The Navier-Stokes Equations

Transcription

1 s University of New Hampshire February 22, 202

2 and equations describe the non-relativistic time evolution of mass and momentum in fluid substances. mass density field: ρ = ρ(t, x, y, z) velocity field: v i = v i (t, x, y, z), i =, 2, 3 We will derive them by using conservation of mass and force laws on a control volume V. The control volume propagates in time, V =. s (We will assume an isothermal continuum, so we don t need to consider energy conservation.)

3 s mass gives a continuity equation. (Demand that the mass flux through any closed surface be the change in total mass, and use divergence theorem) ρ + i (ρv i ) = 0 This provides us with nice simplifications if we later assume incompressibility incompressibility = ρ = 0 = i v i = 0

4 The Forces We start by considering the forces on a control volume V. There are forces on the body, like gravity. Call this force field f i. And there are forces on the surface of the control volume, pressures along normals and viscous forces tangentially. These are described by a stress tensor σ ij. σij can be physically defined by the way it operates on a normal n j to a surface element of area da. It gives the force on that surface element: σ ij n j da σ ij is the stress (the force per area) in the j direction acting on the cube face with normal in the i direction. Thus the diagonal components are pressures and the off-diagonal components are shear stresses. s

5 The Force Law Now we write the force law on a control volume V : d ( ρv i ) d 3 ( x = f i ) d 3 ( x + σ ij ) n j da dt d dt p = F body + F surface Divergence theorem on the surface force term gives: d ( ρv i ) d 3 ( x = f i ) d 3 ( x + j σ ij) d 3 x dt s It now remains to turn the first term into a volume integral. The time derivative cannot simply be pushed in because the volume of integration is time-dependent.

6 Reynold s Transport Theorem We need to deal with the time derivative on d dt α(t) d 3 x for a function α = α(t, x, y, z) and a time-dependent volume of integration. d α(t) = dt ( ) α(t + t) α(t) = t 0 t V (t+ t) ( ) α(t + t) α(t + t) + α(t + t) α(t) = t 0 t V (t+ t) ( ) α α(t + t) α(t + t) + = t 0 t V (t+ t) t ( ) α α(t + t) + t 0 t V t s

7 Reynold s Transport Theorem The integral over the difference in volume, V α(t + t), can be expressed in terms of an integral of the changes in volume over the surface. s The differential volume change at da is v j n j t da.

8 Reynold s Transport Theorem t 0 t 0 t 0 ( t ( t ( t cube at each da V ) α(t) = cube at each da cube at each da α(center of cube) ( V at patch da) α(center of cube)v j n j t da α(center of cube)v j n j da = α(t)v j n j da ) = ) = s

9 And so we have Reynold s transport theorem: d α(t) = α(t)v j n j da + dt We can use it on the force equation: ( ρv i v j n j ) da + ( ρv i) = t α t f i + j σ ij s Then use divergence theorem again, and combine the volume integrals: ( ( j ρv i v j) + ( ρv i) ) f i j σ ij = 0 t

10 Since this holds for any control volume, we get the differential form of the equation. We can also use mass conservation ρ = j (ρv j ) = j ρv j ρ j v j ( ) j ρv i v j + ( ρv i ) f i j σ ij = 0 t j ρv i v j + ρ j v i v j + ρv i j v j + ρv i + ρ v i f i j σ ij = 0 j ρv i v j + ρ j v i v j + ρv i j v j j ρv i v j ρv i j v j + ρ v i f i j σ ij = 0 s And we are finally left with the Cauchy momentum equation: ρ j v i v j + ρ v i f i j σ ij = 0

11 s To get an actual equation we must choose a form for the stress tensor σ ij. ρ j v i v j + ρ v i f i j σ ij = 0