DIY Dehumidifier

Melody,

I have written a few articles on humidity and mold issues that were published. Here are links to some of them.

Crawlspace problems:

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We're in the mold business:
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Humidity Control:
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Stretch Kevin O'Neill

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stretch
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Thank you for explaining who DH is. I was wandering ... Dear Husband or Deceised Husband but did not want to ask. :

Reply to
MG

I measured a new efficient dehumidifier, which produced 1.6 kWh of heat for every kWh consumed, ie 1 kWh from motors plus 0.6 kWh to condense about 2 pints of water, ie it consumed 0.5 kWh per pound of water, ie

5 cents/pint at 10 cents/kWh. A 70 F basement with 60% RH has humidity ratio w = 0.00947 pounds of water per pound of dry air. NREL says w = 0.008 for outdoor air on an average May day in Phila with a 62.9 F 24-hour average temp and 52.7 and 73.1 daily min and max. On an _average_ (vs dry) May day, we might run Lasko's 16" 2470 cfm 90 watt 2155A window fan ($53 from Ace Hardware) for an hour and remove 60x2470x0.075(0.00947-0.008) = 16.34 pounds of water at 90/16.34 = 5.5 Wh/pint, ie $0.1x90/1000/16.33 = 0.055 cents/pint, 91X less than the dehumidifier.

But some days are drier than average. Running the fan longer on dry days vs running it every day would be more efficient. We might invent a nice adaptive algorithm with the help of a simulation using TMY2 weather data.

And the outdoor humidity ratio (vs the RH) doesn't change much over a day, so we might as well run the fan when it's warmer outdoors if the house needs heat or when it's cooler outdoors if the house needs cooling.

Nick

Reply to
nicksanspam

A computer with sensors can. Check out the U Mass team house at the 2005 DOE DC Decathlon contest in October.

Radio Shack sells "data collectors." Here's some "proper software" for an $8 Casio FX-260 calculator: ventilate when Ro < Rie^(9621(1/(460+To)-1/(460+Ti)), ie Ro < Rie^(9621(1/(460+To)-1/530) if it's 70 F indoors. For example, if it's

70 F with 80% RH indoors and 80 F outdoors, ventilate when Ro < 0.8e^-0.336 = 0.57, ie the outdoor RH on the $20 Radio Shack display is less than 57%.

I've noticed that few a/c contractors have strong HVAC backgrounds.

Raleigh looks fairly humid, but a properly controlled exhaust fan might help from October (w = 0.0081) through May (w = 0.0099). July is dampest, with w = 0.0149 on a 78.1 F average day with 68.1 and 88.0 daily min and max. An "airtight" house with 15 cfm of natural air leakage (vs. 2.5 cfm in Canada) and w = 0.0120 indoors would need about 24hx60x15x0.075(0.0149-0.0120) = 4.7 lb/day of dehumidification from air leakage plus about 2 gallons per day from human activities (Andersen's estimate for a family of 4), about 21 lb/day.

On an average July day, 1920 Btu/ft^2 falls on the ground and 750 falls on a south wall in Raleigh. We might have an EPDM rubber liner with a passive greenhouse-type solar still with shallow LiCl lakes separated by dry EPDM beds to act as water collectors and parasitic air heaters, like this, viewed in a fixed font like Courier:

| 2' | carbo poly s nate flat p clear clear a flat c poly nate... e carbo epdm LiCl r lake epdm heater dry bed epdm epdm 2x4 epdmepdmepdm 2x4 epdmepdmepdmepdmepdmepdmepdm 2x4 epdm

--------------------------------------------- top of SIP ------

(What's a good lake to heater area ratio?)

How many square feet of 80 F LiCl solution (precooked to 160 F) are needed to remove 20 pounds of water from 80 F house air with w = 0.012 in 12 hours? Here's a 9-pound calc for Miami, based on some crude assumptions:

1) The LiCl still operates at a constant temp for 12 hours per day. 2) The solar energy that enters the R1 glazing with 90% transmission equals the sensible and latent heat energy needed for concentration. 3) The solution cools to 25 C at night. 4) The solution gains heat like an ASHRAE pool loses heat.

The next step might be a simple TMY2 simulation.

