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wrote:

the numbers are irrelevant and confuse the issue. we are talking about a principle.

you are saying moist air takes less energy to heat than dry air? this goes against what i believe to be true. water takes more energy to heat than a similiar volume of air. add water to the air and its harder to heat.

im not saying im right. im saying id like to see your proof. if im wrong i'll admit it.

randy

nick will never get it he still doesn't understand OP does not want
humidity added and his idea will add moisture. He has no logic or hands
on experience with homes or engineering. His "winter humidification
wastes energy posts" and "flood your basement floor" posts prove that. I
dough`t he has a house or controls anything about his residence. If he
did hands on experiance would have tought him a few things. Nick could
keep this stupid thread going forever, he will never get it.

Numbers often make the difference. For instance, winter humidification might actually save heating energy in a house if a) a 30% humidity increase made 60 vs 69.5 F air as comfortable as 70 F for people, or b) an average house leaked 100X less air than it does, or c) we could evaporate a pound of water with 10 vs 1000 Btu, but the realities are different, and we might never realize that without numbers. And some things look good in principle, as far as that thinking goes, eg powering a house with a rainwater micro-hydro- electric setup at the base of a downspout, until we estimate that it would only produce 1 average watt (about 1 kWh/year) with a capital cost of $1000. Numbers can give a sense of proportion to wacky ideas.

No, but the difference is small and irrelevant, IMO.

About 4000X more. Heating 1 ft^3 of water 1 F takes 62.33 Btu, vs 0.24x0.075 = 0.018 Btu for 1 ft^3 of air.

But we can only add about 1% water vapor by weight. We might heat a 0.075 lb ft^3 of air containing 1% water vapor 1 F with 0.99x0.018+0.01x0.075x1 = 0.01857 Btu, ie 3% more energy.

That's more polite than your previous demand. Let's see... You wrote:

once you pump all that moisture in it is harder to heat later. so your furnace works harder.

Let's assume it's cold outside and the house temp is constant, say 70 F. The dryer heat recovery scheme I have in mind wouldn't add any moisture to the house air, so let's say you have a different scheme, eg merely filtering the lint out of an electric dryer output and putting the warm moist air into the house, adding 12K Btu of house heat while the furnace supplies 12K Btu less. Let's assume the house air is dry, so the water vapor does not condense indoors. It just raises the RH and specific heat of the house air. But wait... if the house air temperature is constant, we don't have to heat it, so its specific heat doesn't matter.

OK, you say, how about "later"? If we fill up the house with humid air at night and a furnace setback lets the house cool to 60 F by morning, (and the house is airtight :-) the house air is (a small ) part of the thermal capacitance of the house, so the amount of energy it gives to the house as it cools equals the amount of energy it takes back when it warms. Unless you believe the myth that temperature setbacks don't save energy, but a man of your perspicacity wouldn't believe that.

OK, you say, suppose the house isn't airtight? If all the moist air leaks out overnight while giving up some of its heat to the house and it's replaced by dry cold outdoor air that's reheated more easily in the morning, we save a little more energy, in relative terms.

We could go on and on, with more scenarios and Actual Numbers, but perhaps this will suffice for now, given your polite request...

...and humble offer.

Nick

So much for reason.

Bye bye.

Nick

I'm afraid this is not Polite Discourse, nor Simple Seeking of Truth.

You might repost this when you are feeling more sincere.

