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Chapter 4 Differential Relations for a Fluid Particle P4.1 An idealized velocity field is given by the formula 24 2 4tx t y xz V i j k Is this flow field steady or unsteady? Is it two- or three-dimensional? At the point (x, y, z) (–1, 1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the acceleration. Solution: (a) The flow is unsteady because time t appears explicitly in the components. (b) The flow is three-dimensional because all three velocity components are nonzero. (c) Evaluate, by laborious differentiation, the acceleration vector at (x, y, z) (1, 1, 0). 2 2 2 2 4 2 2 du u u u uu v w 4x 4tx(4t) 2t y(0) 4xz(0) 4x 16t x dt t x y z dv v v v vu v w 4ty 4tx(0) 2t y( 2t ) 4xz(0) 4ty 4t y dt t x y z dw w w w wu v w 0 4tx(4z) 2t y(0) 4xz(4x) 16txz 16x z dt t x y z 2 4 2dor: (4x 16t x) ( 4ty 4t y) (16txz 16x z) dt V i j k at (x, y, z) (1, 1, 0), we obtain 2 3d 4(1 4t ) 4t(1 t ) 0 (c) dt Ans. V i j k (d) At (–1, 1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans. P4.2 Flow through the converging nozzle in Fig. P4.2 can be approximated by the one-dimensional velocity distribution o 21 0 0xu V w L (a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case Vo 10 ft/s and L 6 in, compute the acceleration, in g’s, at the entrance and at the exit. Fig. P4.2 Chapter 4 Differential Relations for a Fluid Particle 303 Solution: Here we have only the single ‘one-dimensional’ convective acceleration: 221 . (a)oo Vdu u xu V Ans dt x L L 2 o2V x1 L L 22(10) 2 6 10 , 1 400(1 4 ), 6 /12 6 /12o ft du xFor L and V x with x in feet s dt At x 0, du/dt 400 ft/s2 (12 g’s); at x L 0.5 ft, du/dt 1200 ft/s2 (37 g’s). Ans. (b) 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: 2 2 x 2 2 y du u uu v (x y x)(2x 1) ( 2xy y)( 2y) a dt x y dv v vu v (x y x)( 2y) ( 2xy y)( 2x 1) a 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 cos40i sin40j. Then the velocity component along a 40 line is 40V ( 2 6 ) (cos 40 sin 40 ) . (b)Ans 40V n i j i j 5.39 units (c) The maximum velocity is [(-2)2 + (-6)2]1/2 = 5.32 units, at an angle in the third quadrant, θ = 180° + arctan(-6/-2) = 180° + 71.6° = .251.6°. Ans. (c) (d) The maximum acceleration is amax . [182 262]1/2 . 31.6 units at .55.3. Ans. (c, d) _______________________________________________________________________ �P4.4 A simple flow model for a two-dimensional converging nozzle is the distribution (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. 0)1( w L yUv L xUu oo Solutions Manual Fluid Mechanics, Fifth Edition 304 (c) Where is the largest resultant acceleration and its numerical value? Solution: The streamlines are in the x-y plane and are found from the velocities: These may be plotted for various values of the dimensionless constant C, as shown: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 0.2 0.4 0.6 0.8 1 x/L y/L C = 1 C = 0.75 C = .5 C = 0.25 The streamlines converge and the velocity increases to the right. Ans.(a) (b) The accelerations are calculated from Eq. (4.2): ).(:sstreamlinetheFinally )]/1)(/ln[(:,)/ln()/1ln( Cancel /)/1( :integrateor aAns constantLxLyorconstLyLLxL U LyU dy LxU dx v dy u dx o oo x/L1 C L y Chapter 4 Differential Relations for a Fluid Particle 305 (c) Find the resultant of ax and ay from Ans.(b) above and introduce y/L from Ans.(a): 2 2 2 2 21 2 /(1 ) , where / .( )x ya a a C x L Ans c We observe that the resultant acceleration increases with x and is greatest at x = L, where its numerical value is (Uo2/L) [4 + C2/4]1/2. P4.5 The velocity field near a stagnation point (see Example 1.10) may be written in the form o o ou v and are constantsU x U y U LL L (a) Show that the acceleration vector is purely radial. (b) For the particular case L 1.5 m, if the acceleration at (x, y) (1 m, 1 m) is 25 m/s2, what is the value of Uo? Solution: (a) For two-dimensional steady flow, the acceleration components are 2 o o o o 2 2 o o o o 2 U Udu u u x yu v U U (0) x dt x y L L L L U Udv v v x yu v U (0) U y dt x y L L L L Therefore the resultant 2 2 2 2o o(U /L )(x y ) (U /L (purely radial) (a)Ans. a i j )r (b) For the given resultant acceleration of 25 m/s2 at (x, y) (1 m, 1 m), we obtain 2 2 o o o2 2 2 U Um ra 25 2 m, solve for U . (b) s L (1.5 m) Ans m6.3 s ).()/)(/(0 )1(0)/)](/1([ 2 2 bAns L y L ULULyU y vv x vua L x L ULULxU y uv x uua o ooy o oox Solutions Manual Fluid Mechanics, Fifth Edition 306 P4.6 An incompressible plane flow has the velocity components u = 2y, v = 8x, w = 0. (a) Find the acceleration components. (b) Determine if the vector acceleration is radial. (c) Which two streamlines of this flow are straight lines? Solution: (a, b) With no z activity, we can stick to steady two-dimensional formulas: (2 )(0) (8 )(2) 16 (2 )(8) (8 )(0) 16 .( ) 16 16 16( ) 16 , yes. .( ) x y x y u ua u v y x x x y v va u v y x y Ans a x y a a x y x y Ans b a i j i j i j r Radial (c) Find the streamlines in the manner suggested in Chapter 1, Eq. (1.41): 2 2 , or : , 2 8 2 8 Integrate : 2 8 , or : 4 dx dy dx dy y dy x dx u v y x ydy x dx y x const The streamlines are all curved except when the constant of integration is zero, for which there are two straight streamlines: y = 2 x . Ans.(c) Chapter 4 Differential Relations for a Fluid Particle 307 P4.7 Consider a sphere of radius R immersed in a uniform stream Uo, as shown in Fig. P4.7. According to the theory of Chap. 8, the fluid velocity along streamline AB is given by 3 o 3u 1 RU x V i i Fig. P4.7 Find (a) the position of maximum fluid acceleration along AB and (b) the time required for a fluid particle to travel from A to B. Note that x is negative along line AB. Solution: (a) Along this streamline, the fluid acceleration is one-dimensional: 3 3 3 4 3 4 3 7 o o o du uu U (1 R /x )( 3U R /x ) 3U R (x R x ) for x R dt x The maximum occurs where d(ax)/dx 0, or at x –(7R3/4)1/3 –1.205R Ans. (a) (b) The time required to move along this path from A to B is computedfrom R t 3 3 o o3 3 4R 0 dx dxu U (1 R /x ), or: U dt, dt 1 R /x R 4R 2 1 o 2 2 R (x R) R 2x Ror: U t x ln tan 6 x Rx R 3 R 3 It takes an infinite time to actually reach the stagnation point, where the velocity is zero. Ans. (b) P4.8 When a valve is opened, fluid flows in the expansion duct of Fig. P4.8 according to the approximation 1 tanh 2 x UtU L L V i 308 Solutions Manual Fluid Mechanics, Fifth Edition Find (a) the fluid acceleration at (x, t) (L, L/U) and (b) the time for which the fluid acceleration at x L is zero. Why does the fluid acceleration become negative after condition (b)? Fig. P4.8 Solution: This is a one-dimensional unsteady flow. The acceleration is 2 x u u x U Ut x U Uta u U 1 sech U 1 tanh t x 2L L L 2L 2L L 2 2 2U x Ut 1 Ut(1 )[sech tanh ] L 2L L 2 L At (x, t) (L, L/U), ax (U2/L)(1/2)[sech2(1) – 0.5tanh2 (1)] 0.0650 U2/L Ans. (a) The acceleration becomes zero when The acceleration starts off positive, then goes through zero and turns negative as the negative convective acceleration overtakes the decaying positive local acceleration. P4.9 An idealized incompressible flow has the proposed three-dimensional velocity distribution V 4xy2i f(y)j – zy2k Chapter 4 Differential Relations for a Fluid Particle 309 Find the appropriate form of the function f(y) which satisfies the continuity relation. Solution: Simply substitute the given velocity components into the incompressible continuity equation: 2 2 2 2(4 ) ( ) 4 0u v w f dfxy zy y y x y z x y z dy 2 2: 3 . Integrate: ( ) ( 3 )dfor y f y y dy Ans. dy 3y constant P4.10 Consider the simple incompressible plane flow pattern u = U, v = V, and w = 0, where U and V are constants. (a) Convert these velocities into polar coordinate components, vr and v. [HINT: Make a sketch of the velocity components.] (b) Determine whether these new components satisfy the continuity equation in polar coordinates. Solution: This is harder than it looks. Make a sketch of each separate cartesian component: (a) We can resolve each figure into radial and circumferential components. For Figure (a), U has a radial component U cos and a circumferential component (-U sin). For Figure (b), V has a radial component V sin and a circumferential component V cos. Combine these into the result cos sin ; sin cos .( )rv U V v U V Ans a (b) The original (cartesian) distribution, being constant velocity, obviously satisfied continuity. The new version, in polar coordinates, requires some effort. From Eq. (4.9) for incompressible flow, x y x y U V (a) (b) 310 Solutions Manual Fluid Mechanics, Fifth Edition 1 1 1 1( ) ( ) [ ( cos sin )] ( sin cos ) 1 1( cos sin ) ( cos sin ) Yes, satisfied. .( ) rr v v r U V U Vr r r r r r U V U V Ans b r r 0 P4.11 Derive Eq. (4.12b) for cylindrical coordinates by considering the flux of an incompressible fluid in and out of the elemental control volume in Fig. 4.2. Solution: For the differential CV shown, out ind ol dm dm 0t Fig. 4.2 r r z z r z drr d dr dz v r dz d ( v )dr(r dr)dz d v dz dr t 2 r dr dr( v )d dz dr v r d dr ( v ) r d dr 2 z 2 drv r dz d v dz dr v r d dr 0 2 Cancel (d drdz) and higher-order (4th-order) differentials such as (dr d dz dr) and, finally, divide by r to obtain the final result: .Ans r z 1 1rv v v 0 t r r r z Chapter 4 Differential Relations for a Fluid Particle 311 P4.12 Spherical polar coordinates (r, , ) are defined in Fig. P4.12. The cartesian transformations are x r sin cos y r sin sin z r cos Do not show that the cartesian incompressible continuity relation (4.12a) can be transformed to the spherical polar form Fig. P4.12 2 2 1 1 1( ) ( sin ) ( ) 0 sin sinr r r r rr What is the most general form of r when the flow is purely radial, that is, and are zero? Solution: Note to instructors: Do not assign the derivation of this continuity relation, it takes years to achieve, the writer can’t do it successfully. The problem is only meant to acquaint students with spherical coordinates. 2 r2 1If 0, then (r ) 0, so, in general, rr Ans. r 2 1 fcn , r 312 Solutions Manual Fluid Mechanics, Fifth Edition P4.13 For an incompressible plane flow in polar coordinates, we are given 3 2cos sinrv r r Find the appropriate form of circumferential velocity for which continuity is satisfied. Solution: Substitute into continuity, Eq. (4.9), for incompressible flow: 3 2 2 3 2 1 1 1 1( ) ( ) [ ( cos sin )] , 1or : 4 cos 3 sin Integrate : 4 sin 3 cos ( ) . r v r v v r r r r r r r r r v r r r v r r f r Ans We can’t determine the form of the “constant of integration” f(r) without further information. P4.14 For incompressible polar-coordinate flow, what is the most general form of a purely circulatory motion, (r, , t) and r 0, which satisfies continuity? Solution: If vr 0, the plane polar coordinate continuity equation reduces to: (1/r) ∂vθ/∂θ = 0 , or: vθ = fcn(r, t) Ans. P4.15 What is the most general form of a purely radial polar-coordinate incompressible-flow pattern, r r(r, , t) and 0, which satisfies continuity? Solution: If v 0, the plane polar coordinate continuity equation reduces to: r 1 (r ) 0, or: only r r v Ans. r 1 fcn r v Chapter 4 Differential Relations for a Fluid Particle 313 ________________________________________________________________________ P4.16 Consider the plane polar coordinate velocity distribution 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? Solution: (a) Evaluate the incompressible continuity equation (4.12b) in polar coordinates: 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: -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 O 0 zr vr Kv r Cv ).(000)(1)(1)(1)(1 aAns r K rr Cr rr v r vr rr r 314 Solutions Manual 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. P4.17 An excellent approximation for the two-dimensional incompressiblelaminar boundary layer on the flat surface in Fig. P4.17 is 3 4 1/ 2 3 42 2 for , where , constant( ) y y yu U y C x C (a) Assuming a no-slip condition at the wall, find an expression for the velocity component v(x, y) for y . (b) Then find the maximum value of v at the station x = 1 m, for the particular case of airflow, when U = 3 m/s and = 1.1 cm. Solution: (a) With u known, use the two-dimensional equation of continuity to find v: 3 4 2 4 5 3 4 2 4 5 2 4 5 2 4 5 0 2 6 4( ) , 3 2 3 2: 2 ( ) 2 ( ) .( ) 2 4 5 y v u y d y d y dU y x dx dx dx d y y y d y y yor v U dy U Ans a dx dx (b) First evaluate C from the given data at x = 1 m: 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 0.011 (1 ) , hence 0.011 1 1Or, alternately , ( ) 2 2 2 m C m C m d C x x dx xx Substitute this into Ans.(a) above and note that v rises monotonically with y to a maximum at the outer edge of the boundary layer, y = . The maximum velocity v is thus Fig. P4.17 Chapter 4 Differential Relations for a Fluid Particle 315 max 1 3 2 0.011 32 ( ) 2(3 )[ ]( ) .( ) 2 4 5 2(1 ) 20 d m mv U Ans b dx s m m0.0050 s This is slightly smaller than the exact value of vmax from laminar boundary theory (Chap. 7). P4.18 A piston compresses gas in a cylinder by moving at constant speed V, as in Fig. P4.18. Let the gas density and length at t 0 be o and Lo, respectively. Let the gas velocity vary linearly from u V at the piston face to u 0 at x L. If the gas density varies only with time, find an expression for (t). V = constant x = 0 x = L(t) p(t) x Fig. P4.18 Solution: The one-dimensional unsteady continuity equation reduces to o d u x( u) , where u V 1 , L L Vt, (t) only t x dt x L o t oo u V d dtEnter and separate variables: V x L L Vt o o oThe solution is ln( / ) ln(1 Vt/L ), or: .Ans o o L L Vt P4.19 An incompressible flow field has the cylindrical velocity components Cr, z K(R2 – r2), r 0, where C and K are constants and r R, z L. Does this flow satisfy continuity? What might it represent physically? Solution: We check the incompressible continuity relation in cylindrical coordinates: z r v1 1 v(rv ) 0 0 0 0 . r r r z Ans satisfied identically This flow also satisfies (cylindrical) momentum and could represent laminar flow inside a tube of radius R whose outer wall (r R) is rotating at uniform angular velocity. 316 Solutions Manual Fluid Mechanics, Fifth Edition P4.20 A two-dimensional incompressible velocity field has u K(1 – e–ay), for x L and 0 y . What is the most general form of v(x, y) for which continuity is satisfied and v vo at y 0? What are the proper dimensions for constants K and a? Solution: We can find the appropriate velocity v from two-dimensional continuity: ayu [K(1 e )] 0, or: fcn(x) only y x x v v 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–1}. Ans. P4.21 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 To avoid compressibility corrections, we require (Eq. 4.18) that Ma 0.3 or, in this case, the highest velocity (at the exit) should be Ve 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 P4.22 In an axisymmetric flow, nothing varies with ; the only nonzero velocities are vr and vz (see Fig. 4.2 of the text). If the flow is steady and incompressible and vz = Bz, where B is constant, find the most general form of vr which satisfies continuity. Chapter 4 Differential Relations for a Fluid Particle 317 Solution: With no variation and no v, the equation of continuity (4.9) becomes 2 1 1( ) 0 ( ) ( ) , or : ( ) ; Integrate : ( ) 2 ( )Finally, . 2 z r r r r r v r v r v Bz r r z r r z Br v B r r v r f z r B f zv r Ans r The “function of integration”, f(z), is arbitrary, at least until boundary conditions are set. __________________________________________________________________________ P4.23 A tank volume V contains gas at conditions (o, po, To). 