Two immiscible liquid films of equal thickness h are Bowing

Two immiscible liquid films of equal thickness h are Bowing between two fixed plates as shown in the figure. The liquids have equal densities p, but different viscosities mu_1 and mu_2. There is no pressure gradient in the x direction - the liquids flow due to gravity alone. Because the plates are very wide and very long, the flows are approximately axial, that is u notequalto 0 but v= 0. The velocity distributions in liquid 1 and liquid 2 are u_1(y) and u_2(y) respectively. At the interface of. the two liquids, the shear stresses arc the same. That means, at y = 0, mu_1 partial differential_u1(y)/partial differential_y y=o = mu_2 partial differential_u2(y)/partial differential_y y=0 Additionally, because of the velocity continuity, at y = 0, u_1(0) = u_2(0). Assuming steady laminar flows, (a) starting from the Navier-Stokes equations to find the expressions of u_1, (y) and u_2(y). (b) what is the velocity at the interface (i.e., at y = 0)?

Solution

Navier Stokes eqn for incompressible flows is

In x-direction: (del u / del t) + (u delu / delx + vdelu/dely + wdelu/delz) = g_x - (1/rho)(delP/delx) + neu*(del²u/delx² + del²u / dely² + del²u/delz²)

In y-direction: (del v / del t) + (u delv / delx + vdelv/dely + wdelv/delz) = g_y - (1/rho)(delP/dely) + neu*(del²v/delx² + del²v / dely² + del²v/delz²)

From our problem, we have v = w = 0, delu/delt = delu/delx = delu/delz = 0, delP/delx = 0, g_x = g*sin theta, g_y = -g*cos theta

Therefore we get

In x-direction: 0 = g*sin theta + neu*(del²u / dely² )........eqn1

In y-direction: 0 = -g*cos theta - (1/rho)(delP/dely)

Taking eqn1, del²u / dely² = - (g/neu)*sin theta

del²u / dely² = - (rho*g/meu)*sin theta

meu*(del²u / dely² ) = - rho*(g*sin theta)

Integrating it,

meu*delu/dely = - rho*(g*sin theta)*y + C₁

At y = 0. we have u₁ = u₂ and meu₁*delu₁/dely = meu₂*delu₂/dely

C₁ = meu₁*delu₁/dely = meu₂*delu₂/dely

meu*delu/dely = - rho*(g*sin theta)*y + meu₁*delu₁/dely

Integrating again,

meu*u =  - rho*(g*sin theta)*y² /2 + C₁y + C₂

At y = h, -h. we have u₁ = u₂ = 0

0 = - rho*(g*sin theta)*h² /2 + C₁h + C₂

C2 = rho*(g*sin theta)*h² /2 - (meu₁*delu₁/dely)*h........for fluid1

C2 = rho*(g*sin theta)*h² /2 + (meu₂*delu₂/dely)*h........for fluid2

Thus, for fluid 1, meu₁*u₁ =  - rho*(g*sin theta)*y² /2 + (meu₁*delu₁/dely)*y + rho*(g*sin theta)*h² /2 - (meu₁*delu₁/dely)*h

and, for fluid 2, meu₂*u₂ =  - rho*(g*sin theta)*y² /2 + (meu₂*delu₂/dely)*y + rho*(g*sin theta)*h² /2 + (meu₂*delu₂/dely)*h

Thus, for fluid 1, meu₁*u₁ = rho*(g*sin theta)*(h² - y²) /2 + (meu₁*delu₁/dely)*(y - h)

and, for fluid 2, meu₂*u₂ = rho*(g*sin theta)*(h² - y²) /2 + (meu₂*delu₂/dely)*(y + h)

Thus, for fluid 1, u₁ = (rho*g*sin theta / meu₁)*(h² - y²) /2 + (delu₁/dely)*(y - h)

and, for fluid 2, u₂ = (rho*g*sin theta / meu₂)*(h² - y²) /2 + (delu₂/dely)*(y + h)

At interface where y = 0, we get

For fluid 1, u₁ = (rho*g*sin theta / meu₁)*(h² /2) - (delu₁/dely)* h

and, for fluid 2, u₂ = (rho*g*sin theta / meu₂)*(h² /2) + (delu₂/dely)*h