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 3o3u1RUx Vi i 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. Fig. P4.7 Solution: (a) Along this streamline, the fluid acceleration is one-dimensional: 33 34 23 4 37xo o oua u U (1 R /x )( 3U R /x ) 3U R (x R x ) for x Rx 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 Rt33oo334R 0dx dxuU(1R/x),or: Udt,dt1R/x R4R21o22R(xR) R 2xRor: U t x ln tan6xRxR3R3 It takes an infinite time to actually reach the stagnation point, where the velocity is zero. Ans. (b) P4.17 An excellent approximation for the two-dimensional incompressible laminar boundary layer on the flat surface in Fig. P4.17 is 341/2342 2 for , where , constant()yyyuU y CxC (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: 0uvxy 3424 534 2 45245 2 45026 4(),32 32:2 ( ) 2( ) .()245yvu ydydydUy x dx dx dxdyyy dyyyor v U dy U Ans adx dx (b) First evaluate C from the given data at x = 1 m: 1/2 1/21/2 1/21/20.011 (1 ) , hence 0.01111Or, alternately , ( )22 2mCm C mdCx xdx 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 max1 3 2 0.011 32( )2(3)[ ]() .()245 2(1)20dmmvU Ansbdx s m m0.0050s This is slightly smaller than the exact value of vmax from laminar boundary theory (Chap. 7). 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. Compute the value and position of the maximum viscous normal stress along this streamline. Is this also the position 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.Fig. P4.31 Solution: (a) Along this line of symmetry the convective deceleration is one-dimensional: 22 242x23 35ua2a aaau U1 U 2Uxxx xx xdaThis has a maximum deceleration at 0, or at (5/3) a . (a)dxAnsx 1.29a The value of maximum deceleration at this point is 2x,maxa0.372U/a. (b) The viscous normal stress along this line is given by 2xx3u2aU2 2 with a maximum . (b)xxAns max4Uat x aa 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 max3kg kg 4(0.29)(2.0)SAE 30 oil, 917 , 0.29 , . (b)ms (0.06 m)mAns 39 Pa P4.41 As mentioned in Sec. 4.10, the velocity profile for laminar flow between two plates, as in Fig. P4.40, is max24()0uyhyuwh If the wall temperature is Tw at both walls, use 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): 2222maxpp2224udT d T du d Tc k ,or: c(0)k (h2y),or:dt dydy dy h 222 322 22max max124 416 u 16 udT dT 4y(h 4hy 4y ), Integrate: h y 2hy Cdy 3dy kh kh Before integrating again, note that dT/dy 0 at y h/2 (the symmetry condition), so C1 –h3/6. Now integrate once more: 22342max12416 uyyyTh2hCyC233kh If T Tw at y 0 and at y h, then C2 Tw. The final solution is: .Ans2234maxw2348uy y 4y 2yTTk3hh3h3h 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. P4.79 Study the combined effect of the two viscous flows in Fig. 4.12. 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. 4.12 Solution: The solution for Fig. 4.12 (a) is Vyu12h and the solution for Fig. 4.12 (b) is 222hdp yu12dxh . The combined solution is 222Vyhdp yu1 12h2dxh 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). 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, ,hL 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()12dp yuyyhUdx h where h h(x) may be an arbitrary slowly varying gap width. (For further information on lubrication theory, see Ref. 16.) Solution: With u
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