Document 9LQ2nLej8kGqR68jzYRaygvwD

208 CHAPTER 9 1957 Guide Table 16. Determination op U Value Resulting prom Addition op Insulation to Uninsulated Building Sections (Concluded) (For use with Tables ISA and 1SB). PART E. FLAT ROOFS AND CEILINGS WITH ROOF DECK Conductance C of Roof-Deck Insulation U Value oe Roof without Roof-Decs Insulation 0.12 U 0.15 u 0.19 U .0.24 U 0.36 U 0 10 0 15 0 20 0 25 6.30 0 35 0 40 o sn 0 60 0.70 0.05 0.07 0.08 0.08 0.09 0.09 0.09 0.10 0.10 0.10 0.06 0.08 0.09 0.09 0.10 0.10 0.11 0.12 0.12 0.12 0.07 0.08 0.10 0.11 0.12 0.12 0.13 0.14 0.14 0.15 0.07 0.09 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.08 0.11 0.13 0.15 0.16 0.18 0.19 0.21 0.22 0.24 * Interpolation or mild extrapolation may be used. 0.72 U 0.09 0.12 0.16 0.19 0,21 0.24 0.26 0.29 0.33 0.35 Exam-pie 4: Consider the floor-ceiling construction shown in the example at the top of Table 11, insulated with a sheet of aluminum foil, or paper faced on both sides;, with foil, (effective emissivity (E) of air space = 0.05) placed between the joists and;\ dividing the air space into two equal spaces. From Table 11, the U value for the? uninsulated construction is 0.18 for heat flow down for summer, and 0.23 for heat* flow up for winter. Determine the coefficient Ui for: X (a) Heat flowing downward from uncooled room above to cooled space below (# summer condition) and, ~i (b) Heat flowing upward from heated room below to unheated space above (an. winter condition). , X Solution: J[ (a) Use Table 16 Part D, and corresponding to U = 0.18 in Column 1 find, urf Column 11, Ui = 0.049. Use 0.05. -i' (b) Use Table 16 Part C, and corresponding to U = 0.23 in Column 1 find, in^s Column 11, Ifi = 0.124, (by interpolation). Use 0.12. I,3. Special Uses of Table 16 Values of U-, for insulating applications or combinations other than thoseindicated by the headings of Columns 2 to 14 of Table 16 can be ascertained^: if the table is used appropriately. For instance, going horizontally in th||| table from Column 2 to Column 3 is equivalent to adding Yi in. of fibrous* insulation to the construction. Similarly, going from Column 2 to Column* 4 adds IY in. of fibrous insulation to the construction. In the same way; going horizontally from Column 8 to Column 11 is equivalent to adding ! a construction the insulating value of one additional highly reflectiv;: (E = 0.05) air space, and going from Column 6 to Column 12 in effect adds two non-reflective (E = 0.82) air spaces to the construction, etc. Examples 5 and 6 show the combinational use of Table 16. Example 6: Determine the coefficient Ui for the wall of Example S, with TH Ujj of fibrous insulation added on one side of the air space. f Solution: U-, can be determined in several ways, by using Table 16 Part A. f example: (a) By going from Column 1 to Column 2 three times, for a total of VA inches o, insulation, as follows: enter Column 1 at 0.24 and find Uj = 0.166 in Column 2; ent*J Heat Transmission Coefficients of Building Materials 209 Column 1 at 0.166 and find Uj = 0.126 in Column 2; enter Column 1 at 0.126 and find Ui = 0.102 in Column 2. Therefore, for a total of \lA in. of insulation, use 0.10. (b) Move simply, by going from 0.24 in Column 2 to 0.103 in Column 4. Use 0.10. Example 6: Considering again the wall of Example 3, assume that one-inch blanket insulation is to be installed in mid-space, leaving equal air spaces on the two sides. The coefficient U, is desired for two cases in which (a) the blanket has non-reflective surfaces on both sides, so that both air spaces are non-reflective (E = 0.82) and, (b) highly the blanket has reflective (E = one highly 0.05). reflective surface, so that one of the air spaces ib Solution: (a) In Table 16 Part A, when U without added insulation is 0.24, Ui for one inch of fibrous insulation and one non-reflective air space, is 0.127 in Column 3. To add the second non-reflective air space, enter Column 6 at 0.127, and by interpolating find Ui = 0.112 in Column 9. Use 0.11. (b) Ui for one inch of fibrous insulation and one non-reflective air space is 0.127 as in (a). To add one E = 0.05 air space, enter Column 8 at 0.127 and by interpo lating find Ui = 0.092 in Column 11. Use 0.09. Fig. 6. Correction fob Effect of Framing in Insulated Building Sections Uar =* average U value for building section. Ui ** U value for area between framing members. U =* U value for area backed by framing members. S -- Percentage of area backed by framing members. CORRECTION FOR FRAMING Correction for parallel heat flow through framing and insulated areas may be made by use of Fig. 6. Correction for the effect of framing should be applied after final Ui and UB values have been obtained for a given con struction. In many cases this correction may be omitted. Example 7: Consider a frame wall with 2-in. blanket insulation which has a U-, value of 0.08. By calculation it is found that heat loss from the area backed by fram ing members (U.) is 0.13. U,/Ui is 1.63. From Fig. 6 if 15 percent of wall area is backed by framing, the value tl.v/IA -- 1.1- Unv is therefore 1.1 X 0.08 -- 0.088. Combined Ceiling and Roof Coefficients If the attic space between the ceiling and roof is unheated and not venti"tted, the combined coefficient from room air below the ceiling to exterior a,r can be calculated from the following formula: Rt = _ULr. + n_UL, The combined coefficient U is the reciprocal of Rt, or (4)