Unformatted text preview: π ⋅ r ⋅ ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d + = ρ − π ⋅ R 2 ⋅ U 1 2 ⋅ 2 π ⋅ ρ ⋅ u max 2 ⋅ R 2 2 2 R 4 4 R 2 ⋅ ⋅ − R 6 6 R 4 ⋅ + ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ + = ∆ p p 1 p 2 − = ρ − U 1 2 ⋅ 1 3 ρ ⋅ u max 2 ⋅ + = ρ − U 1 2 ⋅ 1 3 ρ ⋅ 2 U 1 ⋅ ( ) 2 ⋅ + = ρ U 1 ⋅ 1 3 2 ( ) 2 ⋅ 1 − ⎡ ⎣ ⎤ ⎦ ⋅ = 1 3 ρ ⋅ U 1 2 ⋅ = Hence ∆ p 1 3 850 × kg m 3 ⋅ 5.66 m s ⋅ ⎛ ⎝ ⎞ ⎠ 2 × N s 2 ⋅ kg m ⋅ × = ∆ p 9.08 kPa ⋅ =...
View
Full Document
 Fall '07
 Lear
 Fluid Mechanics, Flux

Click to edit the document details