MC 2312 Mecánica De Fluidos – Potter (Solucionario Capítulo 5) pdf

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(1)CHAPTER 5. The Differential Forms of the Fundamental Laws 5.1. 0=. ∂ρ dV − + ∂t c. v .. ∫. c. s .. v   ∂ρ v −.  ∂t + ∇ ⋅ (ρV )d V c. v . c. v . Since this is true for all arbitrary control volumes (i.e., for all limits of integration), the integrand must be zero: v ∂ρ v + ∇ ⋅ ( ρV ) = 0. ∂t This can be written in rectangular coordinates as ∂ρ ∂ ∂ ∂ − = (ρu) + ( ρv ) + ( ρw ). ∂t ∂x ∂y ∂z This is Eq. 5.2.2. The other forms of the continuity equation follow. 0=. 5.2. ∂ρ dV − + ∂t c. v .. v. $ . Using Gauss’ theorem: ∫ ρV ⋅ ndA. ∫. v. v. ∫ ∇ ⋅ (ρV )d V−. =. ∫. ∂m element . ∂t ∂   ρv r ( rdθdz) − ρv r + (ρv r )dr( r + dr )dθdz ∂r   ∂   +ρv θdrdz − ρv θ + ( ρv θ )dθdrdz ∂θ   & in − m & out = m. dr ∂ dr ∂ dr    +ρv z  r +  dθdr − ρv z + (ρv z )dz  r +  dθdr = ρ r +  dθdrdz .  2 ∂z 2 ∂t   2    Subtract terms and divide by rdθdrdz : ρv ∂ r + dr ∂ 1 ∂ r + dr / 2 ∂ r + dr / 2 − r − (ρv r ) − (ρvθ ) − ( ρv z ) = ρ . r ∂r r ∂θ r ∂z r ∂t r Since dr is an infinitesimal, (r + dr ) / r = 1 and ( r + dr / 2) / r = 1. Hence, ∂ρ ∂ 1 ∂ ∂ 1 + ( ρv r ) + (ρvθ ) + ( ρv z ) + ρv r = 0. This can be put in various forms. ∂t ∂r r ∂θ ∂z r. 89.

(2) 5.3. ∂m element . ∂t ∂   ρv r ( rdθ) r sin θdφ − ρv r + (ρv r )dr ( r + dr)dθ( r + dr ) sin θdφ ∂r   dr  ∂ dr      +ρv θdr  r +  sin θdφ − ρvθ + ( ρv θ )dθ dr r +  sin θdφ  2 ∂θ 2    & in − m & out = m.    dr  ∂ dr   +ρv φdr  r +  dθ − ρv φ + (ρv φ )dφdr  r +  dθ    2 ∂φ 2  . =. 2  ∂   dr  ρ r + drdθ sin θdφ      ∂t  2 . Because some areas are not rectangular, we used an average length (r + dr / 2). Now, subtract some terms and divide by rdθ dφdr: dr r+ 2 ∂ ( r + dr ) ∂ 2 sin θ −ρv r sin θ − ρv r sin θ − (ρv r ) sin θ − (ρvθ ) ∂r r ∂θ r 2 dr  dr  r +  r+ ∂ 2 2 = ∂ρ  − (ρv φ ) sin θ ∂φ r ∂t r Since dr is infinitesimal (r + dr ) 2 / r = r and (r + dr / 2) / r = 1. Divide by r sinθ and there results ∂ρ ∂ 1 ∂ 1 ∂ 2 + ( ρv r ) + (ρvθ ) + (ρvφ ) + ρv r = 0 ∂t ∂r r ∂θ r sin θ ∂φ r ∂ρ = 0. Then, with v = w = 0 Eq. 5.2.2 yields ∂t ∂ du dρ ( ρu ) = 0 or ρ +u = 0. ∂x dx dx Partial derivatives are not used since there is only one independent variable.. 5.4. For a steady flow. 5.5. Since the flow is incompressible. Dρ = 0. This gives Dt v ∂ p ˆ 1∂ p ˆ 200 ρ  1 200 ρ  ∴∇ p = ir + iθ = 3  2 − cos2θ  iˆr − 3 sin2θ iˆθ ∂r r ∂θ r r  r ∂ρ ∂ρ u +w = 0. ∂x ∂z v v ∂u ∂w Also, ∇ ⋅ V = 0, or + = 0. ∂x ∂z. 90. or.

