Two immiscible, incompressible, viscous fluids having same densities but different viscosities are contained between two infinite horizontal parallel plates, 2 m apart as shown below. The bottom plate is fixed and the upper plate moves to the right with a constant velocity of 3 m/s. With the assumptions of Newtonian fluid, steady, and fully developed laminar flow with zero pressure gradient in all directions, the momentum equations simplify to

                                                         d²u /dy² = 0

If the dynamic viscosity of the lower fluid, μ2, is twice that of the upper fluid, μ1, then the velocity at the interface (round off to two decimal places) is _______ m/s.

The correct answer is 1.

Let us solve this problem step-by-step.

We are given two horizontal parallel plates separated by a distance of 2 m. The bottom plate is fixed, and the top plate moves at a velocity of 3 m/s. The system contains two immiscible fluids. In this standard configuration, the interface lies exactly midway between the plates, meaning the thickness of each fluid layer is:
$h_1=h_2=1\text{m}$
Thus, the bottom plate is at $y=0$, the interface is at $y=1\text{m}$, and the top plate is at $y=2\text{m}$.

For both fluids, the simplified momentum equation is:
$\frac{d^{2}u}{d{y}^2}=0$
Integrating this equation twice for each fluid layer gives a linear velocity profile for each fluid.

Let $u_2\left(y\right)$ be the velocity profile of the lower fluid (fluid 2) for $0\le y\le 1$:
$u_2\left(y\right)=C_{1}y+C_2$

Let $u_1\left(y\right)$ be the velocity profile of the upper fluid (fluid 1) for $1\le y\le 2$:
$u_1\left(y\right)=C_{3}y+C_4$

Now, we apply the boundary conditions to find the constants:

1. At the bottom plate ($y=0$), the fluid is at rest (no-slip condition):
$u_2\left(0\right)=0\Rightarrow C_2=0$
So, the velocity profile in the lower fluid simplifies to:
$u_2\left(y\right)=C_{1}y$

2. At the top plate ($y=2\text{m}$), the fluid moves with the plate velocity of 3 m/s:
$u_1\left(2\right)=3\Rightarrow 2C_3+C_4=3\Rightarrow {\mathrm{C<!--="4"-->}}_{}=3-2{\mathrm{C<!--="3"-->}}_{}$

3. At the interface ($y=1\text{m}$), the velocity must be continuous:
$u_1\left(1\right)=u_2\left(1\right)=u_i$
Substituting $y=1$ into the velocity equations:
${\mathrm{C<!--="1"-->}}_{}\left(1\right)={\mathrm{C<!--="3"-->}}_{}\left(1\right)+3-2{\mathrm{C<!--="3"-->}}_{}$
${\mathrm{C<!--="1"-->}}_{}=3-{\mathrm{C<!--="3"-->}}_{}$
${\mathrm{C<!--="1"-->}}_{}+{\mathrm{C<!--="3"-->}}_{}=3$ (Equation 1)

4. At the interface ($y=1\text{m}$), the shear stress must also be continuous:
${\mathrm{\tau <!--="1"-->}}_{}={\mathrm{\tau <!--="2"-->}}_{}\Rightarrow {\mathrm{\mu <!--="1"-->}}_{}{\left(\frac{d{\mathrm{u<!--="1"-->}}_{}}{dy}\right)}_{y=1}={\mathrm{\mu <!--="2"-->}}_{}{(\frac{d{\mathrm{u<!--="2"-->}}_{}}{dy}}_{})\mathrm{y<!--=1</mrow-->}$

Since the velocity profiles are linear, the gradients are:
$\frac{d{\mathrm{u<!--="1"-->}}_{}}{dy}={\mathrm{C<!--="3"-->}}_{}$
and
$\frac{d{\mathrm{u<!--="2"-->}}_{}}{dy}={\mathrm{C<!--="1"-->}}_{}$

We are given that the dynamic viscosity of the lower fluid, ${\mathrm{\mu <!--="2"-->}}_{}$, is twice that of the upper fluid, ${\mathrm{\mu <!--="1"-->}}_{}$ (${\mathrm{\mu <!--="2"-->}}_{}=2{\mathrm{\mu <!--="1"-->}}_{}$). Substituting this relationship into the shear stress continuity equation:
${\mathrm{\mu <!--="1"-->}}_{}{\mathrm{C<!--="3"-->}}_{}=2{\mathrm{\mu <!--="1"-->}}_{}{\mathrm{C<!--="1"-->}}_{}$
${\mathrm{C<!--="3"-->}}_{}=2{\mathrm{C<!--="1"-->}}_{}$ (Equation 2)

Now, substitute Equation 2 into Equation 1:
${\mathrm{C<!--="1"-->}}_{}+2{\mathrm{C<!--="1"-->}}_{}=3$
$3{\mathrm{C<!--="1"-->}}_{}=3\Rightarrow {\mathrm{C<!--="1"-->}}_{}=1$

Since the velocity at the interface is ${\mathrm{u<!--="i"-->}}_{}={\mathrm{u<!--="2"-->}}_{}\left(1\right)={\mathrm{C<!--="1"-->}}_{}\cdot 1$:
${\mathrm{u<!--="i"-->}}_{}=1\text{m/s}$