re: most efficient refrigerator
1 feb 2004
>> >> http://enews.lbl.gov/science-articles/archive/energy-myths3.html and
>> >> http://enduse.lbl.gov/info/lbnl-45862.pdf
>> they were just summaries of studies. you might look at the studies cited.
>do you have a link for the actual studies?
i seem to recall they were cited on those web pages.
>i suspect the studies were comparing 'spotless' coils with those that are
>only moderately dirty. but they do admit that dirt affects the heat
>transfer. *if* the dirt is built up enough to prevent normal operation,
>then it's past time to clean them.
sounds reasonable, if you mean the fridge runs all the time or it
doesn't get cold enough or it doesn't cool warm foods fast enough.
>> it seems to me that the full unit would have more cold surface to cool
>> air that flows down through it, so the air would emerge cooler, and
>> thus flow faster.
>guess you and i have a different idea of what 'full' is. i've been
>picturing something like my upright side-by-side freezer/fridge. and
>thinking of the freezer side.
i've been thinking about a refrigerator, as in the subject line above.
>you seem to be thinking of a fridge with a few soda bottles and a gallon
something like that. is 120 is a few? :-)
>> i wonder what we can agree on. suppose the 2'x3'x5' empty unit has 62 ft^2
>> of interior walls, and the full one has 5 wire shelves with 24 vertical
>> 3" diameter x 10" tall 1-liter soda bottles on a 6" grid on each shelf.
>> surrounded by 3" of air, the bottles would introduce minimal flow
>> resistance, but they would add 90 ft^2 of cold surface with no insulation
>> behind it, giving the full fridge 152 vs 62 ft^2 of cold surface, about
>> 2.5x more...
>no, i think of a freezer...
that seems like a different situation. btw, the bottle surfaces are much
stiffer cool sources than the fridge walls, no? they are thin and backed
by liquids, vs a 1/16" plastic fridge wall with insulation behind it.
the bottles would be more effective condensers, after a few seconds.
>> >> >those food items that aren't touching form channels with a high
>> >> >hydraulic diameter resulting in laminar conditions and much lower flow.
>> i'd estimate that a 1/2" gap would provide minimal airflow resistance.
what would you estimate, in this case? might be a simple duct calc.
>> if the room is 70 f and the fridge is 36, how much free channel area is
>> needed to make the air velocity near the back of the fridge at least 90%
>> of the air velocity near the front, 2' away?
another duct calc?
>you assume only 4 ft^3 of 'stuff' for a 'full' freezer?
just thinking about a fridge, packed as described above.
maybe some beer as well. and yoghurt, and tofu, and lettuce.
>> i have data. this morning my kitchen was 59 f with 40% rh. i opened
>> the door of my ge tbx18tazaerwh fridge (about 5 years old) and used my
>> recently calibrated raytek raynger st ir thermometer to measure the temp
>> of a spot on the inside wall and the temp of a pickle jar on a shelf:
>> time wall temp pickle jar temp
>> (sec) (f) (f)
>> 0 40.0 39.0
>> 15 41.5 39.0
>> 30 42.0 39.0
>> 45 43.5 39.0
>> 60 44.0 39.5
>> 75 45.0 39.5
>> 90 46.0 39.5
>> 105 46.0 39.5
>> 120 46.0 39.5
>> an hour before, i put a small mirror inside the fridge. when the door was
>> open, i saw no condensation on the mirror. the wall (vs pickle jar) temp
>> would increase a lot faster with condensation, no?
>your data doesn't support or refute any position.
imo, it indicates that a square foot of fridge wall would absorb less heat
than a square foot of bottled liquid, when the door is open.
>the mirror was obviously warmer than the fridge when you put it inside...
an hour before the test...
>so any air (with moisture) that entered the unit while you did this
>condensed on cooler surfaces.
i didn't see or feel any condensation anywhere.
>if it's a combination fzr/fridge, some of the air was circulated into
>the freezer compartment and some moisture was condensed there as frost.
this is a top freezer (frost free) unit. i only opened the fridge door.
as i recall, the fan that circulates air between the compartments stops
when the door is open.
>leave the mirror in the unit for several hours, then open the door and leave
>it open as you have suggested. if the mirror is on the upper shelf, i would
>predict it will fog then.
i'll try again, altho i think 1 hour is several mirror time constants.
i suspect the room air was too cold and dry for condensation to occur.
>> they go on to describe various methods to estimate this drop, including
>> a general method by colburn and hougen (1934), but, alas, they are all
>> beyond my expertise. i'd also like to know how air interferes with licl
>> water vapor absorption.
>i don't have any direct experience with licl, but it may be similar to libr
>(certainly they would be chemically similar, lithium and a member of the
>halide group). lithium bromide has been used in absorption refrigeration
>cycles for many years. those units use water as the 'refrigerant' and it is
>absorbed by the libr solution. the libr units i've worked on benefited
>greatly by having a small quantity of isopropyl alcohol added to the water.
>this made the water 'wetter'. this aids the 'boiling' of the water in the
i like the idea of heating a house with a licl pool that absorbs water vapor
that diffuses upstairs from a damp basement floor, in the presence of air,
and cooling a house with a water fountain or fogger inside and a licl pool
on the roof that absorbs water vapor from below, in the presence of air...
>fwiw, in the texts that i have, betak is defined as the "temperature
>coefficient of thermal conductivity". units of 1/f, used to calculate the
>thermal conductivity of a material when it changes appreciably over the
>temperature range in question.
>k(t) = k0*(1+betak*t)
>would this 'fit' where you see it used?? perhaps it is the "temperature
>coefficient of 'x'", whatever 'x' is??
dunno. maybe x is a distance. the text says
the resulting expression relates the local sherwood number to the local
reynolds number, the schmidt number, and the parameter betax. a second
equation relating these quantities results from the condition that the
interface is impermeable to the noncondensible gas.
sh re = betaxsc/(1-omega), where omega = wg /wg
x x oo i.
way over my head...
>for the heat transfer across a unit area used in dehumidification where the
>partial pressure of vapor is small compared to the total, the heat transfer
>through the liquid film must be equal to the heat transfer through the gas
>film plus the latent heat of condensation. since the exact conditions at
>the gas-liquid film interface are unknown, the following equation must be
>solved iteratively through trial and error.
>hl*(ti - tl) = hg*(tg - ti) + lambda*k*(yg - yi)
>hl => liquid film heat transfer coefficient
>hg => gas film heat transfer coefficient
>tl => bulk liquid layer temperature
>tg=> bulk gas temperature
>ti=> temperature at the gas-liquid interface
>lambda => heat of vaporization/condensation
>k=>gas-phase mass-transfer coefficient
>yg => concentration of vapor in gas (bulk)
>yi=> concentration of vapor in gas at the gas-liquid interface
>(subscript 'g' is in the gas, 'i' is at the gas-liquid interface, and l is
also way over my head. how many pounds of water vapor per hour would
condense on a 1 ft^2 vertical surface at 36 f, if it were exposed to
70 f still air at 50% rh?
tg = 70 f? hl = 60 btu/h-f-ft^2? tl = 36 f? lambda = 1000 btu/lb?
yg = w = 0.00792 pounds of water vapor per pound of dry air?
what are hg and k and yi? use ti = (tl+tg)/2 to start?