2. Fluid-Flow Equations Governing Equations

• Conservation equations for: ‒ mass ‒ momentum ‒ energy ‒ (other constituents)

• Alternative forms: ‒ integral (control-volume) equations ‒ differential equations Integral (Control-Volume) Approach

Consider the budget of any physical quantity in any control volume V

V

TIME DERIVATIVE ADVECTIVE + DIFFUSIVE FLUX SOURCE + = of amount in 푉 through boundary of 푉 in 푉

→ Finite-volume method for CFD Mass Conservation (Continuity)

Mass conservation: mass is neither created nor destroyed

d u (mass) = net inward mass flux d푡 V un A d (mass) + net outward mass flux = 0 d푡

d mass + ෍ mass flux = 0 d푡 faces

Mass in a cell: ρ푉

Mass flux through a face: 퐶 = ρu • A Mass Conservation - Differential Equation

t Conservation statement:

d z mass + net outward mass flux = 0 w n d푡 y b e d s ρ푉 + (ρ푢퐴) − (ρ푢퐴) + (ρ푣퐴) − (ρ푣퐴) + (ρ푤퐴) − (ρ푤퐴) = 0 x d푡 푒 푤 푛 푠 푡 푏 d ρΔ푥Δ푦Δ푧 + [(ρ푢) − ρ푢) Δ푦Δ푧 + [(ρ푣) − ρ푣) Δ푧Δ푥 + [(ρ푤) − ρ푤) Δ푥Δ푦 = 0 d푡 푒 푤 푛 푠 푡 푏

Divide by volume: dρ (ρ푢) − (ρ푢) (ρ푣) − (ρ푣) (ρ푤) − (ρ푤) + 푒 푤 + 푛 푠 + 푡 푏 = 0 d푡 Δ푥 Δ푦 Δ푧 dρ Δ(ρ푢) Δ(ρ푣) Δ(ρ푤) + + + = 0 d푡 Δ푥 Δ푦 Δ푧

Shrink to a point: 휕ρ 휕(ρ푢) 휕(ρ푣) 휕(ρ푤) 휕ρ + + + = 0 + ∇ • (ρu) = 0 휕푡 휕푥 휕푦 휕푧 휕푡 Continuity in Incompressible Flow

t d z (volume) + net outward volume flux = 0 d푡 w n b y s e x

휕푢 휕푣 휕푤 + + = 0 ∇ • u = 0 휕푥 휕푦 휕푧 Momentum Equation

Momentum Principle: force = rate of change of momentum

F

If steady: force = (momentum flux)out – (momentum flux)in

If unsteady: force = d/d푡(momentum inside control volume)

+ (momentum flux)out – (momentum flux)in Momentum Equation

Rate of change of momentum = force

d (momentum) + net outward momentum flux = force d푡 u d mass × u + ෍ (mass flux × u) = F V un d푡 faces A

Momentum of fluid in a cell = mass × u = (ρ푉)u

Momentum flux through a face = mass flux × u = (ρu • A)u Fluid Forces force Surface forces (proportional to area): stress = area • pressure y  휕푢  • viscous force: τ = μ 휕푦

U • reactions from boundaries force Body forces (proportional to volume): force density = z volume

• gravity: −ρ𝑔e푧  g

axis

2 R   R

2  • centrifugal force: ρΩ R r

u −2ρΩ ∧ u • Coriolis force: 

In inertial frame In rotating frame Differential Equation t

Conservation statement: z w n d y b e (momentum) + net momentum flux = force s d푡 x d ρ푉푢 + (ρ푢퐴) 푢 − (ρ푢퐴) 푢 + (ρ푣퐴) 푢 − (ρ푣퐴) 푢 + (ρ푤퐴) 푢 − (ρ푤퐴) 푢 d푡 푒 푒 푤 푤 푛 푛 푠 푠 푡 푡 푏 푏

= 푝푤퐴푤 − 푝푒퐴푒 + viscous and other forces d ρΔ푥Δ푦Δ푧 푢 + [(ρ푢) 푢 − ρ푢) 푢 Δ푦Δ푧 + [(ρ푣) 푢 − ρ푣) 푢 Δ푧Δ푥 + [(ρ푤) 푢 − ρ푤) 푢 Δ푥Δ푦 d푡 푒 푒 푤 푤 푛 푛 푠 푠 푡 푡 푏 푏

