Document M4QgvQvepzdNV9QjNOkJZ1kw7

128 CHAPTER 6: 1949 Guide' at the University of Minnesota calculated and tested .values. a direct comparison can be made between- Example 1. Calculate the coefficient of heat transmission U for wall as shown in Fig. 4. Wall construction consists of two 4-inch concrete walls separated by a 2} in. space filled with insulation; }-in. diameter meted 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. In Fig. 4 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. con2c.reFtreo. mB to C: Two paths; (a) through 1 in. of tie rod and (b) through 1 in. of 3. From C to D: Two paths; (a) through 2| in. of tie rod and (b) through 2\ in. insulation. 4. From D to E: Two paths; (a) through 1 in. of tie rod and (b) through 1 in. of ncrete. 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- Fig. 4. Section of Concrete Wall Having Steel Tie Rods and Insulation 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: 1. The area of eachf-in. diameter tie rod is 0.0036 square feet and as the tie rods are spaced 9 inches vertically, and 12 inches horizontally, there will be 0.00036 X f = 0.00048 square feet of tie rod to each square foot of wall area.- Then from plane B to plane C, the conductance Ci is `' 0.00048 400 0.99952 12 Ct TT x To + "To~ x ru = 0:192 + 11994 - 12186 Heat'Transmission Coefficients of Building' Material. ^9 2. For Tie Rod qnd Insulation from plane C to plane D the conductance C, is 0 00M8 x -- = 0.077 + 0.120 = 0 197 Ct = ~uT 2.5+ 10 2.5 3 For Tie Rod and Concrete from plane D to plane E the conductance C, is _O0OT^ 0 0J99M 12 = 1.0 i1.n0 + 1.0 11..00 U994 12.186 Series Flew. After the conductance values have been determined, the total re sistance and U value for the wall can be determined as follows: 1 1,1 1 1 RT = fi +ki + Ci + C, + C, + I, +f,, * Kx = i + M + _i_ , 1 _1 *0 1 1.65 T 12.0 T 12.186 T 0.197 + 12.186 + 12.0 + M ST = 0.606 + 0250 + 0.0821 + 5.076 + 0.0821 + 0.250 + 0.167 = 6.513 U A. = = 0.153 Btu per (hr) (sq ft) (F deg). St 6.513 The Hot Box test value, from University of Minnesota, for this wall corrected for a 15 mph wind velocity was V = 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 per cent. 0.150 If the effect of the tie rods were omitted from the calculations, the over all V value would be 0.103. Although the percentage of area occupied by the tie rods per square foot of wall area is X 100 = 0.048 per cent the error between the calculated and test values would be 0.150 - 0.103 v 100 = 45 per cent. 0.103 In making the calculations for values of U shown in Tables 5 to 18, the following-conditions have been assumed: . Equilibrium or steady-state heat transfer, eliminating effects of heat capacity. . Surrounding surfaces at ambient air temperatures. Exterior wind velocity of 15 mph. Surface emissivity of ordinary building materials = 0.83. No correction for position or direction of heat flow. (Average coefficients used). Air spaces are l in. or more in width. Variation of conductivity with mean temperature neglected. Corrections for framing made on basis of parallel heat Sow through 2 X 4 in. (nomi nal) studs, 16 in. on centers, the framing covering 15 per cent of wall area. Actual thicknesses of lumber assumed to be as follows: Nominal Actual lin. (S-2-S).................................................................. If in. 11 in. (S-243)................................. .................. ........... 1A in. . 2 in. (S-2-S).................................................................. If in. 2f in. (S-2-S)................................................................ 2} in. 3 in. (S-2-S)................................................................ 2) in. 4 in. (S-2-S)............... ..................................... 3i in. Finish flooring, (maple or oak)................................... 1A in.