Document xjN6x8NO9VEV1DZrmY0dbm4kJ

HEATING VENTILATING AIR CONDITIONING GUIDE 1941 Where reflective insulating materials are involved the possible increase in the emissivity coefficient due to surface coatings or chemical action4 should be studied by the engineer in order to satisfy himself as to the permanence of the reflective surface for the conditions under which this material will be used. In making installations of this material the par titions between air spaces should be tight, particularly at the top and bottom so that air cannot circulate between adjacent spaces. When reflective insulating materials are installed with multiple air spaces, the position (vertical, horizontal or inclined) of the material in the structure must be taken into consideration. For example, the resistance to heat flow upward is about one-third that of downward flow in a hori zontal position in the same construction, as will be apparent from Table I, Section C. However, the difference between upward heat flow through' single horizontal or sloping air spaces and through single vertical air spaces is comparatively small for the same temperature difference. Con sequently the same conductance value (0.46) was used for computing the coefficients in Tables 8 and 12, involving horizontal and sloping air spaces bounded on one side by aluminum foil applied to plasterboard, as for similar vertical air spaces in Tables 3, 4, 5 and 6. As already stated, a conductance value of 1.10 was similarly used in all cases for calculating the coefficients of construction involving vertical, horizontal and sloping air spaces bounded on both sides by ordinary building materials. PRACTICAL COEFFICIENTS For practical purposes it is necessary to have average coefficients that may be applied to various materials and types of construction without the necessity of making tests on the individual material or combination of materials. In Table 2 coefficients are given for a group of materials which have been selected from various sources. Wherever possible the proper ties of material and conditions of tests are given. However, in selecting and applying these values to any construction a reasonable amount of caution is necessary; variations will be found in the coefficients for the same materials, which may be partly due to different test methods used, but which are largely due to variations in materials. The recommended coefficients which have been used for the calculation of over-all coefficients as given in Tables 3 to 12 are marked by an asterisk. It should be recognized in these tables of calculated coefficients that space limitations will not permit the inclusion of all the combinations of materials that are used in building construction and the varied applications of insulating materials to these constructions. Typical examples are given of combinations frequently used, but any special construction not given in Tables 3 to 12 can generally be computed by using the conductivity values given in Table 2 and the fundamental heat transfer formulae. For example, the tabulation of all of the values for multiple layers of insulating materials would present extensive and detailed problems of calculations for the varied application combinations, but the engineer having the fundamental conductivity values can quickly obtain the proper coefficients. Thermal Teat Coefficients of Aluminum Insulation for Buildings, by G. B. Wilkes, F. C. Hechler and E. R. Queer (A.S.H.V.E. Journal Section, Heating, Piping and Air Conditioning, January, 1940. p. 68). . 76 CHAPTER 3. HEAT TRANSMISSION COEFFICIENTS AND TABLES Attention is called to the fact that the conductivity values per inch of thickness do not afford a true basis for comparison between insulating materials as applied, although they are frequently used for that purpose. The value of an insulating material is measured in terms of the coe fficient (Z7i) of the insulated construction as compared to the coefficient ([/) of the construction without insulation. Certain types of blanket installations are designed to be installed between the studs of a frame building in such manner as to give two .air spaces. In order to get the full value of such materials they should be so installed that each air space is approximately 1 in. or more in thickness and the air spaces should be sealed at the top and bottom to prevent the circulation of air from one space to the other. Another common error in installing such a material is to nail the blanket on the outside of the studs underneath the sheathing, in which case one air space is lost and also the thickness of the insulating material is materially reduced at the studs. There are certain other types of insulation which are very porous, allowing air circulation within the material if not properly installed. The architect or engineer must care fully evaluate the economic considerations involved in the selection of an insulating material as adapted to various building constructions. Lack of good judgment in the intelligent choice of an insulating material, or its improper installation, frequently represents the difference between good or unsatisfactory results. Computed Transmission Coefficients Computed heat transmission coefficients of many common types of building construction are given in Tables 3 to 13, inclusive, each con struction being identified by a serial number. For example, the coefficient of transmission (IT) of an 8-in. brick wall and in. of plaster is 0.46, and the number assigned to a wall of this construction is 1-B, Table 3. Example 1. Calculate the coefficient of transmission (/) of an 8-in. brick wall with in. of plaster applied directly to the interior surface, based on an outside wind exposure of 15 mph. It is assumed that the outside course is of hard (high density) brick having a conductivity of 9.20, and that the inside course is of common (low density) brick having a conductivity of 5.0, the thicknesses each being 4 in. The conductivity of the plaster is assumed to be 3.3, and the inside and outside surface coefficients are assumed to average 1.65 and 6.00, respectively, for still air and a 15 mph wind velocity. Solution, k (hard high density brick) = 9.20; x = 4.0 in.; k (common low density brick) = 5.0; x = 4.0 in.; k (plaster) - 3.3; x = H in.;/i = 1.65;/0 = 6.0. Therefore, U= 1 4.0 , 4,0 ,05 . 1 6.0 + 9.20 + 5.0 + 3.3 + 1.65 0.167 + 0.435 + 0.80 + 0.152 + 0.606 = 0.46 Btu per hour per square foot per degree Fahrenheit difference in tempera ture between the air on the two sides. The coefficients in the tables were determined by calculations similar to those shown in Example 1, using Fundamental Formulae 3, 4 and 5 and the values of k (or C),/,,/0 and a indicated in Table 2 by asterisks. In computing heat transmission coefficients of floors laid directly on the ground (Table 10), only one surface coefficient (/i) is used. For example, 77