0
3.6kviews
Prove it is a case of possible steady incompressible fluid flow. Calculate the velocity and acceleration at a point (2, 1, 3)

A flow $U = x^2yi+yz^2j - (2xyz+yz^2)k$

Prove it is a case of possible steady incompressible fluid flow. Calculate the velocity and

acceleration at a point (2, 1, 3)

$$\vec{u}=x^{2} y \hat{\imath}+z y^{2} \hat{\jmath}-\left(2 x y z+y z^{2}\right) \hat{k}$$ For steady and Incompressible Flow, $\Rightarrow \frac{\partial\left(x^{2} y\right)}{\partial x}+\frac{\partial\left(y^{2} z\right)}{\partial y}-\frac{\partial\left(2 x y z+y z^{2}\right)}{\partial z}$ $\Rightarrow 2 x y+2 y z-2 x y-2 y z=0$ $u=x^{2} y=2^{2} \times 1=4 \mathrm{~m} / \mathrm{s}$ $v=y^{2} z=1^{2} \times 3=3 \mathrm{~m} / \mathrm{s}$ $\omega=-\left(2 x y z+y z^{2}\right)=-\left[(2 \times 2 \times 1 \times 3)+\left(1 \times 3^{2}\right)\right]=-21 \mathrm{~m} / \mathrm{s}$ $a_{x}=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+\omega \frac{\partial u}{\partial z}$
$a_{x}=4 \times(2 x \times y)+3\left(x^{2}\right)-21 \times 0=(4 \times 2 \times 1 × 2)+3 \times 2^{2}=28 \mathrm{~m} / \mathrm{s^2}$ $a_{y}=4 \frac{\partial v}{\partial x}+v \frac{\partial v}{r y}+\omega \frac{\partial v}{r_{2}}$ $=(4 \times 0)+3 \times\left(2 y z\right)-21 \times y^{2}=0+(3 \times 2 \times 1 \times 3)-21 \times 1$ $=-3 \mathrm{~m} /\mathrm{s^2}$ \begin{array}{l} a_{z}=u \frac{\partial \omega}{\partial x}+v \frac{\partial \omega}{\partial y}+\omega \frac{\partial \omega}{\partial z}\ =4 \times(-2 y z)+3 x×-\left[2 x z+z^{2}\right]+21 \times[2 x y+2 y z]\ =4 \times(-2 \times 1 \times 3)+3 \times\left[-(2 \times 2 \times 3)+3^{2}\right]+21 \times[2 \times 1+2 \times 1 \times 3]\ =81 \mathrm{~m} / \mathrm{s}^{2};\ Hence, \text { velocity } v=4 \hat{i}+3 \hat{\jmath} - 21 k ;\ \text { acceleration, } 4=28 \hat{\imath}-3 \hat{\jmath}+81 \hat{k}. \end{array}