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Please provide details, no shortcuts, will give 5 stars for good work. Thanks.

The steady, parallel flow of a viscous fluid between two fixed parallel plates (as shown below) is considered. The gap between the plates is h. The flow is fully-developed in the x-direction (flow direction) such as all velocity gradients in x are 0. The plates are large in the z-direction (into the paper), such that w is zero and gradients in z are also 0. The velocity vector may be expressed as v =(w,v,0) . The body force (gravity) per unit mass is to be neglected in the analysis. The fluid is incompressible and Newtonian with constant viscosity. The fluid velocity is low enough such that the advective terms in the momentum equations can be neglected (i.e. creeping or Stokes flow). State the boundary conditions (BCs) on the u velocity component of the flow and state where there are implemented (e.g. BC at y = ...)(cross out extra entries for BCs) BC1: BC2: Show in an unambiguous fashion that the velocity component in the v-direction, v = 0. What can you say about the velocity component u (in the x-direction)? and why (i.e. show)? u = f(x) u = Const. u = f(y) Simplify the momentum equations in x, y and z by removing all negligible terms. It is possible that for some of the equations, you may get a trivial solution . x-momentum: p/ x = y-momentum: p/ y = z-momentum: p/ z = Which statement is most accurate about ? and why (i.e. show)?

Explanation / Answer

a) bcs

u=0 at y=0 (no-slip criterion)

and u=0 at y=h (no-slip criterion)

b) from continuity equation,

(du/dx) +(dv/dy)+ (dw/dz)=0 (all derivatives are partial)

du/dx=0 (fully developed flow)

dw/dz=0 because w=0 for all z (given)

therefore dv/dy=0

v=0 at walls , that is at y=0,h because of no-penetration boundary condition.

as dv/dy=0 therfore v=constant for all y ...as v=0 at y=0,h , therefore, v=0 for all y

c) As, du/dx=0 (fully developed flow),

and du/dz=0 (given properties do not change along z)

So, u is not a function of x,z....It is only a function of y and hence u=f(y)

d) The second navier-stokes equation is,

r((dv/dt) +u(dv/dx) + v(dv/dy)+w(dv/dz))= -(dp/dy)+ (mu)( d^2 v/dx^2 + d^2 v/dy^2+ d^2 v/dz^2) (all of them are partial dervatives)

mu= viscosity

r=density

z-component is zero (given)

w=0

v=0 as proved above....

So, dp/dy=0

Similarly from the third navier-stokes equation with terms in w,

As w=0,

dp/dz=0

So, p is only a function of x..

So, dp/dx is a total derivative, not a partial derivative...

First navier-stokes equation,

r((du/dt) +u(du/dx) + v(du/dy)+w(du/dz))= -(dp/dx)+ (mu)( d^2 u/dx^2 + d^2 u/dy^2+ d^2 u/dz^2)

du/dt =0 (steady state condition)

du/dx=0 (fully developed flow)

du/dz=0 (properties do not change along z)

d^2 u/dx^2 =0 as du/dx=0

d^2 u/dz^2=0 as du/dz=0

So, the equation reduces to,

(dp/dx)=(mu/r)(d^2 u/dy^2)

e) u is only a function of y

right hand side is only a function of y

p is a function of x only

so, left hand side is only a function of x

So, we have,

f1(x)=f2(y)

This is only matematically true when each side is a constant...

hence,

dp/dx=constant

and

(mu/r)(d^2 u/dy^2)=constant...

Rate well...it took a lot of effort....

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