re: efficient european clothes dryers SOLAR.KnowledgePublications.com

re: efficient european clothes dryers
17 dec 2005
http://www.crosslee.co.uk/cl847.html describes

>>>>>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, about 10% more than the latent heat, vs 0 btu/lb
for an indoor clothesline. how often does the drum move and how much of
that energy comes from the motor? us dryers have no yellow energy labels,
but they probably use a lot more, on the order of 5kw x 1 hour.

if we can tumble-dry a load of clothes containing 5 pounds of water in
0.5 hours at 130 f with vapor pressure ps = 4.53" hg near the clothes and
pd = 0.374 "hg(?) in dryer air (70 f at 50% 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).
an indoor clothesline with lots of 70 f airflow might dry clothes in 5/0.374
= 13.6 hours.

drying in 20 hours in a t (f) closet with c cfm of airflow at 0.25 lb/h
makes wd = 0.00788+0.25/(60x0.075c) and pd = 29.921/(1+0.62198/wd)
= (4.2441c+29.921)/(11.338c+1) and ps = pd+0.25 = (7.0583c+30.171)/(11.338c+1).
if heat only comes from room air, (70-t)c = 1000p = 250 btu/h, approximately,
t = 70-250/c = 9621/(17.863-ln(ps))-460, ie c = 250/(530-9621/(17.863-ln(ps))),
using a clausius-clapeyron approximation. 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 = 66.4 f, approximately.

drying in 8 hours in a 130 f closet makes pd = 4.53-0.625 = 4.47 "hg, and
t = 9621/(17.863-ln(pd))-460 = 128 f at 100% rh, so we might condense 5 lb of
water on a 128 f or cooler surface. if p pounds of water starts at 70 f and
warms to 128, (128-70)p = 5000 btu makes p = 86 pounds, eg 20 2-liter 4"x12"
soda bottles in a 16"x20"x12" tall box, adding no water vapor to house air,
with 5000 btu of drying energy from a heater, with perfect closet insulation. 

now suppose we have 2 86 pound heat batteries, a at 70 f and b charged up to
about 130 f with condensation from drying a load of clothes, and we remove
the clothes and put in another load and use b to heat closet air until a
reaches 100 f, then use the heater to warm closet air until a reaches 130
and the clothes are dry, with half the usual energy. then we cool b to 70 f
with room air and repeat the cycle. what can we do with 3 batteries?

is there a continuous- vs discrete-battery process that can dry clothes with
a small fraction of their latent heat? an efficient condensing clothes dryer
might use very little energy to move liquid water from one place to another.

nick