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
TIME DERIVATIVE ADVECTIVE + DIFFUSIVE FLUX SOURCE
+ =
𝑜𝑓 𝑎𝑚𝑜𝑢𝑛𝑡 𝑖𝑛 𝑉 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑉 𝑖𝑛 𝑉
→ Finite-volume method for CFD
Mass Conservation (Continuity)
Mass conservation: mass is neither created nor destroyed
d
(𝑚𝑎𝑠𝑠) = 𝑛𝑒𝑡 𝑖𝑛𝑤𝑎𝑟𝑑 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥
u
d𝑡
V un
d
A
(𝑚𝑎𝑠𝑠) + 𝑛𝑒𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 = 0
d𝑡
d
𝑚𝑎𝑠𝑠 + 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 = 0
d𝑡
𝑓𝑎𝑐𝑒𝑠
Mass in a cell: ρ𝑉
Mass flux through a face: 𝐶 = ρu • A
Mass Conservation - Differential Equation
t
Conservation statement:
d z
𝑚𝑎𝑠𝑠 + 𝑛𝑒𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 = 0 w n
d𝑡 b
y s e
d
ρ𝑉 + (ρ𝑢𝐴)𝑒 − (ρ𝑢𝐴)𝑤 + (ρ𝑣𝐴)𝑛 − (ρ𝑣𝐴)𝑠 + (ρ𝑤𝐴)𝑡 − (ρ𝑤𝐴)𝑏 = 0 x
d𝑡
d
ρΔ𝑥Δ𝑦Δ𝑧 + [(ρ𝑢)𝑒 − ρ𝑢)𝑤 Δ𝑦Δ𝑧 + [(ρ𝑣)𝑛 − ρ𝑣)𝑠 Δ𝑧Δ𝑥 + [(ρ𝑤)𝑡 − ρ𝑤)𝑏 Δ𝑥Δ𝑦 = 0
d𝑡
Divide by volume:
dρ (ρ𝑢)𝑒 − (ρ𝑢)𝑤 (ρ𝑣)𝑛 − (ρ𝑣)𝑠 (ρ𝑤)𝑡 − (ρ𝑤)𝑏
+ + + = 0
d𝑡 Δ𝑥 Δ𝑦 Δ𝑧
dρ Δ(ρ𝑢) Δ(ρ𝑣) Δ(ρ𝑤)
+ + + = 0
d𝑡 Δ𝑥 Δ𝑦 Δ𝑧
Shrink to a point:
𝜕ρ 𝜕(ρ𝑢) 𝜕(ρ𝑣) 𝜕(ρ𝑤) 𝜕ρ
+ + + =0 + ∇ • (ρu) = 0
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡
Continuity in Incompressible Flow
d z
(𝑣𝑜𝑙𝑢𝑚𝑒) + 𝑛𝑒𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑢𝑥 = 0 n
d𝑡 w
y b e
s
x
𝜕𝑢 𝜕𝑣 𝜕𝑤
+ + =0 ∇•u=0
𝜕𝑥 𝜕𝑦 𝜕𝑧
Momentum Equation
Momentum Principle: force = rate of change of momentum
If steady: 𝑓𝑜𝑟𝑐𝑒 = (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥)𝑜𝑢𝑡 – (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥)𝑖𝑛
d
If unsteady: 𝑓𝑜𝑟𝑐𝑒 = (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛𝑠𝑖𝑑𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒)
d𝑡
+ (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥)𝑜𝑢𝑡 – (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥)𝑖n
Momentum Equation
Rate of change of momentum = force
d
(𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚) + 𝑛𝑒𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥 = 𝑓𝑜𝑟𝑐𝑒
d𝑡
u
d
𝑚𝑎𝑠𝑠 × u + (𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × u) = F V un
d𝑡
𝑓𝑎𝑐𝑒𝑠 A
Momentum of fluid in a cell = 𝑚𝑎𝑠𝑠 × u = (ρ𝑉)u
Momentum flux through a face = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × u = (ρu • A)u
Fluid Forces
Surface forces (proportional to area): 𝑓𝑜𝑟𝑐𝑒
𝑠𝑡𝑟𝑒𝑠𝑠 =
𝑎𝑟𝑒𝑎
• pressure y
𝜕𝑢
• viscous force: τ=μ
𝜕𝑦
