2. The Navier-Stokes Equation

Question 1

Fluid Momentum Transport

Newton’s Second Law

Answer 1

-consider a fixed volume V:

rate of change of momentum in V) = (net inward flow of momentum through S) + (net force acting on V

Question 2

Fluid Momentum Transport

Rate of Change of Momentum in V

Answer 2

I = ∫ ∂/∂t (ρu_) dV
-or in suffix notation:
Ii = ∫ ∂/∂t (ρui) dV

Question 3

Fluid Momentum Transport

Net Inward Flow of Momentum Through S

Answer 3

II = - ∫ ∇ . (ρu_u_) dV
-or in suffix notation:
IIi = ∫ ∂/∂xj (ρuiuj) dV

Question 4

Fluid Momentum Transport

Net Force Acting on V

Answer 4

III = ∫ ρg_ dV (+other volume forces) + ∫ f_ dS
-where f_ is the force density
-or in suffix notation:
IIIi = ∫ ρgi dV + ∫ ∂τji/∂xj dV

Question 5

Force Density

Answer 5

fi = nj*τji

-where n is the normal and τ is the total stress tensor

Question 6

Momentum Equation

Answer 6

ρ∂u_/∂t + ρ(u_.∇)u_ = ρg_ + ∇.τ
-where τ is a matrix
-alternatively:
ρDu_/Dt = ρg_ + ∇.τ

Question 7

Constitutive Equations

Case 1: Ideal Fluid

Answer 7

-for an ideal/inviscid fluid, the only surface force is due to the pressure:
f_ = -Pn_ = n_τ
-or in suffix notation:
fi = -Pni = njτji
-sub into the momentum equation to get the Euler equation:
ρ∂u_/∂t + ρ(u_.∇)u_ = ρg_ + ∇P_

Question 8

Constitutive Equations

Case 2: Newtonian Fluid

Answer 8

-total stress tensor:
τij = -P𝛿ij + σij
-where:
σij = μ(∂ui/∂xj + ∂uj/∂xi)
-so:
τij = -P𝛿ij + μ(∂ui/∂xj + ∂uj/∂xi)
-sub this into the momentum equation to obtain the Navier-Stokes equation

Question 9

Navier-Stokes Equation

Answer 9

ρ(∂u_/∂t + (u_.∇)u_

= ρg_ - ∇P + μ∇²u_

Question 10

Navier-Stokes Equation

In the Absence of Flow

Answer 10

-in the absence of flow the Navier-Stokes equation reduces to:
0_ = ρg_ - ∇P 
-which is easily solvable:
Ph  = Po + ρg_.x_
-where Ph is the hydrostatic pressure

Question 11

Navier-Stokes Equation

Flow Not Driven By Gravity

Answer 11

-if the flow is not driven by gravity, the gravity term in N-S can be absorbed into the pressure term:
p = P - Ph
=>
ρ(∂u_/∂t + (u_.∇)u_
= - ∇p + μ∇²u_

Question 12

Boundary Conditions

Solid Boundary

Answer 12

-for a solid impermeable boundary, fluid velocity at the boundary must be equal to the velocity of the boundary
u_ = U
-force density from the boundary wall on the fluid equal force of fluid on the wall:
f_ = n_ . τ_

Question 13

Boundary Conditions

Free Surface - Dynamic Boundary Condition

Answer 13

-if the fluid is in contact with air (or a fluid of much lower viscosity) we can ignore the viscosity term σ and say that the only force exerted on the fluid by the air is the atmospheric pressure:
n_ . τ_ = -Pn_ + n_.σ_
= -Patm n_
-consequently there is no force parallel to the surface
-if the position of the surface is given by f(x_,t)=0 and all points on the surface must remain on the surface in time, then:
Df/Dt = 0
-in particular u_ . ∇f = 0
-and ∇f = n_
=>
u_ . n_ = 0

Question 14

Boundary Between Immiscible Fluids

Answer 14

-at a boundary between two fluids of different viscosities, both the velocity u_ and the force density n_.τ_ must be continuous
-AND is the surface remains fixed in time:
u_.n_ = 0
-consequently the condition that n_.τ_ is continuous reduces to the condition that both P and μ(n_.∇u_) are continuous

Question 15

Steps to Using the Navier Stokes Equation

Answer 15

• write out in component form
• write out the incompressibiity condition if appropriate
• consider boundary conditions
• cancel terms using the boundary conditions
• the original equations should now be much simplified and solvable

