MHM-A1 maser temperature stabilization
Bob Camp writes:
Ok, but do I need > 800 L of the stuff ?
Forget your 250L of water, it is the wrong way to think about this problem!
Here is a calculated example:
We want to house a HP5065, it dissipates 40 {W}
We build the box as a cube with sides 0.75 {m}.
Surface area = 6 * 0.75 {m} * 0.75 {m} = 3.375 {m²}
Aerated Concrete has a lambda of 0.14 {W/mK}, and thickness 0.05 {m}
Heat loss ("U value") of wall: lambda / thickness = 0.14 {W/mK} / 0.05 {m} = 2.8 {W/m²K}
Multiply by surface area: 3.375 {m²} * 2.8 {W/m²K} = 9.45 {W/K}
Divide into power dispation: 40 {W} / 9.45 {W/K} = 4.23 {K}
The inside of the box will be 4.23 Kelvin hotter than the outside.
The density of Aerated concrete is 540 {kg/m³}
Volume of walls = 3.375 {m²} * 0.05 {m} = 0.168 {m³}
Weight of walls = 540 {kg/m³} * 0.168 {m³} = 91 {kg}
Each of the 6 walls weigh around 15 {kg}, that's workable.
Thermal Capacity of Aerated concrete is 1 {kJ/kgK}
Thermal Capacity of walls = 91 {kg} * 1 {kJ/kgK} = 91 {kJ/K}
/This/ is the number that matter: The amount of energy it takes to change the temperature of the walls 1{K}.
If we dive into algebra, we can now calculate how big a 24h temperature
excursion on the outside it takes to meaningfully change the
temperature of the inner walls in the box, but I'll leave that as
an exercise for the reader.
Instead we build the same box, but with 25mm EPS foam board:
lambda = 0.041 {W/mK} thickness 0.025 {m}
Heat loss ("U value") = 0.041 {W/mK} / 0.025 {m} = 1.64 {W/m²K}
Multiply by surface area: 3.375 {m²} * 1.64 {W/m²K} = 5.535 {W/K}
Divide into power disipation: 40 {W} / 5.535 {W/K} = 7.22 {K}
Density: 13 {kg/m³}
Weight of walls: 13 {kg/m³} * 3.375 {m²} * 0.025 {m} = 1.1 {kg}
Thermal Capacity of (E)PS: 1.3 {kJ/kgK}
Thermal Capacity of walls = 1.1 {kg} * 1.3 {kJ/kgK} = 1.43 {kJ/K}
QED:
It takes 63 times /more/ energy to change the temperature of the inner
wall in the 50mm Aerated concrete box, than it does for the foam-box.
/That/ is the "thermal impedance" we need to attenuate the diurnal
and random temperature changes.
... and why we need algebra as soon as we introduce time :-/
Note also that the temperature rise almost twice as much in
the foam box than in the Aerated Concrete box.
--
Poul-Henning Kamp | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG | TCP/IP since RFC 956
FreeBSD committer | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.
sooo, rigid connected foam that will hold the water...
On 2023-01-17 13:01, Bob Camp via time-nuts wrote:
Hi
Yes, a big swimming pool of mercury or water is going to do a lot of
things that
make it really fun to analyze. Folks seem to run circulation pumps to
reduce the
impact. Not on my list right now ….
Water is attractive because it holds a lot of heat per unit mass. One
KG of concrete
soaks up 880 J/(Kg-K). One L of water comes in at 4,181. More fun
numbers at:
https://www.omnicalculator.com/physics/specific-heat
Specific Heat Calculator
omnicalculator.com
Solid concrete weighs about 3X what water does by volume, but water
still wins
the race. If the concrete has air spaces in it, it falls further
behind on a volume basis.
In terms of a practical answer for a fluid, put it in small(er)
containers (or baffle it).
Then you don’t have it doing weird flow stuff. Yes, another rabbit
hole to wander down.
Right now a bunch of 2 L glass jars are sitting over there on the shelf
…..
Coming back to the basic question:
Concrete *or* water, how much do I need?
Not looking for anything past a rough order of magnitude to see if my
magic
math number of > 250 liters of water is anywhere close. ( In concrete
numbers,
is > 400 liters close? )
Just trying to get an idea of the scope of the project. If the answer
is 25 and not 250,
it’s trivial to do. If it’s 250 … gulp … If its 2,500 … yikes ….. As
you might guess from
the reference to the glass jars, empirical data suggests that 25L is
not the answer.
At some point adding this or that makes the enclosure bigger and the
insulation needs
to get thicker to stay at the same net loss value. There’s only so
much room over there :)
Bob
As I do this in my usual hand waving fashion, I come up with hundreds
of liters of water for
the thermal mass. It just goes up if I move from 1C and get closer to
0.1C.
Liquids and gasses are trouble in this context, because you easily
end up with stationary or oscillating circulations which, if nothing
else, really ruins the predictability.
So let us stick to solids for the moment:
A thermal resistance is an insulating material with low thermal
mass: Sheep's wool, mineral wool, foam-boards, extruded foam and
ultimately aerogel.
A thermal capacitance, is a material with high thermal mass and
high thermal conductivity: Ideally silver and copper, but in general
any metal.
