Chapter 2. Review of Hydrodynamics and Vector Analysis
2.1 Taylor series
L
L+ − +
+
− +
− +
=a a x x a x x an x x n
x
f( ) 0 1( 0) 2( 0)2 ( 0)
0 0) (x a
f =
L
L+ − +
+
− +
= 1 2 2( 0) ( 0) −1
) (
' x a a x x nan x x n
f
1 0) (
' x a
f =
L
L+ − − +
+
=2 2 ( 1)( 0) −2
) (
" x a n n x x n
f
2
0) 2
(
" x a
f = ⇒
2 ) (
" 0
2
x a = f
M
L − +L
+ +
− +
− +
=
∴ !
) )(
) ( 2 (
) (
"
) )(
( ' ) ( )
( 0 0
) ( 2
0 0
0 0
0 n
x x x x f
x x x f
x x f x f x f
n n
+Δ = + Δ + Δ +L
) 2 (
"
) ( ' ) ( ) (
2 0 0
0 0
x x f x x f x f x x f
If f(x) is known at x0, then it can be approximated at x0 +Δx by the Taylor series.
On the other hand, −Δ = − Δ + Δ −L ) 2
(
"
) ( ' ) ( ) (
2 0 0
0 0
x x f x x f x f x x f
2.2 Conservation of mass
x -direction:
z x y
x z u
y x u z y x z
y z x y x u z x y
x ⎥⎦⎤Δ Δ
⎢⎣⎡ Δ +
∂
−∂
= Δ
⎟Δ
⎠
⎜ ⎞
⎝
⎛ −Δ
⎟⎠
⎜ ⎞
⎝
⎛ − Δ L
2 ) ) (
, , ( ) , , ( ,
2 , ,
2 ,
ρ ρ ρ
− x y z
x z u
y x u z y x z
y z x y x u z x y
x Δ Δ
⎥⎦⎤
⎢⎣⎡ Δ +
∂ + ∂
= Δ
⎟Δ
⎠
⎜ ⎞
⎝
⎛ + Δ
⎟⎠
⎜ ⎞
⎝
⎛ + Δ L
2 ) ) (
, , ( ) , , ( ,
2 , ,
2 ,
ρ ρ ρ
OUTx
IN [massflux]
] flux mass ) [
( Δ Δ Δ = −
∂
−∂
z x
y x x
ρu
= mass accumulated with time in cube by x -direction
∴ Net accumulation in 3 directions for unit time
z y z x
w y
v x
u ⎥Δ Δ Δ
⎦
⎢ ⎤
⎣
⎡
∂
−∂
∂
−∂
∂
−∂
= (ρ ) (ρ ) (ρ )
On the other hand, mass increase in cube during Δ is t
z y x t t z y x t z y x z y x t t
t z y x
z y x t z y x z y x t t z y x
Δ Δ Δ
∂ Δ
=∂ Δ Δ Δ
− Δ Δ
⎥⎦Δ
⎢⎣ ⎤
⎡ Δ
∂ +∂
=
Δ Δ Δ
− Δ Δ Δ Δ +
ρ ρ ρ ρ
ρ ρ
) , , , ( )
, , , (
) , , , ( )
, , , (
t z y z x
w y
v x
z u y x
t t ⎥Δ Δ Δ Δ
⎦
⎢ ⎤
⎣
⎡
∂
−∂
∂
−∂
∂
−∂
= Δ Δ Δ
∂ Δ
∴∂ρ (ρ ) (ρ ) (ρ )
= accumulation in 3 directions during tΔ
) 0 ( ) ( )
( =
∂ +∂
∂ + ∂
∂ +∂
∂
∴∂
z w y
v x
u t
ρ ρ
ρ
ρ Exact conservation of mass equation
=0
∂ + ∂
∂ + ∂
∂ + ∂
∂ + ∂
∂ + ∂
∂ + ∂
∂
∂
z w w z
y v v y
x u u x
t ρ ρ ρ ρ ρ ρ
ρ
1 0
∂ = + ∂
∂ + ∂
∂ +∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂ + ∂
∂
∂
z w y v x u w z
v y u x
t
ρ ρ
ρ ρ
ρ
1 0
∂ = + ∂
∂ + ∂
∂ +∂
z w y v x u Dt Dρ ρ where
w z v y
u x t Dt
D
∂ + ∂
∂ + ∂
∂ + ∂
∂
= ∂ = total derivative or material derivative
Bulk modulus, 1 1 0
≅
=
⇒
= dt
dp E dt d d
E dp ρ
ρ ρ ρ
2
9 N/m
10 07 .