Nick PS: Here's a start:

Equation 9.30 on page 253 of the 1998 2nd edition of Schaum's Outline by Pitts and Sissom says the heat transfer coefficient for laminar film condensation on vertical tubes with diameter D >> film thickness is h = 1.13(RHOlG(RHOl-RHOv)HfgKl^3)/(MUlL(Tsat-Ts)^0.25 W/m^2, where liquid density RHOl = 999 kg/m^3, gravitational G = 9.8 m/s^2, sat vapor density RHOv = 6 kg/m^3, vaporization enthalpy Hfg = 2.064x10^6 J/kg, conductivity Kl = 0.641 W/mK, viscosity MUl = 4.8x10^-4 kg/m-s, sat temp Tsat = 54 C, surface temp Ts = 24 C, and length L = 0.3048 m, so h = 1.13(999x9.8(999-6) x2.064x10^6x0.641^3)/(4.8x10^-4x0.3048(54-24))^0.25 = 6656 W/m^2, ie 222 W/m^2K, ie 39 Btu/h-F-ft^2, ie R0.026 for the outside of the bottle (25X more than a slow-moving airfilm :-) in series with R0.016 for a still water film, totaling R0.043, so condensing the output of a 5 kW (17.1K Btu/h) dryer with a 27 F temp diff with A ft^2 of bottles with A/R0.043 = 23.2A Btu/h-F of thermal conductance requires that 23.2A27 = 17.1K, ie A = 27.2 ft^2, eg 27.2/1.22 = 22 2-liter bottles with 631 Btu/h-F of air-to-water conductance.

But, as you say, the bottles will warm as the dryer runs, which leads to two problems: 1) warmer bottles will condense a smaller fraction of the water vapor in the dryer output, and 2) we have to get rid of most of the sensible heat in the bottles before the next dryer load.

And it looks like a typical dryer wastes about 30% of its energy heating air (sensible Sheat below.) Reducing the airflow helps, but that raises the dryer temp and risks damaging clothes...

20 DRYPOWP00'dryer power (W) 30 DRYHEAT=3.412*DRYPOW'dryer heat (Btu/h) 40 FOR CFMU TO 65 STEP 5'dryer airflow (cfm) 50 AC 0'clothing area (ft^2, both sides) 60 TIN`'dryer inlet temp (F) 70 TERM1=DRYHEAT/CFM+TIN'terms 80 TERM20*AC/CFM 90 TERM3=TERM1/TERM2 100 TERM4=4.5***.62198***CFM/(.1*AC)
110 T0'initial dryer temp est (F)
120 PW=EXP(17.863-9621/(460+T))'vapor pressure near clothes ("Hg)
130 B=-TERM4-PW'quadratic term
140 C).921'quadratic term
150 PA=(-B-SQR(B^2-4*C))/2'vapor pressure in dryer air ("Hg)
160 TH–21/(17.863-LOG(TERM3+PA-T/TERM2))-460'new dryer temp est (F)
170 IF ABS(TH-T)/T>.0001 THEN T=TH:GOTO 120'iterate to 0.01%
180 PRINT CFM,PW,PA,T
190 SHEATÏM*(T-60)'sensible heat (Btu/h)
200 LHEAT0***AC***(PW-PA)'latent heat (Btu/h)
210 PRINT SHEAT,LHEAT,100*SHEAT/LHEAT
230 NEXT CFM

cfm Pw Pa T

55 4.247575 3.57495 126.0513 3632.82 13452.5 27.00479

60 3.965667 3.302619 123.6099 3816.596 13260.96 28.78068

65 3.721862 3.069051 121.3723 3989.197 13056.24 30.55396

Sheat Lheat %S/L

Most of the sensible heat can be recovered by using the dryer output (Tdo and Wdo below) to heat an incoming airstream (Tlo and Wlo), but it seems hard to recover the latent heat. Maybe the best we can do is to run the dryer output into one side of an air-air heat exchanger, with condensation on the dryer air side, then run the (Tao, Wao) airstream past some cool water containers, then run the (Tlo, Wlo) airstream into the other side of the air-air heat exchanger, like this, viewed in a fixed-font:

-------------- ---------------------- ------- Tdo, Wdo | | Tao, Wao | | | |----->----------->------------->------water containers->- | | | | condensation | | condensation | | | dryer | |--------------| |------------------- | | | |-----<-----------<-------------<------water containers-<- | ------- Tdi, Wdi | | Tlo, Wlo | condensation | -------------- ---------------------- air-air hx condensing hx

This requires finding the dryer air inlet and adding a hose for that. (Nitwits please note: this adds no humidity to the house air :-)

This would be for an electric dryer with no "Co vaper."