(3) ∂ ∂ρ = 0, ≠ 0. Since water can be considered to be incompressible, we ∂t ∂z Dρ ∂ρ ∂ρ demand that = 0. ∴u +w = 0, assuming the x-direction to be in the Dt ∂x ∂z v v ∂u ∂w direction of flow. Also, we demand that ∇ ⋅ V = 0, or + = 0. ∂x ∂z. 5.6. Given:. 5.7. We can use the ideal gas law, ρ =. 5.8. p . Then, the continuity equation RT v v p v v Dρ 1 Dp = − ρ∇ ⋅ V becomes, assuming RT to be constant, =− ∇ ⋅V Dt RT Dt RT v v 1 Dp = −∇ ⋅ V . p Dt. or. a) Use cylindrical coordinates with v θ = v z = 0: 1 ∂ ( rv r ) = 0 r ∂r Integrate: C rv r = C. ∴vr = . r b) Use spherical coordinates with v θ = v φ = 0: 1 ∂ 2 ( r vr ) = 0 r 2 ∂r Integrate: C r 2 v r = C. ∴vr = 2 . r. 5.9. v v  ∂u ∂v  kg Dρ = −ρ∇ ⋅V = −ρ +  = −2.3( 200 × 1 + 400 × 1) = −1380 . Dt m3 ⋅s  ∂x ∂y . 5.10. In a plane flow, u = u( x , y ) and v = v ( x , y ). Continuity demands that If u = const, then. ∂u ∂v = 0 and hence = 0. Thus, v = const also. ∂x ∂y. 91. ∂u ∂v + = 0. ∂x ∂y.

(4) 5.11. 5.12. If u = C1 and v = C2 , the continuity equation provides, for an incompressible flow, ∂u ∂v ∂w ∂w + + = 0. ∴ = 0 and w = C 3 . ∂x ∂y ∂z ∂z The z-component of velocity w is also constant. We also have Dρ ∂ρ ∂ρ ∂ρ ∂ρ =0= +u +v +w Dt ∂t ∂x ∂y ∂z The density may vary with x, y, z and t. It is not, necessarily, constant. ∂u ∂v + = 0. ∂x ∂y. ∴ A+. But, v ( x , o) = 0 = f ( x ). 5.13. ∂u ∂v + = 0. ∂x ∂y. ∴ v (x , y ) = ∫. 5.14. ∴. ∂v = 0. ∂y. ∴ v ( x , y ) = − Ay + f ( x ). ∴ v = − Ay.. ( x 2 + y 2 )5 − 5x( 2 x) 5x 2 − 5y 2 ∂v ∂u =− =− = − ∂y ∂x (x 2 + y 2 )2 (x 2 + y 2 )2. 5y 2 − 5x 2 5y dy + f ( x ) = 2 + f ( x ). 2 2 2 (x + y ) x + y2. f ( x ) = 0.. 1 ∂ 1 ∂ vθ 1 .4  ( rv r ) = − = −  10 + 2  sin θ. r ∂r r ∂θ r r  .4  .4    ∴ rv r = ∫  10 + 2  sin θdr + f (θ ) =  10r −  sin θ + f (θ ).   r  r. From Table 5.1:. .4   .2v r (.2, θ ) =  10 ×.2 −  sin θ + f (θ ) = 0.  .2 . ∴ f (θ ) = 0.. 0.4   ∴ v r =  10 − 2  sin θ .  r  5.15. 1 ∂ 1 ∂vθ −20  1 ( rv r ) = − =  1 + 2  cos θ. r ∂r r ∂θ r  r  1 1   ∴ rv r = ∫ −20  1 + 2  cos θdr + f (θ ) = −20  r −  cos θ + f (θ ).   r  r v r (1,θ ) = −20( 1 − 1)cos θ + f (θ ) = 0. ∴ f (θ ) = 0.. From Table 5.1:. 1  ∴ v r = −20 1 − 2  cos θ .  r . 92. ∴v =. 5y . x + y2 2.

