soln06 1

# soln06 1 - lb 174 . 32 s lb in 12 ft in 2 s 60 min min ft...

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Solution (Pairs Exercise 1) ( 29 2 2 f 2 ft h mi lb 00256 . 0 = = A V C F d F = 0.00256 C d V 2 A where: F = wind force (lb f ) C d = drag coefficient (no units) V = wind velocity (mi/h) A = projected area(ft 2 )

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[T] L] [ s m velocity fluid [L] (m) length pipe ] [L M] [ m kg density [1] less) (dimension factor friction [L] (m) diameter pipe ] [L][T M] [ m·s kg m s kg·m m N (Pa) loss pressure 3 3 2 2 2 2 2 = = = = = = = = = = = = = = = v L f d p ρ d v L f 2 p 2 = ] [L][T M] [ [L] [T] L] [ ] [L [M] L] ][ 1 [ ] [L][T M] [ 2 2 3 2 = = (1) Solution (Pair Exercise 2)
(2) Solution (Pair Exercise 2) d Lv 2f p 2 ρ = ft m 3048 . 0 in 12 ft in 2 s 60 min ft m 3048 . 0 min ft 200 m cm 100 g 1000 kg cm g 1 ft m 0.3048 ft 20 ) 02 . 0 ( 2 2 3 3 × × × × × × × = p = 4955 Pa = 5000 Pa

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( 29 ft ·
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Unformatted text preview: lb 174 . 32 s lb in 12 ft in 2 s 60 min min ft 200 ft cm 48 . 30 g 453.6 lb cm g 1 ft 20 ) 02 . ( 2 m 2 f 2 3 m 3 = p (3) Solution (Pair Exercise 2) d Lv 2f p 2 = 2 f 2 f ft lb 100 ft lb 5 . 103 = = p ( 29 ( 29 3/2 3 3/2 ft ft /s ft LH Q 5.35 = = ( 29 2 / 5 3 in 12 ft min s 60 ft gal 7.4805 ft /s ft 5.35 5/2 3 5/2 in gal/min 4.81 = 3/2 4.81 Q LH = Solution (Pair Exercise 3)...
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## This note was uploaded on 09/01/2009 for the course ENGR 111 taught by Professor Walker during the Spring '07 term at Texas A&M.

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soln06 1 - lb 174 . 32 s lb in 12 ft in 2 s 60 min min ft...

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