Hybrid Car – More Fun with Less Gas

Modeling an air heater with a massy ceiling

register ::  Login Password  :: Lost Password?
Posted by Nick Pine on December 29, 2004, 1:52 pm
 
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


Posted by nicksanspam on January 1, 2005, 2:57 pm
 
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


Posted by Astro on January 1, 2005, 11:29 pm
 

<lots snipped>

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!


Posted by nicksanspam on January 5, 2005, 1:03 pm
 
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


This Thread
Bookmark this thread:
 
 
 
 
 
 
  •  
  • Subject
  • Author
  • Date
please rate this thread