this up. I got the idea from a document that ultimately comes from the Colorado State University extension service (.pdf). What caught my eye was the claim of freeze protection, down to zero degrees, using a space blanket draped over a plastic-covered cold frame.
Notice how interesting that result is? First, zero Farenheit is pretty cold. Second, the cold frame was already wrapped in plastic, so the effect of the space blanket doesn’t have anything to do with keeping the air inside from mixing with the cold air outside. And third, it surely doesn’t have anything to do with insulation, because thin sheet of aluminized mylar (a space blanket) provides essentially zero insulation.
So this has to be about radiation. About the ground losing heat by radiating it away as infrared. And how a space blanket stops that.
Which leads to the obvious question:
Why is radiative heat loss important for a garden bed?
Short answer: A garden bed is like a big open window, looking up into the cold night sky. Heat loss, in that case, is fundamentally different from heat loss from an insulated house.
I didn’t start out to look at infrared radiation and garden beds. I actually started from internet sources that suggested passive solar heating as a way to keep garden beds warm at night. It’s easy, so the internet tells me. Spray-paint a few gallon jugs black and fill them with water. Placed in a garden bed, they’ll warm in the sun during the day and radiate heat at night. That will keep your garden warm.
But is that fact or is that folklore? Can a few gallon jugs of water add enough energy to make a material difference in the nighttime temperature of a garden bed?
Obviously, one way to test this is just to build it, stick a recording thermometer in the bed, and see how it does. And I’ve gotten to that, eventually. But first I wanted to do a little calculation to shed some light on whether or not it’s even remotely feasible.
And so, before I start painting a bunch of jugs black, I decided to do a little arithmetic to see if this is even plausible. And it was in the process of that deep dive that I realized that heat loss through radiation is more important than heat loss through conduction.
What I found was that, from the standpoint of heat loss, a raised garden bed isn’t at all like a house. So all your standard notions of how to stop heat loss, based on insulating your house, give you an incomplete picture of how you actually lose heat from a garden bed.
Houses lose most of their heat by conduction from their outer shell. That’s why you put insulation in the walls and attic. It slows the conduction of heat from the interior to the exterior of the house (in winter), or vice-versa (in summer).
In a garden bed with, say, a piece of plastic on top, that corresponds to the heat that is conducted from the ground to the cold night air, via that plastic. That’s energy lost to using the heat from the soil to heat the still air directly above the garden bed.
The second largest way that houses lose heat is via air exchange, which you can think of as a form of convection. Even in a tightly-sealed modern house, you’re supposed to have no fewer than one full air exchange every three hours. In effect, every three hours, you have to replace your inside air with outside area, heating or cooling that new air in the process.
In a raised garden bed, I guess that corresponds to having the wind blow over the bare soil. It’s not that you’re heating up the still air over the bed, it’s that you’re continuously having to heat up new air.
In my house, based on my calculations, my heat loss is about two-thirds conduction, and about one-third air exchange. But I live in an old and relatively poorly-sealed house.
By far, the least important way that houses lose heat is via radiation. That is, radiating energy away from the house in the form of infrared. If you get a typical home energy audit, energy loss via radiation is not even on the list. (Although, weirdly, people will try to sell you “IR-blocking paint”, under the theory that it helps stem energy losses.) The only place that radiation losses matters for a home is the windows, where low-emissivity (low-e) glass is used to prevent radiant energy from coming into the house in summertime.
But a garden bed is, in fact, just like a window in a house. From an energy-loss standpoint, it’s an open window pointing directly up into the sky. And that’s why radiation matters crucially here, when it hardly matters at all in your house. Your garden bed isn’t a house at all, it’s a window.
The way to stop that loss of heat energy via infrared radiation is to add a radiant barrier. Something that reflects infrared back to where it came from. Hence, the space blanket.
(Finally, a garden bed has yet a fourth way to lose heat, through evaporation or equivalent. I think that’s small enough, in most cases, that I can ignore it for purposes of the next set of calculations.)