U
• reactions from boundaries
𝑓𝑜𝑟𝑐𝑒
Body forces (proportional to volume): 𝑓𝑜𝑟𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
z 𝑣𝑜𝑙𝑢𝑚𝑒
• gravity: −ρ𝑔e𝑧 g
axis
R
2
R
• centrifugal force: ρΩ2 R r
u
• Coriolis force: −2ρΩ ∧ u
In inertial frame In rotating frame
Differential Equation t
z
Conservation statement: n
w
y b
d s e
(𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚) + 𝑛𝑒𝑡 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑓𝑙𝑢𝑥 = 𝑓𝑜𝑟𝑐𝑒
d𝑡 x
d
ρ𝑉𝑢 + (ρ𝑢𝐴)𝑒 𝑢𝑒 − (ρ𝑢𝐴)𝑤 𝑢𝑤 + (ρ𝑣𝐴)𝑛 𝑢𝑛 − (ρ𝑣𝐴)𝑠 𝑢𝑠 + (ρ𝑤𝐴)𝑡 𝑢𝑡 − (ρ𝑤𝐴)𝑏 𝑢𝑏
d𝑡
= 𝑝𝑤 𝐴𝑤 − 𝑝𝑒 𝐴𝑒 + 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑎𝑛𝑑 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
d
ρΔ𝑥Δ𝑦Δ𝑧 𝑢 + [(ρ𝑢)𝑒 𝑢𝑒 − ρ𝑢)𝑤 𝑢𝑤 Δ𝑦Δ𝑧 + [(ρ𝑣)𝑛 𝑢𝑛 − ρ𝑣)𝑠 𝑢𝑠 Δ𝑧Δ𝑥 + [(ρ𝑤)𝑡 𝑢𝑡 − ρ𝑤)𝑏 𝑢𝑏 Δ𝑥Δ𝑦
d𝑡
= 𝑝𝑤 − 𝑝𝑒 Δ𝑦Δ𝑧 + 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑎𝑛𝑑 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
Divide by volume:
d(ρ𝑢) (ρ𝑢𝑢)𝑒 − (ρ𝑢𝑢)𝑤 (ρ𝑣𝑢)𝑛 − (ρ𝑣𝑢)𝑠 (ρ𝑤𝑢)𝑡 − (ρ𝑤𝑢)𝑏 𝑝𝑒 − 𝑝𝑤 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑎𝑛𝑑
+ + + = − +
d𝑡 Δ𝑥 Δ𝑦 Δ𝑧 Δ𝑥 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
d(ρ𝑢) Δ(ρ𝑢𝑢) Δ(ρ𝑣𝑢) Δ(ρ𝑤𝑢) Δ𝑝 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑎𝑛𝑑
+ + + = − +
d𝑡 Δ𝑥 Δ𝑦 Δ𝑧 Δ𝑥 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
Shrink to a point:
𝜕(ρ𝑢) 𝜕(ρ𝑢𝑢) 𝜕(ρ𝑣𝑢) 𝜕(ρ𝑤𝑢) 𝜕𝑝
+ + + =− + μ∇2 𝑢 + 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥
General Scalar
Time derivative + net outward flux = source
ϕ = concentration (amount per unit mass)
u
Amount in a cell: ρ𝑉ϕ (mass concentration) V un
A
Flux through a face:
‒ advection: (ρu • A)ϕ (mass flux concentration)
𝜕ϕ
‒ diffusion: −Γ 𝐴 (diffusivity gradient area)
𝜕𝑛
Source: 𝑆 = 𝑠𝑉 (source density volume)
d 𝜕ϕ
𝑚𝑎𝑠𝑠 × ϕ + (𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × ϕ − Γ 𝐴) = 𝑠 𝑉
d𝑡 𝜕𝑛
𝑓𝑎𝑐𝑒𝑠
𝜕(ρϕ) 𝜕 𝜕ϕ 𝜕 𝜕ϕ 𝜕 𝜕ϕ
+ ρ𝑢ϕ − Γ + ρ𝑣ϕ − Γ + ρ𝑤ϕ − Γ = 𝑠
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧
Momentum Components as General Scalars
Momentum equation:
d 𝜕𝑢
𝑚𝑎𝑠𝑠 × 𝑢 + 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × 𝑢 = (μ 𝐴) + 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
d𝑡 𝜕𝑛
𝑓𝑎𝑐𝑒𝑠 𝑓𝑎𝑐𝑒𝑠
𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒𝑠
d 𝜕𝑢
𝑚𝑎𝑠𝑠 × 𝑢 + 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × 𝑢 − μ 𝐴 = 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠
d𝑡 𝜕𝑛
𝑓𝑎𝑐𝑒𝑠
General scalar-transport equation:
d 𝜕ϕ
𝑚𝑎𝑠𝑠 × ϕ + 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × ϕ − Γ 𝐴 = 𝑆
d𝑡 𝜕𝑛
𝑓𝑎𝑐𝑒𝑠
• Velocity components 𝑢, 𝑣, 𝑤 satisfy individual scalar-transport equations:
‒ concentration, ϕ velocity component
‒ diffusivity, viscosity
‒ source, 𝑆 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 2
(𝑦 ) = 𝑔(𝑥) conservative
d𝑥
d𝑦
2𝑦 = 𝑔(𝑥) non-conservative
d𝑥
Same equation! ... but only the first can be integrated directly
Rate of Change Following the Flow
ϕ ≡ ϕ(𝑡, x)
Total derivative dϕ 𝜕ϕ 𝜕ϕ d𝑥 𝜕ϕ d𝑦 𝜕ϕ d𝑧
(following any path x(𝑡) ≡ + + +
(x(t), y(t), z(t)) d𝑡 𝜕𝑡 𝜕𝑥 d𝑡 𝜕𝑦 d𝑡 𝜕𝑧 d𝑡
Material derivative d𝑥 Dϕ 𝜕ϕ 𝜕ϕ 𝜕ϕ 𝜕ϕ
=u ≡ +𝑢 +𝑣 +𝑤
(following the flow): d𝑡 D𝑡 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
D 𝜕 𝜕 𝜕 𝜕
≡ +𝑢 +𝑣 +𝑤
D𝑡 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
D 𝜕
≡ +u•∇
D𝑡 𝜕𝑡
Non-Conservative Flow Equations
conservative form non-conservative form
𝜕 𝜕 𝜕 𝜕 Dϕ
𝜕𝑡
(ρϕ) +
𝜕𝑥
(ρ𝑢ϕ) +
𝜕𝑦
(ρ𝑣ϕ) + (ρ𝑤ϕ)
𝜕𝑧 → ρ
D𝑡
(mass conservation)
D𝑢 𝜕𝑝
e.g. momentum equation: ρ =− + μ∇2 𝑢
D𝑡 𝜕𝑥
𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒𝑠
Proof: 𝜕(ρϕ) + 𝜕(ρ𝑢ϕ) + 𝜕(ρ𝑣ϕ) + 𝜕(ρ𝑤ϕ)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕ρ 𝜕ϕ 𝜕(ρ𝑢) 𝜕ϕ 𝜕(ρ𝑣) 𝜕ϕ 𝜕(ρ𝑤) 𝜕ϕ
= ϕ+ρ + ϕ + ρ𝑢 + ϕ + ρ𝑣 + ϕ + ρ𝑤
𝜕𝑡 𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧
𝜕ρ 𝜕(ρ𝑢) 𝜕(ρ𝑣) 𝜕(ρ𝑤) 𝜕ϕ 𝜕ϕ 𝜕ϕ 𝜕ϕ
= + + + ϕ + ρ +𝑢 +𝑣 +𝑤
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
=0 𝑏𝑦 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 =Dϕ/D𝑡 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛
Dϕ
=ρ
D𝑡
Example Q1 (Equation Manipulation)
In 2-d flow, the continuity and x-momentum equations can be written in conservative form as
𝜕ρ 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕𝑝
+ (ρ𝑢) + (ρ𝑣) = 0 (ρ𝑢) + (ρ𝑢𝑢) + (ρ𝑣𝑢) = − + μ∇2 𝑢
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥
(a) Show that these can be written in the equivalent non-conservative forms:
Dρ 𝜕𝑢 𝜕𝑣 D𝑢 𝜕𝑝
+ ρ( + ) = 0 ρ =− + μ∇2 𝑢
D𝑡 𝜕𝑥 𝜕𝑦 D𝑡 𝜕𝑥
(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 𝑧-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 𝑝 + ρ𝑔𝑧.
axis
(f) In a rotating reference frame there are additional apparent forces (per unit volume): R
2
R
centrifugal force: ρΩ2 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 𝑧 axis.