It's actually better than that.

Homeowner m might use 4.43 kWh/day for drying and 3.63 for house heating, as well as exhausting 0.88hx60minx60cfmx0.075lb/ft^3 = 238 pounds per day of conditioned house air, eg another 0.24x238(70-30) = 2281 Btu, ie a total of 8.73 kWh/day, or more, with wasteful winter humidification.

Comparing apples to apples, homeowner d might use 3.63 kWh/day for drying and 0 kWh for house heating, totaling 5.1 fewer kWh/day and saving about 51 cents per day at 10 cents/kWh, compared to homeowner m, while feeling ecological******all over*******.

Nick

PS: Rethinking this, it would probably make more sense to put the whole 160' of 4" pipe inside a 4'x8'x2'-tall box without the radiator above and run the cold water line for the house through the pipe.

#### Site Timeline

- posted on March 31, 2005, 7:45 am

wrote:

the numbers are irrelevant and confuse the issue. we are talking about a principle.

you are saying moist air takes less energy to heat than dry air? this goes against what i believe to be true. water takes more energy to heat than a similiar volume of air. add water to the air and its harder to heat.

im not saying im right. im saying id like to see your proof. if im wrong i'll admit it.

randy

- posted on March 31, 2005, 10:32 am

- posted on March 31, 2005, 3:51 pm

Numbers often make the difference. For instance, winter humidification might actually save heating energy in a house if a) a 30% humidity increase made 60 vs 69.5 F air as comfortable as 70 F for people, or b) an average house leaked 100X less air than it does, or c) we could evaporate a pound of water with 10 vs 1000 Btu, but the realities are different, and we might never realize that without numbers. And some things look good in principle, as far as that thinking goes, eg powering a house with a rainwater micro-hydro- electric setup at the base of a downspout, until we estimate that it would only produce 1 average watt (about 1 kWh/year) with a capital cost of $1000. Numbers can give a sense of proportion to wacky ideas.

No, but the difference is small and irrelevant, IMO.

About 4000X more. Heating 1 ft^3 of water 1 F takes 62.33 Btu, vs 0.24x0.075 = 0.018 Btu for 1 ft^3 of air.

But we can only add about 1% water vapor by weight. We might heat a 0.075 lb ft^3 of air containing 1% water vapor 1 F with 0.99x0.018+0.01x0.075x1 = 0.01857 Btu, ie 3% more energy.

That's more polite than your previous demand. Let's see... You wrote:

once you pump all that moisture in it is harder to heat later. so your furnace works harder.

Let's assume it's cold outside and the house temp is constant, say 70 F. The dryer heat recovery scheme I have in mind wouldn't add any moisture to the house air, so let's say you have a different scheme, eg merely filtering the lint out of an electric dryer output and putting the warm moist air into the house, adding 12K Btu of house heat while the furnace supplies 12K Btu less. Let's assume the house air is dry, so the water vapor does not condense indoors. It just raises the RH and specific heat of the house air. But wait... if the house air temperature is constant, we don't have to heat it, so its specific heat doesn't matter.

OK, you say, how about "later"? If we fill up the house with humid air at night and a furnace setback lets the house cool to 60 F by morning, (and the house is airtight :-) the house air is (a small ) part of the thermal capacitance of the house, so the amount of energy it gives to the house as it cools equals the amount of energy it takes back when it warms. Unless you believe the myth that temperature setbacks don't save energy, but a man of your perspicacity wouldn't believe that.

OK, you say, suppose the house isn't airtight? If all the moist air leaks out overnight while giving up some of its heat to the house and it's replaced by dry cold outdoor air that's reheated more easily in the morning, we save a little more energy, in relative terms.

