Respuesta :
The flow is 2-dimensional and incompressible. Its velocity field is
[tex]\vec{v} = u\hat{i}+v\hat{j} \\where \\ u=-2xy \\ v = y^{2}-x^{2}[/tex]
Part (a)
Because the flow is incompressible, conservation of mass requires that
[tex]\bigtriangledown . \vec{v} = 0[/tex]
[tex]\bigtriangledown . \vec{v} = \frac{\partial}{\partial x} (-2xy) + \frac{\partial}{\partial y}(y^{2}-x^{2}) = -2y +2y = 0[/tex]
Answer: The flow satisfies the conservation of mass
Part (b)
Note that
[tex]\bigtriangledown ^{2}u = 0 \\ \bigtriangledown ^{2}v = 0[/tex]
Therefore the pressure field is given by
[tex]- \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[/tex]
Also,
[tex]- \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}[/tex]
Therefore
[tex]-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)[/tex]
Because p(0,0) = ρa, therefore
f(0) = g(0) = -ρa.
[tex]p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )[/tex]
Answer: [tex]p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )[/tex]
[tex]\vec{v} = u\hat{i}+v\hat{j} \\where \\ u=-2xy \\ v = y^{2}-x^{2}[/tex]
Part (a)
Because the flow is incompressible, conservation of mass requires that
[tex]\bigtriangledown . \vec{v} = 0[/tex]
[tex]\bigtriangledown . \vec{v} = \frac{\partial}{\partial x} (-2xy) + \frac{\partial}{\partial y}(y^{2}-x^{2}) = -2y +2y = 0[/tex]
Answer: The flow satisfies the conservation of mass
Part (b)
Note that
[tex]\bigtriangledown ^{2}u = 0 \\ \bigtriangledown ^{2}v = 0[/tex]
Therefore the pressure field is given by
[tex]- \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[/tex]
Also,
[tex]- \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}[/tex]
Therefore
[tex]-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)[/tex]
Because p(0,0) = ρa, therefore
f(0) = g(0) = -ρa.
[tex]p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )[/tex]
Answer: [tex]p(x,y) = \rho a - \rho(x^{2}y^{2} + \frac{x^{4}+y^{4}}{2} )[/tex]
