Passive Solar Design for New Mexico
Class IV
Review of previous classes:
 The Three Pillars of Passive Solar: Gain, Mass, and Insulation
 Guidelines
 Basic energy concepts: BTUs, solar constant, light transfer, heat transfer
 Heat Load Calculations
Thermal Mass:
 density (r), specific heat (c), conductivity (k)
 (Conventional) Heat Capacity (C): Q_{stored}
= V C DT_{_material}, V =
volume, C = heat capacity,
DT = change in temperature of the
material.
 Diurnal Heat Capacity (C_{d}) (See the
article by Doug Balcomb called "Thermal Storage in Passive
Buildings" in the Advanced Passive Solar Design Book  this appears to
have appeared in Supplement One of an ASHRAE publication titled
"Passive Solar Heating Analysis", published in November 1987): The
formula involving diurnal heat capacity is Q_{stored, returned after 6pm} = A C_{d}
_{ }
DT. Note that the material's area is used here, not
volume, in
contrast to the conventional heat capacity. Thickness is still important,
but is taken into account implicitly  the values for C_{d}
must either be for a specified thickness, or for infinite thickness, in
which case the material actually used must be thick enough to approximate
the behavior of infinite thickness. This is possible of course because we
are talking about the harmonic response, not a "dc" (static)
response, wherein heat goes in and out on a daily cycle, so that it doesn't
penetrate too deeply. As the general passive guidelines suggest, four inches
is roughly adequate for materials with a high thermal mass index, and less
for materials with low indices. Also note that the temperature swing in this
case is for the air in the room. This allows one to add up a total diurnal
heat capacity for many materials acting simultaneously in the same
room. In fact, once added up, the temperature swing can be predicted
using DT_{swing }= .61 Q_{solar}/C_{d_total}, where Q_{solar }is the daily solar heat
gain, and C_{d_total} is the total diurnal heat capacity of
all the material surfaces in the room. If the room temperature swing was not
incorporated, for example, if you use the conventional head capacity, you'd
have to determine in some way the swing of each material separately. To
summarize, the use of area instead of volume, and room temperature swing
instead of material temperature swing, allow the diurnal heat capacity
approach to implicitly embody an enormous amount of information about the
transfer of energy from the air to the material, and the propagation of heat
in the material for a 24 hour cycle, making diurnal heat capacity easy to it
use but extremely powerful for architectural design applications.
Material 
density (r) lbm/ft^{3} 
specific heat (c) Btu/lbm^{o}F 
heat capacity (C) Btu/ft^{3}^{o}F 
conductivity (k) Btu/fthr^{o}F 
Thermal Mass Index
(rck=Ck) 
diurnal heat capacity (radiative/convective)
Btu/ft^{2}day^{o}F

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 
1.42.2 
Example: 1000 ft^{2 }house: Assume 120 ft^{2 }south glazing
(12% of floor area); Assume thermal mass of concrete, with an area of six times
south glazing = 720 ft^{2}, thickness of four inches. Volume of mass is
therefore 720/3 = 240 ft^{3}. Heat storage accomplished with DT
= 10 ^{o}F is Q_{stored} = 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 ^{o}F? Area
is 1080 ft^{2}, Volume is 1080/3 = 360 ft^{3}. Q_{stored}
= V C DT = 360 x
30 x 15 = 162,000 Btu = 1.62 therm. This is more than enough (thus we have
verified the 69 times mass area guideline). What if
diurnal heat capacity used instead? Q_{stored} = A C_{d}
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, Q_{stored}
= A C_{d} 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.
 Thermal Wave Propagation: Here we will check the diurnal heat capacity
concept from another perspective  by examining the dynamics of the thermal
wave into concrete over a six hour time scale. Standard heat transport
theory yields that:
Q = rate of heat flux into or out of an infinitely thick wall
= A k (T_{outside}  T_{insideinfinity})
/( p D t)^{ 1/2}_{, }where D =k/(rc)
(diffusion coefficient)
Q_stored = energy stored in an infinitely thick wall
= (2/p^{1/2}) A (rck)^{1/2}
(T_{outside}  T_{insideinfinity}) (t)^{ 1/2}
Note that the thermal mass index directly controls the amount stored.
Example: Concrete, rck = 16.2,
Assume (T_{outside}  T_{insideinfinity}) = 10 ^{o}F,
Q_stored = 45.4 A (t)^{ 1/2};
Time (hours) 
Heat Stored (Btu/ft^{2}) 
1 
45.4 
2 
64.2 
3 
78.6 
4 
90.8 
5 
101.5 
6 
111.2 
The value of 111.2 Btu/ft^{2 }stored after 6 hours is almost exactly
what the diurnal heat capacity would predict to be stored with a temperature
change of 10 ^{o}F. Thus the value reported for the diurnal heat
capacity looks quite reasonable.
Homework:
 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.
 Calculate how many therms are lost through the window over a 24 hour
period, if the average temperature inside and outside are 65 ^{o}F
and 30 ^{o}F, respectively. Assume the window has a U value of .4
Btu/(hr ft^{2} ^{o}F).
 Compare the two answers above: What can you conclude?