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= \u2212 \u2212 = = oSince at y 0 for all x, then it must be that .v v Ans= = o const= =v v The dimensions of K are {K} = {L/T} and the dimensions of a are {L\u20131}. Ans. 4.21 Air flows under steady, approximately one-dimensional conditions through the conical nozzle in Fig. P4.21. If the speed of sound is approximately 340 m/s, what is the minimum nozzle-diameter ratio De/Do for which we can safely neglect compressibility effects if Vo = (a) 10 m/s and (b) 30 m/s? Solution: If we apply one-dimensional continuity to this duct, Fig. P4.21 2 2 2 o o o e e e o e e o o eV D V D , or V V (D /D ) if4 4 \u3c0 \u3c0\u3c1 \u3c1 \u3c1 \u3c1= \u2248 \u2248 To avoid compressibility corrections, we require (Eq. 4.18) that Ma \u2264 0.3 or, in this case, the highest velocity (at the exit) should be Ve \u2264 0.3(340) = 102 m/s. Then we compute 1/2 1/2 e o min o e o o(D /D ) (V /V ) (V /102) if V 10 m/s . (a)Ans= = = =0.31 oif V 30 m/s . (b)Ans= =0.54 268 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition 4.22 A flow field in the x-y plane is described by u = Uo = constant, v = Vo = constant. Convert these velocities into plane polar coordinate velocities, vr and v\u3b8. Solution: Each pair of components must add to give the total velocity, as seen in the sketch at right. The geometry of the figure shows that Ans.\u3b8 v U \u3b8 V \u3b8 v U \u3b8 V \u3b8 r o o o o cos sin ; sin cos = + = \u2212 + Fig. P4.22 4.23 A tank volume V contains gas at conditions (\u3c10, p0, T0). At time t = 0 it is punctured by a small hole of area A. According to the theory of Chap. 9, the mass flow out of such a hole is approximately proportional to A and to the tank pressure. If the tank temperature is assumed constant and the gas is ideal, find an expression for the variation of density within the tank. Solution: This problem is a realistic approximation of the \u201cblowdown\u201d of a high- pressure tank, where the exit mass flow is choked and thus proportional to tank pressure. For a control volume enclosing the tank and cutting through the exit jet, the mass relation is tank exit exit d d(m ) m 0, or: ( ) m CpA, where C constant dt dt \u3c1\u3c5+ = = \u2212 = \u2212 =\ufffd \ufffd o p(t) t o o p o CRT Ap dpIntroduce and separate variables: dt RT p \u3c1 \u3c5 = = \u2212\ufffd \ufffd The solution is an exponential decay of tank density: p = po exp(\u2013CRToAt/\u3c5). Ans. 4.24 Reconsider Fig. P4.17 in the following general way. It is known that the boundary layer thickness \u3b4(x) increases monotonically and that there is no slip at the wall (y = 0). Further, u(x, y) merges smoothly with the outer stream flow, where u \u2248 U = constant outside the layer. Use these facts to prove that (a) the component \u3c5(x, y) is positive everywhere within the layer, (b) \u3c5 increases parabolically with y very near the wall, and (c) \u3c5 is a maximum at y = \u3b4. Solution: (a) First, if \u3b4 is continually increasing with x, then u is continually decreasing with x in the boundary layer, that is, \u2202 u/\u2202 x < 0, hence \u2202 v/\u2202 y = \u2013\u2202 u/\u2202 x > 0 everywhere. It follows that, if \u2202 v/\u2202 y > 0 and v = 0 at y = 0, then v(x, y) > 0 for all y \u2264 \u3b4. Ans. (a) Chapter 4 \u2022 Differential Relations for a Fluid Particle 269 (b) At the wall, u must be approximately linear with y, if \u3c4w \u2265 0: u df df v u dfNear wall: u y f(x), or y , where 0. Then y x dx dx y x dx \u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \ufffd \ufffd \u2248 = < = \u2212 = \ufffd \ufffd\ufffd \ufffd 2 0 df df yThus, near the wall, v y dy . (b) dx dx 2 y Ans\ufffd \ufffd \ufffd \ufffd\u2248 \u2248\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd\ufffd Parabolic (c) At y = \u3b4, u \u2192 U, then \u2202 u/\u2202 x \u2248 0 there and thus \u2202 v/\u2202 y \u2248 0, or v = vmax. Ans. (c) 4.25 An incompressible flow in polar coordinates is given by 2v cos 1r bK r \u3b8 \ufffd \ufffd= \u2212\ufffd \ufffd\ufffd \ufffd 2v sin 1 bK r \u3b8 \u3b8 \ufffd \ufffd = \u2212 +\ufffd \ufffd\ufffd \ufffd Does this field satisfy continuity? For consistency, what should the dimensions of constants K and b be? Sketch the surface where vr = 0 and interpret. Fig. P4.25 Solution: Substitute into plane polar coordinate continuity: r 2 v1 1 1 b 1 b(rv ) 0 K cos r K sin 1 0 r r r r r r r r Satisfied\u3b8\u2202\u2202 \u2202 \u2202\u3b8 \u3b8\u2202 \u2202\u3b8 \u2202 \u2202\u3b8 ? \ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd + = = \u2212 + \u2212 + =\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd \ufffd \ufffd The dimensions of K must be velocity, {K} = {L/T}, and b must be area, {b} = {L2}. The surfaces where vr = 0 are the y-axis and the circle r = \u221ab, as shown above. The pattern represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8). 