Declared/published performance of the 847 A Class machine is:
>With test conditions as stipulated in EN 61121- European standard.
>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.
>
>Please note that modern washing machines spin dry at higher spinning speeds
>than when the standard was written so better performance can be expected...
... 2.5kWhx3412/7.7lb = 1108 Btu/lb. Given the long drying time, this might be vented indoors... 7.7/8 = 0.96 lb/h, so it would only raise the indoor humidity ratio of an average house in Phila with 224 cfm of air leakage in January to 0.0025+0.96/(60x224x0.075) = 0.00345, ie 22% RH at 70 F.
But is there a non-heat-pump machine that dries with less than
1000 Btu/lb of electrical energy, like an indoor clothesline?
Yeah. An indoor clothesline. It is very common in my area to use a folding wire rack, sold expressly for that purpose. Works especially well if you have direct sunlight coming in your window.
I would only go back to using mechanical dryers if I were living in an extremely small space, and/or had to use a public laundromat, where I didn't want to haul wet clothes back home.
Cost: One-time $15 for the rack, plus zero ongoing.
Why do people have such an attachment to those mechanical power-hogs?
Electric dryer? In any case while I would agree that good filters can stop most of the lint, I personally would prefer not to have any and to vent it outside.
I can't answer your question, but since we're on the topic, for interest I just monitored one of our typical laundry loads.
Machine is a Splendide 2000, vented model. Capacity advertised at 5kg washing, 3kg washing/drying. Set for regular wash, warm/cold, high spin, half heat, 30 minutes drying.
Load was a mix of shirts, T-shirts, hand towels, washcloths, socks and underwear. 5lb according to an electronic bath scale which is too crude for the purpose. IMO the machine would be quite happy with a 50% larger load. But since my wife disagrees and I don't need any more chores... ;-)
Assuming I didn't miss anything (I didn't stand over it the whole time), it filled once and spent about 45 minutes washing. In this mode it spends about 12 seconds turning one way, rests for 5, reverses. About 180W while turning according to a KillaWatt, on which the display never really stabilized due to the constantly changing consumption.
After the wash cycle, the machine drains and refills for rinse, short wash action, drains and spins for a short time. Repeats, two rinses total.
Next it enters spin mode, perhaps 10 minutes of various tumbling, spin speed ramp-up and pumping. Power consumption tops out at about 400W until it goes into scary-spin mode, about 500W steady for another 10 minutes.
70 minutes to this point, .21kWh. Removed clothes and weighed, about
1.5lb gain. Too wet to hang up IMO. Clothes would probably drip, and dry crispy.
Replaced clothes for drying, 700W steady for 20 minutes. Vent air is very gentle flow, barely warmer than room air. Last 10 minutes is by design tumbling with heat off. Consumption down to about 100W when drum is turning, about 150W with pump combined.
Total time 1:45. Total consumption .45kWh. No detectable weight loss since spinning, but my scale was too useless to tell. Clothes were definitely drier though, plenty dry enough to hang up. Shirts clammy but wearable in a pinch on a summer day. Although I'm sure that any self-respecting yuppie would leave them in at least another half hour.
I didn't stand over it, so it may have rested longer at times. And I'd expect that the other programs would include less washing, perhaps less rinsing as well. A chart in the manual seems to confirm that. For the "knits and delicates" program it shows "S" agitate speed, and 650 spin speed. But the energy for agitating on our machine is perhaps
60Wh per load, so there's not a lot to save regardless.
Perhaps they should be rated in Gs to account for drum diameter. :-) Anyway, I looked it up for you - 1100 RPM max. How does it compare?
More correctly, the load weighed that much after spinning. My scale wasn't accurate enough to detect the weight loss after drying so I described it as best I could. We have one of those chintzy baby scales as well, so I may try using that on another load for interest. Since the thinner items were relatively dry, I expect that most of the moisture still remaining was in the towels and washcloths. I don't generally pay this much attention to the washing, but in my new role as Iaundry critic, I noticed this morning that those towels are definitely crispy to the touch. I suppose for maximum efficiency one could remove the lighter items after a half-hour and leave the heavy ones in longer. But I will *not* be making any recommendations. :-)
It wasn't an experiment, merely some quick and dirty observations to give an idea of how one machine works and how we use it. Perhaps I'll post similar notes on the dishwasher as well since these per-use energy consumption figures with setting details seem to be hard to come by.
ditto. I was a very obvious joke. (LOL may indicate this)
That's a pretty fast spin cycle. It's a good thing they are only off balance when they first start up. Mine has danced across the rom a few times before the pinball tilt switch picks up on it and shuts it down.
... 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.
They are, on some sites.
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%
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%.
... 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.
800 rpm spin is basic, 1000 so so, top end ones do 1400 now. At 800 clothes come out saturated, at 1300 they come out no more than damp. Theres really no need for a dryer if youve got a fast spin washer, but for some reason driers have become common anyway.
(The old twintubs used to do around 2000-2500 rpm, but that was with a much smaller drum.)
Machine design is an ongoing issue, getting them to do high spin speeds without jumping about, pummelling other nearby appliances or dancing across the floor. This is a common fault mode. Modern machines need to have off balance load sensors so they dont try to spin up when not distributed correctly, If they did, the drum housing would break. As you can guess, heavy vibration is normal.
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