Document 4JOrMyQ0Q3QmXkJGxaXeo4ENj

276 CHAPTER 12 1950 Guide at 2:00 pjn. during July for an outdoor design temperature 95 F, and an inside tem perature 80 F. Solution: From Table 14 in 2 p.m. column for 6 in. concrete plus 2 in. insulation, find the total equivalent temperature differential 34 deg. The overall heat trans mission coefficient for summer is taken from Table 16 and is found to be 0.13. The heat flow rate equals 34 X 0.13 -- 4.42 Btu per (hr) (sq ft). Example 7: For the conditions of Example 6, find the rate of heat flow into build ing at 2:00 pjn. during July for design temperatures of 105 F (outdoor) and 78 F (in door). Dady range of temperature 30 degrees, i.e., outdoor temperature minimum of 75 F which occurs at 4:00 or 5:00 a.m.; this being 30 deg less than the maximum. Solution: Make correction in equivalent temperature differential in accordance with Note 5 in Table 14 as follows: The correction for 27 deg design temperature difference is (27 -- 15) = + 12. The correction for 30 deg daily range is Net total correction is + 12 -- 5 + 7. The heat flow rate at 2:00 pjn. therefore is (34 + 7) X 0.13 = 5.32 Btu per (hr) (sq ft). A method of determining heat flaw rates, when structure is not given in Tables 14 or 15 is illustrated in Example 8. Example 8: A 4 in. stone concrete roof covered with an average depth of 4 in. cin der concrete (k = 4.9) on which is placed a } in. thick felt roof with 1 in. pitch and slag surface, is exposed to the sun. The location is the central part of the United States. Design temperatures are: outdoor 95 F: daily range 20 deg; indoor temperature 80 F. Find the heat flow rate at 2:00 pjn. fora day in July. Solution: For the purpose of selecting the equivalent temperature differential, this construction is assumed to be equal approximately to an uninsulated 6 in. concrete roof, for which the equivalent temperature is found to be 38 deg in the 2:00 pjn. column of Table 14. Calculate the overall heat transmission coefficient U of the roof as follows: U =---------------------- ----------------------- = 0.33. 1 4 4 0.375 0.60 1 -- -I------------ 1--------------1------------------------------------ ----------- 1.2 12 4.9 1.33 1.00 4.0 The heat flow rate is then 38 X 0.33 equals 12.5 Btu per (far) (sq ft). GLASS AREAS--DESIGN TABLES . In order to clearly set forth the principles involved in calculating heat, transfer through glass areas, the general instantaneous heat-balance rela-` tion will be presented. Fig. 5 shows this schematically. The net heat gain for the indoor space is the result of several contributing phenomena.. In discussing the heat gain through glass') there are three dimensionless quantities which require definition: a = absorptivity of glass, or fraction of incident radiation intensity which is ab sorbed within the glass itself. r = transmissivity of glass, or fraction of incident radiation intensity which is transmitted through the glass. r = reflectivity of glass, or fraction of incident radiation intensity which is reflected at the surface. It is necessary that o+t+r-1 (9) These quantities vary with wave length and angle of incidence, prima rily; and they are determined by the properties and thickness of the glass material concerned. Data are obtainable from glass, manufacturers for their various products. The dependence of these quantities upon wave Cooling Load 277 Table 16. Summer Coefficients of Heat Transmission U of Flat Roofs Covered With Built-Up Roofing* Btu per (hour) (square foot) (F deg difference between the air on the two sides) " Nominal thickness of wood is specified but actual thiekneBS was used in calculations. If corkboard insulation is used, the coefficient U may be decreased 10 per cent. length has important practical consequences; for example, common window glass transmits a large portion of incident solar radiation, whereas it transmits outward only a very small portion of the indoor radiation