Question 16

Plane-Poiseuille Flow

Answer 16

• consider the stationary flow of fluid along a channel driven by a pressure gradient
• we define Cartesian coordinates with x along the channel in the direction of the flow and y across the channel with boundaries at y=±h
• the fluid velocity satisfies no-slip boundary conditions at the walls, no wall-normal flow, unidirectional flow in the direction of the pressure gradient, invariance in the spanwise direction
• the velocity has a parabolic profile with a shear-stress at the wall

Question 17

Hagen-Poiseuille Flow

Answer 17

-the axisymmetric equivalent of plane-poiseuille flow where a fluid flows along a cylindrical pipe

Question 18

Flow Down an Inclined Plane

Answer 18

• a plane inclined at an angle α is coated with a layer of fluid thickness h
  -define Cartesian coordinates with x down the slope and y perpendicular to the slope
  -using symmetries, assume flow of the form:
  u_ = (u(y), 0 , 0)
  -since gravity is driving the flow, consider the full pressure an include the body force due to gravity
  -the flow profile is parabolic and corresponds to the flow in the bottom half of the channel

Question 19

Taylor-Couette Flow

Answer 19

-consider a flow between two concentric cylinders of radii a and b respectively where the inner cylinder is rotating at angular velocity Ω
-define the cylindrical polar coordinates about the axis of the cylinders so that:
u_ = (0, v(r) , 0) with boundary conditions v(a)=aΩ and v(b)=0
-the pressure increases increases with the distance to the center which is why the free-surface dips near a rotating rod
-can obtain an equation for the torque required to rotate the inner cylinder in terms of viscosity, length, angular velocity, a, and b
-since all of these values are known, measurable or calculable; this can be used to experimentally measure viscosity

Question 20

The Reynolds Number

Dynamic Similarity

Answer 20

-consider a flow pattern generated by an obstacle, in this case a sphere of size D in a uniform flow of speed U in a fluid density ρ and viscosity µ
-start by defining a new system of units bases upon independent units for mass, M, length, L, time, T:
L = D
T = D/U
-since both ρ and µ involve mass we can choose either:
M1 = ρD³
OR
M2 = µD²/U
-we can divide these for a single independent dimensionless group:
Re = M1/M2 = ρUD/µ
-Re is the Reynold’s number

Question 21

What is the Reynold’s number?

Answer 21

Re = ρUD/µ

• the Reynold’s number indicates the balance between inertia and viscous forces
• flows with the same Reynold number display the same flow pattern and are thus dynamically similar

Question 22

Flow Past a Cylinder

Re<1

Answer 22

• for small values of Re the flow is nearly fore-aft symmetric
• however this flow pattern is distinct from the potential flow solution as it satisfies u_=0_ on the cylinder surface

Question 23

Flow Past a Cylinder

1

Answer 23

• as Re increases, the flow loses its fore-aft symmetry and two recirculating vortices appear on the downstream side of the cylinder
• these cells grow in size as Re increases
• although the flow is no longer fore-aft symmetric it remains steady

Question 24

Flow Past a Cylinder

46>Re

Answer 24

• above Re of around 46, the flow is no longer steady
• the vortices behind the cylinder become unsteady and are shed alternately from the two sides forming a double line of vortices known as a von Karmen vortex street
• as Re increases further the flow in this wake region behind the cylinder becomes chaotic

Question 25

The Reynold’s Number and the Navier-Stokes Equation

Description

Answer 25

-we can obtain the Reynold’s number from the Navier-Stokes equation by non-dimensionalising it, i.e choosing units based on the natural length and time scales

Question 26

The Reynold’s Number and the Navier-Stokes Equation

Small Re

Answer 26

-a small Reynold’s number suggests that the inertia terms are small compared to other terms in the Navier-Stokes Equation and might be neglected
-in the limit of small Re, we obtain Stokes Equation:
0 = -∇p + µ∇²u_

Question 27

The Reynold’s Number and the Navier-Stokes Equation

Large Re

Answer 27

-a large Reynold’s number suggests that the viscosity terms are small compared to other terms in the Navier-Stokes Equation and might be neglected
-in the limit of large Re, we obtain the Euler Equation:
ρ (∂u_/∂t + (u_.∇)u_) = -∇p