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Evaporation through HDPE - Page 4

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Posted by David Delaney on October 4, 2003, 2:27 am
Well, I finally found that the usual language for loss of water vapor
through plastic is "water vapor transmission rate" or "WVTR". Google
then allowed me to find several claims for different HDPE products,
all reasonably close to HDPE WVTR = 1 g*mil/(100in^2*24hr) at 100F and
100% RH.

A standard 5 gallon HDPE bucket is about 12 inches in diameter and 15
inches high, for a total surface area of 790 in^2.  The thickness of
the walls and lid is approximately 90 mil.

total WVTR for the bucket, WVTRt,
WVTRt = WVTR*(790/100)*365/90
      = 3.2 g/year

WVTR increases with temperature by a fair amount, lets guess a factor
of 10, going from 1 to 10 when the temperature increases from 100F to
190F (the hot fill limit of HDPE buckets), for a total of 32 g/year.
Over thirty years this would give a total loss of 960 g, say 1 kg, out
of the 19 kg in the bucket.

If nearly correct, this says that evaporation through a cracked seal,
or some other perforation of the bucket, gives a greater threat of
water loss. Anyone have a view of this probability? Does anyone have a
better view? Anyone know if the level of water in a "natural" HDPE
bucket can be seen if you shine a strong light on the bucket?

David Delaney, Ottawa

On Tue, 30 Sep 2003 15:00:46 GMT, ddelaney@sympatico.ca (David
Delaney) wrote:

Posted by Cosmopolite on October 4, 2003, 4:06 pm

David Delaney wrote:

Your water loss calculation is based on 190 F., does this mean that you
to run this system at that temp?

Posted by David Delaney on October 6, 2003, 12:50 pm
 On Sat, 04 Oct 2003 16:06:28 GMT, Cosmopolite

Nope. Just playing with extremes.  A collector
operating by natural convection might be located
below the thermal mass. The collector would likely
have a nonselective absorber. I don't think
temperatures would exceed 125-145F (50-60C).
Temperatures higher than this should be prevented
by the losses of the collector and the guaranteed
availability (since there are no fans to fail) of
the thermal mass to cool the collector air.

The pails of water might be stacked on industrial
steel shelving of the kind used to store pallets
of goods. Such shelving is pretty cheap. It's open
front and back, and comes with wire mesh decking.
(The industrial shelf industry calls a shelf a
"deck".) It comes rated for far heavier loads than
you need to stack pails of water. Air could
circulate freely in any direction around pails in
such a stack, and enter the stack at any level.
The stack could accept air rising convectively
from a collector below it at any temperature above
the temperature of the lowest pails in the stack.
To maintain thermal stratification, the stack
could be arranged so that the hot air would rise
beside the stack from the collector below,
entering the stack at the level at which the
buoyancy of the rising air becomes neutral and its
temperature matches the temperature of the air in
the stack.

David Delaney, Ottawa

Posted by John on October 8, 2003, 12:19 pm
 You could fill each pail to a specific weight and simply weigh them every X
number of years and refill as necessary.

With all that steel, how much heat will the steel take in?

Posted by David Delaney on October 8, 2003, 4:08 pm
 On Wed, 8 Oct 2003 08:19:05 -0400, "John"

The water on one bay of shelves would weigh 1200
kg, the steel dedicated to the bay would not weigh
more than, say, 400 kg. The specific heat of steel
is 1/9 the specific heat of water, so a weight of
steel holds 1/9 the heat of the same weight of
water at the same temperature, and the steel of
the bay would store at most (400/9)/(400/9 + 1200)
~ 3.6%  of the heat stored by the steel and water
of the bay, a little more for a stratified store,
because the temperature of the steel would be much
less stratified.

David Delaney, Ottawa

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