* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download zero. Ans. (b) P4.8 When a valve is opened, fluid flows in the
Survey
Document related concepts
Lift (force) wikipedia , lookup
Classical mechanics wikipedia , lookup
Routhian mechanics wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Centripetal force wikipedia , lookup
Equations of motion wikipedia , lookup
Classical central-force problem wikipedia , lookup
Biofluid dynamics wikipedia , lookup
Flow conditioning wikipedia , lookup
Bernoulli's principle wikipedia , lookup
History of fluid mechanics wikipedia , lookup
Transcript
Chapter 4 307 Differential Relations for a Fluid Particle 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 R3 i x3 ui Uo 1 V 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: du dt u u x U o (1 R 3 /x 3 )( 3U o R 3 /x 4 ) 3U o R 3 (x 4 R 3 x 7 ) for x R 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 computed from u or: U o t R dx dt 3 3 U o (1 R /x ), or: 4R x R (x R)2 ln 2 6 x Rx R 2 dx 1 R 3 /x 3 t U o dt, 0 R R 3 tan 1 2x R R 3 4R 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 V iU 1 x Ut tanh 2L L 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 ax u t u u x U 1 x U Ut sech 2 2L L L U2 x Ut (1 )[sech 2 L 2L L At (x, t) (L, L/U), ax U 1 x 2L U Ut tanh 2L L 1 Ut tanh 2 ] 2 L (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 Solution: With no variation and no v , the equation of continuity (4.9) becomes vz z 1 (r vr ) r r or : 317 Differential Relations for a Fluid Particle r (r vr ) 0 1 (r vr ) r r B r ; Integrate : r vr Finally, vr B r 2 z ( Bz ) , B 2 r 2 f ( z) r f ( z) Ans. 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 highpressure 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 d d (m tank ) mexit 0, or: ( ) mexit C p A, where C constant dt dt Introduce p RTo p(t) and separate variables: The solution is an exponential decay of tank density: p po dp p CRTo A t dt o 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 u ( A Bx) (h 2 y 2 ) , 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 v y u x Integrate : v ( B)(h 2 B (h 2 Fluid Mechanics, Fifth Edition y2 ) y 2 )dy If v 0 at y 0, then f ( x) B(h2 y v 0. y3 ) 3 f ( x) y3 ) 3 B(h 2 y Ans.(a ) (b) Just simply introduce this boundary condition into the answer to part (a): v( y h) vw B ( h3 h3 ) , 3 hence B P4.25 An incompressible flow in polar coordinates is given by b vr K cos 1 2 r b v K sin 1 2 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. 3 vw Ans.(b) 2 h3 Fig. P4.25 Solution: Substitute into plane polar coordinate continuity: 1 1 v (rv r ) r r r 0 1 K cos r r r b r 1 r K sin 1 b r2 0 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). Solution for problem 6). ∂() ∂y a). According to the continuity equation (two-d flow, = 0) ∂u ∂w + =0 ∂x ∂z we have ! " #$ 2z ∂u ∂ z2 √ − 2 =− U0 ∂x ∂x C x C x # " " # 1 1 U0 z 2 ∂ 2U0 z ∂ √ + = − C ∂x C 2 ∂x x x U0 z 2 −2 U0 z − 3 x 2− x = C C2 Solving the above partial differential equation we have ∂w ∂z = − 3 U0 x− 2 z 2 U0 x−2 z 3 − + C1 (x) 2C 3C 2 Due to the no-slip boundary condition, w(x, z = 0) = 0. Subsitute it into the above general solution we have C1 (x) = 0, hence the expression for vertical velocity is w(x, z) = 3 U0 x− 2 z 2 U0 x−2 z 3 w(x, z) = − 2C 3C 2 b). As assumed by this question, we have δ(x = 1) = 0.011 m, thus 0.011 = C × 11/2 =⇒ C = 0.011 m1/2 At x = 1 we can define w1 (z) = w(x = 1, z), such that w1 (z) = Thus we have U0 z 3 U0 z 2 − 2C 3C 2 dw1 U0 z U0 z 2 = − dz C C2 2 2U0 z d w1 U0 − = dz 2 C C2 and U0 % z& dw1 = 0 =⇒ z 1− =0 dz C C Therefore w1 (z) have maximum or minimum values at z = 0 and z = C. Since we have ' d2 w1 '' U0 >0 = ' 2 dz z=0 C ' d2 w1 '' U0 U0 U0 −2 =− <0 = ' 2 dz z=C C C C Thus the maximun value of w1 (z) is w1 (z = C), hence w(x = 1, z)max U0 C U0 C − 2 3 = w(x = 1, z = C) = = 3 × 0.011 U0 C = = 0.0055 m/s 6 6 1 326 Solutions Manual p Fluid Mechanics, Fifth Edition 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. υθ (r) r r=a No slip r=b Fig. P4.35 Solution: The preliminary work for this problem is identical to Prob. 4.32 on an earlier page. That is, there are two possible solutions for purely circulating motion (r), hence v C1r C2 , subject to v (a) 0 r This requires C1 C2 0, or v C1a C2 /a and v (b) 0 0 (no steady motion possible between fixed walls) Ans. y g 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 C1b C2 /b 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. h (y) θ x Fig. P4.36 Solution: There is atmospheric pressure all along the surface at y h, hence p/ x The x-momentum equation can easily be evaluated from the known velocity profile: u u x u y v p x 2 gx u, or: 0 g sin 2 