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Fig. 5.4 Forces acting on an element of fluid.
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respectively denoted by F1 and G. Thus we can write
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(F1 + G) dX 1 dX 2 dx 3
Therefore Newton's second law for a fluid element takes the form
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)U = F1+ G
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Thus the derivation of the Navier-Stokes equation reduces to the derivation of a definite expression for G. Let uS choose a coordinate system such that the fluid element under consideration is a cube with edges along the three coordinate axes, as shown in Fig. 5.4. The six faces of this cube are subjected to forces exerted by neighboring fluid elements. The force on each face is such that its direction is determined by the direction of the normal vector to the face. That is, its direction depends on which side of the face is considered the "outside." This is physically obvious if we remind ourselves that this force arises from hydrostatic pressure and viscous drag. Let Ti be the force per unit area acting on the face whose normal lies along the x i axis. Then the forces per unit area acting on the two faces normal to the Xi axis are, respectively (see Fig. 5.4),
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The total force acting on the cube by neighboring fluid elements is then given by
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We denote the components of the vectors T1 , T z, T3 as follows:
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(P11' P1Z , P13 ) (PZ1 ' Pzz , P23 ) (P31 , P32 , P33 )
G=-V' p
With this, (5.104) becomes
p(~ at
+ u
V')U =
V' . P
which is of the same form as (5.22) if we set F1 = pF1m, where F is the external force per molecule and m is the mass of a molecule. To derive the Navier-Stokes equation, we only have to deduce a more explicit form for Pij We postulate that (5.110) is valid, whatever the coordinate system we choose. It follows that Pi} is a tensor. We assume the fluid under consideration to be isotropic, so that there can be no intrinsic distinction among the axes Xl' Xz, x 3 Accordingly we must have
PH = Pzz = P33 == P (5.111) where P is by definition the hydrostatic pressure. Thus Pi} can be written in the form (5.112) Pi} = ~i}P + Pi}
where Pi} is a traceless tensor, namely,
" pI ..., II
This follows from the fact that (5.113) is true in one coordinate system and that the trace of a tensor is independent of the coordinate system. Next we make the physically reasonable assumption that the fluid element under consideration, which is really a point in the fluid, has no intrinsic angular momentum. This assumption implies that Pij , and hence Pi}' is a symmetric tensor: (5.114) To see this we need only remind ourselves of the meaning of, for example, P1z . A glance at Fig. 5.5a makes (5.114) obvious. Finally we incorporate into Pi} the empirical connection between the shear force applied to a fluid element and the rate of deformation of the same fluid element. A shear force F per unit area acting parallel to a face of a cube of fluid
Fig. 5.5. Nonrotation of fluid element implies P{2
Pli .
Fig. 5.5b Deformation of fluid element due to shear
tends to stretch the cube into a parallelopiped at a rate given by R' = p.(dcf>/dt), where p. is the coefficient of viscosity and cf> is the angle shown in Fig. 5.5b. Consider now the effect of P12 on one fluid element. It can be seen from Fig. 5.5c, where P{2 is indicated in its positive sense in accordance with (5.105), that
P2l =
_p.(dcf>l + dcf>2) dt dt
p. ( x ~
_p.(au 1 + au l ) aX l aX 1
In general we have
au.) + a x~
To make Pi} traceless we must take
-Jl [(
au au) ax~ + ax~
3~ij\7 u
Fig. 5.5e P{2 as shear force.
p . = l).p IJ IJ
au [( ax.
2 -l)'\7
which is identical in form to (5.75). This completes the phenomenological derivation, which makes it plausible that the Navier-Stokes equation is valid for dilute gas and dense liquid alike.
To illustrate the mathematical techniques of dealing with the equations of hydrodynamics (5.110)-(5.102), we consider two examples of the application of the Navier-Stokes equation to a liquid.