MC 2312 Mecánica De Fluidos – Potter (Solucionario Capítulo 3) pdf

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(1)CHAPTER 3. Introduction to Fluids in Motion 3.1 pathline streamline. 3.2. Pathlines:. streakline. Release several at an instant in time and take a time exposure of the subsequent motions of the bulbs.. Sreakline: Continue to release the devises at a given location and after the last one is released, take a snapshot of the “line” of bulbs. Repeat this for several different release locations for additional streaklines. 3.3 streakline pathline t =0. hose time t. boy. 3.4 y. streakline at t = 3 hr pathline t = 2 hr. streamlines t = 2 hr x. 34.

(2) 3.5. dx = 2t + 2 dt x = t 2 + 2t + c1. a) u =. v=. dy = 2t dt y = t 2 + c2. y. streamlines t=5s (27, 21). = y +2 y. 39.8o. ∴parabola.. ∴ x − 2 xy + y = 4 y 2. 2. (35, 25). x. b) x = t 2 + 2t + c 1 . ∴ c 1 = −8 , and c 2 = −4. = y + 4 + 2( ± y + 4 ) − 8 ∴parabola.. ∴ x 2 − 2xy + y 2 + 8 x − 12 y = 0.. 3.6 3.7. v V = uiˆ + vjˆ + wkˆ.   v dr = dxiˆ + dy ˆj + d z kˆ . using iˆ × jˆ = kˆ , ˆj × iˆ = −kˆ.. Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian:. 3.8. v v (V × dr ) z = udy − vdx. Several college students would be positioned at each intersection and quantities would be recorded as a function of time.. a) At t = 2 and (0 ,0,0 ) V = 2 2 = 2 m / s . At t = 2 and (1, −2 ,0) V = 3 2 + 2 2 = 3.606 m / s. b) At t = 2 and (0 ,0, 0) V = 0. At t = 2 and ( 1,−2 ,0 ) V = ( −2) 2 + ( −8 ) 2 = 8.246 m / s. c) At t = 2 and (0 ,0,0 ) V = (−4 ) 2 = 4 m / s. At t = 2 and (1, −2 ,0). 3.9. (D). V = 2 2 + ( −4 ) 2 + ( −4) 2 = 6 m / s.. ( −51.4 × 10−5 ˆj ). A simultaneous solution yields n x = 4/5 and n y = 3/5. (They must both have the same sign.. 3.10. v a) cos α = V ⋅ i$ / V = (1 + 2)/ v V ⋅ n$ = 0.. ∴nx =. 3 2 + 2 2 = 0.832.. ∴ α = 33.69o. 3 ny = − nx 3n x + 2n y = 0  2 (3 $i + 2 $j ) ⋅ ( n x i$ + n y $j ) = 0.  ∴ 9 2 n x2 + n y2 = 1  2 n x + nx = 1 4 2 3 1 , ny = − or n$ = ( 2i$ − 3 $j ). 13 13 13. 35.

(3) v b) cos α = V ⋅ i$ / V = −2 / ( −2) 2 + ( −8) 2 = −0.2425. ∴ α = 104 o v −2 n x − 8 n y = 0 n x = −4n y V ⋅ n$ = 0. (−2i$ − 8 $j ) ⋅ (n x $i + n y $j ) = 0. ∴  2 2 n x + ny = 1  16n 2y + n y2 = 1 1 4 , nx = − or n$ = 17 17. ∴ ny =. v c) cos α = V ⋅ i$ / V = 5 / v V ⋅ n$ = 0.. 5 2 + ( −8) 2 = 0.6202.. ( 5$i − 8 $j ) ⋅ ( n x i$ + n y $j ) = 0.. ∴ α = −51.67 o. 8 nx = ny 5n x − 8n y = 0 5  ∴ 64 n 2x + n y2 = 1  2 n y + n 2y = 1 25. 5 8 1 , nx = or n$ = ( 8$i + 5 $j ). 89 89 89. ∴ny = 3.11. 1 ( −4i$ + $j ). 17. [(x + 2)i$ + xtj$] × (dxi$ + dyj$) = 0.. v v a) V × dr = 0.. ∴ ( x + 2) dy − xtdx = 0 or t. xdx = dy . x+2. xdx = dy . t [ x − 2ln x + 2 ] = y + C. x +2 ∫ 2(1 − 2ln 3) = −2 + C. ∴ C = 0.8028. t [ x − 2ln x + 2 ] = y + 0.8028. Integrate: t ∫. v v b) V × dr = 0.. [ xyi$ − 2 y $j ] × (dxi$ + dyj$) = 0. 2. dy 2dx =− . x y Integrate: 2 lnx = −ln ( y / C). 2ln(1) = − ln( −2 / C). ∴ xydy + 2 y 2 dx = 0 or ∴ C = −2. v v c) V × dr = 0.. lnx 2 = − ln( y / −2).. [(x. 2. ∴ x 2 y = −2.. ]. + 4)i$ − y 2 tj$ × ( dxi$ + dyj$ ) = 0.. ( x 2 + 4) dy + y 2 tdx = 0 or. dy tdx =− 2. 2 x +4 y. t x 2 1 1  1   tan −1 + C = .  tan −1 + C = − .  y  2 2 2 2 2 x   ∴ C = −0.9636. yt  tan −1 − 0.9636 = 2   2 Integrate:. 36.

