Sunday, July 28, 2013

Revisiting Dan's PA greenhouse

Daniel Baker <baker_daniel@...> wrote:

>... the calculations don't support what I was hoping to accomplish.
> Is that about right?

I think so. For starters, it seems to me that the greenhouse needs more insulation at night and on cloudy days, and the heat store needs to move more heat into greenhouse air at night.

Filling the space between 2 layers of polyethylene film with R20 soap bubble foam at night would reduce a 500 ft^2 cover conductance to 25 Btu/h-F, so the greenhouse would only need (64-26.6)25 = 935 vs 18.7K Btu/h to stay 64 F on a 26.6 F night, or 5x24x935 = 112.2K Btu for 5 cloudy days in a row.

This could come from N 55 gallon drums with 25N ft^2 of surface and a 1.5x25N Btu/h-F conductance to slow-moving air and a Tm = 64+935/(1.5x25N) F minimum water temp.

Drums in an insulated box (a solar closet (tm)) with an R1 transparent south wall over an insulated south wall (a passive thermosyphoning air heater) under a bench that gain and lose 0.9x0.8x830 = 598 Btu/ft^2 = 6h(Ta-82)1ft^2/R1 in a greenhouse that's 82 F for 6 hours on an  average December day would have temperature Ta = 182 F.

A 55 gallon drum has 55x8.33 = 458 Btu/F of thermal capacitance. Storing 112.2K Btu = (182-Tm)458N = (182-64-24.93/N)458N makes N = 2.28. N = 4 makes Tm = 70.2 F, storing 205K Btu, enough heat for 219 cloudy hours, ie 9.1 days, with an approximate 1-2^-9.1 = 0.998 solar heating fraction.

If the drum tops and bottoms are not exposed to air, that lowers the heat transfer surface to about 75 ft^2, with a 1.5x75 = 112 Btu/h-F airfilm conductance, which raises Tm to 64+935/112 = 72.3 F. And the drum box needs a 62 F thermostat and a fan, eg a 665 cfm fan, which further raises Tm to 64+935/(1/112+1/665) = 73.7 F. And with finite insulation, the box will lose some heat at night, making the water cooler than the surrounding air during the day, which lowers Ta.

And a 9'x10' tomato greenhouse needs about 0.4x9'x10' = 36 lb/day of winter dehumidification, which normally comes from daytime ventilation with dry outdoor air, which precludes beneficial CO2 enrichment with compost, which can also generate heat. Air at 82 F and 70% RH has vapor pressure Pa = 0.7e^(17.863-9621/(460+82)) = 0.783 "Hg and humidity ratio wa = 0.62198/(29.921/Pa-1) = 0.01671 pounds of water per pound of dry air.

NREL's Blue Book says an average 33.6 F December day in Harrisburg has a 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The average December humidity ratio wo = 0.0028, and air weighs about 0.075 lb/ft^2, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wo) = 36lb, with a C = 96 cfm fan. With an average (33.6+40.6)/2 = 37.1 F daytime temp, this costs about 6h(82-37.1)96 = 25.9K Btu/day of heat.

The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F. If we cool it to Tg < 71 (F) with condensation, it will have vapor pressure Pg = e^(17.863-9621/(460+Tg)) "Hg at 100% RH and humidity ratio wg = 0.62198/(29.921/Pg-1). For example, cooling it to 43 F makes Pg = 0.282 "Hg and wg = 0.0059, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wg) = 36lb, with a C = 124 cfm fan, at a cost of about 6h(82-43)124 = 29K Btu of sensible heat + 36K Btu for the latent heat of condensation, totaling 65K Btu/day.

If the condensation happens on the inside of 500 ft 2 of outer glazing film with a 3 Btu/h-F-ft^2 conductance to 37.1 F outdoor air, the glazing temp Tg will be 37.1+65K/6h/500/3 = 44.3 F, which is warmer than 43 F, so let's try cooling the air to (44.3+43)/2 = 43.7. Then Pg = 0.290 "Hg and wg = 0.0061 and C = 1.33/(0.01671-0.0061) = 125.4 cfm and the glazing gains 6h(82-43.7)125.4+6K = 64.8K Btu and Tg = 37.1+64.8K/6h/500/3 = 44.3 F. That's closer.

With tall indeterminate tomato plants and an 8' height and 80% solar transmission, the greenhouse could gain 0.8x10'(9'x520+8'x830') = 90.6K Btu/day. Dehumidifcation with 64.9K Btu/day leaves 90.6K-64.9K = 25.7K Btu for 18 hours of nighttime heat and more efficient dehumidification by night ventilation. With 18h(64-33.6)(500ft^2/R20+C) = 25.7K, C = 22 cfm, which can remove 18hx60x22x0.075(0.01671-0.0028) = 25 lb/day of water vapor. We could reduce the dew point of
greenhouse air to 64 F by condensation before night ventilation to avoid dripping water on plants.

