Velocity Field near a Rotating Disk A hand-held surface grinder (see Figure 1.11) of radius R is rotating at a constant angular velocity Ω in the z-direction. Assume that the density of air is constant and that the flow is steady. Gravitational effects are also neglected. Assume a velocity traverse

Question

Velocity Field near a Rotating Disk
A hand-held surface grinder (see Figure 1.11) of radius R is rotating at a constant angular velocity Ω in the z-direction. Assume that the density of air is constant and that the flow is steady.
Gravitational effects are also neglected. Assume a velocity traverse has been made and indicates that only the tangential velocity component [latex](U_0)[/latex] is important; the velocity components in the radial and axial directions are negligible. The velocity field is given by:

- Region I: [latex]U_θ[/latex] = Ωr for r < R [latex]\left(\frac{U_ heta }{\Omega R}=\frac{r}{R} for \frac{r}{R} \lt 1 ight)[/latex]

- Region II: [latex]U_θ=\frac{ΩR^2}{r}[/latex] for r > R [latex]\left(\frac{U_ heta }{\Omega R}=\frac{R}{r} for \frac{r}{R} > 1 ight)[/latex]

A sketch of the flow field is shown in Figure E7.1a. The tangential velocity and radius have been normalized (non-dimensionalized) for convenience.
To do: To illustrate the previous fundamental concepts, analyze the flow field with respect to the following:
(a) vorticity and rotationality
(b) the pressure field needed to satisfy conservation of momentum
(c) applicability of the Bernoulli equation

Accepted Answer

(a) Vorticity - In cylindrical coordinates, the velocity vector is [latex]\vec{U} = ( U_r,U_θ,U_z )[/latex] for unit vector [latex](\hat{r} ,\hat{θ} ,\hat{z} )[/latex]. The vorticity, Eq. (7-1), is

[latex] \vec{\zeta } =\vec{ riangledown } imes \vec{U} =curl (\vec{U})[/latex] (7-1)

[latex]\vec{\zeta } =\vec{ riangledown } imes \vec{U} =\hat{r} \left(\frac{1}{r}\frac{∂U_z}{∂ heta }-\frac{∂U_ heta }{∂z} ight) +\hat{ heta } \left(\frac{∂U_r}{∂z}-\frac{∂U_z}{∂r} ight) +\hat{z} \frac{1}{r} \left(\frac{∂(rU_ heta )}{∂r} -\frac{∂U_r}{∂ heta } ight)[/latex]

Applying this to Region I,

[latex]\vec{\zeta } =\hat{z} \frac{1}{r}\left(\frac{∂(r^2\Omega )}{∂r} ight) =2\Omega \hat{z}[/latex]

and to Region II,

[latex]\vec{\zeta } =\hat{z} \frac{1}{r}\left(\frac{∂(\Omega R^2)}{∂r} ight) =0[/latex]

Thus the flow is irrotational in the outer region (Region II) and rotational with constant vorticity in the inner region (Region I). In addition, since the vorticity is constant in Region I, the rotational flow field is said to undergo solid body rotation, i.e. to rotate as a solid body such as a rotating compact disk (CD). The analysis illustrates clearly that just because air revolves about a center of rotation, it is not necessarily rotational; it can be either rotational or irrotational, that is to say it may or may not have vorticity. All curvilinear flows revolve about a center of rotation, but irrotation or rotation in the fluid dynamic sense depends on the vorticity.

(b) Pressure distribution - To find the pressure distribution consistent with the conservation equations, the continuity and momentum equations are applied to the velocity profile. Equation (7-8) becomes

[latex]\vec{ riangledown } .\vec{U} =0[/latex] (7-8)

[latex]\vec{ riangledown } .\vec{U} = \frac{1}{r}\frac{∂(rU_r)}{∂r}+\frac{1}{r}\frac{∂U_ heta }{∂ heta }+\frac{∂U_z}{∂z} =0[/latex]

Substituting the observed velocity distributions, it is seen that the continuity equation is identically satisfied in both Region I and Region II. While the continuity equation is a scalar equation, the momentum equation (Navier-Stokes equation) is a vector equation. The radial (r-direction) and tangential (6-direction) components of Eq. (7-7) are