(4) 3.12. v v v v ∂V ∂V ∂V v ∂V a= +u +v +w = 2 xy (2 yiˆ) − y 2 (2 xiˆ − 2 yjˆ ) = −16iˆ − 8iˆ + 16 jˆ. ∂t ∂x ∂y ∂z. (C). ∴ a = 82 + 162 = 17.89 m/s 3.13. v v v v v DV ∂V ∂V ∂V ∂V a) =u +v +w + =0. Dt ∂x ∂y ∂z ∂t v v v v ∂V ∂V ∂V ∂V b) u +v +w + = 2x ( 2i$ ) + 2 y ( 2 $j ) = 4 xi$ + 4 yj$ = 8i$ − 4 $j ∂x ∂y ∂z ∂t v v v v ∂V ∂V ∂V ∂V c) u +v +w + = x 2t ( 2xti$ + 2 ytj$) + 2 xyt( 2 xtj$ + 2ztk$ ) + x 2 i$ + 2 xyj ∂x ∂y ∂z ∂t +2 yzk = 68i$ − 100 $j − 54 k$ v v v v ∂V ∂V ∂V ∂V d) u +v +w + = x( $i − 2 yzj$) − 2 xyz( −2 xzj$ ) + tz( −2 xyj$ + tk$ ) + zk$ ∂x ∂y ∂z ∂t = xi$ − (2 yz − 4 x 2 yz 2 + 2 xyzt ) $j + ( zt 2 + z) k$ = 2i$ − 114 $j + 15 k$. 3.14. v 1  ∂w ∂v  $ 1  ∂u ∂w  $ 1  ∂v ∂u  $ Ω=  − i +  −  j +  −  k. 2  ∂y ∂z  2  ∂z ∂x  2  ∂x ∂y  v 1 ∂u $ a) Ω = − k = 20 yk$ = −20 k$ 2 ∂y v 1 1 1 b) Ω = ( 0 − 0)i$ + (0 − 0) j$ + ( 0 − 0) k$ = 0 2 2 2 v 1 1 1 c) Ω = ( 2 zt − 0)i$ + ( 0 − 0) $j + ( 2 yt − 0) k$ = 6i$ − 2 k$ 2 2 2 v 1 1 1 d) Ω = ( 0 + 2 xy )$i + (0 − 0) $j + ( −2 yz − 0)k$ = −2i$ + 3k$ 2 2 2. 3.15. v v The vorticity ω = 2Ω. Using the results of Problem 3.7: v v v v a) ω = −40i$ b) ω = 0 c) ω = 12i$ − 4k$ d) ω = −4 i$ + 6k$. 3.16. a) ε xx =. ε xy = ε yz. ∂u = 0, ∂x. ε yy =. ∂v = 0, ∂y. ε zz =. ∂w = 0. ∂z. 1  ∂u ∂v  1  ∂u ∂w   +  = −20 y = 20, ε xz =  +  = 0, 2  ∂y ∂x  2  ∂z ∂x . 1  ∂v ∂w  =  +  = 0. 2  ∂z ∂y .  0 20 0  ∴ rate - of strain = 20 0 0   0 0 0  37.

