Document 442xYVQvV7Ddgx6nbLR5MYypx

158 CHAPTER 9 1948 Guide openings have a flow coefficient slightly greater than that of a square-, edged orifice. If the openings are not advantageously placed with respect to the wind, the flow per unit area of the openingswill be less and, if unusually well placed, the flow will be slightly more than that given by the formula. Inlets should be placed to face directly into the prevailing wind, while outlets should be placed in one of the five places listed: ' 1. -On the side of the building directly opposite the direction of the prevailing wind. 2. On the roof in the low pressure area caused by the jump of the wind (see Fig. 1). Natural Ventilation 159 ' where >. Q = air flow, cubic feet per minute. . '- . A J= free area of inlets or outlets (assumed equal), square feet, ft = height from inlets to outlets, feet. I = average temperature of indoor air in height ft, Fahrenheit degrees. t0 = temperature of outdoor air, Fahrenheit degrees. 9.4 = constant of proportionality, including a value of 65 per cent for effectiveness of openings. This should be reduced to 50 per cent (constant = 7.2) if conditions .. are.not favorable. HEAT REMOVAL In problems of heat removal, knowing the amount of heat to be re moved and, having selected a desirable temperature difference, the amount Fig. 1. The Jump op Wind prom Windward Face op Building. (A--Length op Suction Area; B--Point of Maximum Intensity of Suction; C--Point op Maximum Pressure) 3. On. the sides adjacent to the windward face where low pressure areas occur. 4. In a monitor on the side opposite from the wind. 5. In roof ventilators or stacks. TEMPERATURE DIFFERENCE FORCES2 The stack effect produced within a building when the outdoor tempera ture is lower than the indoor temperature is due to the difference iri weight of the warm column of air within the building and cooler air outside. The flow due to stack effect is proportional to the square root of the draft head, or approximately; Q = 9.4 A y h (t - to) (2) Fig. 2. Increase in Flow,Caused by Excess of One Opening Over Another of air to be passed through the building per minute to maintain this tem perature difference can be determined by means of Equation, 3. H = 0.0175 Q (t - to) where H = heat removed, Btu per minute. : .' Q = air flow, cubic feet per minute. t--t0= inside^outside temperature difference, Fahrenheit degrees.' (3) EFFECT OF UNEQUAL OPENINGS. The largest flow per unit area of openings is obtained when inlets and outlets .are equal, and the equations given previously are based on this condition. Increasing outlets over inlets, or vice-versa, will increase the air flow, but not in proportion to the added area. When solving.prdblems having an unequal distribution of openings, use the smaller area, either inlet or outlet, in the equations and add the increase as determined from Fig. 2. COMBINED FORCES OF WIND AND TEMPERATURE Equations for determining the air flow due to temperature difference and wind have already been given. It must be remembered that-when