Question

2) Figure 2 shows a typical duct layout. Design the duct system using equal friction method. Take the velocity of air in the main duct (A) as 9 m/s. Estimate the static pressure in the index run of the duct network. There is a pressure drop of 15 Pa each at the outlet grilles. Assume the resistance of the fittings at 10 Pa for elbow and 15 Pa for Tee. (10 marks) 2 m/s 2 1 mºls 3 D F -15 m 12 m 18 m 6 m с E A B 6 m Fan 1 m/s 1 Figure 2

Answer 1

Given data Velocity fair ús the main diect = 9mis Pressure drop of 15pa retr'o tance of the fitting at top elbow 2m3 Ls Im3ls © Tee = lspa 18 m Lm kism to 12. PA = -1-f2+1 - Gmis VA=ants DPezet = 15pa Opelbow = lopa OP Tee - ispa Fan Em PA -AV A = Q = 4 Imuls DA= 0.7522 m F Here the equal ferchon method forichom presure drop per cenit length in any part of the duct is same SE JA - sef) OPA where Apf = 0.022243 x Q 2.852 XL 24.973 Apf GRE LA = DPF Vio 1.852 1-853 DAC 4.973 D24.993 For duct A DA= 0.7522 m 4.933 & DB= [(14)1-052 1992) yo C0.752294.973 DB= 0.4488m

A like Dc = 0.61 SS) Dos 0.5510m DE 0.4488 m DF = 0.4488 m E la = 0.022243 X PA DA 4.983 1.852 0.022243 XC4) (0.7522) 9.923 OPELLA 1.1945 pr. 1065 cen pipe A-B DP) A-B = DPF DAT DPF) seppelbow TDP exit =(1-1945X15) + (1-1945X6 ) +10+15 DPF) A-B- 17.91 +7-167 + 10 +15 App) A-B = 50.67 pa EC - DP) A-C-D Oppla tapp)c Topfot spree to rellow TAP exet = 1.1945 (15+1246)+15+10415 AP) AC-D = 79.42pQ

DPTee tapree + DP ellow top exit op) A-(-E- PAPE) A +APF) CFDPF), DPF) + 1.1945( 15+12+18 +6) +15+15 +10+15 DP)A-C-E-F 115.9229 Accordingly index san ces A-C-E-F To FTP = 115 92 pq

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