(5) 5.16. 1 ∂ 2 1 ∂ (r v r ) = − ( v θ sin θ). 2 r ∂r r sin θ ∂θ. From Table 5.1, spherical coordinates:. 1 ∂ 2 1  40  (r v r ) =  10 + 3  2 sin θ cos θ. 2 r ∂r r sin θ  r  40  80    2 2 ∴ r v r = ∫ r  10 + 3  2 cos θdr + f (θ ) =  10r −  cos θ + f (θ )   r  r  80   2 4v r ( 2, θ ) =  10 × 2 −  cos θ + f (θ ) = 0. ∴ f (θ ) = 0.  2 ∴. 80   ∴ v r =  10 − 3  cos θ .  r  ∂ du dρ ( ρu ) = 0. ∴ρ +u = 0. ∂x dx dx slug p 18 × 144 du 526 − 453 ρ= = = 0.00302 . = = 219 fps / ft. 3 RT 1716 × 500 ft dx 2 × 2 / 12 dρ ρ du .00302 ∴ =− =− × 219 = −0.00136 slug / ft 4 . dx u dx 486. 5.17. Continuity:. 5.18. ∂u ∂v + = 0. ∂x ∂y. ∂ −20( 1 − e − x ) = −20 e − x ∂x. [. ]. Hence, in the vicinity of the x-axis: ∂v = 20e − x and v = 20ye − x + C. ∂y But v = 0 if y = 0. ∴ C = 0. v = 20 ye − x = 20( 0.2)e −2 = 0.541 m / s. 5.19. 1 ∂ ∂v ∂ ( rv r ) + z = 0. −20(1 − e − z ) = −20 e − z r ∂r ∂z ∂z Hence, in the vicinity of the z-axis: 1 ∂ r2 ( rv r ) = 20 e − z and rv r = 20 e − z + C. r ∂r 2 But v r = 0 if r = 0. ∴ C = 0.. [. ]. v r = 10 re − z = 10( 0.2) e −2 = 0.271 m / s. 5.20. The velocity is zero at the stagnation point. Hence, 40 0 = 10 − 2 . ∴R = 2 m R ∂u ∂v ∂u The continuity equation for this plane flow is + = 0. Using = 80 x − 3 , ∂x ∂y ∂x. 93.

(6) we see that. ∂v = −80 x −3 near the x-axis. Consequently, for small ∆y , ∂y. ∆v = −80x −3 ∆y. 5.21. so that v = −80( −3 ) −3 (0.1) = 0.296 m / s.. The velocity is zero at the stagnation point. Hence 40 0 = 2 − 10. ∴R = 2 m R 1 ∂ 2 1 ∂ 20 r vr ) = 2 ( 40 − 10r 2 ) = − . ( 2 r ∂r r ∂r r Near the negative x-axis continuity provides us with 1 ∂ 20 vθ sin θ) = . ( r sin θ ∂θ r Integrate, letting θ = 0 from the y-axis: v θ sin θ = −20 cos θ + C Since v θ = 0 when θ = 90 o , C = 0. Then, with α = tan −1 v θ = −20. 0.1 = 1.909 o , 3. cos θ cos 88.091 0.0333 = −20 = −20 = 0.667 m / s sin θ sin 88.091 0.999. ∂u ∂v ∆v ∆u 13.5 − 11.3 m/s + = 0. ∴ =− =− = −220 . ∂x ∂y ∆y ∆x 2 ×.005 m ∴ ∆v = v − 0 = −220∆y . ∴ v = −220 ×.004 = −0.88 m / s. ∂u b) a x = u = 12.6 × ( +220) = 2772 m /s 2 . ∂x. 5.22. Continuity:. 5.23. ΣFy = may . For the fluid particle occupying the volume of Fig. 5.3:. ∂τ yy dy  ∂τ zy dz  ∂τ xy dx      τ yy +  dxdz +  τ zy +  dxdy +  τ xy +  dydz ∂y 2  ∂z 2  ∂x 2     ∂τ yy dy  ∂τ zy dz  ∂τ xy dx     − τ yy −  dxdz −  τ zy −  dxdy −  τ xy −  dydz ∂y 2  ∂z 2  ∂x 2     +ρg y dx dy dz = ρdx dy dz. Dividing by dx dy dz , and adding and subtracting terms: ∂τ xy ∂τ yy ∂τ zy Dv + + + ρg y = ρ . ∂x ∂y ∂z Dt 5.24. Check continuity: ∂u ∂v ∂w ( x 2 + y 2 )10 − 10 x( 2 x) ( x 2 + y 2 )10 − 10y (2 y ) + + = + = 0. ∂x ∂y ∂z (x 2 + y 2 )2 ( x2 + y2 ) 2. 94. Dv Dt.

