Question

Consider a steady, two-dimensional, incompressible flow of a newtonian fluid in which the velocity field is known: u=-2xy, v=y 2 - x 2 , w=0. (a) does this flow satisfy conservation of mass? (b) find the pressure field, p(x, y) if the pressure at the point (x=0, y=0) is equal to pa.

Answer 1

The flow is 2-dimensional and incompressible. Its velocity field is
$\vec{v} = u\hat{i}+v\hat{j} \\where \\ u=-2xy \\ v = y^{2}-x^{2}$

Part (a)
Because the flow is incompressible, conservation of mass requires that
$\bigtriangledown . \vec{v} = 0$
$\bigtriangledown . \vec{v} = \frac{\partial}{\partial x} (-2xy) + \frac{\partial}{\partial y}(y^{2}-x^{2}) = -2y +2y = 0$

Answer: The flow satisfies the conservation of mass

Part (b)
Note that
$\bigtriangledown ^{2}u = 0 \\ \bigtriangledown ^{2}v = 0$
Therefore the pressure field is given by
$- \frac{\partial p}{\partial x} \\ =\rho (u \frac{\partial u}{\partial x} +v \frac{\partial u}{\partial y} ) \\ =\rho[-2xy(-2y)+(y^{2}-x^{2})(-2x)] \\ = 2xy^{2}+2x^{3$
Also,
$- \frac{\partial p}{\partial y} \\ = \rho[u \frac{\partial v}{\partial x} +v \frac{\partial v}{\partial y} ) \\ = \rho [-2xy(-2x)+(y^{2}-x^{2})(2y)] \\ = 4x^{2}y+2y^{3}-2x^{2}y \\ = 2x^{2}y + 2y^{3}$

Therefore
$-p = \rho(x^{2}y^{2}+ \frac{x^{4}}{2} )+f(y) \\ or \\ -p = \rho(x^{2}y^{2}+ \frac{y^{4}}{2} ) + g(x)$

Because p(0,0) = ρa, therefore
f(0) = g(0) = -ρa.
$p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )$

Answer: $p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )$