Explanation Given Information: Pipe diameter, d = 6 c m = 0.06 m Length of perforated section, L = 2 m Number of holes in perforated section = 500 Hole diameter, d h = 1 m m Air pressure at section, P 1 = 105 k P a = 105000 P a Air flow rate, Q 1 = 110 m 3 / h = 0.0305 m 3 / s At 20 ° C air temperature, Air density, ρ a = 1.25 k g / m 3 Air viscosity, μ a = 1.8 × 10 − 5 k g / m . s For holes, assume, β = 0 ; C d = 0.61 Pipe velocity is given by v 1 = Q 1 A = 0.0305 ( Π / 4 ) × d 2 = 0.0305 ( Π / 4 ) × 0.06 2 = 10.8 m / s Re = ρ v 1 d μ = 1.25 × 10.8 × 0.06 1.8 × 10 − 5 Re = 45000 We know, f = 0.0214 τ w = f 8 ρ v 1 2 = 0.0214 8 × 1.25 × 10.8 2 τ w = 0.39 P a We can write, Q o r i = 500 C d A o r i 2 Δ P ρ = 500 × 0.61 × Π 4 × 0.001 2 × 2 × ( 105000 − 101350 ) 1

8th Edition

ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 6, Problem 6.147P

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The value of pressure and flow rate at section 2.

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Chapter 6, Problem 6.146P

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