The surface of a nominally quiescent pool of water is deflected symmetrically downward near R = 0 because of vortical flow within the pool as shown in Figure 5.4. In cylindrical coordinates, the water surface profile is z = h(R). Assume the velocity field in the water has only an angular component,

Question

The surface of a nominally quiescent pool of water is deflected symmetrically downward near R = 0 because of vortical flow within the pool as shown in Figure 5.4. In cylindrical coordinates, the water surface profile is z = h(R). Assume the velocity field in the water has only an angular component, [latex]\mathbf{u}=u_{\varphi}(R)\mathbf{e}_{\varphi}[/latex], and determine [latex]u_{\varphi}(R)[/latex] when the pressure on the water surface is [latex]p_{o}[/latex].

Accepted Answer

Integrating the equivalents of (5.5a) and (5.5b) in cylindrical coordinates, leads to:

[latex]- ho u^{2}_{ heta}/r=-\partial p/\partial r[/latex], and [latex]0=-\partial p/\partial z= ho\mathrm{g}[/latex]. (5.5a, 5.5b)

[latex]p(R,z)-p_{o}=\overset{R}{\underset{0}{\int}} ho\frac{u^{2}_{\varphi}}{R}dR+f(z)[/latex], and [latex]p(R,z)-p_{o}=- ho\mathrm{g}z+\mathrm{g}(R)[/latex],

where ƒ and g are single-variable functions of integration. These two results are consistent when:

[latex]p(R,z)-p_{o}=\overset{R}{\underset{0}{\int}} ho\frac{u^{2}_{\varphi}}{R}dR- ho\mathrm{g}z[/latex].

On the water surface the pressure is [latex]p_{o}[/latex], that is p(R,h) = [latex]p_{o}[/latex], so this pressure equation implies:

[latex]\overset{R}{\underset{0}{\int}} ho\frac{u^{2}_{\varphi}}{R}dR= ho\mathrm{g h}[/latex], or [latex] ho\frac{u^{2}_{\varphi}}{R}= ho\mathrm{g}\frac{dh}{dR}[/latex], which leads to: [latex]u_{\varphi}(R)=\pm\sqrt{\mathrm{g}R(dh/dR)} [/latex].

In this case, the vortical flow in the water may swirl with either sense of rotation to produce the surface shape h(R).

Question 5.1: The surface of a nominally quiescent pool of water is deflec...

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