CHAPTER 3 Introduction to Fluids in Motion FE-tpe Eam Review Problems: Problems 3- to 3-9 nˆ 0 ( n ˆi+ n ˆj) (3ˆi 4 ˆj) 0 or 3n 4n 0 3. (D) 3. (C) 3.3 (D) 3.4 (C) 3.5 (B) 3.6 (C) Also n n n + since ˆn is a unit vector. A simultaneous solution ields 4 / 5 and n 3/ 5. (Each with a negative sign would also be OK) ˆ a + u + v + w ( i) (ˆi ˆj) 6ˆi+ 8ˆi+ 6ˆj t z a ( 8) + 6 7.89 m/s u u u u u 0 0(4 ) a + u + v + w u t z (4 ) 0 0( )( )(4 ) 0 0 6.5 m/s (4 ) 4 8 3 The onl velocit component is u(). We have neglected v() since it is quite small. If v() were not negligible, the flow would be two-dimensional. p waterh 980 0.800 3 m/s.3 air p + + 0.00 0.600 9.8 0.400.80 m/s g g g The manometer reading h implies: 3.7 (B) or (60 0.) 9.39 m/s + p p +.3 The temperature (the viscosit of the water) and the diameter of the pipe are not needed. 9 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
3.8 (A) g p p + + g 800,000 40 m/s 980 9.8 3.9 (D) p ( ) ( ) 90 30 5 304,400 Pa Flow Fields d d a) u t+ v t dt dt streamlines t 5 s 3.4 t + t + c t + c (7, ) (35, 5) + 39.8 o + 4 parabola 3.6 Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian: Several college students would be positioned at each intersection and quantities would be recorded as a function of time. a) ˆi + cosα 0.83 α 33.69 3 + 3n + n 0 n ˆ 0 (3ˆ ˆ) ( ˆ ˆ n i+ j ni+ nj) 0 n 3 n 9 + n n + n 4 3.8 3 n, or ˆ (ˆ 3 ˆ n n i j) 3 3 3 c) ˆi 5 cosα 0.60 α 5.67 5 + ( 8) 5n 8n 0 n ˆ 0 (5ˆ 8 ˆ) ( ˆ ˆ n i j ni+ nj) 0 n 5 8 n 5 64 + n n + n 30 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
5 8 n, or ˆ (8ˆ 5 ˆ n n i+ j) 89 89 89 3.0 b) u + v + w + ( ˆi) + ( ˆj) 4ˆi+ 4ˆj 8ˆi 4ˆj z t 3. The vorticit ω Ω. a) ω 40ˆi c) ω ˆi 4k ˆ a) a r 40 80 40 sinθ 40 0 cosθ cosθ 0 ( sin θ) 3 + r r r r r 40 0 + sin θ ( 0.5)( ). 5( ) 9.375 m/s r r 3.4 40 80 θ aθ θ θ θ r r + + 40 r sin r + 40 0 cos sin 0 0 3 cos r 600 00 sin θcosθ 0 since sin(80 ) 0 4 r r a φ 0 3.6 a + u t + v u + w ˆi z t For stead flow u/ t 0 so that a 0 3.8 b) t /0 u ( 0.5 )( e ).875 m/s at t 3.30 D 4 4 3000 0 u + v + w + 0(.3 0 e ) Dt z t 4 3 9. 0 kg/m s D 3.3 u 4 (0.0) 0.04 kg/m 3 s Dt 3 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
u a + u t v 3.34 a + v a + ( ) t t w az + w t π ˆ 5 ˆ Ω k 7.7 0 krad/s 4 60 60 5( 0.707ˆi 0.707 kˆ) 3.535ˆi 3.535km/s ˆ A Ω + Ω ( Ω r ) 5ˆ ˆ ˆ 5 7.7 0 k ( 3.535i 3.535 k) + 7.7 0 k ˆ 3.36 7.7 0 kˆ (6.4 0 )( 0.707i+ 0.707 k ˆ) 5 0 ˆj+ 0.04 ˆi m/s 5 6 ˆ 5 Note: We have neglected the acceleration of the earth relative to the sun since it is quite small (it is / d S dt ). The component ( 5.4 0 5ˆ j ) is the Coriolis acceleration and causes air motions to move c.w. or c.c.w. in the two hemispheres. Classification of Fluid Flows 3.38 Stead: a, c, e, f, h Unstead: b, d, g 3.4 a) inviscid c) inviscid e) viscous inside the boundar laers and separated regions. g) viscous. L 3.46 Re 0.8 0.75.4 0 ν 5 9640 Turbulent 3.48 a) D. 0.0 Re 795 ν 5.5 0 Alwas laminar Assume the flow is parallel to the leaf. Then 3 0 5 T/ν 3.50 5 5 4 3 0 ν / 3.5 0.4 0 / 6 8.7 m T The flow is epected to be laminar 3 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
