Fluid Mechanics-Frank M White Solution Ch4

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Fluid Mechanics-Frank M White Solution Ch4에 대한 보고서 자료입니다.

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P4.1 An idealized velocity field is given by the formula

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). du dt = ∂ u ∂ t +u ∂ u ∂ x+v ∂ u ∂ y+w ∂ u ∂ z =4x+4tx(4t)−2t2y(0)+4xz(0)=4x+16t2x dv dt = ∂ v ∂ t +u ∂ v ∂ x+v ∂ v ∂ y+w ∂ v ∂ z =−4ty+4tx(0)−2t2y(−2t2)+4xz(0)=−4ty+4t4y dw dt = ∂ w ∂ t +u ∂ w ∂ x +v ∂ w ∂ y +w ∂ w ∂ z =0+4tx(4z)−2t2y(0)+4xz(4x)=16txz+16x2z
or: dV dt =(4x+16t2x)i+(−4ty+4t4y)j+(16txz+16x2z)k at (x, y, z) = (−1, +1, 0), we obtain (d) At (–1, +1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans.

P4.2 A three-dimensional flow field has the velocity given by V=4x2i+3xyj+5t(x+y)zk Find the local and total accelerations in terms of x, y, z, and t. Solution
Velocity  V=(4x2, 3xy, 5t(x+y)z) Local acceleration∂  V ∂t =(0, 0, 5(x+y)z)
Convective acceleration ax=u ∂ u ∂ x+v ∂ u ∂ y+w ∂ u ∂ z =(4x2)(8x)+(3xy)(0)+5t(x + y)z(0) =32x3

2

ay=u
∂ v ∂ x+v
∂ u ∂ y+w
∂ u ∂ z =(4x2)(3y)+(3xy)(3x)+5t(x + y)z(0) =12x2y+9x2y =21x2y az=u ∂ w ∂ x +v ∂ w ∂ y +w ∂ w ∂ z =(4x2)(5tz)+(3xy)(5tz)+5t(x + y)z(5t)(x + y) =20tx2z+15txyz+25t2(x+y)2z

Total acceleration = Local accel + convective accel ∂  v ∂ t =(32x3, 21x2y, 20tx2z + 15txyz + 25t2(x+y)2z+5(x+y)z)

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