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re: refrigerator -keep it full?
14 may 1997
theresa mesa wrote:
>you just lost every one of us intelligent readers whose eyes glaze over
>when we see all those numbers interspersed with all those letters and
>symbols, indicating an area of math with which we are unfamiliar,
arithmetic? :-)
>and choose to remain so.
i see :-)
>can you say it in plain english??
i'll try again. (the numbers have changed a bit, btw.)
doug said:
>...once the water gets frozen, the cycles return to a typical on-off pattern
>needed to keep the water frozen...
so we are talking about what happens in the long run, after the ice has
frozen, and the freezer compressor, etc, is keeping the contents cold, by
turning itself on to cool the contents a little bit (say when the interior
temperature rises to 2 f), and off when the freezer is "cold enough" (say
when the inside temperature is 0 f) automatically, under the control of a
thermostat. this makes a cycle that repeats over and over, and the "cycle
time" is the time between complete repetitions of the cycle, if you time it
from similar instant to similar instant, say from one turning-on to the next.
for instance, if you hear the freezer turn on at noon, and it turns off at
12:30, then back on at 1 pm, the cycle time is one hour...
in the us, the flow of heat is typically measured in btu (british thermal
units, no longer used in britain :-) a btu is roughly the amount of heat
in a kitchen match. one btu raises 1 pound of water 1 degree f, ie water
has a heat capacity or thermal mass equivalent to 1 btu/f per pound or a
"specific heat" of 1. so for instance, if you want to make an 8 oz. cup of
tea by raising water from 62 f to boiling (212 f), you need to supply
1/2lb(212-62) = 75 btu, which you could do by burning about 75 kitchen
matches under the cup. or you could put the water on a 750 btu/hour stove
burner, in which case it would take about 75/750 = 0.1 hour or 6 minutes to
boil. if you were heating water for 10 cups of tea in a very-well-insulated
pot on the same 750 btu/hr burner, it would take 10 times longer, about
750/750 = 1 hour to boil.
the more thermal mass, the more heat it takes to warm it up, and the longer
it takes to warm it up, if everything else is the same, eg the stove burner.
a typical freezer might allow about 500 btu per hour to pass from a 70 f room
through the (not very well insulated) freezer walls into the 0 f contents
within, which warms up the contents like a stove burner. the more thermal
mass in the freezer, the longer it takes to warm up, like more water for tea.
a 20 ft^3 freezer absorbing 500 btu/hr of room heat through the walls, with
a 1,000 btu/hr capacity cooler will run 12 hours per day, consuming about
the same energy, whether it's full or empty. the interior temperature extremes
will be the same whether it's full or empty, but the peaks and valleys (within
the narrow range of the thermostat) will happen more often if it is empty. if
the cooling unit needs to supply 24x500 btu/hour = 12,000 btu/day of cooling,
and we assume it is capable of supplying 1,000 btu/hour, then it has to run
12 hours per day to supply 12,000 btu/day, n'est-ce pas?
if it's empty, with say, 50 btu/f of inherent thermal mass (ie the walls and
shelves and such of the empty freezer have a thermal mass equivalent to 50
pounds of water (see teacup example 1.4.3)) and the cooler turns on at 2 f
and off at 0 f (ie the freezer thermostat has a "deadband" of 2 f), it will
take 50x2/(1000-500) = 0.2 hours to cool the inside from 2 to 0 f, and the
room will warm it from 0 to 2 f in 50x2/500 = 0.2 hours, so the cycle time
will be 0.4 hours or 24 minutes, 12 minutes on, and 12 minutes off.
if the freezer contains 20 ft^3 of ice with a heat capacity of 640 btu/f
(somewhat irrelevant physics fact: ice has about half the heat capacity of
water), the cooler will take 640x2/(1000-500) = 2.56 hours to lower the ice
temperature from 2 to 0 f, and the room will warm the ice from 0 to 2 f in
640x2/500 = 2.56 hours, so the cycle time will be 5.12 hours.
each time the freezer turns on, it consumes about 1 kw for about 1 second.
in the first case it turns on 60 times per day, in the second, about 5 times
per day, so the difference in energy used for starting is about 55 btu/day,
ie about 0.016 kwh/day, worth about 0.0016 cents at 10 cents/kwh, a penny
or so every 2 years.
is that better? (i wonder if this would make sense without any numbers...)
nick
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