Question

A frictionless, incompressible steady flow field is given by V = 2xyi − y2j (2) in arbitrary units. Let the density be rhoo = constant and neglect gravity. Find expressions for the pressure gradients in both the x and y directions.

Answer 1

Answer:

pressure gradient in x direction:

$\frac{\partial p}{\partial x} = -\rho 2xy^{2}$

pressure gradient in y direction:

$\frac{\partial p}{\partial y} = \rho 2y^{3}$

Explanation:

Here the gravity is neglected and the field velocities is time independent, so we can use a simplify equation to Navier-Stokes.

$\frac{DV}{Dt} = -\frac{1}{\rho} \nabla p$                                        

V: Field Flow

ρ: Density

p: pressure

Before finding the pressure, let's define the components of the field vector.

$V= 2xyi - y^{2}j$

$u=2xy$    $v = y^{2}$

Now, the x component pressure gradient will be:

$u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partiala x}$

$4xy^{2}-2xy^{2} = -\frac{1}{\rho} \frac{\partial p}{\partial x}$

$\frac{\partial p}{\partial x} = -\rho 2xy^{2}$

We can apply the same analyze to find the y component of the pressure gradient. We just need to take the partial derivative from v.

$u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partial y}$

$0-2y^{3} = -\frac{1}{\rho} \frac{\partial p}{\partial x}$

$\frac{\partial p}{\partial y} = \rho 2y^{3}$

I hope it helps you! :)