We could go on and on, with more scenarios and Actual Numbers, but perhaps this will suffice for now, given your polite request...

...and humble offer.

Nick

- posted on March 31, 2005, 9:54 pm

thats exactly what i thought. if you cant dazzle em with brilliance, baffle
em with bullshit.

you are scrambling. and not very well.

<ploink>

randy

you are scrambling. and not very well.

<ploink>

randy

- posted on April 1, 2005, 1:27 am

So much for reason.

Bye bye.

Nick

- posted on April 2, 2005, 6:26 pm

snipped-for-privacy@ece.villanova.edu wrote:

Okay. Since you asked so politely and didn't ask for "Actual Numbers".

St. Nick's Handy Dandy Rain Barrel Boondoggle Part Deux

You said to fill up the 55 gallon barrel with 2 liter bottles. Ignoring your packing miracle of fitting more than 10% more volume into said barrel, you pointed out that hex tiling of the bottles is on the order of 10% less efficient. Fine, let's work with that.

10% of your larger-than-possible barrel...I'm feeling generous, so let's figure 6 gallons of free air volume in the barrel (but outside of the 2 liter bottles). With me so far?

6 gallons of volume is 0.80 cubic feet.

Your average clothes dryer moves ~150 cubic feet a minute. In other words that miracle barrel experiences ~190 air changes per minute.

To remove 12 pounds of water during the 40 minute dryer cycle requires an average of 5 ounces of condensation per minute.

So your system needs to condense at least (not a linear system, is it?) 5 ounces of water in 20 seconds to operate. Does that sound right to you?

You're not even within an order of magnitude, dude. Probably less than that.

This is why you are a boob. You like the numbers to the point that you don't understand what they mean. You refine numbers based on a bad model. If you ran across an "Actual Number" you wouldn't recognize it.

It's pitiful, in the truest sense of the word.

R

Okay. Since you asked so politely and didn't ask for "Actual Numbers".

St. Nick's Handy Dandy Rain Barrel Boondoggle Part Deux

You said to fill up the 55 gallon barrel with 2 liter bottles. Ignoring your packing miracle of fitting more than 10% more volume into said barrel, you pointed out that hex tiling of the bottles is on the order of 10% less efficient. Fine, let's work with that.

10% of your larger-than-possible barrel...I'm feeling generous, so let's figure 6 gallons of free air volume in the barrel (but outside of the 2 liter bottles). With me so far?

6 gallons of volume is 0.80 cubic feet.

Your average clothes dryer moves ~150 cubic feet a minute. In other words that miracle barrel experiences ~190 air changes per minute.

To remove 12 pounds of water during the 40 minute dryer cycle requires an average of 5 ounces of condensation per minute.

So your system needs to condense at least (not a linear system, is it?) 5 ounces of water in 20 seconds to operate. Does that sound right to you?

You're not even within an order of magnitude, dude. Probably less than that.

This is why you are a boob. You like the numbers to the point that you don't understand what they mean. You refine numbers based on a bad model. If you ran across an "Actual Number" you wouldn't recognize it.

It's pitiful, in the truest sense of the word.

R

- posted on April 3, 2005, 1:42 am

I'm afraid this is not Polite Discourse, nor Simple Seeking of Truth.

You might repost this when you are feeling more sincere.