(7) Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with g x = g y = 0: ∂u ∂u ∂p ρu + ρv =− . ∂x ∂y ∂x. ∂p 10 x 10 y2 − 10 x 2 10 y −20 xy 100( x 2 + y 2 ) y ∴ = −ρ 2 −ρ 2 =ρ ∂x x + y 2 ( x 2 + y 2 )2 x + y 2 ( x 2 + y 2 )2 ( x2 + y2) 3 ∂v ∂v ∂p ρu + ρv =− . ∂x ∂y ∂y ∂p 10 x −20 xy 10 y 10 x 2 − 10 y 2 100( x 2 + y 2 ) y = −ρ 2 − ρ = ρ ∂y x + y 2 (x 2 + y 2 )2 x 2 + y 2 ( x 2 + y 2 )2 (x 2 + y 2 )3 v ∂pˆ ∂p ˆ 100 xρ ˆ 100 yρ ˆ 100ρ ∴∇ p = i + j= i+ j= ( xiˆ + yjˆ ). ∂x ∂y (x 2 + y 2 )2 ( x2 + y 2 ) 2 ( x 2 + y 2 )2 ∴. 5.25. Check continuity (cylindrical coord from Table 5.1): 1 ∂ 1 ∂vθ 10  1 −10  1 ( rv r ) + =  1 + 2  cos θ +  1 + 2  cos θ = 0. ∴It is a possible r ∂r r ∂θ r  r  r  r  flow. For Euler’s Eqs. (let ν = 0 in the momentum eqns of Table 5.1) in cylindrical coord: 2 vθ2 vθ ∂ vr 100 ρ  ∂p ∂ vr 1  1  =ρ − ρ vr −ρ = 1 + 2  sin 2 θ − 10 ρ 1 − 2  ∂r r ∂r r ∂θ r  r   r 10 ρ  1 10  2  −  1 + 2  sin θ  10 − 2  .  r  r  r  vv ∂v v ∂v 1∂p 100 ρ  1  = − ρ r θ − ρ vr θ − ρ θ θ = 1 − 4  sin θ cos θ  r ∂θ r ∂r r ∂θ r  r .  2  20   cos θ  3   r . 2. 1 20 100ρ  1 −10 ρ  1 − 2  cos θ sin θ  3  −  1 + 2  sin θ cos θ .  r  r  r  r  v ∂ p ˆ 1∂ p ˆ 200 ρ  1 200 ρ  ∴∇ p = ir + iθ = 3  2 − cos2θ  iˆr − 3 sin2θ iˆθ ∂r r ∂θ r r  r. 5.26. This is an involved problem. Follow the steps of Problem 5.25. Good luck! v θ2 + v φ2 ∂p v ∂v r ∂v =ρ − ρv r r − ρ θ ∂r r ∂r r ∂θ (v v ) ∂v v ∂v 1 ∂p = −ρ r θ − ρv r θ − ρ θ θ r ∂θ r ∂r r ∂θ. (. ). 95.