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 “blowdown” 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 C p A, where C constant dt dt o p(t) t o o p o CRT Ap dpIntroduce and separate variables: dt RT p The solution is an exponential decay of tank density: p po exp(–CRToAt/V ). Ans. P4.24 For incompressible laminar flow between parallel plates (see Fig. 4.12b), the flow is two-dimensional (v 0) if the walls are porous. A special case solution is 2 2( ) ( )u A Bx h y , where A and B are constants. (a) Find a general formula for velocity v if v = 0 at y = 0. (b) What is the value of the constant B if v = vw at y = +h? Solution: (a) Use the equation of continuity to find the velocity v: 318 Solutions Manual Fluid Mechanics, Fifth Edition 2 2 3 2 2 2 3 2 ( )( ) Integrate : ( ) ( ) ( ) 3 If 0 0, then ( ) 0. ( ) .( ) 3 v u B h y y x yv B h y dy B h y f x yv at y f x v B h y Ans a (b) Just simply introduce this boundary condition into the answer to part (a): 3 3 3 3 ( ) ( ) , .( ) 3 2 w w vhv y h v B h hence B Ans b h P4.25 An incompressible flow in polar coordinates is given by 2v cos 1r bK r 2v sin 1 bK r 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 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 = b, as shown above. The pattern represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8). Chapter 4 Differential Relations for a Fluid Particle319 P4.26 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’s frictionless momentum equation (4.36) in streamline coordinates becomes / ( / ) (1/ ) ( / ) sV t V V s p s g (1) 2 1 n V pV g t R n (2) Fig. P4.26 Further show that the integral of Eq. (1) with respect to s is none other than our old friend Bernoulli’s equation (3.76). Solution: This is 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 Integrate from 1 to 2: (Bernoulli) .Ans 2 22 2 2 1 2 1 1 1 V V dpV ds g z z 0 t 2 P4.27 A frictionless, incompressible steady-flow field is given by V 2xyi – y2j in arbitrary units. Let the density be o constant and neglect gravity. Find an expression for the pressure gradient in the x direction. 320 Solutions Manual Fluid Mechanics, Fifth Edition Solution: For this (gravity-free) velocity, the momentum equation is 2 ou v p, or: [(2xy)(2y ) ( y )(2x 2y )] px y V V i i j 2 3 oSolve for p (2xy 2y ), or: .Ans 2opi j 2xyx P4.28 Consider the incompressible flow field of Prob. P4.6, with velocity components u = 2y, v = 8x, w = 0. Neglect gravity and assume constant viscosity. (a) Determine whether this flow satisfies the Navier-Stokes equations. (b) If so, find the pressure distribution p(x, y) if the pressure at the origin is po. Solution: In Prob. P4.6 we found the accelerations, so we can proceed to Navier-Stokes: 2 2 ( ) [0 (8 )(2)] 0 0; 16 ( ) [(2 )(8) 0] 0 0; 16 x y u u p p pu v x g u x x y x x x v v p p pu v y g v y x y y y y Noting that 2 /( ) 0 in both cases, we conclude .( )p x y Ans a Yes, satisfies Navier - Stokes. (b) The pressure gradients are simple, so we may easily integrate: 2 2 2 2 , or : 16 16 8 ( ) If (0,0) , 8 ( ) .( )o o p pdp dx dy p x dx y dy x y const x y p p then p p x y Ans b This is an exact solution, but it is not Bernoulli’s equation. The flow is rotational. P4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid with the velocity field u –2xy, v y2 – 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. Chapter 4 Differential Relations for a Fluid Particle 321 Solution: Evaluate and check the incompressible continuity equation: ( )0 2 2 0 0 au v w y y x y z 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 uu v w or x y z x x y z 2 2 2 3[ 2 ( 2 ) ( )( 2 )] (0 0 0), : 2 ( )p pxy y y x x or xy x x x and, similarly for the y-momentum relation, 2 2 2 2 2 2 , : v v v p v v vu v w or x y z y x y z 2 2 2 3[ 2 ( 2 ) ( )(2 )] ( 2 2 0), : = 2 ( )p pxy x y x y or x y y y y The two gradients p/x and p/y may be integrated to find p(x, y): 2 2 4 2 ( ), : 2 4y Const p x y xp dx f y then differentiate x 2 2 3 3 42 ( ) 2 ( ), 2 , : ( ) 2 p df dfx y x y y whence y or f y y C y dy dy 2 2 4 4: (2 ) ( , ) (0,0), : 2 a Thus p x y x y C p at x y or aC p Finally, the pressure field for this flow is given by (b)Ans. 2 2 4 4ap p (2x y x y ) P4.30 For the velocity distribution of Prob. P4.4, determine if (a) the equation of continuity and (b) the Navier-Stokes equation are satisfied. (c) If the latter is true, find the pressure distribution p(x,y) when the pressure at the origin equals po. Neglect gravity. Solution: Recall that we were given u = Uo(1+x/L) and v = -Uo y/L. (a) Test continuity: 322 Solutions Manual Fluid Mechanics, Fifth Edition (b) Now substitute these velocities into the x- and y- Navier-Stokes equations: Solve for the two pressure gradients and cross-differentiate to see if they agree: Thus, before finding p(x,y), we know this is an exact solution to Navier-Stokes. Ans.(b) (c) Integrate the two pressure gradients to find the pressure distribution: This is the same as Bernoulli’s equation, but that is a bit hard to see. P4.31 According to potential theory (Chap. 8) for the flow approaching a rounded two- dimensional body, as in Fig. P4.31, the velocity approaching the stagnation point is given by u U(1 – a2/x2), where a is the nose radius and U is the velocity far upstream. ).(.satisfiedOK,0)()]1([ aAns L U L U L yU yL xU xy v x u oo oo 011))(()0()1( 011)0)(()1( 2 2 y pv y p L U L yU L xU y vv x vu x pu x p L yU L U L xU y uv x uu o oo o o o bothfor0Check)1( 222 yx p L y L U y p L x L U x p oo ).() 22 ( 2 ,Then;)() 2 ( 222 2222 cAnsp L y L xx L Up const L y L Uf dy df y pyf L xx L Udx x pp o o oo Chapter 4 Differential Relations for a Fluid Particle 323 Compute the value and position of the maximum viscous normal stress along this streamline. Is this also the position Fig. P4.31 of maximum fluid deceleration? Evaluate the maximum viscous normal stress if the fluid is SAE 30 oil at 20°C, with U 2 m/s and a 6 cm. Solution: (a) Along this line of symmetry the convective deceleration is one-dimensional: 2 2 2 4 2 x 2 3 3 5 u a 2a a aa u U 1 U 2U x x x x x xdaThis has a maximum deceleration at 0, or at (5/3) a . (a) dx Ans x 1.29a The value of maximum deceleration at this point is 2x,maxa 0.372U /a. (b) The viscous normal stress along this line is given by 2 xx 3 u 2a U2 2 with a maximum . (b) x x Ans max 4 U at x a a Thus maximum stress does not occur at the same position as maximum deceleration. For SAE 30 oil at 20°C, we obtain the numerical result max3 kg kg 4(0.29)(2.0)SAE 30 oil, 917 , 0.29 , . (b) m s (0.06 m)m Ans 39 Pa P4.32 The answer to Prob. 4.14 is f(r) only. Do not reveal this to your friends if they are still working on Prob. 4.14. Show that this flow field is an exact solution to the Navier-Stokes equations (4.38) for only two special cases of the function f(r). Neglect gravity. Interpret these two cases physically. 324 Solutions Manual Fluid Mechanics, Fifth Edition Solution: Given v f(r) and vr vz 0, we need only satisfy the -momentum relation: 2 r 2 2 2 v v v v v v1 p 1 1v r , r r r r r r r r 2 1 d df for: (0 0) 0 r 0 , or: r dr dr r 2 1 1f f f 0 r r This is the ‘equidimensional’ ODE and always has a solution in the form of a power-law, f Crn. The two relevant solutions for these particular coefficients are n = 1: f1 C1r (solid-body rotation); f2 C2/r (irrotational vortex) Ans. P4.33 Consider incompressible flow at a volume rate Q toward a drain at the vertex of a 45 wedge of width b, as in Fig. P4.33. Neglect gravity and friction and assume purely radial inflow. (a) Find an expression for vr(r). (b) Show that the viscous term in the r-momentum equation is zero. (c) Find the pressure distribution p(r) if p = po at r = R. Solution: (a) Assume one-dimensional, steady, radial inflow. Then, at any radius r, 4, .