3.5 D u + v + w + 0 Dt z t For a stead, plane flow / t 0 and w 0 Then u + v 0 Bernoulli s Equation 3.54 3.56 3.58 p Use. kg/m 3 a) v p / 000 /. 60 m/s p + 0 p 000 57.0 m/s.3 p U p + + b) Let r rc: pt U 3 d) Let θ 90 : p90 U p U p + + 3.60 a) p ( U u ) 0π 0 0 + 50 + π 50 + c) p ( U u ) 60π 30 30 + 450 + π 450 + 33 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
Assume the velocit in the plenum is zero. Then 3.6 or (60 0.) 9.39 m/s + p p +.3 We found 3. kg/ m 3 in Table B.. Bernoulli from the stream to the pitot probe: Manometer: pt + H HgH h p h p T + p 3.64 Then, + p+ H Hg H p Hg ( H) (3.6 )9800 a) ( 0.04) 000 3.4 m/s (3.6 )9800 b) ( 0.) 000 4.97 m/s 3.66 The pressure at 90 from Problem 3.58 is p90 3 U /. The pressure at the stagnation point is pt U /. The manometer provides: p T H p90 3.04U 9800 0.04.04 U. U.76 m/s Bernoulli: p + g p + g 3.68 Manometer: + + Hg + p z H H z p g Substitute Bernoulli s into the manometer equation: p + ( Hg ) H + p g b) Use H 0.05 m: Substitute into Bernoulli: 9800 (3.6 )9800 0.05 3.56 m/s 9.8 0 3 56 p. 9800 93,600 Pa g 98. 34 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
Write Bernoulli s equation between points and along the center streamline: p z p z + + + + Since the flow is horizontal, z z and Bernoulli s equation becomes 3.70 0.5.5 p+ 000 p + 000 From fluid statics, the pressure at is p h 980 0.5 45 Pa and at, using p H, Bernoulli s equation predicts 0.5.5 45 + 000 980H+ 000 H 0.98 m or 9.8 cm Assume incompressible flow ( < 00 m/s) with point outside the wind tunnel where p 0 and 0. Bernoulli s equation gives 3.7 p 0 + p p 90 3 a).39 kg/m p.39 00 695 Pa RT 0.87 53 p 9 3 c).094 kg/m p.094 00 5470 Pa RT 0.87 93 Bernoulli across nozzle: + p p + p / 3.74 Bernoulli to ma. height: a) p g p + + h p g / 700,000 /000 37.4 m/s + + h h p / h p / 700,000/9800 7.4 m b) p /, 400,000 /000 5.9 m/s h p /,400,000/9800 4.9 m 3.76 + p p + p 00,000 Pa, the lowest possible pressure. 35 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.
a) 600,000 00,000 37.4 m/s 000 000 3.78 b) 300, 000 00, 000 8.3 m/s 000 000 b) p ( ) ( ) 90 0 43,300 Pa d) p ( ) ( ).3 0 59.0 Pa Appl Bernoulli s equation between the eit (point ) where the radius is R and a point in between the eit and the center of the tube at a radius r less than R: 3.80 Since + + p p p <, we see that p is negative (a vacuum) so that the envelope would tend to rise due to the negative pressure over most of its area (ecept for a small area near the end of the tube). 3.8 3.84 A burr downstream of the opening will create a region that acts similar to a stagnation region thereb creating a high pressure since the velocit will be relativel low in that region. The higher pressure at B will force the fluid toward the lower pressure at A, especiall in the wall region of slow moving fluid, thereb causing a secondar flow normal to the pipe s ais. This results in a relativel high loss for an elbow. stagnation region 36 0 Cengage Learning. All Rights Reserved. Ma not be scanned, copied or duplicated, or posted to a publicl accessible website, in whole or in part.