Modeling an air heater with a massy ceiling

Consider an 8' cube with walls and ceiling with R-value R and an Ac ft^2 air heater with a screen absorber that keeps 70 F air near R1 glazing with 90% solar transmission, as in the Barra system. If 1000 Btu/ft^2 of sun falls on the south wall over 6 hours on an average 30 F January day in Philadelphia, we might store overnight heat with water in pipes just under the ceiling at about 60 F min and 140 F max, averaging about 100 F on an average day. If 0.9x1000Ac = 6h(70-30)Ac/R1 + 2h(70-30)64x64/R + 16h(50-30)4x64/R + 24h(100-30)64/R, Ac = 318/R. An R10 cube might have a 4'x8' collector.
With L' of 3" pipe with pipe surface A = Pi3"/12"L = 0.785L ft^2 and pipe- air conductance Gp = 1.5A = 1.18L Btu/h-F and capacitance Cp = 3.1L Btu/F and cube conductance G = 5x64/R10 = 32 Btu/h-F at night and dawn heat loss Qdawn = (50-30)32 = 640 Btu/h and heat storage requirement Ed = 2h(70-30)32 +16h(50-30)32 = 12.8K Btu/day, the minimum usable pipe water temp Twmin = 50+Qdawn/Gp = 50+542/L, eg 59 F with L = 60' of pipe.
T (F) air near the ceiling over 6 hours and a 4'x8' R1 air heater with 90% solar transmission and 70 F air near the glazing makes an hourly gain of 0.9x1000x32/6h = 4800 Btu/h = (70-34)32ft^2/R1[glazing]+(70-34)3.5x64ft^2/R10 [3.5 walls]+(T-Tw)1.5A [into the pipes]+(T-34)64ft^2/R10 [ceiling], which makes T = (2597+TwL)/(L+5.44).
If dTw/dt = (T-Tw)Gp/Cp = 989/(L+5.44)+0.381Tw(L/(L+5.44)-1), dTw/dt+CTw=d and Twmax = d/c+(Tmin-d/c)e^-(6c) = 477+(Tmin-477)e^-(6x0.381(1-L/(L+5.44))). (Twmax-Twmin)Cp = Ed makes L = (L-1.27)e^(-12.46/(L+5.44))+11. L = 50 on the right makes L = 49.92 on the left, then 49.85, and 49.77, so 50' of pipe will do, with Tmin = 60.8 and Tmax = 144.5 and Cp = 155 pounds of water.
Notes:
1. It may not be easy to keep 70 F room air between the screen absorber and the cold glazing by natural convection. A fan might help.
2. A system like this with room air entering the heater may need to control collector airflow to avoid room overheating. If (T-70)cfm = (70-30)3.5x64/10, cfm = 896/(T-70), which will decrease as T increases over an average day.
3. Starting the day with 60.8 F vs warmer water keeps the air heater cool and allows solar collection to be more efficient.
4. To avoid overheating the room by radiation at 144 F, the ceiling needs a low-e surface facing the floor.
5. A shed or cathedral ceiling could be more efficient than a horizontal one, since it would trap warm air near a smaller surface under the ridge, with less heat loss to the outdoors.
5. A slow ceiling fan with an occupancy sensor and a room temp thermostat could bring down warm air from the ceiling as required. Keeping warm air trapped under the ceiling requires controlling air leaks and convection currents arising from glazing and insulation differences on walls. Room air would tend to rise near a warmer wall and sweep across the ceiling (undesirably mixing with hot air) and fall near a cooler wall.
6. A lower-mass structure could cool more quickly to 50 F at the end of the day, making it more efficient than a higher-mass structure.
7. A less passive system with higher performance might have a pump and an underfloor tank to store heat for 5 cloudy days in a row. More insulation would decrease the required tank size.
8. Two pumps and two tanks could be more efficient, with cool water in one and warm in the other, and a hose between them. We could pump cool water up into the pipes at the end of the day to cool the cube to 50 F sooner.