( ) ( / 4)r Q Q C Qv where C Ans a area r b r b The velocity is negative because the flow is inward. (b) The r-momentum equation is not written out in Chapter 4; it is Eq. (D.5) of Appendix D. The viscous term is 2 2 2 2 2 2 3 3 3 2 1 2( ) [ ( ) ] 1 ( / ) 1[ ( ( )) 0] [ ( ) ] ( ) .( ) r r r r v vv v v v r r r rr r r r C C r C C C Cr Ans b r r r r r r rr r r r 0 r Q = /4 Drain Fig. P4.33 Chapter 4 Differential Relations for a Fluid Particle 325 (c) With the viscous term zero, the r-momentum equation reduces to 2 2 3 2 2 2 1 1 12 2 2 2 2 2 2 ( ) ( )( ) , or : Integrate : ; , , 2 2 2 4Finally, ( ) , .( ) 2 r r o o o v CC C p pv r r r rr r C C Cp C at r R p p C C p r R R QC Cp p where C Ans c bR r The two terms in parentheses are the velocities-squared at r = R and r = r, respectively. In other words, it integrates to Bernoulli’s equation because the viscous term is zero (irrotational flow). P4.34 A proposed three-dimensional incompressible flow field has the following vector form: V Kxi Kyj – 2Kzk (a) Determine if this field is a valid solution to continuity and Navier-Stokes. (b) If g –gk, find the pressure field p(x, y, z). (c) Is the flow irrotational? Solution: (a) Substitute this field into the three-dimensional incompressible continuity equation: . ( ) ( ) ( 2 ) 2 0 (a) u v w Kx Ky Kz x y z x y z K K K Ans. Yes, satisfied (b) Substitute into the full incompressible Navier-Stokes equation (4.38). The laborious results are: 2x momentum: ( 0 0) (0 0 0)pK x x 2y momentum: (0 0) (0 0 0)pK y y z momentum: {0 0 ( 2 )( 2 )} ( ) (0 0 0)pKz K g z Integrate each equation for the pressure and collect terms. The result is 326 Solutions Manual Fluid Mechanics, Fifth Edition p p(0,0,0) – gz – (/2)K2(x2 y2 4z2) Ans. (b) Note that the last term is identical to (/2)(u2 v2 w2), in other words, Bernoulli’s equation. (c) For irrotational flow, the curl of the velocity field must be zero: V i(0 – 0) j(0 – 0) k(0 – 0) 0 Yes, irrotational. Ans. (c) P4.35 From the Navier-Stokes equations for incompressible flow in polar coordinates (App. E for cylindrical coordinates), find the most general case of purely circulating motion (r), r z 0, for flow with no slip between two fixed concentric cylinders, as in Fig. P4.35. Solution: The preliminary work for this r No slip r = b r = a υ (r)θ Fig. P4.35 problem is identical to Prob. 4.32 on an earlier page. That is, there are two possible solutions for purely circulating motion (r), hence 2 1 1 2 1 2 Cv C r , subject to v (a) 0 C a C /a and v (b) 0 C b C /b r This requires C1 C2 0, or v 0 (no steady motion possible between fixed walls) Ans. P4.36 A constant-thickness film of viscous liquid flows in laminar motion down a plate inclined at angle , as in Fig. P4.36. The velocity profile is u Cy(2h – y) v w 0 Find the constant C in terms of the specific weight and viscosity and the angle . Find the volume flux Q per unit width in terms of these parameters. (y) g y x h θ Fig. P4.36 Solution: There is atmospheric pressure all along the surface at y h, hence p/x 0. The x-momentum equation can easily be evaluated from the known velocity profile: 2 x u u pu v g u, or: 0 0 gsin + ( 2C) x y x Solve for . (a)Ansg sinC 2 The flow rate per unit width is found by integrating the velocity profile and using C: h h 3 0 0 2Q udy Cy(2h y)dy Ch per unit width . (b) 3 Ans 3gh sin3 P4.37 A viscous liquid of constant density and viscosity falls due to gravity between two parallel plates a distance 2h apart, as in the figure. The flow is fully developed, that is, w w(x) only. There are no pressure gradients, only gravity. Set up and solve the Navier-Stokes equation for the velocity profile w(x). Solution: Only the z-component of Navier- Stokes is relevant: Fig. P4.37 2 20 , : , ( ) ( ) 0 ( ) dw d w gg or w w h w h no-slip dt dx The solution is very similar to Eqs. (4.142) to (4.143) of the text: .w Ans 2 2g (h x ) 2 P4.38 Show that the incompressible flow distribution, in cylindrical coordinates, 00 znr vrCvv 328 Solutions Manual Fluid Mechanics, Fifth Edition where C is a constant, (a) satisfies the Navier-Stokes equation for only two values of n. Neglect gravity. (b) Knowing that p = p(r) only, find the pressure distribution for each case, assuming that the pressure at r = R is po. What might these two cases represent? Solution: (a) The important direction here is the -momentum equation, Eq. (D.6): Cancel C and and rn-2. These terms equal zero only if n2 = 1, or n = 1. Ans.(a). (b) Find the respective pressure distributions for n = 1 and n = -1. Use Eq. (D.5), which reduces simply to p/r = v2/r. Try this for each distribution, n = 1: Case 1, v = Cr, is solid-body rotation. Case 2, v = C/r, is an irrotational potential vortex. 0)(])(1[ :or,]))((1[)0(0000 :or,)2(1 r 1 )( 22221 22 2 22 2 nnnn n n r CrrCnCrrnCr rr r CrCr r r rrr vv v rr vvp r vvv t v V ).()11( 2 :or;)/(:1,2 ).()( 2 :or;)(:1,1 222 222 1 22 222 bAns rR Cppdr r rCdpnCase bAnsRrCppdr r rCdpnCase o r R p p o r R p p o o Chapter 4 Differential Relations for a Fluid Particle 329 P4.39 Reconsider the angular- momentum balance of Fig. 4.5 by adding a concentrated body couple Cz about the z axis [6]. Determine a relation between the body couple and shear stress for equilibrium. What are the proper dimensions for Cz? (Body couples are important in continuous media with microstructure, such as granular materials.) Solution: The couple Cz has to be per unit volume to make physical sense in Eq. (4.39): Fig. 4.5 2 xy yx 2 2 xy yx z 2 1 1 1 ddx dy dx dy dz C dx dy dz dx dy dz(dx dy ) 2 x 2 y 12 dt Reduce to third order termsand cancel (dx dy dz): yx xy Cz Ans. The concentrated couple allows the stress tensor to have unsymmetrical shear stress terms. P4.40 For pressure-driven laminar flow between parallel plates (see Fig. 4.12b), the velocity components are u = U(1– y2/ h2), v = 0, and w = 0, where U is the centerline velocity. In the spirit of Ex. 4.6, find the temperature distribution T(y) for a constant wall temperature Tw. Solution: There are no variations with x or z, so the energy equation (4.53) reduces to 2 2 2 2 2 2 2 22 3 2 2 12 2 4 4 0 ( ) ( ) , 4 42: ( ) ( ) ; Integrate : ( ) 3 p T T u d T duc u k k x y dyy dy U Ud T Uy dT yor y C k dydy h k h k h 330 Solutions Manual Fluid Mechanics, Fifth Edition The condition T = Tw at h is equivalent to dT/dy = 0 at y = 0. Thus C1 = 0. Integrate again: 2 2 24 4 2 2 24 4 4 4( ) ; at : ( ) , 12 12 3 w wU U Uy hT C y h T T C C T kk h k h The final solution for T(y) is, like Ex. 4.6, a quartic polynomial: 2 4 4( ) (1 ) .3w U yT y T Ans k h P4.41 As mentioned in Sec. 4.10, the velocity profile for laminar flow between two plates, as in Fig. P4.40, is max 2 4 ( ) 0u y h yu w h If the wall temperature is Tw at both walls, use Fig. P4.41 the incompressible-flow energy equation (4.75) to solve for the temperature distribution T(y) between the walls for steady flow. Solution: Assume T T(y) and use the energy equation with the known u(y): 2 22 2 max p p2 2 2 4udT d T du d Tc k , or: c (0) k (h 2y) , or: dt dydy dy h 2 22 3 2 2 2 2max max 12 4 4 16 u 16 ud T dT 4y(h 4hy 4y ), Integrate: h y 2hy C dy 3dy kh kh Chapter 4 Differential Relations for a Fluid Particle 331 Before integrating again, note that dT/dy 0 at y h/2 (the symmetry condition), so C1 –h3/6. Now integrate once more: 2 2 3 4 2max 1 24 16 u y y yT h 2h C y C 2 3 3kh If T Tw at y 0 and at y h, then C2 Tw. The final solution is: .Ans 2 2 3 4 max w 2 3 4 8 u y y 4y 2yT T k 3h h 3h 3h This is exactly the same solution as Problem P4.40 above, except that, here, the coordinate y is measured from the boo tom wall rather than the