Answer:
pressure gradient in x direction:
[tex] \frac{\partial p}{\partial x} = -\rho 2xy^{2} [/tex]
pressure gradient in y direction:
[tex] \frac{\partial p}{\partial y} = \rho 2y^{3} [/tex]
Explanation:
Here the gravity is neglected and the field velocities is time independent, so we can use a simplify equation to Navier-Stokes.
[tex] \frac{DV}{Dt} = -\frac{1}{\rho} \nabla p [/tex]
V: Field Flow
ρ: Density
p: pressure
Before finding the pressure, let's define the components of the field vector.
[tex] V= 2xyi - y^{2}j [/tex]
[tex] u=2xy [/tex] [tex] v = y^{2} [/tex]
Now, the x component pressure gradient will be:
[tex] u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partiala x} [/tex]
[tex] 4xy^{2}-2xy^{2} = -\frac{1}{\rho} \frac{\partial p}{\partial x} [/tex]
[tex] \frac{\partial p}{\partial x} = -\rho 2xy^{2} [/tex]
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.
[tex] u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partial y} [/tex]
[tex] 0-2y^{3} = -\frac{1}{\rho} \frac{\partial p}{\partial x} [/tex]
[tex] \frac{\partial p}{\partial y} = \rho 2y^{3} [/tex]
I hope it helps you! :)