Meanwhile, page 1 of says "on average, yields of crops should increase by 33% with a doubling of C02 concentration in the earth's atmosphere," ie 700 vs 350 ppm, and page 549 of Joe Hanon's Greenhouses book (CRC, 1998) says a 1 kg dry weight compost loss can provide 1.5 kg of CO2 and suggests 7-14 kg/m^2 of wet compost to provide sufficient CO2 for 20 days, eg 88 kg (193 lb) for a 9'x10' 8.4 m^2 greenhouse. Some growers spread straw between crops in windrows and keep it moist and turn it every 2-3 days, but it seems more efficient to keep compost in a closed and insulated container with an oxygen sensor and a blower to maintain at least 1% O2 and a humidity sensor and a solenoid valve to maintain a 50% moisture content, or less, if less CO2 is required. says:

>... about 0.50–0.60 kg of CO2/hr/100 m2 must be added in a ‘standard’ glass greenhouse to maintain 1,300 ppm. For double-polyethylene houses supplementation is 0.25–0.35 kg of CO2/hr/100 m2. For glass houses, supplementation is primarily used to offset the dilution due to air infiltration, while for double-poly houses the amount of CO2 required is about equal for the natural air exchange and photosynthesis.

Years ago, I visited a prizewinning orchid grower in Pennsylvania. He was a retired chemist. His greenhouse was rather dark and completely closed in, with a huge air conditioner and a natural gas heater with a thermostat and mister nozzles with a humidistat and a solenoid valve to maintain 90 F at 90% RH for 24 hours per day and alcohol lamps for CO2 enrichment and powerful fans to make plants move for better gas diffusion absorption by stomata...


Friday, July 26, 2013

A PA greenhouse solar heating project

Dan Baker writes:

>I am building a backyard greenhouse about 10 x 9 and am thinking of trying to extend my season as long as possible.  I live in S. Central PA.

NREL's Blue Book says an average 33.6 F December day in Harrisburg has 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The 30-year record December low is -8 F. The deep ground temp is 52.9.

The Harrisburg TMY2 EPW weather data stat file at says the soil is 5.0 C (41.0 F) at a 0.5 meter depth in December.

> My plan after viewing many YouTube videos is to dig out a 4 x 4 x 8 trench in the greenhouse, line it with 1" reflective coated rigid board insulation, place 2 55 gallon (plastic) barrels connected with PVC and cover it all with sand.

Here's an electrical equivalent circuit on an average night, with 128 ft^2 of R5 board insulation and an average 2' sand depth at R0.44 per inch (an R21 vertical bed resistance) and a 1.5x4x8 = 48 Btu/h-F bed surface conductance and 500 ft^2 of R1 greenhouse glazing:

                     Ts                 Tg
      5/128  10.5/32 |   10.5/32  1/48  |  1/500
          -->        |
           I        --- Cs = 916+113ft^3x18Btu/ft^3-F = 2956 Btu/F

With no collector, I = (41.0-26.6)/0.718 = 20.1 Btu/h, which makes
the Thevenin equivalent temp Tt = 41.0-20.1(5/128+10.5/32) = 33.6 F.

To find the Thevenin resistance, ground the 41.0 and 26.6 sources and calculate the resistance from Ts to ground: 0.367 "ohms" to the left in parallel with 0.351 on the right, ie 0.179 "ohms."
So the circuit above is equivalent to:

       |         |
      --- 2956  --- 33.6 F
      ---        -
       |         |
       -         -

with an RC = 0.179F-h/Btux2956Btu/F = 529 hour time constant.

> The collector would be a 4x4 insulated panel using corrugated metal and running copper piping from bottom to top and sealing it with glass or something that would help hold the heat.  Then water would be circulated by a swimming pool pump and go into the storage in the ground.  It would circulate all day and then at night I would turn off the pump and let the heat dissipate into the sand which would radiate into the greenhouse through the night. says 2.75 kWh/m^2 (872 Btu/ft^2) of sun falls on a south wall with a 60 degree tilt on an average December day, so a 4'x4' panel with R1 glazing with 90% solar transmission and perfect back and side insulation and lots of water flow for 6 hours per day would have this equivalent circuit:

   0.9x16ft^2x872/6h = 2093 Btu/h
         ---      |
      R1/16ft^2   |
33.6 -----www-----

which is equivalent to:

  --- 33.6+2093/16 = 164.4 F.

Adding the first equivalent circuit and a switch...      
          1/16   Ts   0.179
    ------www---c  d---www---  
   |         --> \     -->   |
  --- 164.4 F Ic | Ts   Id  --- 33.6 F
   -            ---          -
   |            ---          |
   -             |           -

to charge the sand bed for 6 hours and discharge it
for 18 hours on an average day...

6Ic = 18Id makes 6(164.4-Ts)16 = 18(Ts-33.6)/0.179,
ie the sand bed temp Ts = 97.5 F.

And (97.5-26.6)/(10.5/32+1/48+1/500) = 202 Btu/h flows from the top of the sand bed into the greenhouse at night, which raises the greenhouse air temp to 26.6+202/500 = 27.0 on an average night.


Wednesday, July 24, 2013

Should we insulate hot water pipes?

Malcolm Shealy discusses this at

The SRCC OG-300 solar water heater test procedure uses 6 hourly 3 gpm 10.7 gallon 135 F bursts in a 67.5 F environment...

With no insulation, 1' of 135 F 1/2" horizontal bare copper pipe with a 0.625" OD in 67.5 F still air would have a 0.27((135-67.5)/0.625"/12))^0.25 = 1.62 Btu/h-F-ft^2 convective thermal conductance, using equation 2.20US for laminar free convection from a horizontal pipe in Kreider and Rabl's 1994 Heating and Cooling of Buildings book. With pipe conductance Gb = Pi0.625"/12x1'x1.62 = 0.265 Btu/h-F-ft, it would lose (135-67.5)0.265 = 17.9 Btu/h per linear foot. A 2 gpm (1000 Btu/h-F) water stream would lose 2 F with a 2000 Btu/h pipe loss after flowing through 2000/17.9 = 112 feet of bare pipe.

How much will a $1.18 piece of R2.3 6' pipe insulation save in 1 year, if we heat water with electricity at $0.10/kWh and use it in 6 10.7 gallon 3 gpm hourly bursts each day?

With "R2.3 insulation" (ie a system R-value after installation, including the outer air film), Gi = Pi0.625"/12x1'/R2.3 = 0.071 Btu/h-F-ft, so the pipe would lose (135-67.5)0.071 = 4.8 Btu/h per linear foot. A 2 gpm insulated water stream would be 2 F warmer than a bare pipe stream after flowing through 2000/(17.9-4.8) = 153 feet of insulated pipe.

Type M 1/2" copper pipe weighs 0.933lb/5' = 0.1866 lb/foot, with a 0.0915x0.1866 = 0.0171 Btu/F thermal capacitance. Its 0.5" ID holds 62.33Pi(0.25/12)^2x1'= 0.085 lb of water, making the total (tiny) capacitance C = 6x0.1021 = 0.6126 Btu/F.

If hot water uses are 3.6 minutes long, with (60-3.6)/60 = 0.94 hours between uses, bare pipe with a C/Gb = 0.39 hour time constant would cool from 135 to 67.5+(135-67.5)e^(-0.94/0.39) = 73.4 F before the next use during the day, and 67.5 F overnight. Reheating 6' of water to 135 F would take (5((135-73.4)+(135-67.5)0.6126 = 169 Btu/day, or 61644 Btu/year, worth about $0.10x61644/3412 = $1.81/year.

Insulated pipe with a C/Gi = 1.44 hour time constant would cool to 67.5+(135-67.5)e^(-0.94/1.44) = 102.6 F during the day, and 67.5 F overnight. Reheating it  takes (5((135-102.6)+(135-67.5)0.6126 = 141 Btu/day, or 51341 Btu/year, worth about $0.10x51341/3412 = $1.50/year.The reheating difference is 31 cents per year.

So this $1.18 investment has a 100x$0.31/$1.18 = 26% tax free return and a $1.18/$0.31 = 3.8 year simple payback, if the pipe is in an unheated space and labor is free, with hourly hot water uses.

Pipe insulation within a few feet of a water heater makes even more sense, but plumbing heat traps don't seem to make sense, given pipe insulation, since they only slow convection inside a single pipe, unless there's a hot-cold connection above. And a 4 watt mechanical timer that turns off a well-insulated water heater at night would use more electrical power than it saves.

How about a simple greywater heat exchanger? Home Depot sells 300' of 1" HDPE 100 psi NSF plastic pipe for $96.64...

With a 1.049" ID and a 1.119" OD, and a 0.035" wall thickness and still clean water on the inside and some crud on the outside, 1' of pipe could have a 1'xPi1.049/12x20 = 5.50 Btu/h-F conductance and a Pi(1.049/24)^2x62.33 = 0.374 Btu/F conductance and an RC = C/G = 0.374/5.50 = 0.068 hour (4.1 minute) time constant, so a 300' pressurized pipe coil near the top of a tall unpressurized stratified tank with a much larger volume and non-conductive walls and a drain from the bottom (eg a 4' diameter x 8' tall culvert or ferrocement or thin bent plywood tank in a basement with an insulated lining and a greywater dip tube to the bottom) could heat Pi(1.049/24)^2x7.48x200 = 13.5 gallons of 60 F water to 105-(105-60)e^(-0.94/0.068) = 104.9999554 F between hourly uses, with a 105 F shower drain temp.

Every piece of NSF pipe is tested to 5 times the nominal pressure rating at 73 F, and it loses about 10% of its strength for each 10 F temperature increase above 73 F.

"Scott L" <shiva@...> wrote from Fairbanks:

>... we insulate our cold and hot water pipes is to keep them from freezing as fast if the heat quits.

NREL says the average temp is -10.1 F in January in Fairbanks, with a -18.5 min and a -61.0 30-year record low and a 0.0006 humidity ratio. The deep ground temp is 26.9 F.

