Page 31 - Introduction to Computational Fluid Dynamics

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INTRODUCTION
stresses are given by May 20, 2005 12:20
∂u

σ x =−p + σ =−p + q + τ xx =−p + q + 2µ , (1.7)
x
∂x
∂v

σ y =−p + σ =−p + q + τ yy =−p + q + 2µ , (1.8)
y
∂y
∂w

σ z =−p + σ =−p + q + τ zz =−p + q + 2µ . (1.9)
z
∂z

In these normal stress expressions, σ is called the deviotoric stress and the
significance of quantity q in its definition requires elaboration. Schlichting [65]
and Warsi [86], for example, define a space-averaged pressure p as
1
p =− (σ x + σ y + σ z ). (1.10)
3
Now, an often overlooked requirement of the Stokes’s relations is that, in a
continuum, p must equal the point value of pressure p and the latter, in turn, must
equal the thermodynamic pressure p th . Thus,
2
p = p = p th = p − q − µ · V. (1.11)
3
In the context of this requirement, we now consider different flow cases to derive
the significance of q.
1. Case 1 (V = 0): In this hydrostatic case,

p = p − q. (1.12)
But in this case, p can only vary linearly with x, y, and z and, therefore, the point
value of p exactly equals its space-averaged value p in both continuum as well
as discretised space and hence q = 0 exactly.
2. Case 2 (µ = 0or · V = 0): Clearly when µ = 0 (inviscid flow) or . V = 0
(constant-density incompressible flow) Equation 1.12 again holds. But, in this
case, since fluid motion is considered, p can vary arbitrarily with x, y, and z
and, therefore, p may not equal p in a discrete space. To understand this matter,
consider a case in which pressure varies arbitrarily in the x direction, whereas
its variation in y and z directions is constant or linear (as in a hydrostatic case).
Such a variation is shown in Figure 1.3. Now consider a point P. According to
Stokes’s requirement p P must equal p in a continuum. However, in a discretised
P
space, the values of pressure are available at points E and W only, and if these
points are equidistant from P then p = 0.5(p W + p E ). Now, this p will not
P
P
equal p P , as seen from the figure, and therefore the requirement of the Stokes’s
relations is not met.
However, without violating the continuum requirement, we may set

q = λ 1 (p − p), (1.13)