Document gbao7Qx5yvn3E6Vj4je1Mbr4q

160 CHAPTER 9 1948'Guide 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 heads acting on that opening. . ' When the two heads are about equal in value 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 head acting independently under condition^ ideal to it. This percentage decreases rapidly as one head increases over the other and the larger will predominate. . The wind velocity and direction, the outdoor temperature, or the indoor distribution, cannot be predicted with certainty, and refinement in calculations is not justified; consequently, a simplified method can be' 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 that of the inlet is 30 ft. Oil-fuel, of 18,000 Btu per pound is used in this shop at the rate of 15 gai per hour. (7.75 .1b 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? - ' Solutionr,for ,Temperat.u..re Difference Only. The heat H = 15 X 7~76^0 X. 18'000 = 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: ' .. . ' .. Natural Ventilation 161 H Q = 0.0175 (1 - to) 34,875: = 199,286 cfm. 0.0175 X 10 This is equal to about 20 air changes per hour. From Equation 2 the inlet (or outlet) opening area should be: VA '= 9.4 h.(t -- (o) 199,286 = 1224 sq. ft.' 9*4 -^30 X 10 The flow per square foot of inlet or outlet would be 199,286. -S- .1224 = 163'cfm with all windows-open. Solution for Wtnd 'Only^With 1,224 sq ft of inlet , openings distributed around the 'sidewalls, there will be"about 410 sq ft in each long side and 202 sq ft in each end. The outlet area will be equally distributed on the two sides of the monitor, or 612 sq ft on each side. With the wind perpendicular to.the long side, there will.be 410 sq ft.of opening in its path for inflow and 612 in the lee side of the monitor for outflow with the windward side closed. The air flow, as calculated by Equation 1, will be: . ' Q = 0.60 X 410 X 704 = 173,200 cfm. This gives 17.3 air changes per hour, which should be more than ample when {here is no heat to be removed. Solution for Combined Heads. Since the windward side of the monitor is closed when the wind is blowing, the flow due to temperature difference must be calculated for this condition, using Fig. 2. This chart shows that when inlets are twice the size of the outlets, in this case 1,224 sq ft in the sidewalls and 612 sq ft in the monitor, the flow will be increased 26.5 per cent over that produced by equal openings. Using the smaller opening and the flow per square foot obtained previously, the calculated amount for this condition will be: 612 X 163 X 1.265 = 126,200 cfm.. Adding the two computed flows: Temperature Difference = 126,200 = 42 per cent. Wind = 173,200 = 58 per cent. Total 299,400 = 100 per cent. From Fig. 3, it is determined that when the flow, due to temperature difference, is 42 per cent of the total, the actual flow, due to the combined forces, will be about 1.6 times that calculated for temperature difference alone, or 201,920 cfm. The original flow, due to temperature difference alone, was 199,286 cfm with all openings in use. The effect of the wind is to increase this to 201,920 cfm even though half of the outlets are closed. A. factor of judgment is necessary in the location of the openings in a building, especially those in the roof, where heat, smoke and fumes are to be removed. Usually windward monitor openings should be closed, but if the wind is low enough for the temperature head to overcome it, all windows may be opened. TYPES OF OPENINGS Types of openings may be classified as: (1) windows, doors, monitor openings and skylights, (2) roof ventilators, (3) stacks connecting to registers, and (4) specially designed inlet or outlet openings. Windows, Doors and Skylights Windows have the advantage of transmitting light, as well as providing ventilating area when open. Their movable parts are arranged to open in various ways; they may open by sliding either vertically or horizon tally, by tilting on horizontal pivots at or near the center, or by swinging' on pivots at the top, bottom or side. Regardless of their design, the air flow per square foot of opening may be considered to be the same under the same conditions. The type of pivoting should receive consideration