Document 93GDbENpvjxjORZKMD2N3qJdq

812 CHAPTER 31 . ,1958 Guide * Based on Equation 13. Fig. 14. Static Regain Chart* : For Determining Velocity in Ducts Carrying Q_ to 3000. cfm . 4. Loss in Section F. = 0.13 X 0.225 = 0.03 in. water. 5. Since the operating pressure of Outlet 1 is to be the same as that of Outlet 6, the loss in Section B must equal the loss in Section F. Using the Static Regain Chart, Fig. 15, size Section B for a net loss of 0.03 in. water. ii The procedure for using these charts is indicated by arrow heads and dashed lines on Fig. 15. Proceed as follows: a. Locate the velocity of the preceding (upstream) duct section along-the velocity scale on the left margin (1500 fpm). b. Proceed horizontally to the air volume abscissa (6000 cfm). c. Proceed parallel to the curved lines to intersect the diagonal base line. d. Go vertically to the net loss desired (0.03 net loss). e. Go horizontally to the air velocity base line. f. Proceed parallel to the curved lines to intersect ordinate for 25 ft duct length. g. Move horizontally to the air velocity scale and read the velocity in the downstream section of duct (1500 fpm). 6. This procedure is repeated for Section C except that the no gain or loss line is used in previous Step 5 d. Outlet No. 2 will then nave the same static pressure be hind it as Outlet No. 1, thus fulfilling the problem condition of the same operating pressure for all outlets. 7. At this point, the equivalent length of the elbow in Section D should be esti mated. Its width .will be somewhat less than that of Section C. Assume W = 26 in.; hence H/W = 16/26 = 0.6. With a typical radius ratio of 1.25, from Fig. 8, * Based on Equation 13. Fig. 15. Static Regain Chart* For Determining Velocity in Ducts Carrying 3000 to 30,000 cfm L/W -- 6; therefore L -- 6 X (26/12) = 13 ft. Adding this to the actual length of Section D results in a total equivalent length of 26 ft. 8. Using the no yom or loss line in Fig. 14, size Sections D, E and G; Results are listed in Table 8. ' r ?' total pressure loss of the system is the sum of the losses in Sections A and B (or F) and the outlet operating pressure: Loss in Section A = 0.13 X 0.40 = 0.05 Loss in Section B (or F) = 0.03 Outlet Operating Pressure = 0.12 Total Pressure Loss = 0.20 in. Section A B C D E F G Table 8. Tabulation of Results (Example 7) Aib Volume cfm Equiva lent Length ft Velocity fpm Rectangular Duct in. Diam. in. Friction Per 100 - FT in. HsO Net Pres sure Loss in. HtO 8000 40 1500 6000 25 1500 4500 15 1300 3000 26* 1040 1500 15 860 2000 22.5 1200 1000 15 900 48 x.16 36 x 16 31 x 16 . 26 x 16 16 x 16 15 x 16 10 x 16 29.2 -- --- -- -- 17. -- ' 0.13 -- ___ ___ -- 0.13 .-- .05 .03 0 0 0 .03 0