Solve for C 0 g sin + ( 2C) Ans. (a) The flow rate per unit width is found by integrating the velocity profile and using C: h Q h u dy 0 gh 3 sin 3 2 3 Ch 3 Cy(2h y) dy 0 per unit width Ans. (b) 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 NavierStokes is relevant: dw dt 0 Fig. P4.37 d2w , or: w dx 2 g g , w( h ) w( h) 0 (no-slip) The solution is very similar to Eqs. (4.142) to (4.143) of the text: w P4.38 g 2 (h 2 x 2 ) Ans. Show that the incompressible flow distribution, in cylindrical coordinates, vr 0 v C rn vz 0 0. Chapter 4 331 Differential Relations for a Fluid Particle Before integrating again, note that dT/dy 0 at y h/2 (the symmetry condition), so C1 –h3/6. Now integrate once more: 16 u2max 2 y 2 h 2 kh 4 T If T Tw at y 0 and at y T Tw h, then C2 2h y3 3 y4 3 C1y C2 Tw. The final solution is: 8 u 2max y k 3h y2 h2 4y 3 3h 3 2y 4 3h 4 Ans. 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 conditions are V(r, z, t). The initial condition is: V(r, z, 0) 0. The boundary 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. : p patm, rz At the free surface, z 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 u ( h) 0 Ans. (b) (b) At the surface, y h, no shear stress, 0, or y y Solution: With vz fcn(r) only, the Navier-Stokes z-momentum relation is dvz dt or: 1 d dvz r r dr dr g p z 0 vz , gr 2 4 , Integrate twice: vz vz (b) r gb2 r ln 2 a The final solution is v z b The flow rate is Q 2 g The proper B.C. are: u(a) 0 (no-slip) and ga 4 8 vz 2 r dr a b a where g 4 3 u(y, t) and p/ x u u u t x or: u t v u 0 0 t C1 ln(r) C2 0 (no free-surface shear stress) 4 r2 1 4 a2 2 Ans. 4 4 ln , Ans. P4.85 A flat plate of essentially infinite width and breadth oscillates sinusoidally in its own plane beneath a viscous fluid, as in Fig. P4.85. The fluid is at rest far above the plate. Making as many simplifying assumptions as you can, set up the governing differential equation and boundary conditions for finding the velocity field u in the fluid. Do not solve (if you can solve it immediately, you might be able to get exempted from the balance of this course with credit). Solution: Assume u 361 Differential Relations for a Fluid Particle Chapter 4 Fig. P4.85 0. The x-momentum relation is u y 2 p x u x2 gx 2 u , y2 2 0 0 u , or, finally: y2 0 2 u y2 subject to: u(0, t) U o sin t and u ,t 0. Ans. 368 Solutions Manual 1 d dv (r ) r dr dr As r , v v r 2 , Fluid Mechanics, Fifth Edition Solution : v 0 , hence C1 C1 r C2 r 0 C2 R2 ; C2 R 2 ; Finally, v Ans. R r Rotating a cylinder in a large expanse of fluid sets up (eventually) a potential vortex flow. At r R, v R ________________________________________________________________________ *P4.95 Two immiscible liquids of V equal thickness h are being sheared h y 2, 2 between a fixed and a moving plate, 1, h as in Fig. P4.95. Gravity is neglected, 1 x Fixed and there is no variation with x. Fig. P4.95 Find an expression for (a) the velocity at the interface; and (b) the shear stress in each fluid. Assume steady laminar flow. Solution: Treat this as a Ch. 4 problem (not Ch. 1), use continuity and Navier-Stokes: u v v 0 0 ; thus v const 0 for no slip at the walls x y y This tells us that there is no velocity v, hence we need only consider u(y) in Navier-Stokes: Continuity : 1, 2 (u u x u v ) y p x Thus 2 1, 2 ( u u 2 x2 a y by u ) or : 0 2 0 0 1, 2 (0 d 2u dy 2 ) 369 Differential Relations for a Fluid Particle Chapter 4 The velocity profiles are linear in y but have a different slope in each layer. Let uI be the velocity at the interface. (a) The shear stress is the same in each layer: 1 uI h V uI 2 Solve for u I h 2 1 V Ans.(a ) 2 (b) In terms of the upper plate velocity, V, the shear stress is 1 ( 2 1 ) 2 V h Ans.(b) ________________________________________________________________________ P4.96 Reconsider Prob. P1.44 and calculate (a) the inner shear stress and (b) the power required, if the exact laminar-flow formula, Eq. (4.140) is used. (c) Determine whether this flow pattern is stable. [HINT: The shear stress in (r, ) coordinates is not like plane flow.] Solution: The exact laminar-flow velocity is Eq. (4.140), and the shear stress is Eq. (D.9): v i ri [ ( r (ro / r ) ( r / ro ) ] (ro / ri ) (ri / ro ) dv dr v ) r [ i ri (ro / ri ) (ri / ro ) ] 2ro r2 Recall the data from Prob. P1.44: ri = 5 cm, ro = 6 cm, L = 120 cm, = 0.86 kg/m-s (SAE 50W oil), and i = 900 rev/min = 94.25 rad/s. At the inner cylinder, inner [ i ri (ro / ri ) (ri / ro ) ] 2ro ri2 (0.86)[ 94.25(0.05) 2(0.06) ] 0.06 / 0.05 0.05 / 0.06 (0.05) 2 531 Pa Ans.(a) The moment and power required are M i2 ri2 L Power (531) 2 (0.05)2 (1.20) i M 10.0 N (94.25 rad / s)(10.0 N m) m 943 watts Ans.(b)