nick pine

Mike Jay... what would happen if instead of 55 gallon drums of water, you had a 55 gallon drum of glycol?  It phase changes at 63 degrees which could do some interestingly beneficial things.  Getting heat into the drum could require a circulating heat transfer liquid.  [/quote wrote:

This, which melts at 27.7 C? https://www.sciencedirect.com/science/article/pii/S0927024811000365 Seems complicated and expensive, compared to water.

Nick

Two layers of greenhouse polyethylene film over a half-cylindrical 8’x16’ x8’-tall frame would make the curved greenhouse south wall/roof area 4’Pi16’ = 200 ft^2 with a 200ft^2/R2 = 100 Btu/h-F thermal conductance. With R20 insulation, the north wall conductance would be about 200ft^2/R20 = 10 Btu/h-F. R20 endwalls would add Pi8’^2/R20 = 10, making the total greenhouse conductance 120 Btu/h-F.

Where I live near Philadelphia, 1000 Btu/ft^2 of sun falls on a south wall over an average 30 F January day in about 6 hours. With 80% solar transmission, the 128 ft^2 south wall projection would transmit 128ft^2x0.8x1000Btu/ft^2/6h = 17067 Btu/h. With no thermal mass, the greenhouse air temp would be about 30+17067/120 = 172 F for 6 hours/day before it suddenly fell back to 30 F, theoretically.

LOTS of thermal mass and mass surface and 102400/24h = 4267 Btu/h on an average January day would make the greenhouse a constant 30 + 4267/120 = 66 F, 24 hours per day.

Each of 2 55 gallon water-filled drums would have about 450 Btu/F of thermal mass and 25 ft^2 of mass surface and 1.5x25 = 40 Btu/h-F of air-water surface conductance with a thermal equivalent circuit which makes more sense if viewed in a fixed font like Courier New after downloading. Tg is the greenhouse air temperature:

 17067 Btu/h  Tg
    --        |     1/120 F-h/Btu
|---|—>|-------*------vvv--- 30 F
    --        |
              <
              < 1/80
              <
              |
             ---
             --- 900 Btu/F
              |
              -  

with this Thevenin equivalent circuit https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ :

      1/120
   ----www------------ Tg
  |           |
  |           <
  | Vt        < 1/80       Vt = 30 + 17067/120 = 172 F.
 ---          <
  -           |                
  |          ---          
  |          --- 900 Btu/F    
  |           |                
  -           -  
                                           
which is equivalent to:  
 
      1/48 = 1/(1/120+1/80)                              
   ----www----                  
  |           |
 --- 172     ---                
  -          --- 900 Btu/F    
  |           |                
  -           -              


with a 900/48 = 19 hour RC time constant.


137 F  |         Tg peak air temp, which

      |                 falls quickly at dusk.
      |     Tg
      |         Tg
      |  Tg        
 84   |      T T T peak water temp, which does not.    
      |Tg  T      T        
      |  T      Tg    T
      | T                 T
      |Tg               Tg   T            
      |Tg                       T
      |T                           T
      |Tg                        Tg    T
      |T                                   T
 54    ------------------------------------------
                6h                         24h


If T = Td just before dusk and T = Tn just before dawn, 18 hours later,

Tn = 30 + (Td-30)e^-(18/19) = 18.4 + 0.388Td, and
Td = 172 + (Tn-172)e^-(6/19) = 46.6 + 0.729Tn, then
Tn = 18.4 + 0.388(46.6+0.729Tn) = 36.5 + 0.283Tn, so
Tn = 36.5/(1-0.283) = 50.9 F.

Just before dawn, (50.9-30)48 = 1004 Btu/h flows out of the drums, warming the greenhouse air to 30 + 1004/120 = 38.4 F.

Just before dusk, Td = 46.6 + 0.729Tn = 83.7 F. The greenhouse air would be hotter: (172-83.7)48 = 4238 Btu/h flows into the drums, adding 4238/80 = 53 F to the drum temp to make 137 F air, if I did that right.  

