To determine Determine the screen drag force. Explanation Express the cross-sectional area of the duct in section 1. A 1 = π 4 D 2 (I) Here, the diameter of the duct in section 1 is D 1 . As the diameter of the duct remains the same, the cross-sectional area of the duct remains constant at sections (1) and (2). A 2 = A 1 Express the velocity in section 2. u 2 = u max ( r / R ) Here, the maximum velocity in section 2 is u max , the variable radius is r and the maximum radius of the duct is R . Express the maximum velocity at section 2 using principle of conservation of mass. m ˙ 1 = m ˙ 2 ρ V 1 A 1 = ρ u 2 A 2 ρ V 1 A 1 = ρ ∫ 0 R u max ( r / R ) ( 2 π r d r ) V 1 A 1 = u max ∫ 0 R 2 π r 2 R d r u max = V 1 A 1 ∫ 0 R 2 π r 2 R d r = V 1 A 1 2 π R 2 / 3 (II) Here, the velocity in section 1 is V 1 . Express the screen drag force by applying the linear momentum principle to the flow along the axial direction. − ρ V 1 2 A 1 + ∫ 0 R ρ u 2 2 ( 2 π r d r ) = p 1 A 1 − p 2 A 2 − R x R x = ρ V 1 2 A 1 − ∫ 0 R ρ u 2 2 ( 2 π r d r ) + p 1 A 1 − p 2 A 2 (III) Here, the density of the air is ρ , upstream pressure is p 1 , downstream pressure is p 2 and screen drag force is R x . Substitute u max ( r / R ) for u 2 in Equation (III). R x = ρ V 1 2 A 1 − ∫ 0 R ρ [ u max ( r / R ) ] 2 ( 2 π r d r ) + p 1 A 1 − p 2 A 2 = ρ V 1 2 A 1 − 2 π ρ u max 2 R 2 [ r 4 4 ] 0 R + p 1 A 1 − p 2 A 2 = ρ V 1 2 A 1 − 2 π ρ u max 2 R 2 R 2 + p 1 A 1 − p 2 A 2 (IV) Conclusion: Substitute 2 ft for D 1 in Equation (I)

8th Edition

ISBN: 9781119571490

Author: GERHART

Publisher: WILEY

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Chapter 5.2, Problem 66P

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Determine the screen drag force.

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