Document rewGayg1m74wnjD8dNK4rrnka

CHAPTER 13 1956 Guide 296 Table 11. Sommeb 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) * The summer coefficients are considered temporary, and have been calculated with an outdoor wind velocity of 8 znph. For summer an inside surface conductance of 1.2 has been used instead of the regular 1.55 value. In all of these roofs a 4 ply felt roof has been assumed f in. thick, thermal conductivity -- 1.33* Fitch and slag have been assumed as an additional thickness of 4 in. which has been assigned thermal con ducbti8v7i4type=rc1e.0n.t gIynpsbuomth, c1a2s4epsethrceernmt awlocoodnfdibuecrt.iviTtyhircekfenresstsoinodneiciantcehdtihnicclkundeesss4. in* gypsum board. This is aepNouormedinraolothf.ickness of wood is specified, but actualthickness was used in calculations. d If corkboard insulation is used, the coefficient V may be decreased 10 percent. A method of determining heat flaw rates, when structure is not given to Tables 9 or 10, is illustrate in Example 7. 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 | in. thick felt roof with ) in. pitch and slag surface, is exposed to the sun. The location is the central part of the United States. Cooling Load 297 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 U (see Equa tion 3 of Chapter 9) of the roof as follows: U =------- :----- --------------------------------- = 0.33. J_ 4 0375 0.50 I 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 (hr) (sq ft). SOW ssx_ INCIDENT OUTDOOR SURROUNDING^ / AREFLECTED -Fh (WAVE LENGTHS UNCHANGajT* TRANSMITTED INDOOR RADIATION (wave lengths unchanged) INCIDENT INDOOR RADIATION REFLECTED I indoor OUTDOOR CONVECTION *< v THERMAL CAPACITANCE OF CLASS INDOOR CONVECTION V<U V>u gIITTEO OUTDOOR RADIATION irrERENT DISTRIBUTION 'OF ENERGY VS. WAVE LENGTH THEN TRANSMITTER -ft1--tr--7T- EMITTED INDOOR RADIATION _ ^DIFFERENT DISTRIBUTION OF ENERGY VS.WAVE LENGTH THEN TRANSMITTED) OUTDOOR AIR TEMPERATURE > tgota OUTDOOR CLASS - SURFACE TEMPERATURE tl INDOOR AIR TEMPERATURE INDOOR. GLASS- SURFACE TEMPERATURE Fig. 2. Instantaneous Heat-Balance Conditions on a Glass Section 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