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e <br /> 11-3 SINGLE-PASS (INTERMITTENT) PACKED-BED FILTERS 737 <br /> a. The headloss in pipe without orifices is determined from Eq. (6-5): <br /> Q �_a5 . <br /> hfp = 10.5(Lf_„)�Q D-a.87 <br /> = <br /> �� <br /> 10.5(15.5) <br /> 4.73)' 0.930-4.57 = 0.39 ft <br /> b. The beadloss in the lateral pipe with orifices is determined from Eq. (11-6b): <br /> hfdp = !hfp = Ah(,—n) , <br /> = 3(0.39) = 0.13 ft <br /> 8. Determine the difference in the discharge between the first and last orifice in each <br /> lateral. <br /> F <br /> a. The head on the first orifice,from Eq. (11-9), is: <br /> Ah(,-n) = h, - h„ <br /> - hi = hr + Ahtj-n� <br /> b. Determine the value of m from Eq. (11-8): <br /> h„ = m2 h, <br /> �5.0 <br /> 5 �. <br /> m = V 0+ 0 13 = 0.987 <br /> The difference in the discharge between the first and last orifice in each lateral is <br /> about 1.3 percent[(I—0.987)x 1001,which is below the 5 percent value specified. <br /> If the orifices are drilled with a hand drill in the field,differences in the discharge <br /> between orifices as high as 5 percent can be expected. Improved accuracy is ob- <br /> tained with a drill press. <br /> 9. Estimate the total dynamic head.Assume a nominal'1.5-in(38-mm)diameter plastic <br /> pipe [actual ID = 1.720 in(44 mm)] will be used to supply the filter laterals. <br /> a. The headloss in the 1:5-in pipe from the septic tank to the sand filter; computed <br /> by Eq. (6-5), is: <br /> 1.85 <br /> hfp = 10.5 (L)(9) <br /> D-4.87 <br /> L = 60ft <br /> ! <br /> QT = 52.0 gal/min <br /> hfp = 10.5 (60) 52.01"=55(1.720)-4-" = 6.3 ft <br /> ,. 150 J <br /> i <br /> I b. Estimate the headloss in the 1.5-in distribution manifold used to connect the ]at- <br /> erals: <br /> Length of distribution manifold = 15 ft <br /> Flow in the distribution manifold = 52.0 gal/min <br /> Headloss in distribution manifold without laterals: <br /> 52.0 f.$5 - <br /> hfp = .10.5(15) T50 (1.720)74' 1.6 ft 3 <br />