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 http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/state.html 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 http://apps1.eere.energy.gov/buildings/energyplus/cfm/weather_data3.cfm/region=4_north_and_central_america_wmo_region_4/country=1_usa/cname=USA 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
41.0---www-----www---*-----www----www-------www---26.6
--> |
I --- Cs = 916+113ft^3x18Btu/ft^3-F = 2956 Btu/F
---
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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:
0.179
Ts-------www---
| |
--- 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.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi 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:
1/16
------www----->
|
--- 33.6+2093/16 = 164.4 F.
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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
- --- -
| --- |
- | -
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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.
Nick
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