Document 2N9VNG89rLRrx245DN2xq19L

: ! 290 CHAPTER 13 1955 < Cooling Load 291 mission coefficient is different, provided the structure has thermal and properties similar to one of those listed in Tables 9 and 10. 3. Adjustments can be made, according to instructions given in the footno room and outdoor conditions different from those on which the tables are ba Examples of Use of Equivalent Temperature Tables Example 5: Given: A roof is constructed of 6 in. of stone concrete with 2 in>5of. insulating board and tar felt roofing in. thick, and is exposed to the sun. The locaV tion is the central part of the United States. Find the rate of heat flow into building^ at 2:00 p.m. during July for an outdoor design temperature 95 F, and an inside'tem- perature 80 F. .A - Solution: From Table 9 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- Example 7: 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 f 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 p.m. for a 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 p.m. column of Table 9. Calculate the overall heat transmission coefficient!/ (see Equa tion 3 of Chapter 9) of the roof as follows: .-11-.2--1-14-2---1-44-.-9-- 1-01-..-33-37-5-- 1-10-..-05-00-- 1-14--.0- The heat flow rate is then 38 X 0.33 equals 12.5 Btu per (hr) (sq ft). SURROUNQlNGg / (Wave lengths unchange^^ JRANSMITTEO INDOOR RADIATION 0*AVC LENGTHS UNCHANGED) vINCIDENT INDOOR RAOIATION \i THERMAL capacitance or GLASS V< u V>`i EMITTED __________ _ (different distribution or ENERGY vs. WAVE LENGTH THEN TRANSMITTER) -ft1--tt1--yy- EMITTED INDOOR RADIATION _ CDIFFERENT DISTRIBUTION OF ENERGY VS.WAVE LENGTH THEN TRANSMITTED) to OUTDOOR AIR TEMPERATURE tg,, m OUTDOOR GLASS - SURFACE TEMPERATURE ti -- INDOOR AIR TEMPERATURE INDOOR GLASS-- SURFACE TEMPERATURE Fig. 2. Instantaneous Heat-Balance Conditions on a Glass Section mission coefficient for summer is taken from Table 11 and is found to be 0.13. The heat flow rate equals 34 X 0.13 = 4.42 Btu per (hr) (sq ft). Example 6: For the conditions of Example S, find the rate of heat flow into build ing at 2:00 p.m. during July for design temperatures of 105 F (outdoor) and 78 F (in door). Daily range of temperature 30 deg, 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 9 as follows: The correction for 27 deg design temperature difference is (27 -- 15) =+ 12. (- 3--0_---20\ ) -- ~ 5. Net total correction is + 12 -- 5 => + 7. The heat flow rate at 2:00 p.m. therefore is (34 + 7) X 0.13 = 5.32 Btu per (hr) (sq ft). A method of determining heat flow rates, when structure is not given $ Tables 9 or 10, is illustrated in Example 7 which follows. TABLES FOR CALCULATING SOLAR HEAT GAIN THROUGH GLASS AREAS Basic Principles In order to set forth the principles involved in calculating heat flow through glass areas, the general instantaneous heat-balance relation will be presented. It will be shown schematically in Fig. 2. The net heat gain for the indoor space is the result of several contributing phenomena. Some observations concerning the behavior of glass with respect to radiant energy will lead to a better understanding of the heat-balance relation. To various degrees glass transmits radiation having wave lengths between 0.29 and 4.75 microns. Of the portion not transmitted, part is absorbed, and the remainder is reflected. Outside these limits glass is opaque, absorbing approximately 94 percent and reflecting 6 percent. Only a negligible amount of radiant energy from a surface at 450 F has a wave length shorter than 4.75 microns. It is therefore convenient to treat all forms of solar radiant energy separately from radiant energy from other sources, so long as the temperature of these sources is not over approxi mately 450 F. The complete heat-balance for a glass section can be expressed for a unit time interval as follows: ]Total heat flow, ~1 __ FTransmitted, [^through glass section -- | solar radiation THeat flow by convective and radiative exchanges at Lthe indoor surface (2a) The second term of the right side of Equation 2a can also be expressed by a heat balance equation as follows: Heat flow by convective-) rAbsorbed-) ["Radiative exchanges be- ") And radiative exchanges I = solar I tween outer surface of glass | Lat the indoor surface J Lradiation J Land outdoor surroundings J ("Convective exchanges "1 ["Heat storage"! J between outer surface of I =fc I within the j (2b) Lglass and outdoor air J Lglass sectionJ Equations 2a and 2b can be combined and expressed in symbolic terms y Equation 2c. Tabular values of the two bracketed terms of Equation -a are presented later in this section for various types of glass for specific design conditions. (9/A) tT 1)7D 4" Td/d] 4" [ced/d 4" Otdld 4- <goEa -- <g1.7!Ro /co (fgo to) 51, Btu per (hr) (sq ft) (2c) }' {