We know of no materials which act as pure thermal inductances: It
would be a material which conducts heat well, but resists change
to the heat-flow. Certain crystaline semiconductors behave a little
bit like a thermal inductance under certain circumstances, but it
is not useful in practice.
So we are more or less limited to RC filters.
We can make a "lumped" RC with foamboard insulation and
a lot of metal inside.
This is what metrology-labs do for their resistance standards:
Typically a slap of aluminium roughly 1'x2'x2' with holes for the
individual resistors (+oil) insulated with 4" of foam/mineral wool
or similar.
But that concept, as your own calculation also showed you, scales
up badly: A big box of foam-board and lots of metal (or water),
is both expensive and unpractical in so many ways.
Fortunately almost all geology is a distributed RC thermal filter:
Limited heat conductivity combined with some thermal mass.
A "box" built from 2" aerated concrete, cinderblocks or pretty much
any geology you might have at hand, is cheaper, much more practical,
and almost certain to give better results.
I mention 2" aerated concrete specifically, because if you cut the
slabs to size and paint them to bind the dust, they are very
handy building blocks: You can stack them around your equipment
when you want to, and remove again when you need to access it.
--
Poul-Henning Kamp | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG | TCP/IP since RFC 956
FreeBSD committer | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by
incompetence.
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Dr. Don Latham AJ7LL
PO Box 404, Frenchtown, MT, 59834
VOX: 406-626-4304
Hi
Ok, here’s some numbers:
In the original example, the box heats 4C with 100W of power.
Each degree of “damping” means you have done something with 25W of power.
If we want to turn a 2C room swing into a 1C box swing, we “damp” 1C.
We started out talking about 24 hours so let’s go with that first
25 W over 24 hours is 24 * 60 * 60 * 25 = 2.16x10^6 Joules.
My mass rises 1 degree when you put 2.16 MJ into it.
If I use concrete at 880 J / (KG-K), that gets me to 2,454 KG
(there’s a range of about 750 to 960 depending on this or that)
If I use water at 4,186 J / (KG-K) that gets me to 518 KG
(yes, there could be issues if it’s not baffled in some way)
Drop back to 12 hours and the numbers come down by 50%.
Swing the room 4C and they go up by 3X.
Control the box at 0.5C, they go up by 2X.
To me, they are into the crazy end of things.
Bob
Ok, but do I need > 800 L of the stuff ?
Forget your 250L of water, it is the wrong way to think about this problem!
Here is a calculated example:
We want to house a HP5065, it dissipates 40 {W}
We build the box as a cube with sides 0.75 {m}.
Surface area = 6 * 0.75 {m} * 0.75 {m} = 3.375 {m²}
Aerated Concrete has a lambda of 0.14 {W/mK}, and thickness 0.05 {m}
Heat loss ("U value") of wall: lambda / thickness = 0.14 {W/mK} / 0.05 {m} = 2.8 {W/m²K}
Multiply by surface area: 3.375 {m²} * 2.8 {W/m²K} = 9.45 {W/K}
Divide into power dispation: 40 {W} / 9.45 {W/K} = 4.23 {K}
The inside of the box will be 4.23 Kelvin hotter than the outside.
The density of Aerated concrete is 540 {kg/m³}
Volume of walls = 3.375 {m²} * 0.05 {m} = 0.168 {m³}
Weight of walls = 540 {kg/m³} * 0.168 {m³} = 91 {kg}
Each of the 6 walls weigh around 15 {kg}, that's workable.
Thermal Capacity of Aerated concrete is 1 {kJ/kgK}
Thermal Capacity of walls = 91 {kg} * 1 {kJ/kgK} = 91 {kJ/K}
/This/ is the number that matter: The amount of energy it takes to change the temperature of the walls 1{K}.
If we dive into algebra, we can now calculate how big a 24h temperature
excursion on the outside it takes to meaningfully change the
temperature of the inner walls in the box, but I'll leave that as
an exercise for the reader.
Instead we build the same box, but with 25mm EPS foam board:
lambda = 0.041 {W/mK} thickness 0.025 {m}
Heat loss ("U value") = 0.041 {W/mK} / 0.025 {m} = 1.64 {W/m²K}
Multiply by surface area: 3.375 {m²} * 1.64 {W/m²K} = 5.535 {W/K}
Divide into power disipation: 40 {W} / 5.535 {W/K} = 7.22 {K}
Density: 13 {kg/m³}
Weight of walls: 13 {kg/m³} * 3.375 {m²} * 0.025 {m} = 1.1 {kg}
Thermal Capacity of (E)PS: 1.3 {kJ/kgK}
Thermal Capacity of walls = 1.1 {kg} * 1.3 {kJ/kgK} = 1.43 {kJ/K}
QED:
It takes 63 times /more/ energy to change the temperature of the inner
wall in the 50mm Aerated concrete box, than it does for the foam-box.
/That/ is the "thermal impedance" we need to attenuate the diurnal
and random temperature changes.
... and why we need algebra as soon as we introduce time :-/
Note also that the temperature rise almost twice as much in
the foam box than in the Aerated Concrete box.
--
Poul-Henning Kamp | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG | TCP/IP since RFC 956
FreeBSD committer | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.