Insulation doesn't help much with small pipes. At -10.1, a 1/2" pipe with R2.3 insulation and a 1.43 hour time constant would cool from 70 to 32 F in -1.43ln((32-(-10))/(70-(-10)) = 0.92 hours and freeze solid after another 0.085x144/((32-(-10))0.071) = 4.1 hours. A battery or generator could run (32-(-10))0.071/3.412 = 0.87 watts/foot of heat tape under pipe insulation when the heat quits. A 12V 200 Ah battery could keep 200' of R2.3 pipe from freezing for 13.7 hours.

>The cold water pipes sweat and drip if left uninsulated and exposed to a heated area with our cold water.

The air would need significant moisture for that to happen. Air with a w = 0.0006 humidity ratio has vapor pressure P = 29.921/(0.62198/w+1) = 0.0288 "Hg and dew point Tdp = 9621/(17.863-ln(P))-460 = -10.6 F.

> Lots of us haul or have our water delivered. $80 for 800 gallons in the town area and easily twice that if you live outside the city area by much. You can not afford to run the water until it gets hot over a long distance.

We could insulate both hot and cold pipes and use a motion detector and pump to circulate water back through the heater before hot water uses. That could also help avoid frozen pipes when the heat quits. Cooling a 50 gallon 135 F tank to 32 F with 200' of R2.3 pipe in -10 F air would take (135-32)50x8.33Btu/(200x(32-(-10))0.071) = 72 hours.


Saturday, July 20, 2013

Personal cooling

This $184 Polar system has an icewater cooler and a pump and a U-shaped bladder and an optional lumbar compression wrap for the bladder.

We might circulate antifreeze through an optional 12.5"x22.5" rectangular bladder under ice-cube trays in a small freezer in series with the U-bladder, in order to cool the cooler without adding ice.

This system has adjustable flow rate temperature control, ie we can vary the pump flow rate manually to vary the cooling. A more advanced system could have an adjustable bladder thermostat to run the pump as needed and measure the dew point of the surrounding air  and limit the lower bladder temp to avoid condensation. Air at T (F) (absolute temp 460+T R) and RH% has an approximate (T+460)/(1-ln(RH/100)(T+460)/9621))-460 F dewpoint. For example, 70 F air at 50% RH has a 530/(1-ln(0.5)530/9621))-460 = 50.5 F dewpoint.

If it's humid, air movement helps. The Berkeley online comfort calculator says 25.4 C (77.7 F) air with a 25 C mean radiant (wall) temp and a 0.15 m/s air speed and a 50% RH and a 1.2 metabolic rate and 0.5 clo clothing is comfortable (PMV = 0.01 on a scale of -3 (very cold) to +3 (very hot)). Increasing the RH to 60% makes the PMV = 0.08 (slightly warm.) Increasing the air speed to 0.17 m/s makes things comfortable again, with PMV = -0.01. So does removing some clothing, with PMV = -0.02 at 0.45 clo. See also and and

Losing weight also helps, allometrically-speaking, ie increasing our heat-losing-surface-to-heat-generation-volume ratio. The ASHRAE Handbook of Fundamentals says a W-pound animal generates P = 6.6W^0.75 Btu/h of heat, and a W pound x H inch tall person has A = 0.108W^0.425H^0.725 ft^2 of DuBois surface area, eg P = 289 Btu/h and A = 19.6 ft^2 for a 154 lb 68" tall ASHRAE-standard human, who might be dT = P/(1.5A) = 9.8 F warmer than surrounding still air if naked, with no sweating or shivering or other adaptations. Raising W to 200 lb makes P =  351 Btu/h and A = 21.9 ft^2 and dT =  10.7 F, with H = 68". A proportional height gain to H = 12.7x200^(1/3) = 74.2" raises A to 23.3 ft^2 and lowers dT to 10.0 F.


Thursday, July 18, 2013

Recycle a freezer into a heat storage tank

This 24.9 ft^3 chest freezer could hold 24.9x7.48 = 186 gallons of water in a 10'x12' folded EPDM liner...

It uses 568 kWh/year, eg 3x568x3412Btu/kWh/365d/24h = 664 Btu/h with a COP of 3 at 0 F in a 70 F room, with a 664Btu/h/70F = 9.5 Btu/h-F thermal conductance and about 80 ft^2 of surface with an R-value of 80ft^2/9.5Btu/h-F = R8.4 ft^2-F-h/Btu, which could be raised with fiberglass or foamboard. R10 foamboard over the top and sides would reduce the conductance to 65ft^2/R18.4+15ft^2/R8.4 = 5.3 Btu/h-F.

It could live above a 4'x16' bare fin 2.56 gpm DIY collector made from $67 worth of 0.018" brown- or black-painted aluminum flashing as described at

The collector could have 128 ft of 1/2" PEX-Al-PEX tubing on 6" centers, with 2 Ts to make 2 64' 1.28 gpm flow paths. says 64' of 1/2" PEX tubing with a 0.475" ID would have a 1.28 gpm flow with a 3.7' pressure drop, and an $85 Grundfos 15-58C pump can deliver 2.56 gpm with a 8' pressure drop on speed 1, using 60 watts.

