Two viscous, incompressible, immiscible fluids of same density (=ρ) but different viscosities (viscosity of the lower fluid layer = μ1 and that of the upper fluid layer = μ2

Question

Two viscous, incompressible, immiscible fluids of same density [latex] (=\rho) [/latex] but different viscosities (viscosity of the lower fluid layer [latex] =\mu_{1} [/latex] and that of the upper fluid layer [latex] =\mu_{2}<\mu_{1} [/latex] ) flow in separate layers between parallel boundaries located at [latex] y=\pm H [/latex] , as shown in the Fig. 8.6. Thickness of each fluid layer is identical and their interface is flat. The flow is driven by a constant favourable pressure gradient of [latex] \frac{d p}{d x} [/latex] . Derive expressions for the velocity profiles in the fluid layers. Also, make a sketch of the velocity profiles. Assume the flow to be steady and the plates to be of infinitely large width.

Accepted Answer

Large lateral width of the plate renders the basic flow consideration to be two-dimensional, for which the continuity equation under incompressible flow conditions reads:

[latex] \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 [/latex]

For fully developed flow, [latex] \frac{\partial u}{\partial x}=0 [/latex]

Hence, [latex] \frac{\partial v}{\partial y}=0 [/latex]

[latex] \Rightarrow v eq v(y)[/latex]

Since [latex] v=0 [/latex] at [latex] y=\pm H [/latex] as a result of the no-penetration at the walls, [latex] v [/latex] is identically equal to zero for all y, i.e.,

[latex] herefore \quad v=0 ext { at }-H \leq y \leq H [/latex]

Now, considering x momentum equation, we have, for steady flow,

[latex] ho\left[u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} ight]=-\frac{d p}{d x}+\mu\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} ight) [/latex]

Using derivations identical to those presented in Sec. 8.3.1, it follows that

[latex] \frac{d^{2} u}{d y^{2}}=\frac{1}{\mu} \frac{d p}{d x}=\mathrm{constant} [/latex]

[latex]\underbrace{ }_{Function of y only}[/latex]

[latex] \underbrace{ }_{Function of x only} [/latex]

For [latex] -H \leq y \leq 0 [/latex]

[latex] \frac{d^{2} u_{1}}{d y^{2}}=\frac{1}{\mu_{1}} \frac{d p}{d x} [/latex]

Integrating the above equation, we have

[latex] \frac{d u_{1}}{d y} =\frac{1}{\mu_{1}} \frac{d p}{d x} y+C_{1} [/latex]

[latex] u_{1} =\frac{1}{\mu_{1}} \frac{d p}{d x} \frac{y^{2}}{2}+C_{1} y+C_{2}[/latex] (8.39a)

where [latex] C_{1} [/latex] and [latex] C_{2} [/latex] are constants of integration.

For [latex] 0 \leq y \leq H [/latex]

[latex] \frac{d^{2} u_{2}}{d y^{2}}=\frac{1}{\mu_{2}} \frac{d p}{d x} [/latex]

Integrating the above equation, we have

[latex] \frac{d u_{2}}{d y} =\frac{1}{\mu_{2}} \frac{d p}{d x} y+C_{3} [/latex]

[latex] u_{2} =\frac{1}{\mu_{2}} \frac{d p}{d x} \frac{y^{2}}{2}+C_{3} y+C_{4}[/latex] (8.39b)

where [latex] C_{3} [/latex] and[latex] C_{4} [/latex] are constants of integration.

Equations (8.39a) and (8.39b) are subjected to the following boundary conditions.

