Forced and Free Vortices The flow pattern in a liquid stirred in a large closed cylindrical tank can be approximated by a forced vortex of radius a and angular velocity ω (corresponding to the location of the stirrer), surrounded by a free vortex that extends indefinitely radially outward, as shown

Question

Forced and Free Vortices

The flow pattern in a liquid stirred in a large closed cylindrical tank can be approximated by a forced vortex of radius [latex]a[/latex] and angular velocity [latex]\omega[/latex] (corresponding to the location of the stirrer), surrounded by a free vortex that extends indefinitely radially outward, as shown in Fig. E7.1.1.

If the pressure is [latex]p_{\infty}[/latex] for large values of [latex]r[/latex], derive and plot expressions for the tangential velocity [latex]v_{ heta}[/latex] and pressure [latex]p[/latex] as functions of radial location [latex]r[/latex]. If the top of the tank is removed, so that the liquid is allowed to have a free surface, comment on its shape. Also, discuss how the free vortex is generated in this example.

Accepted Answer

First, consider the tangential velocity [latex]v_{ heta}[/latex] in the two regions:

Forced [latex]\operatorname{vortex}(r \leq a): v_{ heta}=r \omega ; \quad[/latex] free vortex [latex](r \geq a): v_{ heta}=\frac{c}{r}.[/latex] (E7.1.1)

At the junction between the two regions, where [latex]r=a[/latex], the velocities must be identical, so that [latex]\left(v_{ heta} ight)_{r=a}=a \omega=c / a[/latex]; thus, the constant is [latex]c=\omega a^{2}[/latex] and the velocities in the two regions are:

Forced vortex [latex]: v_{ heta}=r \omega ; \quad[/latex] free vortex [latex]: v_{ heta}=\frac{\omega a^{2}}{r}[/latex], (E7.1.2)

representative profiles being shown in Fig. E7.1.2.

Second, since there is zero vorticity in the free vortex, Bernoulli's equation applies, it being noted that the velocity declines to zero for large [latex]r[/latex] :

Free vortex [latex]: \frac{p}{ ho}+\frac{v_{ heta}^{2}}{2}=\frac{p_{\infty}}{ ho}, \quad[/latex] or [latex]\quad p=p_{\infty}-\frac{ ho \omega^{2} a^{4}}{2 r^{2}}[/latex], (E7.1.3)

so that the pressure at the junction between the two vortices is:

[latex]p_{a}=p_{\infty}-\frac{1}{2} ho \omega^{2} a^{2}.[/latex] (E7.1.4)

For the forced vortex, the radial variation of pressure is already given in Eqn. (1.44):

[latex]\frac{\partial p}{\partial r}= ho r \omega^{2},[/latex] (1.44)

[latex]\frac{d p}{d r}= ho r \omega^{2},[/latex] (E7.1.5)

a result that can also be obtained from the full [latex]r[/latex] momentum balance in cylindrical coordinates, Eqn. (5.75), by making appropriate simplifications and substituting [latex]v_{ heta}=r \omega .[/latex] Integration of Eqn. (E7.1.5), noting that [latex]p=p_{a}[/latex] at [latex]r=a[/latex], gives:

In terms of velocities, for a Newtonian fluid with constant [latex] ho[/latex] and [latex]\mu[/latex] :

[latex]\begin{aligned} ho\left(\frac{\partial v_{r}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{r}}{\partial heta}-\frac{v_{ heta}^{2}}{r}+v_{z} \frac{\partial v_{r}}{\partial z} ight) \&=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial\left(r v_{r} ight)}{\partial r} ight)+\frac{1}{r^{2}} \frac{\partial^{2} v_{r}}{\partial heta^{2}}-\frac{2}{r^{2}} \frac{\partial v_{ heta}}{\partial heta}+\frac{\partial^{2} v_{r}}{\partial z^{2}} ight]+ ho g_{r}, \ ho\left(\frac{\partial v_{ heta}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{ heta}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{ heta}}{\partial heta}+\frac{v_{r} v_{ heta}}{r}+v_{z} \frac{\partial v_{ heta}}{\partial z} ight) \&=-\frac{1}{r} \frac{\partial p}{\partial heta}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial\left(r v_{ heta} ight)}{\partial r} ight)+\frac{1}{r^{2}} \frac{\partial^{2} v_{ heta}}{\partial heta^{2}}+\frac{2}{r^{2}} \frac{\partial v_{r}}{\partial heta}+\frac{\partial^{2} v_{ heta}}{\partial z^{2}} ight]+ ho g_{ heta}, \ ho\left(\frac{\partial v_{z}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{z}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{z}}{\partial heta}+v_{z} \frac{\partial v_{z}}{\partial z} ight) \&=-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial r} ight)+\frac{1}{r^{2}} \frac{\partial^{2} v_{z}}{\partial heta^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}} ight]+ ho g_{z}. (5.75)\end{aligned}[/latex]

Forced vortex : [latex]p=p_{a}-\frac{1}{2} ho \omega^{2}\left(a^{2}-r^{2} ight)=p_{\infty}- ho \omega^{2}\left(a^{2}-\frac{1}{2} r^{2} ight) .[/latex] (E7.1.6)

A representative complete pressure distribution is given in Fig. E7.1.3. Note that the overall pressure changes in the two vortices are identical [latex]\left( ho \omega^{2} a^{2} / 2 ight)[/latex]. If the liquid has a free surface, the pressure toward the bottom of its container must be balanced hydrostatically by a column of liquid that extends to the free surface. Thus, the free surface must have a shape that resembles the curve in Fig. E7.1.3, and this is substantiated by experiment.

Regarding the generation of the free vortex, consider the liquid in the region [latex]r>a[/latex] to be initially at rest when the stirrer is turned on. With appropriate simplifications, the [latex] heta[/latex] momentum balance becomes:

[latex] ho \frac{\partial v_{ heta}}{\partial t}=\mu \frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial\left(r v_{ heta} ight)}{\partial r} ight),[/latex] (E7.1.7)

and will govern how [latex]v_{ heta}[/latex] builds up to its final value as shown in Fig. E7.1.2, at which stage the right-hand side of Eqn. (E7.1.7) is zero. Somewhat paradoxically in this example, viscosity is important in determining the final velocity profile, even though the resulting free vortex region has zero vorticity.

Question 7.1: Forced and Free Vortices The flow pattern in a liquid stirre...

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