A $100 differential thermostat controller or an Arduino controller could run the pump when the collector is less than 40 F to avoid freezing or when the tank is warmer than 160 F to avoid overheating the EPDM liner. The tank could could contain a 1"x300' pressurized PEX coil as a heat exchanger, or it could be a gravity-pressurized hot water supply with a float valve. (Rich Komp has a gravity hot water supply with a 3' head (a 55 gallon drum) for a shower with a rainwater supply in his off-grid house in Maine. The flow is adequate, but it won't knock you back against the wall like a shower in a hotel.)

Where I live near Philadelphia (not an easy climate for solar heating), 1000 Btu/ft^2 of sun falls on a south wall on a 30 F average January day with a 34 F daytime temp. If 0.8x8'x16'x1000 = 102.4K Btu enters 8'x16' of R2 twinwall glazing with 80% solar transmission over 6 hours and 51.2K falls on a 4'x16' 160 F bare collector which also gains 6h(Ta-160)4'x16'x3 Btu from both sides from Ta (F) air inside the glazing and

102.4K = 51.2K + 6h(Ta-160)4'x16'x3 + 6h(Ta-34)8'x16'/R2,

then Ta = 161.8 F, with 51.2K + 6h(Ta-160)4'x16'x3 = 53.3K Btu of water heating on an average January day, ie 30% more than the 41K Btu/day SRCC 0G-300 spec, in colder, cloudier weather, eg 53.3K/(110-60)/8.33 = 128 gallons of 110 F water heated from 60 F.

On the 1st cloudy day after a long string of average days, the tank can supply 53.3K Btu and cool to 160-53.3K/(186x8.33) = 125.6 F, with no backup heat requirement.

On the 2nd cloudy day, as it cools from 125.6 to 110 F, the tank can supply (125.6-110)186x8.33 = 24.2K Btu, enough to heat 24.2K/(110-60)/8.33 = 58 gallons of water from 60 to 110 F. Adding 128-58 = 70 gallons at 60 F will reduce the tank temp to (60x70+110x186)/(70+186) = 96.3 F, with an average (110+96.3)/2 = 103.2 exit temp, and heating that 70 gallons to 110 requires (110-103.2)70x8.33 = 3986 Btu, making the backup fraction 3986/53.3K = 0.075 after the 2nd cloudy day.

On the 3rd cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+96.3x186)/(128+186) = 81.5, with an average (96.3+81.5)/2 = 88.9 exit temp, and heating that 128 gallons to 110 requires (110-88.9)128x8.33 = 22.5K Btu, making the backup fraction 22.5K/53.3K = 0.422 after the 3rd day.

On the 4th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+81.5x186)/(128+186) = 72.7, with an average (81.5+72.7)/2 = 77.1 exit temp, and heating that 128 gallons to 110 requires (110-77.1)128x8.33 = 35.1K Btu, making the backup fraction 35.1K/53.3K = 0.634 after the 4th day.

On the 5th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+72.7x186)/(128+186) = 67.5, with an average (72.7+67.5)/2 = 70.1 exit temp, and heating that 128 gallons to 110 requires (110-70.1)128x8.33 = 42.5K Btu, making the backup fraction 42.5K/53.3K = 0.769 after the 5th cloudy day.

(If I did that right. I should learn how to use spreadsheets :-)

Assuming cloudy days are like coin flips, the expected backup heating fraction is approximately

0x2^-1+0x2^-2+0.075x2^3+0.422x2^-4+0.634x2^-5+0.769x2^-6 = 
0         +0        +9.4E-3      +26.4E-3     +19.8E-3    +12.0E-3       = 0.0676,

so, ignoring the tank heat loss, the expected solar heating fraction is 1-0.0676 = 93% in January.  


Now that all of us are plagued with the pollution resulting from the overabundance of devices we have purchased, perhaps government or church groups should sponsor a series of "you don't need it" commercials. Instead of the bright uniformed "service personnel" of the Ace Air Conditioning Company briskly delivering and installing the latest gadgets, the commercials would show the expensive equipment misused: a bored housewife growing geraniums in her new dishwashing machine; a small child casually dismantling a TV-stereo combo with a claw hammer...

from the book "Sunspots," by Steve Baer, 1979.

Sunday, July 14, 2013

DIY air-air heat exchangers

"yreysa" <gary@...> mentions Shurcliff's example:

It has about 36x65"x19"/12^2 = 309 ft^2 of heat exchange surface with an approximate U = 0.75 Btu/h-F-ft^2 2-sided slow airfilm conductance, so AU = 309x0.75 = 232 Btu/h-F, with NTU = 232/100 = 2.32 and effectiveness E = NTU/(NTU+1) = 70% at 100 cfm and 82% at 50 cfm.

