Question

The velocity vector in a fluid flow is given by V=4x3i-10x2yj+2tk.find the velocity and acceleration of a fluid particle at (2,1,3)at time t=1

Answer 1

Step-by-step Explanation:

Given: the velocity vector $\vec{v} = 4x^{3} \hat{i} - 10x^{2} y \hat{j} + 2t \hat{k}$, coordinates (2,1,3), time t = 1s

To Find: velocity and acceleration of the fluid particle

Solution:

• The velocity of the fluid particle at (2,1,3) and t = 1s

We have the velocity vector $\vec{v} = 4x^{3} \hat{i} - 10x^{2} y \hat{j} + 2t \hat{k}$, therefore, at $(x,y,z,t) = (2,1,3,1)$

$\Rightarrow \vec{v} \ |_{(2,1,3,1)} = 4(2)^{3} \hat{i} - 10 (2)^{2} (1) \hat{j} + 2(2) \hat{k}$

$\Rightarrow \vec{v} \ |_{(2,1,3,1)} = 32 \hat{i} - 40 \hat{j} + 4 \hat{k} \ \ \cdots \cdots \cdots (1)$

• The acceleration of the fluid particle at (2,1,3) and t = 1s

To find acceleration, let $\vec{v} = \vec{v} (x,y,z,t)$ such that,

$d\vec{v} = \Big( \frac{\partial \vec{v}}{\partial x} \Big)_{(y,z,t)} dx + \Big( \frac{\partial \vec{v}}{\partial y} \Big)_{(x,z,t)} dy + \Big( \frac{\partial \vec{v}}{\partial z} \Big)_{(y,x,t)} dz + \Big( \frac{\partial \vec{v}}{\partial t} \Big)_{(x,y,z)} dt$

$\Rightarrow \vec{a} = \frac{d\vec{v}}{dt} = \Big( \frac{\partial \vec{v}}{\partial x} \Big)_{(y,z,t)} \frac{dx}{dt} + \Big( \frac{\partial \vec{v}}{\partial y} \Big)_{(x,z,t)} \frac{dy}{dt} + \Big( \frac{\partial \vec{v}}{\partial z} \Big)_{(y,x,t)} \frac{dz}{dt} + \Big( \frac{\partial \vec{v}}{\partial t} \Big)_{(x,y,z)}$

$\Rightarrow \vec{a} = \frac{d\vec{v}}{dt} = \Big( \frac{\partial \vec{v}}{\partial x} \Big)_{(y,z,t)} v_x + \Big( \frac{\partial \vec{v}}{\partial y} \Big)_{(x,z,t)} v_y + \Big( \frac{\partial \vec{v}}{\partial z} \Big)_{(y,x,t)} v_z} + \Big( \frac{\partial \vec{v}}{\partial t} \Big)_{(x,y,z)} \ \ \cdots \cdots \cdots (2)$

The above expression (2) will give the acceleration of the fluid particle

Since we have $\vec{v} = 4x^{3} \hat{i} - 10x^{2} y \hat{j} + 2t \hat{k}$  such that $v_x = 4x^{3}$, $v_y = -10x^{2}y$ and $v_z = 2t$. Therefore,

$\frac{\partial \vec{v}}{\partial x} = 12x^{2} \hat{i} - 20xy \hat{j} \Rightarrow v_x \frac{\partial \vec{v}}{\partial x} = 48x^{5} \hat{i} - 80x^{4}y \hat{j} \ \cdots \cdots (3)$

$\frac{\partial \vec{v}}{\partial y} = - 10x^{2} \hat{j} \Rightarrow v_y \frac{\partial \vec{v}}{\partial y} = 100x^{4}y \hat{j} \ \cdots \cdots (4)$

$\frac{\partial \vec{v}}{\partial z} = 0 \Rightarrow v_z \frac{\partial \vec{v}}{\partial z} = 0 \ \cdots \cdots (5)$

$\frac{\partial \vec{v}}{\partial t} = 2 \hat{k} \ \cdots \cdots (6)$

Substituting (3), (4), (5), and (6) in (2), we get;

$\vec{a} = \frac{d\vec{v}}{dt} = 48x^{5} \hat{i} +20x^{4}y \hat{j} +2 \hat{k}$

$\Rightarrow \vec{a} = \frac{d\vec{v}}{dt} \ \Big{|}_{(2,1,3,1)} = 48(2)^{5} \hat{i} +20(2)^{4}y \hat{j} +2 \hat{k}$

$\Rightarrow \vec{a} \ |_{(2,1,3,1)} = 1536 \hat{i} + 320 \hat{j} + 2 \hat{k} \ \cdots \cdots \cdots (7)$

Hence the velocity and acceleration of the fluid particle are $\vec{v} \ |_{(2,1,3,1)} = 32 \hat{i} - 40 \hat{j} + 4 \hat{k}$ and $\vec{a} \ |_{(2,1,3,1)} = 1536 \hat{i} + 320 \hat{j} + 2 \hat{k}$ respectively.