Flow Between Parallel Plates Fig. E6.1.1 shows a fluid of viscosity μ that flows in the x direction between two rectangular plates, whose width is very large in the z direction when compared to their separation in the y direction. Such a situation could occur in a die when a polymer is being extrud

Question

Flow Between Parallel Plates

Fig. E6.1.1 shows a fluid of viscosity [latex]\mu[/latex] that flows in the [latex]x[/latex] direction between two rectangular plates, whose width is very large in the [latex]z[/latex] direction when compared to their separation in the [latex]y[/latex] direction. Such a situation could occur in a die when a polymer is being extruded at the exit into a sheet, which is subsequently cooled and solidified. Determine the relationship between the flow rate and the pressure drop between the inlet and exit, together with several other quantities of interest.

Accepted Answer

Simplifying assumptions. The situation is analyzed by referring to a cross section of the duct, shown in Fig. E6.1.2, taken at any fixed value of [latex]z[/latex]. Let the depth be [latex]2 d(\pm d[/latex] above and below the centerline or axis of symmetry [latex]y=0)[/latex] and the length [latex]L[/latex]. Note that the motion is of the Poiseuille type, since it is caused by the applied pressure difference [latex]\left(p_{1}-p_{2} ight)[/latex]. Make the following fairly realistic assumptions about the flow:

As already stated, it is steady and Newtonian, with constant density and viscosity. (These assumptions will often be taken for granted, and not restated, in later problems.)
Entrance effects can be neglected, so that the flow is fully developed, in which case there is only one nonzero velocity component --- that in the direction of flow, [latex]v_{x}[/latex]. Thus, [latex]v_{y}=v_{z}=0[/latex].
Since, in comparison with their spacing, [latex]2 d[/latex], the plates extend for a very long distance in the [latex]z[/latex] direction, all locations in this direction appear essentially identical to one another. In particular, there is no variation of the velocity in the [latex]z[/latex] direction, so that [latex]\partial v_{x} / \partial z=0[/latex].
Gravity acts vertically downward; hence, [latex]g_{y}=-g[/latex] and [latex]g_{x}=g_{z}=0[/latex].
The velocity is zero in contact with the plates, so that [latex]v_{x}=0[/latex] at [latex]y=\pm d[/latex].

Continuity. Start by examining the general continuity equation, (5.48):

[latex]\frac{\partial ho}{\partial t}+\frac{\partial\left( ho v_{x} ight)}{\partial x}+\frac{\partial\left( ho v_{y} ight)}{\partial y}+\frac{\partial\left( ho v_{z} ight)}{\partial z}=0,[/latex] (5.48)

which, in view of the constant-density assumption, simplifies to Eqn. (5.52):

[latex]\frac{\partial v_{x}}{\partial x}+\frac{\partial v_{y}}{\partial y}+\frac{\partial v_{z}}{\partial z}=0.[/latex] (5.52)

But since [latex]v_{y}=v_{z}=0[/latex] :

[latex]\frac{\partial v_{x}}{\partial x}=0,[/latex] (E6.1.1)

so [latex]v_{x}[/latex] is independent of the distance from the inlet, and the velocity profile will appear the same for all values of [latex]x[/latex]. Since [latex]\partial v_{x} / \partial z=0[/latex] (assumption 3), it follows that [latex]v_{x}=v_{x}(y)[/latex] is a function of [latex]y[/latex] only.

Momentum balances. With the stated assumptions of a Newtonian fluid with constant density and viscosity, Eqn. (5.73) gives the [latex]x, y[/latex], and [latex]z[/latex] momentum balances:

[latex]\begin{aligned} ho\left(\frac{\partial v_{x}}{\partial t}+v_{x} \frac{\partial v_{x}}{\partial x}+v_{y} \frac{\partial v_{x}}{\partial y}+v_{z} \frac{\partial v_{x}}{\partial z} ight) &=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^{2} v_{x}}{\partial x^{2}}+\frac{\partial^{2} v_{x}}{\partial y^{2}}+\frac{\partial^{2} v_{x}}{\partial z^{2}} ight)+ ho g_{x}, \ ho\left(\frac{\partial v_{y}}{\partial t}+v_{x} \frac{\partial v_{y}}{\partial x}+v_{y} \frac{\partial v_{y}}{\partial y}+v_{z} \frac{\partial v_{y}}{\partial z} ight) &=-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^{2} v_{y}}{\partial x^{2}}+\frac{\partial^{2} v_{y}}{\partial y^{2}}+\frac{\partial^{2} v_{y}}{\partial z^{2}} ight)+ ho g_{y}, \ ho\left(\frac{\partial v_{z}}{\partial t}+v_{x} \frac{\partial v_{z}}{\partial x}+v_{y} \frac{\partial v_{z}}{\partial y}+v_{z} \frac{\partial v_{z}}{\partial z} ight) &=-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^{2} v_{z}}{\partial x^{2}}+\frac{\partial^{2} v_{z}}{\partial y^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}} ight)+ ho g_{z} . (5.73)\end{aligned}[/latex]

With [latex]v_{y}=v_{z}=0[/latex] (from assumption 2), [latex]\partial v_{x} / \partial x=0[/latex] [from the simplified continuity equation, [latex](\mathrm{E} 6.1 .1)], g_{y}=-g, g_{x}=g_{z}=0[/latex] (assumption 4), and steady flow (assumption 1), these momentum balances simplify enormously to:

[latex]\begin{aligned}\mu \frac{\partial^{2} v_{x}}{\partial y^{2}} =\frac{\partial p}{\partial x}, \ \end{aligned}[/latex] (E6.1.2)

[latex]\begin{aligned}\frac{\partial p}{\partial y} =- ho g, \ \end{aligned}[/latex] (E6.1.3)

[latex]\begin{aligned}\frac{\partial p}{\partial z} =0.\end{aligned}[/latex] (E6.1.4)

Pressure distribution. The last of the simplified momentum balances, Eqn. (E6.1.4), indicates no variation of the pressure across the width of the system (in the [latex]z[/latex] direction), which is hardly a surprising result. When integrated, the second simplified momentum balance, Eqn. (E6.1.3), predicts that the pressure varies according to:

[latex]p=- ho g \int d y+f(x)=- ho g y+f(x) .[/latex] (E6.1.5)

Observe carefully that since a partial differential equation is being integrated, we obtain not a constant of integration, but a function of integration, [latex]f(x)[/latex].

Assume --- to be verified later --- that [latex]\partial p / \partial x[/latex] is constant, so that the centerline pressure [latex]([/latex] at [latex]y=0)[/latex] is given by a linear function of the form:

[latex]p_{y=0}=a+b x .[/latex] (E6.1.6)

The constants [latex]a[/latex] and [latex]b[/latex] may be determined from the inlet and exit centerline pressures:

[latex]x=0: \quad p=p_{1}=a,[/latex] (E6.1.7)

[latex]x=L: \quad p=p_{2}=a+b L[/latex], (E6.1.8)

leading to:

[latex]\quad a=p_{1}, \quad b=-\frac{p_{1}-p_{2}}{L}.[/latex] (E6.1.9)

Thus, the centerline pressure falls linearly from [latex]p_{1}[/latex] at the inlet to [latex]p_{2}[/latex] at the exit:

[latex]f(x)=p_{1}-\frac{x}{L}\left(p_{1}-p_{2} ight),[/latex] (E6.1.10)

so that the complete pressure distribution is:

[latex]p=p_{1}-\frac{x}{L}\left(p_{1}-p_{2} ight)- ho g y .[/latex] (E6.1.11)

That is, the pressure declines linearly, both from the bottom plate to the top plate and also from the inlet to the exit. In the majority of applications, [latex]2 d \ll L[/latex], and the relatively small pressure variation in the [latex]y[/latex] direction is usually ignored. Thus, [latex]p_{1}[/latex] and [latex]p_{2}[/latex], although strictly the centerline values, are typically referred to as the inlet and the exit pressures, respectively.

