[time-nuts] Re: MHM-A1 maser temperature stabilization

[Bob Camp]

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



> On Jan 17, 2023, at 5:30 PM, Poul-Henning Kamp <phk at phk.freebsd.dk> wrote:
> 
> --------
> 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 at 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.