Question

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.656y v=0.754−2.18x−2.05y . Calculate the acceleration field (find expressions for acceleration components ax and ay), and calculate the acceleration at the point (x,y) = (-1, 5).

Answer 1

1) $a_x=4.287+2.772x\\a_y=-5.579+2.772y$

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

$u=1.85+2.05x+0.656y$

$v=0.754-2.18x-2.05y$

The x-component and y-component of the acceleration field can be found using the following equations:

$a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}$

$a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}$

The derivatives in this problem are:

$\frac{du}{dt}=0$

$\frac{dv}{dt}=0$

$\frac{du}{dx}=2.05$

$\frac{du}{dy}=0.656$

$\frac{dv}{dx}=-2.18$

$\frac{dv}{dy}=-2.05$

Substituting, we find:

$a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x$

And

$a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y$

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

$a_x=4.287+2.772x\\a_y=-5.579+2.772y$

And so we find:

$a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281$

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

$a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418$