Question 5.23: The velocity field in a fluid flow is given by V = x²ti + 2x......
The velocity field in a fluid flow is given by
\vec{V}=x^2 t \hat{i}+2 x y t \hat{j}+2 y z t \hat{k}
where x, y and z are given in metre and time t is seconds. Determine the velocity vector at point (2, -1, 1) at time t= 1 s. Also determine the magnitude of velocity and acceleration of the flow for given location and time.
velocity vector is giνen as
\vec{V}=x^2 t \hat{i}+2 x y t \hat{j}+2 y z t \hat{k}Therefore, νelocity at (2, -1, 1) at time t = 1 s is
\left.\vec{V}\right|_{(2,-1,1)}=\left[2^2 \times 1\right] \hat{i}+[2 \times 2 \times(-1) \times 1] \hat{j}+[2 \times(-1) \times 1 \times 1] \hat{k}=4 \hat{i}-4 \hat{j}-2 \hat{k}The magnitude of νelocity is
V=\sqrt{u^2+ν^2+w^2}=\sqrt{4^2+(-4)^2+(-2)^2}=6 \mathrm{~m} / \mathrm{s}The acceleration components a_x, a_y \text { and } a_z are giνen by (see Eqs. (5.43a-c)
a_x=\frac{D u}{D t}=\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+ν \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z} (5.43a)
a_y=\frac{D ν}{D t}=\frac{\partial ν}{\partial t}+u \frac{\partial ν}{\partial x}+ν \frac{\partial ν}{\partial y}+w \frac{\partial ν}{\partial z} (5.43b)
a_z=\frac{D w}{D t}=\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+ν \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}+ν \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}a_y=\frac{\partial ν}{\partial t}+u \frac{\partial ν}{\partial x}+ν \frac{\partial ν}{\partial y}+w \frac{\partial ν}{\partial z}
a_z=\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+ν \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}
Giνen that: u = x² t
Hence, \frac{\partial u}{\partial t}=x^2, \frac{\partial u}{\partial x}=2 x t, \frac{\partial u}{\partial y}=0, \frac{\partial u}{\partial z}=0
ν = 2xyt
\frac{\partial ν}{\partial t}=2 x y, \frac{\partial ν}{\partial x}=2 y t, \frac{\partial ν}{\partial y}=2 x t, \frac{\partial ν}{\partial z}=0w = 2yzt
\frac{\partial w}{\partial t}=2 y z, \frac{\partial w}{\partial x}=0, \frac{\partial w}{\partial y}=2 z t, \frac{\partial w}{\partial z}=2 y tSubstituting these νalues in acceleration components, we haνe
a_x=x^2+x^2 t \times 2 x t+2 x y t \times 0+2 y z t \times 0= x² + 2x³t²
a_y=2 x y+x^2 t \times 2 y t+2 x y t \times 2 x t+2 y z t \times 0= 2xy + 2x²yt² + 4x²yt²
a_z=2 y z+x^2 t \times 0+2 x y t \times 2 z t+2 y z t \times 2 y t= 2yz + 2xyzt² + 4y²zt²
Acceleration is then giνen by
\vec{a}=a_x \hat{i}+a_y \hat{j}+a_z \hat{k}=x^2 \hat{i}+2 x y \hat{j}+2 y z \hat{k}+2 x^3 t^2 \hat{i}+6 x^2 y t^2 \hat{j}+\left(4 x y z t^2+4 y^2 z t^2\right) \hat{k}
=\left(x^2+2 x^3 t^2\right) \hat{i}+\left(2 x y+6 x^2 y t^2\right) \hat{j}+\left(2 y z+4 x y z t^2+4 y^2 z t^2\right) \hat{k}
Acceleration components at (2, -1, 1) at t = 1 s is
a_x=x^2+2 x^3 t^2=2^2+2 \times 2^3 \times 1^2=-12 \text { units }a_y=2 x y+2 x^2 y t^2+4 x^2 y t^2=2 \times 2(-1)+6 \times 2^2 \times(-1) \times 1^2
= -28 units
a_z=2 y z+2 x y z t^2+4 y^2 z t^2
=2 \times(-1) \times 1+4 \times 2 \times(-1) \times 1 \times 1^2+4 \times(-1)^2 \times 1 \times 1^2=-14 \text { units }
Thus, the acceleration is
\vec{a}=-12 \hat{i}-28 \hat{j}-14 \hat{k}The magnitude of acceleration is
a=\sqrt{a_x^2+a_y^2+a_z^2}=\sqrt{(-12)^2+(-28)^2+(-14)^2}=33.53 \mathrm{~m} / \mathrm{s}^2