We could limit the air temp to about 80 F max by opening a greenhouse vent whenever the indoor air rises to 80 F using an unpowered vent opener, or a thermostat and a 24V 2W motorized damper actuator or a fan. The vent needs to dump about 17067 – (80-30)120 = 11067 Btu/h. With an A ft^2 open area for upper and lower vents and an 8’ height difference, A = 11067/(16.6sqrt((80-30)^1.5) = 0.67 ft^2 min, eg 1 ft^2 vents, which could vent up to 16600 Btu/h at 332 cfm and 80 F.

Ventilation can also reduce wintertime humidity. Air at 80 F (540 R) and 60% RH (an upper limit to avoid mold) has a water vapor pressure Pw = 0.6e^(17.863-9621/540) = 0.628 “Hg. The humidity ratio wa = 0.62198/(29.921/Pw-1) = 0.0133 pounds of water per pound of dry air. NREL says the average outdoor humidity ratio w = 0.0024 in January in Philadelphia (although the current online Blue Book is missing Phila data.) A 100 cfm vent airflow would remove 60m/hx100x0.075lb/ft^3(0.0133-0.0024) = 4.9 lb/h of water vapor from the greenhouse. Some people  say that green plants can evaporate 1 lb of water per day per square foot of greenhouse floor space.

With 4 drums:

      1/120
   ----www------------ Ta
  |           |
  |           <
  | Vt        < 1/160       Vt = 30 + 17067/120 = 172 F.
 ---          <
  -           |
  |          ---

  |          --- 1800 Btu/F
  |           |  
  -           -


which is equivalent to:

      1/69
   ----www----
  |           |
 --- 172 F   ---
  -          --- 1800 Btu/F
  |           |
  -           -

with an 1800/69 = 26 hour RC time constant.

If T = Td just before dusk and T = Tn just before dawn,

Tn = 30 + (Td-30)e^-(18/26) = 15.0 + 0.500Td, and
Td = 172 + (Tn-172)e^-(6/26) = 35.4 + 0.794Tn, then
Tn = 15.0 + 0.500(35.4+0.794Tn) = 32.7 + 0.397Tn, so
Tn = 32.7/(1-0.397) = 54.2 F.


Just before dawn, (54.2-30)69 = 1670 Btu/h flows out of the drums, warming the house air to 30 + 1670/120 = 43.9 F. More drums help. We put 200 55 gallon water drums as plant pallet supports into a 20’x96’ single cover greenhouse in PA, and it never froze in wintertime.

The drum heat would last longer if the greenhouse air were kept at a constant 40 F all night instead of gradually cooling from the peak to the minimum temp, but that would require some sort of control.

With more controls, 8 drums in an R20 4x4x8’ insulated box with an R1 90% transparent south wall could store more heat, ie 3600(132-44) = 317K Btu, vs 8 cooler drums exposed to greenhouse air storing 3600(58-44) = 50K Btu. To reduce heat loss at night, a 500 cfm high temp 23 watt fan, eg https://m.grainger.com/mobile/product/4WT44?cm_mmc=PPC:+Google+PLA&s_kwcid=AL!2966!3!166593068073!!!g!102431315157!&ef_id=WGOu3wAAAH9I1Syw:20180705181606:s could circulate warm air through an R20 partition during the day, with no airflow at night. The 4’x8’ vertical partition could separate the south wall and the vertical drums in a single 2x4 layer under a 4’x8’ bench would have an equivalent circuit like this during the day:

 3888 Btu/h     1/500 fan  R20/96ft^2 box            
    --        
|---|—>|----------vvv---------vvv--- 80 F
    --       |          |
      R1/32  |          <
80 F ---vvv---           < 1/320
                        <
                        |  T
                       ---
                       --- 3600 Btu/F
                        |
                        -  

which is equivalent to this:


         1/32  1/500       1/4.8
   -------www---vvv---------vvv---- 80 F
  |                     |
  |                     <
  | Vt                  < 1/320       Vt = 80 + 3888/32 = 202 F.
 ---                    <
  -                     |                
  |                    ---          
  |                    --- 3600 Btu/F    
  |                     |                
  -                     -  

and this:

      1/34.8
   -----vvv-----
  |             |      
 --- 185       ---
  -            --- 3600                

  |             |    
  -             -  

with a daytime time constant RC = 3600Btu/F/34.8Btu/h-F = 103 hours.

