Creeping viscous flow in a semi-infinite channel

2 Creeping viscous flow in a semi-infinite channel 

As a very simple example of the use of the Navier-Stokes equation under creeping flow conditions, consider the flow of a Newtonian liquid in a parallel-sided, semi-infinite channel in the absence of body forces. Let the boundaries be planes located at y = /z, each lying in an xz plane and let the liquid flow be in the z direction. The motion is assumed to be so slow that a creeping solution is obtained, the liquid is assumed to be incompressible and r is treated as a constant. From the symmetry of the problem the liquid flow velocity is a function of y only, so that the three components of the Navier-Stokes equation are  

Equations [6.2b] and [6.2c] show that p is a function of z only so that the gradient of p may be written dp/dz. Further, b u ldy is a function of y only, and may be written d u /dy it follows that both dp/dz and rjid u /dy ) must be constants. 

By symmetry, du ldy = 0 when y = 0. This makes the constant in equation [6.3] equal to zero, but notice that du /dy is not zero at the walls. 

The condition of no slip at the solid boundary gives = 0 when y = h, so that the constant in equation [6.4] is equal to Therefore  

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