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re: clothes drying rack?
16 nov 1998
hvacman wrote:
> any suggestions on where to purchase a durable clothes drying rack?
>i'm tired of the rickity wooden type, and refuse to run the electric
>dryer if i don't have to...
how about a more ephemeral indoor clothesline on a retractable spool?
or an insulated closet with a $4.95 humidistat at the top and a few
100 watt bulbs near the floor, or a 200 f sauna closet, with a cheap
heater and a $2.95 thermostat...
dave.garland@wizinfo.com (dave garland) writes:
>...aren't you overlooking the fact that you'll have to be moving air
>through the dryer closet to get rid of the moisture?
ne> yes. how much air? if the closet door is open, we lose lots
ne> of heat. if the closet is perfectly airtight, the moisture
ne> never leaves the closet. there must be an optimum airflow
ne> that dries clothes with minimum energy.
>probably would require some experiment. you need both to circulate air
>within the closet, and pump moisture-laden air out. the typical largish
>muffin fan pumps 30-40 cfm...
this would probably work without a fan--70 f house air at 50% relative
humidity and 13.5 ft^3/lb contains 0.5x0.0158 = .0079 pounds of water
per pound of dry air, and 108 f closet air at 80% humidity (with the
humidistat setpoint higher than house humidity so the bulbs turn off
when the clothes are dry--we could use a differential humidistat) has
0.8x0.0558 = 0.0446 water by weight. the difference is 0.0367, so
replacing closet air with 13.5 ft^3 of room air removes 0.0367 pounds
of water. removing 4 pounds of water (20% of a 20 pound washload)
requires that 4 pounds = 0.0367/13.5xcfmx12hoursx60min, or cfm = 2.
using an empirical chimney formula, 2 = 16.6 av sqrt(8'(108-70)), so
this airflow requires a hole at the top and bottom of the closet with
area av = 0.0069 ft^2 or 1 in^2, eg a 1"x1" slot.
in an airtight house, the top and bottom holes might be connected by a
4" pvc pipe outside of the insulation to make a thermosyphoning condensing
air-air heat exchanger with a drain and bucket at the bottom. that would
work almost as quickly, since 70 f air at 100% humidity contains 0.0158
pounds of water per pound of dry air, vs 0.8x0.0558 = 0.0446 for 80% 108 f
closet air, a smaller difference of 0.0288. we could speed up the airflow
by increasing the hole size...
10 u=224/25'thermal conductance of dryer room (btu/h-f)
20 for power=1500 to 1500 step 100'dryer power (watts)
30 heat=3.412*power'dryer heat (btu/h)
40 rhr=100'relative humidity of room air (%)
50 wr=.015832*rhr/100'room air humidity ratio (water/dry air by weight)
60 d=1/13.5'air density (pounds/ft^3)
70 p=4/2'initial drying rate (pounds of water/hour)
80 for cfm=1 to 5 step 1'airflow (cfm)
90 w=wr+p/(60*cfm*d)'dryer air humidity ratio (water/dry air by weight)
100 pr=29.921/(1+.62198/w)'partial pressure of water in dryer ("hg)
110 td=9621/(17.863-log(pr))-459.57'clausius-clapeyron dew point (f)
120 latheat=60*cfm*d*(w-wr)*1000'latent heat loss (btu/h) [1000=144 for am]
130 t=70+(heat-latheat)/(u+cfm)'dryer air temperature (f)
140 ps=exp(17.863-9621/(t+459.57))'saturation pressure ("hg)
145 rhd=100*pr/ps'relative humidity in dryer (%)
150 pe=(t-td)*100*1.5/1000'estimated drying rate (pounds of water per hour)
160 if abs(pe-p)/p > .01 then p=.99*p+.01*pe: goto 90'iterate to 1%
170 drytime=4/p'drying time (hours)
180 kwh=power*drytime/1000
190 print cfm;t,td,rhd,drytime,kwh
200 next:next
run
cfm temp (f) dew point rh drying time kwh
1 206.7377 181.4805 56.61469 1.064936 1.597405
2 192.7962 167.3976 55.02187 1.060402 1.590603
3 182.529 157.5558 54.53419 1.060402 1.590603
4 174.8427 150.0256 53.93522 1.064045 1.596068
5 168.6737 143.9713 53.43038 1.069371 1.604057
it looks like that 4x8x8' tall r25 box with a 1500 watt heater works
better at sauna than "warm room" temperatures, with minimum clothes
drying energy between 2 and 3 cfm, with a condensing heat exchanger.
nick
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