(5) 2 0 0 rate-of strain = 0 2 0 0 0 0. ε xx = 2, ε yy = 2, ε zz = 0. b) ε xy = 0, ε xz = 0, ε yz = 0.. c) ε xx = 2xt = 8 , ε yy = 2 xt = 8, ε zz = 2 yt = −4. 1 1 1 ε xy = ( 2 yt ) = −2, ε xz = ( 0) = 0, ε yz = ( 2 zt ) = 6. 2 2 2  8 −2 0  rate-of strain = −2 8 6   0 6 −4  d) ε xx = 1, ε yy = −2xz = −12, ε zz = t = 2. 1 ( −2 yz ) = 3 , ε xz = 2 1 3 rate-of strain = 3 −12 0 2. ε xy =. 3.17. 1 1 (0) = 0, ε yz = ( −2xy ) = 2. 2 2 0 2 2. 40 80 40 sin θ  40  a) ar =  10 − 2  cos θ  3  cos θ −  10 + 2   1 − 2  ( − sin θ )  r   r  r  r  r  2. 1 40 −  10 + 2  sin 2 θ = (10 − 2.5)(−1)1.25( −1) = 9.375 m/s2 . r r  40  40  sin θ  40    80   aθ =  10 − 2  cos θ  3  sin θ +  10 + 2   10 + 2  cos θ  r   r  r  r  r  1 1600  since sin 180° = 0. −  100 − 4  sin θ cos θ = 0 r r  aφ = 0. 1 40 1 40 b) ω r = 0, ω θ = 0, ω z =  −10 + 2  sin θ −  10 − 2  ( − sin θ ) = 0. r r  r r  v v At (4, 180°) ω =0 since ω = 0 everywhere.. 3.18. 80 240 80 sin θ 80   a) ar =  10 − 3  cos θ  4  cos θ −  10 + 3  ( − sin θ ) 10 − 3    r    r  r  r r . 80 sin 2 θ − 10 + 3  = 8.75( −1)(.9375)(−1) = 8.203 m/s2  r  r aθ = 0 since sin 180° = 0. aφ = 0 since v φ = 0. 2. b) ω r = 0, ω θ = 0, ω φ = 0,. since sin 180° = 0.. 38.

(6) 3.19. v v v v ∂V ∂V ∂V ∂ u ˆ v ∂V v a= +u +v +w = i . For steady flow ∂u / ∂t = 0 so that a = 0. ∂t ∂x ∂y ∂z ∂ t. 3.20. Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate equations of Table 3.1 and recognizing that v θ = 0 and ∂ / ∂θ = 0: ∂v ∂v ∂u ∂u ar = v +u ax = v +u ∂r ∂x ∂r ∂x. 3.21. a) u = 2(1 − 0)(1 − e − t/ 10 ) = 2 m / s at t = ∞. ∂u  1  − t/ 10 2 ax = = 2(1 − 0)  e = 0.2 m / s at t = 0.  10  ∂t 2 b) u = 2(1 − 0.5 )(1 − e − t/ 10 ) = 1.875 m / s at t = ∞.. (. ). 2 2  1 − t / 10  2 ax = 2(1 − 0.5 / 2 ) e  = 0.0125 m / s at t = 0.  10 . c) u = 2(1 − 2 2 / 2 2 )(1 − e − t/ 10 ) = 0 for all t . 2 2  1 − t / 10  ax = 2(1 − 2 / 2 ) e  = 0 for all t .  10  3.22. DT ∂T ∂T ∂T ∂T πt π  π  =u +v +w + = 20(1 − y 2 )  − sin = − × 0.5878  Dt ∂x ∂y ∂ z ∂t 5  100  100 = −0.3693 °C/s.. 3.23. −4 kg Dρ ∂ρ ∂ρ ∂ρ ∂ρ =u +v +w + = 10( −1.23 × 10 −4 e −3000 ×10 ) = −9.11 × 10 −4 3 . Dt ∂x ∂y ∂z ∂t m ⋅s. 3.24. Dρ ∂ρ ∂ρ ∂ρ ∂ρ 1000  =u +v +w + = 10 −   Dt ∂x ∂y ∂z ∂t 4 . 3.25. Dρ ∂ρ =u = 4 × (.01) = 0.04 kg/m3 ⋅s Dt ∂x. 3.26. (D). 3.27. = −2500. kg . m3 ⋅s. ∂u ∂u ∂u ∂u ∂u 10 ∂ +u +v +w =u = [10(4 − x )−2 ] 2 ∂t ∂x ∂y ∂z ∂x (4 − x ) ∂x 10 10 1 = 10( −2)(−1)(4 − x) −3 = × 20 × = 6.25 m/s 2 . 2 4 8 (4 − x) v v v ∂ D = V ⋅∇ + observing that the dot product of two vectors A = A x i$ + A y $j + A z k$ Dt ∂t v v v $ and B = Bx i + B y $j + Bz k$ is A ⋅ B = AxB x + AyB y + Az Bz . ax =. 39.

