# Fluid mechanics: calc the acceleration of a particle at a point.

## Homework Statement

A velocity field is given by $\vec{V}$= [Ax$^{3}$ + Bxy$^{2}$]$\hat{i}$ + [Ay$^{3}$ + Bx$^{2}$y]$\hat{j}$; A=0.2 m$^{-2}$s$^{-1}$, B is a constant, and the
coordinates are measured in meters. Determine the value and
units for B if this velocity field is to represent an incompressible
flow. Calculate the acceleration of a fluid particle at point
(x, y)=(2, 1). Evaluate the component of particle acceleration
normal to the velocity vector at this point.

## Homework Equations

u=$\frac{\partial\Psi}{\partial y}$ v=-$\frac{\partial\Psi}{\partial x}$

## The Attempt at a Solution

I used the above equation to get value equations u and v, there is something I'm missing, a bit of reasoning that has to be made using the given information to determine a constraint that exists since the fluid is incompressible. This should link the rates of change to each other. But that's where I get stuck, having trouble hanging on to all the concepts..

u = [2Bxy]$\hat{i}$ + [.6y$^{2}$ + Bx$^{2}$]$\hat{j}$
v = -[.6x$^{2}$ + By$^{2}$]$\hat{i}$ - [2Bxy]$\hat{j}$

Assumptions:
1. Incompressible flow
2. B is constant

Last edited:

mfb
Mentor
"Incompressible" links the derivatives of the velocities in both directions.
This gives you one constraint and allows to calculate B.

so du = -dv? increases of u lead to decreases of v.

mfb
Mentor
Not du = -dv, but du/dx = -dv/dy (or the other way round?) should work.

when I try $\frac{du}{dx}$ = -$\frac{dv}{dy}$ I get

[2By]$\hat{i}$ + [2Bx]$\hat{j}$ = [2By]$\hat{i}$ + [2Bx]$\hat{j}$

which doesn't help solve for B, unless I'm messing up somewhere or forgetting something.
The same happens when I take $\frac{du}{dy}$ = -$\frac{dv}{dy}$

[2Bx]$\hat{i}$ + [1.2y]$\hat{j}$ = [1.2x]$\hat{i}$ + [2By]$\hat{j}$

mfb
Mentor
The second equation can be solved for B.

I'm getting B = 1.2/2
does that look right?

mfb
Mentor
I think so. It solves the equations for i and j at the same time, which looks good.