Chapter4_SM

# Chapter4_SM - Chapter 4 Differential Relations for a Fluid...

This preview shows pages 1–4. Sign up to view the full content.

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 du u u u u u v w 4x 4tx(4t) 2t y(0) 4xz(0) 4x 16t x dt t x y z ∂∂∂ = +++ = + + = + dv v v v v u v w 4ty 4tx(0) 2t y( 2t ) 4xz(0) 4ty 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 d or: (4x 16t x) ( 4ty 4t y) (16txz 16x z) dt + + + + V i 2 j k at (x, y, z) = ( 1, + 1, 0), we obtain 23 d 4(1 4t ) 4t(1 t ) 0 (c) dt Ans. + + V ij 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 x uV w L υ ⎛⎞ ≈+ ⎜⎟ ⎝⎠ 0 (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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 4 Differential Relations for a Fluid Particle 303 Solution: Here we have only the single ‘one-dimensional’ convective acceleration: 2 2 1. o o V du u x u V Ans dt x L L ⎡⎤ ⎛⎞ == + = ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ o 2V x 1 LL 2 + ( a ) 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 ax and ay , (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 du u u u v (x y x)(2x 1) ( 2xy y)( 2y) a ∂∂ =+= + + + = y 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 ax = 18 i and ay = 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 ) Ans ° ° + ° 40 V n i j i j 5.39 units ° . ( b ) (c) The maximum acceleration is amax = [18 2 + 26 2 ] 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 0 ) 1 ( = = + = w L y U v L x U u o o ( 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.
Solutions Manual Fluid Mechanics, Fifth Edition 304 ( c ) Where is the largest resultant acceleration and its numerical value?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/24/2011 for the course MECH 122 at San Jose State.

### Page1 / 68

Chapter4_SM - Chapter 4 Differential Relations for a Fluid...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online