(7) ∂u v v  + V ⋅ ∇u  ∂t v v v v ∂v v v  v ∂V ay = + V ⋅ ∇v  ∴ a = + ( V ⋅ ∇ )V ∂t ∂t  ∂w v v  az = + V ⋅ ∇w  ∂t ax =. 3.28. 3.29. 3.30. Using Eq. 3.2.12: v v v d 2 sv v v v v v dΩ v a) A = a + 2 + 2Ω × V + Ω × (Ω × r ) + ×r dt dt = 2( 20 k$ × 4i$ ) + 20 k$ × ( 20 k$ × 1.5$i ) = 160 $j − 600i$ m 2 / s v v v v v v b) A = 2Ω × V + Ω × (Ω × r ) = 2( 20k$ × −20 cos 30 o $j ) + 20 k$ × ( 20 k$ × 3i$ ) = −507 $i v 2π Ω= k$ = 7 .272 × 10 −5 k$ rad/s. 24 × 60 × 60 v v = 5( −.707 i$ −.707 k$ ) = −3.535i$ − 3.535k$ m/s. v v v v v v A = 2Ω × V + Ω × ( Ω × r ) = 2 × 7 .272 × 10 −5 k$ × (−3.535i$ − 3.535 k$ ) + 7.272 × 10 −5 k$ × [ 7.272 × 10 −5 k$ × 6 × 10 6 (−.707i$ +.707 k$ )] = −51.4 × 10 −5 $j + 0.0224 $i m / s 2 . Note:. We have neglected the acceleration of the earth relative to the sun since it is quite small. v 2 (it is d s / dt ). The component 2. ( −51.4 × 10−5 ˆj ) is the Coriolis acceleration and causes air. motions to move c.w. or c.c.w. in the two hemispheres.. 3.31. a) two-dimensional (r, z) c) two-dimensional (r, z) e) three-dimensional (x, y, z) g) two-dimensional (r, z). b) two-dimensional (x, y) d) two-dimensional (r, z) f) three-dimensional (x, y, z) h) one-dimensional (r). 3.32. Steady:. Unsteady:. 3.33. b. It is an unsteady plane flow.. 3.34. a). 3.35. f, h. a, c, e, f, h. d). e). 40. b, d, g.

(8) 3.36. a) inviscid. b) inviscid. c) inviscid. d) viscous inside the boundary layer. e) viscous inside the boundary layers and separated regions. f) viscous. g) viscous. h) viscous.. 3.37. d and e. Each flow possesses a stagnation point.. 3.38 3.39. (C). 3.40. Re = V L / ν = 2 × .015/.77 × 10-6 = 39 000.. 3.41. Re =. 3.42. 3.43. 3.44. The only velocity component is u(x). We have neglected v(x) since it is quite small. If v(x) in not negligible, the flow would be two-dimensional. ∴Turbulent.. VL = .2 × .8/1.4 × 10-5 = 11 400. ν. ∴Turbulent.. VL 4 ×.06 = = 14 100. ν 1.7 × 10 −5 Note: We used the smallest dimension to be safe! Re =. ∴Turbulent.. a). Re =. VD 1.2 × 0.01 = = 795. ν 1.51 × 10 −5. Always laminar.. b). Re =. VD 1.2 × 1 = = 79 500. ν 1.51 × 10 −5. May not be laminar.. VxT . ν a) T = 223 K or −50°C. Re = 3 × 105 =. ν = µ/ ρ. ∴ µ = 1.5 × 10 − 5 N ⋅ s / m 2 . ∴ν =. 3 × 10 5 =. where µ = µ(T ).. 900 × 1000 x T . 3600 × 2.5 × 10 −5. 1.5 × 10 −5 = 2.5 × 10 − 5 m 2 /s. .3376 × 1.23 ∴xT = 0.03 m. b) T = −48°F. ∴µ = 3.3 × 10−7 lb-sec/ft2 . ν = 3 × 10 5 =. 600 × 5280x T . 3600 × 3.7 × 10 −4. or. 3 cm. 3.3 × 10 −7 = 3.7 × 10 −4 ft2 /sec. .00089. ∴xT = 0.13' or 1.5". 41.

(9) 3.45. Assume the flow is parallel to the leaf. Then 3 × 105 = Vx T / ν . ∴ x T = 3 × 10 5 ν / V = 3.5 × 10 5 × 1.4 × 10 −4 / 6 = 8.17 m . The flow is expected to be laminar.. 3.46. 3.47. V 100 = = 0.325. For accurate calculations the flow is c 1.4 × 287 × 236 compressible. Assume incompressible flow if an error of 4%, or so, is acceptable. V 80 b) M = = = 0.235. ∴Assume incompressible. c 1.4 × 287 × 288 V 100 c) M = = = 0.258. ∴Assume incompressible. c 1.4 × 287 × 373 Dρ ∂ρ ∂ρ ∂ρ ∂ρ =u +v +w + = 0. For a steady, plane flow Dt ∂x ∂y ∂z ∂t ∂ρ / ∂t = 0 and w = 0. Then ∂ρ ∂ρ u +v = 0. ∂x ∂y a) M =. 3.48. Dρ ∂ρ ∂ρ ∂ρ ∂ρ =u +v +w + = 0. Dt ∂x ∂y ∂ z ∂t. 3.49. (B). 3.50. V2 p = . 2 ρ. ∴incompressible.. V2 p γ h 9810 × 0.800 = = water = . 2 ρ ρair 1.23. ∴V = 113 m/s.. Use ρ = 0.0021 slug/ft3 .. a) v =. 2p / ρ =. 2×.3 × 144/ .0021 = 203 ft/sec.. b) v =. 2p / ρ =. 2×.9 × 144/ .0021 = 351 ft/sec.. c) v =. 2p / ρ =. 2×.09 × 144/ .0021 = 111 ft/sec.. V2 120 × 1000  = 1.23  / 2 = 683 Pa.  2 3600  ∴F = pA = 683 π × 0.0752 = 12.1 N. 2. 3.51. 3.52. p= ρ. V2 p + = 0. 2 ρ. ∴V =. −2 p = ρ. 2 × 2000 = 57.0 m/s 1.23. 42.

(10) 3.53. (C). V12 p V22 + = . 2g γ 2g. V12 + 0.200 = 0.600. 2g. 3.54. (B). The manometer reading h implies:. ∴V = 2 × 9.81 × 0.400 = 2.80 m/s.. V12 p1 V22 p2 2 + = + or V22 = (60 −10.2). ∴V2 = 9.39 m/s The 2 ρ 2 ρ 1.13 temperature (the viscosity of the water) and the diameter of the pipe are not needed.. 3.55. 3.56. a). V 2 p V02 p + = + o. 2 ρ 2 ρ. (−10 x )2 p po + = . 2 ρ ρ. b). V 2 p V02 p + = + o. 2 ρ 2 ρ. (10 y ) 2 p po + = . 2 ρ ρ. ∴ p = po − 50 x 2 ρ ∴ p = po − 50 y 2 ρ. V 2 p U ∞2 p + = + ∞ . 2 ρ 2 ρ. a) v θ = 0 and θ = 180 o : v r = U ∞ (1 − rc2 / r 2 )( −1). 4 ρ 2 ρ 2  rc2  rc   2 ∴ p = U ∞ − v r = U ∞  2 2 −   .  r   2 2  r. (. b) Let r = rc :. pT =. ρ 2 U∞ 2. c) v r = 0 and r = rc : v θ = −U ∞ 2 sin θ . ∴ p = d) Let θ = 90 o :. 3.57. ). p 90 = −. ρ ρ U 2∞ − v θ2 = U ∞2 1 − 4 sin 2 θ 2 2. (. ). [. 3 ρU ∞2 2. V 2 p U ∞2 p + = + ∞ . 2 ρ 2 ρ 3 6 ρ 2 ρ 2   rc   rc   2 a) v θ = 0 and θ = 180 : p = U ∞ − v r = U ∞ 2  −   .  r   2 2   r . (. o. ). 1 ρU ∞2 . 2 ρ ρ c) v r = 0 and r = rc : p = U 2∞ − v θ2 = U ∞2 1 − 4 sin 2 θ 2 2 3 d) Let θ = 90 o : p 90 = − ρU ∞2 2 b) Let r = rc :. pT =. (. ). 43. [. ]. ].

