Document rB43Rr9L0dGaZ8koXNZXrNgYE

688 CHAPTER 31 1952 Guide locities. Balancing is] obtained by use of dampers. The method is illus trated in Example 4. inSCCTIOM---A iO"5.E..C..-.B.-- O? ttCrC O} **rr o*rr 20-ft SEC-D11 o* 2o-ft Ffo. 10. Duct Layout fob Example 4 Example 4: (Velocity Reduction Method). A duct layout is shown in Fig. 10. The fan delivers 8000 cfm. Four outlets deliver 2000 cfm each. Find duct dimen sions and total pressure loss. Solution: Select velocity for Section A (2200 fpm) and reduce velocity arbitrarily along run. Find duct areas by using Equation 17. For selection of circular equiva-. lents of rectangular ducts refer to Table 2, and: for determination of friction loss in duct refer to Fig. 2 (See Example.!). Results are tabulated in Table 5. Table 5. Tabulation of Results (Example 4) Section Ant Volume cfm Velocity fpm Area sq ft Area sq in. . Duct Size in. . . Diam in. Frict pee ' 100 FT in. HjO Filer Loss in. HiO A - . 8000 2200 3.64 B 6000 2000 3.00 C 4000 1800 2.22 D 2000 1600 1.25 524 432 320. , 180 28x20 22x20 20 x 16 12...x 16. . 24.8 0.25 22.9 ' 0.23 19.5 0.24 15.1: .0.24 0.10 0.05 0.05 0.05 Total resistance, 0.25 2. Equal Friction Method When the equal friction method of design is used, the duct system is designed for equal friction per foot of length. This prevents one section of the duct from having an excessive resistance compared with another. The usual procedure in this method is to select the main , duct velocity to be consistent, with good practice from a standpoint of noise for a particular type of building. This velocity should be less than the fan outlet velocity. All ducts are then sized for equal friction per unit length by the use of Figs. 1 or 2 and Table 2. The equal friction method has the advantage of auto matically reducing the velocities in the various sections of the system, and . also of allowing a quick check of the total duct resistance. In cases where the fan or factory assembled air conditioning unit can operate against only a limited external resistance, it is necessary to divide the permissible total resistance by the total equivalent length of the longest or most complicated run of duct to determine the design resistance per 100 ft, and then to size all ducts at this resistance value. This will auto matically determine the duct velocities and give the desired. total duct resistance. A further refinement,.which is sometimes used in large systems, is to size each branch duct so that it has a resistance equal, to the resistance of the main system at the point of juncture. Even when this refinement is added, regulating dampers are recommended in each branch. Example 5: (Equal Friction Method). A duct layout is shown in Fig. 11. The fan delivers 2500 cfm. Outlets No.- l and 2 deliver 750 cfm each and outlet No. 3 de livers 1000 cfm'. Trunk velocity is assumed as 1560 fpm; the area will be 1.67 sq ft (240 sq in.); and the size will be 20 x 12 in. Determine sizes of ducts for sections B, Cy D and E and find the total pressure loss. Solution: The equivalent round diameter of a 20 x 12 in;. rectangular duct is 16.8 in-. (from.Table 2).. Referring to Friction Chart, Fig. 2, a volume of 2500 cfm through Air. Duct Design 689 2-0 SECTION-A 20-FT ' SEC-0 O-FT SEC-C IO-FT , sec-e 15-FT SCC-B O-FT > 3-FT <U tS-rr. 3-6 Fig. 11. Duct Layout fob Example 5 a 16.8 in. duct gives a resistance of 0.2 in. per 100 ft. The amount of air to be handled by each section is known, and the corresponding round duct sizes with equal pressure drop for these values can be located on the 0.2 m. friction line. The 'equivalent rec tangular duct sizes are then selected from Table 2. Results are tabulated in Table 6. Table 6. Tabulation of Results (Example 5) Section Air Friction -Volume PER .100 FT. rfm in. Diam. in. Veloc ity fpm Rectan gular Duct . Friction PER 100 FT. in. in. Diam. in. Veloc ity fpm Rectan gular .Duct in. A' B C D E. 2500 750 1750 750 1000 0.2 16.8- 1620 20 x 12 0.2 17 1600 20 x 12 0.2 10.7 1190 10 x 9 0.286 10 1350 10 x 8 0.2 14.8 1400 15 x 12 0.2 14.5 1450 15 x 12 0.2 10.7 1190 10 x 9 0.4 9.8 1350 10 x 8 0.2 12 1300 10 x 12 0.2 12 1300 10 x 12 The total pressure loss in the longest run is the friction loss.in Sections (A + C + E), plus the loss in one,elbow and the loss through the'outlet (3h The additional pressure' loss in the elbow wifi be assumed as kf = 12 (Fig. 5); thus, the additional ' w equivalent length of duct is 10 ft, and the design loss will be 0.02 in. , Friction loss (A + C + E).. 0.12 (Du ct length = 20 ft + 10 ft +, 15 ft + 15 ft) Elbow loss.............................. 0.02- . Loss through outlet............. 0.12 . Total pressure loss in duct.................................. = 0.26 in.. The pressure required at the beginning of the main run'is therefore 0.26.in. The fan selected for the duct system must not only deliver the required volume of air against this los8,_but also against the losses in all air conditioningapparatus such as washers or spray chambers, heating or cooling coils and filters. The static head required of the1fan for the usual air conditioning installation.is between 1 and 1.5 in. of water. About one-third of this represents losses in the duct system.' The ldsses in the air conditioning apparatus can be obtained from manufacturers' catalogs. Resizing of Ducts In order to equalize the pressure drop in the system, the following addi tional procedure is recommended: : Assume ASi, &Hi, AH, to be the total pressure loss through ducts (1), (2) and (3);r,, Tb, rc, rd, re the friction losses in the straight sections of the system; n*, rde, r,, the elbow losses, and r,, r,, r, the loss through the outlets. Then, AH, = ri + Tb + 2fbe -f- Ti AH, = r + r,, + rd + rd + rs AH, = r + r0 + r,, + Tee +Tl If AH, = AH, = AH, = AH n> + 2rbi "AH -- r. -- n rd + rd, = AH -- r. -- r. -- r.