2 ×
=
E for water (incompressible fluid) For incompressible fluid,
=0
∂ +∂
∂ + ∂
∂
∂
z w y v x
u Continuity equation
2.3 Hydrostatic pressure
Newton’s 2nd law:
flow) no (
0 Q
=
= am F
0 0
sin ΔΔ = Δ Δ − Δ Δ = ⇒ − =
− Δ Δ
∑
Fx = px y z pn θ l y px y z pn z y px pn2 0 1
2 0 1 2
cos 1
= Δ
−
−
⇒
= Δ Δ Δ
− Δ Δ
− Δ Δ
= Δ Δ Δ
− Δ Δ
− Δ Δ
∑
=z g p
p
y z x g y x p y x p y z x g y l p
y x p F
n z
n z
n z
z
ρ
ρ ρ
θ
As the prism becomes smaller and smaller, or Δx,Δz,W →0, we have px = pz = pn
Pressure is a scalar. But the pressure force acting on a submerged body acts in the direction normal to the surface.
z y x x p
z x y
x z p y x p z x y
x z p y x p
z y z x y x p z y z x y x p Fx
Δ Δ
∂ Δ
−∂
=
Δ
⎥⎦Δ
⎢⎣ ⎤
⎡ Δ +
∂ + ∂
− Δ
⎥⎦Δ
⎢⎣ ⎤
⎡ Δ +
∂
−∂
=
Δ
⎟Δ
⎠
⎜ ⎞
⎝
⎛ +Δ
− Δ
⎟Δ
⎠
⎜ ⎞
⎝
⎛ −Δ
=
∑
=L
L ( , , ) 2
) 2 , , (
, 2 , ,
2 , 0
) , ( 0 p C1 y z x
p = ⇒ =
∂
∴∂ (1)
Similarly,
) , ( 0
0 p C2 x z
y
Fy p = ⇒ =
∂
⇒ ∂
∑
= (2) From (1) and (2),) (z p
p= (3) In z -direction,
z y x g y z x z y x p y z x z y x p
Fz ⎟Δ Δ − Δ Δ Δ
⎠
⎜ ⎞
⎝
⎛ +Δ
− Δ
⎟Δ
⎠
⎜ ⎞
⎝
⎛ − Δ
=
∑
=0 , , 2 , , 2 ρ=0 Δ Δ Δ
− Δ Δ
∂ Δ
−∂ z x y g x y z z
p ρ
) ,
3(x y C gz p
z g
p =− ⇒ =− +
∂
∴∂ ρ ρ (4)
From (3) and (4), C3 =constant. Because p=0 at z=0, we have C3 =0. Finally p=−ρgz
2.4 Equation of motion
Surface forces:
zz yy
xx σ σ
σ , , = normal stresses
zx xz
xy τ τ
τ , , , etc = shear stresses
The 1st subscript indicates the plane, while the 2nd subscript indicates the direction.
Sign convention: Stresses are positive if positive direction on positive plane or negative direction on negative plane.
Body force:
) (X Y Z z
y
xΔ Δ + + ρΔ
where X , Y, and Z are body forces per unit mass in x , y, and z directions, respectively.
Net forces acting on the cube of mass ρΔxΔyΔz are
x : X x y z
z y
Fx xxx yx zx ⎟⎟⎠Δ Δ Δ
⎜⎜ ⎞
⎝
⎛ +
∂ +∂
∂ + ∂
∂
= ∂σ τ τ ρ
y: Y x y z
z x
Fy yyy xy zy ⎟⎟⎠Δ Δ Δ
⎜⎜ ⎞
⎝
⎛ +
∂ +∂
∂ +∂
∂
= ∂σ τ τ ρ
z : Z x y z
y x
Fz zzz xz yz ⎟⎟⎠Δ Δ Δ
⎜⎜ ⎞
⎝
⎛ +
∂ +∂
∂ + ∂
∂
= ∂σ τ τ ρ
On the other hand, Newton’s 2nd law gives
( )
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂ + ∂
∂ Δ ∂ Δ Δ
= Δ
Δ Δ
=
× Δ Δ Δ
= z
w u y v u x u u t z u y Dt x
zDu y x a
z y x
Fx ρ x ρ ρ
( )
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂ + ∂
∂ Δ ∂ Δ Δ
= Δ Δ Δ
=
× Δ Δ Δ
= z
w v y v v x u v t z v y Dt x
z Dv y x a
z y x
Fy ρ y ρ ρ
( )
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂ + ∂
∂ Δ ∂ Δ Δ
= Δ
Δ Δ
=
× Δ Δ Δ
= z
w w y v w x u w t z w y Dt x
z Dw y x a
z y x
Fz ρ z ρ ρ
Note: Newton’s 2nd law applies to a fluid particle consisting of the same molecules or a system, so the total acceleration (or local acceleration plus convective accelerations) should be used.
z X y
x Dt
Du σxx τyx τzx ρ
ρ +
∂ +∂
∂ + ∂
∂
=∂
z Y x
y Dt
Dv σyy τxy τzy ρ
ρ +
∂ +∂
∂ +∂
∂
=∂
y Z x
z Dt
Dw σzz τxz τyz ρ
ρ +
∂ +∂
∂ + ∂
∂
=∂
Introduce pressure (scalar):
3
zz yy
p σxx +σ +σ
−
= positive as compression
Define
xx p
xx =σ +
τ
yy p
yy =σ +
τ
zz p
zz =σ +
τ
Then
z X y
x x p Dt
Du τxx τyx τzx ρ
ρ +
∂ +∂
∂ +∂
∂ +∂
∂
−∂
=
z Y y
x y p Dt
Dv τxy τyy τzy ρ
ρ +
∂ + ∂
∂ +∂
∂ +∂
∂
−∂
=
z Z y
x z p Dt
Dw τxz τyz τzz ρ
ρ +
∂ + ∂
∂ + ∂
∂ +∂
∂
−∂
=
For u=v=w=0 and τxx =τxy =L=0,
0 )
( + + =
∂ +
−∂
∂
−∂
∂
−∂ X Y Z
z p y p x
p ρ
If z is positive vertically upwards, ∂p/∂x=∂p/∂y=0. If gravity is the only body
force, X = Y =0 and Z =−g. Then
z g p =−ρ
∂
∂
gz
p=−ρ ← hydrostatic pressure
Assume all the shear stresses are zero (inviscid fluid), that is valid for most water wave problems. Then
z g p Dt
Dw
y p Dt
Dv
x p Dt
Du
∂ −
− ∂
=
∂
− ∂
=
∂
− ∂
=
ρ ρ ρ
1 1
1
Euler equation (Equation of motion for inviscid fluid)
Consider angular momentum:
ω&
I M =
∑
where M = moment, I = moment of inertia, and ω& = angular acceleration.
( )
ωτ ρ τ τ
τ
τ τ τ τ
&
2 2
12 1 2 2
2 2
2 2
2 2
z x z y x x
y x z x y z
z x z
y x x z x y z
z x z
xz xz
zx zx
xz xz
zx zx
Δ + Δ Δ Δ Δ Δ =
Δ Δ
⎟⎠
⎜ ⎞
⎝
⎛ Δ
∂
−∂ Δ −
Δ Δ
⎟⎠
⎜ ⎞
⎝
⎛ Δ
∂
−∂ +
Δ Δ Δ
⎟⎠
⎜ ⎞
⎝
⎛ Δ
∂ +∂ Δ −
Δ Δ
⎟⎠
⎜ ⎞
⎝
⎛ Δ
∂ +∂
( )
ωρ τ
τ 2 2 &
12
1 x z
xz
zx − = Δ +Δ ∼O
( )
Δx2 ω&As the cube becomes smaller and smaller, ω& increases, that implies fast spinning of water particle. Therefore, τzx −τxz must be zero. In general,
) , , ,
(i j x y z
ji
ij =τ =
τ
For laminar flow, shear stress is proportional to rate of strain, so that
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ +∂
∂
= ∂
= x
v y u
yx
xy τ μ
τ
⎟⎠
⎜ ⎞
⎝
⎛
∂ +∂
∂
= ∂
= x
w z u
zx
xz τ μ
τ
x u
xx ∂
= μ ∂
τ 2
where μ = coefficient of viscosity. Substituting into x -direction momentum equation,
z X w y v x u z x
u y
u x
u x
p
x X w z
u z x v y
u y x u x
x p Dt
Du
ρ μ
ρ μ
μ μ
μ μ
ρ
⎥+
⎦
⎢ ⎤
⎣
⎡ ⎟⎟
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
∂
∂ + ∂
∂ +∂
∂ +∂
∂ + ∂
∂
−∂
=
⎟+
⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
∂
∂ + ∂
∂
−∂
=
2 2 2 2 2 2
2
Using the continuity equation, we obtain the Navier-Stokes equation, equation of motion for incompressible, Newtonian fluid:
z Z w y
w x
w z
p Dt
Dw
z Y v y
v x
v y
p Dt
Dv
z X u y
u x
u x
p Dt
Du
⎟⎟+
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ +∂
∂ + ∂
∂
− ∂
=
⎟⎟+
⎠
⎜⎜ ⎞
⎝
⎛
∂ +∂
∂ + ∂
∂ + ∂
∂
− ∂
=
⎟⎟+
⎠
⎜⎜ ⎞
⎝
⎛
∂ +∂
∂ +∂
∂ + ∂
∂
− ∂
=
2 2 2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2
1 1 1
ρ μ ρ
ρ μ ρ
ρ μ ρ
2.5 Review of vector analysis
k a j a i a
a= x + y + z
where i, j,k = unit vectors in x,y,z directions. The length or magnitude of the vector a is given by
2 2 2
z y
x a a
a
a = + +
2.5.1 Dot product
a b b
a b
a⋅ = cosθ = ⋅ (commutative)
where θ = angle between the two vectors. For unit vectors, we have
k k j j i
i⋅ =1×1×cos0°=1= ⋅ = ⋅ k j k i j
i⋅ =1×1×cos90°=0= ⋅ = ⋅
Using the above relation,
z z y y x x z
y x z y
xi a j a k b i b j b k a b a b a b
a b
a⋅ =( + + )⋅( + + )= + +
Note: 1) Dot product is a scalar.
2) If a ≠0, b ≠0, but a⋅ b=0, then a⊥ (b Qcosθ =0) 3) Projection of a onto b = a⋅b/b
2.5.2 Cross product
a b b
a b
a× = sinθ ≠ × (not commutative)
Cross product of vectors is a vector (magnitude + direction). The magnitude is given by the above expression, and the direction is given by the right-hand rule.
For unit vectors,
) 0 sin (
0 =
=
×
=
×
=
×i j j k k Q θ
i
k i j k j
i× = , × =− , j×k=i, k× j=−i, k×i= j, i×k =−j
k b a b a j b a b a i b a b a b b b
a a a
k j i b
a y z z y z x x z x y y x
z y x
z y
x =( − ) +( − ) +( − )
=
×
Note: If a ≠0, b ≠0, but a× b=0, then a // (b Qsinθ =0)
2.5.3 Vector differential operator
zk y j
xi ∂
+ ∂
∂ + ∂
∂
= ∂
∇
z k y j
x i ∂
+∂
∂ +∂
∂
=∂
=
∇φ gradient ofφ (scalar) φ φ φ
Gradient indicates the spatial rate of change of a scalar. Gradient is a vector whose direction indicates the maximum rate of change.
( )
z u y u x u
k u j u i u zk y j xi
u u
y z x
z y x
∂ +∂
∂ + ∂
∂
=∂
+ +
⎟⎟⋅
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
= ∂
=
⋅
∇ divergenceof
If u=ui+vj+wk,
z w y v x u u
∂ +∂
∂ + ∂
∂
= ∂
⋅
∇
=0
⋅
∇ u continuity equation in vector notation
Laplacian operator:
2
2 2 2 2 2 2
z y k x
j z i y
k x j z i y
x ∂
+∂
∂ +∂
∂
=∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ +∂
∂ + ∂
∂
⋅ ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
= ∂
∇
⋅
∇
=
∇ φ φ φ φ φ φ φ φ
( )
y k u x j u x u z i u z u y u
u u u
z y x
k j i
k u j u i u zk y j xi
u u
y x z
y x z
z y x
z y x
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
−∂
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂
−∂
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂
−∂
∂
= ∂
∂
∂
∂
∂
∂
= ∂
+ +
⎟⎟×
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂ + ∂
∂
= ∂
=
×
∇ curlof
Useful vector identities:
Divergence of curl is zero, ∇⋅
( )
∇×u =0Curl of gradient is zero, ∇×∇φ =0
2.6 Rotation of fluid particle
At time t → At time t+Δt
x t w x
x t x w w x x w w
∂ Δ
= ∂ Δ
⎥Δ
⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
⎛ Δ
∂
−∂
⎟−
⎠
⎜ ⎞
⎝
⎛ Δ
∂ + ∂
=
≅ 2 2
tanα α
x w
t ∂
=∂
∂
∴ ∂α
← angular velocity (or rate of rotation)
z
u
t ∂
−∂
∂ =
∂β
The mean angular velocity, or vorticity is given by
⎟⎠
⎜ ⎞
⎝
⎛
∂
−∂
∂
= ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂
= ∂
z u x w t
t 2
1 2
1 α β
ω
For irrotational flow, ω=0. Note that in an irrotational flow, the overall shape of the fluid particle can be distorted, but the mean angular velocity (or vorticity) must be zero.
2.7 Velocity potential
Definition (3-D):
If ω =∇×u=(∂w/∂y−∂v/∂z)i+(∂u/∂z−∂w/∂x)j+(∂v/∂x−∂u/∂y)k =0, then the flow is irrotational.
Theorem:
=0
×
∇
= u
ω if and only if there exists a scalar field φ such that u=∇φ.
Proof (Greenberg, 1978. Foundations of Applied Mechanics, Prentice-Hall, 170-171):
1) Assume that there is a scalar φ such that u=∇φ. 2) Then ω =∇×u=∇×
( )
∇φ =0 (Q curl of gradient = 0)3) We must show that ∇ u× =0 implies the existence of a scalar φ such that φ
∇
=
u .
By Stokes theorem,
) 0 (
0 ∇× =
=
×
∇
⋅
=
∫
Cu⋅dr∫
Sn udσ Q uHowever,
∫
⋅ =0=∫
1⋅ +∫
− 2⋅ =∫
1⋅ −∫
2⋅C C
C C
Cu dr u dr u dr u dr u dr
Therefore,
∫
Cu1⋅dr=∫
Cu2⋅dr ← independent of path of integrationDefine
( , , ) ( , , )
) , ,
(x0y0z u0 dr x y z
z y
x ⋅ ≡φ
∫
Then the velocity potential φ(x,y,z) is uniquely defined by the point P(x,y,z), not by the path of integration.
Using
u=ui+vj+wk dr=dxi+dyj+dzk we have
=
∫
⋅ =∫
+ + +∫
(( ,, ,,)) + +) , , (
) , , ( )
, , (
) , ,
( 1
1 0 0 0 0
0 0
) (
) (
) , ,
( xyz
z y x z
y x
z y x z
y x
z y
x u dr udx vdy wdz udx vdy wdz
z y φ x
Using the fundamental theorem of integral calculus,
f( )d f(x) dx
d x
a =
∫
ξ ξwe have
udx vdy wdz u
x x
z y x
z y
x + + =
∂
= ∂
∂
∂φ
∫
((1,,,,))( )In the same way,
w
v z
y =
∂
= ∂
∂
∂φ φ
,
Therefore,
u=∇φ
This is the end of the proof.
Considering the above diagram, physically the flow must be in positive direction. But according to the definition of the velocity potential, we have
<0
∂
=∂ u φx
or the flow is in negative direction. Therefore, we redefine the velocity potential as
φ φ
φ φ
−∇
=
→
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
∂
−∂
=
∂
−∂
=
∂
−∂
=
u
w z v y u x
so that the flow occurs in the direction from high potential to low potential.
2.8 Stream function
Consider 2D x− plane z∫
⋅= Ψ ( ,)
) , ( 0 0
z x
z
x u ndr ← flow rate across the line connecting P0 and P
( ) ( )
∫
∫
+
−
=
+
−
⋅ +
=
) , (
) , (
) , (
) , ( 0 0
) (
z x
z x
z x
z x
wdx udz
k dx i dz k w i u
0 0
For conservation of mass, is independent of path of integration. For this, the integrand must be an exact differential,
Ψ
dΨ. This requires that
wdx udz z dz
x dx
d =− +
∂ Ψ +∂
∂ Ψ
=∂ Ψ
or
u z w x
∂ Ψ
−∂
∂ = Ψ
=∂ ,
Using
z x z w
∂
∂ Ψ
= ∂
∂
∂ 2
and
x z x u
∂
∂ Ψ
−∂
∂ =
∂ 2
we have
=0
∂ +∂
∂
∂ z w x u
which is 2-D continuity equation. Therefore, a stream function exists for 2-D incompressible flow.
orthogonal are
across and integral line
:
along integral line
: → Ψ
⎭⎬
⎫
Ψφ φ
C C
Finally, φ and satisfies the Cauchy-Riemann conditions: Ψ
x z
z x
∂ Ψ
−∂
∂ =
∂
∂ Ψ
= ∂
∂
∂ φ φ
2.9 Bernoulli equation
Euler equation in x -direction isx p z
w u x u u t u
∂
− ∂
∂ = + ∂
∂ + ∂
∂
∂
ρ 1
Using 2-D irrotational flow condition, ∂u/∂z=∂w/∂x,
x p w
x u
x t u
∂
− ∂
⎟⎟=
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
ρ 1 2
2
2 2
Similarly, in z - direction, we have
z g p w
z u
z t
w −
∂
− ∂
⎟⎟=
⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
∂
ρ 1 2
2
2 2
Introducing φ and u=−∂φ/∂x, w=−∂φ/∂z,
2 0
) / ( ) / : (
2 0
) / ( ) / : (
2 2
2 2
⎥=
⎦
⎢ ⎤
⎣
⎡ ∂ ∂ + ∂ ∂ + +
∂ +
−∂
∂
∂
⎥=
⎦
⎢ ⎤
⎣
⎡ + ∂ ∂ + ∂ ∂ +
∂
−∂
∂
∂
p gz z
x t
z z
p z x
t x x
ρ φ
φ φ
ρ φ
φ φ
Integration gives
) , 2 (
) / ( ) / : (
) , 2 (
) / ( ) / : (
2 2
2
1 2
2
t x C p gz
z x
z t
t z p C z x
x t
+
−
=
∂ +
∂ +
∂ + ∂
∂
−∂
=
∂ +
∂ +
∂ + ∂
∂
−∂
ρ φ
φ φ
ρ φ
φ φ
The above two equations give
) ( ) , ( )
, ( )
,
( 2 2
1 z t gz C x t C x t C t
C =− + ⇒ =
Therefore,
) 2 (
) / ( ) /
( 2 2
t C p gz
z x
t + ∂ ∂ + ∂ ∂ + + =
∂
−∂
ρ φ
φ φ
which is the Bernoulli equation for unsteady flow, giving relationship between pressure field and flow kinematics (u,w).
Using
t t t f
C ∂
=∂ ( ) )
(
the Bernoulli equation becomes
[ ]
02 ) 1 (
2 2
= +
⎥+
⎥⎦
⎤
⎢⎢
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂ +
− ∂ p gz
z t x
t f ρ
φ φ φ
Defining φ'=φ+ f(t), we have
' 0 '
2 1
' 2 2 + + =
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝
⎛
∂ + ∂
⎟⎠
⎜ ⎞
⎝
⎛
∂ + ∂
∂
−∂ p gz
z x
t ρ
φ φ
φ
The Bernoulli term C(t) is included in φ . However, the flow kinematics are the same ' for φ and φ : '
w
z u z
x
x =
∂
= ∂
∂
= ∂
∂
=∂
∂
∂φ φ φ' φ
' ,
