Document KJbXMkraD7K0MKMKyrK3Gj5qK

188 CHAPTER 9 1951 Guide nomic 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 unsatisfactory results. Computed Heat Transmission Coefficients Computed overall heat transmission coefficients of many common types of building construction are given in Tables 6 to 20, inclusive, each coeffi cient being identified by a serial number, except in Tables 19 and 20. For example, the coefficient U of a brick veneer, frame wall with wood sheath ing and 3-in. of plaster on gypsum lath is 0.27 (Wall No. 28-C in Table 6) and with 2 inches of blanket or bat insulation, the coefficient would be 0.097 (No. 49-B in Table 7). 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 deter mine the total resistance for the wall, the conductance must be known. The importance of this analysis cannot be over-emphasized. This is especially true in wall constructions in which there are parallel paths of 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 1 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 1. Calculate the coefficient of heat transmission U for wall as shown in Fig. 5. Wall construction consists of two 4-in. concrete walls separated by a 2$-in. space filled with insulation; J-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. 3. From C to D: Two paths, (a) through 2J in. of tie rod, and (b) through 2} in. 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 as one. If equilibrium or steady state heat transfer is assumed, there will exist a tem perature difference between the metal tie rod and the concrete, and also between the metal tie rod and the insulating material. The rate of heat transfer between these materials is dependent upon their conductivity values and the temperature difference. As the conductivity of the metal tie rods is considerably higher than that of the concrete or insulating material, it cannot be assumed that the same rate of heat transfer takes place for all parallel paths. Likewise, an appreciable error would be made by assuming that no heat transfer takes place between the metal tie rod and the surrounding materials. Although the pattern of the isotherms is unknown, the following method of calculation does partially take into account the heat flow be tween the metal tie rods and its bounding materials. ' Parallel Flow. The conductances through the areas of parallel heat flow may be determined as follows: /l. The area of each i-in. diameter tie rod is 0.00036 sq ft, and as the tie rods are Heat Transmission Coefficients of Building Materials 189 .,,j a jn vertically,rand 12 in. horizontally, there will be 0.00036 X } = 0.00048 sq ft of tie rod to each square footmf wall area. Then from plane B to plane C, the conductance Ci is 0.00048 129 + '92252 x -- = 0.192 + 11.994 = 12.186 Cl 1.0 .1 0T 1.0 10 2. For tie rod and insulation from plane C to plane D the conductance Ct is Ct' 0.00048 1.0 - 400 XF5 + 0.999521.0 wX0--.-3-0 2.5 = 0.077 + 0.120 = 0.197 3. For tie rod and concrete from plane D to plane E the conductance C is C, + = 0.192 + 11.994 = 12.186 L0 Fio. 5. Section of Concbete Wall Havino Steel Tie Rods and Insulation Series Flow. After the conductance values have been determined, the total re sistance and XJ value for the wall' can be determined as follows: ,, 1 ii 1 1 , 1 1 Kt ~ A + fc, + C,+ C,+ C,+ k, +h . _ J_ JI0 1 1 1 3.0 1 fiT " 1.65 + 12.0 + 12.186 + 0.197 + 12.186 + 12.0 + 6.0 Rt = 0.606 + 0.250 + 0.0821 + 5.076 + 0.0821 + 0.250 + 0.167 = 6.513 7 = -- = ----r = 0.153 Btu per (hr) (sq ft) (F deg). Rt 6.513 . The Hot Box test-value, from University of Minnesota, for this wall, corrected for a 15 mph wind velocity, was U = 0.150 Btu per (hr) (sq ft) (F deg). The error between the calculated and test values would be . 0.153 - 0.150 X 100 = 2 percent. 0.150 If the effect of the tie rods were omitted from the calculations, the over all U value would be 0.103. Although the percentage of area occupied