Question

Heated air at 1 atm, 100 ° F, and 23% relative humidity is to be transported in a 490 ft long circular plastic duct at a rate of 740 cfm (ft 3 /min). If the head loss in the duct is not to À A 11 - Arial (com exceed 790 in. of air, determine the minimum diameter of the duct. (Hint: initial guess value for the diameter is 8 inch). For air at 1 atm and 100 · F, p = 0.07088 lb/ft3 and j = 1.281 x 10-5 lb/(ft s) = 0.0007686 To calculate the friction factor for turbulent flow, you may use Colebrook Equation: -2log 2.51 ED RealF 3.71

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Answer 1

mo Green that, Heated air at lain and G=100°F Relatices humiwilij = 234 C=490lb Giren guess ss Value diameter 5inch = 0.6674 Va grabe g flowdr 740CFMCft3/min) 212.33 ft3/sec Gireen for salon and 100f pe 0.07088061 ft' and M =1281X1oublfts Head loss in duct mustrit enceed 790 in 265.83FE ais 12,33416 0 69014 20.23 foi relative humidity Dynamic kisicosity change Monin qv+(1-91ua where, MVr 9.22+0.02869-273,15) x16€ (where t' in Kelrein]

129.22+0.028(310.77-273.15) (100°F 2310.776) - 10.273x10udrtis So, Horin = 0.23x10.273x16+61-0.23) X1 281x165 11222x1óulitis Remin - for umin 0.0708880.667XV 11222X105 Remin = 0.0386xVx105 We know that Ve XC 12.33 M (0.673 4 E 12.33 31141% 0.4489 35i27ft/sec Remin = 0.0386X35.27 는 1.36X105 for friction facto we use Colebrook 금 -zlog peber Equation e + E10 REJF 371 fol smooth flow, Elo-O Thus / SF = -210g/251 = -210g 136x18F 2:51

3 = 2log 05-25) SF = 0.0616 F2 00616 F -0.00379 The head loss in air duct can be Calculated as hy of 4ox v/29 65.83= 0.00379x 490 190 x (35.2707 2x32.15 Dz0.546ft on 6.552 in Mininum Diameter duct. of the 20.54ft, on 6.552 in: