Session 6-1 Continuity equation
Contents
6.1 Continuity Equation
6.2 Stream Function in 2-D, Incompressible Flows 6.3 Rotational and Irrotational Motion
6.4 Equations of Motion
6.5 Examples of Laminar Motion 6.6 Irrotational Motion
6.7 Frictionless Flow 6.8 Vortex Motion
Objectives
- Derive 3D equations of continuity and motion
- Derive Navier-Stokes equation for Newtonian fluid - Study solutions for simplified cases of laminar flow
- Derive Bernoulli equation for irrotational motion and frictionless flow - Study solutions for vortex motions
∙ Derivation of 3-D Eq.
conservation of mass → continuity eq.
conservation of momentum → eq. of motion → Navier-Strokes eq.
x y z
∆ ∆ ∆
Consider infinitesimal control volume ( ) → point form [Cf] Finite control volume – arbitrary CV → integral form
Apply principle of conservation of matter to the CV
→ sum of net flux = time rate change of mass inside C.V.
1) mass flux per unit time (mass flow)
mass vol
Q u A
time ρ time ρ ρ
= = = = ∆
∙ net flux through face perpendicular to x -axis
= flux in –flux out
∙ net flux through face perpendicular to y -axis
∙ net flux through face perpendicular to z -axis
( u ) ( u )
u y z u x y z x y z
x x
ρ ρ
ρ ρ ∂ ∂
= ∆ ∆ − + ∂ ∆ ∆ ∆ = − ∂ ∆ ∆ ∆
( v )
x y z y
ρ
= − ∂ ∆ ∆ ∆
∂
( w )
x y z z
ρ
= − ∂ ∆ ∆ ∆
∂
(A)2) time rate change of mass inside C.V.
( x y z ) t ρ
= ∂ ∆ ∆ ∆
∂
(B)Thus, equating (A) and (B) gives
( ) ( ) ( )
( ) u v w
x y z x y z x y z x y z
t x y z
ρ ρ ρ
∂ ρ ∆ ∆ ∆ = − ∂ ∆ ∆ ∆ − ∂ ∆ ∆ ∆ − ∂ ∆ ∆ ∆
∂ ∂ ∂ ∂
( ) ( )
LHS x y z x y z x y z
t t t
ρ ρ ρ
∂ ∂ ∂
= ∆ ∆ ∆ = ∆ ∆ ∆ + ∆ ∆ ∆
∂ ∂ ∂
Since C.V. is fixed →
( ) x y z 0
t
∂ ∆ ∆ ∆ =
∂ LHS x y z
t ρ
∴ = ∆ ∆ ∆ ∂
∂
Cancelling terms makes
( ) ( ) ( )
u v w 0
t x y z
ρ ρ ρ ρ
∂ + ∂ + ∂ + ∂ =
∂ ∂ ∂ ∂
0 t q
ρ ρ
∂ + ∇ ⋅ =
∂
→ Continuity Eq. for compressible fluid in unsteady flow (point form) (6.1)
The 2 term of Eq. (6.1) can be expressed as
( ρ q ) q ρ ρ q
∇ ⋅ = ∇ + ∇ ⋅
I II
gradient divergence
(I): q ( ui vj wk ) i j k
x y z
ρ ρ ρ
ρ ∂ ∂ ∂
∇ = + + ∂ + ∂ + ∂
u v w
x y z
ρ ρ ρ
∂ ∂ ∂
= + +
∂ ∂ ∂
(II): u v w
q x y z
ρ ∇ ⋅ = ρ ∂ ∂ + ∂ ∂ + ∂ ∂
( ) q u v w u v w
x y z
x y z
ρ ρ ρ ρ
ρ ∂ ∂ ∂ ∂ + ∂ + ∂
∴∇ ⋅ = ∂ + ∂ + ∂ + ∂ ∂ ∂
(i)Substituting (i) into Eq (6.1) yields
u v w 0
u v w
t x y z x y z
ρ ρ ρ ρ ρ
∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ =
d dt
ρ
d u v w 0
dt ρ ρ + ∂ ∂ x + ∂ ∂ y + ∂ ∂ z = ( ) 0
d q
dt ρ ρ + ∇ ⋅ =
(6.2a)
(6.2b)
0 t q
ρ ρ
∂ + ∇ ⋅ =
∂
(6.1)
[Re] Total derivative (total rate of density change)
d dx dy dz
dt t x dt y dt z dt ρ = ∂ ρ + ∂ ρ + ∂ ρ + ∂ ρ
∂ ∂ ∂ ∂
u v w
t x y z
ρ ρ ρ ρ
∂ ∂ ∂ ∂
= + + +
∂ ∂ ∂ ∂
1) For steady-state conditions
0 t
→ ∂ ρ
∂ =
Then (6.1) becomes
( ) ( ) ( )
( ) 0
u v w
x y z q
ρ ρ ρ ρ
∂ + ∂ + ∂ = ∇ ⋅ =
∂ ∂ ∂
(6.3)2) For incompressible fluid (whether or not flow is steady)
0 d
dt
→ ρ =
Then (6.2) becomes
0 u v w
x y z q
∂ + ∂ + ∂ = ∇ ⋅ =
∂ ∂ ∂
(6.5)(6.4)
[Re] Continuity equation derived using a finite CV method Eq. (4.5a):
CV
dV
CSq dA 0
t
ρ ρ
∂ + ⋅ =
∫ ∂ ∫
(4.5)→ volume-averaged (integrated) form
● Gauss' theorem:
volume integral ↔ surface integral
– reduce dimensions by 1 (3D → 2D)
( )
v
∇ ⋅ X dV =
AX dA ⋅
∫ ∫
Thus,
2 ( )
CS CV
nd term = ∫ ρ q dA ⋅ = ∫ ∇ ⋅ ρ q dV
( ) ( ) 0
CV
dV
CVq dV
CVq dV
t t
ρ ρ ρ ρ
∂ ∂ + ∇ ⋅ = ∂ ∂ + ∇ ⋅ =
∫ ∫ ∫
Eq. (4.5) becomes
Since integrands must be equal to zero.
( q ) 0 t
ρ ρ
∂ + ∇ ⋅ =
∂
→ same as Eq. (6.1) → point form
(A)
where V = cross-sectional average velocity
1D Continuity equation
( ) : u 0
A dA dA
t x
ρ ρ
∂ + ∂ =
∂ ∂
∫ ∫
0
dA udA
t ρ x ρ
∂ + ∂ =
∂ ∫ ∂ ∫
For incompressible fluid flow
0
dA udA
t x
ρ ∂ + ρ ∂ =
∂ ∫ ∂ ∫
A VA
where q = flow through σ
A VA 0
t x
∂ ∂
∴ + =
∂ ∂
Consider lateral inflow/outflow
A VA
t x
σqd σ
∂ + ∂ =
∂ ∂ ∫
For steady flow;
A 0 t
∂ =
∂ VA 0
x
∴ ∂ =
∂
.
VA = const = Q
[Re] Continuity equation in polar (cylindrical) coordinates
radial
, - u r
azimuthal
, - v θ
axial
, - w z
For compressible fluid of unsteady flow
1 ( ) 1 ( ) ( )
ur v w 0
t r r r z
ρ ρ ρ ρ
θ
∂ + ∂ + ∂ + ∂ =
∂ ∂ ∂ ∂
For incompressible fluid
1 ( ) 1
ur v w 0
r r r θ z
∂ + ∂ + ∂ =
∂ ∂ ∂
For incompressible fluid and flow of axial symmetry
( )
0, 0, v 0
t r z
ρ ρ ρ ρ ρ
θ θ
∂ = ∂ = ∂ = ∂ = ∂ =
∂ ∂ ∂ ∂ ∂
1 ( )
ur w 0
r r z
∂ ∂
∴ + =
∂ ∂
→ 2-D boundary layer flow Example: submerged jetr, u
z, w
[Re] Green's Theorem
1) Transformation of double integrals into line integrals
( )
2 1
1 2
C R
F F
dxdy F dx F dy
x y
∂ − ∂ = +
∂ ∂
∫∫ ∫
( )
CR
curl F ⋅ kdxdy = F dr ⋅
∫∫ ∫
1 2
F = F i + F j
2) 1st form of Green's theorem
(
2)
T S
f g grad f grad g dV f g dA n
∇ + ⋅ = ∂
∫∫∫ ∫∫ ∂
where n = outer unit normal vector of S 3) 2nd form of Green's theorem
(
2 2)
T S
g f
f g g f dV f g dA
n x
∂ ∂
∇ + ∇ = ∂ − ∂
∫∫∫ ∫∫
[Re] Divergence theorem of Gauss
→ transformation between volume integrals and surface integrals
T S
divF dV = F n dA ⋅
∫∫∫ ∫∫
1 2 3
F = F i + F j + F k
cos cos cos n = α i + β j + γ k
1 2 3
T
F F F
dxdydz
x y z
∂ + ∂ + ∂
∂ ∂ ∂
∫∫∫
(
1cos
2cos
3cos )
S
F α F β F γ dA
= ∫∫ + +
By the way
(
1 2 3)
S S
F ndA ⋅ = F dydz + F dzdx + F dxdy
∫∫ ∫∫
( )
1 2 3
1 2 3
T
S
F F F
dxdydz
x y z
F dydz F dzdx F dxdy
∂ ∂ ∂
∴ ∂ + ∂ + ∂
= + +
∫∫∫
∫∫
(6.6)In 2D flowfield, define stream function ψ(x, y) as
2-D incompressible continuity eq. is
u v 0
x y
∂ + ∂ =
∂ ∂
(6.7)u y
ψ
= − ∂
∂
v x
ψ
= ∂
∂ (6.9)
(6.8)
• We can a simplified equation by having to determine only one unknown function.