Velocity profile. Since, from Eqn. (E6.1.1), [latex]v_{x}[/latex] does not depend on [latex]x[/latex], [latex]\partial^{2} v_{x} / \partial y^{2}[/latex] appearing in Eqn. (E6.1.2) becomes a total derivative, so this equation can be rewritten as:

[latex]\mu \frac{d^{2} v_{x}}{d y^{2}}=\frac{\partial p}{\partial x},[/latex] (E6.1.12)

which is a second-order ordinary differential equation, in which the pressure gradient will be shown to be uniform between the inlet and exit, being given by:

[latex]-\frac{\partial p}{\partial x}=\frac{p_{1}-p_{2}}{L} .[/latex] (E6.1.13)

A minus sign is used on the left-hand side, since [latex]\partial p / \partial x[/latex] is negative, thus rendering both sides of Eqn. (E6.1.13) as positive quantities.

Equation (E6.1.12) can be integrated twice, in turn, to yield an expression for the velocity. After multiplication of both sides by [latex]d y[/latex], a first integration gives:

[latex]\begin{aligned}\int \frac{d^{2} v_{x}}{d y^{2}} d y=\int \frac{d}{d y}\left(\frac{d v_{x}}{d y} ight) d y &=\int \frac{1}{\mu}\left(\frac{\partial p}{\partial x} ight) d y, \ \frac{d v_{x}}{d y} &=\frac{1}{\mu}\left(\frac{\partial p}{\partial x} ight) y+c_{1}. (E6.1.14)\end{aligned}[/latex]

A second integration, of Eqn. (E6.1.14), yields:

[latex]\begin{aligned}\int \frac{d v_{x}}{d y} d y &=\int\left[\frac{1}{\mu}\left(\frac{\partial p}{\partial x} ight) y+c_{1} ight] d y, \v_{x} &=\frac{1}{2 \mu}\left(\frac{\partial p}{\partial x} ight) y^{2}+c_{1} y+c_{2}. (E6.1.15)\end{aligned}[/latex]

The two constants of integration, [latex]c_{1}[/latex] and [latex]c_{2}[/latex], are determined by invoking the boundary conditions:

[latex]\begin{array}{ll}y=0: & \frac{d v_{x}}{d y}=0, \ \end{array}[/latex] (E6.1.16)

[latex]\begin{array}{ll}y=d: & v_{x}=0,\end{array}[/latex] (E6.1.17)

leading to:

[latex]c_{1}=0, \quad c_{2}=-\frac{1}{2 \mu}\left(\frac{\partial p}{\partial x} ight) d^{2} .[/latex] (E6.1.18)

Eqns. (E6.1.15) and (E6.1.18) then furnish the velocity profile:

[latex]v_{x}=\frac{1}{2 \mu}\left(-\frac{\partial p}{\partial x} ight)\left(d^{2}-y^{2} ight),[/latex] (E6.1.19)

in which [latex]-\partial p / \partial x[/latex] and [latex]\left(d^{2}-y^{2} ight)[/latex] are both positive quantities. The velocity profile is parabolic in shape and is shown in Fig. E6.1.2.

Alternative integration procedure. Observe that we have used indefinite integrals in the above solution and have employed the boundary conditions to determine the constants of integration. An alternative approach would again be to integrate Eqn. (E6.1.12) twice, but now to involve definite integrals by inserting the boundary conditions as limits of integration.

Thus, by separating variables, integrating once, and noting from symmetry about the centerline that [latex]d v_{x} / d y=0[/latex] at [latex]y=0[/latex], we obtain:

[latex]\mu \int_{0}^{d v_{x} / d y} d\left(\frac{d v_{x}}{d y} ight)=\frac{\partial p}{\partial x} \int_{0}^{y} d y,[/latex] (E6.1.20)

or:

[latex]\frac{d v_{x}}{d y}=\frac{1}{\mu}\left(\frac{\partial p}{\partial x} ight) y.[/latex] (E6.1.21)

A second integration, noting that [latex]v_{x}=0[/latex] at [latex]y=d[/latex] (zero velocity in contact with the upper plate - the no-slip condition), yields:

[latex]\int_{0}^{v_{x}} d v_{x}=\frac{1}{\mu}\left(\frac{\partial p}{\partial x} ight) \int_{d}^{y} y d y .[/latex] (E6.1.22)

That is:

[latex]v_{x}=\frac{1}{2 \mu}\left(-\frac{\partial p}{\partial x} ight)\left(d^{2}-y^{2} ight),[/latex] (E6.1.23)

in which two minus signs have been introduced into the right-hand side in order to make quantities in both pairs of parentheses positive. This result is identical to the earlier Eqn. (E6.1.19). The student is urged to become familiar with both procedures before deciding on the one that is individually best suited.

