The velocity of a viscous fluid flowing between the parallel surfaces in Fig. 7–10a is defined by U = 0.002[1 - 10(10³)y²] m/s, where y is in meters. Determine the vorticity and shear-strain rate of a fluid element located at y = 5 mm within the flow.

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Fluid Description. We have nonuniform steady one-dimensional flow of a real fluid.

Vorticity. We must apply Eq. 7–7, where u = 0.002 [1 - 10(10³)y²] m/s and [latex]v = 0[/latex].

[latex] \zeta=\frac{\Gamma}{A}=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y} [/latex] (7-7)

[latex] \begin{aligned} \zeta & =\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y} \ & =0-0.002\left[0-10\left(10^3 ight)(2 y) ight] \mid_{y = 0.005 m} rad / s =0.200 rad / s \end{aligned} [/latex]

This vorticity is a consequence of the fluid’s viscosity, and since [latex] \zeta eq 0 [/latex], we have rotational flow. Actually, the element has a rate of rotation of [latex] \omega_z=\zeta / 2=0.1 [/latex] rad/s, Fig. 7–10b. It is positive (counterclockwise) because at y = 0.005 m, the velocity profile shows that the top of the fluid element is moving slower than its bottom, Fig. 7–10a.

Shear-Strain Rate. Applying Eq. 7–4,

[latex] \dot{\gamma}_{x y}=\dot{\alpha}+\dot{\beta}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} [/latex] (7-4)

[latex] =0+0.002\left[0-10\left(10^3 ight)(2 y) ight]\mid_{y = 0.005 m} rad / s =-0.200 rad / s [/latex]

The result is negative because β is positive clockwise, and the shear strain is defined as the difference in the angle 90° - (90° + β) = -β, Fig. 7–10c.

Question 7.2: The velocity of a viscous fluid flowing between the parallel...

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