Flow Through an Annular Die Following the discussion of polymer processing in the previous section, now consider flow through a die that could be located at the exit of the screw extruder of Example 6.6. Consider a die that forms a tube of polymer (other shapes being sheets and filaments). In the d

Question

Flow Through an Annular Die

Following the discussion of polymer processing in the previous section, now consider flow through a die that could be located at the exit of the screw extruder of Example 6.6. Consider a die that forms a tube of polymer (other shapes being sheets and filaments). In the die of length [latex]D[/latex] shown in Fig. E6.7, a pressure difference [latex]p_{2}-p_{3}[/latex] causes a liquid of viscosity [latex]\mu[/latex] to flow steadily from left to right in the annular area between two fixed concentric cylinders. Note that [latex]p_{2}[/latex] is chosen for the inlet pressure in order to correspond to the extruder exit pressure from Example 6.5. The inner cylinder is solid, whereas the outer one is hollow; their radii are [latex]r_{1}[/latex] and [latex]r_{2}[/latex], respectively. The problem, which could occur in the extrusion of plastic tubes, is to find the velocity profile in the annular space and the total volumetric flow rate [latex]Q[/latex]. Note that cylindrical coordinates are now involved.

Accepted Answer

Assumptions and continuity equation. The following assumptions are realistic:

There is only one nonzero velocity component, namely, that in the direction of flow, [latex]v_{z}[/latex]. Thus, [latex]v_{r}=v_{ heta}=0[/latex].
Gravity acts vertically downward, so that [latex]g_{z}=0[/latex].
The axial velocity is independent of the angular location; that is, [latex]\partial v_{z} / \partial heta=0[/latex]. To analyze the situation, again start from the continuity equation, (5.49):

[latex]\frac{\partial ho}{\partial t}+\frac{1}{r} \frac{\partial\left( ho r v_{r} ight)}{\partial r}+\frac{1}{r} \frac{\partial\left( ho v_{ heta} ight)}{\partial heta}+\frac{\partial\left( ho v_{z} ight)}{\partial z}=0,[/latex] (5.49)

which, for constant density and [latex]v_{r}=v_{ heta}=0[/latex], reduces to:

[latex]\frac{\partial v_{z}}{\partial z}=0,[/latex] (E6.7.1)

verifying that [latex]v_{z}[/latex] is independent of distance from the inlet and that the velocity profile [latex]v_{z}=v_{z}(r)[/latex] appears the same for all values of [latex]z[/latex].

Momentum balances. There are again three momentum balances, one for each of the [latex]r, heta[/latex], and [latex]z[/latex] directions. If explored, the first two of these would ultimately lead to the pressure variation with [latex]r[/latex] and [latex] heta[/latex] at any cross section, which is of little interest in this problem. Therefore, we extract from Eqn. (5.75) only the [latex]z[/latex] momentum balance:

In terms of velocities, for a Newtonian fluid with constant ρ and μ:

[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]

[latex]\begin{aligned} 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}. (E6.7.2)\end{aligned}[/latex]

With [latex]v_{r}=v_{ heta}=0[/latex] (from assumption 1), [latex]\partial v_{z} / \partial z=0[/latex] [from Eqn. (E6.7.1)], [latex]\partial v_{z} / \partial heta=0[/latex] (assumption 3), and [latex]g_{z}=0[/latex] (assumption 2), this momentum balance simplifies to:

[latex]\mu\left[\frac{1}{r} \frac{d}{d r}\left(r \frac{d v_{z}}{d r} ight) ight]=\frac{\partial p}{\partial z},[/latex] (E6.7.3)

in which total derivatives are used because [latex]v_{z}[/latex] depends only on [latex]r[/latex].

