Page 351 - Handbook of Energy Engineering Calculations

(0.65)(54.387) = 35,352 Btu (37.3 kJ). This step is not required for the design
of thermal-storage wall systems since the storage system is integrated within
the collector.

5. Compute the volume of thermal storage material required
For a direct-gain system, use the formula V = i /(d)(c )(Δt )(C ), where V M
M
p
S
S
S
3
3
= volume of thermal storage material, ft (m ); d = density of storage
3
3
material, lb/ft (kg/m ); c = specific heat of the material, Btu/(lb · °F)
p
[kJ/(kg · K)]; Δt = temperature increase of the material, °F (°C); and, C =
S
S
fraction of insolation absorbed by the material due to color, expressed as a
decimal.
Select concrete as the thermal storage material. Entering Table 11, we find
3
3
the density and specific heat of concrete to be 144 lb/ft (2306.7 kg/m ) and
0.22 Btu/(lb · °F) [0.921 kJ/(kg · K)], respectively.
TABLE 11 Properties of Thermal Storage Materials

A suitable temperature increase of the storage material in a direct-gain

systems is Δt = + 15°F (+8.3°C). A range of +10 to +20°F (+5.6 to 11.1°C)
S
can be used with smaller increases being more suitable. Select from Table 12.

In this space, thermal energy will be stored in floors and walls, resulting in a
weighted average of C = 0.60. Thus, V = 35,352/(144)(0.22)(15) (0.60) =
S
M
3
3
124 ft (3.5 m ).
TABLE 12 Insolation Absorption Factors for Thermal Storage
Material Based on Color