chpt_4_suggested_qns

chpt_4_suggested_qns - Chapter 4 Differential Relations for...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 4 Differential Relations for Fluid Flow P4.1 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). 2 2 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 2 2 2 2 dv v v v v u v w 4ty 4tx(0) 2t y( 2t ) 4xz(0) 4ty 4t y dt t x y z d w w w w w u v w 4tx(4z) 2t y(0) 4xz(4x) 16txz 16x z dt t x y z 2 4 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 2 3 d 4(1 4t ) 4t(1 t ) (c) dt Ans. V i j k (d) At (1, 1, 0) there are many unit vectors normal to d V /dt. One obvious one is k . Ans. P4.2 Solution: Here we have only the single one-dimensional convective acceleration: 2 2 1 . o o V du u x u V A n s dt x L L o 2V x 1 L L (a) 2 2 2(10) 2 6 10 , 1 400(1 4 ), 6 /12 6 /12 o ft du x For L and V x with x in feet s d t At x 0, du/dt 400 ft/s 2 (12 gs); at x L 0.5 ft, du/dt 1200 ft/s 2 (37 gs). Ans. (b) P4.3 Solution: (a) Do each component of acceleration: 2 2 x y du u u u v (x y x)(2x 1) ( 2xy y)( 2y) a dt x y 2 2 dt x y dv v v u v (x y x)( 2y) ( 2xy y)( 2x 1) a 212 Solutions Manual Fluid Mechanics, Seventh Edition 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 ) (cos40 sin 40 ) 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) _______________________________________________________________________ P4.5 Solution: (a) For two-dimensional steady flow, the acceleration components are 2 o o o o 2 U U du u u x y u v U U ( ) x y 2 o o o o 2 dt x y L L L L U U dv v v x y u v U ( ) U dt x y L L L L Therefore the resultant 2 2 2 2 o o (U /L )(x y ) (U /L (purely radial) (a) Ans. a i j ) r (b) For the given resultant acceleration of 25 m/s...
View Full Document

Page1 / 10

chpt_4_suggested_qns - Chapter 4 Differential Relations for...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online