Document 15dbKXbO6zLgk93r9MEY0YKLa

228 CHAPTER 11 ; 1954 Guide distribution, cannot, be predicted with certainty, and refinement, in calcu lations 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 co ordination of the forces. Then, by determining the ratio of the flow pro duced by temperature difference to the sum of the two flows, the actual flow due to the combined forces can be approximated from Fig. 4. . 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 lb is used in this shop at the rate of 15 gph (7.75 lb per gal). Desired summer temperature difference is 10 deg, and the prevail ing 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? Infiltration and Ventilation 229 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 3, will.be: Q = 0.60 X 410 X 704 = 173^00 cfm. This gives 17.3 air changes per hour, which should be more than ample when there is no heat to be removed. ' Solution to Combined ffeads: 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. 3. This chart shows that, when inlets are twice the size of the outlets, in this case 1,224 sq ft in the sidewalls ahd:612.sq>ft:in the monitor, the flow will be increased 26.5 percent over that produced by equal openings. Using Fio. 3. Increase in Flow Caused by Excess of One Opening Over Another Solution for Temperature Difference Only: The heat H = ^ .X . 34,875 Btu per min. * 18,000 60 :ii By Equation 5, the air flow required to remove this-heat with an average temper'i ature difference of 10 deg is H____________ 34,875 0.0175(!j - U,) ~ 0.0175 X 10 199,286 cfm. This is equal to about 20 air changes per hour. From Equation 4, the inlet (or outlet); opening area should be if Q____________199,286 A= = 1224 sq ft. 9.4Vh(li - to) ~ 9.4V30 X 10 r The flow per square foot of inlet or outlet would be 199,286 -+ 1224 = 163 cfm, with all5-' windows open. ; :U.(J Solution for Wind Only: With 1,224 s<) ft of inlet openings distributed around the), sidewalls, there will be about 410 sq ft in each long side and 202 Bq ft in each end.; The outlet area will be equally distributed on the two sides of the monitor, or 612scjt. ft on each side. With the wind perpendicular to the long side, there will be 410 sqV Fig. 4. Determination of Flow Caused by Combined Forces of Wind and Temperature Difference 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 percent. Wind = 173,200 = 58 percent. Total 299,400 = 100 percent. From Fig. 4 it is determined that, when the flow due to temperature difference is 42 percent 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 "alf of the outlets are closed. i ^,factor of judgment is necessary in the location of the openings in a uiiding, especially those in the roof, where heat, smoke; and fumes are to removed. Usually, windward monitor Openings should be closed, but if