= 푝푤 − 푝푒 Δ푦Δ푧 + viscous and other forces Divide by volume:

d(ρ푢) (ρ푢푢)푒 − (ρ푢푢)푤 (ρ푣푢)푛 − (ρ푣푢)푠 (ρ푤푢)푡 − (ρ푤푢)푏 푝푒 − 푝푤 viscous and + + + = − + d푡 Δ푥 Δ푦 Δ푧 Δ푥 other forces

d(ρ푢) Δ(ρ푢푢) Δ(ρ푣푢) Δ(ρ푤푢) Δ푝 viscous and + + + = − + d푡 Δ푥 Δ푦 Δ푧 Δ푥 other forces

Shrink to a point: 휕(ρ푢) 휕(ρ푢푢) 휕(ρ푣푢) 휕(ρ푤푢) 휕푝 + + + = − + μ∇2푢 + other forces 휕푡 휕푥 휕푦 휕푧 휕푥 General Scalar Time derivative + net outward flux = source

ϕ = concentration (amount per unit mass) u

Amount in a cell: ρ푉ϕ (mass  concentration) V un A Flux through a face:

‒ advection: (ρu • A)ϕ (mass flux  concentration) 휕ϕ ‒ diffusion: −Γ 퐴   휕푛 (diffusivity gradient area)

Source: 푆 = 푠푉 (source density  volume) d 휕ϕ mass × ϕ + ෍ (mass flux × ϕ − Γ 퐴) = 푠 푉 d푡 휕푛 faces

휕(ρϕ) 휕 휕ϕ 휕 휕ϕ 휕 휕ϕ + ρ푢ϕ − Γ + ρ푣ϕ − Γ + ρ푤ϕ − Γ = 푠 휕푡 휕푥 휕푥 휕푦 휕푦 휕푧 휕푧 Momentum Components as General Scalars Momentum equation: d 휕푢 mass × 푢 + ෍ mass flux × 푢 = ෍ ( μ 퐴) + other forces d푡 휕푛 faces faces viscous forces d 휕푢 mass × 푢 + ෍ mass flux × 푢 − μ 퐴 = other forces d푡 휕푛 faces

General scalar-transport equation: d 휕ϕ mass × ϕ + ෍ mass flux × ϕ − Γ 퐴 = 푆 d푡 휕푛 faces

• Velocity components 푢, 푣, 푤 satisfy individual scalar-transport equations: ‒ concentration, ϕ  velocity component, 푢, 푣, 푤 ‒ diffusivity,   viscosity, μ ‒ source, 푆  non-viscous forces

• Differences: ‒ momentum equations are non-linear ‒ momentum equations are coupled ‒ the velocity field also has to be mass-consistent Differential Equations For Fluid Flow

Forms of the equations in primitive variables may be:

• Conservative ‒ can be integrated directly to give “net flux = source” • Non-conservative ‒ can’t be integrated directly

Other forms of the equations include those for:

• Derived variables ‒ e.g. velocity potential Example

d (푦2) = 𝑔(푥) conservative d푥

d푦 2푦 = 𝑔(푥) non-conservative d푥

Same equation! ... but only the first can be integrated directly Rate of Change Following the Flow

ϕ ≡ ϕ(푡, x)

Total derivative dϕ 휕ϕ 휕ϕ d푥 휕ϕ d푦 휕ϕ d푧 ≡ + + + (following any path x(푡) (x(t), y(t), z(t)) d푡 휕푡 휕푥 d푡 휕푦 d푡 휕푧 d푡

Material derivative d푥 Dϕ 휕ϕ 휕ϕ 휕ϕ 휕ϕ = u ≡ + 푢 + 푣 + 푤 (following the flow): d푡 D푡 휕푡 휕푥 휕푦 휕푧

D 휕 휕 휕 휕 ≡ + 푢 + 푣 + 푤 D푡 휕푡 휕푥 휕푦 휕푧 D 휕 ≡ + u • ∇ D푡 휕푡 Non-Conservative Flow Equations

conservative form non-conservative form 휕 휕 휕 휕 Dϕ (ρϕ) + (ρ푢ϕ) + (ρ푣ϕ) + (ρ푤ϕ) → ρ 휕푡 휕푥 휕푦 휕푧 D푡 (mass conservation)