How much does radiative heat loss matter to garden beds?
Lots. On a cold, clear night, if you were to cover your raised garden bed with simple piece of plastic sheet, you’d lose a lot of heat through conduction. A plastic sheet is a poor insulator, and heat would flow from the ground into the cold air, right through that sheet.
Here’s the kicker: As much heat as you might lose to the cold air, via conduction, you could easily lose three times that much via radiation. That’s the arithmetic I’m going to show later. And that’s why, if you’re going to cover your beds with a sheet of plastic on a cold night, it’s well worth your while to make that sheet a space blanket.
This is the section where I do the arithmetic and/or just look up some numbers that somebody has already calculated, to show this. This was news to me, and I figure it might be news to a lot of other people. Because, again, this is not the way it works for keeping your house warm.
An apology to the more reasonable parts of the world, and clarifying a common confusion about BTU, BTUH, KW, KWH.
Before we can do the math, we have to define the units we’re going to use. Hence the title, because those of you who use the metric system are going to get a chuckle over how we go about that in the U.S.A.
Joules? We don’t need no stinkin’ Joules. We spit upon the Joule. In the U.S.A., this calculation is done in good old all-American British Thermal Units (BTUs). A BTU is the energy required to raise one pound of water by one degree Farenheit.
(That’s antique, but it’s not the most ridiculous unit we use to discuss heating and cooling. That honor goes to the ton. Ton? Ton of what? A one-ton air conditioner is a unit which, over the course of 24 hours, can extract enough heat to freeze one ton of ice, at 32F. That works out to be 12,000 BTU per hour. Which is then colloquially termed a “12,000 BTU” air conditioner, causing physicists the world over to splutter as they try to explain, to the unwashed masses, the difference between energy (BTUs) and power (BTUs per hour, or BTUH). I will get into that below.)
I need to drag in one more unit definition. The insulating value of building materials in the U.S. — their resistance to heat flow — is expressed in R-values. The R-value of a material is the inverse of the number of BTUs per hour required to maintain a one degree Farenheit difference across one square foot of that material.
A single-pane window is typically R-1. You’d need to inject one BTU of energy, per hour, to the area on one side of the glass, to maintain a one degree F difference between the two sides of that window pane. By contrast, standard 2×4 walls with fiberglass insulation are typically R-12. You’d only need one-twelfth of a BTU, per hour, to maintain that same one-degree F temperature difference. (Or, equivalently, a one BTU per hour energy input would maintain a 12 F temperature difference across the R-12 wall.)
(All that last paragraph, per square foot. So an R-1 surface requires one BTU per hour heat input, on one side of the surface, per square foot of surface, in order to maintain a one degree F temperature difference between the two sides of that surface.)
A brief public service announcement: Energy is not power.
If you know this, skip it. But the terms we use to describe energy and power are a) confusing, and b) frequently confused. I’m going to take two minutes to clarify.
Think of energy as being, say, a quantity of fuel. Gallons of gasoline. By contrast, power is the rate at which you burn that fuel/use that energy. If energy is gasoline, then power is gallons of gasoline burned per hour.
Here’s where it gets confusing.
A watt is a unit of power. It tells you how fast you are using energy. A 100-watt incandescent bulb consumes energy ten times as fast as a 10-watt bulb. And it’s ten times brighter.
A watt-hour is a unit of energy. If you run a 100 watt bulb for an hour, or run a 10 watt bulb for ten hours, you’ll have used 100 watt-hours of energy. Use 1000 watt-hours, and that’s a kilowatt-hour (KWH).
A BTU is a unit of energy. I’ve heard it said that it’s about as much energy as you get by burning one large wooden kitchen match.
A BTU per hour (BTU/H often written BTUH) is a unit of power. It tells you how fast an appliance is using or producing energy. If a burner on a stove can output 10,000 BTUs of heat in an hour, that’s a 10,000 BTUH burner. And a 10,000 BTUH burner will heat things ten times faster than a 1,000 BTUH burner.