Room air at 70 F and 50% RH has 0.00787 pounds of water per pound of dry air and weighs 0.075 lb/ft^3. Freezing all the water out of a 30 cfm airstream would make 24hx60m/hx30cfmx0.075lb/ft^3x0.00787lb/ft^3 = 25.5 pounds of ice per day. Melting ice takes 144 Btu/lb. If the exhaust fan runs without the intake fan for T hours per day to melt ice and T(70-32)30 = (24-T)60x30x0.075x0.00787x144, T =  2.84 hours per day, with a 12% duty cycle. Condensation and ice could form in the outgoing airstream between Coroplast flat plates, with incoming cold air inside the plate corrugations.

But (IMO) this is so rarely needed in the US that it makes more sense to just run an exhaust fan with a humidistat if the RH reaches say, 60% in an "airtight house." Swedish houses have 0.025 ACH. A SIP house might have 0.1 or 0.2. An average US house has about 1 ACH. On an average day, a 4,000 ft^2 house can provide ASHRAE's standard 15 cfm for a full-time occupant with a natural air leakage of only 15x60/(4Kx8') = 0.028 ACH, ie about 0.56 ACH on a 50 Pa blower door test.

Bill Shurcliff also suggested a "lung" with an external bellows that could turn all the cracks and crevices in a house envelope into very efficient bidirectional latent and sensible air-air heat exchangers. ORNL's preliminary experiments used a bellows that was too small to exhaust all the air from a 2x6 stud cavity.

To avoid that restriction, a 90 watt 2470 cfm Lasko 2155a reversible window fan could be in an interior wall that divides the house in two partitions with a humidistat in series with a repeat cycle timer like Grainger's $83.90 2A179 (with its $4.26 5X582 socket) which can reverse the fan airflow with adjustable off and cycle times from 1.2 seconds to 300 hours.

If every 10'x10' envelope section contains a 4x10'x6"x1/32" crack and 30 cfm flows through 4,096 ft^2 of envelope and each section has a "duct" with 40/32/12 = 0.104 ft^2 of cross-sectional area, the entire envelope has 4096/100x0.104 = 4.27 ft^2, and the average air velocity is 2x30/4.27 = 14 fpm. Each wall section has 40x0.5x2 = 40 ft^2 of heat exchange surface, so Cmin = 30 and NTU = AU/Cmin = 4096/100x40x1.5/30 = 82 and E = 1-exp(-82), ie 100%. Not bad :-)

The cycle time should be long enough that stale house air clears the path to the outdoors before the fan reverses. If it isn't, we'll know, because the humidity and CO2 concentration in the house won't decrease much as the fan runs.

This could also dehumidify a house... says Energy Star dehumifidifiers typically remove 4 pounds of water per kWh (and add heat to a house if not part of an air-conditioner.) Philadelphia has an average 76.7 F temp and a 67.2 daily min and 0.0133 humidity ratio in July. ASHRAE says 80 F with a 0.0120 humidity ratio is comfortable.

A Lasko fan with a smart ventilation controller could remove 60x2470x0.075(0.0120-0.0100) = 22.2 pints per hour of water vapor from house air with a 0.0120 humidity ratio when the outdoor humidity ratio drops to 0.0100 using 90Wh, at 22.3/0.090 = 247 pints/kWh, using 988 times less energy than an Energy Star dehumidifier.

And this could happen at night in summertime, when outdoor air is cooler, since house materials can store dryness as well as coolth. Concrete weighs about 150 lb/ft^3 and stores about 25 Btu/F-ft^3 and 1% of its weight by volume as the RH of the surrounding air rises from 40 to 60%.   


Low-flow showers

The March/April 2013 issue of Sierra (the club magazine) says wind is now
cheaper than any other form of electrical energy generation, at $48-95/MWH, vs
$61-89 for a natural gas combined cycle plant with 60% efficiency or coal at
$62-141 or $200-231 for an industrial gas turbine (eg Capstone?), IF we remove
all subsidies for all forms of generation. But conservation and energy
efficiency are still cheaper, at $0-50/MWH.

I use a fully-enclosed shower with a continuous 1/4 gpm brass nozzle head and
swivel fitting from McMaster-Carr in parallel with occasional 1 gpm sprays from a hose
with a hand valve...\
Drilling a hole
through the plastic disk shutoff valve in the sprayer T will keep the 1/4 gpm
shower head going when you are using the sprayer to wash off soap. describes the benefits of a fully-enclosed shower.

This $9.17 nozzle uses 0.2 gpm at 40 psi with a 60 degree spray cone...
Here's a $15.60 swivel fitting...

This uses about 5 times less water than a 1.5 gpm "low-flow" shower head...\

McGill U's "Water Conservation and the Mist Experience" pamphlet suggests using a garden sprayer, eg this $8.57 1-gallon sprayer...

Bucky Fuller's Dymaxion pressure mist shower used 1 pint per hour, ie 0.002 gpm, ie 2 milligpm, 750 times less than a "low-flow showerhead."


Wednesday, July 10, 2013

A tomato greenhouse in Nashville

"brad.howard308" <bradhoward@...> wrote:

>I am trying to design a solar heating system for a new 32 ft x 96 ft greenhouse on my farm.

