Page 348 - Handbook of Energy Engineering Calculations
P. 348
+ 0.20 = 0.21, where 0.24 and 0.20 are the ratios at 48°N and 36°N,
respectively; 12 is a constant derived from 48 − 36; and 40 is the latitude for
which a ratio is sought.
Next, find the collector area by using the relation A = (g)(A ), where A =
C
F
C
2
2
collector area, ft (m ); g = ratio of collector area to floor area, expressed as a
2
2
decimal; and A = floor area, ft (m ). Therefore, A = (0.21)(2.10) = 44 ft 2
C
F
2
(4.1 m ).
To compute the conductive heat loss through a surface, use the general
relation H = UA Δt, where H = conductive heat loss, Btu/h (W); U =
C
C
2
overall coefficient of heat transmission of the surface, Btu/(h · ft · °F)
2
2
2
[W/(m · K)]; A = area of heat transmission surface, ft (m ); and Δt =
temperature difference, °F = 65 − t (°C = 18.33 − t ), where t = average
o
o
o
monthly temperature, °F (°C). The U values of materials can be found in
ASHRAE and architectural handbooks.
Since a direct-gain system was selected, the total area of glazing is the sum
2
2
2
2
of the collector and noncollector glazing 44 ft (4.1 m ) + 12 ft (1.1 m ) = 56
2
2
ft (5.2 m ). Double glazing is recommended in all passive solar designs and
2
2
is found to have a U value of 0.42 Btu/(h · ft · °F) [2.38 W/(m · K)] in
winter. Thus, the conductive heat loss through the glazing is H = UA Δt =
C
(0.42)(56)(65 – 32) = 776 Btu/h (227.4 W).
The area of opaque wall surface subject to heat loss can be estimated by
multiplying the wall height by the total wall length and then subtracting the
estimated glazed areas from the total exterior wall area. Thus, the opaque
2
2
wall area of this space is (8)(15 + 14) – 56 = 176 ft (16.3 m ). Use the same
general relation as above, substituting the U value and area of the wall. Thus,
2
2
2
2
U = 0.045 Btu/(h · ft · °F) [0.26 W/(m · K)], and A = 176 ft (16.3 m ).
Then H = UA Δt = (0.045)(176)(65 – 32) = 261 Btu/h (76.5 W).
C
To determine the conductive heat loss of the roof, use the same general
relation as above, substituting the U value and area of the roof. Thus, U =
2
2
2
2
0.029 Btu/(h · ft · °F) [0.16 W/(m · K)] and A = 210 ft (19.5 m ). Then H c
= UA Δt = (0.029)(210)(65 – 32) = 201 Btu/h (58.9 W).
To calculate infiltration heat loss, use the relation H = Vn Δt/55, where V =
i
3
3
volume of heated space, ft (m ); n = number of air changes per hour,