(8) 5.27.  2µ v v  2µ v v ∴ p = p − + λ ∇ ⋅V . ∴ p − p = − + λ ∇ ⋅V .  3   3  ∂ s$ ∆$s ∆αn$ n$ ≅ =− =− . ∂ s ∆s R∆α R ∂ s$ ∆$s n$ ∆θ ∂θ ≅ = = n$ . ∂t ∆ t ∆t ∂t v DV  ∂ V ∂ V   ∂θ V 2  ∴ = +V sˆ +  V −  nˆ. Dt  ∂ t ∂ s   ∂ t R   V2 ∂V For steady flow, the normal acc. is  − .  , the tangential acc. is V ∂s  R . 5.28. 5.29. v For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to Ω. Thus, Euler’s equation becomes v v v v v v v dΩ v  v  DV v ρ + 2Ω × V + Ω × (Ω × r ) + × r  = − ∇ p − ρ g. dt  Dt . v v ∂u + λ ∇ ⋅ V = −30 psi. ∂x = − p = −30 psi.. τ xx = − p + 2µ τ yy = τ zz. ∂u ∂v  .1  −5  −5 τ xy = µ  +  = 10 30 − 1440 ×  = 18 × 10 psf. 12   ∂ y ∂ x  τ xy 18 × 10 −5 τ xz = τ yz = 0. = = 4.17 × 10 −8 . τ xx 30 × 144 5.30. ∂v ∂u 16 y 16 y 2 8 y2 16 y 3 =− = − . ∴ v ( x, y ) = 9 / 5 − 2 13/5 + f ( x ). ∂y ∂ x C x 9 / 5 C 2 x13/5 Cx 3C x 4/ 5 v ( x , o) = 0. ∴ f ( x ) = 0. 8 = C 1000 . ∴ C = 0.0318. ∴ u( x , y ) = 629 yx − 4/ 5 − 9890 y 2 x −8/ 5 .. τ xx. v ( x , y ) = 252 y 2 x −9/ 5 − 5270 y 3 x −13/ 5 . ∂u = − p + 2µ = −100 + 0 = −100 kPa. ∂x τ yy = τ zz = − p = −100 kPa.. ∂u ∂v  −5 −4 / 5  τ xy = µ  + = 5.01 ×10 −5 Pa.  = 2 ×10  629 × 1000  ∂ y ∂ x   τ xz = τ yz = 0.. 96.

(9) 5.31. v v Du ∂ u  ∂ ∂ ∂  = + u +v +w u = ( V ⋅ ∇) u.  Dt ∂t  ∂ x ∂y ∂z v v Dv ∂ v  ∂ ∂ ∂  = + u +v +w v = ( V ⋅ ∇ ) v.  Dt ∂t  ∂ x ∂y ∂ z v v Dw ∂ w  ∂ ∂ ∂  = +u +v +w w = ( V ⋅ ∇)w  Dt ∂t  ∂ x ∂y ∂z v v v v DV Du ˆ Dv ˆ Dw ˆ v v ˆ ˆ ∴ = i+ j+ k = V ⋅ ∇ (ui + vj + wkˆ ) = (V ⋅ ∇ )V . Dt Dt Dt Dt. 5.32. Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible effects: v v v µ ∂ v v µ ∂ v v µ ∂ v vˆ DV v ρ = − ∇p + ρ g + µ∇ 2V + ∇ ⋅ Viˆ + ∇ ⋅ Vjˆ + ∇ ⋅ Vk Dt 3∂x 3∂y 3∂z v v µ ∂ ∂ $ ∂ $ v v v = −∇p + ρg + µ∇ 2V +  i$ + j+ k ∇ ⋅V 3  ∂x ∂y ∂z  v v v µ v v v DV v ∴ρ = −∇p + ρg + µ∇ 2V + ∇( ∇ ⋅ V ). Dt 3. 5.33. If u=u(y), then continuity demands that. ∂v = 0. ∂y. ∴ v = C.. But, at y=0 (the lower plate) v=0. ∴ C = 0 , and v ( x , y ) = 0. ∴ρ.  ∂ 2u ∂ 2u ∂ 2 u  Du ∂u ∂u ∂u ∂u ∂p = ρ +u +v +w = − + ρ g + µ  2 + 2 + 2  .  x Dt ∂x ∂y ∂z ∂x ∂y ∂z   ∂t ∂ x ∴0 = −. ρ ρ. 5.34. ∂p ∂x. +µ. ay 2. .. Dv ∂p =0= − . Dt ∂y. ∂p Dw =0=− + ρ( − g ). Dt ∂z. Continuity:. ∂2u. ∴0 = −. ∂p ∂z. − ρg .. ∂ ( rvr ) = 0. ∴ rvr = C. At r = 0, vr ≠ ∞. ∂r Dvr 1∂p =0=− . Dt ρ ∂r. Dvθ 1 ∂p =0= − . Dt ρ r ∂θ. 97. ∴ C = 0..

