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re: question - calculation of convection forces
25 mar 2002
daniel matthews wrote:

> say i have a container [a] with a hole at the top and bottom.

say each hole is a ft^2, and the container is h feet tall.

> the top is open to the atmosphere [b].
> the bottom connects to another larger container [c]...
> i have 10mj/s of energy radiating into [a] via one wall.

you might get 10mw of power from 10,000 m^2 exposed to the sun, eg a 2.5
acre square 328 feet on a side... 10mwx3.412btu/h-w = 34120k btu/h, like
341 home oil burners...

> [a] is 1 m^3 in volume, with the dimensions of the radiant wall being
> [ten times the size of the box depth. i.e. it is a flat box with all the
> other sides insulated and reflective...

so d = 0.1 l, and lxlxd = 0.1l^2 = 1, ie l = 3.16 m (10.4') and d = 0.316 m
and maybe the open top and bottom vent areas a = ld = 1 m^2 (10.8 ft^2.)

>the radiant side is black as possible.

on the outside?

>the 10mj/s is after the losses from the radiant wall's [b]
>side been calculated.

and you are saying the only way heat leaves the a box is via warmer air... 

>... [c] temp = 25 deg c. [b] temp = 30-45 deg c.

say 77 and 100 f.

>what is the best way of calculating the air flow into [b] from [c] in
>litres per second?

here's an empirical chimney formula: cfm = 16.6asqrt(h(tt-tb)), where
tt is the temp (f) at the top of the chimney and tb is the cooler temp
at the bottom, and the air inside the chimney balances an equal height
tb column of air outside. you can find more accurate formulas in the
second edition of passive solar energy by anderson and wells (brick house,
1994) and heating and cooling of buildings by kreider and rabl (mcgraw
hill, 1994) and schaum's outline on heat transfer by pitts and sissom
(mcgraw hill, 1998, a bargain at $14.95.) if you want to go crazy, try
looking at the handbook of heat transfer by rohsenow, harnett and cho
(mcgraw hill, 1998, a non-bargain at $150.)

>[c] is indirectly connected to [b] such that all spaces form
>a convective loop.

if the 100 f air is cooled to 77 f near the top of the chimney, eg by
an air conditioner in the attic, each cfm of airflow with a temp diff dt
(f) moves about dt btu/h, so heatflow cfmxdt = 16.6(10.8)sqrt(10.4)dt^1.5
= 34,120k btu/h and dt = 1518 f (wow), so cfm = 16.6(10.8)sqrt(10.4x1518)
= 22,529 cfm or 10,632 l/s. that's probably way outside the temperature
range of applicability of the formula.

if the 100 f air is cooled to 77 f near the chimney bottom, eg by an earth
tube, the air column inside the chimney is unbalanced by a lighter 100 f
vs 77 f column outside the chimney, so we'd expect less flow.

we might crudely model this as a chimney with a preheater below, with
a 100 f chimney inlet and a t (f) outlet and an external 100 f column,
so cfm = 16.6(10.8)sqrt(10.4(t-100)), but we also have to heat the inlet
air from 77 to 100 f, so 34120k = cfm(100-77)+cfm(t-100) = cfm(t-77)
= 578sqrt(t-100)(t-77), ie t^3-254t^2+21329t-3.48x10^9 = 0, which has
a solution t = 1600 which makes cfm = 22,386, or 10,564 l/s.

nick




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