centerline. P4.42 Suppose that we wish to analyze the rotating, partly-full cylinder of Fig. 2.23 as a spin-up problem, starting from rest and continuing until solid-body-rotation is achieved. What are the appropriate boundary and initial conditions for this problem? Solution: Let V V(r, z, t). The initial condition is: V(r, z, 0) 0. The boundary conditions are Along the side walls: v(R, z, t) R, vr(R, z, t) 0, vz(R, z, t) 0. At the bottom, z 0: v(r, 0, t) r, vr(r, 0, t) 0, vz(r, 0, t) 0. At the free surface, z : p patm, rz z 0. P4.43 For the draining liquid film of Fig. P4.36, what are the appropriate boundary conditions (a) at the bottom y 0 and (b) at the surface y h? Fig. P4.36 Solution: The physically realistic conditions at the upper and lower surfaces are: (a) at the bottom, y 0, no-slip: u(0) 0 Ans. (a) u(b) At the surface, y h, no shear stress, 0, or . (b) y Ans u ( ) 0 y h 332 Solutions Manual Fluid Mechanics, Fifth Edition P4.44 Suppose that we wish to analyze the sudden pipe-expansion flow of Fig. P3.59, using the full continuity and Navier-Stokes equations. What are the proper boundary conditions to handle this problem? Solution: First, at all walls, one would impose the no-slip condition: ur uz 0 at all solid surfaces: at r r1 in the small pipe, at r r2 in the large pipe, and also on the flat- faced surface between the two. Fig. P3.59 Second, at some position upstream in the small pipe, the complete velocity distribution must be known: u1 u1(r) at z z1. [Possibly the paraboloid of Prob. 4.34.] Third, to be strictly correct, at some position downstream in the large pipe, the complete velocity distribution must be known: u2 u2(r) at z z2. In numerical (computer) studies, this is often simplified by using a “free outflow” condition, u/ z 0. Finally, the pressure must be specified at either the inlet or the outlet section of the flow, usually at the upstream section: p p1(r) at z z1. P4.45 For the sluice gate problem of Example 3.10, list all the boundary conditions needed to solve this flow exactly by, say, Computational Fluid Dynamics (CFD). 1 2 3 3 2 4 Chapter 4 Differential Relations for a Fluid Particle 333 Solution: There are four different kinds of boundary conditions needed, as labeled. (1) Known velocity V1 upstream, and of course the depth y1 must be known. (2) Known pressure patm at both the upstream and downstream free surfaces. (3) No-slip (V = 0) all along the bottom and on the gate inner wall. (4) The downstream flow is complicated because we don’t know V2 or y2 and therefore cannot specify them. What CFD modelers do is to have an adjustable upper boundary and specify that the exit flow is “smooth”, or “zero gradient”, that is, V/x = 0. P4.46 Fluid from a large reservoir at temperature To flows into a circular pipe of radius R. The pipe walls are wound with an electric-resistance coil which delivers heat to the fluid at a rate qw (energy per unit wall area). If we wish to analyze this problem by using the full continuity, Navier-Stokes, and energy equations, what are the proper boundary conditions for the analysis? Solution: Letting z 0 be the pipe entrance, we can state inlet conditions: typically uz(r, 0) U (a uniform inlet profile), ur(r, 0) 0, and T(r, 0) To, also uniform. At the wall, r R, the no-slip and known-heat-flux conditions hold: uz(R, z) ur(R, z) 0 and k( T/ r) qw at (R, z) (assuming that qw is positive for heat flow in). At the exit, z L, we would probably assume ‘free outflow’: uz/ z T/ z 0. Finally, we would need to know the pressure at one point, probably the inlet, z 0. P4.47 Given the incompressible flow V 3yi 2xj. Does this flow satisfy continuity? If so, find the stream function (x, y) and plot a few streamlines, with arrows. 334 Solutions Manual Fluid Mechanics, Fifth Edition Solution: With u 3y and v 2x, we may check u/ x v/ y 0 0 0, OK. Find the streamlines from u / y 3y and v –/x 2x. Integrate to find .Ans 2 23 2 y x Set 0, 1, 2, etc. and plot some streamlines at right: flow around corners of half-angles 39 and 51. 39° 51° x y Fig. P4.47 P4.48 Consider the following two-dimensional incompressible flow, which clearly satisfies continuity: u Uo constant, v Vo constant Find the stream function (r, ) of this flow, that is, using polar coordinates. Solution: In cartesian coordinates the stream function is quite easy: u / y Uo and v –/x Vo or: Uoy – Vox constant But, in polar coordinates, y rsin and x rcos. Therefore the desired result is (r, ) Uor sin – Vor cos constant Ans. P4.49 Investigate the stream function K(x2 – y2), K constant. Plot the streamlines in the full xy plane, find any stagnation points, and interpret what the flow could represent. Solution: The velocities are given by u 2Ky; v 2Kx y x This is also stagnation flow, with the stream-lines turned 45 from Prob. 4.48. Fig. P4.49 P4.50 In 1851, George Stokes (of Navier-Stokes fame) solved the problem of steady incompressible low-Reynolds-number flow past a sphere, using spherical polar coordinates (r, ) – [Ref. 5, page 168]. In these coordinates, the equation of continuity is 2( sin ) ( sin ) 0rr v r vr (a) Does a stream function exist for these coordinates? (b) If so, find its form. Solution: Two velocity components and two continuity terms. Yes, exists! Ans.(a) (b) The stream function should be defined such that continuity takes the form 2 2 2 0 , or : 1 1; .( ) sin sin r r r v v Ans b r r r P4.51 The velocity profile for incompressible pressure-driven laminar flow between parallel plates (see Fig. 4.12b) has the form u = C(h2 – y2), where C is a constant. (a) Determine if a stream function exists. (b) If so, determine a formula for the stream function, Solution: (a) A stream function exists, for a single velocity component u, if u/x = 0, which it certainly is, since u is a function only of y. Yes, exists. Ans.(a) (b) Finding the stream function is just a matter of direct integration: 3 2 2 2 0 , hence is a function only of ( ) ; Integrate : ( ) .( ) 3 v y x yu C h y C h y constant Ans b y P4.52 A two-dimensional, incompressible, frictionless fluid is guided by wedge-shaped walls into a small slot at the origin, as in Fig. P4.52. The width into the paper is b, and the volume flow rate is Q. At any given distance r from the slot, the flow is radial inward, with constant velocity. Find an expression for the polar-coordinate stream function of this flow. Fig. P4.52 Solution: We can find velocity from continuity: r Q Q 1v from Eq. (4.101). Then A ( /4)rb r .Ans 4Q constant b θπ This is equivalent to the stream function for a line sink, Eq. (4.131). P4.53 For the fully developed laminar-pipe-flow solution of Eq. (4.137), find the axisymmetric stream function (r, z). Use this result to determine the average velocity V Q/A in the pipe as a ratio of umax. Solution: The given velocity distribution, vz umax(1 – r2/R2), vr 0, satisfies continuity, so a stream function does exist and is found as follows: 2 4 2 2 z max max 2 1 r rv u (1 r /R ) , solve for u f(z), now use in r r 2 4R r 1 df v 0 0 , thus f(z) const, . r z dz Ans 2 4 max 2 r ru 2 4R We can find the flow rate and average velocity from the text for polar coordinates: 2 4 2 1-2 2 1 0-R max max max2 R RQ 2 ( ), or: Q 2 u u (0 0) R u 2 24R Chapter 4 Differential Relations for a Fluid Particle 337 2 2 avg pipe maxThen V Q/A [( /2)R u /( R )] .Ans max1 u2 338 Solutions Manual Fluid Mechanics, Fifth Edition P4.54 An incompressible stream function is defined by 2 3 2( , ) (3 ) Ux y x y y L where U and L are (positive) constants. Where in this chapter are the streamlines of this flow plotted? Use this stream function to find the volume flow Q passing through the Fig. E4.7 rectangular surface whose corners are defined by (x, y, z) (2L, 0, 0), (2L, 0, b), (0, L, b), and (0, L, 0). Show the direction of Q. Solution: This flow, with velocities u / y 3U/L2(x2 – y2), and v –/ x –6xyU/L2, is identical to Example 4.7 of the text, with “a” 3U/L2. The streamlines are plotted in Fig. E4.7. The volume flow per unit width between the points (2L, 0) and (0, L) is 2 3 2 2 U UQ/b (2L, 0) (0, L) (0 0) [3(0) L L ] UL, or: . L L Ans Q ULb Since at the lower point (2L, 0) is larger than at the upper point (0, L), the flow through this diagonal plane is to the left, as per Fig. 4.9 of the text. P4.55 For the incompressible plane flow of Prob. P4.6, with velocity components u = 2y, v = 8x, w = 0, determine (a) if a stream function exists. (b) If so, determine the form of the stream function, and (c) plot a few representative streamlines. Solution: (a) Check to see is two-dimensional continuity is satisfied: (2 ) (8 ) 0 0 0 .