NREL says 730 Btu/ft^2 of sun falls on the ground and 1030 falls on a south wall on an average 36.2 F January day in Nashville, with a 26.5 and 45.9 low and high and a (36.2+45.9)/2 = 41 average daytime temp and a wa = 0.0035 humidity ratio (0.0035 pounds of water per pound of dry air.) The deep ground temp is 59.1. The 30-year record low is -17 F. The average July temp is 79.3, with a 68.9 and 89.5 low and high and 1980 Btu/ft^2 of sun on the ground and 770 on a south wall and a 0.0152 humidity ratio.

>I need to try and keep temps in the greenhouse between 50F - 90F with the optimum range 60F - 80F. recommends a 70 to 82 F day temperature at 60-70% RH and a 64 F minimum night temperature, with nutritional deficiencies below 60 F and fruit damage below 57 F. says a tomato greenhouse can evaporate 0.4 lb/ft^2 per day of water vapor, about 0.4x32'x96' = 1229 lb for a 32'x96' greenhouse.

Air at 82 F and 70% RH has vapor pressure Pg = 0.7e^(17.863-9621/(82+460)) = 0.783 "Hg and humidity ratio wg = 0.62198/(29.921/Pg-1) = 0.0167, and it weighs about 0.075 lb/ft^3, so it looks like maintaining a 70% RH requires 1229/(6hx60m/h/0.075lb/ft^3(wg-wa)) = 3448 cfm of ventilation for 6 hours per day. The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F.  

>Idea 1... a south facing 4 ft x 96 ft screen collector tilted at about 50 degrees, laying on its side down the length of the greenhouse blowing into 4 inch corrugated plastic non-perforated drain pipe buried 3 ft under the green house with 2 ft perimeter insulation.

Why not omit the screen and its plant shading and just blow warm air down from the peak of the greenhouse through perforated pipes, to allow any condensation to drain from the pipes (altho there won't be any condensation if the ground is warmer than 71 F)... suggests something like this, but I'd still worry about mold and mildew in the pipes infecting the plants.

The Kissock equation says A ft^2 of soil with Rf floor insulation and Rp perimeter insulation would lose Qf = UA(Tg-Ts) Btu/h, with U = 0.11/(4+Rf+Rp)+0.89/(16+Rf), where Tg is the greenhouse air temp and Ts is the deep ground temp. With A = 32'x96' and Rf = 0 and Rp = 10 and Tg = 59.1, Qf = 205(Tg-59.1) Btu/h.

With 2 R1 polyethylene film covers, the greenhouse thermal conductance would be about 32'x96'/R2 = 1536 Btu/h-F.

With tall indeterminate tomato plants and a 12' height and 80% solar transmission, it could gain about 0.8x96'(32'x730+1030x12') = 2.74 million Btu/day.

With lots of thermal mass and airflow and a constant Tg (F) temp 24 hours per day,

2.64M = 6(Tg-41)3448+24(Tg-36.2)1536+24(Tg-59.1)205
      = 20688Tg-848208+36864Tg-1334477+4920Tg-290722

makes Tg = 83.5, but the greenhouse could be 70 F during the day and 64 at night, so there's plenty of solar heat on an average day.

Keeping the greenhouse 64 F on an average night requires (64-26.5)1536 = 57.6K Btu/h. Soil with a 32x96x1.5 = 4608 slow-moving airfilm conductance could provide this at a 64+57.6K/4608 = 76 F surface temp.

Keeping the greenhouse 64 F on a 24-hour average 36.2 F day requires (64-36.2)1536 = 42.7K Btu/h... 16 96' x 4" pipes would have 1608 ft^2 of surface with a 1608x1.5 = 2413 Btu/h-F airfilm conductance... says 55 cfm flowing through 48' of 4" duct would have a 0.1 H20 pressure drop, so 16 pipes fed from the center would have about 0.1 "H20 with a 16x2x55 = 1760 cfm flow. With a combined surface and pipe conductance of 4608+1/(1/2413+1/1760) = 5626 Btu/h-F, the soil can provide 42.7K Btu/h at a 64+42.7K/5626 = 72 F temp.

The greenhouse needs 5dx24hx57.6K = 6.9 million Btu to stay 64 F for 5 cloudy 36.2 F days in a row. This could come from 6.9M/(82-72) = 690K Btu/F of mass cooling from 82 to 72 F, eg 690K/30Btu/F-ft^3 = 23K ft^3 of soil with a 30 Btu/ft^3-F specific heat by volume, eg a 23K/32'x96' = 7.5' deep x 32'x96' greenhouse floor. That's very deep, because the temperature swing of this mass is only 10 F, and the floor probably needs more than 1 level of pipes, given the R1 per foot soil thermal resistance.

>Idea 2... a south facing 4 ft x 96 ft PEX or CPVC type collector in the same configuration but use a "tank", basically an insulated hole lined with pond liner as a heat storage tank. Use pex buried in the ground to heat the ground up and inside the greenhouse.

