Question 5.24: A three-dimensional velocity field is given by V= — x² yi — ......

Velocity vector is given as

\vec{V}=-x^2 y \hat{i}-y^2 z \hat{j}+\left(2 x y z+y z^2\right) \hat{k}

Therefore, velocity at (1, 1, 2) is

\left.\vec{V}\right|_{(1,1,2)}=-\left[1^2 \times 1\right] \hat{i}-\left[1^2 \times 2\right] \hat{j}+\left[2 \times 1 \times 1 \times 2+1 \times 2^2\right] \hat{k}=-\hat{i}-2 \hat{j}+8 \hat{k}

The acceleration components a_x, a_y \text { and } a_z are given by (see Eqs. (5.43a—c)

a_x=\frac{D u}{D t}=\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}       (5.43a)

a_y=\frac{D v}{D t}=\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}     (5.43b)

a_z=\frac{D w}{D t}=\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}     (5.43C)

a_x=\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}

a_y=\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}

a_z=\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}

Given that:                      = —x²y

Hence,                  \frac{\partial u}{\partial t}=0, \frac{\partial u}{\partial x}=-2 x y, \frac{\partial u}{\partial y}=-x^2, \frac{\partial u}{\partial z}=0

v = —y²z

\frac{\partial v}{\partial t}=0, \frac{\partial v}{\partial x}=0, \frac{\partial v}{\partial y}=-2 y z, \frac{\partial v}{\partial z}=-y^2

w= 2xyz + yz²

\frac{\partial w}{\partial t}=0, \frac{\partial w}{\partial x}=2 y z, \frac{\partial w}{\partial y}=\left(2 x z+z^2\right), \frac{\partial w}{\partial z}=(2 x y+2 y z)

Substituting these values in acceleration components, we have

a_x=0-x^2 y \times(-2 x y)-y^2 z \times\left(-x^2\right)+\left(2 x y z+y z^2\right) \times 0

= 2x³y² + x²y²z

a_y=0-x^2 y \times 0-y^2 z \times(-2 y z)+\left(2 x y z+y z^2\right) \times\left(-y^2\right)

= 2y³z² — 2xy³z — y3z² = y³z² — 2xy³z

a_z=0-x^2 y \times(2 y z)-y^2 z \times\left(2 x z+z^2\right)+\left(2 x y z+y z^2\right) \times(2 x y+2 y z)

= —2x² y²z — 2xy²z² — y²z³ + 4x² y²z + 4xy²z²+ 2xy²z² + 2y²z³
= 2x² y²z + 4xy²z² + y²z³

Acceleration is then given by

\vec{a}=a_x \hat{i}+a_y \hat{j}+a_z \hat{k}

=\left(2 x^3 y^2+x^2 y^2 z\right) \hat{i}+\left(y^3 z^2-2 x y^3 z\right) \hat{j}+\left(2 x^2 y^2 z+4 x y^2 z^2+y^2 z^3\right) \hat{k}

Acceleration at (1, 1, 2) is

\left.\vec{a}\right|_{(1,1,2)}=\left[2 \times 1^3 \times 1^2+1^2 \times 1^2 \times 2\right] \hat{i}+\left[2 \times 1^3 \times 2^2-2 \times 1 \times 1^3 \times(2)\right] \hat{j}

+\left[2 \times 1^2 \times 1^2 \times 2+4 \times 1 \times 1^2 \times 2^2+1^2 \times 2^3\right] \hat{k}

=4 \hat{i}+28 \hat{k}