[latex]\frac{∂\vec{U} }{∂t} +(\vec{U}.\vec{ riangledown } )\vec{U} =-\frac{1}{ ho } \vec{ riangledown } P+\vec{g} +v riangledown ^2\vec{U}[/latex] (7-7)

[latex](\vec{U}.\vec{ riangledown } )U_r -\frac{1}{r} U_ heta ^2=-\frac{1}{ ho } \frac{∂P}{∂r}[/latex]

and

[latex](\vec{U}.\vec{ riangledown } )U_ heta +\frac{1}{r} U_r U_ heta =-\frac{1}{ ho r} \frac{∂P}{∂ heta }[/latex]

Note that the flow is steady, gravity is neglected, and the viscous terms have dropped out. In Region I, although the fluid is viscous, viscosity does not affect solid body rotation. In Region II, the flow is irrotational and thus the viscous terms drop out as discussed previously. Upon substituting the observed velocity distribution, it is found that in both Region I and Region II,

[latex]\frac{∂P}{∂r}=\frac{ ho U_ heta ^2}{r} [/latex]

in the radial direction, and

[latex]\frac{∂P}{∂ heta }=0 [/latex]

in the tangential direction. Thus the pressure is only a function of r. The r-momentum equation becomes an ordinary differential equation in both Region I and Region II, and the partial derivative operator in the above equation can be replaced by the total derivative operator,

[latex]\frac{dP}{dr} =\frac{ ho U_ heta ^2}{r}[/latex]

First consider Region II (r > R). Substituting [latex]U_ heta= ΩR²/r[/latex], and integrating the above equation between r and infinity where [latex]U_ heta[/latex] is zero and the pressure is the ambient pressure [latex]P_0[/latex], one obtains

[latex]P(r)=P_0- ho \Omega ^2(R^2-\frac{r^2}{2} ) for r\lt R \left(\frac{P_0-P(r)}{ ho \Omega ^2R^2}=1-\frac{1}{2}\frac{R^2}{r^2} for \frac{r}{R} \lt 1 ight)[/latex]

At the interface between Region I and Region II (r = R) the pressure must be continuous, thus P(R) can be found by evaluating the above equation at r = R,

[latex]P(R)=P_0-\frac{ ho \Omega^2 R^2}{2}[/latex] at r=R

Now consider Region I (r < R). Substituting [latex]U_0 = Ωr[/latex], and integrating between r and R,

[latex]P(R)=P_0 - ho \Omega^2\left(R^2-\frac{r^2}{2} ight) \quad r

Figure E7.1b is a graph of the pressure distribution, and shows that the pressure at the center of the flow is one half the value at the edge of the rotational region (Region I). Again, for convenience, the pressure has been normalized.

(c) Bernoulli equation - The Bernoulli constant (C) from Eq. (7-17) or (7-18) is equal to

[latex]\frac{P}{ ho }+\frac{U^2}{2} +gz=C= ext{constant everywhere}[/latex] (7-17)

[latex]\frac{P}{ ho }+\frac{U^2}{2} +gz=C= ext{constant along streamlines}[/latex] (7-18)

[latex]C=\frac{P}{ ho }+\frac{U^2}{2}[/latex]

since gravitational effects have been neglected. Since the velocity and pressure distributions have been computed for Region I and Region II, the Bernoulli constant can be found. In Region I,

[latex]C=\frac{P_0}{ ho } -\Omega ^2(R^2-r^2) for r\lt R[/latex]

which is obviously not constant. In Region II,

[latex]C=\frac{P_0}{ ho } for r\gt R[/latex]

which is constant. Thus Bernoulli's equation is valid everywhere in the irrotational region (Region II), but is only valid along streamlines in the rotational region (Region I). In Region I the constant has a unique value for each radius .

Discussion: The above results are consistent with the assumptions made at the beginning of the analysis. If a more detailed analysis were desired, engineers would have to consider the three dimensional features of the flow. Viscous effects are also significant close to the rotating disk where a boundary layer exists. Secondly, the predicted low pressure region in the center of the disk initiates an axial flow of air toward the disk which in turn produces a radially outward flow of air. Under these conditions, [latex]U_z[/latex] and [latex]U_r[/latex] are no longer zero and the full set of Navier-Stokes equations would have to be solved.

Question 7.1: Velocity Field near a Rotating Disk A hand-held surface grin...

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