The velocity field V = { -6xi + 6yj }m/s defines the two-dimensional ideal fluid flow in the horizontal plane between two walls, Fig. 7–14. Determine the volumetric dilatation rate and the rotation of a fluid element located at point B(1 m, 2 m). If the pressure at point A(1 m, 1 m) is 250 kPa,

Question

The velocity field V = { -6xi + 6yj }m/s defines the two-dimensional ideal fluid flow in the horizontal plane between two walls, Fig. 7–14. Determine the volumetric dilatation rate and the rotation of a fluid element located at point B(1 m, 2 m). If the pressure at point A(1 m, 1 m) is 250 kPa, what is the pressure at point B? Take ρ = 1200 kg/m³.

Accepted Answer

Fluid Description. Since the velocity is not a function of time, the flow is steady. The fluid is an ideal fluid.

Volumetric Dilatation Rate. Applying Eq. 7–1, where u = (-6x) m/s, [latex]v[/latex] = (6y) m/s, and w = 0, we have

[latex] \frac{\delta \sout{V} / d \sout{V}}{\partial t}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} [/latex] (7-1)

= -6 + 6 + 0 = 0

The result confirms that there is no change in the volume of the fluid element at B, because an ideal fluid is incompressible.

Rotation. The angular velocity of the fluid element at B is defined by Eq. 7–2.

[latex] \omega_z=\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y} ight) [/latex] (7-2)

[latex] =\frac{1}{2}(0-0)=0 [/latex]

Therefore, the fluid element will not rotate about the z axis. Actually, the above two results apply at all points since they are independent of x and y. This is to be expected since we have an ideal fluid.

Pressure. Since the flow is irrotational and steady, we can apply the Bernoulli equation at two points not located on the same streamline, Fig. 7–14. The velocities at A and B are

[latex] \begin{aligned} & V_A=\sqrt{[-6(1)]^2+[6(1)]^2}=8.485 m / s \ & V_B=\sqrt{[-6(1)]^2+[6(2)]^2}=13.42 m / s \end{aligned} [/latex]

Since motion is in the horizontal plane, we have

[latex] \frac{p_A}{\gamma}+\frac{V_A^2}{2 g}+z_A=\frac{p_B}{\gamma}+\frac{V_B^2}{2 g}+z_B [/latex]

[latex] \frac{250\left(10^3 ight) N / m ^2}{\left(1200 kg / m ^3 ight)\left(9.81 m / s ^2 ight)}+\frac{(8.485 m / s )^2}{2\left(9.81 m / s ^2 ight)}+0=\frac{p_B}{\left(1200 kg / m ^3 ight)\left(9.81 m / s ^2 ight)}+\frac{(13.42 m / s )^2}{2\left(9.81 m / s ^2 ight)}+0 [/latex]

[latex] p_B=185 kPa [/latex]

Question 7.3: The velocity field V = { -6xi + 6yj }m/s defines the two-dim...

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