daestrom wrote:

The ice box might be 2'x2'x8' tall, with a 55 gallon drum packed with
100 2 liter water bottles surrounded by an antifreeze solution
thermosyphoning through an old 800 Btu/h-F auto radiator above with
an 8' chimney above. Cold air might enter the radiator and flow out
at the top of the chimney.

With 800 Btu/h-F of conductance from 32 F water to outdoor air at temp T,
(32-T)800 = 3456 Btu makes T = 32-3456/800 = 27.7 for 1-hour per day
cooling and T = 32-3456/1600 = 29.8 for 2-hour cooling, and so on.

With a radiator pump and fan and a differential thermostat and a fridge pump
and a fridge thermostat and a compact ice box with the drum on the ground.
All low power stuff, but can we omit some? What's a good compromise?

The radiator might have a chimney, and it might stay primed for a long time,
with both supply and return pipes underwater, and the fridge loop might stop
working before it freezes the fridge...

How do antifreeze or salt water thermosyphon when cold?

I figured cold water weighs about 62.67 - 0.003T lb/ft^3...

You mentioned glycol, so assuming a 50/50 mix, the density change is more
like 0.01884 lb/ft^3-F...

Hey, 6X better for thermosyphoning :-)

Even weak solutions aren't very helpful, using just 10% glycol would give
about 0.001 lb/ft^3-F. A bit of a trade off, higher mixtures give better
density change per degree, but hurt viscosity and heat capacitance.
Probably for a given operating temperture, an 'optimum' mixture could be
determined.

....50/50 sounds good, if the viscosity doesn't increase much. We might
not care much what happens much below 32 F, when all the water bottles
are frozen. Do we need water bottles, vs slush?

The density difference caused by the temperature difference between up and
down pipes causes a pressure difference proportional to the height of the
water column, making a flow through the resistance of the pipe loop. With
8' of height and a dT temperature difference, dP = 0.024dT lb/ft^2.

...dP = 0.01884dT * H, for dP = 0.15072dT for 50/50 glycol mix running 8'.

OK.

A pipe with radius r and length L in feet and pressure diff dP has laminar
flow Q = Pir^4dP/(8MuL) ft^3/s. Viscosity Mu is 6x10^-7 lb-s/ft^2 at 32 F.
With 16' of 1/2" pipe, 8' up and down, Q = Pi(0.25/12)^4dP/(8x6x10^-7x16')
= 0.0077dP ft^3/s or 1729dT lb/h,

No Way!!!

Hmmm.

Check some references, viscosity of water is more like 3.7x10^-5 lbf-s/ft^2
at 32 F (about 1.8 centipoise). Since your number is about 1/32 of the
right value, I suspect your's is missing a 'g sub c' division or something
(slugs??) (damned imperial units ;-).

Thanks for the correction. How many legs in a centipoise?

50/50 glycol is much worse, about 1.88x10^-3 lbf-s/ft^2 (about 9 centipoise).
For laminar flow... I use Q = dP*(d^4)/(.000273*mu*L) (Q in gpm, dP in psi,
d in inches, mu in centipoise and L in ft) (Crane TP 410 pg 3-2).

Where do we get a Crane TP 410 pg 3-2?

That gives Q = (0.15072/144 * dT)*(.5^4)/ (.000273*9*16) for about .001664dT
gpm for 50/50 glycol.

50/50 glycol is about 0.746 BTU/lbm-F so you would get 0.001664*dT*9*0.746*60
BTU/hr or 0.67 BTU/hr /F.

A gallon of 50/50 glycol weighs 9 pounds?

The flow is 0.898dT pounds per hour, right? And each pound moves 0.746dT Btu,
so the flow moves 0.898dTx0.746dT = 0.67dT^2 Btu/h.

Replace the 1/2 pipe with 2" (always good idea to
use large diameter pipe with thermosyphoning :-)

Thanks for the improvement. The radiator might have 2" hoses...

and you can get about 171 BTU/hr/F.

....171dT^2 Btu/h?

At that, you would need about 4.7 F delta T to get your target 800 BTU/hr.

Or maybe sqrt(800/171) = 2.2. Then again, maybe it's -26 F outdoors.

Nick