(11) 3.58. V 2 p U ∞2 p + = + ∞ . 2 ρ 2 ρ. a) p =. 2 2   ρ 2 ρ 2  20π   1  2 U ∞ − u = 10 −  10 +   = 50 ρ 1 −  1 +    2 2 2π x   x    2 1  = − 50ρ  + 2  x x . (. ). b) u = 0 when x = −1.. p− 1 = −50 ρ( −2 + 1) = 50 ρ. 2   1 2  ρ 2 ρ 2  60π   2 1 2 c) p = U ∞ − u = 30 −  30 +  = 450 ρ 1 −  1 +   = −450 ρ + 2   x x  2 2  2π x   x      d) u = 0 when x = −1. p− 1 = −450 ρ ( − 2 + 1) = 450 ρ. (. 3.59. 3.60. ). V12 p 1 V 22 p2 + = + . V1 = 0 and p1 − p 2 = 20 kPa. 2 ρ 2 ρ 2 2 V22 = ( p1 − p 2 ) = ( 20 000) = 40. ∴ V 2 = 6.32 m / s ρ 1000 Assume the velocity in the plenum is zero. Then V12 p1 V22 p2 2 + = + or V22 = (60 −10.2). 2 ρ 2 ρ 1.13 We found ρ = 113 . kg / m 3 in Table B.2.. 3.61. ∴V2 = 9.39 m/s. Bernoulli from the stream to the pitot probe:. pT = ρ. V2 + p. 2. Manometer: pT + γ H − γ Hg H − γ h = p − γ h. Then, ρ. V2 + p + γ H − γ Hg H = p . 2. ∴V 2 =. (13.6 − 1)9800 ( 2 × 0.04). 1000 (13.6 − 1)9800 b) V 2 = ( 2 × 0.1). 1000 (13.6 − 1)62.4 c) V 2 = ( 2 × 2 / 12). 1.94 (13.6 − 1)62. 4 d) V 2 = (2 × 4 / 12 ). 1.94. a) V 2 =. 44. γ Hg − γ ρ. (2 H ). ∴ V = 3.14 m / s ∴ V = 4.97 m / s ∴ V = 11.62 fps ∴ V = 16.44 fps.

(12) 3.62. The pressure at 90° from Problem 3.56 is p90 = −3ρU ∞2 /2. The pressure at the. 3.63. stagnation point is pT = ρU∞2 /2. The manometer provides: pT − γH = p 90 1 3 × 1.204U ∞2 − 9800 × 0.04 = − × 1.204U ∞2 . ∴ U ∞ = 12.76 m/s 2 2 The pressure at 90° from Problem 3.57 is p90 = −3ρU ∞2 /2. The pressure at the stagnation point is pT = ρU∞2 /2. The manometer provides: pT − γH = p 90 1 3 × 1.204U ∞2 − 9800 × 0.04 = − × 1.204U ∞2 . ∴ U ∞ = 12.76 m/s 2 2. 3.64. Assume an incompressible flow with point 1 outside in the room where p1 = 0 and v 1 = 0. The Bernoulli’s equation gives, with p2 = γ w h2 , V12 p1 V22 p2 + = + . 2 ρ 2 ρ V 22 −9800 × 0.02 + . 2 1.204 V 2 −9800 × 0.08 b) 0 = 2 + . 2 1.204 V 22 −62.4 × 1 / 12 c) 0 = + . 2 0.00233 V 2 −62.4 × 4 / 12 d) 0 = 2 + . 2 0.00233 a) 0 =. 3.65. ∴ V 2 = 36.1 m / s ∴ V 2 = 66.8 fps ∴ V 2 = 133.6 fps. Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where p1 = 0 and V 1 = 0. Bernoulli’s equation gives 0=. V 22 p2 + . 2 ρ. ∴ p2 = −. p 90 = = 1.239 kg / RT 0.287 × 253 p 95 b) ρ = = = 1.212 kg / RT 0.287 × 273 p 92 c) ρ = = = 1.094 kg / RT 0.287 × 293 p 100 d) ρ = = = 1.113 kg / RT 0.287 × 313. a) ρ =. 3.66. ∴ V 2 = 18.04 m / s. (A). V12 p V2 p + 1 = 2 + 2. 2g γ 2g γ. 1 ρV 2 2 2. 1 × 1.239 × 100 2 = −6195 Pa 2 1 m 3 . ∴ p 2 = − × 1.212 × 100 2 = −6060 Pa 2 1 m 3 . ∴ p2 = − × 1.094 × 100 2 = −5470 Pa 2 1 m 3 . ∴ p2 = − × 1113 . × 100 2 = −5566 Pa 2 m 3 . ∴ p2 = −. 800000 V22 = . 9810 2 × 9.81. 45. ∴V2 = 40 m/s..

