{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch08solutions

ch08solutions - Problem 8.1[1 Given Find Solution Air...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 8.1 [1] Given: Air entering duct Find: Flow rate for turbulence; Entrance length Solution: The governing equations are Re V D ν = Re crit 2300 = Q π 4 D 2 V = The given data is D 6 in = From Table A.9 ν 1.62 10 4 × ft 2 s = L laminar 0.06 Re crit D = or, for turbulent, L turb = 25D to 40D Hence Re crit Q π 4 D 2 D ν = or Q Re crit π ν D 4 = Q 2300 π 4 × 1.62 × 10 4 × ft 2 s 1 2 × ft = Q 0.146 ft 3 s = For laminar flow L laminar 0.06 Re crit D = L laminar 0.06 2300 × 6 × in = L laminar 69.0 ft = For turbulent flow L min 25 D = L min 12.5 ft = L max 40 D = L max 20 ft =
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Problem 8.2 [2]
Image of page 2