( )u v y x Ans a x y x y Yes, ψ exists. (b) Find the stream function by relating velocities to derivatives of : 2 22 ; 8 ; Integrate : 4 .( )u y v x y x constant Ans b y x Chapter 4 Differential Relations for a Fluid Particle 339 (c) Plot a few streamlines, that is, plot y2 = 4x2 + C for various C. Here are the results: 0 1 2 3 4 5 6 -3 -2 -1 0 1 2 3 y2 = 4x2 + C We are showing only the upper half plane, which is the mirror image of the lower half. P4.56 Investigate the velocity potential = Kxy, K = constant. Sketch the potential lines in the full xy plane, find any stagnation points, and sketch in by eye the orthogonal streamlines. What could the flow represent? Solution: The potential lines, constant, are hyperbolas, as shown. The streamlines, Fig. P4.56 sketched in as normal to the lines, are also hyperbolas. The pattern represents plane stagnation flow (Prob. 4.49) turned at 45. P4.57 A two-dimensional incompressible flow field is defined by the velocity components 2 2x y yu V v V L L L 340 Solutions Manual Fluid Mechanics, Fifth Edition where V and L are constants. If they exist, find the stream function and velocity potential. Solution: First check continuity and irrotationality: 2 2 0 20 0 u v V V x y L L v u Vx x y L exists does not exist ; V k k To find the stream function , use the definitions of u and v and integrate: 2 2 , 2 ( ) 2 x y xy yu V V f x y L L L L 2 2Evaluate Thus 0 and . Vy df Vyv x L dx L df Ans dx xy yV const L L 22 P4.58 Show that the incompressible velocity potential in plane polar coordinates (r,) is such that 1 r r r Finally show that as defined satisfies Laplace’s equation in polar coordinates for incompressible flow. Solution: Both of these things are quite true and easy to show from the definition of the gradient vector in polar coordinates. Ans. P4.59 Consider the two-dimensional incompressible velocity potential xy x2 – y2. (a) Is it true that 2 0, and, if so, what does this mean? (b) If it exists, find the stream function (x, y) of this flow. (c) Find the equation of the streamline which passes through (x, y) (2, 1). Solution: (a) First check that 2 0, which means that incompressible continuity is satisfied. 2 2 2 2 2 0 2 2 0x y Yes Chapter 4 Differential Relations for a Fluid Particle 341 (b) Now use to find u and v and then integrate to find . 2 2 , hence 2 ( ) 2 yu y x xy f x x y 2 2 2 , hence ( ) 2 df xv x y y f x const y x dx The final stream function is thus . (b)Ans 2 21 2 2 y x xy const (c) The streamline which passes through (x, y) (2, 1) is found by setting a constant: 2 21 3 5( , ) (2, 1), (1 2 ) 2(2)(1) 4 2 2 2 At x y Thus the proper streamline is . (c)Ans 2 21 52 2 2 y x xy P4.60 Liquid drains from a small hole in a tank, as shown in Fig. P4.60, such that the velocity field set up is given by r 0, z 0, R2/r, where z H is the depth of the water far from the hole. Is this flow pattern rotational or irrotational? Find the depth zc of the water at the radius r R. Solution: From Appendix D, the angular velocity is patm z r z = 0 r = R z = H Fig. P4.60 z 1 1(rv ) (v ) 0 (IRROTATIONAL) r r r Incompressible continuity is valid for this flow, hence Bernoulli’s equation holds at the surface, where p patm, both at infinity and at r R: 2 2 atm r atm r R c 1 1p V gH p V gz 2 2 2 2 r r RIntroduce V 0 and V R to obtain .2C Rz H Ans g 342 Solutions Manual Fluid Mechanics, Fifth Edition P4.61 For the incompressible plane flow of Prob. P4.6, with velocity components u = 2y, v = 8x, w = 0, determine (a) if a velocity potential exists. (b) If so, determine the form of the velocity potential, and (c) plot a few representative potential lines. Solution: (a) A velocity potential exists if the vorticity is zero. Here, for plane flow in (x, y) coordinates, we need only evaluate rotation around the z axis: 2 8 2 6 0 Rotational , does not exist. .( )z z v u Ans a x y (b, c) There is no velocity potential – no plot, no formula. The flow has constant vorticity. P4.62 Show that the linear Couette flow between plates in Fig. 1.6 has a stream function but no velocity potential. Why is this so? Solution: Given u Vy/h, v 0, check continuity: Fig. 1.6 ?u v 0 0 0 Find from x y Satisfied therefore exists . Vyu , v 0 , solve for . h y x Ans 2V y const 2h Now check irrotationality: ? z v u V2 0 0 0! . x y h Ans Rotational, does not exist. Chapter 4 Differential Relations for a Fluid Particle 343 P4.63 Find the two-dimensional velocity potential (r,) for the polar-coordinate flow pattern r Q/r, K/r, where Q and K are constants. Solution: Relate these velocity components to the polar-coordinate definition of : r Q K 1v , v ; solve for . r r r r Ans Q ln(r) K const P4.64 Show that the velocity potential (r, z) in axisymmetric cylindrical coordinates (see Fig. 4.2 of the text) is defined by the formulas: r zr z Further show that for incompressible flow this potential satisfies Laplace’s equation in (r, z) coordinates. Solution: Both of these things are quite true and are easy to show from their definitions. Ans. P4.65 A two-dimensional incompressible flow is defined by 2 2 2 2 Ky Kxu x y x y where K constant. Is this flow irrotational? If so, find its velocity potential, sketch a few potential lines, and interpret the flow pattern. Fig. P4.65 Solution: Evaluate the angular velocity: 2 2 z 2 2 2 2 2 2 2 2 2 2 v u K 2Kx K 2Ky2 . x y x y (x y ) x y (x y ) Ans 0 (Irrotational) Introduce the definition of velocity potential and integrate to get (x, y): 2 2 2 2 Ky Kxu ; v , solve for . x yx y x y Ans 1 yK tan K x The lines are plotted above. They represent a counterclockwise line vortex. 344 Solutions Manual Fluid Mechanics, Fifth Edition P4.66 A plane polar-coordinate velocity potential is defined by cos constK K r Find the stream function for this flow, sketch some streamlines and potential lines, and interpret the flow pattern. Solution: Evaluate the velocities and thence find the stream function: Fig. P4.66 r 2 2 Kcos 1 1 Ksinv ; v , r r r rr r solve .Ans Ksin r The streamlines and potential lines are shown above. This pattern is a line doublet. Chapter 4 Differential Relations for a Fluid Particle 345 P4.67 A stream function for a plane, irrotational, polar-coordinate flow is ln and constC K r C K Find the velocity potential for this flow. Sketch some streamlines and potential lines, and interpret the flow pattern. Solution: If this problem is given early enough (before Section 4.10 of the text), the Fig. 4.14 students will discover this pattern for themselves. It is a line source plus a line vortex, a tornado-like flow, Eq. (4.134) and Fig. 4.14 of the text. Find the velocity potential: r 1 C K 1v ; v , solve . r r r r r r Ans C ln(r) K The streamlines and potential lines are plotted above for negative C (a line sink). P4.68 For the velocity distribution of Prob. P4.4, (a) determine if a velocity potential exists and, if it does, (b) find an expression for (x,y) and sketch the potential line which passes through the point (x, y) = (L/2, L/2). Solution: Recall the given flow, u = Uo(1+x/L) and v = Uo(y/L). (a) Calculate if the flow is irrotational. For plane flow, only one term of the curl(V) is needed: Therefore a velocity potential does exist. Ans.