10 A1=12.7409'LiCl vapor pressure constants from the 1993 Hawlader paper 20 A2=-.065536 30 A3=-8.2416E-04 40 B1=-4675.4 50 B2=+29.31 60 B3=+.66911 70 C1=372690! 80 C2=-1689.8 90 C3=-187.1 100 TA=82.8'average ambient August temperature in Miami (F) 110 SG=1770'average August sun on ground in Miami (Btu/ft^2-day) 120 H=12'distillation day length (hours) 130 W=.0176'average ambient August humidity ratio in Miami 140 PV=25.4*29.921/(1+.62198/W)'ambient vapor pressure (mmHg) 150 P=9'dehumidification load (lb H2O/day) 160 FOR TC=60 TO 90 STEP 10'solution temp (C) 170 TK=273.1+TC'solution temp (K) 180 C=A1+B1/TK+C1/TK^2-LOG(PV)/LOG(10) 190 B=A2+B2/TK+C2/TK^2 200 A=A3+B3/TK+C3/TK^2 210 CONC=(-B-SQR(B^2-4*A*C)/(2*A))'equilibrium soln conc (wt%) 220 TF=1.8*TC+32'solution temp (F) 230 CONCSURF=1000*P/(.9*SG-H*(TF-TA))'LiCl surf needed for conc (ft^2) 240 TK=298.1'solution temp (25 C) 250 AP=A1+A2*CONC+A3*CONC^2 260 BP=B1+B2*CONC+B3*CONC^2 270 CP=C1+C2*CONC+C3*CONC^2 280 PVC=10^(AP+BP/TK+CP/(TK^2))'vapor pressure at 25 C (mmHg) 290 PVI=29.921/(1+.62198/.012)'indoor vapor pressure ("Hg) 300 PVL=PVC/25.4'LiCl vapor pressure ("Hg) 310 DRYRATE=.1*(PVI-PVL)'lb/h/ft^2 H2O (like an ASHRAE pool) 320 DRYSURF=P/(12*DRYRATE)'LiCl surface needed to dry P lb H2O in 12 h (ft^2) 330 PRINT TC,CONC,PVC,DRYSURF,CONCSURF 340 NEXT

still solution LiCl Pv drying concentrating temp (C) conc (wt%) (mmHg) surf (ft^2) surf (ft^2)

60 39.15389 5.493444 21.42437 9.927201 70 45.89019 2.75522 16.3801 13.03215 80 52.33653 1.244954 14.49746 18.96333 90 58.57794 .5091767 13.72873 34.80277

If the still temp is too low, it looks like we need lots of drying surface. If it's too high, we need lots of concentrating surface. A 70 or 80 C still temp seems good...

Nick

Reply to
nicksanspam

Reply to
unknown user

Possibly. Are you a nubile senior with similar interests? :-)

That's a Clausius-Clapeyron approximation.

Nick

Reply to
nicksanspam

Nick wrote: I measured a new efficient dehumidifier, which produced 1.6 kWh of heat

for every kWh consumed, ie 1 kWh from motors plus 0.6 kWh to condense about 2 pints of water, ie it consumed 0.5 kWh per pound of water, ie

5 cents/pint at 10 cents/kWh.

Nick,

How can a dehumidifier produce mor heat than the equivalent of the electricity it uses. A heat pump does that because it extracts heat from outside. A dehumidifier is usually self contained...the whole thing is all in the same space. All electricity consumed should be converted to heat. Even though it converts the water vapor to liquid water, the total heat is still there because it never left the room, only changed form.

Stretch

Reply to
stretch

It's an indoor heat pump that converts latent to sensible heat, with a COP of 1.6, in the case above.

Not always. The dehum above is 60% more efficient than electric resistance house heating in wintertime. Bill Shurcliff suggests air conditioning damp basements in wintertime. The AC might be mounted in a stairwell, with the hot coils in the living space.

It is, but beyond that, each pint of water condensed adds about 1000 Btu of sensible heat to the room air. Evaporating water requires heat energy. Condensing water releases heat energy.

Nick

Reply to
nicksanspam

Have you been voted most boring nerd on campus yet?

=======================

A computer with sensors can. Check out the U Mass team house at the 2005 DOE DC Decathlon contest in October.

Radio Shack sells "data collectors." Here's some "proper software" for an $8 Casio FX-260 calculator: ventilate when Ro < Rie^(9621(1/(460+To)-1/(460+Ti)), ie Ro < Rie^(9621(1/(460+To)-1/530) if it's 70 F indoors. For example, if it's

70 F with 80% RH indoors and 80 F outdoors, ventilate when Ro < 0.8e^-0.336 = 0.57, ie the outdoor RH on the $20 Radio Shack display is less than 57%.

I've noticed that few a/c contractors have strong HVAC backgrounds.

