Fluid Dynamics Sol ch4

Fluid Dynamics Sol ch4 - Chapter 4 Differential Relations...

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Chapter 4 Differential Relations for a Fluid Particle 4.1 An idealized velocity field is given by the formula 2 42 4 tx t y xz =− + Vi jk 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). 22 4 d u uuu u u v w 4x 4tx(4t) 2t y(0) 4xz(0) 4x 16t x dt t x y z d v vvv v u v w 4ty 4tx(0) 2t y( 2t ) 4xz(0) 4t y dt t x y z dw w w w w u v w 0 4tx(4z) 2t y(0) 4xz(4x) 16txz 16x z dt t x y z ∂∂∂ = +++ = + + = + = + + + + + + =+ + + = + + = + 24 2 d or: (4x 16t x) ( 4ty 4t y) (16txz 16x z) dt + + + + V ij k at (x, y, z) = ( 1, + 1, 0), we obtain 23 d 4(1 4t ) 4t(1 t ) 0 (c) dt Ans. + + V k (d) At (–1, + 1, 0) there are many unit vectors normal to d V /dt. One obvious one is k . Ans. 4.2 Flow through the converging nozzle in Fig. P4.2 can be approximated by the one-dimensional velocity distribution o 2 10 0 x uV w L υ ±² ≈+ ³´ µ¶ (a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case V o = 10 ft/s and L = 6 in, compute the acceleration, in g ’s, at the entrance and at the exit. Fig. P4.2
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Chapter 4 Differential Relations for a Fluid Particle 259 Solution: Here we have only the single ‘one-dimensional’ convective acceleration: 2 2 1. ( a ) o o V du u x uV A n s dt x L L ±² ³´ ³ ´ == + = µ¶ µ ¶ ·¸ ¹º ¹ º »¼ 2 o 2V x 1 LL 2 + 2 2(10) 2 6 10 , 1 400(1 4 ), 6/12 o ft du x For L and V x with x in feet sd t = + = + ′′ At x = 0, du/dt = 400 ft/s 2 (12 g’s); at x = L = 0.5 ft, du/dt = 1200 ft/s 2 (37 g’s). Ans. (b) 4.3 A two-dimensional velocity field is given by V = ( x 2 y 2 + x ) i – (2 xy + y ) j in arbitrary units. At ( x , y ) = (1, 2), compute (a) the accelerations a x and a y , (b) the velocity component in the direction θ = 40 ° , (c) the direction of maximum velocity, and (d) the direction of maximum acceleration. Solution: (a) Do each component of acceleration: 22 x y du u u u v (x y x)(2x 1) ( 2xy y)( 2y) a dt x y dv v v u v (x y x)( 2y) ( 2xy y)( 2x 1) a dt x y ∂∂ =+= + + + = + + = At (x, y) = (1, 2), we obtain a x = 18 i and a y = 26 j Ans. (a) (b) At (x, y) = (1, 2), V = –2 i – 6 j . A unit vector along a 40 ° line would be n = cos40 ° i + sin40 ° j . Then the velocity component along a 40 ° line is 40 V( 2 6 ) ( c o s 4 0 s i n 4 0 ) . ( b ) Ans ° ° + ° 40 V n i j i j 5.39 units ° (c) The maximum acceleration is a max = [18 2 + 26 2 ] 1/2 = 31.6 units at 55.3 ° Ans. (c, d) 4.4 Suppose that the temperature field T = 4x 2 – 3y 3 , in arbitrary units, is associated with the velocity field of Prob. 4.3. Compute the rate of change dT/dt at (x, y) = (2, 1). Solution: For steady, two-dimensional flow, the rate of change of temperature is 2 ( )(8 ) ( 2 )( 9 ) dT T T uv x y x x x y y y dt x y + + At ( x , y ) = (2, 1), dT/dt = (5)(16) – 5(–9) = 125 units Ans.
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Solutions Manual Fluid Mechanics, Fifth Edition 260 4.5 The velocity field near a stagnation point (see Example 1.10) may be written in
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Fluid Dynamics Sol ch4 - Chapter 4 Differential Relations...

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