Nick PS: Here's a start:

Equation 9.30 on page 253 of the 1998 2nd edition of Schaum's Outline by Pitts and Sissom says the heat transfer coefficient for laminar film condensation on vertical tubes with diameter D >> film thickness is h = 1.13(RHOlG(RHOl-RHOv)HfgKl^3)/(MUlL(Tsat-Ts)^0.25 W/m^2, where liquid density RHOl = 999 kg/m^3, gravitational G = 9.8 m/s^2, sat vapor density RHOv = 6 kg/m^3, vaporization enthalpy Hfg = 2.064x10^6 J/kg, conductivity Kl = 0.641 W/mK, viscosity MUl = 4.8x10^-4 kg/m-s, sat temp Tsat = 54 C, surface temp Ts = 24 C, and length L = 0.3048 m, so h = 1.13(999x9.8(999-6) x2.064x10^6x0.641^3)/(4.8x10^-4x0.3048(54-24))^0.25 = 6656 W/m^2, ie 222 W/m^2K, ie 39 Btu/h-F-ft^2, ie R0.026 for the outside of the bottle (25X more than a slow-moving airfilm :-) in series with R0.016 for a still water film, totaling R0.043, so condensing the output of a 5 kW (17.1K Btu/h) dryer with a 27 F temp diff with A ft^2 of bottles with A/R0.043 = 23.2A Btu/h-F of thermal conductance requires that 23.2A27 = 17.1K, ie A = 27.2 ft^2, eg 27.2/1.22 = 22 2-liter bottles with 631 Btu/h-F of air-to-water conductance.

But, as you say, the bottles will warm as the dryer runs, which leads to two problems: 1) warmer bottles will condense a smaller fraction of the water vapor in the dryer output, and 2) we have to get rid of most of the sensible heat in the bottles before the next dryer load.

And it looks like a typical dryer wastes about 30% of its energy heating air (sensible Sheat below.) Reducing the airflow helps, but that raises the dryer temp and risks damaging clothes...

20 DRYPOWP00'dryer power (W) 30 DRYHEAT=3.412*DRYPOW'dryer heat (Btu/h) 40 FOR CFMU TO 65 STEP 5'dryer airflow (cfm) 50 AC 0'clothing area (ft^2, both sides) 60 TIN`'dryer inlet temp (F) 70 TERM1=DRYHEAT/CFM+TIN'terms 80 TERM20*AC/CFM 90 TERM3=TERM1/TERM2 100 TERM4=4.5

cfm Pw Pa T

55 4.247575 3.57495 126.0513 3632.82 13452.5 27.00479

60 3.965667 3.302619 123.6099 3816.596 13260.96 28.78068

65 3.721862 3.069051 121.3723 3989.197 13056.24 30.55396

Sheat Lheat %S/L

Most of the sensible heat can be recovered by using the dryer output (Tdo and Wdo below) to heat an incoming airstream (Tlo and Wlo), but it seems hard to recover the latent heat. Maybe the best we can do is to run the dryer output into one side of an air-air heat exchanger, with condensation on the dryer air side, then run the (Tao, Wao) airstream past some cool water containers, then run the (Tlo, Wlo) airstream into the other side of the air-air heat exchanger, like this, viewed in a fixed-font:

-------------- ---------------------- ------- Tdo, Wdo | | Tao, Wao | | | |----->----------->------------->------water containers->- | | | | condensation | | condensation | | | dryer | |--------------| |------------------- | | | |-----<-----------<-------------<------water containers-<- | ------- Tdi, Wdi | | Tlo, Wlo | condensation | -------------- ---------------------- air-air hx condensing hx

This requires finding the dryer air inlet and adding a hose for that. (Nitwits please note: this adds no humidity to the house air :-)

- posted on April 3, 2005, 2:49 am

Neither can you; because the numbers refute your allegation. Used
correctly, that sort of a system can create a noticeable difference, plus
maintain normal humidity levels during the coldest seasons.
Also it isn't necessary for the humidity to all go into the house with
a properly installed, albeit more expensive than a flap door system. I have
one that uses a transfer chamber and it works quite well. Now if only I
could make it refrigerate in the summer ...

Pop

Pop

--

Let someone else do it

I\'m retired!

Let someone else do it

I\'m retired!

Click to see the full signature.

- posted on April 6, 2005, 5:28 pm

Several companies make ventless dryers that condense water vapor from
dryer air with a dehumidifier or cool water which gets discarded or
a large fan and an air-air heat exchanger, but they are expensive and
often waste sensible heat, and people complain they are noisy and use
more energy and take longer than conventional dryers...