RC = 3600(20/128+1/320) = 574 hours at night:

                          R20/128ft^2 box
                         -----vvv----- 40F            
                        |
                        <
                        < 1/320
                        <
                        | T
                       ---
                       --- 3600 Btu/F
                        |
                        -

Tn = 40 + (Td-40)e^-(18/574) = 1.2 + 0.969Td, and
Td = 185 + (Tn-185)e^-(6/103) = 10.5 + 0.943Tn, then
Tn = 1.2 + 0.969(10.5+0.943Tn) = 11.4 + 0.914Tn, so
Tn = 11.4/(1-0.914) = 132.3 F.
Td = 10.5+0.943x132.3 = 135.2 F.

If we insulate the entire greenhouse (including the south wall) with R20 soap bubble foam at night and fill the 6 triangular spaces between 2 drums and a 2’x4’x4’ tall R20 box with an R1 transparent south wall with 90% solar transmission, the 6 spaces (totaling 2’x4’- 2Pi(1’^2) = 1.72 ft^2 could hold 8 4”x3’ vertical PVC perforated pipes (to reduce airflow resistance) surrounded by about 300 pounds of 2” diameter rocks, which would add about 48 Btu/F to the drum thermal capacitance and 108 Btu/h-F to the drum thermal conductance, with an equivalent circuit like this, during the day:

 1944 Btu/h     1/500 fan  R20/48ft^2 box            
    --        
|---|—>|----------vvv---------vvv--- 80 F
    --       |          |
      R1/16  |          <
80 F ---vvv---           < 1/189
                        <
                        |  T
                       ---
                       --- 948 Btu/F
                        |
                        -  

which is equivalent to this:


         1/16  1/500       1/2.4
   -------www---vvv---------vvv---- 80 F
  |                     |
  |                     <
  | Vt                  < 1/189       Vt = 80 + 1944/16 = 202 F.
 ---                    <
  -                     |                
  |                    ---          
  |                    --- 948 Btu/F    
  |                     |                
  -                     -  

and this:

      1/16.4
   -----vvv-----
  |             |      
 --- 186       ---
  -            --- 948                

  |             |    
  -             -  

with a daytime time constant RC = 948Btu/F/16.4Btu/h-F = 58 hours.

RC = 948(20/64+1/189) = 301 hours at night:

                          R20/64ft^2 box
                         -----vvv----- 40F            
                        |
                        <
                        < 1/189
                        <
                        | T
                       ---
                       --- 948 Btu/F
                        |
                        -

Tn = 40 + (Td-40)e^-(18/301) = 2.3 + 0.942Td,
Td = 186 + (Tn-186)e^-(6/58) = 18.3 + 0.902Tn, then
Td = 18.3 + 0.902(2.3+0.942Td) = 20.4 + 0.850Td, so
Td = 20.4/(1-0.850) = 136 F.
Tn = 2.3+0.942x136 = 130 F.

Foaming the greenhouse at night lowers the heat requirement to (40-30)30 = 300 Btu/h. The rocks lower the min usable water temp to 300/189 = 42 F. The 2 130 F drums can keep the greenhouse air 40 F for 948(130-42)/(24hx300) = 11 30 F cloudy days in a row.