This seems more promising. Water has twice the thermal mass by volume of soil, and it could have a higher temp on an average day and a larger temperature swing. At 140 F, a 4'x96' water tray with an R1 cover under the ceiling would collect 4'x96'(0.8x730-6h(140-82)/R1) = 90.6K Btu/day. With 4'x96'x2x1.5 = 1152 Btu/h-F of airfilm conductance and a 10K cfm ceiling fan and a 1/(1/1152+1/10K) = 1033 Btu/h-F total conductance, tray water at 64+57.6K/1033 = 120 F could keep the greenhouse 64 F on a 26.5 F night.

That's a high minimum usable tray water temp, so the greenhouse needs a large water volume to store enough heat for 5 cloudy days in a row: 6.9M/(140-120)/62.33 = 5469 ft^3 of water, eg an 18' cube. Adding 8 1000 Btu/h-F car radiators and fans would raise the total conductance to 9033 Btu/h-F with a minimum tray water temp of 64+57.6K/9033 = 70 F and a 6.9M/(140-70)/62.33 = 1581 ft^3 tank, eg a 12' cube, but it seems more practical to fill the space between 2 poly film covers with soap bubble foam on cloudy days.

R20 foam would reduce the heat requirement to (64-26.5)32'x96'/R20 = 5.76K Btu/h on an average night, and the tray and ceiling fan could provide that with a minimum tray water temp of 64+5.76K/1033 = 70 F. The greenhouse would only need 5dx24hx5.76K = 690K Btu to stay 64 F for 5 cloudy 36.2 F days in a row, eg a 690K/(140-70)/62.33 = 158 ft^3 tank, eg a 6' cube or a 4' wide x 3' tall x 14' long plywood box with a 10'x20' folded EPDM liner under a bench.

Zelon’s Swedish patent (US No. 3672184, June, 1972) described insulating shop windows at night with soap bubble foam. Professor John Groh at U. Arizona measured US R3 per inch for soap bubble greenhouse insulation in 1968. Professor Otho Wells at U. New Hampshire did later greenhouse experiments. In 1995, Bill Sturm built a 12,000 ft^2 tomato greenhouse with a soap bubble foam roof in Calgary, Alberta and measured an 84% propane energy savings with and without foam on alternate nights at 20 F below zero.

Bill showed me a simple foam generator with a shop vac blower connected to a horizontal 2" PVC pipe full of holes in a shallow trough near the ground containing a 10% detergent solution which expands by a factor of 300 when foamed. He suggests some window screen to push on a microswitch in an air return at the top of the glazing cavity to turn off the shop vac for all but a few seconds every half-hour at night, along with a standard greenhouse inflation blower above the trough water line to fill the space between the 2 films with air during the day, with a check valve to reduce blower power...

How about summer cooling?


Monday, July 1, 2013

Improving indoor evaporative cooling

ASHRAE says 80 F air with a 0.0120 max humidity ratio (pounds of water per pound
of dry air) is comfortable. Vapor pressure Pi = 29.921/(0.62198/0.0120+1) =
0.5663 "Hg, and RH = 0.5663/e^(17.863-9621/(460+80)) = 0.54, ie 54% max.

In a dry climate, a portable indoor evaporative cooler (eg\
sr_1_2?s=appliances&ie=UTF8&qid=1372588973&sr=1-2&keywords=evaporative+cooler )
can be efficient if a line-voltage thermostat (\
QAAEAE ) runs the cooler when the room air temp rises to 80 F and a humidistat ( ) runs an exhaust fan (\
ef=sr_1_1?ie=UTF8&qid=1372590555&sr=8-1&keywords=lasko+2155a ) when or if the
room RH rises to 54%.

If the outdoor air temp is Ta = 86.7 F with a wa = 0.008 humidity ratio (an
average July afternoon in Billings MT) and the cooler evaporates 10 pounds of
water per hour and the exhaust fan moves C cfm of 0.075lb/ft^3 air and 10 =
60C0.075(wi-wa), C = 555 cfm, and the net cooling is 10x1000 - (86.7-80)555 =
6278 Btu/h, like a typical window AC with about 10X less electrical power.

Cool exhaust air could desirably fill a house attic ("up ducts") with a one-way
plastic film damper to prevent warm outdoor air from flowing down through the
house when the fan isn't running and allow cooler outdoor air to flow up through
the house when it is available. If the house has enough natural air leakage, the
fan may never run.

A cooler near an outdoor air inlet can evaporate more water than one handling
cooler and moister house air. A supply plenum could enclose the exhaust fan and
the cooler inlet, with a plastic film damper that allows house air to enter the
plenum when the exhaust fan is not running. The exhaust fan could become a
2-watt motorized damper, raising the COP.

A solenoid valve scrounged from an old washing machine and some misters that
spray water on a rock pile in a tub could replace the indoor evaporative cooler
and lower the cost and raise the SEER to about 6278Btu/h/(90Wx555/2470) = 310 vs
10 for a window AC, using the 90 watt 2470 cfm Lasko fan...\

Fancier controls could turn off evaporation and run conventional AC when the
weather is unsuitable.