Two immiscible liquid films of equal thickness h are Bowing between two fixed pl

Question

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)?

Explanation / Answer

Navier Stokes eqn for incompressible flows is

In x-direction: (del u / del t) + (u delu / delx + vdelu/dely + wdelu/delz) = gx - (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) = gy - (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, gx = g*sin theta, gy = -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

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