If cloudy days are like coin flips, a greenhouse that can store enough heat to keep itself 40 F for N 30 F cloudy days in a row would have a Solar Heating Fraction (SHF) of 1-2^-N, eg 0.5 for 1 cloudy day, 0.75 for 2, 1-2^-3 = 0.875 for 3, and so on. For instance, the 8 drums in the box would store 317K Btu. Keeping the greenhouse 40 F for a day requires 24h(40-30)120 = 28.8K Btu, so N = 317K/28.8K = 11 days, with an SHF = 1-2^-11 = 0.9995, ie 99.95%, ie solar-heated in all but 4 hours per year of 12 months of January weather.

In summary:                                      
                                        min
# of   C      RC      Twp    Twm   dT    usable
drums  Btu/F  hours   peak   min   F     temp   SHF

0      0      0       172    30    142   -      0
2      900    19      87     51    36    55     0
2+foam 948    58/301  136    130   6     42     0.9995
4      1800   26      78     54    24    48     0.23
8      3600   41      75     58    16    44     0.70
8-box  3600   103/574 135    132   3     44     0.9995
oo     oo     oo      66     66    0     40     1

Nick

Re: https://permies.com/t/82794/Moderating-temperature-hoop-house

Grace Gierucki writes from Southern Michigan:

>I have a small hoop house 8’x14’x7’ at the peak.

Say 8’x16’x8’, in rounder numbers.

>It’s relatively well sealed from air leaks but, naturally, the temperature drops ultra fast at night.

With no thermal mass, it would quickly drop to about 30 F at night, where I live near Phila. Southern Michigan is colder and cloudier, with a 20.9 vs 30 F average temp and 750 vs 1000 Btu/ft^2 of sun on an average January day in Lansing, based on NREL’s Blue Book.

>I’m just trying to figure out how to keep it from dropping below about 40 degrees...I have lots of 2” rigid foam panels...

R8? Rectangular? How about R19 fiberglass batts between two layers of north greenhouse polyethylene film?

That would make the greenhouse thermal conductance about 8’x16’/R2 = 64 Btu/h-F for 2 layers of south glazing + 12’x16’/R20 = 10 for the roof + 6’x16’/R20 = 5 for the lower north wall + 2x7’x12’/R20 = 8 for the endwalls, totaling 87 Btu/h-F.

Where I live, a south wall receives 1000 Btu/ft^2 of sun on a 30 F January day, in about 6 hours. With 80% solar transmission, the 128 ft^2 south wall would transmit 128ft^2x0.8x1000Btu/ft^2/6h = 17067 Btu/h, raising the house temp to 30+17067/87 = 226 F for 6 hours/day before it drops back to 30 F, theoretically.

With LOTS of thermal mass and mass surface, the south wall would transmit 102,400 Btu of sun on an average January day, ie 102400/24h = 4267 Btu/h, making the greenhouse a constant 30 + 4267/87 = 69 F.

>I also have 2 55 gallon barrels I could fill and use to replace my current work bench area.

Each drum would have about 500 Btu/F of thermal mass and 40 ft^2 of mass surface and 60 Btu/h-F of air-water surface conductance with a thermal equivalent circuit like this, viewed in a fixed font (Tg is the house air temperature):

 17067 Btu/h Tg
    --       |      1/87
|---|—>|------*------vvv--- 30 F
    --       |
             |
             <
             < 1/120  (How do we post in a fixed font?)
             <
             |
             |
            ---
            --- 1000
             |
             |
             -  

with a Thevenin equivalent circuit like this:

      1/87
   ----www----
  |           |
  |           |
  |           <
  | Vt        < 1/120       Vt = 30 + 17067/87 = 226 F.
 ---          <
  -           |
  |           |
  |          ---
  -          --- 1000 Btu/F
              |
              |
              -  

which is equivalent to:

      1/50
   ----www------- T
  |           |
  | 226 F     |
 ---         ---
  -          --- 1000 Btu/F
  |           |
  |           |
  -           -
           
with water temp T and a 1000/50 = 20 hour RC time constant.

