Document oDmY4YZyaQe17G0GZ8jov1EBD

144 CHAPTER 11 1959 Guide -- average temperature of indoor air in height A, Fahren heit. t, = temperature of outdoor air, Fahrenheit. 9.4 - constant of proportionality, including a value of 65 percent for effectiveness of openings. This should be reduced to 50 percent (constant = 7.2) if conditions are not favorable. HEAT REMOVAL In problems of heat removal, knowing the amount of heat to be removed and having selected a desirable tempera ture difference, the amount of air to be passed through the building per minute, to maintain this temperature difference, can be determined by means of Equation 5. Hg Q " C, X p X 60(4 - U) " -1.08(4 - 4) fig- 2 - -.. The Jump of Wind from Windward Face of Building. (A--Length of Suction Area; 6--Point of Maximum Intensity of Suction; C--Point of Maximum Pressure) 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 openings will 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 out lets 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. 2). 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 roef ventilators or stacks. where Q = air removed, cubic feet per minute, g = heat removed, Btu per hour. Cr -- specific heat of air at constant pressure, 0.24. p a density of standard air, 0.075 pounds per cubic foot. ti -- tt indoor-outdoor temperature differences, Fahren heit. EFFECT OF UNEQUAL OPB41NGS The largest flow per unit area of openings is obtained . when inlets and outlets are equal, and the preceding equa tions are based on this condition. Increasing outlets over in lets, or vice-versa, will increase the air flow, but not in pro portion 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. 3. COMBINED FORCES OF WIND * AND TEMPERATURE Equations have already been given for determining the air flow due to temperature difference and wind. 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 TEMPERATURE DIFFERENCE FORCES10 The stack effect produced within a building, when the outdoor temperature is lower than the indoor temperature, is due to the difference in weight of the warm column of air within the building and cooler air outdoors. The flow due to stack effect is proportional to the square root of the draft head, or approximately where Q = 9.4A VA. - <) (4) 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. fig. 3 .... Increase in flow Caused by Excess of One Opening Over Another (6) infiltration and Ventilation 145 Solution for Wind Only: With 1,190 sq ft of inlet openings , distributed around the side walla, there will be about 400 sq It in each long side and 195 sq ft in each end. The outlet area will be equally distributed on the two sides of the monitor, or 595 sq ft on each side. With the wind perpendicular to the long side, there will be 400 sq ft of opening in its p&tb for inflow, and 595 in the lee side of the monitor for outflow with the wind ward side closed. The air flow, as calculated by Equation 3. will be: Q - 0.60 X 400 X 704 - 168,960 cfm. This gives 16.9 air changes per hour, which should be more than ample when there 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 Sow due to temperature difference must be calculated for this condition, using Fig. 3. This chart shows that, when inlets are twice the rise of the outlets, in this case 1,190 sq ft in the ride walls and 595 sq ft in the monitor, the flow will be increased 26.5 percent over that produced by equal openings. Using the smaller open ing and the flow per square foot obtained previously, the cal culated amount for this condition will be fig. 4 .... Determination of Flow Caused by Combined Forces of Wind and Temperature Difference 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 percent greater than that produced by either head acting in dependently under conditions ideal to it. This percentage decreases rapidly as one head increases over the-other. The effect of the larger bead will predominate. The wind velocity and direction, the outdoor temperature, or the indoor distribution, cannot be predicted with- cer tainty, and refinement in calculations is not justified; con sequently,, a simplified method can be used. This may -be done by using the equations and calculating the flows pro duced by.each force separately, under conditions of openings best suited for coordination of the forces. Then, by determin ing 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. 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 aii 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 dif ference is 10 deg, and the prevailing wind is 8 mph perpendicu lar to the long, dimension,. What is the necessary area for the inlets and outlets, and wE&t is.the rate of air flow through the building? Solution for Temperature Difference Only: The heat H -- 15 X 7.75 x 18,000 = 2,092,500. Btu per hr. By Equation 5, the air flow required to remove this heat with an average temperature difference of 10 deg is Q H l.oga- - o 2.092,600 1.08 X-10 = 193,750 cfm. This is equal to about 20 air changes per hour. From Equa tion 4, the inlet (or outlet), opening area should be Q________ 193,750 9.4vT(t, - 4) = 9.4%/sO X 10 1190 sq ft 595 X 163 X 1-265 - 122,700 cfm. Adding the two computed flows: Temperature Difference = 122,700 = 42 percent. Wind = 168,960 = 58 percent. Total 291,660 = 100 percent. From Fig. 4 it is determined that, when the flow due to tem perature difference is 42 percent of the total, the actual flow due to the combined forces will be about 16 times that cal culated for temperature difference alone, or 196,300 cfm. The original flow due to temperature difference alone was 193,750 cfm with all openings in use. The effect of the wind is to increase this to 196,300 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, wind ward 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 de signed inlet or outlet openings. The various types and principles of operation are discussed in following paragraphs. 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 horizontally, by tilting on horizontal pivots at or near the center, or by swinging on pivots at the top, bottom, ot side. Beardless 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 from the standpoint of weather protection, and certain types may be advantageous in controlling the distribution of incoming air. Deflectors are sometimes used for the mma purpose and should be considered a part of the ventilation system. Roof Ventilators The flow per square foot of inlet or outlet would be 193,750 -T-.1190 1625 cfm, with all windows open. The function of a roof ventilator is to provide a storm and weatherproof air outlet. They are actuated by the