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