
re: hair brained solar cistern???
7 jan 2003
daestrom wrote:
>using a threetermseries expansion... i get some numbers
>that are not far from your estimate...
erf(a) = 2/sqrt(pi)(aa^3/(3x1!)+a^5/(5x2!)...)?
erf(a) = a works pretty well for a < 0.5.
>1) the time it takes to reach ~63% of the way from 70 to 120 is only 218
>hours. this would imply a basic timeconstant of 218 hours. trying to
>calculate the time to reach 119 degrees is about as useless as trying to
>calculate the time to reach 120...
estimating the time for the layer 1' in to reach 119 f tells us how long
we have to wait for equilibrium within 1 f in an experiment or simulation.
>...i think the 'more correct' approach is to calculate the time to reach
>the ~63% point (as done with exponential curves).
there's an ac skin depth analogy, but it seems more natural to think about
halving the difference between the original body temp 1' in and the new
surface temp, which happens in 1/alpha = 95.1 hours for ashrae sand.
>2) the 'schaum outline' is for a body extending infinitely deep away from
>the surface. any realworld application would be finite and have insulation
>or a heat sink at the boundary.
they talk about that, saying
a useful criterion for the semiinfinite solution to apply to a body
of finite thickness (slab) subjected to onedimensional heat transfer
is x > 3.65sqrt(alphat) [x > 0.92' for t = 6 h with ashrae sand]...
where 2x is the thickness of the body. this criteron results in (tts)
greater than 99% of (tits) for any value of the halfthickness x
greater than 3.65sqrt(alphat), and the slab meeting this criterion
is effectively infinite in thickness.
>you or toby can say that heat travels at some 'speed' through sand...
not me... the floor temp hardly varies. imo, the main problem is still the
high sand bed thermal resistance. moisture helps... googling curves, the
diffusivity of sand increases with water content up to a max (at 20%?)
where conductivity seems to increase more slowly than capacity.
problem 4.12 from page 110 of the outline:
at what depth should a water pipe be buried in wet soil (alpha
= 7.75x10^7 m^2/s) initially at a uniform soil temp of 5 c, for
the surrounding soil temp to remain above 2 c if the soil surface
temp suddenly drops to 20 c and remains at this value for 10 hours?
...if the temp will just reaches 2 c after 10 h, (t(x,t)ts)/(tits)
= (2(20))/(5(20)) = 0.88 = erf(x/sqrt(4alphat), ie x/sqrt(alphat)
= 1.1, so x = 1.1sqrt[4(7.75x10^7 m^2/s)(10h)3600s/h] = 0.37 m [15"].
nick