At [latex] y=-H, u_{1}=0 [/latex]

At [latex] y=H, u_{2}=0 [/latex]

At [latex] y=0, u_{1}=u_{2}[/latex] (continuity of flow velocity)

At [latex] y=0, \quad \mu_{1} \frac{d u_{1}}{d y}=\mu_{2} \frac{d u_{2}}{d y}[/latex] (continuity of shear stress)

From the boundary conditions at [latex] y=-H, u_{1}=0 [/latex] , and at [latex] y=H, u_{2}=0 [/latex] , we get

[latex] C_{2}=C_{1} H-\frac{1}{\mu_{1}} \frac{d p}{d x} \frac{H^{2}}{2} [/latex]

[latex] C_{4}=-C_{3} H-\frac{1}{\mu_{2}} \frac{d p}{d x} \frac{H^{2}}{2}[/latex]

From the boundary condition at [latex] y=0, \mu_{1} \frac{d u_{1}}{d y}=\mu_{2} \frac{d u_{2}}{d y} [/latex] , we have

[latex] \mu_{1} C_{1}=\mu_{2} C_{3} [/latex]

[latex] \Rightarrow [/latex] [latex] C_{1}=\frac{\mu_{2}}{\mu_{1}} C_{3} [/latex]

From the boundary condition at [latex] y=0, u_{1}=u_{2} [/latex] , one can write

[latex] C_{2}=C_{4} [/latex]

Or [latex] C_{1} H-\frac{1}{\mu_{1}} \frac{d p}{d x} \frac{H^{2}}{2}=-C_{3} H-\frac{1}{\mu_{2}} \frac{d p}{d x} \frac{H^{2}}{2} [/latex]

Or [latex] \left(C_{1}+C_{3} ight) H=\frac{d p}{d x} \frac{H^{2}}{2}\left[\frac{1}{\mu_{1}}-\frac{1}{\mu_{2}} ight] [/latex]

Or [latex] \left(\frac{\mu_{2}}{\mu_{1}}+1 ight) C_{3}=\frac{d p}{d x} \frac{H}{2}\left[\frac{1}{\mu_{1}}-\frac{1}{\mu_{2}} ight] [/latex]

Or [latex] C_{3}=\frac{d p}{d x} \frac{H}{2} \frac{\left(\mu_{2}-\mu_{1} ight)}{\left(\mu_{2}+\mu_{1} ight) \mu_{2}} [/latex]

[latex] C_{1}=\frac{d p}{d x} \frac{H}{2} \frac{\left(\mu_{2}-\mu_{1} ight)}{\left(\mu_{2}+\mu_{1} ight) \mu_{1}}[/latex]

Substituting the values of [latex] C_{1}, C_{2}, C_{3} [/latex] and [latex] C_{4} [/latex] in the Eqs (8.39a) and (8.39b), we have

[latex] u_{1}=\frac{1}{\mu_{1}} \frac{d p}{d x} \frac{1}{2}\left[y^{2}-H^{2}+\frac{\left(\mu_{2}-\mu_{1} ight)}{\left(\mu_{2}+\mu_{1} ight)} H(y+H) ight] [/latex]

[latex] u_{2}=\frac{1}{\mu_{2}} \frac{d p}{d x} \frac{1}{2}\left[y^{2}-H^{2}+\frac{\left(\mu_{2}-\mu_{1} ight)}{\left(\mu_{2}+\mu_{1} ight)} H(y-H) ight][/latex]

The velocity profiles are shown in Fig. 8.7.

Note from the above figure that since [latex] \mu_{2}<\mu_{1},\left.\frac{d u_{2}}{d y} ight|_{y=0}>\left.\frac{d u_{1}}{d y} ight|_{y=0} [/latex]

(so as to satisfy [latex] \left.\mu_{1} \frac{d u_{1}}{d y} ight|_{y=0}=\left.\mu_{2} \frac{d u_{2}}{d y} ight|_{y=0}[/latex])

Or [latex] \left.\frac{d y}{d u_{1}} ight|_{y=0}>\left.\frac{d y}{d u_{2}} ight|_{y=0} [/latex]

or equivalently, [latex] an heta_{1}> an heta_{2} [/latex] .

Question 8.2: Two viscous, incompressible, immiscible fluids of same densi...

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