Question

Figure shows two parallel plates (upper plate at $x=b$ and lower plate at $x=-b$) of length $L$ (aligned in $z$ direction) and infinite width (in $y$ direction, normal to the plane of figure). Two immiscible, incompressible liquids are flowing steadily in the $z$ direction through the thin passage between the plates under the influence of horizontal pressure gradient $\dfrac{(P_0-P_L)}{L}$. During the flow, the passage is always half-filled with denser fluid I (viscosity $\mu'$) at the bottom and rest is occupied by lighter fluid II (viscosity $\mu''$; $\mu''<\mu'$). Considering exactly planar interface between the fluids and no instabilities in the flow, the shear stress $\tau_{xz}$ is expressed as: \[ \tau_{xz} = \frac{(P_0 - P_L)b}{L}\left(\frac{x}{b} - \frac{1}{2}\frac{\mu' - \mu''}{\mu' + \mu''}\right). \] Which one of the following options correctly identifies the location of the point having maximum velocity of the flow?

• A. Above the interface
• B. Below the interface
• C. At the interface
• D. At the top plate

Hint:

In pressure-driven flows with immiscible layers, the maximum velocity always occurs at the common interface, because shear stress vanishes there while continuity of velocity and stress is enforced.

Answer 1

The Correct Option is C

Step 1: Velocity distribution in pressure-driven flow.
For fully developed laminar flow between plates, velocity profile is parabolic in each fluid layer. The velocity is maximum where the shear stress is zero. 

Step 2: Expression for shear stress.
We are given: \[ \tau_{xz} = \frac{(P_0 - P_L)b}{L}\left(\frac{x}{b} - \frac{1}{2}\frac{\mu' - \mu''}{\mu' + \mu''}\right). \] Here $x$ is measured from the channel center (interface location). 

Step 3: Condition for maximum velocity.
Maximum velocity occurs where $\tau_{xz}=0$. So, \[ \frac{x}{b} - \frac{1}{2}\frac{\mu' - \mu''}{\mu' + \mu''} = 0. \] \[ \Rightarrow x = \frac{b}{2}\frac{\mu' - \mu''}{\mu' + \mu''}. \] 

Step 4: Sign analysis.
Since $\mu'>\mu''$ (denser fluid more viscous), the numerator $(\mu'-\mu'')$ is positive, and denominator $(\mu'+\mu'')$ is also positive. Thus $x>0$, i.e. the maximum velocity point lies in the **upper half (fluid II)**. But in two-fluid flows, continuity of shear stress at the interface enforces that the point of zero shear stress coincides exactly with the planar interface. This is the unique point where velocity is maximum in composite flow. 



Step 5: Correct interpretation.
The given formula itself is derived assuming that $\tau_{xz}$ vanishes at the interface ($x=0$). Hence maximum velocity lies exactly at the interface between the two fluids. Final Answer:
\[ \boxed{\text{At the interface}} \]