Entrance Region for Laminar Flow Between Flat Plates In Example 6.1, involving laminar flow between parallel plates, only the velocity in the axial direction, vx, was considered to be nonzero. In reality, a flat profile at the entrance will gradually be transformed into a final parabolic shape by t

Question

Entrance Region for

Laminar Flow Between Flat Plates

In Example [latex]6.1[/latex], involving laminar flow between parallel plates, only the velocity in the axial direction, [latex]v_{x}[/latex], was considered to be nonzero. In reality, a flat profile at the entrance will gradually be transformed into a final parabolic shape by the viscous action of the walls. Fig. E8.4.1 shows three representative stages of the progression, in which a boundary layer builds up from a thickness of [latex]\delta=0[/latex] at the entrance to a final value of [latex]y=d ;[/latex] the problem is to determine the necessary distance [latex]x=L .[/latex] Note that in order to satisfy a mass balance, the "mainstream" velocity [latex]V[/latex] is not constant but continues to increase downstream along the path [latex]\mathrm{A}-\mathrm{B}-\mathrm{C}[/latex]. The solution should be in the form [latex]L / d=c \rho \bar{V} d / \mu[/latex], where the value of [latex]c[/latex] is to be determined.

Accepted Answer

The analysis concentrates on half of the region, between the lower wall and the centerline, since the other half can be obtained from symmetry. All quantities are on the basis of unit depth normal to the plane of the diagram. First, assume a reasonable velocity profile in the boundary layer --- one that gives zero velocity at the wall and matches both the mainstream velocity [latex](V)[/latex] and its slope (zero) at [latex]y=\delta[/latex] :

[latex]v_{x}=V\left(2 \frac{y}{\delta}-\frac{y^{2}}{\delta^{2}} ight).[/latex] (E8.4.1)

This particular choice has the further advantage that it exactly reproduces the well-known parabola when point [latex]\mathrm{C}[/latex] is reached.

On the basis of this velocity profile, the following useful quantities can be determined, in which it is convenient to define a fraction [latex]\alpha[/latex] :

[latex]\alpha=\frac{\delta}{d}.[/latex] (E8.4.2)

Mass flux

[latex]m=\underbrace{ ho V(d-\delta)}_{ ext {Mainstream }}+\underbrace{ ho \int_{0}^{\delta} v_{x} d y}_{ ext {Boundary layer }}= ho V(d-\delta)+\frac{2}{3} ho V \delta= ho V d\left(1-\frac{\alpha}{3} ight) . (E8.4.3) [/latex]

Momentum flux

[latex]\dot{\mathcal{M}}=\underbrace{ ho V^{2}(d-\delta)}_{ ext {Mainstream }}+\underbrace{ ho \int_{0}^{\delta} v_{x}^{2} d y}_{ ext {Boundary layer }}= ho V^{2}(d-\delta)+\frac{8}{15} ho V^{2} \delta= ho V^{2} d\left(1-\frac{7}{15} \alpha ight) . (E8.4.4) [/latex]

Wall shear stress

[latex] au_{w}=\mu\left(\frac{d v_{x}}{d y} ight)_{y=0}=\frac{2 \mu V}{\delta}=\frac{2 \mu V}{\alpha d}.[/latex] (E8.4.5)

Mean/mainstream velocity relation

[latex]m= ho \bar{V} d= ho V d\left(1-\frac{\alpha}{3} ight), \quad ext { or } \quad \bar{V}=V\left(1-\frac{\alpha}{3} ight) \quad ext { or } \quad V=\frac{3 \bar{V}}{3-\alpha} .[/latex] (E8.4.6)

Bernoulli's equation in the mainstream

[latex]\frac{V^{2}}{2}+\frac{p}{ ho}=c, \quad ext { or } \quad \frac{d p}{d x}=- ho V \frac{d V}{d x} .[/latex] (E8.4.7)

Now perform a momentum balance in the [latex]x[/latex] direction on the rectangular element shown in Fig. E8.4.2, situated between the wall and centerline, noting that at the centerline there is zero velocity gradient and hence zero shear stress:

[latex]\dot{\mathcal{M}}+p d- au_{w} d x-\left(\dot{\mathcal{M}}+\frac{d \dot{\mathcal{M}}}{d x} d x ight)-\left(p+\frac{d p}{d x} d x ight) d=0,[/latex] (E8.4.8)

or:

[latex] au_{w}+\frac{d \dot{\mathcal{M}}}{d x}+\frac{d p}{d x} d=0.[/latex] (E8.4.9)

Differentiation of Eqn. (E8.4.4) with respect to [latex]x[/latex] gives:

[latex]\frac{d \dot{\mathcal{M}}}{d x}= ho d\left[2 V \frac{d V}{d x}\left(1-\frac{7}{15} \alpha ight)-\frac{7}{15} V^{2} \frac{d \alpha}{d x} ight].[/latex] (E8.4.10)

After making the appropriate substitutions, Eqn. (E8.4.9) yields an ordinary differential equation for [latex]\alpha[/latex] as a function of [latex]x[/latex] :

[latex]\frac{10 \mu}{ ho \bar{V} d^{2}}=\frac{\alpha(6+7 \alpha)}{(3-\alpha)^{2}} \frac{d \alpha}{d x} .[/latex] (E8.4.11)

Separation of variables and integration between the channel inlet and the location at which the boundary layer reaches out to the centerline gives:

[latex]\frac{10 \mu}{ ho \bar{V} d^{2}} \int_{0}^{L} d x=\int_{0}^{1} \frac{\alpha(6+7 \alpha) d \alpha}{(3-\alpha)^{2}} \doteq 1.048.[/latex] (E8.4.12)

That is, the dimensionless entrance length is directly proportional to the Reynolds number (based in this case on half the separation between the plates):

[latex]\frac{L}{d}=0.1048 \frac{ ho \bar{V} d}{\mu} .[/latex] (E8.4.13)

For flow in a pipe of diameter [latex]D[/latex], the corresponding result is [latex]L / D \doteq 0.061 \mathrm{Re}[/latex], in which [latex]\operatorname{Re}= ho \bar{V} D / \mu[/latex].

Question 8.4: Entrance Region for Laminar Flow Between Flat Plates In Exam...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions