Document 3ejQkxVXk8OJK0kewLEJ1NNkJ

284 CHAPTER 12 1951 Guide Table 14. USummeb Coefficients of Heat Transmission of Plat Rah*. Covered With Built-Up Roofing* " Btu per {hour) {square foot) {F deg difference between Ike air on the two sides) Ttpb or Roof Deck Ceiling not shown Flat Metal Roof Deck Thickness of Roof Deck (Inches) 4 Ply Felt Roof Ditto + $ in. Blag Insulation on Top of Deck (Cowered With Built-Up Roofing) No Ceiling--Underside of Roof Exposed Furred Ceiling w th Air Space, Metal Lath and Plaster No In sula tion Insulating Board4 Thickness, In. 4 1 n2 No In sula tion Insulating Board4 Thickness, In. h I 1 ] U 1 s, 0.73 0.3f 0.23 0.17 0.13 0.40 0.23 O.lf 0.14 0.11 0.54 0.30 0.2C o.ie 0.13 0.84 0.22 0.1(1 0.13 o.n Preeast Cement Tile 4 Ply Felt Roof Ditto . + $in. Blag U 1| 0.67 0.33 0.22 0.17 0.13 0.38 0.24 0.11 0.14 0.12 0.60 0.28 0.20 0.15 0.12 0.32 0.21 0.17 0.13 0.11 Concrete 4 Ply Felt Roof 2 4 6 Ditto 2 4 + $ in. Slag 6 0.65 0.69 0.54 0.49 0.40 0.42 0.33 0.22 0.16 0.13 0.37 0.31 0.21 0.16 0.13 0.36 0.30 0.20 0.16 0.18 0.33 0.28 0.27 0.20 0.20 0.19 0.19 0.15 0.16 0.14 0.12 0.12 0.12 0.31 0.30 0.29 0.24 0.18 0.14 0.13 0.23 0.22 0.17 0.17 0.18 0.12 0.18 0.11 0.21 0.16 0.13 0.11 0.21 0.16 0.13 0.11 0.20 0.16 0.18 0.10 Gypeum and Wood Fiber1* on $" Gypsum Board 4 Ply Felt Roof 2$ 3$ Ditto 2$ + $ in. Slag 3$ 0.34 0.28 0.23 0.17 0.13 0.12 0.25 0.20 0.15 0.12 0.11 0.21 0.18 0.16 0.14 0.13 0.12 0.097 0.11 0.094 0.29 0.25 0.20 0.16 0.18 0.11 0.22 0.18 0.14 0.12 0.10 0.19 0.16 0.16 0.13 0.13 0.11 0.093 0.10 0.090 poop MaSqS I Wood* 4 Ply Felt Roof 1 0.43 0.26 1$ 0.33 0.22 2 0.29 0.20 3 0.22 0.16 0.16 0.12 0.29 0.13 0.11 0.24 0.13 0.11 0.22 0.11 0.09 0.17 0.20 0.15 0.13 0.11 0.18 0.14 0.12 9.097 0.16 0.13 0.11 9.094 0.18 0.12 0.10 9.035 . Ditto l| + $ in. Slag 2 3 0.35 0.29 0.26 0.20 0.23 0.17 0.14 0.11 0.25 0.20 0.15 0.12 0.10 0.21 0.19 0.14 0.12 0.10 0.20 0.15 0.12 0.10 0.09 0.16 0.13 0.14 0.12 MO 0.17 0.13 o.n 9.093 0.16 0.13 0.13 0.11 0.10 >.090 0.09 9.031 * The summer coefficients are considered temporary, and bare been calculated with an outdoor wind velocity of 8 mph. For summer an inside surface conductance of 1.2 has been used instead of the regular 1.65 value. In all of these roofs a 4 ply felt roof has been assumed t in. thick, thermal conductivity **1.33. Pitch and elag have been assumed as an additional thickness of 1 in. which has been assigned thermal conductivity =* 1.0. Inboth cases thermal conductivity refers to one inch thickness. b 87$ percent gypsum, 12$ percent wood fiber. Thickness indicated indudes $ in. gypsum board. This is a poured roof. ! Nominal thickness of wood ia specified, but actual thickness was used in calculations. If corkboard insulation ia need, the coefficient V may be, decreased 10 percent. sky which is not included in the Mackey and Wright method. The tem perature differentials for roofs were based on an inside surface conductance of 1.65 because the charts prepared by Mackey and Wright11 used this Cooling Load ^5 value and it was not considered practicable to repeat their work using a different film coefficient. An examination of the values given in their paper indicates that the temperature differential would be changed very little even if a value 1.20 were used instead of 1.65. But to obtain the heat flow rates through roofs, more accurate values will be obtained if the overall heal trans mission coefficient is calculated using 1.2 as the inside film conductance of heat transfer in summer. The roof coefficients of transmission for summer shown in Table 14 are based on surface conductances /,,,, of 4.0 for an outside roof surface and 1.20 for an inside ceiling surface. The outside conductance 4.0 is used for summer because it corresponds to a wind velocity of approximately 7.5 mph averaged for rough and smooth surfaces, and is more representative of summer wind velocities. Also, the lower wind velocity should be used in order to be on the safe side in determining the sol-air temperature. The inside conductance 1.20 is used because the convective portion of the film conductance factor of downward heat flow from a horizontal surface is appreciably less than the winter conductance, which applies when heat is flowing upward. Since there is little difference in wall transmission coefficients for summer, based on conductances of 4.0 and 1.65, and the winter coefficients, based on 6.0 and 1.65, it is recommended that the overall coefficient V, for walls, be taken directly from the tables in Chapter 9 in which they are based on an outside film conductance of 6.0, corresponding to a 15 mph wind velocity. Advantages of Equivalent Temperature Differential Method The advantages of the equivalent temperature differential method of determining the total heat transmission are given in following paragraphs, and are apparent from Examples 10 to 12. 1. The total sensible heat flow is obtained by multiplying the overall heat trans mission coefficient, U, and the equivalent temperature differential indicated in Tables 12 and 13. 2. The temperature differentials listed for a few representative types of con struction may be used on all classes of walls and roofs, even though the overall heat transmission coefficient is different, provided the structure has thermal and physical properties similar to one of those listed in Tables 12 and 13. 3. Adjustments can be made, according to instructions given in the footnotes for room and outdoor conditions different from those on which the tables are based Examples of Use of Equivalent Temperature Tables Example 10. Given: A roof is constructed of 6 in. of stone concrete with 2 in. of insulating board and tar felt roofing f in. thick, and is exposed to the sun. The loca tion is the central part of the United States. Find the rate of heat flow into building at 2:00 pun. during July for an outdoor design temperature 95 F, and an inside tem perature 80 F. Solution: From Table 12 in 2 pjn. 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 14 and is found to be.0.13. The heat flow rate equals 34 X 0.13 = 4.42 Btu per (hr) (sq ft). Example It: For the conditions of Example 10, find the rate of heat flow into build ing at 2:00 p.rn. 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 6 in Table 12 as follows: The correction for 27 deg design temperature difference is (27 -- 15) = + 12.