2. The Navier-Stokes Equation Flashcards
fluid momentum transport, constitutive equation, the navier-stokes equation, hydrostatic and dynamic pressure, boundary conditions, one dimensional flow examples, the reynolds number
Fluid Momentum Transport
Newton’s Second Law
-consider a fixed volume V:
rate of change of momentum in V) = (net inward flow of momentum through S) + (net force acting on V
Fluid Momentum Transport
Rate of Change of Momentum in V
I = ∫ ∂/∂t (ρu_) dV
-or in suffix notation:
Ii = ∫ ∂/∂t (ρui) dV
Fluid Momentum Transport
Net Inward Flow of Momentum Through S
II = - ∫ ∇ . (ρu_u_) dV
-or in suffix notation:
IIi = ∫ ∂/∂xj (ρuiuj) dV
Fluid Momentum Transport
Net Force Acting on V
III = ∫ ρg_ dV (+other volume forces) + ∫ f_ dS
-where f_ is the force density
-or in suffix notation:
IIIi = ∫ ρgi dV + ∫ ∂τji/∂xj dV
Force Density
fi = nj*τji
-where n is the normal and τ is the total stress tensor
Momentum Equation
ρ∂u_/∂t + ρ(u_.∇)u_ = ρg_ + ∇.τ
-where τ is a matrix
-alternatively:
ρDu_/Dt = ρg_ + ∇.τ
Constitutive Equations
Case 1: Ideal Fluid
-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_
Constitutive Equations
Case 2: Newtonian Fluid
-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
Navier-Stokes Equation
ρ(∂u_/∂t + (u_.∇)u_
= ρg_ - ∇P + μ∇²u_
Navier-Stokes Equation
In the Absence of Flow
-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
Navier-Stokes Equation
Flow Not Driven By Gravity
-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_
Boundary Conditions
Solid Boundary
-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_
Boundary Conditions
Free Surface - Dynamic Boundary Condition
-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
Boundary Between Immiscible Fluids
-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
Steps to Using the Navier Stokes Equation
- 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
Plane-Poiseuille Flow
- 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
Hagen-Poiseuille Flow
-the axisymmetric equivalent of plane-poiseuille flow where a fluid flows along a cylindrical pipe
Flow Down an Inclined Plane
- 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
Taylor-Couette Flow
-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
The Reynolds Number
Dynamic Similarity
-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
What is the Reynold’s number?
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
Flow Past a Cylinder
Re<1
- 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
Flow Past a Cylinder
1
- 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
Flow Past a Cylinder
46>Re
- 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
The Reynold’s Number and the Navier-Stokes Equation
Description
-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
The Reynold’s Number and the Navier-Stokes Equation
Small Re
-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_
The Reynold’s Number and the Navier-Stokes Equation
Large Re
-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
