Page 348 - Handbook of Energy Engineering Calculations
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+  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,
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