Actually, I don't know about you, but I've literally forgotten more about physics than most people will ever know, insofar that I've forgotten pretty much everything I ever knew. And I used to know... some.
So here is the thing with the winter preparation. We posit that my trees are gonna be hardy to zone 5 and I will leave them outside until the end of December. This means that they're allowed to drop to -30 C over three months. Therefore, how much insulation do they need?
The answer, in reality, would be complicated, because the temperature outside the insulation will drop gradually. But let's assume the outside temperature is -30 C the whole time. Then, we need to know how much energy is in the pots to begin with, which if I recall correctly, would be something like heat capacity times volume times temperature. I'll have to google it, but it sounds about right. We assume the starting temperature is 0 C, and then this would give us something in Joules. Then we divide it by the 7.8 million seconds the trees have to survive the cold, and that would give us something in Watts. From there we get the RSI, which is in units of K.m2/W. We have the Watts, we have the temperature difference (the K), and I determine the surface area. And an inch of rigid extruded polystyrene has an RSI of 0.88.
Therefore, in the unlikely event that I'm not mistaken, the answer would be:
Inches of rigid = (30 K) (2.23 m2) (7.8 x 10^6 s) / (heat capacity of soil) / (3.1 x 10^-2 m3 of soil) / (273 K) / (0.88 K.m2.s/J)
Anyone want to check the units? I don't feel like it just this instant.
Oh, fine, I'll do it. What are the units of heat capacity? J / m3 / K? I don't remember. But other than that:
K x m2 x s x m-3 x K-1 x K-1 x m-2 x s-1 x J x m3 x J-1 x K
K's cancel out.
s's cancel out.
m's cancel out.
J's cancel out.
Yep, looks about right.
Now I don't even have a preliminary guess on the heat capacity of soil, but seeing that there is a multiplier of a million in there, I'm gonna have to hope that it's a really huge number.
In the quite plausible event that I can't insulate enough to achieve this temperature curve, I can compensate by filling the hot box with ice.
That's right, ice. Why would I fill my hot box with ice? Because it has a high heat capacity, therefore, it's gonna be a reservoir of heat. If we start with more heat inside, we can afford to lose more of it. Ice usually comes out of the freezer at a temperature of approximately -18 C, but I could heat it up to near zero before putting it in the box. Even better, I could fill the hot box with water, but then, first of all I'd have to make my hot box waterproof, and second I'd have to allow for the expansion of the water as it slowly freezes over three months. The much bigger problem I'd then have is how to break my pots out of the ice, so that a) I don't have to try and drag a huge block of ice into the house, and b) I don't have a huge block of ice melting in my house. Maybe I need a box within a box.
Who knew gardening was so complicated?
So here is the thing with the winter preparation. We posit that my trees are gonna be hardy to zone 5 and I will leave them outside until the end of December. This means that they're allowed to drop to -30 C over three months. Therefore, how much insulation do they need?
The answer, in reality, would be complicated, because the temperature outside the insulation will drop gradually. But let's assume the outside temperature is -30 C the whole time. Then, we need to know how much energy is in the pots to begin with, which if I recall correctly, would be something like heat capacity times volume times temperature. I'll have to google it, but it sounds about right. We assume the starting temperature is 0 C, and then this would give us something in Joules. Then we divide it by the 7.8 million seconds the trees have to survive the cold, and that would give us something in Watts. From there we get the RSI, which is in units of K.m2/W. We have the Watts, we have the temperature difference (the K), and I determine the surface area. And an inch of rigid extruded polystyrene has an RSI of 0.88.
Therefore, in the unlikely event that I'm not mistaken, the answer would be:
Inches of rigid = (30 K) (2.23 m2) (7.8 x 10^6 s) / (heat capacity of soil) / (3.1 x 10^-2 m3 of soil) / (273 K) / (0.88 K.m2.s/J)
Anyone want to check the units? I don't feel like it just this instant.
Oh, fine, I'll do it. What are the units of heat capacity? J / m3 / K? I don't remember. But other than that:
K x m2 x s x m-3 x K-1 x K-1 x m-2 x s-1 x J x m3 x J-1 x K
K's cancel out.
s's cancel out.
m's cancel out.
J's cancel out.
Yep, looks about right.
Now I don't even have a preliminary guess on the heat capacity of soil, but seeing that there is a multiplier of a million in there, I'm gonna have to hope that it's a really huge number.
In the quite plausible event that I can't insulate enough to achieve this temperature curve, I can compensate by filling the hot box with ice.
That's right, ice. Why would I fill my hot box with ice? Because it has a high heat capacity, therefore, it's gonna be a reservoir of heat. If we start with more heat inside, we can afford to lose more of it. Ice usually comes out of the freezer at a temperature of approximately -18 C, but I could heat it up to near zero before putting it in the box. Even better, I could fill the hot box with water, but then, first of all I'd have to make my hot box waterproof, and second I'd have to allow for the expansion of the water as it slowly freezes over three months. The much bigger problem I'd then have is how to break my pots out of the ice, so that a) I don't have to try and drag a huge block of ice into the house, and b) I don't have a huge block of ice melting in my house. Maybe I need a box within a box.
Who knew gardening was so complicated?
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