9. We might have fin-tube vs PVC pipes near the ceiling.
10. And more fin-tube or a flat poly film water duct under the floor, with a central thermal chimney, instead of the slow ceiling fan.
Nick
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It looks like we might store overnight heat for an 8' R10 cube near Phila with a 4'x8' air heater that keeps 70 F air near the glazing using 16' of fin-tube under the ceiling and 8' under the floor. The ceiling tube might heat 50 lb/h (0.1 gpm) of 53 F water to 96 during the day. The floor tube might cool 15 lb/h of 96 F water back to 53 F at night, with a 36 gallon unpressurized stratified heat storage tank under the floor. Eric Hawkins of Powertech in the UK says has found that the fin-tubes can stay primed for years if the ends of the supply and return pipes are underwater.
We might keep the room air 70 F over an average solar collection day with varying sun and outdoor temp by pumping more or less water through the ceiling tubes. A ceiling fan might do away with the floor tubes (or poly film water duct under the floor) and heat the cube faster after a setback.
The tank must be larger for 5 cloudy days in a row. More insulation (R10 isn't much) or an 8'x8' vs 4'x8' air heater would raise the water temp and reduce the tank size. It might make water for showers with a coil in the unpressurized tank and a flat tank under the floor for a greywater-to-air heat exchanger. The 8'x8' air heater might be 2 4'x8' pieces of 0.020 inch polycarbonate film over a picture frame over a dark south wall, with 2 4'x8' pieces of dark window screen for the mesh absorber.
10 RV'wall and ceiling US R-value (ft^2-h-F/Btu) 20 GC=5*8*8/RV'cube to outdoor conductance (Btu/h-F) 30 TA0'outdoor temp (F) 40 TSP'setback temp (F) 50 EDS=2*(70-TA)*GC+16*(TS-TA)*GC'daily heat storage requirement (Btu) 60 EDSHS/6'daily heat storage rate (Btu/h) 70 SSUN00'daily sun on south wall (Btu/ft^2) 80 AHA=4*8'air heater area (ft^2) 90 NAHG=.9*SSUN*AHA-6*(70-TA)*AHA/1'net air heater gain (Btu/day) 100 EDW=6*(70-TA)*(4*64-AHA)/RV'6-hour heat loss from walls (Btu) 110 TC=TA+(NAHG-EDS-EDW)/(6*64/RV)'average daytime air temp under ceiling (F) 120 LU'ceiling fin-tube length (feet) 130 GU=5*LU'ceiling fin-tube to air conductance (Btu/h-F) 140 QDP'daytime water flow rate (Btu/h-F) 150 TIND=TC-EDSH/(QD*(1-EXP(-GU/QD)))'daytime input temp (F) 160 TOUD=TIND+EDSH/QD'daytime output and nighttime input temp (F) 170 DAWNHEAT=(TS-TA)*GC'heat requirement at dawn (Btu/h) 180 QNWNHEAT/(TOUD-TIND)'nighttime water flow rate (Btu/h-F) 190 LL=-(LOG(1-((DAWNHEAT/QN)/(TOUD-TS)))*QN)/5'floor fin-tube length (feet) 200 PRINT TS,TIND,TOUD,TC'Ts,Tcold,Thot,Tc (F) 210 PRINT QD,QN'day and night water flow rates (lb/hour) 220 PRINT LU,LL,LU+LL'upper, lower, total fin tube length (feet) 230 CTANKS/(TOUD-TIND)'stratified tank volume (pounds of water) 240 PRINT CTANK,CTANK/8.33,CTANK/(8.33*8)
Setback Tcold Thot Tceiling 50 53.2066 95.87326 106.6667 F
daytime nighttime water flow water flow 50 15 lb/hour
ceiling floor total fin-tube fin-tube fin-tube 16 7.982017 23.98202 feet
Stratified tank size 300 lb 36.01441 gal 4.501801 ft^3
Happy new year,
Nick
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Hey Nick (or anybody else that might be able to answer this question),
how about modelling a geothermal loop system to see how much heat can be supplied? I emailed you directly but I'm not sure if the email address was proper.