If T = Td just before dusk and T = Tn just before dawn, 18 hours later,

Tn = 30 + (Td-30)e^-(18/20) = 17.8 + 0.407Td, and

Td = 226 + (Tn-226)e^-(6/20) = 58.6 + 0.741Tn, then

Tn = 17.8 + 0.407(58.6+0.741Tn) = 41.7 + 0.302Tn, so

Tn = 41.7/(1-0.302) = 59.7 F.

Just before dawn, (59.7-30)50 = 1484 Btu/h flows out of the drums, warming the house air to 30 + 1484/87 = 47.1 F.

Just before dusk, Td = 58.6 + 0.741Tn = 102.8 F. And the greenhouse air will be hotter: (226-102./(1/50) = 6160 Btu/h flows into the drums, adding 6160/120 = 51 F to the drum temp to make 154 F air, if I did that right. Too hot.

Suppose we limit the drum temp to 80 F max by opening a greenhouse vent whenever the greenhouse air temperature reaches 80 F using an unpowered vent opener or a thermostat and a 24V 2W motorized damper actuator or a fan. The vent needs to dump about 17067 – (80-30)87 = 12717 Btu/h. with an A ft^2 open area for upper and lower vents and an 8’ height difference, 16.6A = sqrt 8 (80-30)^1.5 = 12717 makes A = 0.77 ft^2, eg a 1 ft^2 vent. Ventilation will also desirably reduce wintertime humidity.

With water temp T = 80 just before dusk and T = Tn just before dawn, 18 hours later,

Tn = 30 + (80-30)e^-(18/20) = 50.3 F.

Just before dawn, (50.3-30)50 = 1016 Btu/h flows out of the drums, warming the house air to 30 + 1016/87 = 41.7 F.

Mr Ken Koch wrote:

>Come on Nick! Where is the optimism factor?

As an optimistic alternative, the Winnipeg house could have R32 SIP walls and an R48 roof with a 1024/48+2048/32 = 85 Btu/h-F conductance and a frugal 300 kWh/mo (1421 Btu/h) of indoor electrical use and a 6.6F+1421Btu/h/(85Btu/h-F) = 23.3 F effective outdoor temp (see how nicely the units work out?), so it would only need 24h(70-23.3)85 = 95.2K Btu/day of solar heat.

At 80 F, a 1 ft^2 twinwall air heater would gain 0.8x992-6h(80-6.6)1ft^2/R2 = 573 Btu on an average December day, so the house could have about 95.2K/573 = 166 ft^2 of air heater glazing, eg an 8'x20' south wall.

An 8500 Btu/F 4"x1024ft 2 radiant floorslab and 7K Btu/F of walls and furnishings would make time constant RC = 15,500Btu/F/(85Btu/h-F) = 182 hours. If the house is 72.3 at 3 PM on an average December day, it would be 23.3+(72.3-23.3)e^(-18/182) = 67.7 F by 9 AM the next morning, with no water flow in the floor.

On a cloudy day, with a 1Btu/h-F-ft^2x1024ft^2 slow airfilm conductance in series with a 1000 Btu/h-F 2 gpm water flow conductance, the floor could provide (70-23.3)85 = 3970 Btu/h of house heat with 70+3970/2000 = 72 F water.

The house needs 5x24x3970 = 476K Btu for 5 cloudy days in a row, eg a 476K/(140-72)/62.33 = 112 ft^3 tank cooling from 140 to 72 F, eg a 4'x10'x3'-tall plywood box with a 10'x16' folded EPDM liner or a 4'-tall x 6' diameter tank with 132 ft^2 of surface.

With R20 insulation, the round tank would lose 24h(140-70)132/20 = 11.1K Btu on an average December day. We could keep it 140 F with a separate 29 ft^2 twinwall air heater and a 1000 Btu/h-F car radiator and a 1000 cfm fan and a 2 gpm pump, or timeshare a slightly larger basic house air heater or a smaller deep-mesh air heater http://www.builditsolar.com/Experimental/DeepMeshCol/120116Test.htm

A larger air heater and tank could also provide hot water for showers.

