Document ppN55Y9qKXm2V5Qpz37wB389D

HEATINC VENTILATING AIR CONDITIONING GUIDE 1942 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 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 CHAPTER 42. NATURAL VENTILATION 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 F is: y H 0.0175 (l - to) 34,875 .0.0175 X 10 199,286 cfm. This is equal to about 20 air changes per hour. From Equation 2 the inlet (or outlet) opening area should be: 199,286 1224 sq ft. 9.4 (t - t0) 9.4 ^30 X 10 Fig. 2. Increase in Flow Caused by Excess of One Opening Over Another percentage decreases rapidly as one force increases over the other and the larger force 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 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 gal per hour (7.75 lb per gall- Desired summer temperature difference is 10 F 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? 764 Fig. 3. Determination of Flow Caused by Combined Forces of Wind and Temperature Difference The flow per square foot of inlet or outlet would be 199,286 + 1224 = 163 cfm with all windows open. Solution for Wind Only. With 1,224 sq ft of inlet openings distributed around the sidewalls, there would 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 there is no heat to be removed. Solution for Combined Forces. 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 765