(a) (b) To find , integrate from u and v: 0 )curl(Yes,;0002 V y u x v z 346 Solutions Manual Fluid Mechanics, Fifth Edition For a potential line to pass through (L/2, L/2), we must have /(Uo/L) = [1/2 + {(1/2)2 - (1/2)2}/2] = ½. For convenience let the const = 0. Thus we are to plot this potential line: The result is plotted (red) in the graph below, along with the (blue) line, which has the analytic form = Uo(y + xy/L) = 3UoL/4. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x/L y/L /UoL /UoL 2 2 2 2 2 ; (1 ) ( ) ( ) 2 0 , or : 2 Thus ( )( ) .( ) 2 o o o o o x xu Thus u dx U dx U x f y x L L df y yv U f U constant y dy L L x yxU L const Ans b L L 2 22 22 1 L yx L x LUo Chapter 4 Differential Relations for a Fluid Particle 347 P4.69 A steady, two-dimensional flow has the following polar-coordinate velocity potential: where C and K are constants. Determine the stream function (r, ) for this flow. For extra credit, let C be a velocity scale U, let K = UL, and sketch what the flow might represent. Solution: Write out the and expressions for polar-coordinate velocities: Extra credit: Plot a typical streamline for C = U and K = UL: -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 O All the streamlines are logarithmic spirals coming out from the origin in every direction. rKrC lncos .sin,)0(sin1 )(sinhence,1cos AnsconstantKrChence r KC r v rfKrC rr KC r vr 348 Solutions Manual Fluid Mechanics, Fifth Edition P4.70 A CFD model of steady two-dimensional incompressible flow has printed out the values of stream function (x, y), in m2/s, at each of the four corners of a small 10cm-by-10cm cell, as shown in Fig. P4.70. Use these numbers to estimate the resultant velocity in the center of the cell and its angle with respect to the x axis. Solution: Quick analysis:the values are higher on the top than the bottom, therefore u is to the right. The values are higher on the right than the left, therefore v is down. There are several ways to estimate the center velocities. One simple way is to compute average values of on the sides: x = 1.5 m x = 1.6 m y = 1.0 m y =1.1 m ? V ? = 1.9552 m2/s 2.0206 1.79781.7308 m2/s Fig. P4.70 u v 1.9879 1.9092 1.7643 1.8430 Chapter 4 Differential Relations for a Fluid Particle 349 Then ucenter /y = (1.9879-1.7643 m2/s)/(0.1m) = 2.236 m/s to the right. And vcenter /x = (1.9092-1.8430 m2/s)/(0.1m) = 0.662 m/s down. The resultant and its angle are The values in this problem are in fact taken from an exact solution, V = 2.3315 m/s, = 16.505. P4.71 Consider the following two-dimensional function f(x, y): (a) Under what conditions, if any, on (A,B,C,D) can this function f be a steady, plane-flow velocity potential? (b) If you find a (x, y) to satisfy part (a), also find the associated stream function x, y), if any, for this flow. Solution: (a) If f is to be a plane-flow velocity potential, it must satisfy Laplace’s equation: (b) To find , use to get u and v and work backwards to get the stream function: .) 236.2 662.0(tan;)662.0()236.2( 122 AnsdownV o16.5m/s2.332 V 0where,223 ADxCyxBxAf ).(3ispotentialvelocityThe 0and3026 23 2 aAnsDxyAxA CABifCBxAxf ).(3,Finally ,66 )(3,33 32 3222 bAnsconstyAyxA constf dx dfAxy x Axy y v xfyAyxA y yAxA x u 350 Solutions Manual Fluid Mechanics, Fifth Edition P4.72 Water flows through a two-dimensional narrowing wedge at 9.96 gal/min per meter of width into the paper. If this inward flow is purely radial, find an expression, in SI units, for (a) the stream function, and (b) the velocity potential of the flow. Assume one-dimensional flow. The included angle of the wedge is 45. Solution: The wedge angle equals /4 radians. At any given position r, the inward flow equals We have already been advised that v = 0. (a) Work from radial velocity to stream function: Note that r must be in meters. (b) Work from radial velocity to obtain velocity potential: r Q Drain Fig. P4.72 3 3 4 /( ) ( / 4) 4(9.96 / min/ )(6.309 5 / / )where 4( / ) / 0 00080 r Q bQ Qv A rb r gal m E m s gpm mQ b . s m 30.00080 / 1 ; Solve 0.00080 .( ) r m s mv Ans a r r 30.00080 ; Solve 0.00080 ln( ) .( ) r m s mv r Ans b r r Chapter 4 Differential Relations for a Fluid Particle 351 P4.73 A CFD model of steady two-dimensional incompressible flow has printed out the values of velocity potential (x, y), in m2/s at each of the four corners of a small 10cm-by10cm cell, as shown in Fig. P4.73. Use these numbers to estimate the resultant velocity in the center of the cell and its angle with respect to the x axis. Solution: Quick analysis: the values are lower on the left than the right, therefore u is to the right. The values are lower on the top than the bottom, therefore v is down. There are several ways to estimate the center velocities. One simple way is to compute average values of on the sides: x = 1.5 m x = 1.6 m y = 1.0 m y = 1.1 ? V ? = 4.8338 m2/s 5.0610 5.12364.9038 m2/s Fig. P4.73 352 Solutions Manual Fluid Mechanics, Fifth Edition Then ucenter /x = (5.0923-4.8688 m2/s)/(0.1m) = 2.235 m/s to the right. And vcenter |/y| = (5.0137-4.9474 m2/s)/(0.1m) = 0.663 m/s down. The resultant and its angle are The values in this problem are in fact taken from an exact solution, V = 2.3315 m/s, = 16.505 down. P4.74 Consider the two-dimensional incompressible polar-coordinate velocity potential where B is a constant and L is a constant length scale. (a) What are the dimensions of B? (b) Locate the only stagnation point in this flow field. (c) Prove that a stream function exists and then find the function (r, ). u v 4.9474 m/s 5.0923 5.0137 4.8688 .) 235.2 663.0(tan;)663.0()235.2( 122 AnsV o16.5m/s2.331 V LBrB cos Chapter 4 Differential Relations for a Fluid Particle 353 Solution: (a) To give its correct dimensions of {L2/T}, the constant B must have the dimensions of velocity, or {L/T}. Ans.(a) (b) Calculate velocities in polar coordinates: At first it doesn’t look as if we can find a stagnation point, but indeed there is one: As discussed later in Chap. 8, this is the velocity potential of a Rankine half-body. (c) With the velocities known, check the continuity equation: Continuity is satisfied. Find the stream function from the definition of (r, ): P4.75 Given the following steady axisymmetric stream function: valid in the region 0 r R and 0 z L. (a) What are the dimensions of the constant B? (b) Show whether this flow possesses a velocity potential and, if so, find it. (c) What might this flow represent? [HINT: Examine the axial velocity vz.] r BLB r vB r vr sin1;cos ).(0,0:180, bAns L BLBvvLr r satisfiedYes,0coscos01)(1 r B r Bv r vr rr r ).(lnsin:Integrate sin;cos1 cAnsconstrLBrB r BLB r vB r vr constantsareandwhere,) 2 ( 2 2 4 2 RB R rrB 354 Solutions Manual Fluid Mechanics, Fifth Edition Solution: (a) From the definition of (r, z) in Eqs. (4.105), the dimensions of are {L3/T}. Thus B has velocity dimensions, {B} = {L/T}. Ans.(a) (b) To test for irrotationality, first find the velocity components from Eqs. (4.106): Now evaluate the curl of the velocity, which has only one possible non-zero component. From Appendix D, Eq. (D.11), (c) The interpretation of the flow follows immediately from the velocity components. The velocity profile is a paraboloid of revolution and represents Poiseuille pipe flow, Eq. (4.137). Ans.