<< Chapter < Page Chapter >> Page >
F p = - s p n d S = - v p d V = - ρ f g V

where ρ f is the (constant) density of the fluid. Thus, the upward force on the object due to pressure equals the weight of an equivalent volume of liquid; this is Archimedes' law . An important implication of this equation is that F p is independent of the absolute pressure (provided that the density is independent of pressure).

The buoyancy force is the net force on the object due to the same static pressure variation and gravity. Evaluating the gravitational force on the solid body by setting ρ = ρ f , the buoyancy force is found to be

F B = ( ρ i - ρ f ) g V .

Principal axes of stress and the notion of isotropy

The diagonal terms T 11 , T 22 , T 33 of the stress tensor are sometimes called the direct stresses and the terms T 12 , T 21 , T 31 , T 13 , T 23 , T 32 the shear stresses . When there are no external or stress couples, the stress tensor is symmetric and we can invoke the known properties of symmetric tensors. In particular, there are three principal directions and referred to coordinates parallel to these, the shear stresses vanish. The direct stresses with these coordinates are called the principal stresses and the axes the principal axes of stress .

An isotropic fluid is such that simple direct stress acting in it does not produce a shearing deformation. This is an entirely reasonable view to take for isotropy means that there is no internal sense of direction within the fluid. Another way of expressing the absence of any internally preferred direction is to say that the functional relation between stress and rate of deformation must be independent of the orientation of the coordinate system. We shall show in the next section that this implies that the principal axes of stress and rate of deformation coincide.

The stokesian fluid

The constitutive equation of a non-elastic fluid satisfying the hypothesis of Stokes is called a Stokesian fluid. This fluid is based on the following assumptions.

  1. The stress tensor T i j is a continuous function of the rate of deformation tensor e i j and the local thermodynamic state, but independent of other kinematical quantities.
  2. The fluid is homogeneous, that is, T i j does not depend explicitly on x .
  3. The fluid is isotropic, that is, there is no preferred direction.
  4. When there is no deformation ( e i j = 0 ) the stress is hydrostatic, ( T i j = - p δ i j ) .

The first assumption implies that the relation between the stress and rate of strain is independent of the rigid body rotation of an element given by the antisymmetric kinematical tensor Ω i j . The thermodynamic variables, for example, pressure and temperature, will be carried along this discussion without specific mention except where it is necessary for emphasis. We are concerned with a homogeneous portion of fluid so the second assumption is that the stress tensor depends only on position through the variation of e i j and thermodynamic variables with position. The third assumption is that of isotropy and this implies that the principal directions of the two tensors coincide. To express this as an equation we write T i j = f i j e p q , then if there is no preferred direction T i j is the same function f i j of e p q as T i j is of e p q . Thus

T ¯ i j = f i j e ¯ p q .

The fourth assumption is that the tensor

P i j = T i j + p δ i j

vanishes when there is no motion. P i j is called the viscous stress tensor.

Constitutive equations of the stokesian fluid

The arguments for the form of the constitutive equation for the Stokesian fluid is given by Aris and is not repeated here. The equation takes the form

T i j = - p δ i j + β e i j + γ e i k e k j , T = - p I + β e + γ e e

which insures that T i j reduces to the hydrostatic form when the rate of deformation vanishes.

The newtonian fluid

The Newtonian fluid is a linear Stokesian fluid, that is, the stress components depend linearly on the rates of deformation. Aris gives two arguments that deduce the form of the constitutive equation for a Newtonian fluid.

T i j = ( - p + λ Θ ) δ i j + 2 μ e i j where Θ = e i j = ν i , i = v

Interpretation of the constants λ And μ

Consider the shear flow given by

v 1 = f ( x 2 ) , v 2 = v 3 = 0

For this we have all the e i j zero except

e 12 = e 21 = 1 2 f ' ( x 2 ) = 1 2 v 1 x 2

Thus

P 12 = P 21 = μ f ' ( x 2 ) = μ v 1 x 2

and all other viscous stresses are zero. It is evident that μ is the proportionality constant relating the shear stress to the velocity gradient. This is the common definition of the viscosity, or more precisely the coefficient of shear viscosity of a fluid.

For an incompressible, Newtonian fluid the pressure is the mean of the principal stresses since this is

1 3 T i i = - p + λ Θ + 2 3 μ Θ = - p

For a compressible fluid we should take the pressure p as the thermodynamic pressure to be consistent with our ideas of equilibrium. Thus if we call - p ¯ the mean of the principal stresses,

p ¯ - p = - ( λ + 2 3 μ ) Θ = - ( λ + 2 3 μ ) v = - ( λ + 2 3 μ ) 1 ρ D ρ D t

Since p , the thermodynamic pressure, is in principle known from the equation of state p ¯ - p is a measurable quantity. The coefficient in the equation is known as the coefficient of bulk viscosity . It is difficult to measure, however, since relatively large rates of change of density must be used and the assumption of linearity is then dubious. Stokes assumed that p = p ̲ and on this ground claimed that

λ + 2 3 μ = 0

supporting this from an argument from the kinetic theory of gases.

Assignment 5.1

Assume a Newtonian fluid and calculate the stress tensor for the flow fields of assignment 4.2. Evaluate the force due to the stress on the surface y = 0 .

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Transport phenomena' conversation and receive update notifications?

Ask