And there’s the confusion. The H in KWH is hours, the H in BTUH is per hour. And so:
- BTU is analogous to KWH is a unit of energy
- BTUH (really, BTU/H) is analogous to W is a unit of power.
When somebody describes an air conditioner or a stove burner as “12,000 BTU”, physicists cringe. The only way that an air conditioner is “12,000 BTU” is if you could burn it and so release 12,000 BTUs of heat. Properly said, both those devices are rated for a power output of 12,000 BTU/H.
Conductive and radiative loss through thin plastic sheet: How many water jugs do I need to keep my raised bed warm?
First question: How many BTUs of energy do I need, to maintain a 10-degree F temperature difference across a 4′ x 16′ polyethelene sheet, for an eight-hour night.
In other words, if I’m going to cover my raised bed with a sheet of plastic, and add enough heat below the plastic to keep the bed 10 degrees F warmer than the surrounding air, for an eight-hour night, say, how much heat do I need?
A single layer of plastic sheet has an R-value of about 0.85. (Per this reference, but I’ve seen that value before, so I’m pretty sure that’s an agreed-upon number.) This bed, 16′ x 4′, will therefore require (16′ x 4′ x 10 degrees x (1/.85) BTU per degree per square foot per hour x 8 hours =) ~ 6000 BTUs.
That’s just to offset the conductive losses. Over the course of a night, I’m going to need about 6000 BTUs to keep the inside of that thin plastic sheet ten degrees warmer than the outside.
That ignores all other energy losses.
Converting units, that’s 1.75 Kilowatt-hours (KWH) of energy. That a little over 200 watts of power, running for eight hours.
Let me say this next statement precisely. If I had no other source of heat energy, I’d have to run two old-fashioned incandescent 100 watt bulbs, all night, to keep that garden 10 degrees above ambient temperature, with just that thin plastic sheet for insulating the top.
The part in italics is key. Because, as it turns out, the soil itself stores quite a bit of heat. Not a huge amount. But enough to matter.
Second question: How large are the energy losses via radiation, in that same situation. This calculation is far more loosey-goosey.
For one thing, the net loss from radiating infrared into the sky depends strongly on atmospheric conditions. Losses are much greater under a clear sky than under a humid or cloudy sky. (Not because the bed radiates less, but because the sky radiates more “down-welling long-wave” back toward the ground.) At 50F, this source shows that soil might lose anywhere from 15 watts to almost 100 watts per square meter, depending on the state of the atmosphere, from the warm earth radiating energy into the cold sky.
Taking 50 watts per square meter as a plausible value, over the course of an 8-hour night, my 64-square-foot bed would lose (8 hours x 50 watts per square meter x 64/10.75 square meters =) about 2.4 KWH of energy. Just by radiating it away into the night sky. And if it’s a clear, low-humidity night, you could radiate double that amount, or 4.8 KWH of energy.
The point is, in this real-world example — trying to keep a raised bed 10 degrees F warmer than the surrounding air, using a thin sheet of plastic — I’d lose more heat from radiation than from conduction through the sheet. And if it’s a clear, low-humidity night, I’d be losing almost three times as much energy from radiation as from conduction.
To me, this was completely unexpected. I am really only familiar with energy loss in houses. There, you never even see radiation mentioned as an issue. Here, it’s the principal source of energy loss. This only happens because, in effect, my raised bed is a house consisting of a 64-square-foot open window.
In theory, covering the open bed with something will reduce that radiation loss. But polyethylene film is almost completely transparent to infrared. In that case, using a simple sheet of clear plastic isn’t going to do much to reduce losses due to radiation of infrared from the garden bed.
Third question: With that as context, how many gallon jugs of sun-warmed water do I need to keep this bed warm?
