Solutions11

Solutions11 - Problem 8.173 [2] Given: Flow through a...

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Problem 8.173 [2] Given: Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!) Find: Flow rate Solution: Basic equation m actual CA t 1 β 4 2 ρ p 1 p 2 () = t 1 β 4 2 ρ Δ p = Note that m actual is mass flow rate (the software cannot render a dot!) For Re D1 > 2 x 10 5 , 0.980 < C < 0.995. Assume C = 0.99, then check Re β D t D 1 = β 3 6 = β 0.5 = Also Δ p ρ Hg g Δ h = SG Hg ρ g Δ h = Then Q m actual ρ = t ρ 1 β 4 2 ρ Δ p = π C D t 2 4 ρ 1 β 4 2 ρ SG Hg ρ g Δ h = π C D t 2 41 β 4 2SG Hg g Δ h = Q π 4 1 0.5 4 × 0.99 × 1 4 ft 2 × 2 13.6 × 32.2 × ft s 2 1 × ft × = Q 1.49 ft 3 s = Hence V Q A = 4Q π D 1 2 = V 4 π 1 1 2 ft 2 × 1.49 × ft 3 s = V 7.59 ft s = At 75 o F,(Table A.7) ν 9.96 10 6 × ft 2 s = Re D1 VD 1 ν = Re D1 7.59 ft s 1 2 × ft s 9.96 10 6 × ft 2 × = Re D1 3.81 10 5 × = Thus Re D1 > 2 x 10 5 . The volume flow rate is Q 1.49 ft 3 s =
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Problem 9.1 [2] Given: Model of riverboat Find: Distance at which transition occurs Solution: Basic equation Re x ρ U x μ = Ux ν = and transition occurs at about Re x 51 0 5 × = For water at 10 o C ν 1.30 10 6 × m 2 s = (Table A.8) and we are given U 3.5 m s = Hence x p ν Re x U = x p 0.186m = x p 18.6 cm =
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This note was uploaded on 08/25/2011 for the course AME 331 taught by Professor Zohar during the Fall '08 term at Arizona.

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Solutions11 - Problem 8.173 [2] Given: Flow through a...

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