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re: efficient european clothes dryers
6 dec 2005
wmbjk wrote:
http://www.crosslee.co.uk/cl847.html
>>>>>dry cotton load = 5kg, room temperature 20c
>>>>>energy to dry the above load in a class mode = 2.5 kwh
>>>>>wetted condition of test load = 70% so 3.5kg of water or 7.7 lbs
>>>>>time to dry the test load under these conditions = 8hrs.
>>>>
>>>>... 2.5kwhx3412/7.7lb = 1108 btu/lb.
... of electrical energy, vs 0 btu/lb for an indoor clothesline.
i wonder how often the drum moves and how much of that energy
comes from the motor.
>>scary-spin mode sounds like fun. uk machines are rated by rpm.
>
>perhaps they should be rated in gs to account for drum diameter. :-)
they are, on some sites.
>anyway, i looked it up for you - 1100 rpm max. how does it compare?
i'm not sure. one us doe site lists "remaining moisture content" standards
after spins...
warm spin cold spin
15 min 4 min 15 min 4 min
100 gs ~45% ... 50%
...
500 gs 24% ... 30%
>>>total time 1:45. total consumption .45kwh. no detectable weight loss
>>>since spinning, but my scale was too useless to tell.
digital scales are getting cheaper...
~~~~~
if we can tumble-dry a load of clothes containing 5 pounds of water in
0.5 hours at 130 f with ps = 4.53" hg and pd = 0.374 (?) (70 f at 50% rh
with wd = 0.00788, approximately) and 0.1a(ps-pd)0.5 = 5, using an ashrae
swimming pool formula, we might say their equivalent area a = 24 ft^2.
let's arbitrarily reduce this to 10 ft^2, with no tumbling, which makes
the numbers easy: drying time = 5/(ps-pd).
so an indoor clothesline in free air at 70 f and 50% rh might dry in 5/0.374
= 13.3 h at an approximate twb = 9621/(22.47-ln(460+70+37.4-twb). plugging in
510 r (50 f) on the right makes twb = 522 on the right, then 516, 519, 517.3,
518.5, 517.8, 518.2, and 518.0 r (58.0 f)
if we dry clothes in 20 hours in a closet with c cfm of airflow at 0.25 lb/h,
pd = 29.921/(1+0.62198/(0.00788+0.25/(4.5c)) = (4.2441c+29.921)/(11.338c+1)
and ps = e^(17.863-9621/(460+t)) = pd + 0.25 = (7.0583c+30.171)/(11.338c+1).
if the only heat comes from room air, (70-t)c = 1000p = 250, so t = 70-250/c
= 9621/(17.863-ln(ps))-460, ie c = 250/(530-9621/(17.863-ln(ps))). plugging
in c = 100 on the right makes c = 60 cfm on the left, then 74, 67, 70, 68.5,
69.0, and 68.8, which makes t = 70-250/68.9 = 66.4 f, approximately.
if we speed this up with closet insulation and heat, 10 hours at 100 f makes
ps = 1.979 "hg, pd = ps - 0.5 = 1.479, wd = 0.62198/(29.921/pd-1) = 0.03234
= 0.00788 + 0.5/(4.5c), and c = 4.54 cfm (not much), with 10h(100-70)4.54
= 1363 btu of heat, about 0.4 kwh, only 27% of the water's latent heat :-)
with good insulation, longer drying times and higher temps and less airflow
minimize the electrical energy needed for drying: 5 hours at 120 f make ps
= 3.579 "hg, pd = ps - 1 = 2.579, wd = 0.62198/(29.921/pd-1)) = 0.05867, and
c = 4.38, with 5h(120-70)4.38 = 1094 btu, ie 0.32 kwh... 10 hours makes pd
= ps - 0.5 = 3.079, wd = 0.62198/(29.921/pd-1)) = 0.07134, and c = 1.75 cfm,
with 10h(120-70)1.75 = 875 btu, ie 0.26 kwh. this might come from a holmes
hfh111 1500 w fan space heater ($12.88 at wal-mart) with its thermostat set
to 130 f (if that's below the upper temp limit) running 100x260wh/(1500wx10h)
= 1.7% of the time.
if we dislike stiff clothes and don't mind extra labor, we might put
a dryer inside the closet with a sequencer that only turns it on for
1 out of every 10 minutes and a humidistat and fan that circulates
room air through the closet when the rh reaches 80%.
nick
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