D푢 휕푝 ρ = − + μ∇2푢 e.g. momentum equation: D푡 휕푥 mass × acceleration forces

휕(ρϕ) 휕(ρ푢ϕ) 휕(ρ푣ϕ) 휕(ρ푤ϕ) Proof: + + + 휕푡 휕푥 휕푦 휕푧

휕ρ 휕ϕ 휕(ρ푢) 휕ϕ 휕(ρ푣) 휕ϕ 휕(ρ푤) 휕ϕ = ϕ + ρ + ϕ + ρ푢 + ϕ + ρ푣 + ϕ + ρ푤 휕푡 휕푡 휕푥 휕푥 휕푦 휕푦 휕푧 휕푧

휕ρ 휕(ρ푢) 휕(ρ푣) 휕(ρ푤) 휕ϕ 휕ϕ 휕ϕ 휕ϕ = + + + ϕ + ρ + 푢 + 푣 + 푤 휕푡 휕푥 휕푦 휕푧 휕푡 휕푥 휕푦 휕푧 =0 by continuity =Dϕ/D푡 by definition

Dϕ = ρ D푡 Example Q1 (Equation Manipulation) In 2-d flow, the continuity and x-momentum equations can be written in conservative form as 휕ρ 휕 휕 휕 휕 휕 휕푝 + (ρ푢) + (ρ푣) = 0 (ρ푢) + (ρ푢푢) + (ρ푣푢) = − + μ∇2푢 휕푡 휕푥 휕푦 휕푡 휕푥 휕푦 휕푥 (a) Show that these can be written in the equivalent non-conservative forms: Dρ 휕푢 휕푣 D푢 휕푝 + ρ( + ) = 0 ρ = − + μ∇2푢 D푡 휕푥 휕푦 D푡 휕푥 (b) Define carefully what is meant by the statement that a flow is incompressible. To what does the continuity equation reduce in incompressible flow?

(c) Write down conservative forms of the 3-d equations for mass and x-momentum.

(d) Write down the 푧-momentum equation, including the gravitational force.

(e) Show that, for constant-density flows, pressure and gravity can be combined in the momentum equations via the piezometric pressure 푝 + ρ𝑔푧.

axis (f) In a rotating reference frame there are additional apparent forces (per unit volume): 2 R   R 2 centrifugal force: ρΩ R  r Coriolis force: −2ρΩ ∧ u

where Ω is the angular velocity of the reference frame, u is the fluid velocity in that frame, r is the position vector and R is its projection perpendicular to the axis of rotation. By writing the centrifugal force as the gradient of some quantity show that it can be subsumed into a modified pressure. Also, find the components of the Coriolis force if rotation is about the 푧 axis. In 2-d flow, the continuity and x-momentum equations can be written in conservative form as 휕ρ 휕 휕 휕 휕 휕 휕푝 + (ρ푢) + (ρ푣) = 0 (ρ푢) + (ρ푢푢) + (ρ푣푢) = − + μ∇2푢 휕푡 휕푥 휕푦 휕푡 휕푥 휕푦 휕푥 (a) Show that these can be written in the equivalent non-conservative forms: Dρ 휕푢 휕푣 D푢 휕푝 + ρ( + ) = 0 ρ = − + μ∇2푢 D푡 휕푥 휕푦 D푡 휕푥

Continuity: Momentum:

휕ρ 휕 휕 휕 휕 휕 휕푝 + (ρ푢) + (ρ푣) = 0 (ρ푢) + (ρ푢푢) + (ρ푣푢) = − + μ∇2푢 휕푡 휕푥 휕푦 휕푡 휕푥 휕푦 휕푥 휕ρ 휕푢 휕ρ 휕ρ 휕푢 휕ρ 휕푣 LHS = 푢 + ρ + 푢 + ρ + 푣 + ρ = 0 휕푡 휕푡 휕푡 휕푥 휕푥 휕푦 휕푦 휕(ρ푢) 휕푢 + 푢 + ρ푢 휕푥 휕푥 휕(ρ푣) 휕푢 + 푢 + ρ푣 휕ρ 휕ρ 휕ρ 휕푢 휕푣 휕푦 휕푦 + 푢 + 푣 + ρ + = 0 휕푡 휕푥 휕푦 휕푥 휕푦 휕ρ 휕(ρ푢) 휕(ρ푣) 휕푢 휕푢 휕푢 LHS = + + 푢 + ρ + 푢 + 푣 휕푡 휕푥 휕푦 휕푡 휕푥 휕푦 Dρ 휕푢 휕푣 =0 by mass conservation =D푢ΤD푡 + ρ + = 0 D푡 휕푥 휕푦 D푢 휕푝 ρ = − + μ∇2푢 D푡 휕푥 (b) Define carefully what is meant by the statement that a flow is incompressible. To what does the continuity equation reduce in incompressible flow?

