Solutions11 - Problem 8.173[2 Given Flow through a venturi...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 C A t 1 β 4 2 ρ p 1 p 2 ( ) = C A 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 = C A 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 4 1 β 4 2 SG 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 = 4 Q 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 V D 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 =
Image of page 1

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

View Full Document Right Arrow Icon