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HW3+solutions - Chapter 4 303 Differential Relations for a...

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Unformatted text preview: Chapter 4 303 Differential Relations for a Fluid Particle Solution: Here we have only the single of California Davis University ‘one-dimensional’ convective acceleration: Department of Mechanical o 2Vo2 du u 2 x 2V and Aerospace Engineering x dt For L 6 u x Vo 1 L L L 1 L Ans. (a) E du 2(10)2 ft NG 103 Fluid Mechanics 2x 10 , 1 400(1 4x ), with x in feet s dt 6 /12 6 /12 Fall 2011 Professor Kennedy 2 (12 g’s); at x L 0.5 ft, du/dt 1200 ft/s2 (37 g’s). Ans. (b) 400 ft/s Homework 3 Solutions and Vo At x 0, du/dt P4.3 A two-dimensional velocity field is given by V (x2 – y2 x)i – (2xy y) j in arbitrary units. At (x, y) (1, 2), compute (a) the accelerations ax and ay, (b) the velocity component in the direction 40 , (c) the direction of maximum velocity, and (d) the direction of maximum acceleration. Solution: (a) Do each component of acceleration: du u u u v (x 2 y 2 x)(2x 1) ( 2xy y)( 2y) a x dt x y dv v v u v (x 2 y 2 x)( 2y) ( 2xy y)( 2x 1) a y dt x y At (x, y) (1, 2), we obtain ax 18i and ay 26j Ans. (a) (b) At (x, y) (1, 2), V –2i – 6j. A unit vector along a 40 line would be n sin40 j. Then the velocity component along a 40 line is V40 V n 40 ( 2 i 6 j) (cos 40 i sin 40 j) 5.39 units cos40 i Ans. (b) (c) P4.4 A simple flow model for a two-dimensional converging nozzle is the distribution x y _______________________________________________________________________ u U o (1 ) v Uo w0 L L (a) Sketch a few streamlines in the region 0<x/L<1 and 0<y/L<1, using the method of Section 1.11. (b) Find expressions for the horizontal and vertical accelerations. Chapter 4 313 Differential Relations for a Fluid Particle ________________________________________________________________________ P4.16 Consider the plane polar coordinate velocity distribution C K v vz 0 r r where C and K are constants. (a) Determine if the incompressible equation of continuity is satisfied. (b) By sketching some velocity vector directions, plot a single streamline for C = K. What might this flow field simulate? vr Solution: (a) Evaluate the incompressible continuity equation (4.12b) in polar coordinates: 1 (r v r ) rr 1 r (v ) 1 C (r ) rr r 1 r ( K ) r 0 0 0 Ans.(a ) Incompressible continuity is indeed satisfied. (b) For C = K, we can plot a representative streamline by putting in some velocity vectors and sketching a line parallel to them: 1.2 1.0 0.8 0.6 0.4 0.2 0.0 O - 0.2 -0.2 314 -0.1 Solutions Manual 0.0 0.1 0.2 0.3 0.4 Fluid Mechanics, Fifth Edition The streamlines are logarithmic spirals moving out from the origin. [They have axisymmetry about O.] This simple distribution is often used to simulate a swirling flow such as a tornado. 4.27 P4.17 An excellent approximation for the two-dimensional incompressible laminar boundary layer on the flat surface in Fig. P4.17 is u U (2 y 2 y3 y4 3 4 ) for y , where C x1/ 2 , C constant Fig. P4.17 320 Solutions Manual Fluid Mechanics, Fifth Edition Solution: For this (gravity-free) velocity, the momentum equation is u V x v Solve for 334 V y p, or: 2 p Solutions o (2xy i Manual o [(2xy)(2yi) 3 ( y 2 )(2xi 2yj)] p x 2y ), or: FluidjMechanics, Fifth Edition o 2xy 2 p Ans. 4.47 P4.28 Consider the incompressible flow field of Prob. P4.6, with velocity components y u = 2y, v = 8x, w = 0. Neglect gravity and assume constant viscosity. (a) Determine whether Solution: With u 3y and v 2x, we this flow satisfies the Navier-Stokes equations. (b) If so, find the pressure distribution p(x, y) if v/ y 0 0 0, OK. may check u/ x tFind the streamlines from u o. he pressure at the origin is p / y 3y and v – / x 2x. Integrate to find Solution: In Prob. P4.6 we found the accelerations, so we can proceed to Navier-Stokes: 51° 32 39° y x 2 Ans. x u u2 p p p 2 v) gx u (u [ 0 (8 x)(2)] 0 0; 16 x Fig. P4.47 x y x x x Set 0, 1, 2, etc. and plot some v v p p 2 streamlines at right: flowyaround cornerspof g v) v (u [(2 )(8) 0] 0 0; 16 y y x y y y half-angles 39 yand 51 . QNoting that .5 P2 p /( xCy ) 4.48 onsider the following two-dimensional incompressible flow, which clearly satisfies 0 in both cases, we conclude Yes, satisfies Navier - Stokes. Ans.(a ) continuity: (b) The pressure gradients areUo constant, may easilycintegrate: u simple, so we v Vo onstant Find the stream function (r, ) of this flow, that is, using polar coordinates. p p dp dy , coordinates16 xtream function dy quite easy: y 2 ) const 16 y is 8 ( x2 Solution:dx In cartesian or : p the s dx x y u / y Uo and v – / x Vo or: 2 Uoy – Vox constant y2 ) Ans.(b) If p (0, 0) po , then p po 8 ( x But, in polar coordinates, y rsin and x rcos . Therefore the desired result is This is an exact solution, but it isrnsinBernoulli’s equationstant flow is rotational. (r, ) Uo ot – Vor cos c on. The Ans. P4.29 Consider a the stream function P4.49 Investigate steady, two-dimensional, incompressible flow of a newtonian fluid with – y2), K constant. u –2xy, v y2 – x2, and w 0. (a) Does this flow satisfy K(x2 the velocity field Plot the streamlines cn the full xy mass? (b) Find the stagnation i onservation of plane, find any pressure field p(x, y) if the pressure at point (x 0, y 0) ipoints, and ainterpret what the flow could s equal to p . represent. Solution: u y The velocities are given by 2Ky; v x 2Kx This is also stagnation flow, with the streamlines turned 45 from Prob. 4.48. Fig. P4.49 ...
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