The velocity field of a flow is given by V= 2x2ti+[4y(t-1)+2x2t]j m/s, where x and y are in...

Question:

The velocity field of a flow is given by {eq}V= 2x^2t \ \hat{i}+[4y(t-1)+2x^2t]\hat{j} \ m/s {/eq}, where x and y are in meters and t is in seconds. For fluid particles on the x-axis, determine the speed and direction of flow.

Vector:

Vector are characterized by its magnitude and direction such as velocity, acceleration,force and momentum.Vector quantity is only depends on magnitude and direction of the particle.

In this question we have to find the direction and magnitude of particles about x-axis.

Given data:

• The field flow is given by {eq}{\rm{V = 2}}{{\rm{x}}^2}t\mathop i\limits^ \wedge + [4y\left( {t - 1} \right) + 2{x^2}t]\mathop j\limits^ \wedge {\rm{m/s}} {/eq}

Where x and y are in meters and t is in seconds.

• The general expression of the vector in 2D-plane.

V ={eq}A\mathop i\limits^ \wedge + B\mathop j\limits^ \wedge {/eq}

Where {eq}\mathop i\limits^ \wedge{/eq} and{eq}\mathop j\limits^ \wedge {/eq} represent the unit vector in x and y direction.

• The expression of the magnitude of vector V.

{eq}\left| {\rm{V}} \right| = \sqrt {{A^2} + {B^2}} {/eq}

• Expressions for direction vector V from x-axis .

{eq}\theta = {\tan ^{ - 1}}\left( {\dfrac{B}{A}} \right) {/eq}

Where {eq}\theta{/eq} is the angle of the vector V from the x-axis.

Part 1 :

• We have given that the fluid particle is on the x axis,

So y = 0.

• Now the expressions for calculating the velocity of particles along x-axis is,

{eq}{\rm{V = 2}}{{\rm{x}}^2}t\mathop i\limits^ \wedge + [4y\left( {t - 1} \right) + 2{x^2}t]\mathop j\limits^ \wedge {/eq}

Here y=0

Substitute the values in above equation,

{eq}\begin{align*} {V }&= 2{{\rm{x}}^2}t\mathop i\limits^ \wedge + [4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t]\mathop j\limits^ \wedge \\ {V}&= 2{x^2}t\mathop i\limits^ \wedge + 2{x^2}t\mathop j\limits^ \wedge ......\left( 1 \right) \end{align*} {/eq}

Now expression for calculating the magnitude of vector V is,

{eq}\left| {\rm{V}} \right| = \sqrt {{A^2} + {B^2}} {/eq}

From equation 1 we get A ={eq}2{x^2}t and {\rm{B = 2}}{{\rm{x}}^2}t {/eq}

{eq}\begin{align*} \left| {\rm{V}} \right| &= \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\ \left| {\rm{V}} \right| &= (2\sqrt 2 ){x^2}t \end{align*} {/eq}

Thus the velocity of the fluid particles on the x axis is{eq}(2\sqrt 2 ){x^2}tm/s. {/eq}

• Part 2:
• Now the expression for calculating the directional flow is,

{eq}\theta = {\tan ^{ - 1}}\left( {\dfrac{B}{A}} \right) {/eq}

Substitute the values of A and B is ,

{eq}\begin{align*} \theta &= {\tan ^{ - 1}}\left( {\dfrac{{2{x^2}t}}{{2{x^2}t}}} \right)\\ \theta &= {\tan ^{ - 1}}\left( 1 \right)\\ \theta &= 45^\circ \end{align*} {/eq}

Thus the direction of flow of particle from positive x-axis is{eq}\ 45^\circ . {/eq}