Document V3JBD2R4v0yzEBXVXx3ajjKwK

650 CHAPTER 30 1952 Guide Iv ' Throw Equations'for the throw of straight flow side wall outlets have been'<fe-! veloped- on the! basis of the momentum theory. -Equation:`4 States the throw in terinsof the'area of the outlet and the primitry airVoliimel* ' " : ; where L = 0.82 Qi Vh (4) L = throw, feet: Ai = effective outlet area, in square inches = (gross measured area) X (percen tage of free area/100) X (discharge coefficient). : The discharge coefficient is. approximately 0.8. ' Equation 4 has been, developed.under the assumption that the tempera ture of the supply air is the same as the temperature of the room air. It applies only to straight:flow: outlets with aspect ratios less than 16. Equation 5 for the performance of straight flow outlets evolved from research1 allows the calculation of the maximum residual velocity at any distance perpendicular to the ,outlet-face. It applies for aspect ratios up to 50. ': ; : where rVVr:^- :~x~:`~,,AlvQTi....l... .(B) V, = maximum residual velocity in air stream, t.e., the highest maintained ve locity at the given cross section in'the room, feet per minute. Vi = average initial velocity across outlet, feet per. minute. K = constant of proportionality. Ai = effective outlet area in square feet = (gross measured area) X .(percentage of free area/100) X (discharge coefficient). ! X = normal distance from outlet face, feet. Equation 5 together with Equation 6 (which reduces to Equation 7 if the jet angle is 20 deg) for the entrainment ratio,"' where Entrainment Ratio 0.785 KrYi/3`Z+2Xtan|y - RXy/A,i\y .0.785 . .. 2/ (6) , R =, ratio of maximum residual velocity to average residual velo'city. 0 = jet angle or spread angle in degrees. Entrainment Ratio = (i /.4'-- + 0.35 X ) --' 1' ` 20:deg jet angle)- RXyfiAi .0.785 ,/ . - i (7) has been used to develop, charts3 which provide the graphical solution of probieins involving the determination of the throw of air from slots and.jets, the residual velocity,, and .the size of openings. (See Figs. 2 and 3). The charts,apply, only to air discharging into room air of same temperature as the stream.' They can be used to determine the throw of. air and entrainment Air-' Distribution 651 ratios up to 40:1 with initial velocities of 1000 to 6000 fpm, and with resid: ual velocities of-100 to 1000 fpm. The charts furthermore are for use with sharphedged orifices or slots, and include.the coefficient of discharge. If .air is discharged from an orifice with a well-rounded entrance or from a length of straight duct, the coefficient of discharge is unity and the .'actual area of the opening is the effective area. For such rectangular openings the ef fective diameter is the diameter of a circle with an area equal to the actual withian initial velocity of 2000 fpm.. Determine the maximum residual velocity and the entrainment ratio at a distance-of 15 ft from the slot. - From Fig. 3 the effective diameter = 6.2 in. = 0.52 ft.1 The number of effective diameters in 15 ft f 15/0.52 = 28.8. ': . From Fig. 2 at 2000 ft initial velocity read entrainment ratio = 6.6 and maximum residual velocity = 390 fpm. From tests.it has been shown that the average residual velocity may be taken as 1 of the numnum.or 130 fpm in this ease. Example 2: Using the data from Example 1 determine the distance at which the maximum residual velocity will be 150 fpm. . From Fig. 2 at V, = 2000 and Vr = i50, the number of effective diameters-is read directly as 73 and the throw of the air is therefore 73 x.0.52 = 38 ft. ..; Example S: Air issues from a round orifice plate with an initial average velocity of 4000 fpm. It is to have a maximum residual velocity of 400 fpm at a distance of 30 ft from the opening. Calculate the size of the opening required and the entrainment ratio. On Fig. 2 at the intersection of the curve of.4000 fpm, the entrainment ratio is. read directly as 15 and the effective diameters of throw = 55.