stagnation

[logistic_guy]

A steady, two-dimensional velocity field is given by

\(\displaystyle \bold{V} = (u,v) = (-0.781 - 4.67x)\bold{i} + (-3.54 + 4.67y)\bold{j}\)

Calculate the location of the stagnation point.

 

[logistic_guy]

The stagnation point is \(\displaystyle (x,y)\) when \(\displaystyle \bold{V} = 0\).

Let's calculate.

\(\displaystyle u = -0.781 - 4.67x\)

\(\displaystyle 0 = -0.781 - 4.67x\)

This gives:

\(\displaystyle x = -\frac{0.781}{4.67} = -0.16724 \ \text{m}\)

And

\(\displaystyle v = -3.54 + 4.67y\)

\(\displaystyle 0 = -3.54 + 4.67y\)

This gives:

\(\displaystyle y = \frac{3.54}{4.67} = 0.75803 \ \text{m}\)

Then, the location of the stagnation point is:

\(\displaystyle (x,y) = \textcolor{blue}{(-0.16724 \ \text{m},0.75803 \ \text{m})}\)