Shortly, we shall prove that the pressure gradient is uniform between the die inlet and exit, being given by:

[latex]-\frac{\partial p}{\partial z}=\frac{p_{2}-p_{3}}{D},[/latex] (E6.7.4)

in which both sides of the equation are positive quantities. Two successive integrations of Eqn. (E6.7.3) may then be performed, yielding:

[latex]v_{z}=-\frac{1}{4 \mu}\left(-\frac{\partial p}{\partial z} ight) r^{2}+c_{1} \ln r+c_{2} .[/latex] (E6.7.5)

The two constants may be evaluated by applying the boundary conditions of zero velocity at the inner and outer walls,

[latex]r=r_{1}: v_{z}=0, \quad r=r_{2}: v_{z}=0,[/latex] (E6.7.6)

giving:

[latex]c_{1}=\frac{1}{4 \mu}\left(-\frac{\partial p}{\partial z} ight) \frac{r_{2}^{2}-r_{1}^{2}}{\ln \left(r_{2} / r_{1} ight)}, \quad c_{2}=\frac{1}{4 \mu}\left(-\frac{\partial p}{\partial z} ight) r_{2}^{2}-c_{1} \ln r_{2} .[/latex] (E6.7.7)

Substitution of these values for the constants of integration into Eqn. (E6.7.5) yields the final expression for the velocity profile:

[latex]v_{z}=\frac{1}{4 \mu}\left(-\frac{\partial p}{\partial z} ight)\left[\frac{\ln \left(r / r_{2} ight)}{\ln \left(r_{2} / r_{1} ight)}\left(r_{2}^{2}-r_{1}^{2} ight)+\left(r_{2}^{2}-r^{2} ight) ight],[/latex] (E6.7.8)

which is sketched in Fig. E6.7. Note that the maximum velocity occurs somewhat before the halfway point in progressing from the inner cylinder to the outer cylinder.

Volumetric flow rate. The final quantity of interest is the volumetric flow rate [latex]Q[/latex]. Observing first that the flow rate through an annulus of internal radius [latex]r[/latex] and external radius [latex]r+d r[/latex] is [latex]d Q=v_{z} 2 \pi r d r[/latex], integration yields:

[latex]Q=\int_{0}^{Q} d Q=\int_{r_{1}}^{r_{2}} v_{z} 2 \pi r d r.[/latex] (E6.7.9)

Since [latex]r \ln r[/latex] is involved in the expression for [latex]v_{z}[/latex], the following indefinite integral is needed:

[latex]\int r \ln r d r=\frac{r^{2}}{2} \ln r-\frac{r^{2}}{4},[/latex] (E6.7.10)

giving the final result:

[latex]Q=\frac{\pi\left(r_{2}^{2}-r_{1}^{2} ight)}{8 \mu}\left(-\frac{\partial p}{\partial z} ight)\left[r_{2}^{2}+r_{1}^{2}-\frac{r_{2}^{2}-r_{1}^{2}}{\ln \left(r_{2} / r_{1} ight)} ight].[/latex] (E6.7.11)

Since [latex]Q, \mu, r_{1}[/latex], and [latex]r_{2}[/latex] are constant throughout the die, [latex]\partial p / \partial z[/latex] is also constant, thus verifying the hypothesis previously made. Observe that in the limiting case of [latex]r_{1} ightarrow 0[/latex], Eqn. (E6.7.11) simplifies to the Hagen-Poiseuille law, already stated in Eqn. (3.12).

[latex]\begin{aligned} Q &=\int_{0}^{Q} d Q=2 \pi \int_{0}^{a} \underbrace{\frac{1}{4 \mu}\left(-\frac{d p}{d z} ight)}_{ ext {Constant }}\left(a^{2}-r^{2} ight) r d r \ &=\frac{\pi a^{4}}{8 \mu}\left(-\frac{d p}{d z} ight)=\frac{\pi a^{4}}{8 \mu} \frac{p_{1}-p_{2}}{L}, (3.12)\end{aligned}[/latex]

This problem may also be solved by performing a momentum balance on a shell that consists of an annulus of internal radius [latex]r[/latex], external radius [latex]r+d r[/latex], and length [latex]d z[/latex].

Question 6.7: Flow Through an Annular Die Following the discussion of poly...

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