→ Thus, continuity equation is satisfied.
1) Apply stream function to the equation for a stream line in 2-D flow Eq. (2.10):
Substitute (6.8) into (6.11)
(6.10)
2 2
=0
u v
x y x y y x x y x y
ψ ψ ψ ψ
∂ ∂ + ∂ ∂ = ∂ ∂ − ∂ ∂ + ∂ ∂ ∂ ∂ = − ∂ ∂ ∂ + ∂ ∂ ∂
0
vdx − udy =
(6.11)= =0
dx dy d
x y
ψ ψ ψ
∂ ∂
∂ + ∂
(6.12)ψ = constant
2) Apply stream function to the law of conservation of mass
Substitute (6.8) into (6.14)
(6.13)
→ The stream function is constant along a streamline.
qdn udy vdx
− = − +
(6.14)qdn dy dx d
y x
ψ ψ ψ
∂ ∂
− = + =
∂ ∂
(6.15)→ Change in ψ (dψ) between adjacent streamlines is equal to the volume rate of flow per unit width.
• Stream function in cylindrical coordinates
radial
v
rr ψ
θ
= − ∂
∂ v azimuthal
θ
r
ψ
= ∂
∂
Assume the rate of rotation of fluid element ∆x and ∆y about Z -axis is positive when it rotates counterclockwise.
- time rate of rotation of ∆x -face about z -axis 6.3.1 Rotation and vorticity
1
v v x v t
x v
t x x
+ ∂ ∆ − ∆
∂ ∂
= =
∆ ∆ ∂
- time rate of rotation of ∆y -face about z - axis
net rate of rotation = average of rotations of ∆x -and ∆y -face
1
u u y u t
y u
t y y
+ ∂ ∆ − ∆
∂ ∂
= − = −
∆ ∆ ∂
1
z
2
v u
x y
ω = ∂ ∂ − ∂ ∂
1
x
2
w v
y z
ω = ∂ ∂ − ∂ ∂
1
y
2
u w
z x
ω = ∂ ∂ − ∂ ∂
(6.16a)1) Rotation
1 1 1
2 2 2
w v u w v u
i j k
y z z x x y
ω = ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂
( )
1 1
2 q 2 curl q
= ∇ × =
(6.16b)
2 2 2
x y z
ω = ω + ω + ω
a) Ideal fluid → irrotational flow
0
∇ × = q
x y z
0
ω = ω = ω =
, ,
w v u w v u
y z z x x y
∂ = ∂ ∂ = ∂ ∂ = ∂
∂ ∂ ∂ ∂ ∂ ∂
(6.17)b) Viscous fluid → rotational flow
0
∇ × ≠ q
2) Vorticity
2 curl q q
ζ = = ∇ × = ω
Rotation in cylindrical coordinates
1 1 2
z r
v v
r z
ω
θθ
∂ ∂
= ∂ − ∂
1 2
v
rvz
z r
ω
θ= ∂ ∂ − ∂ ∂ 1 1
2
r z
v v v
r r r
θ θ
ω θ
∂ ∂
= − ∂ + + ∂
rotational flow
irrotational flow
rotational flow
Γ = line integral of the tangential velocity component about any closed contour S
6.3.2 Circulation
Γ = ∫ q ds ⋅
(6.19)– take line integral from A to B, C, D, A ~ infinitesimal CV
2 2 2 2
u dy v dx u dy v dx
d u dx v dy u dx v dy
y x y x
∂ ∂ ∂ ∂
Γ ≅ − ∂ + + ∂ − + ∂ − − ∂
v u
x y dxdy
∂ ∂
= ∂ − ∂
v u
d dxdy
x y
∂ ∂ Γ ≅ ∂ − ∂
( )
2
z zA A A
v u
dA dA q dA
x y ω
∂ ∂
Γ = ∫∫ ∂ − ∂ = ∫∫ = ∫∫ ∇ ×
(6.20)For irrotational flow,
circulation Γ = 0 (if there is no singularity vorticity source).
[Re] Fluid motion and deformation of fluid element translation
Motion
rotation
linear deformation Deformation
angular deformation
(1) Motion
1) Translation: ξ, η
, d u dt u
dt ξ = = ξ
, d v dt v
dt η = = η
2) Rotation ← Shear flow
1
x
2
v u
x y
ω = ∂ ∂ − ∂ ∂
1) Linear deformation – normal strain
x
x
ε = ∂ ξ
∂
y
y
ε = ∂ η
∂
i) For compressible fluid, changes in temperature or pressure cause change in volume.
ii) For incompressible fluid, if length in 2-D increases, then length in another 1-D decreases in order to make total volume unchanged.
2) Angular deformation– shear strain
xy