I'm interested in learning two key things: 1) how much heat can be supplied to a 4 loop system, 100' bores, 4 1/2" diameter, bores 6' apart (square pattern), sand grouted in solid granite. By supplied, I mean where is the balance point where the loops can sustain so many thousand BTUs without dropping the ground temerature more than, say, 1F/week, which would allow the system to run through the cooling season before the ground loops drop to a temperature at which the system output is significantly diminished, say 35F. Starting ground temperature, Philly 51F. Assume a system COP of 3.5.
2) How many 100' loops, spaced at 15', would be required to supply a 4 ton system.
More parameters: direct exchange geothermal - copper tubing running refrigerant directly from the compressor system. System run-time percentage based - 100%, 75%, 50%
I know this is underspecified. Just ask and I'll give you additional design parameters.
If there's anybody interested in tackling this modeling problem, let me know. I'm looking for a qualified consultant to work with me on this.
Thanks!
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I got your email and replied at length...

I suggested that you 1) concentrate on resistance ie steady-state long-term performance and ignore capacitance, and 2) work out an arm's-length test with the contractor. If he agreed to deliver a 4-ton system, test it at 4 tons for a while. Your technical efforts to help make the system work with your theories and instrumentation may irritate him and subtly shift that burden from him to you.
Good luck,
Nick
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Looks like a 4.2'x8' air heater will do, with half the previous water flow, ie 0.05 gpm during the day and 0.02 (from thermosyphoning) at night.
10 RV'wall and ceiling US R-value (ft^2-h-F/Btu) 20 GC=5*8*8/RV'cube to outdoor conductance (Btu/h-F) 30 TA0'outdoor temp (F) 40 TSP'setback temp (F) 50 EDS=2*(70-TA)*GC+16*(TS-TA)*GC'daily heat storage requirement (Btu) 60 EDSHS/6'daily heat storage rate (Btu/h) 70 SSUN00'daily sun on south wall (Btu/ft^2) 80 AHA=4.2*8'air heater area (ft^2) 90 NAHG=.9*SSUN*AHA-6*(70-TA)*AHA/1'net air heater gain (Btu/day) 100 EDW=6*(70-TA)*(4*64-AHA)/RV'6-hour heat loss from walls (Btu) 110 TC=TA+(NAHG-EDS-EDW)/(6*64/RV)'average daytime air temp under ceiling (F) 120 LU'ceiling fin-tube length (feet) 130 GU=5*LU'ceiling fin-tube to air conductance (Btu/h-F) 140 QD&.5'daytime water flow rate (Btu/h-F) 150 TIND=TC-EDSH/(QD*(1-EXP(-GU/QD)))'required daytime input temp (F) 160 TOUD=TIND+EDSH/QD'daytime output and nighttime input temp (F) 170 DAWNHEAT=(TS-TA)*GC'heat requirement at dawn (Btu/h) 180 QNWNHEAT/(TOUD-TIND)'nighttime water flow rate (Btu/h-F) 190 LL=-(LOG(1-((DAWNHEAT/QN)/(TOUD-TS)))*QN)/5'floor fin-tube length (feet) 200 PRINT TS,TIND,TOUD,TC'Ts,Tcold,Thot,Tc (F) 210 PRINT QD,QN'day and night water flow rates (lb/hour) 220 PRINT LU,LL,LU+LL'upper, lower, total fin tube length (feet) 230 ECS=8*(70-TA)*GC+16*(50-TA)*GC'cloudy day heat requirement (Btu) 240 CTANK=(EDS+5*ECS)/(TOUD-TIND)'stratified tank volume (pounds of water) 250 PRINT CTANK,CTANK/8.33,CTANK/(8.33*8)
Setback Tcold Thot Tceiling 50 50.52835 131.0315 135.1666 F
daytime nighttime water flow water flow 26.5 7.95 lb/hour
ceiling floor total fin-tube fin-tube fin-tube 16 8.002228 24.00223 feet
Stratified tank size 1431 lb 171.7887 gal 21.47359 ft^3.
Nick
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