(c) *P4.76 A two-dimensional incompressible flow has the velocity potential where K and C are constants. In this discussion, avoid the origin, which is a singularity (infinite velocity). (a) Find the sole stagnation point of this flow, which is somewhere in the upper half plane. (b) Prove that a stream function exists and then find (x, y), using the hint that dx/(a2+x2) = (1/a)tan-1(x/a). )1() 2 42( 2 11;01 2 2 2 3 R rB R rrB rrr v zr v zr ).(.existnotdoes,0202 2 bAnsR Br r v z v zr Rotational )ln()( 2222 yxCyxK Chapter 4 Differential Relations for a Fluid Particle 355 Solution: (a) Find the velocity components and see where they both equal zero: For positive K and C, u cannot be zero anywhere except at x = 0. Then v = 0 if (b) First check the velocities to see if continuity is satisfied: The algebra is messy but, indeed, continuity is satisfied, exists. Ans.(b) – part 1. Now integrate the velocity components to find the stream function : P4.77 Outside an inner, intense-activity circle of radius R, a tropical storm can be simulated by a polar-coordinate velocity potential (r, ) = Uo R , where Uo is the windvelocity at radius R. (a) Determine the velocity components outside r = R. (b) If, at R = 2222 22;22 yx CyKy x v yx CxKx x u ).(0:,22 aAns K Cyandxatstagnationor y CKy 0] )( 422[] )( 422[ 222 2 22222 2 22 yx Cy yx CK yx Cx yx CK y v x u ).()(tan22obtaintoIntegrate 2222 1 2222 bAnsconst x yCKxy yx CyKy x vand yx CxKx y u 356 Solutions Manual Fluid Mechanics, Fifth Edition 25 mi, the velocity is 100 mi/h and the pressure 99 kPa, calculate the velocity and pressure at r = 100 mi. Solution: (a) First, convert Uo = 100 mi/h = 44.7 m/s and R = 25 mi = 40,200 m. The velocities are calculated from , as requested in Prob. P4.58: 1( ) 0 ; ( ) .( )or o o U R v U R v U R Ans a r r r Outside the “intense” region, the wind is simulated as a circulating “potential vortex” whose velocity drops off inversely as the radius. (b) The flow is irrotational, otherwise would not exist. Thus Bernoulli’s equation applies outside r = R, with no elevation change at the ocean surface. Take surface air density to be sea-level standard, = 1.225 kg/m3. 44.7At 4 100 , 11.2 4 4 4 o oU R U mr R mi v R s 2 2 1 1 2 2Bernoulli : , or :2 2 p V p V 2 2 2 10099,000 (1.225 / 2)(44.7) (1.225 / 2)(11.2) , Solve .( )mip p Pa Ans b 100,150 The pressure far from the storm is approximately sea-level standard pressure. P4.78 An incompressible, irrotational, two-dimensional flow has the following stream function in polar coordinates: sin( ) , where and are constants.nAr n A n Find an expression for the velocity potential of this flow. Chapter 4 Differential Relations for a Fluid Particle 357 Solution: Use to find the velocity components, then integrate back to find . 1 1 1 cos( ) ; Integrate : cos( ) ( ) 1 1sin( ) [ sin( )] Compare : 0 , and thus cos( ) constant . n n r n n n v n Ar n Ar n f r r dfv n Ar n nAr n r r r d df Ar n Ans d P4.79 Study the combined effect of the two viscous flows in Fig. 4.16. That is, find u(y) when the upper plate moves at speed V and there is also a constant pressure gradient (dp/dx). Is superposition possible? If so, explain why. Plot representative velocity profiles for (a) zero, (b) positive, and (c) negative pressure gradients for the same upper-wall speed V. Fig. 4.16 Solution: The combined solution is 2 2 2 V y h dp yu 1 1 2 h 2 dx h The superposition is quite valid because the convective acceleration is zero, hence what remains is linear: p 2V. Three representative velocity profiles are plotted at right for various (dp/dx). Fig. P4.79 358 Solutions Manual Fluid Mechanics, Fifth Edition P4.80 An oil film drains steadily down the side of a vertical wall, as shown. After an initial development at the top of the wall, the film becomes independent of z and of constant thickness. Assume that w w(x) only that the atmosphere offers no shear resistance to the film. (a) Solve Navier-Stokes for w(x). (b) Suppose that film thickness and [ w/ x] at the wall are measured. Find an expression which relates to this slope [ w/ x]. Solution: First, there is no pressure gradient p/ z because of the constant-pressure atmosphere. The Navier-Stokes z-component is (d2w/dx2) g, and the solution requires w 0 at x 0 and (dw/dx) 0 (no shear at the film edge) at x . The solution is: . (a) NOTE: is negative (down)Ans wgxw x 2 )2 The wall slope is dw/dx / , rearrange [ / | ] . (b)/wall wallg : g dw dx Ans P4.81 Modify the analysis of Fig. 4.17 to find the velocity v when the inner cylinder is fixed and the outer cylinder rotates at angular velocity o. May this solution be added to Eq. (4.146) to represent the flow caused when both inner and outer cylinders rotate? Explain your conclusion. Solution: We apply new boundary condi- tions to Eq. (4.145) of the text: 1 2 i 1 i 2 i v C r C /r; At r r , v 0 C r C /r Fixed Fluid: ρ, μ ro ri υ r Ω θ Fig. 4.17 o o o 1 o 2 oAt r r , v r C r C /r 1 2Solve for C and C . The final result: .Ans i i o o o i i o r/r r /rv r r /r r /r Chapter 4 Differential Relations for a Fluid Particle 359 This solution may indeed be added to the inner-rotation solution, Eq. (4.146), because the convective acceleration is zero and hence the Navier-Stokes equation is linear. P4.82 A solid circular cylinder of radius R rotates at angular velocity in a viscous incompressible fluid which is at rest far from the cylinder, as in Fig. P4.82. Make simplifying assumptions and derive the governing differential equation and boundary conditions for the velocity field v in the fluid. Do not solve unless you are obsessed with this problem. What is the steady-state flow field for this problem? Fig. P4.82 Solution: We assume purely circulating motion: vz vr 0 and / 0. Thus the remaining variables are v fcn(r, t) and p fcn(r, t). Continuity is satisfied identically, and the -momentum equation reduces to a partial differential equation for v: subject to v (R, t) R and v ( , t) 0 .Ans 2 v v v1 r t r r r r I am not obsessed with this problem so will not attempt to find a solution. However, at large times, or t , the steady state solution is v R2/r. Ans. P4.83 The flow pattern in bearing lubrication can be illustrated by Fig. P4.83, where a viscous oil (, ) is forced into the gap h(x) between a fixed slipper block and a wall moving at velocity U. If the gap is thin, ,h L it can be shown that the pressure and velocity distributions are of the form p p(x), u u(y), w 0. Neglecting gravity, reduce the Navier-Stokes equations (4.38) to a single differential equation for u(y). What are the proper boundary conditions? Integrate and show that 21 ( ) 1 2 dp yu y yh U dx h where h h(x) may be an arbitrary slowly varying gap width. (For further information on lubrication theory, see Ref. 16.) 360 Solutions Manual Fluid Mechanics, Fifth Edition Fig. P4.83 Solution: With u u(y) and p p(x) only in the gap, the x-momentum equation becomes 2 2 2 2 du dp u d u 1 dp0 , or: constant dt dx dxy dy 2 1 2 1 dp yIntegrate twice: u C y C , with u(0) U and u(h) 0 dx 2 With C1 and C2 evaluated, the solution is exactly as listed in the problem statement: .Ans 21 dp yu y yh) U 1 2 dx h P4.84 Consider a viscous film of liquid draining uniformly down the side of a vertical rod of radius a, as in Fig. P4.84. At some distance down the rod the film will approach a terminal or fully developed draining flow of constant outer radius b, with z z(r), r 0. Assume that the atmosphere offers no shear resistance to the film motion. Derive a differential equation for z, state the proper boundary conditions, and solve for the film velocity distribution. How does the film radius b relate to the total film volume flow rate Q? Fig. P4.84 Chapter 4 Differential Relations for a Fluid Particle 361
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