Incompressible: flow-induced changes to pressure (or temperature) do not cause significant changes in density

Dρ = 0 D푡

Dρ 휕푢 휕푣 + ρ + = 0 D푡 휕푥 휕푦

휕푢 휕푣 + = 0 휕푥 휕푦 휕ρ 휕 휕 2-d continuity: + (ρ푢) + (ρ푣) = 0 휕푡 휕푥 휕푦 휕 휕 휕 휕푝 휕 휕 2-d x-momentum: (ρ푢) + (ρ푢푢) + (ρ푣푢) = − + μ∇2푢 ∇2≡ + 휕푡 휕푥 휕푦 휕푥 휕푥2 휕푦2

(c) Write down conservative forms of the 3-d equations for mass and x-momentum. 휕ρ 휕 휕 휕 3-d continuity: + (ρ푢) + (ρ푣) + (ρ푤) = 0 휕푡 휕푥 휕푦 휕푧 휕 휕 휕 휕 휕푝 휕 휕 휕 3-d x-momentum: (ρ푢) + (ρ푢푢) + (ρ푣푢) + (ρ푤푢) = − + μ∇2푢 ∇2≡ + + 휕푡 휕푥 휕푦 휕푧 휕푥 휕푥2 휕푦2 휕푧2 (d) Write down the 푧-momentum equation, including the gravitational force. 휕 휕 휕 휕 휕푝 3-d z-momentum: ρ푤 + ρ푢푤 + ρ푣푤 + ρ푤푤 = − − ρ𝑔 + μ∇2푤 휕푡 휕푥 휕푦 휕푧 휕푧

(e) Show that, for constant-density flows, pressure and gravity can be combined in the momentum equations via the piezometric pressure 푝 + ρ𝑔푧.

휕푝 휕 휕푝∗ Pressure + gravity forces: 푥: − = − (푝 + ρ𝑔푧) = − 휕푥 휕푥 휕푥 휕푝 휕 휕푝∗ 푦: − = − (푝 + ρ𝑔푧) = − 휕푦 휕푦 휕푦 휕푝 휕 휕푝∗ 푧: − − ρ𝑔 = − (푝 + ρ𝑔푧) = − 푝∗ ≡ 푝 + ρ𝑔푧 휕푧 휕푧 휕푧 (f) In a rotating reference frame there are additional apparent forces (per unit volume): centrifugal force: ρΩ2R axis 2 Coriolis force: −2ρΩ ∧ u R   R where Ω is the angular velocity of the reference frame, u is the fluid velocity in that frame, r is the position vector and R is its projection perpendicular to the axis of  rotation. r

By writing the centrifugal force as the gradient of some quantity show that it can be subsumed into a modified pressure. Also, find the components of the Coriolis force if rotation is about the 푧 axis.

1 1 ρΩ2R = ρΩ2 푥, 푦, 0 = ∇ ρΩ2 푥2 + 푦2 = ∇( ρΩ2푅2) 2 2

1 −∇푝 + ρΩ2R = −∇ 푝 − ρΩ2푅2 2 L Coriolis modified pressure force pressure force H 0 푢 −Ω푣 2ρΩ푣 −2ρΩ ∧ u = −2ρ 0 ∧ 푣 = −2ρ Ω푢 = −2ρΩ푢 Ω 푤 0 0 Northern hemisphere Example Q2 (Exact Solutions)

The 푥-component of the momentum equation is given by D푢 휕푝 ρ = − + μ∇2푢 D푡 휕푥

Using this equation, derive the velocity profile in fully-developed, laminar flow for:

(a) pressure-driven flow between stationary parallel planes (“Plane Poiseuille flow”);

(b) constant-pressure flow between stationary and moving planes (“Couette flow”).