Also, the reader who is troubled by the assumption of symmetry of [latex]v_{x}[/latex] about the centerline (and by never using the fact that [latex]v_{x}=0[/latex] at [latex]y=-d[/latex] ) should be reassured by an alternative approach, starting from Eqn. (E6.1.15):

[latex]v_{x}=\frac{1}{2 \mu}\left(\frac{\partial p}{\partial x} ight) y^{2}+c_{1} y+c_{2} .[/latex] (E6.1.24)

Application of the [latex]t w o[/latex] boundary conditions, [latex]v_{x}=0[/latex] at [latex]y=\pm d[/latex], gives

[latex]c_{1}=0, \quad c_{2}=-\frac{1}{2 \mu}\left(\frac{\partial p}{\partial x} ight) d^{2},[/latex] (E6.1.25)

leading again to the velocity profile of Eqn. (E6.1.19) without the assumption of symmetry.

Volumetric flow rate. Integration of the velocity profile yields an expression for the volumetric flow rate [latex]Q[/latex] per unit width of the system. Observe first that the differential flow rate through an element of depth [latex]d y[/latex] is [latex]d Q=v_{x} d y[/latex], so that:

[latex]Q=\int_{0}^{Q} d Q=\int_{-d}^{d} v_{x} d y=\int_{-d}^{d} \frac{1}{2 \mu}\left(-\frac{\partial p}{\partial x} ight)\left(d^{2}-y^{2} ight) d y=\frac{2 d^{3}}{3 \mu}\left(-\frac{\partial p}{\partial x} ight).[/latex] (E6.1.26)

Since from an overall macroscopic balance [latex]Q[/latex] is constant, it follows that [latex]\partial p / \partial x[/latex] is also constant, independent of distance [latex]x[/latex]; the assumptions made in Eqns. (E6.1.6) and (E6.1.13) are therefore verified. The mean velocity is the total flow rate per unit depth:

[latex]v_{x m}=\frac{Q}{2 d}=\frac{d^{2}}{3 \mu}\left(-\frac{\partial p}{\partial x} ight),[/latex] (E6.1.27)

and is therefore two-thirds of the maximum velocity, [latex]v_{x \max }[/latex], which occurs at the centerline, [latex]y=0[/latex].

Shear-stress distribution. Finally, the shear-stress distribution is obtained by employing Eqn. (5.60):

[latex] au_{y x}=\mu\left(\frac{\partial v_{x}}{\partial y}+\frac{\partial v_{y}}{\partial x} ight).[/latex] (5.60)

By substituting for [latex]v_{x}[/latex] from Eqn. (E6.1.15) and recognizing that [latex]v_{y}=0[/latex], the shear stress is:

[latex] au_{y x}=-y\left(-\frac{\partial p}{\partial x} ight) .[/latex] (E6.1.28)

Referring back to the sign convention expressed in Fig. 5.12, the first minus sign in Eqn. (E6.1.28) indicates for positive [latex]y[/latex] that the fluid in the region of greater [latex]y[/latex] is acting on the region of lesser [latex]y[/latex] in the negative [latex]x[/latex] direction, thus trying to retard the fluid between it and the centerline and acting against the pressure gradient. Representative distributions of pressure and shear stress, from Eqns. (E6.1.11) and (E6.1.28), are sketched in Fig. E6.1.3. More precisely, the arrows at the left and right show the external pressure forces acting on the fluid contained between [latex]x=0[/latex] and [latex]x=L[/latex].

Question 6.1: Flow Between Parallel Plates Fig. E6.1.1 shows a fluid of vi...

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