Nick

Sean Rauch wrote:

-We're planning the basic south facing windows with thermal mass inside to collect as much winter sun as possible.

> That could work. If the house has no internal heat gains and it's 75 F on an average day and 65 after 5 cloudy days in a row (with a 2^-5 probability), 65 = 6.6 + (75-6.6)e^(-5x24/RC) makes time constant RC = 759 hours. With no air leaks or floor heat loss and US R30 walls and an R60 ceiling and R30 night- and cloudy-day window insulation and a 1/R = G = 1024/60+2048/30 = 85 Btu/h-F cloudy-day conductance, it would need C = 759hx85Btu/h-F = 64.8K Btu/F of room temp thermal mass, eg 64.8K/25 = 2591 ft^3 of concrete or 4048 ft^3 of cylindrical rock gabions, about 25% of the house volume, or more, with less night window insulation.

A 1 ft^2 R2 south window with 80% solar transmission and an R28 shutter (6" Styrofoam inserts that double as tables for 6 hours per day?) would gain 0.8x992-6h(75-6.6)1ft^2/R2 = 588 Btu/day, so you could heat the house with about 119K/588 = 202 ft^2 of such windows.

Windows with weighted insulated R3 curtains rolled up every morning by a solar panel driven 12v motor and then released when the sun went down would gain 588-18h(75-6.6)1ft^2/R5 = 342 Btu/day, so you could heat the house with about 119K/342 = 348 ft^2 of them. On a cloudy day, with the windows shuttered, the house conductance would be about 1024/60+348/5+(2048-348)/30 = 143 Btu/h-F, so it would need C = 759hx143Btu/h-F = 108790 Btu/F of room temp thermal mass, eg 108790/16 = 6800 ft^3 of cylindrical rock gabions, about 42% of the house volume.

A ft^2 of R4 south windows with 50% solar transmission would gain 0.5x992A = 24h(75-6.6)(1024/60+A/4+(2048-A)/30) Btu/day, which makes A = 998 ft^2 (eg a 16'x64' south window wall on a 16'x64'x16'-tall house) with G = 1024/60+1024/4+(2560-1024)/30 = 324 Btu/h-F, so it would need C = 759hx324Btu/h-F = 246118 Btu/F of room temp thermal mass, eg 246118/16 = 15382 ft^3 of cylindrical rock gabions, about 94% of the house volume.

With no shutters, A ft^2 of R2 south windows with 80% solar transmission would gain 0.8x992A = 24h(75-6.6)(1024/60+A/2+(2048-A)/30) Btu/day, which makes A = 5090 ft^2...

Nick

allen lumley writes:

>... have you seen the Kalwal and Lumira Aerogel window panels from> www.kalwal.com/aerogel.htm ?

They look expensive, with 22% solar transmission, compared to $1.30/ft^2 twinwall from my local greenhouse supplier (Nolt's Produce in Leola, PA.)

Nick

Sean Rauch wrote:

-We live in Manitoba Canada and plan to build just outside of Winnipeg

http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/International/pvwattsv1_intl.cgi says 3.13 kWh/m^2 (992 Btu/ft^2) of sun falls on a south wall on an average December day in Winnipeg.

The stat file at http://apps1.eere.energy.gov/buildings/energyplus/cfm/weather_data3.cfm/region=4_north_and_central_america_wmo_region_4/country=3_canada/cname=CANADA says the average December temp is -14.1 C (6.6 F) with a -12.1 high.

-I'm working with a model right now that has 2000 square feet of living space spread out over two floors

... eg 32'x32'x16'-tall, with a 1024 ft^2 ceiling and 2048 ft^2 of walls.

-We're planning the basic south facing windows with thermal mass inside to collect as much winter sun as possible.