In 2-d flow, the continuity and x-momentum equations can be written in conservative form as
𝜕ρ 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕𝑝
+ (ρ𝑢) + (ρ𝑣) = 0 (ρ𝑢) + (ρ𝑢𝑢) + (ρ𝑣𝑢) = − + μ∇2 𝑢
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥
(a) Show that these can be written in the equivalent non-conservative forms:
Dρ 𝜕𝑢 𝜕𝑣 D𝑢 𝜕𝑝
+ ρ( + ) = 0 ρ =− + μ∇2 𝑢
D𝑡 𝜕𝑥 𝜕𝑦 D𝑡 𝜕𝑥
Continuity: Momentum:
𝜕ρ 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕𝑝
+ (ρ𝑢) + (ρ𝑣) = 0 (ρ𝑢) + (ρ𝑢𝑢) + (ρ𝑣𝑢) = − + μ∇2 𝑢
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥
𝜕ρ 𝜕𝑢
𝜕ρ 𝜕ρ 𝜕𝑢 𝜕ρ 𝜕𝑣 LHS = 𝑢+ρ
+ 𝑢+ρ + 𝑣+ρ =0 𝜕𝑡 𝜕𝑡
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕(ρ𝑢) 𝜕𝑢
+ 𝑢 + ρ𝑢
𝜕𝑥 𝜕𝑥
𝜕(ρ𝑣) 𝜕𝑢
+ 𝑢 + ρ𝑣
𝜕ρ 𝜕ρ 𝜕ρ 𝜕𝑢 𝜕𝑣 𝜕𝑦 𝜕𝑦
+𝑢 +𝑣 +ρ + =0
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
𝜕ρ 𝜕(ρ𝑢) 𝜕(ρ𝑣) 𝜕𝑢 𝜕𝑢 𝜕𝑢
LHS = + + 𝑢+ρ +𝑢 +𝑣
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦
Dρ 𝜕𝑢 𝜕𝑣 =0 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 =D𝑢ΤD𝑡
+ρ + =0
D𝑡 𝜕𝑥 𝜕𝑦
D𝑢 𝜕𝑝
ρ =− + μ∇2 𝑢
D𝑡 𝜕𝑥
(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
D𝑡
Dρ 𝜕𝑢 𝜕𝑣
+ρ + =0
D𝑡 𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕𝑣
+ =0
𝜕𝑥 𝜕𝑦
2-d continuity: 𝜕ρ 𝜕 𝜕
+ (ρ𝑢) + (ρ𝑣) = 0
𝜕𝑡 𝜕𝑥 𝜕𝑦
𝜕 𝜕 𝜕 𝜕𝑝 𝜕 𝜕
2-d x-momentum: (ρ𝑢) + (ρ𝑢𝑢) + (ρ𝑣𝑢) = − + μ∇2 𝑢 ∇2 ≡ +
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2
(c) Write down conservative forms of the 3-d equations for mass and x-momentum.
𝜕ρ 𝜕 𝜕 𝜕
3-d continuity: + (ρ𝑢) + (ρ𝑣) + (ρ𝑤) = 0
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕 𝜕 𝜕 𝜕 𝜕𝑝 𝜕 𝜕 𝜕
3-d x-momentum: (ρ𝑢) + (ρ𝑢𝑢) + (ρ𝑣𝑢) + (ρ𝑤𝑢) = − + μ∇2 𝑢 ∇2
≡ + +
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
(d) Write down the 𝑧-momentum equation, including the gravitational force.
𝜕 𝜕 𝜕 𝜕 𝜕𝑝
3-d z-momentum: ρ𝑤 + ρ𝑢𝑤 + ρ𝑣𝑤 + ρ𝑤𝑤 = − − ρ𝑔 + μ∇2 𝑤
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑧
(e) Show that, for constant-density flows, pressure and gravity can be combined in the momentum
equations via the piezometric pressure 𝑝 + ρ𝑔𝑧.
𝜕𝑝 𝜕 𝜕𝑝∗
Pressure + gravity forces: 𝑥: − = − (𝑝 + ρ𝑔𝑧) =−
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕𝑝 𝜕 𝜕𝑝∗
𝑦: − = − (𝑝 + ρ𝑔𝑧) =−
𝜕𝑦 𝜕𝑦 𝜕𝑦
𝜕𝑝 𝜕 𝜕𝑝∗
𝑧: − − ρ𝑔 = − (𝑝 + ρ𝑔𝑧) =− 𝑝∗ ≡ 𝑝 + ρ𝑔𝑧
𝜕𝑧 𝜕𝑧 𝜕𝑧
(f) In a rotating reference frame there are additional apparent forces (per unit volume):
centrifugal force: ρΩ2 R axis
Coriolis force: −2ρΩ ∧ u R R
2
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 𝑧 axis.
1 2 2 1
ρΩ2 R = ρΩ2 𝑥, 𝑦, 0 = ∇ ρΩ 𝑥 + 𝑦 2 = ∇( ρΩ2 𝑅 2 )
2 2
1
−∇𝑝 + ρΩ2 R = −∇ 𝑝 − ρΩ2 𝑅 2
2
L Coriolis
force
modified pressure pressure
force
H
0 𝑢 −Ω𝑣 2ρΩ𝑣
−2ρΩ ∧ u = −2ρ 0 ∧ 𝑣 = −2ρ Ω𝑢 = −2ρΩ𝑢
Ω 𝑤 0 0
Northern hemisphere
Example Q2 (Exact Solutions)
The 𝑥-component of the momentum equation is given by
D𝑢 𝜕𝑝
ρ =− + μ∇2 𝑢
D𝑡 𝜕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”).
Assume flow in the x direction, and bounding planes 𝑦 = 0 and
𝑦 = ℎ. The velocity is then (𝑢(𝑦), 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 𝑈𝑤 .
The 𝑥-component of the momentum equation is given by
D𝑢 𝜕𝑝 y u(y) h
ρ =− + μ∇2 𝑢
D𝑡 𝜕𝑥 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”).
D𝑢
Fully-developed: =0
D𝑡
2
d2 𝑢 𝜕𝑝 d2 𝑢
𝑢 a function of 𝑦 only: ∇ 𝑢→ 2 𝑥 − momentum: 0=− +μ 2
d𝑦 𝜕𝑥 d𝑦
𝜕𝑝
𝑦 − momentum: 0=−
𝜕𝑦
𝜕𝑝/𝜕𝑥 must be constant, say −𝐺
d2 𝑢 𝐺 1𝐺
= − (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) Case (a): 𝑢 = 0 on 𝑦 = ℎ: 𝐴= ℎ
d𝑦 2 μ 2μ
1𝐺 2 𝐺
𝑢=− 𝑦 + 𝐴𝑦 + 𝐵 𝑢= 𝑦(ℎ − 𝑦)
2μ 2μ
𝑢 = 0 on 𝑦 = 0, so 𝐵 = 0 Case (b): 𝐺 = 0 and 𝑢 = 𝑈𝑤 on 𝑦 = ℎ: 𝑈𝑤 = ℎ𝐴
1𝐺 𝑦
𝑢 = 𝑦(𝐴 − 𝑦) 𝑢 = 𝑈𝑤
2μ 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 (𝐿0 ), velocity (𝑈0 ) and density (0 ) scales:
𝐿0 ∗
x = 𝐿0 x∗ , 𝑡= 𝑡 , u = 𝑈0 u∗ , ρ = ρ0 ρ∗ , 𝑝 = 𝑝𝑟𝑒𝑓 + ρ0 𝑈02 𝑝∗ , 𝑒𝑡𝑐.
𝑈0
Substitute into the governing equations:
D𝑢 𝜕𝑝 ρ0 𝑈02 ∗ D𝑢∗ ρ0 𝑈02 𝜕𝑝∗ μ𝑈0 ∗ 2 ∗
ρ =− + μ∇2 𝑢 ρ =− + 2 ∇ 𝑢
D𝑡 𝜕𝑥 𝐿0 D𝑡 ∗ 𝐿0 𝜕𝑥 ∗ 𝐿0
D𝑢 ∗ 𝜕𝑝 ∗ μ
ρ∗ ∗ = − ∗ + ∇∗ 2 𝑢 ∗
D𝑡 𝜕𝑥 ρ0 𝑈0 𝐿0
∗ ∗
D𝑢 𝜕𝑝 1
ρ∗ ∗ = − ∗ + ∇ ∗ 2 𝑢 ∗
D𝑡 𝜕𝑥 Re
ρ0 𝑈0 𝐿0 D𝑢 𝜕𝑝 1 2
Identify important dimensionless groups: Re = ρ =− + ∇ 𝑢
μ D𝑡 𝜕𝑥 Re
Common Dimensionless Groups
ρ𝑈𝐿
Re ≡ Reynolds number (viscous flow; μ = dynamic viscosity)
μ
𝑈
Fr ≡ Froude number (open-channel flow; 𝑔 = gravity)
𝑔𝐿
𝑈
Ma ≡ Mach number (compressible flow; 𝑐 = speed of sound)
𝑐
𝑈
Ro ≡ Rossby number (rotating flows; Ω = angular velocity of frame)
Ω𝐿
ρ𝑈 2 𝐿
We ≡ Weber number (free-surface flows; σ = surface tension)
𝜎
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
