EBK FUNDAMENTALS OF AERODYNAMICS
EBK FUNDAMENTALS OF AERODYNAMICS
6th Edition
ISBN: 9781259681486
Author: Anderson
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 4, Problem 4.13P

In Section 3.15 we studied the case of the lifting flow over a circular cylinder. In real life, a rotating cylinder in a flow will produce lift; such real flow fields are shown in the photographs in Figures 3.34(b) and (c). Here, the viscous shear stress acting between the flow and the surface of the cylinder drags the flow around in the direction of rotation of the cylinder. For a cylinder of radius R rotating with an angular velocity w in an otherwise stationary fluid, the viscous flow solution for the velocity field obtained from the Navier-Stokes equations (Chapter 15) is V θ = R 2 w r where V θ is the tangential velocity along the circular srcamlines and r is the radial distance from the center of the cylinder. (See Schlichting, Boundary-Layer Theory, 6th ed., McGraw-Hill, 1968, page 81.) Note that V9 varies inversely with r and is of the same form as the inviscid flow velocity for a point vortex given by Equation (3.105). If the rotating cylinder has a radius of 1 m and is flying at the same velocity and altitude as the airfoil in Problem 4.11, what must its angular velocity be to produce the same lift as the airfoil in Problem 4.11? (Note: You can check your results with the experimental data for lift on rotating cylinders in Hoerner, Fluid-Dynamic Lift, published by the author, 1975, pp. 2l, Fig. 5.)

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Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/s
Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4
Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…
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