(13) 3.67. a) p A = γh = 9800 × 4 = 39 200 Pa, V A = 0. VA2 p V2 p + A + hA = 2 + 2 + h2 . 2g γ 2g γ. Using hA = h2 ,. p2 = p A −. V22 γ 2g. = 39 200 −. 14 2 × 9800 = −58 700 Pa 2 × 9.81. b) p B = 0 and VB = 0. Bernoulli’s eq. gives, with the datum through the pipe, 2 2 VB pB V2 p2 + + hB = + + h2 . 2g γ 2g γ. 3.68.  14 2  p2 =  4 −  9800 = −58 700 Pa 2 × 9.81 . V22 p V2 p + 2 = 1 + 1 2g γ 2g γ. Bernoulli:. V22 γ + p2 2g Substitute Bernoulli’s into the manometer equation:. Manometer: p1 + γ z + γ Hg H − γ H − γ z =. V12 γ + p1. 2g V12 × 9800 a) Use H = 0.01 m: = (13.6 − 1)9800 × 0.01 ∴ V1 = 1.572 m / s 2 × 9.81 Substitute into Bernoulli: V 22 − V12 20 2 − 1.572 2 p1 = γ = × 9800 = 198 600 Pa 2g 2 × 9.81. (. ). p1 + γ Hg − γ H =. V12 × 9800 b) Use H = 0.05 m: = (13.6 − 1)9800 × 0.05 ∴ V1 = 3.516 m / s 2 × 9.81 Substitute into Bernoulli: V 22 − V12 20 2 − 3.516 2 p1 = γ = × 9800 = 193 600 Pa 2g 2 × 9.81 V12 × 9800 c) Use H = 0.1 m: = (13.6 − 1) 9800 × 0.1 ∴ V1 = 4.972 m / s 2 × 9.81 Substitute into Bernoulli: V 22 − V12 20 2 − 4.972 2 p1 = γ = × 9800 = 187 400 Pa 2g 2 × 9.81. 46.

(14) 3.69. Bernoulli across nozzle:. V12 p V2 p + 1 = 2 + 2. 2 ρ 2 ρ. Bernoulli to max. height:. V12 p V2 p + 1 + h1 = 2 + 2 + h2 . 2g γ 2g γ. ∴ V2 = 2 p1 / ρ ∴ h2 = p1 / γ .. a) V 2 = 2 p 1 / ρ = 2 × 700 000 / 1000 = 37.42 m / s h2 = p1 / γ = 700 000 / 9800 = 71.4 m b) V 2 = 2 p 1 / ρ = 2 × 1 400 000 / 1000 = 52.92 m / s h2 = p1 / γ = 1 400 000 / 9800 = 142.9 m c) V 2 = 2 p 1 / ρ = 2 × 100 × 144 / 1.94 = 121.8 fps. h2 = p1 / γ = 100 × 144 / 62.4 = 231 ft d) V 2 = 2 p1 / ρ = 2 × 200 × 144 / 1.94 = 172.3 fps. h2 = p1 / γ = 200 × 144 / 62.4 = 462 ft 3.70. a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream flow: V12 p V2 p + 1 + h1 = 2 + 2 + h2 . 2g γ 2g γ. ∴V 2 = 2g ( H − h). b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the bottom of the downstream flow: V12 p V2 p + 1 + h1 = 2 + 2 + h2 . 2g γ 2g γ Using p1 = γH , p 2 = γh and h1 = h2 ,. 3.71. V 2 = 2 g( H − h). V12 p V2 p + 1 = 2 + 2. p2 = −100 000 Pa, the lowest possible pressure. 2 ρ 2 ρ 600 000 V 22 100 000 a) = − . ∴ V2 = 37.4 m/s. 1000 2 1000 300 000 V 22 100 000 b) = − . 1000 2 1000. ∴ V2 = 28.3 m/s.. 47.