That could work. If the house has no internal heat gains and it's 75 F on an average day and 65 after 5 cloudy days in a row (with a 2^-5 probability), 65 = 6.6 + (75-6.6)e^(-5x24/RC) makes time constant RC = 759 hours. With US R30 walls and an R60 ceiling and R30 night- and cloudy-day window insulation and a 1/R = G = 1024/60+2048/30 = 85 Btu/h-F cloudy-day conductance, it would need C = 759hx85Btu/h-F = 64.8K Btu/F of room temp thermal mass, eg 64.8K/25 = 2591 ft^3 of concrete or 4048 ft^3 of cylindrical rock gabions, about 25% of the house volume, or more, with less night window insulation.

-The house wont be off grid but I don't want to rely on the grid for any of the climate control demands of the house.

It might have a shiny massy ceiling heated by passive air heaters, with a thermostat and a slow low-power ceiling fan to mix down hot ceiling air as needed to keep the house exactly 70 F for 5 cloudy days, or 70 F during the day and 60 at night. Here's a nice air heater: http://www.builditsolar.com/Projects/SpaceHeating/solar_barn_project.htm With hot ceiling air and colder air outdoors, you might use R2 twinwall polycarbonate with 80% solar transmission instead of a single layer in balmy Montana. Gary Reysa and I tried this scheme in 2010, with one big mistake, an uninsulated partition wall that turned the air heater into an air cooler at night... http://www.builditsolar.com/Experimental/dCube/Barra/BaraBox.htm

With no windows (use flat screen TVs and outdoor cameras), the house would need about (65-6.6)85 = 5K Btu/h or 119K Btu/day (collect 20K Btu/h for 6 hours) or 596K Btu for 5 cloudy days in a row. At (say) 160 F, a 1 ft^2 twinwall air heater would gain 0.8x992-6h(160-6.6)1ft^2/R2 = 460 Btu on an average day, so the house could have 119K/460 = 260 ft^2 of air heater glazing, eg an 8'x32' south wall. A 1024 ft^2 ceiling with a 3 Btu/h-F-ft^2 slow 2-sided airfilm conductance would be about 20KBtu/h/3KBtu/h-F = 7 F cooler than the hot air around it when collecting heat on an average day, and a 5.5"x32' air heater flow path with A = 14.7 ft^2 and H = 8' would make the air heater air (20K/(16.6x14.7sqrt(8)))^(2/3) = 9.4 F warmer than the air near the ceiling, so the ceiling would be 160-9.4-7 = 144 F on an average day.

A 1000 cfm ceiling fan could keep the house 70 F with a 70+5KBtu/h(1/3K+1/1K) = 77 F ceiling, and (144-77)C = 596K makes C = 4.3K Btu/F for 5 cloudy days in a row, ie 4.2 psf, eg 0.8 inches of water, or deeper water in trays that cover less ceiling area.

Have fun :-)

Nick

>One of the problems with any convective heat is that you end up with a whole bunch of heat near the ceiling and very little near the floor. It gets layered. Air at the ceiling could be 20 degrees warmer than the air at the floor.

This might be a myth. http://www.builditsolar.com/Experimental/StratificationTest/stratificationtest.htm shows tiny air temp increases with height. Rich Komp says gravity vs diffusion can make a positive difference over thousands of feet, but it seems to me that a room with a heat source near the floor would have a cooler ceiling in wintertime, as in a thermal chimney...

cfm = 16.6Asqrt(HdT), right? So Q = 16.6Asqrt(H)dT^1.5 Btu/h, approximately. Hence, an 8' 70 F R16 cube with a 20 Btu/h-F conductance above the floor would need (70-30)20 = 800 Btu/h on a 30 F day. If air from the floor rises up to the ceiling in a 4 ft^2 central column and 800 = 16.6x4xsqrt(8)dT^1.5, the ceiling could be dT = 2.6 F cooler than the floor, in the center of the room. OTOH, the air near the floor might be cooler than the ceiling near a wall, after it cools and falls in a square toroidal convective loop.

Nick