Passive Solar Design for New Mexico

Class IV

Review of previous classes:

Thermal Mass:

Material density (r) lbm/ft3  specific heat (c) Btu/lbm-oF heat capacity (C) Btu/ft3-oF conductivity (k) Btu/ft-hr-oF Thermal Mass Index

(rck=Ck)

diurnal heat capacity (radiative/convective)

Btu/ft2-day-oF

water 62.5 1 62.5 0.35 21.9

 

concrete 143 .21 30.0 .54 16.2 10.71
adobe 106 .24 25.44 0.3 7.63 5.52
         wood          40 .5 20 0.09 1.8 1.4-2.2
 

Example: 1000 ft2 house: Assume 120 ft2 south glazing (12% of floor area); Assume thermal mass of concrete, with an area of six times south glazing = 720 ft2, thickness of four inches. Volume of mass is therefore 720/3 = 240 ft3.  Heat storage accomplished with DT = 10 oF is Qstored = V C DT = 240 x 30 x 10 = 72,000 Btu = .72 therm. This is a little low to provide adequate heat, but not awful. How about with thermal mass area of nine times south glazing with DT = 15 oF? Area is 1080 ft2, Volume is 1080/3 = 360 ft3. Qstored = V C DT = 360 x 30 x 15 = 162,000 Btu = 1.62 therm. This is more than enough (thus we have verified the 6-9 times mass area guideline). What if  diurnal heat capacity used instead? Qstored = A Cd DT = 720 x 10.7 x 10 = 77,040 Btu = .77 therm. This is very close to the result of .72 therm calculated above using a 4 inch thickness! For the larger surface area and temperature swing,  Qstored = A Cd DT = 1080 x 10.7 x 15 = 173,340 = 1.73 therm. Again, this is close to the corresponding value of 1.62 therm calculated above. Thus we have verified the four inch mass rule using the diurnal heat capacity concept. In fact, note from above that the diurnal heat capacity of concrete is 10.71 per square foot, which is also exactly one third of the heat capacity of concrete per cubic foot, 30. This makes sense, because a slab consisting of the first four inches of a cubic foot contains exactly one third of the mass of the cubic foot.   

Q = rate of heat flux into or out of an infinitely thick wall

= A k (Toutside - Tinside-infinity) /( p D t) 1/2, where D =k/(rc) (diffusion coefficient)

Q_stored = energy stored in an infinitely thick wall

= (2/p1/2) A (rck)1/2 (Toutside - Tinside-infinity) (t) 1/2

Note that the thermal mass index directly controls the amount stored.

Example: Concrete, rck = 16.2, Assume (Toutside - Tinside-infinity) = 10 oF,

Q_stored = 45.4 A (t) 1/2

 Time (hours) Heat Stored (Btu/ft2)
1 45.4
2 64.2
3 78.6
4 90.8
5 101.5
6 111.2

The value of 111.2 Btu/ft2 stored after 6 hours is almost exactly what the diurnal heat capacity would predict to be stored with a temperature change of 10 oF. Thus the value reported for the diurnal heat capacity looks quite reasonable.

Homework:

  1. Calculate how many Btus and then therms of energy are provided by solar gain over a clear six hour period through a 200 square foot, southfacing window. Assume total transmission and misalignment losses of 20% through the window.
  2. Calculate how many therms are lost through the window over a 24 hour period, if the average temperature inside and outside are 65 oF and 30 oF, respectively. Assume the window has a U value of .4 Btu/(hr ft2 oF).
  3. Compare the two answers above: What can you conclude?