(10)  ∂ 2v  ∂v Dvz ∂ v z vθ ∂ vz ∂ vz  ∂p 1 ∂vz 1 ∂ 2vz ∂ 2vz z z  =− ρ = ρ + vr + + vz +µ + + +  ∂t  ∂ r 2 r ∂ r r 2 ∂θ 2 Dt ∂r r ∂θ ∂z  ∂z ∂ z2    ∴0 = −. 5.35. Continuity:. ∂p  ∂2 v 1 ∂v z  + µ 2z + . ∂z r ∂r   ∂r. 1 ∂ 2 ( r v r ) = 0. r 2 ∂r. ∴ r 2 v r = C.. At r = r1 , v r = 0.. ∴ C = 0.. vθ2 ∂p  2v  − ρ=− + µ  − 2θ cot θ . r ∂r  r . 5.36. 0=−. 1∂ p  1 ∂  2 ∂ vθ +µ r 2 r ∂θ ∂r r ∂ r . 0= −. 1 ∂p . r sin θ ∂φ. vθ   − 2 2   r sin θ . v v For an incompressible flow ∇ ⋅ V = 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and 5.3.3: Du ∂  ∂u  ∂  ∂u ∂v  ∂  ∂u ∂w  ρ = µ  +  + µ +  −p + 2µ  +  + ρg x . Dt ∂x  ∂x  ∂y  ∂y ∂x  ∂z  ∂z ∂x  =− ∴ρ. ρ. ∂p ∂2u ∂2u ∂2u ∂  ∂u ∂v ∂w  +µ 2 +µ 2 +µ 2 +µ  + +  + ρg x ∂x ∂x ∂y ∂z ∂x  ∂x ∂y ∂z . ∂p  ∂2u ∂2u ∂2u Du =− + µ  2 + 2 + 2  + ρg x . Dt ∂x ∂y ∂z   ∂x. Dv ∂  ∂ u ∂ v  ∂  ∂ v  ∂ ∂ v ∂ w  = µ + µ + +  − p + 2µ +  + ρgy. Dt ∂ x  ∂ y ∂ x  ∂ y  ∂ y ∂ z ∂ z ∂ y  ∂p ∂2v ∂2v ∂ 2v ∂  ∂u ∂v ∂w  =− + µ 2 +µ 2 +µ 2 +µ  + +  + ρg y ∂y ∂x ∂y ∂z ∂y  ∂x ∂y ∂z . ∂p  ∂2v ∂2v ∂2 v Dv ∴ρ =− + µ  2 + 2 + 2  + ρg y . Dt ∂y ∂y ∂z   ∂x. ρ. Dw ∂  ∂ u ∂ w  ∂  ∂ v ∂ w  ∂  ∂w = µ + µ + + +  − p + 2µ  + ρ gz Dt ∂ x  ∂ z ∂ x  ∂ y  ∂ z ∂ y  ∂ z  ∂z . 98.    .