Assume flow in the x direction, and bounding planes 푦 = 0 and 푦 = ℎ. The velocity is then (푢(푦), 0,0), y u(y) h

In part (a) both walls are stationary. In part (b) the upper wall x slides parallel to the lower wall with velocity 푈푤. The 푥-component of the momentum equation is given by D푢 휕푝 y u(y) h ρ = − + μ∇2푢 D푡 휕푥 x Using this equation, derive the velocity profile in fully-developed, laminar flow for: (a) pressure-driven flow between stationary parallel planes (“Plane Poiseuille flow”); (b) constant-pressure flow between stationary and moving planes (“Couette flow”). D푢 Fully-developed: = 0 D푡 d2푢 휕푝 d2푢 푢 a function of 푦 only: ∇2푢 → 푥 − momentum: 0 = − + μ d푦2 휕푥 d푦2 휕푝 푦 − momentum: 0 = − 휕푦 휕푝/휕푥 must be constant, −퐺 say

d2푢 퐺 1 퐺 = − (constant) Case (a): 푢 = 0 on 푦 = ℎ: 퐴 = ℎ d푦2 μ 2 μ 1 퐺 퐺 푢 = − 푦2 + 퐴푦 + 퐵 푢 = 푦(ℎ − 푦) 2 μ 2μ

푢 = 0 on 푦 = 0, so 퐵 = 0 Case (b): 퐺 = 0 and 푢 = 푈푤 on 푦 = ℎ: 푈푤 = ℎ퐴 1 퐺 푦 푢 = 푦(퐴 − 푦) 푢 = 푈푤 2 μ ℎ Non-Dimensionalisation - Advantages

• All dynamically-similar problems (same Re, Fr etc.) can be solved with a single computation

• The number of parameters is reduced

• It indicates the relative size of different terms in the governing equations: in particular, which might be neglected

• Computational variables are similar size, yielding better numerical accuracy Non-Dimensionalisation

Form non-dimensional variables using length (퐿0), velocity (푈0) and density (0) scales: ∗ 퐿0 ∗ ∗ ∗ 2 ∗ x = 퐿0x , 푡 = 푡 , u = 푈0u , ρ = ρ0ρ , 푝 = 푝ref + ρ0푈0 푝 , 푒푡푐. 푈0

Substitute into the governing equations: D푢 휕푝 ρ 푈2 D푢∗ ρ 푈2 휕푝∗ μ푈 ρ = − + μ∇2푢 0 0 ρ∗ = − 0 0 + 0 ∇∗2푢∗ D푡 휕푥 ∗ ∗ 2 퐿0 D푡 퐿0 휕푥 퐿0 ∗ ∗ ∗ D푢 휕푝 μ ∗2 ∗ ρ ∗ = − ∗ + ∇ 푢 D푡 휕푥 ρ0푈0퐿0

D푢∗ 휕푝∗ 1 ρ∗ = − + ∇∗2푢∗ D푡∗ 휕푥∗ Re

ρ 푈 퐿 D푢 휕푝 1 Identify important dimensionless groups: Re = 0 0 0 ρ = − + ∇2푢 μ D푡 휕푥 Re Common Dimensionless Groups

ρ푈퐿 Re ≡ Reynolds number (viscous flow; μ = dynamic viscosity) μ

푈 Fr ≡ Froude number (open-channel flow; 𝑔 = gravity) 𝑔퐿

푈 Ma ≡ Mach number (compressible flow; 푐 = speed of sound) 푐

푈 Ro ≡ Rossby number (rotating flows; Ω = angular velocity of frame) Ω퐿

ρ푈2퐿 We ≡ Weber number (free-surface flows; σ = surface tension) 휎 Summary (1)

• The fluid-flow equations are conservation equations for: ‒ mass ‒ momentum ‒ energy ‒ (additional constituents)

• The equations can be written in equivalent integral (control-volume) or differential forms

• The finite-volume method is a direct discretisation of the control-volume equations

• Differential forms of the flow equations may be conservative or non-conservative

• For any conserved property and arbitrary control volume: time derivative + net outward flux = source Summary (2)

• There are really just two canonical equations to solve: ‒ mass conservation (continuity) ‒ a generic scalar-transport equation

• Each momentum component satisfies its own scalar-transport equation ‒ concentration  velocity component ‒ diffusivity  viscosity ‒ source  non-viscous forces

• However, the momentum equations are: ‒ non-linear ‒ coupled ‒ also required to be mass-consistent

• Non-dimensionalisation: ‒ solves dynamically-similar (Re, Fr, Ro, …) flows with a single computation ‒ reduces the number of parameters ‒ identifies relative importance of different terms in equations ‒ maintains numerical variables of similar size