Document dYgMDvNYQ2KMapV93JgZJ1YLq

.728 Chapter 42 __________ ______ - 1945 Guide_ where Q = air flow, cubic feet per minute. A = free area of inlets or outlets (assumed equal), square feet. H = height from inlets to outlets, feet. ' t = average temperature of indoor air in height H, degrees'Fahrenheit, to = temperature of outdoor air, degrees Fahrenheit. 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 ,_Ndtu ralJKentila tion both forces are acting together, even without interference, the resulting air flow is not equal to the sum of the two estimated quantities. The flow through any opening is proportional to the square root of the sum of the forces acting on that opening. When the two forces are about equal in intensity and the ventilating openings are operated so as to coordinate them, the total air flow through the building is about 10 per cent greater than that produced by either force acting independently under conditions ideal to that force. This percentage decreases rapidly as one force increases over the other and the. larger force will predominate. The wind, velocity and direction, the outdoor temperature, of the indoor distribution, cannot be predicted with certainty, and refinement in calculations is not justified; consequently, a simplified method can be 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. where ,H = 0.0175 Q (I - fo) (3) H = heat removed, Btu per minute. Q = air flow, cubic feet per minute. <--lo= inside-outside temperature difference, degrees Fahrenheit. ' 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 problems 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 AS PER CENT OF TOTAL Fig. 3. Determination of Flow Caused by Combined Forces of Wind and Temperature Difference used. This may be done by using the equations and calculating the flows produced by each force separately under conditions of openings best suited for coordination of the forces. Then, by determining, as a per centage, the ratio of the flow produced by temperature difference to the sum of the two flows, the actual flow due to the combined forces can be approximated from Fig. 3. ~ Example 1. Assume a drop forge shop, 200 ft long, 100 ft wide, and 30 ft high. The cubical content is 600,000 cu ft, and the height of the air outlet over {hat of the inlet is 30 ft. Oil fuel of 18,000 Btu per pound is used in this shop at the rate of 15 gal per hour (7.75 lb per gal). Desired summer temperature difference is 10 deg and the prevailing wind is 8 mph perpendicular to the long dimension. What is the necessary area for the inlets and outlets, and what is the rate of air flow through the building? Solution for Temperature Difference Only. The. heat H = 15 X 7.75 X 18,000 60 34, 875 Btu per minute. By Equation 3, the air flow required to remove this heat with an average temperature difference of 10 deg is: .