(11) =−. ∂p ∂2w ∂2w ∂2 w ∂  ∂u ∂v ∂w +µ 2 +µ 2 +µ 2 +µ  + +  + ρg z ∂z ∂x ∂y ∂z ∂z  ∂x ∂y ∂z . ∂p  ∂2w ∂2w ∂2w  Dw ∴ρ = − + µ  2 + 2 + 2  + ρg z . Dt ∂z ∂y ∂z   ∂x. 5.37. If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with µ = µ ( x , y , z ) we arrive at.  ∂ 2u ∂ 2u ∂ 2u  Du ∂p ∂µ ∂ u ∂µ  ∂ u ∂ v  ∂µ  ∂ u ∂ w  ρ =− + ρ gx + µ  2 + 2 + 2  + 2 + + +  +     Dt ∂x ∂ x ∂ x ∂ y∂ y ∂ x  ∂ z ∂ z ∂ x  ∂y ∂z  ∂ x. 5.38. If plane flow is only parallel to the plate, v = w = 0. Continuity then demands that ∂u / ∂x = 0. The first equation of (5.3.14) simplifies to.  ∂ 2u ∂ 2u ∂ 2u ∂u ∂u ∂u ∂u ∂p  ρ +u +v +w = − + ρ g + µ + +  x  ∂ x2 ∂ y2 ∂ z2 ∂x ∂y ∂z ∂x ∂t  ρ.    . ∂u ∂ 2u =µ ∂t ∂ y2. We assumed g to be in the y-direction, and since no forcing occurs other than due to the motion of the plate, we let ∂p / ∂x = 0. 5.39. 5.40. τ xx + τ yy + τ zz. v v 2µ  ∂u ∂v ∂w   + +  − λ ∇ ⋅ V. 3 3  ∂x ∂y ∂z   2µ v v  2µ v v ∴ p = p − + λ ∇ ⋅V . ∴ p − p = − + λ ∇ ⋅V .  3   3  From Eqs. 5.3.10, −. = p−. v v  ∂ ∂ ∂ (V ⋅ ∇ )V =  u + v + w  (uiˆ + vjˆ + wkˆ ) ∂y ∂z   ∂x v v  ∂  ∂w ∂w ∂ w  ∂  ∂v ∂v ∂v   ∇ × (V ⋅ ∇ )V =   u +v +w −  u + v + w   iˆ  ∂y ∂z  ∂ z  ∂x ∂y ∂ z   ∂y  ∂x.  ∂  ∂u ∂u ∂u  ∂  ∂w ∂w ∂w   +  u +v + w  − u + v + w   ˆj ∂y ∂z  ∂ x  ∂x ∂y ∂z    ∂z  ∂x  ∂  ∂v ∂v ∂v  ∂  ∂u ∂u ∂u   +  u + v + w  − u + v + w   kˆ ∂y ∂ z  ∂ y  ∂x ∂y ∂z   ∂x  ∂x ∂w ∂v ˆ ∂u ∂w ˆ ∂v ∂u ˆ v Use the definition of vorticity: ω = ( − )i + ( − ) j + ( − )k ∂y ∂z ∂z ∂x ∂x ∂y. 99.

(12) v  ∂w ∂v ∂ ∂u ∂w ∂ ∂ v ∂x ∂  v (ω ⋅ ∇ )V =  ( − ) + ( − ) + ( − )  ( uiˆ + vjˆ + wkˆ)  ∂y ∂z ∂x ∂z ∂x ∂y ∂x ∂y ∂z  v ∂ ∂   ∂ w ∂v ∂u ∂w ˆ ∂v ∂u ˆ  v  ∂ (V ⋅ ∇ )ω = u + v + w  ( − )iˆ + ( − ) j + ( − )k  ∂y ∂ z   ∂y ∂z ∂z ∂x ∂x ∂y   ∂x Expand the above, collect like terms, and compare coefficients of iˆ , jˆ, and kˆ. 5.41. Studying the vorticity components of Eq. 3.2.21, we see that ω z = −∂ u / ∂ y is the only vorticity component of interest. The third equation of Eq. 5.3.24 then simplifies to Dω z = ν∇ 2ω z Dt ∂ 2ω z =ν ∂ y2 since changes normal to the plate are much larger than changes along the plate, ∂ω z ∂ω z i.e., >> . ∂y ∂x. 5.42. If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces to Dω z =0 Dt that is, there is no change in vorticity (along a streamline) between sections 1 and 2. Since (see Eq. 3.2.21), at section 1, ∂v ∂u ωz = − = −10 ∂x ∂y we conclude that, for the lower half of the flow at section 2, ∂u = 10. ∂y. This means the velocity profile at section 2 is a straight line with the same slope of the profile at section 1. Since we are neglecting viscosity, the flow can slip at the wall with a slip velocity u0 ; hence, the velocity distribution at section 2 is u 2 ( y ) = u 0 + 10 y . Continuity then allows us to calculate the profile: V1 A 1 = V 2 A 2 1 (10 × 0.04 )(0.04w ) = (u 0 + 10 × 0.02 / 2 )(0.02w ). ∴ u 0 = 0.3 m / s. 2 Finally, u 2 ( y ) = 0.3 + 10 y. 100.

