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American. Society of Heating and Ventilating Engineers Guide, 1932 walls, ceilings and floors next to cold or unheated spaces are found, of course, by taking the inside dimensions of such areas, measured on the heated side. CALCULATIONS FOR TRANSMISSION LOSSES The calculations for heat transmission losses are made by multiplying the area A in square feet of wall, glass, roof or floor through which the loss takes place, by the proper coefficient U for such construction (Tables 9 to 33, or by computation as described under Transmission Coefficients by Computation) and by the temperature difference between the inside air temperature t at the proper level (in many cases not the breathing line) and the outside air temperature ta. Therefore, where Ht = AU (t - I,,) (12) Ht = Btu per hour transmitted through the material of the wall, glass, roof or floor. A = area in square feet of wall, glass, roof or floor, taken from building plans or actually measured. (Use the net inside or heated surface dimensions in all cases). I -- to = temperature-difference between inside and outside air, in which t must always be taken at the proper level. Note that t may not be the breathing-line temperature in many cases. Table 33. Coefficients of Transmission (V) of Doors, Windows and Skylights Note.--These coefficients aee based on a wind exposure of 15 MILES per HOUR, AND ARB EX PRESSED IN Btu per hour, per square foot, per deg fahr difference in temperature between THE AIR INSIDE AND OUTSIDE OF THE DOOR, WINDOW OR SKYLIGHT. A. Windows and Skylights Single............................. .................... ...................... Double....................................................................... Triple...,................................................................... V 1 13a-c 0 45s 0.281a B. Solid Wood Doorsb-C Nominal Thickness Inches > Actual Thickness Inches V i i*A 1 2 2K 3 % IMe IMa m IK 2K 2K 0.563. 0.485 . 0.432 0,-421 0,382- , 0.321 0.277 ' ' ^ See page 212, Volume I, Mechanical Equipment of Buildings, by Harding and Willard, second edition. ^Computed using C <= 1.0 for wood;/i = 1.34'and/0 *= 4.02. clt is sufficiently accurate to use the same coefficient of transmission for doors containing thin wood panels, as that of single panes of glass, namely. 1.13 Btu per hour per square foot per degree difference' between inside and outside air temperature. \ 66 Chapter 3--Heat Transfer Through Materials and Constructions CONDENSATION ON BUILDING SURFACES* Condensation on the interior surfaces of buildings is often a serious problem. Water dripping from a ceiling may cause irreparable damage to manufactured articles and machinery. It often results in short-cir cuiting of electric power and lighting systems, necessitating shut-downs and incurring costly repairs. It also causes rotting of wood roof struc tures, corrosion of metal roofs, and spalling and disintegration of gypsum and other types of roof decks not properly protected. Condensation is caused by the contact of the warm humid air in a building with surfaces below the dew-point temperature, and can be remedial in two ways, (1) by increasing the temperature of such surfaces above the dew-point temperature, or (2) by lowering the humidity. Dehumidification, of course, is not permissible where a high relative humidity is necessary for manufacturing processes. Hence, the only alter native is to increase the surface temperature by decreasing the inside surface resistance. This can be accomplished by increasing the velocity of air passing over the surface, or by increasing the over-all resistance of the wall or roof by installing a sufficient thickness of insulation. The latter method is generally used, and the thickness of insulation is determined by ascertaining the amount of resistance to be added to increase the temperature of the interior surface above _ the dew-point temperature for the maximum conditions involved. This in turn is based on the fundamental principle that the drop in temperature is proportional to the resistance. Condensation Chart The chart (Fig. 3) can be used for approximating the thickness of insulation required to prevent condensation on the interior wall or roof surfaces of a building. Although this chart is intended primarily for roofs, it can be used for walls by taking the dry-bulb temperature and the cor responding relative humidity near the walls at the point which will neces sitate the maximum heat resistance to prevent condensation, instead of using the temperature and humidity near the ceiling. Example 1. Determine the thickness of insulation required to prevent ceiling con densation for the following conditions: Dry-bulb temperature near ceiling, 85 F; Relative humidity, 70 per cent; Lowest outside temperature, -- 10 F; Construction of uninsulated roof, 1 in. yellow pine sheathing and built-up roofing; Coefficient of trans mission of roof, 0.485; Conductivity of insulation to be used, 0.30. Solution. The solution of this problem is indicated on the chart (Fig. 3) by the dotted line: 1. Locate the inside dry-bulb temperature of 85 F on scale A, and draw a line hori zontally to the 70 per cent relative humidity curve, indicated on scale B. 2. Draw line 2 vertically downward from the intersection located as per paragraph 1. 3. Locate on scale D the temperature difference of 95 F between the ceiling tem perature of 85 F and the lowest outside temperature of -- 10 F, and draw a line hori zontally until it intersects with line 2. 4. From the point of intersection of lines 2 and'S, draw a line to the point P. `For additional information on this subject see. Preventing Condensation on Interior Building Surfaces, by Paul D. Close (A.S.H.V.E. Transactions, Vol. 36, 1930). 67