(15) 3.72. 80 × 144 V 22 14.7 × 144 c) = − . 1.94 2 1.94. ∴ V2 = 118.6 ft/sec.. 40 × 144 V22 14.7 × 144 d) = − . 1.94 2 1.94. ∴ V2 = 90.1 ft/sec.. A water system must never have a negative pressure, since a leak could ingest impurities. ∴ The least pressure is zero gage. V12 p 1 V2 p + + gz 1 = 2 + 2 + gz 2 . V1 = V 2 . Let z 1 = 0, and p2 = 0. 2 ρ 2 ρ 500 000 = 9.81 z 2 . ∴ z2 = 51.0 m. 1000 ρ 2 ρ b) p1 = 2 ρ c) p1 = 2 ρ d) p1 = 2. a) p1 =. 3.73. (V. 2 2. ). (V22 − V12 ) (V22 − V12 ) (V22 − V12 ). (. ). ( ( (. ) ) ). 1000 2 2 − 10 2 = −48 000 Pa 2 902 2 = 2 − 10 2 = −43300 Pa 2 680 2 = 2 − 10 2 = −32600 Pa 2 1.23 2 = 2 − 102 = −59.0 Pa 2. − V12 =. (. ). (. ). ρ 2 1.23 2 V2 − V12 = 2 − 82 = −36.9 Pa 2 2. 3.74. V12 p 1 V 22 p 2 + = + . 2 ρ 2 ρ. 3.75. (D). 3.76. Apply Bernoulli’s equation between the exit (point 2) where the radius is R and a point 1 in between the exit and the center of the tube at a radius r less than R: V12 p 1 V 22 p 2 V 2 − V 12 + = + . ∴ p1 = ρ 2 . 2 ρ 2 ρ 2 Since V2 < V1 , we see that p1 is negative (a vacuum) so that the envelope would tend to rise due to the negative pressure over most of its area (except for a small area near the end of the tube).. 3.77. VD . For air ν ≅ 1.5 × 10 −5 . Use reasonable dimensions from your ν experience!. p1 =. (. p1 =. ). (. ). ρ 2 902 V2 − V12 = 30 2 − 152 = 304400 Pa 2 2. Re =. 48.

(16) 20 × 0.03 = 4 × 10 4 . ∴Separate 1.5 × 10 −5 20 × 0.005 b) Re = = 6700. ∴Separate 1.5 × 10 −5 20 × 2 c) Re = = 2.7 × 10 6 . ∴Separate −5 1.5 × 10 5 × 0.002 d) Re = = 670. ∴Separate 1.5 × 10 −5 20 × 2 e) Re = = 2.7 × 10 6 . ∴Separate 1.5 × 10 −5 100 × 3 f) Re = = 2 × 10 7 . 1.5 × 10 −5 ∴It will tend to separate, except streamlining the components eliminates separation.. a) Re =. 3.78. 3.79. A burr downstream of the opening will create a region that acts similar to a stagnation region thereby creating a high pressure since the velocity will be relatively low in that region.. V2 10 2 ∆n = 1000 × 0.02 = 40 000 Pa R 0.05 expect VA > 10 m / s and VB < 10 m /s. ∆p = ρ. stagnation region. B. Along AB, we A VA. 3.80. 3.81. The higher pressure at B will force the fluid toward the lower pressure at A, especially in the wall region of slow moving fluid, thereby causing a secondary flow normal to the pipe’s axis. This results in a relatively high loss for an elbow. V12 p1 V 22 p2 Refer to Bernoulli’s equation: + = + 2 ρ 2 ρ p A > pB since VA < VB. pC < pD. since VC > V D. pB > p D. since VD > V B. 49. VB.

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