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 V ui Uo 1 3 i x 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 ax u u U o 1 R 3 x 3 3U o R 3 x 4 3U o2 R 3 x 4 R 3 x 7 for x R 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 computed from dx u U o 1 R 3 x 3 or dt R 4R t dx U o dt 1 R 3 x 3 0 R R x R 2 R 1 2x R or U o t x ln 2 tan 6 x Rx R 2 3 R 3 4R 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 u U 2 y 2 y3 3 y4 4 for y where C x1 2 C constant 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 u v 0 x y v u 2 y d 6 y3 d 4 y 4 d U 2 4 5 y x dx dx dx or v 2U d dx y 0 y 2 3 y3 4 2 y4 5 dy 2U d y 2 3 y 4 2 y5 2 4 5 dx 2 4 5 Ans a b First evaluate C from the given data at x 1 m 0 011 m C 1 m 1 2 hence C 0 011 m1 2 d 1 1 1 2 1 2 Or alternately dx 2 Cx 1 2 2 x x 2x 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 vmax 2U d 1 3 2 m 0 011m 3 m Ans b 2 3 0 0050 dx 2 4 5 s 2 1 m 20 s 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 twodimensional 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 ax u a2 2a2 a2 a 4 u U 1 2 U 3 2U2 3 5 x x x x x This has a maximum deceleration at da x 0 or at x 5 3 a 1 29a dx Ans a The value of maximum deceleration at this point is a x max 0 372U 2 a b The viscous normal stress along this line is given by xx 2 2a 2 U u 4 U at x a Ans b 2 3 with a maximum max x a x 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 SAE 30 oil 917 kg kg 4 0 29 2 0 39 Pa 0 29 max 3 m s 0 06 m m Ans b P4 41 As mentioned in Sec 4 10 the velocity profile for laminar flow between two plates as in Fig P4 40 is u 4umax y h y h2 w 0 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 2 2 du dT d2T d2T 4u h 2y or cp k 2 or cp 0 k 2 max 2 dt dy dy h dy 2 2 2 16 u max 2 d T dT 16 u max 2 4y3 2 2 C1 h 4hy 4y Integrate h y 2hy 2 4 4 dy 3 dy 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 T 16 u2max 2 y 2 y3 y 4 h 2h C1y C2 4 2 3 3 kh If T Tw at y 0 and at y h then C2 Tw The final solution is y y 2 4y 3 2y 4 2 3 4 Ans 3h 3h 3h h 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 T Tw 8 u 2max k 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 u V y 1 2 h and the solution for Fig 4 12 b is u h 2 dp y 2 1 2 dx h 2 The combined solution is V y h 2 dp y 2 u 1 1 2 h 2 dx h 2 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 h L …
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