It looks like most of the sensible heat can be recovered by heating an incoming airstream (Tlo and Wlo below), with a dryer output (Tdo and Wdo), but it seems hard to reduce the latent heat. Maybe the best we can do is to run the dryer output into one side of an air-air heat exchanger with condensation on the dryer air side, then run the (Tao, Wao) airstream past some cool water pipes, then run the (Tlo, Wlo) airstream into the other side of the air-air hx, like this, viewed in a fixed-font:

125 0.08 -------------- 113 0.07 ---------------------- ------- Tdo, Wdo | | Tao, Wao | | | |----->----------->------------->------water containers->- | | | | condensation | | condensation | | | dryer | |--------------| |------------------- | | | |-----<-----------<-------------<------water containers-<- | ------- Tdi, Wdi | | Tlo, Wlo | condensation | 117 0.01 -------------- 67 0.01 ---------------------- air-air hx condensing hx

This requires finding the dryer air inlet and adding a hose for that...

The air-air hx could be 32 2'x4' vertical plastic film layers on 3/4" centers inside half of a 4'x4'x2'-tall box. The condensing hx could be 16 4"x4' thinwall PVC pipes in the other half of the box, plumbed close in a 4x4 array with 3/4" PVC male adapters as bulkhead fittings, with 96' of 4" pipe above the box in 2 2'4x4' pipe shelves with 4 pipe posts and a back to add more thermal mass and act as a passive radiator with 160' of pipe containing 870 pounds of water with a 5.8 hour external time constant. The box would contain enough pipe surface (about 70 ft^2) to make its condensing heat exchange effectiveness close to 100%. If one load of laundry per day adds 12K Btu, it might warm from 60.26 to 12K/870 = 74.05 by the end of the load and cool back to 60.26 23 hours later...

