re: simple solar heating problems SOLAR.KnowledgePublications.com

re: simple solar heating problems
26 may 1996
i wrote: 

 here's another litmus test for an energy expert: what will the average water 
 temperature be inside a 4' cube full of water surrounded by 5 r20 foam walls,
 sitting outside in december, when the air has an average 24 hour temperature
 of 36 f and the sun puts 1,000 btu/ft^2/day of heat into the r1 glazed side
 of the box? and how will that change over time if we shade the sunny wall?

which russ schmidt nicely solved :-)

nobody has solved the one below yet, after a few days, so i'll try...

 bonus problem (another $10): how does all this change if a december day
 is 6 hours long, and during each 18 hour night, that r1 glazed wall is
 somehow insulated to r20, like the other walls, and it stays r20 for
 24 hours a day during the dark time.

this would be more interesting if the cube were part of a high-performance
solar house, with the south face of the cube as the back wall of a sunspace,
and the rest of the cube walls inside the house. below are two solutions for
the outdoor cube, in a simple program that compares their results. one is an
iterative arithmetic solution, and another uses exponentials.

the steady state temp is determined by ein = 16k btu/day = eout, where

eout = 6 hr (t-36) 16 ft^2/r1      daytime south wall loss with damper open
    + 18 hr (t-36) 16 ft^2/r20     nightime south wall with damper closed
    + 24 hr (t-36) 5*16 ft^2/r20   other walls, 24 hours a day
     = (96+14.4+96)(t-36), so

t = 36 + 16k/206.4 = 113.5 f, after a long string of average december days.
insulating the glazed south wall at night makes this cube warmer than the
cube where the water lost heat through the r1 glazing all night... 

the thermal resistance of the cube with its south damper or door closed is 
r = r20/(6*16ft^2) = 0.20833, and the 4,000 pounds of water inside has a
thermal mass of c = 4,000 btu/f, so the time constant rc of the closed cube
is rc = (0.20833)(4000) = 833 hours or 35 days :-) so...

t(t) = 36 + (t(0)-36)exp(-t/35), where t(0) is the water temperature at the
start of a cloudy week and t is in days. the program below calculates this,
and compares it with an arithmetic solution that finds a new water temp each
day, as the heat loss from the water cools its own thermal mass, assuming
the water temp is constant over a day. whole-day time steps work well here,
because the rc time constant of the closed-up cube is so big.

10 'heating and cooling a 4' cube of water surrounded by r20 insulation,
20 '   sitting outdoors on a 36 f day, with 1,000 btu/ft^2/day of sun
30 '   passing into one r1 side wall for 6 hours a day
40 '
50 ta = 36'ambient temp
60 v=4^3'volume of water (ft^3)
70 c=64*v'thermal mass of water (btu/f)
80 uday = 16/1 + 5*16/20'daytime thermal conductance of cube
90 unight = 6*16/20'nightime thermal conductance of cube
100 sun=16*1000'solar input to sunny side of cube (btu/day)
110 'solar heating during an average week
120 tss=36+sun/(6*uday+18*unight)'estimate steady state water temp
130 tw = tss'initial water temp
140 for day=1 to 7'simulate for a week, day by day
150 eout=(6*uday+18*unight)*(tw-ta)'energy flowing out of the cube
160 tw=tw+(sun-eout)/c'new water temp
170 print day;ta;tss;tw
180 next day
190 'passive cooling during a cloudy week
200 twi=tw'initial water temp
210 print
220 rc=c/unight'time constant of closed and shaded cube in hours
230 for day=1 to 7'simulate for a week, day by day
240 eout=24*unight*(tw-ta)'energy flowing out of the cube in an hour
250 tw=tw-eout/c'new water temp, calculated arithmetically
260 texp=ta+(twi-ta)*exp(-24*day/rc)' calculated exponentially
270 print day;ta;texp;tw
280 next day

 1  36  113.5194  113.5194  once the cube reaches this temperature,
 2  36  113.5194  113.5194  after a long string of average days,
 3  36  113.5194  113.5194  the temperature does not change...
 4  36  113.5194  113.5194
 5  36  113.5194  113.5194  the south wall damper is open for 6 hours  
 6  36  113.5194  113.5194  a day in this case...
 7  36  113.5194  113.5194

this is an average water temperature. the water would be about
1 f higher at dusk and 1 f lower at dawn, just coasting... :-)

 1  36  111.3695  111.3391  here is how the water temperature of 
 2  36  109.2793  109.2202  the cube changes, with the south wall
 3  36  107.247   107.1609  damper closed all the time, during a week 
 4  36  105.2711  105.1595  without sun, when the air temp is 36 f...
 5  36  103.35    103.2144
 6  36  101.4822  101.324   there's only a small difference between
 7  36  99.66614  99.48676  the arithmetic and exponential solutions...

the outdoor cube temperature decreases slowly during a cloudy week.

how long will it take to reach 80 f?

if t(t) = 80 = 36 + (113.5-36)exp(-t/35), then 44 = 77.5exp(-t/35), 
exp(-t/35) = 0.567, -t/35 = ln(0.567), and t = -35 ln(0.567) = 19.8 days :-)

if the south face of the cube is the back wall of a sunspace which is 80 f
when the sun is shining, ie warmer than the 36 f outdoor temp, and the rest
of the cube walls are inside the 70 f living space of a house, ie they lose
less heat every day than if they lost heat to 36 f outdoor air, the steady
state temperature goes up, so we can store more heat in the cube, ie

ein = 16k btu/day, but

eout = 6 hr (t-80) 16 ft^2/r1      daytime south wall loss with damper open
    + 18 hr (t-36) 16 ft^2/r20     nightime south wall with damper closed
    + 24 hr (t-70) 5*16 ft^2/r20   other walls, 24 hours a day
     = (96+14.4+96)t - 7680 - 518 - 6720, and 

16k + 14,918 = 206.4 t, so t = 30,918/206.4 = 149.8 f, after a long string
of average december days in phila.

putting a ground reflector in front of this solar closet, eg snow at
60% reflectivity, raises the solar input above the number given with 
20% ground reflectivity, so ein = 21.33k and t = 36,251/206.4 = 175.6 f.

how long will this indoor closet take to reach 80 f?

80 = 70 + (175.6-70)exp(-t/35) ==> t = -35 ln(10/105.6) = 31.8 days :-)

i'll offer another $10 to the first person to calculate how the steady state
temp would change, if at this point we add another layer of glazing to the
solar closet, using these assumptions, or any other reasonable assumptions...

bigger closets have bigger time constants, increasing as their dimension,
squared, so an 8' cube might stay warm for about 4 months indoors :-) 

nick