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, 4). The acceleration components are ax = + x ay = + y Acceleration components at (-1, 4) are ax = ay =

Answer 1

Answer:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

Explanation:

The acceleration field is obtained by deriving the components in function of the time. That is to say:

$\frac{du}{dt}=2.05\cdot \frac{dx}{dt}+0.656\cdot \frac{dy}{dt}\\\frac{dv}{dt} = -2.18\cdot \frac{dx}{dt} -2.05\cdot \frac{dy}{dt}$

Where $\frac{dx}{dt} = u$ and $\frac{dy}{dt} = v$.

The velocity components at given point are, respectively:

$u = 2.424$

$v=-5.266$

Lastly, the acceleration components are found:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$