20 DRYPOWP00'dryer power (W) 30 DRYHEAT=3.412*DRYPOW'dryer heat (Btu/h) 40 CFM`'dryer airflow (cfm) 50 AC 0'clothing area (ft^2, both sides) 60 TDI6.842'initial dryer inlet temp est. (F) 70 WDI=.01443'initial dryer inlet hum rat est. 80 PRINT "Tdi =";TDI,"Wdi =";WDI 90 ALPHA=.62198***4.5***CFM'terms
100 BETA=4.5***CFM***WDI
110 GAMMA=.1***AC
120 TERM1=(DRYHEAT+TDI***CFM)/(100*AC)
130 TERM2ÏM/(100*AC)
140 TDO0'initial dryer temp est (F)
150 PW=EXP(17.863-9621/(460+TDO))'vapor pressure near clothes ("Hg)
160 A=GAMMA
170 B=-(ALPHA+BETA+GAMMA***(29.921+PW))'quadratic term
180 C).921***(BETA+GAMMA*PW)'quadratic term
190 PAO=(-B-SQR(B^2-4***A***C))/(2*A)'vapor pressure in dryer air ("Hg)
200 TDOH–21/(17.863-LOG(TERM1+PAO-TERM2*TDO))-460'new dryer temp est (F)
210 IF ABS(TDOH-TDO)/TDO>.0001 THEN TDO=TDOH:GOTO 150'iterate to 0.01%
220 LHEAT0***AC***(PW-PAO)'latent heat removal rate (Btu/h)
230 DRYTIME000/LHEAT'drying time (hours)
240 DRYENERGY=5*DRYTIME'drying energy (kWh)
250 WDO=.62198/(29.921/PAO-1)'dryer output humidity ratio
260 PRINT "Tdo =";TDO,"Wdo =";WDO
270 TROOM`'basement air temp (F)
280 LIPIPE=4***16'length of 4" pipe inside box (feet)
290 LEPIPE–'length of 4" pipe outside box (feet)
300 CCHXb.33***3.14159*(4/12/2)^2*(LIPIPE+LEPIPE)'heat cap (Btu/F)
310 SPIPE=3.14159*4/12*LEPIPE'exposed pipe surface (ft^2)
320 RCÌHX/(1.5*SPIPE)'condensing hx time constant (hours)
330 DT000/CCHX'temp rise during drying (F)
340 TMIN=(TROOM+(DT-TROOM)*EXP(-23/RC))/(1-EXP(-23/RC))'min cond hx temp (F)
350 TLO=TMIN+DT/2'average condensing hx output temp (F)
360 PLO=EXP(17.863-9621/(460+TLO))'vapor pressure after cond hx ("Hg)
370 WLO=.62198/(29.921/PLO-1)'hum rat after cond hx
380 N$/.75'number of films in air-air hx
390 A=N***2***4'air-air hx area (ft^2)
400 NTU=1.5*A/CFM'air-air hx NTU
410 E=NTU/(NTU+1)'air-air hx effectiveness
420 TDI=TLO+E***(TDO-TLO)'dryer input air temp (F)
430 AAHEATÏM***(TDI-TLO)'air-air hx heat (Btu/h)
440 TMX=TDO+1000***60***.075*WDO-AAHEAT/CFM'terms
450 TMY=TMX+4500*.62198
460 TERM3.863-LOG(29.921)
470 TAO0'initial Tao est (F)
480 TAOH–21/(TERM3-LOG((TMX-TAO)/(TMY-TAO)))-460
490 IF ABS(TAOH-TAO)/TAO > .0001 THEN TAO=TAOH: GOTO 480
500 WAO=(TMX-TAO)/4500'hum rat
510 PRINT "Tao =";TAO,"Wao =";WAO
520 PRINT "Tlo =";TLO,"Wlo =";WLO
530 WDI=WLO'no condensation in cold stream of air-air hx
540 PRINT "Tdi =";TDI,"Wdi =";WDI
550 PRINT "Troom =";TROOM,,"Tmin =";TMIN
560 PRINT"Drytime =";DRYTIME,"Dryenergy =",DRYENERGY

Tdi = 116.842 F Wdi = .01443 humidity ratio Tdo = 124.6056 Wdo = 7.579565E-02 Tao = 113.006 Wao = 6.733185E-02 Tlo = 67.15541 Wlo = 1.443278E-02 Tdi = 116.8421 Wdi = 1.443278E-02

Troom = 60 Tmin = 60.26116

Drytime = .7242576 hours Dryenergy = 3.621288 kWh

We can simulate the dryer alone by changing line 60 above to

60 TDI`'initial dryer inlet temp est. (F)

Drytime = .8868306 hours Dryenergy = 4.434153

So this box might save 60(0.887-0.724) = 10 minutes of drying time and put 3.62x3412 = 12350 Btu of heat (but no water vapor) into the house, vs exhausting 4.43x3412 = 15115 Btu of warm moist air from the house with the dryer alone.

At today's energy prices, this might save 2(4.43-3.62)10 = 16 cents/day :-)

Nick

It looks like most of the sensible heat can be recovered by heating an incoming airstream (Tlo and Wlo below), with a dryer output (Tdo and Wdo), but it seems hard to reduce the latent heat. Maybe the best we can do is to run the dryer output into one side of an air-air heat exchanger with condensation on the dryer air side, then run the (Tao, Wao) airstream past some cool water pipes, then run the (Tlo, Wlo) airstream into the other side of the air-air hx, like this, viewed in a fixed-font:

125 0.08 -------------- 113 0.07 ---------------------- ------- Tdo, Wdo | | Tao, Wao | | | |----->----------->------------->------water containers->- | | | | condensation | | condensation | | | dryer | |--------------| |------------------- | | | |-----<-----------<-------------<------water containers-<- | ------- Tdi, Wdi | | Tlo, Wlo | condensation | 117 0.01 -------------- 67 0.01 ---------------------- air-air hx condensing hx

This requires finding the dryer air inlet and adding a hose for that...

