Explanation Write the expression to calculate the area of pipe. A = π 4 D 2 (I) Here, the diameter of pipe is D . Write the equation for Reynolds number. Re = V D ν (III) Here, the velocity of the fluid is V and the kinematic viscosity of fluid is ν . Write the relation for the friction factor in turbulent flow. 1 f = − 1.8 log [ ( ε / D 3.7 ) 1.11 + 6.9 R e ] (IV) Write the expression for the pressure drop without the straws in the pipe. Δ p no straw = f ( l D ) ( 1 2 ρ V 2 ) (V) Here, length of the straw in l . Calculate the number of straws by consider the cross section of straw is 0.25 in 2 . N = 4 A c ( A p i p e ) b N = 4 A c ( π 4 D 2 ) (VI) Here, area of the straw is A c and diameter of the pipe is D . Write the relationship between the diameter and discharge. Q g a p Q s t r a w = ( D g ) 4 d 4 (VII) Here, diameter of the straw is d . Write the expression for the velocity of flow in the straw. Q = V s t r a w ( N A s t r a w ) V s t r a w = Q N A s t r a w (VIII) Write the expression for the friction factor for laminar flow: f = 64 Re (IX) Write the expression for the pressure drop when straws in the pipes. Δ p straw = f ( l d ) ( 1 2 ρ V 2 ) (X) Write the expression for the pressure drop between the point (1) and (2). Δ p = Δ p straw − Δ p no straw (XI) Conclusion: Substitute 12 in. for D in equation (I). A = π × ( 12 in . 2 ) 2 A = π × ( 12 in × 1 ft 12 in 2 ) 2 = 0.7857 ft 2 Write the expression for the discharge. Q = A V (II) Here, velocity of the air is V . Substitute 5.0 ft 3 /s for Q and 0.7857 ft 2 for A in equation (II). 5.0 ft 3 /s = ( 0.7857 ft 2 ) V V = 5 0.7857 = 6.37 ft/s Refer to Table 1.7, “Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (BG Units)”, ν = 1.57 × 10 − 4 ft 2 /s Substitute 6.37 ft/s for V , 12 in . for D , and 1.57 × 10 − 4 ft 2 /s for ν in equation (III). Re = 6.37 ft/s × 12 in . 1.57 × 10 − 4 ft 2 /s Re = 6.37 ft/s × 12 in . × 1 ft 12 in . 1.57 × 10 − 4 ft 2 /s = 40549 > 2100 Thus, the flow is turbulent. Substitute 40549 for Re , 0 for ε , and 12 in. for D in equation (IV). 1 f = − 1.8 log [ ( ε / D 3.7 ) 1.11 + 6.9 Re ] 1 f = − 1.8 log [ ( 0 3

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ISBN: 9781119547990

Author: HOCHSTEIN

Publisher: JOHN WILEY+SONS INC.

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Chapter 8.4, Problem 57P

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The extra pressure drop between points (1) and (2) caused by these straws.

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Chapter 8.4, Problem 56P

Chapter 8.4, Problem 58P

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EBK MUNSON, YOUNG AND OKIISHI'S FUNDAME

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