A gallon contains about 8.5 pounds of water. By the BTU definition, that stores 8.5 BTUs for every degree over ambient temperature. Suppose that on a sunny spring day, I can get that gallon up to 85F. (Not implausible, that’s what my plastic-wrapped water jugs hit today.) If I want to maintain a temperature of 50F, that gallon jug of warm water has a potential to supply (85 – 50)*8.5 = around 300 BTUs of heat energy.
But I’ll never be able to extract that all, because the rate of heat loss falls as it nears ambient temperature. In the morning, the jug will still be somewhat warmer than the 50 degree bed. In my last experiment, the jugs ended up at 60F. If that’s typical then I can only extract (85 – 60)*8.5 = a bit over 200 BTUs per sun-warmed gallon of water, over the course of the night.
The upshot is that with the thin polyethylene sheet, if there is no other source of heat energy, I’d need 30 gallon jugs just to offset the conductive losses through the plastic sitting atop the 4′ x 16′ bed.
And if I’ve done the math right on radiative losses, if there is no other source of heat energy, I’d need another 30 and up to offset the radiative losses through the polyethelene sheet.
In short, I’d need something like 60+ gallons of warm water to keep this bed at 50F, on a 40F evening, if the bed is covered with polyethelene sheet. If there is no other source of heat energy.
Step 1: Replace the polyethylene sheet with a space blanket. In theory, I can eliminate almost all my radiative losses by using a radiant barrier in place of the polyethelene sheet. Space blankets (aluminized mylar sheets) are said to reflect about 95% of the infrared directed at them. That’s equivalent to having a 5% emissivity, or in other words, this would reduce the infrared emissions from the bed by about 95%.
Step 2: But wait, there’s more. I keep using this phrase “if there is no other source of heat energy”. But that’s absurd, because the raised bed itself stores some heat.
How much heat energy is available from the bed itself, and how does that compare to the energy in a gallon of warm water?
This is a tricky question, because even if there’s a lot of heat energy there, it will only travel slowly through the soil. It’s the same issue as the water jug that doesn’t fully cool overnight. I’m limited in the rate at which I can extract energy from the soil.
But just to get a handle on it, ignore that for now. For the purposes of this calculation, how much heat is stored in the 4′ x 16′ x 1′ garden bed, with an average soil temperature of (say) 60 degrees?
The specific heat of wet soil is about one-third that of water (calculated from this reference). The raised bed holds 64 cubic feet of soil, and there are about 7.5 gallons per cubic foot. And with 60F soil, where I want to maintain the space above the bed at 50F, the total BTUs available work out to be something like (8.5 BTU/gallon of water x .33 (specific heat of wet soil relative to water) x 64 cubic feet x 7.5 gallons per cubic feet x 10 degrees = ) roughly 13,500 BTUs.
That’s interesting because that’s not really a huge amount. If I could extract it all, it would be a lot. But in fact, heat moves relatively slowly through soil. And so, while there’s a lot of potential there, my guess is, the actual amount you can bleed off in one night is not all that much.
The upshot is that yeah, you probably do want to use that sun-warmed water. Without the aid of those sun-warmed gallons, the bed probably will not be able to remain as far above ambient air temperature as it would, with a significant number of water jugs installed.
And the upshot is that it’s the combination of the readily-available energy from the warm water, and the radiant barrier, that keeps the bed surface warm. And so, I expect I will end up painting a bunch of water jugs black after all.
Step 3: Final addendum. But if I put aside all practical issues, there is one building material that would give me the best of both worlds: foil-covered foam board. This material is made to provide both an insulating layer against conduction, and a radiant barrier against radiation. Except for the fact that it would be incredibly awkward to use, that would probably be the ideal insulator for the garden bed.
As it stands, if I can maintain my required 10 degrees F of additional warmth, with just some water jugs and space blankets, I think I’ll stick with that.
I’m still going to test all this, in the next few days. But based on the math, just a handful of warm water jugs probably make a material difference. And at 89 cents each (with free distilled water inside), they’re a bargain.