(13) 5.43. No. The first of Eqs. 5.3.24 shows that, neglecting viscous effects, Dω x ∂u ∂u ∂u = ωx + ωy + ωz Dt ∂x ∂y ∂z so that ω y , which is nonzero near the snow surface, creates ω x through the term ω y ∂ u / ∂ y, since there would be a nonzero ∂u / ∂y near the tree.. 5.44. v $ = ∫ k∇T ⋅ ndA. c .s .. ∫. p v ∂ V 2   V2 ~ $ + gz + u ) ρ d V − + + gz + ~ u +  ρV ⋅ ndA   ∫c.v. ∂t  2 ∫   2 ρ  c . s.. v v ∇⋅ ( k∇ T )d V − =. c .v .. ∂ V 2  ∫c.v. ∂t  2 + gz + ~u ρd V− +. v v V2 p ∇ ∫c.v. ⋅ ρV 2 + gz + ~u + ρ d V−. v V 2  p  ∂  V2  v 2 ~ ∴ ∫ − k∇ T +  ρ + ρgz + ρu  + ∇ ⋅ ρV  + gz + ~ u +  d V − = 0. ∂t  2   2 ρ  c. v .  v v v V2 v v ∂ V v v ∇p v  ∂ V2 v p  V 2  ∂ρ v ρ + ∇ ⋅ ρV  + gz +  = + ∇ ⋅ ρ V  + ρV ⋅  + V ⋅ ∇V + + g ∇z  = 0.  ∂t 2 ρ  2  ∂t ρ ∂t    2. continuity ~ v v ∂ Du ∴ − k∇ 2T + ρu~ + ρV ⋅ ∇~ u = 0. ∴ρ = k∇ 2T . ∂t Dt 5.45. Divide each side by dxdydz and observe that. ∂T ∂x. − x + dx. ∂T ∂x. x. dx Eq. 5.4.5 follows. 5.46. momentum. =. ∂ T 2. ∂x 2. ,. ∂T ∂y. − y + dy. ∂T ∂y. dy. y. =. ∂ T 2. ∂x2. ,. ∂T ∂z. − z + dz. dz. ∂T ∂z. z. =. ∂ 2T ∂z 2. v Du% D( h − p / ρ ) Dh Dp p D ρ Dh Dp p =ρ =ρ − + =ρ − +  − ρ∇ ⋅ V  Dt Dt Dt Dt ρ Dt Dt Dt ρ v where we used the continuity equation: D ρ / Dt = − ρ∇ ⋅ V . Then Eq. 5.4. 9 becomes v v Dh Dp p ρ − +  − ρ∇ ⋅ V  = K ∇2T − p∇ ⋅ V Dt Dt ρ ρ. which is simplified to Dh Dp ρ = K ∇2T + Dt Dt. 101.

(14)  ∂T ∂T ∂T ∂T  ∴ ρc  +u +v + w  = k∇ 2T . ∂x ∂y ∂z   ∂t ∂T Neglect terms with velocity: ρc = k∇ 2 T. ∂t. 5.47. See Eq. 5.4.9: ~ u = cT .. 5.48. The dissipation function Φ involves viscous effects. For flows with extremely large velocity gradients, it becomes quite large. Then DT ρcp =Φ Dt DT and is large. This leads to very high temperatures on reentry vehicles. Dt. 5.49. 5.50. ∂u = −2r × 10 5 . ( r takes the place of y ) ∂r  1  ∂u  2  From Eq. 5.4.17, Φ = 2µ     = µ 4 r 2 × 10 10 .  2  ∂y   At the wall where r = 0.01 m , Φ = 1.8 × 10 −5 × 4×.012 × 10 10 = 72 N / m 2 ⋅ s . ∂u At the centerline = 0 so Φ = 0. ∂r At a point half-way: Φ = 1.8 × 10 −5 × 4×.005 2 × 10 10 = 18 N / m 2 ⋅ s . u = 10( 1 − 10 000 r 2 ).. (a) Momentum:. ∴. ∂u ∂2u =ν 2 ∂t ∂y 2.  ∂u  ∂T ∂ 2T Energy: ρc = K 2 + µ  . ∂t ∂y  ∂y . (b) Momentum: ρ. ∂u ∂ 2u ∂µ ∂u =µ 2 + ∂t ∂y ∂y ∂y 2.  ∂u  ∂T ∂ 2T Energy: ρc = K 2 + µ  . ∂t ∂y  ∂y . 102.

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