Consider incompressible, fully-developed, steady, two-dimensional flow in a channel of height H driven by a given, constant, pressure gradient dp/dx . Unlike the problem done in class, the channel has two fluid layers, one over the other. Each layer has its own density and viscosity. Find the velocity profiles in the two fluids, assuming that the interface is at height ha. (The boundary conditions at the two walls are no-slip. At the interface between the fluids, the velocity and shear stress are continuous.)

u(y)/U = [(y/H) - ((H²/2μ)(dp/dx)(y/H))] × [(1 - y/H)]

Explanation:

The flow is steady, compressible and planar. Thus the incompressible the continuity equation is given as;

(δu/δx) + (δv/δy) = 0

The velocity(v) in the vertical direction would be zero at both boundaries as well as everywhere in the flow.

This means the continuity equation will dictate that:

∂u

/∂x = 0

It means that u is only just a function of y i.e. u = u(y).

Thus, The Navier-Stokes equation in the y-direction will now be reduced to:

∂p

/∂y = 0

This means the pressure can only then be a function of x.

The Navier-Stokes equation in the x-direction would be;

ρ[(∂u

/∂t) + u(∂u/∂x) + v(∂u/∂y) + w(∂u/∂z)] = -dp/dx + μ(∂²u/∂x²) + v(∂²u/∂y²) + w(∂²u/∂z²)

Recall that v = 0 and u = u(y).

Thus, the Navier-Stokes equation in the x-direction would now become;

∂²u/∂y² = (1/μ)(dp/dx)

We now Integrate twice with respect to y to give;

u(y) = (1/2μ)(dp/dx)y² + c1•y + c2)

At boundary condition of y = 0, C2 will be zero.

Thus;

u(y) = (1/2μ)(dp/dx)y² + c1•y)

At height of y = H, we have;

U = (1/2μ)(dp/dx)H² + c1•H) - - - (eq 1)

Making C1 the subject gives;

c1 = (U/H) - (H/2μ)(dp/dx)

Putting that for c1 in (eq 1) and rearranging to simplify gives us;

u(y)/U = [(y/H) - ((H²/2μ)(dp/dx)(y/H))] × [(1 - y/H)]

answer:

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explanation:

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