4.26 Curvilinear, or streamline, coordinates are defined in Fig. P4.26, where n is normal to the streamline in the plane of the radius of curvature R. Show that Euler\u2019s frictionless momentum equation (4.36) in streamline coordinates becomes 1 s pV VV g t s s \u2202\u2202 \u2202 \u2202 \u2202 \u3c1 \u2202+ = \u2212 + (1) 2 1 n V pV g t R n \u2202\u3b8 \u2202 \u2202 \u3c1 \u2202\u2212 \u2212 = \u2212 + (2) Fig. P4.26 270 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition Further show that the integral of Eq. (1) with respect to s is none other than our old friend Bernoulli\u2019s equation (3.76). Solution: This a laborious derivation, really, the problem is only meant to acquaint the student with streamline coordinates. The second part is not too hard, though. Multiply the streamwise momentum equation by ds and integrate: s 2 V dp dp dpds V dV g ds gsin ds g dz t \u2202 \u3b8\u3c1 \u3c1\u3c1\u2202 + = \u2212 + = \u2212 \u2212 = \u2212 \u2212 Integrate from 1 to 2: (Bernoulli) .Ans\ufffd \ufffd \u2202 \u3c1\u2202 2 22 2 2 1 2 1 1 1 V V dpV ds g z z 0 t 2 \u2212 + + + ( \u2212 ) = 4.27 A frictionless, incompressible steady-flow field is given by V = 2xyi \u2013 y2j in arbitrary units. Let the density be \u3c10 = constant and neglect gravity. Find an expression for the pressure gradient in the x direction. Solution: For this (gravity-free) velocity, the momentum equation is 2 ou v p, or: [(2xy)(2y ) ( y )(2x 2y )] p x y \u2202 \u2202\u3c1 \u3c1\u2202 \u2202 \ufffd \ufffd + = \u2212 + \u2212 \u2212 = \u2212\ufffd \ufffd\ufffd \ufffd V V i i j\u2207 \u2207 2 3 oSolve for p (2xy 2y ), or: .Ans\u3c1= \u2212 + 2o pi j 2xy x \u2207 = \u2212\u2202 \u3c1\u2202 4.28 If z is \u201cup,\u201d what are the conditions on constants a and b for which the velocity field u = ay, \u3c5 = bx, w = 0 is an exact solution to the continuity and Navier-Stokes equations for incompressible flow? Solution: First examine the continuity equation: ?u v w 0 (ay) (bx) (0) 0 0 0 for all and x y z x y z a b\u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u2202 \u2202 \u2202 \u2202 \u2202 \u2202+ + = = + + = + + Given gx = gy = 0 and w = 0, we need only examine x- and y-momentum: 2 2 2 2 u u p u u p u v [(ay)(0) (bx)(a)] (0 0) x y x xx y \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ \u3c1 µ\u2202 \u2202 \u2202 \u2202\u2202 \u2202 \ufffd \ufffd\ufffd \ufffd + = \u2212 + + = + = \u2212 + +\ufffd \ufffd\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd 2 2 2 2 v v p v v p u v [(ay)(b) (bx)(0)] (0 0) x y y yx y \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ \u3c1 µ\u2202 \u2202 \u2202 \u2202\u2202 \u2202 \ufffd \ufffd\ufffd \ufffd + = \u2212 + + = + = \u2212 + +\ufffd \ufffd\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd Chapter 4 \u2022 Differential Relations for a Fluid Particle 271 p pSolve for abx and aby, or: x y Ans.\u2202 \u2202\u3c1 \u3c1\u2202 \u2202= \u2212 = \u2212 2 2p ab(x y ) const 2 = \u2212 + + \u3c1 The given velocity field, u = ay and v = bx, is an exact solution independent of a or b. It is not, however, an \u201cirrotational\u201d flow. 4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid with the velocity field u = \u20132xy, v = y2 \u2013 x2, and w = 0. (a) Does this flow satisfy conservation of mass? (b) Find the pressure field p(x, y) if the pressure at point (x = 0, y = 0) is equal to pa. Solution: Evaluate and check the incompressible continuity equation: ( )0 2 2 0 0 au v w y y x y z \u2202 \u2202 \u2202 \u2202 \u2202 \u2202+ + = = \u2212 + + \u2261 Yes! Ans. (b) Find the pressure gradients from the Navier-Stokes x- and y-relations: 2 2 2 2 2 2 , : u u u p u u u u v w or x y z x x y z \u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ\u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \ufffd \ufffd\ufffd \ufffd + + = \u2212 + + +\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd 2 2 2 3[ 2 ( 2 ) ( )( 2 )] (0 0 0), : 2 ( )p pxy y y x x or xy x x x \u2202 \u2202\u3c1 µ \u3c1\u2202 \u2202\u2212 \u2212 + \u2212 \u2212 = \u2212 + + + = \u2212 + and, similarly for the y-momentum relation, 2 2 2 2 2 2 , : v v v p v v v u v w or x y z y x y z \u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ\u2202 \u2202 \u2202