Hybrid Car – More Fun with Less Gas

Re: Opinions on power for rural property

register ::  Login Password  :: Lost Password?
Posted by nicksanspam on March 31, 2006, 8:46 am
 


Maybe 40'x48', with 10 8'x24' SIP roof panels and 8 wall panels and
1920 ft^2 of floorspace and 156 ft^2 of R4 windows with 50% solar
transmission and 0.1 natural air changes per hour (26 cfm.)


NREL says January is the hardest month for solar heating in Albuquerque,
when 1010 Btu/ft^2 of sun falls on the ground and 1640, 690, 240, and 700
fall on SWNE walls on an average 34.2 F day with a 46.8 daily max. Say it's
7 F cooler at 7000 feet, ie 27 F, and 0.5(80x1640+32x690+12x240+700x32)
= 89.3K Btu of solar heat enters windows on an average day.

With no other form of heat and a 60 F average daily indoor temp (say 70 F
for 12 hours per day and 50 at night) the max house thermal conductance
G = 89.3K/(24h(60-27)) = 113 Btu/h-F. Subtracting 156/4 = 39 Btu/h-F for
windows and 26 for air leaks leaves 48 for 3172 ft^2 of SIPs, so they need
a min 3172/48 = 66 US R-value, eg 17" R4/inch EPS SIP walls...

A single 128 ft^2 layer of COEX or Dynaglas polycarbonate "solar siding"
over the 48'x8' south wall might collect an additional 0.92x128ft^2x1640
= 193.1K Btu and lose 6h(100-33)128ft^2/R1 = 51.5K, for a net gain of 142K,
or 231K, counting the windows, which makes Gmax = 231K/(24h(60-27)) = 292
(not counting the warmer south wall during the day and the added R1 cover
at night) and lowers the minimum SIP R-value to 3172/227 = 14, but that
would require a lot of thermal mass for heat storage.

R24 (6") SIPs make G = 3172/24+26+39 = 197, so the house needs about
(60-27)197 = 6.5K Btu/h or 156.2K Btu/day, or 781K for 5 cloudy days.
If the siding collects 193.1K Btu/day at an average daily air temp T
and we need 156.2K/day of space heat plus 50K Btu/day of hot water and
193.1K = 156.2K+50K-89.3K+6h(T-33)128, T = 99 F. If 64 ft^2 of Dynaglas
air heater glazing inside the siding collects 1388x64 = 88.4K Btu/day
and provides 50K Btu/day of hot water at temp Tw (F) from fin-tube pipe
and loses 6h(T-99)64ft^2/R1 from the air heater glazing, Tw = 200 F.
At 170 F, we might store cloudy day heat in 781K/(170-80) = 8678 pounds
(139 ft^3) of water in a 3'x4'x12' EPDM-lined plywood tank on the ground,
with a pressurized copper coil heat exchanger to heat water for showers.
A greywater heat exchanger (eg a $0 1"x300' PE pipe coil in another 4'
of tank) could reduce the size of the main tank.

If the sunspace keeps the house warm for 6 hours per day, we need to
store about 105K Btu of overnight heat. If that comes from an overhead
thermal mass cooling from 95 to 55 F, we need 105K/(95-55) = 2625 pounds
of water in poly film ducts on shallow shelves or 24 20'x4" thinwall PVC
water pipes on a foil-covered platform under a foil-covered ceiling.
(The foil prevents overheating the room by radiation.)

At night and on cloudy days, we could bring down warm air as needed with
a room temp thermostat and a slow ceiling fan and an occupancy sensor.
After 5 cloudy days, we might pump 6.5KBtu/(80F-75F) = 1300 lb/h (2.7 gpm)
of tank water up through the pipes under the ceiling, with supply and
return pipes under the tank water level and a partial vacuum aloft to
minimize pump power.

Nick


Posted by nicksanspam on March 31, 2006, 1:10 pm
 


Well, 21 pipes should be enough, or fewer, with more insulation...

20 SW:WW2:NW:EW2'window areas (ft^2)
30 AWIND=SW+WW+NW+EW'total window area (ft^2)
40 RWIND=4'US window R-value
50 G=AWIND/RWIND'window thermal conductance (Btu/h-F)
60 W@:LH'house dimensions (feet)
70 RCEIL$'ceiling R-value (US)
80 G=G+L*W/RCEIL'add ceiling conductance
90 RWALL$'wall R-value
100 G=G+(2*(L+W)*8-AW)/RWALL'add wall conductance
110 ACH=.1'natural air changes per hour
120 CFMH*W*L*8/60'air leakage (cfm)
130 G=G+CFM'add air leakage conductance
140 HHEAT=(60-27)*G'average hourly heat (Btu/h)
150 DHEAT$*HHEAT'daily heat
160 CHEAT=5*DHEAT'heat for 5 cloudy days
170 WHEAT=.5*(SW*1640+WW*690+NW*240+EW*700)'window heat (Btu/day)
180 AS8'sunspace area (ft^2)
190 GS=.92*AS*1640'sunspace gain (Btu/day)
200 DHWP000!'water heating (Btu/day)
210 TS3+(GS-DHEAT+WHEAT-DHW)/6/AS'sunspace temp (F)
220 AP *3.1416*4/12'pipe area (ft^2)
230 VP *3.1416*(2/12)^2'pipe volume (ft^3)
240 CPb.33*VP'pipe thermal capacitance (Btu/F)
250 RCP=2/3/AP*CP'pipe time constant (hours)
260 OHEAT=DHEAT*18/24'approx overnight heat (Btu)
270 DHEAT=(50-14.7)*G'dawn heat (Btu/h)
280 NP 'number of 20'x4" ceiling pipes
290 TPMINP+DHEAT/(NP*AP*1.5)'dawn pipe temp (F)
300 TPMAX=TS+(TPMIN-TS)*EXP(-6/RCP)'dusk pipe temp (F)
310 IF (TPMAX-TPMIN)*NP*CP<OHEAT THEN NP=NP+1:GOTO 290
320 PRINT NP,TPMIN,TPMAX
330 AAHd'air heater area (ft^2)
340 GAHH*.92*.92*1640'air heater gain (Btu/day)
350 TAH=TS+(GAH-DHW)/6/AAH'air heater temp (F)
360 TW0'water temp (F)
370 QFi0*9.6865E-04*(TAH-TW)^1.4172'fin-tube pipe gain (Btu/h-ft)
380 LFT=DHW/QF/6'fin-tube pipe length (feet)
390 PRINT TS,TAH,TW,LFT
400 MHEAT=(70-14.7)*G'morning heat (Btu/h)
410 TPMINCp+MHEAT/(NP*AP*1.5)'min cloudy-day pipe temp (F)
420 CTANK=CHEAT/(TW-TPMINC)'tank capacitance (Btu/F)
430 LTANK=CTANK/62.33/3/4'tank length (feet)
440 PRINT TPMINC,LTANK

number         min pipe      max pipe
of pipes       temp (F)      temp (F)

 21            60.87604      114.4807

sunspace       air heater    water         fin-tube
air temp (F)   temp (F)      temp (F)      length (ft)

 125.9938      227.1348      170           40.35743

morning pipe   tank length
temp (F)       (ft)

 87.0381       12.97192

Nick


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