Raleigh looks fairly humid, but a properly controlled exhaust fan might help from October (w = 0.0081) through May (w = 0.0099). July is dampest, with w = 0.0149 on a 78.1 F average day with 68.1 and 88.0 daily min and max. An "airtight" house with 15 cfm of natural air leakage (vs. 2.5 cfm in Canada) and w = 0.0120 indoors would need about 24hx60x15x0.075(0.0149-0.0120) = 4.7 lb/day of dehumidification from air leakage plus about 2 gallons per day from human activities (Andersen's estimate for a family of 4), about 21 lb/day.

On an average July day, 1920 Btu/ft^2 falls on the ground and 750 falls on a south wall in Raleigh. We might have an EPDM rubber liner with a passive greenhouse-type solar still with shallow LiCl lakes separated by dry EPDM beds to act as water collectors and parasitic air heaters, like this, viewed in a fixed font like Courier:

| 2' | carbo poly s nate flat p clear clear a flat c poly nate... e carbo epdm LiCl r lake epdm heater dry bed epdm epdm 2x4 epdmepdmepdm 2x4 epdmepdmepdmepdmepdmepdmepdm 2x4 epdm

--------------------------------------------- top of SIP ------

(What's a good lake to heater area ratio?)

How many square feet of 80 F LiCl solution (precooked to 160 F) are needed to remove 20 pounds of water from 80 F house air with w = 0.012 in 12 hours? Here's a 9-pound calc for Miami, based on some crude assumptions:

1) The LiCl still operates at a constant temp for 12 hours per day. 2) The solar energy that enters the R1 glazing with 90% transmission equals the sensible and latent heat energy needed for concentration. 3) The solution cools to 25 C at night. 4) The solution gains heat like an ASHRAE pool loses heat.

The next step might be a simple TMY2 simulation.

10 A1=12.7409'LiCl vapor pressure constants from the 1993 Hawlader paper 20 A2=-.065536 30 A3=-8.2416E-04 40 B1=-4675.4 50 B2=+29.31 60 B3=+.66911 70 C1=372690! 80 C2=-1689.8 90 C3=-187.1 100 TA=82.8'average ambient August temperature in Miami (F) 110 SG=1770'average August sun on ground in Miami (Btu/ft^2-day) 120 H=12'distillation day length (hours) 130 W=.0176'average ambient August humidity ratio in Miami 140 PV=25.4*29.921/(1+.62198/W)'ambient vapor pressure (mmHg) 150 P=9'dehumidification load (lb H2O/day) 160 FOR TC=60 TO 90 STEP 10'solution temp (C) 170 TK=273.1+TC'solution temp (K) 180 C=A1+B1/TK+C1/TK^2-LOG(PV)/LOG(10) 190 B=A2+B2/TK+C2/TK^2 200 A=A3+B3/TK+C3/TK^2 210 CONC=(-B-SQR(B^2-4*A*C)/(2*A))'equilibrium soln conc (wt%) 220 TF=1.8*TC+32'solution temp (F) 230 CONCSURF=1000*P/(.9*SG-H*(TF-TA))'LiCl surf needed for conc (ft^2) 240 TK=298.1'solution temp (25 C) 250 AP=A1+A2*CONC+A3*CONC^2 260 BP=B1+B2*CONC+B3*CONC^2 270 CP=C1+C2*CONC+C3*CONC^2 280 PVC=10^(AP+BP/TK+CP/(TK^2))'vapor pressure at 25 C (mmHg) 290 PVI=29.921/(1+.62198/.012)'indoor vapor pressure ("Hg) 300 PVL=PVC/25.4'LiCl vapor pressure ("Hg) 310 DRYRATE=.1*(PVI-PVL)'lb/h/ft^2 H2O (like an ASHRAE pool) 320 DRYSURF=P/(12*DRYRATE)'LiCl surface needed to dry P lb H2O in 12 h (ft^2) 330 PRINT TC,CONC,PVC,DRYSURF,CONCSURF 340 NEXT

still solution LiCl Pv drying concentrating temp (C) conc (wt%) (mmHg) surf (ft^2) surf (ft^2)

60 39.15389 5.493444 21.42437 9.927201 70 45.89019 2.75522 16.3801 13.03215 80 52.33653 1.244954 14.49746 18.96333 90 58.57794 .5091767 13.72873 34.80277

If the still temp is too low, it looks like we need lots of drying surface. If it's too high, we need lots of concentrating surface. A 70 or 80 C still temp seems good...

Nick

Reply to
Gideon

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