Document 7O4w5p2kMOyw305YoMGOVvZM8

180 CHAPTER 9 1955 Guide i ! in accordance with the manufacturer's specification. The engineer mustij evaluate carefully the economic considerations involved in the selection':'; of an insulating material as adapted to various building constructions. | Lack of proper evaluation, or improper- installation may lead to unsafe- factory results. Special attention must be given to vapor barriers as out- , lined in Chapter 10. Water-soaked insulation loses its effectiveness as insulation. Computed Heat Transmission Coefficients Computed overall heat transmission coefficients of many common types of building construction are given in Tables 7 to 21, inclusive, each coeffi cient being identified by a serial number, except in Tables 19, 20, and 21. Table 6. Coefficients of Transmission (U) of Frame Walls and Roofs with' Insulation Between Framing" Coefficients are expressed in Btu per (hour) (square foot) (Fahrenheit degree difference tn temperature between - the air on the.two irfe), and are bated on an outside wind velocity of IS mph. COEFFICIENT with NO INSULATION BETWEEN FRAMING Q.ll 0.13 0.15 0.17 0.19 . 0.21 0.23 0.25 0.27 0.29 0.31 0.33 . 0.35 0.37 0.39 0.41 0.43 COEFFICIENT with insulation between framing Mineral Wool or Vegetable Fibers in Blanket ob Bat Form* (Thickness below) 1 IN. A 0.078 0.088 0.097 0.10 0.11 0.12 0.12 0.13 0.14 0.14 0.14 0.15 0.15 0.16 0.16 0.16 0.17 2 IN. B 0.063 0.070 0.075 0.080 0.084 0.088 0.091 0.094 0.097 0.10 0.10 0.10 . Q.ll 0.11 0.11 0.11 o.n 3 IN. C 0.054 0.058 0.062 0.066 0.069 0.072 0.074 0.076 0.078 0.080 0.081 0.083 0.084 0.085 0.086 0.037 0.088 3} IN. Mineral Wool Framing6 D 0.051 0:055 0.059 0.062 0.065 0.067 0.069 0.071 0.073 0.075 0.076 0.077 0.076 0.080 0.081 0.082 0.082 tae ' Da z- 33 3539 41., 43 ' 45 49 61 ,f 53 > 55 57 ` 01. r 65 Coefficients corrected for 2 x 4 framing, 16 in. on centers--15 percent of surface area. b Based on one air space between framing. 6 No air space. For example, the coefficient U of a brick veneer, frame wall with woofj: sheathing and 5-in. of plaster on gypsum lath is 0.27 (Wall No. 28-C in. Table 7) and with 2 inches of blanket or bat insulation, the coefficient would be 0.097 (No. 49-B in Table 6). i In the analysis of any wall construction for the purpose of calculating the overall coefficient of heat transmission U, it is first necessary to deter mine the paths of heat flow, that is, whether they are parallel or series, or a combination of both. This is in accordance with the basic laws of heat transfer which state that in parallel flow the conductances are additive, while in series flow the resistances are additive. Likewise, in order to determine the total resistance for the wall, the conductance must be known. The importance of this analysis cannot be over-emphasized. This i|, especially true in wall constructions in which there are parallel paths 0*. Heat Transmission Coefficients of Building Materials 181 heat flow, and one path has a high heat transfer, while others have a low heat transfer. The method of making this calculation can best be shown by Example 2 and Fig. 5. As this wall was tested by the hot box method at the University of Minnesota, a direct comparison can be made between calculated and tested values. Example 2: Calculate the coefficient of heat transmission U for a wall shown in Fig. 5. Wall construction consists of two 4-in. concrete walls separated by a 2}^-in. space filled with insulation; 34-in. diameter metal tie rods are imbedded a distance of 1 in. in each 4-in. concrete wall, and spaced 9 in. vertically and 12 in: horizontally. Values of k are: insulation 0.30, concrete 12.00, tie rods 400.00. Solution: In Fig. 5 the following paths of heat flow from plane A to plane F will be noted: " 1. From A to B: One path through 3 in. of concrete. 2. From B to C: Two paths, (a) through 1 in. of tie rod, and (b) through 1 in. of concrete. - -' Fig. 5. Section of Concrete Wall Having Steel Tie Rods and Insulation 3- From CtoD: Two paths, (a) through 2% in. of tie rod, and (b) through 2J4 m. of insulation. 4. From D to E: Two paths, (a) through 1 in. of tie rod, and (b) through 1 in. of concrete. 5. From E to F: One path through 3 in. of concrete. It will be noted that items 2 and 4 are paths of similar flow, and could be treated te ne ^ equilibrium or steady state heat transfer is assumed, there will exist a t mpcrature difference between the metal tie rod and the concrete, and also between thp He ro^ and ^h insulating material. The rate of heat transfer between diff 6 ma^er*a^8 i dependent upon their conductivity values and the temperature of flrence- A8 the conductivity of the metal tie rods is considerably higher than that heat t concre^e or insulating material, it cannot be assumed that the same rate of be place for all parallel paths. Likewise, an appreciable error would the aC*6 ^ as8uming that no heat transfer takes place between the metal tie rod and foIIoU-rUnding mal;erials' Although the pattern of the isotherms.is unknown, the method of calculation does partially take into account the heat flow ben the metal tie rods and its bounding materials. Flow. The conductances through the areas of parallel heat flow may be Renamed as follows: - v-: . 1- The area of each 3^-in. diameter tie rod is 0.00036 sq ft. and as the tie rods are