Math 575-Lecture 2 1 Conservation of Momentum and Cauchy Stress Tensor

Home , Body force, Cauchy stress tensor, Force density, Stress (mechanics)

Math 575-Lecture 2

1 Conservation of momentum and Cauchy stress tensor

Consider a ﬂuid parcel with volume V . The force acting on it includes the body force and surface force F = F v + F S The body force usually comes from external ﬁeld. For example, if there is only gravitational force, then Z F v = ρgdV. V The interesting part is the surface force, which is the forces exerted across the boundary of the volume by the neighboring material, called the stress. Let T be the stress vector (force per unit area, also called traction vector), the total surface force is given by Z F S = T dS ∂V

Now, consider the material volume Rt. By conservation of momentum, we have d Z Z Z ρudV = ρbdV + T dS (1.1) dt Rt Rt ∂Rt where b is some general body force density. By the second convection identity (if you feel confused, consider each component) to the ﬁrst term: d Z Z Z Du ρudV = (ρu)t + ∇ · (u ⊗ (ρu))dV = ρ dV dt Rt Rt Rt Dt Du This actually makes perfect sense. Dt is the acceleration and thus the right hand side is the integrals of the forces of ﬂuid elements, which should be total force. As a result, we have Z Du Z ρ − ρbdV = T dS. Rt Dt ∂Rt As long as we have this, it is clear that this equality should be true for any volume V , since any volume can be regarded as some material value at time t. Let V to be a tetrahedron with one vertex at x and the three edges origin from x are parallel to coordinate axes. The traction T should depend on x, t and the direction n. Consider we scale V by length L. Then, the equation is written as Du X (ρ − ρb)VL3 = T (x, t, n )S L2 + O(L3) Dt i i i

1 As L → 0, we ﬁnd that, we have X T (x, t, ni)Si = 0. i Hence, 3 X Sj T (x, t, n) = − T (x, t, −e ) i S i=1 4

It can be seen easily that Sj/S4 = nj. Hence,

T (x, t, n) = n · σ = niσij where σij = −Tj(x, t, −ei). σ is called the Cauchy stress tensor. By an argument of conservation of angular momentum (which means the net torque acting on an infinitesimal fluid element is zero, and may not be true in some rare cases), σ is symmetric. In this course, we consider all fluids so that σ is symmetric. Now, we can rewrite the Equation (1.1) as d Z Z Z ρudV = ρbdV + n · σdS dt Rt Rt ∂Rt Applying the Reynolds’ transport theorem and divergence theorem, we have Z Z Z Z Z (ρu)t + ∇ · (ρu ⊗ u)dV = (ρu)tdV + n · (ρu ⊗ u)dS = ρbdV + n · σdS Rt Rt S Rt ∂Rt Hence, Z Z Z (ρu)tdV = ρbdV − n · (ρuu − σ)dS. (1.2) Rt Rt ∂Rt ρuu − σ is then called the total momentum flux. This relation yields, (ρu)t = ρb − ∇ · (ρuu − σ). Meanwhile, as we have shown, Z Du Z Z ρ dV = ρbdV + n · σdS Rt Dt Rt ∂Rt Applying the divergence theorem, we have Du ρ = ρb + ∇ · σ Dt This equation can be interpretated as F = ma.

2 Pressure and ideal ﬂuid

We can decompose the Cauchy stress tensor into the isotropic part and anisotropic part:

σ = −pI + τ (1.3) where trace(τ) = 0. p is a scalar depending on x and t only. It is called the ‘pressure’. Usually, τ depends on the rate of strain, or in other words, the spatial derivative of velocity. In the case of Newtonian ﬂuid,

τ = µ(∇u + ∇uT ) which we will come back later in detail, the equation is reduced to Du ρ = ρ(u + u · ∇u) = −∇p + µ∆u + ρb Dt t Here, µ is called the viscosity.

Remark 1. In the compressible case, τ is not traceless. We allow τ = λ(∇ · u)I + µ(∇u + ∇uT ). See C.M. book P32 but (Equation 1.3.1 has a sign error).

An ideal fluid is that with τ = 0, which in the Newtonian fluid case corresponds to zero viscosity, or there is no energy dissipated in the fluid. Hence, we have the equations derived for ideal fluids:

ρt + ∇ · (ρu) = 0, (1.4) ρ(ut + u · ∇u) = −∇p + ρb

You can see that we have unknowns ρ, u, p(5 unknowns) but only 4 equations. The system is not closed. There are two directions:

• If the ﬂuid is compressible, then, the pressure p depends on the density and entropy (and maybe other quantities like temperature). In this case, we may also consider conservation of energy and the equation of state. This is usually the case we do for ideal gas. The resulted equations will be called the compressible Euler equations. If pressure is a function of density only in the isentropic case, the two equations are closed, which we study in detail next lecture.

3 Dρ • If the ﬂuid is incompressible, then Dt = 0 or ∇ · u = 0. The ﬁrst equation can be dropped and we have a closed equation

ρ(ut + u · ∇u) = −∇p + ρb (1.5) ∇ · u = 0.

This is called the compressible Euler equations. For incompressible ﬂuid, the pressure p can not be determined through the state of the system. You can imagine that for water, the pressure can change a lot with a little change of volume. In the ideal case, the liquid becomes incompressible, then the pressure is undetermined. It is now a Lagrangian multiplier of the constraint ∇ · u = 0 and must be solved from the equations.

Then, we will mention the gas dynamics briefly and then focus on the incompressible case in other lectures. Boundary conditions for ideal fluids? Usually, we impose that the condition so that the fluid particle initially at the wall will stay at the wall. The normal component of the fluid velocity should be the same as the normal component of a material point on the wall:

u · n = w · n. (1.6)

Alternatively, if the wall is given by an equation Σ(x, t) = 0. Then,

Σt|x + u(x, t) · ∇Σ = 0.

Mathematically, there is only ﬁrst order spatial derivative in the PDEs, we can only impose the normal component. If you impose the full velocity at the boundary, you get too many boundary conditions.

Conservation of energy for incompressible ﬂuids

Even though we do not need the conservation of energy for the system to be closed, we can still check that the energy satisfies the conservation property. For incompressible fluids, there is no exchange between internal energy and kinetic energy. We define the total energy to be Z 1 E = ρ|u|2dx. Rt 2

4 By the second convection identity, we have

d Z 1 Z 1 D Z D Z ρ|u|2dV = ρ |u|2dV = ρu · udV = u · (−∇p + ρb)dV dt Rt 2 Rt 2 Dt Rt Dt Rt Z Z Z = −∇ · (pu)dV + ρu · b)dV = − pu · ndS + ρu · bdV Rt ∂Rt Rt These two terms are work done by the stress and body force.