The air-air hx could be 32 2'x4' vertical plastic film layers on 3/4" centers inside half of a 4'x4'x2'-tall box. The condensing hx could be 16 4"x4' thinwall PVC pipes in the other half of the box, plumbed close in a 4x4 array with 3/4" PVC male adapters as bulkhead fittings, with 96' of 4" pipe above the box in 2 2'4x4' pipe shelves with 4 pipe posts and a back to add more thermal mass and act as a passive radiator with 160' of pipe containing 870 pounds of water with a 5.8 hour external time constant. The box would contain enough pipe surface (about 70 ft^2) to make its condensing heat exchange effectiveness close to 100%. If one load of laundry per day adds 12K Btu, it might warm from 60.26 to 12K/870 = 74.05 by the end of the load and cool back to 60.26 23 hours later...

20 DRYPOWP00'dryer power (W) 30 DRYHEAT=3.412*DRYPOW'dryer heat (Btu/h) 40 CFM`'dryer airflow (cfm) 50 AC 0'clothing area (ft^2, both sides) 60 TDI6.842'initial dryer inlet temp est. (F) 70 WDI=.01443'initial dryer inlet hum rat est. 80 PRINT "Tdi =";TDI,"Wdi =";WDI 90 ALPHA=.62198

Tdi = 116.842 F Wdi = .01443 humidity ratio Tdo = 124.6056 Wdo = 7.579565E-02 Tao = 113.006 Wao = 6.733185E-02 Tlo = 67.15541 Wlo = 1.443278E-02 Tdi = 116.8421 Wdi = 1.443278E-02

Troom = 60 Tmin = 60.26116

Drytime = .7242576 hours Dryenergy = 3.621288 kWh

We can simulate the dryer alone by changing line 60 above to

60 TDI`'initial dryer inlet temp est. (F)

Drytime = .8868306 hours Dryenergy = 4.434153

So this box might save 60(0.887-0.724) = 10 minutes of drying time and put 3.62x3412 = 12350 Btu of heat (but no water vapor) into the house, vs exhausting 4.43x3412 = 15115 Btu of warm moist air from the house with the dryer alone.

At today's energy prices, this might save 2(4.43-3.62)10 = 16 cents/day :-)

Nick

- posted on April 6, 2005, 6:18 pm

Where does the gas vaper -Co go, inside?. 16c a day , pretty good, I do
1 load a week. Whats the payback 100 years. or in actuality from
interest lost, never.

- posted on April 6, 2005, 9:19 pm

snipped-for-privacy@ece.villanova.edu wrote:

Please do, and be sure to let us know when you're done.*

R

*Hint - There'll be an odd sensation, a plop! and you'll be able to hear again.

Please do, and be sure to let us know when you're done.*

R

*Hint - There'll be an odd sensation, a plop! and you'll be able to hear again.

- posted on April 6, 2005, 11:57 pm

This would be for an electric dryer with no "Co vaper."

It's actually better than that.

Homeowner m might use 4.43 kWh/day for drying and 3.63 for house heating, as well as exhausting 0.88hx60minx60cfmx0.075lb/ft^3 = 238 pounds per day of conditioned house air, eg another 0.24x238(70-30) = 2281 Btu, ie a total of 8.73 kWh/day, or more, with wasteful winter humidification.

Comparing apples to apples, homeowner d might use 3.63 kWh/day for drying and 0 kWh for house heating, totaling 5.1 fewer kWh/day and saving about 51 cents per day at 10 cents/kWh, compared to homeowner m, while feeling ecological

Nick

PS: Rethinking this, it would probably make more sense to put the whole 160' of 4" pipe inside a 4'x8'x2'-tall box without the radiator above and run the cold water line for the house through the pipe.

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