Selected Problems of Real
F lu i ds
Advanced Fluid
M ech anics
1
U n steady B ern ou lli E q u ati on
UNSTEADY BERNOULLI
EQUATION
2.1. UNSTEADY LIQUID FLOWS IN PIPES
Assumptions: rigid pipe boundaries, incompressible fluid and
flow along a streamline
frictionless Euler equation
DU
1
= F − ∇p
Dt
ρ
( 2. 1 )
Let us convert the above vector equation into the scalar form by
taking the dot product with d s - the element of distance along a
streamline
DU
1
o d s = F o d s − ∇p o d s
( 2.2 )
Dt
ρ
I
II
III
Term I
DU
DU
 ∂U ∂U ds 
o ds =
⋅ ds = 
+
⋅  ds =
Dt
Dt
∂
t
∂
s
dt 

∂U
∂U
∂U
=
⋅ ds + U
⋅ ds =
⋅ ds + UdU
( 2. 3 )
∂t
∂s
∂t
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
2
U n steady B ern ou lli E q u ati on
Term II
Assuming the potential character of the mass forces field
 ∂Φ
∂Φ
∂Φ 
F = ∇Φ = 
i+
j+
k
∂
x
∂
y
∂
z


( 2. 4 )
Thus
 ∂Φ
∂Φ
∂Φ 
F o ds = 
i+
j+
k  o (dx ⋅ i + dy ⋅ j + dz ⋅ k ) =
∂
x
∂
y
∂
z


∂Φ
∂Φ
∂Φ
=
dx +
dy +
dz = dΦ
( 2.5 )
∂x
∂y
∂z
For gravitational mass forces field
∂Φ
∂Φ
∂Φ
= Fx = 0 ;
= Fy = 0 ;
= Fz = − g
∂x
∂y
∂z
So finally
F o d s = dΦ =
∂Φ
⋅ dz = − g ⋅ dz
∂z
( 2.6 )
( 2. 7 )
Term III
 ∂p
∂p
∂p 
∇p o d s =  i +
j + k  o (dx ⋅ i + dy ⋅ j + dz ⋅ k ) =
∂y
∂z 
 ∂x
∂p
∂p
∂p
= dx + dy + dz = dp
∂x
∂y
∂z
( 2. 8 )
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
3
U n steady B ern ou lli E q u ati on
Utilising expressions (2.3), (2.7) and (2.8) leads to the following
form of Euler equation
U 2 
∂U
dp
ds + d 
 = − g dz −
∂t
ρ
 2 
( 2. 9 )
Integration along a streamline from point 1 to point 2 leads to
Bernoulli equation for unsteady flows
∫1
2
2 dp
U 22 − U 12
∂U
ds +
+ g ( z 2 − z1 ) + ∫1
=0
∂t
2
ρ
( 2.10 )
For incompressible fluid ρ=const
2 ∂U
U 12 p1
U 22 p2
+
+ g ⋅ z1 =
+
+ g ⋅ z 2 + ∫1
ds
2
ρ
2
ρ
∂t
( 2.11 )
example
A long pipe is connected to a large open reservoir that is initially
filled with water to a depth of H. The pipe is D in diameter and L
long. As a first approximation, friction may be neglected.
Determine the flow velocity leaving the pipe as a function of time
after a cap is removed from its free end. The reservoir is large
enough so that the change in its level may be neglected.
cross section A
cross section B
zA = H
pA = pa
UA = 0
zB = 0
pB = pa
UB = ?
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
4
U n steady B ern ou lli E q u ati on
Bernoulli equation along a streamline from A to B
B ∂U
U B2
gH =
+ ∫A
ds
2
∂t
( 2.12 )
assumption: velocity in reservoir may be neglected except for a
small region near the inlet to the tube
⇓
∫A
B
B ∂U
∂U
ds ≈ ∫0
ds
∂t
∂t
( 2.13 )
from continuity equation implies that in the tube U=UB
everywhere, so that
∫0
B
B dU
dU B
∂U
B
ds = ∫0
ds = L
∂t
dt
dt
( 2.14 )
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
5
U n steady B ern ou lli E q u ati on
Substitution gives
U B2
dU B
gH =
+L
2
dt
( 2.15 )
dU B
dt
=
2 gH − U B2 2 L
( 2.16 )
Separating variables
and integrating
∫0
UB
=
we have
dU B
U1
1
-1 

=
tgh
 2 gH
2 gH − U B2
2 gH

 UB
1
tgh -1 
2 gH
 2 gH
 UB
tgh −1 
 2 gH
 t dt
 = ∫
0
2L




UB
=
0
( 2.17 )

2 gH ⋅ t
 =
2L

( 2.18 )
and finally
 2gH ⋅ t
U B = 2 gH tgh
2L




( 2.19 )
Selected Problems of Real
F lu i ds
Advanced Fluid
M ech anics
dU B
dt
=
t =0
6
U n steady B ern ou lli E q u ati on
gH
⋅
L

cosh

1
2gH ⋅ t 

2L 
=
gH
L
( 2.20 )
t =0
the equation of the tangent to the UB distribution for t≈0
UB(t ) =
gH
t
L
( 2.21 )
the time constant of the dynamic system
T=
2L
2 gH
( 2.22 )
applying T the expression describing the time evolution of the
velocity UB may be written in simplified form
U B = 2 gH tgh(t / T )
( 2.23 )
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
7
U n steady B ern ou lli E q u ati on
2.2. FLOW OF COMPRESSIBLE LIQUID THROUGH
THE ELASTIC PIPE – "WATER HAMMER"
Continuity equation for the pipe with varying cross sectional area A
∂( ρA ) ∂
+ ( ρUA ) = 0
∂t
∂s
for compressible liquid
dp = − k
dv
dρ
=k
v
ρ
( 2.24 )
( 2.25 )
where k is the modulus of volume elasticity of liquid
assuming linearity
p − p0 = k
ρ − ρ0
ρ0
p − p0 

⇒ ρ = ρ0  1 +

k


( 2.26 )
cross sectional area of the elastic pipe
p − p0 D 

A = A0  1 +
⋅ 
E
δ

( 2.27 )
where:
A0 - cross sectional area under reference pressure p0
E - modulus of volume elasticity of pipe wall
D - pipe diameter
δ - pipe wall thickness
let us rewrite continuity equation into the form
∂( ρA ) ∂p ∂( ρUA ) ∂p
⋅ +
⋅ =0
∂p
∂t
∂p
∂s
( 2.28 )
Selected Problems of Real
F lu i ds
Advanced Fluid
M ech anics
8
U n steady B ern ou lli E q u ati on
substituting (2.26) and (2.27) into the above and neglecting the
terms of the second order we obtain
ρ 0 A0  +
1
k
1 D  ∂p
 1 1 D  ∂p
⋅  + ρ 0 A0U  + ⋅  +
E δ  ∂t
 k E δ  ∂s
∂U
+ ρ 0 A0
=0
∂s
( 2.29 )
and finally
∂p
∂p
+U
+ ρ0
∂t
∂s
∂U
=0
1
1
D
∂s
ρ 0  + ⋅ 
k E δ 
1
⋅
( 2.30 )
let us introduce the speed of small disturbance propagating in
elastic pipe and compressible liquid
c2 =
1
ρ 0  +
1
k
1 D
⋅ 
E δ
( 2.31 )
the system of equations governing the flow (continuity and Euler
equations)






∂ p 
∂ p 
∂U

 + U 
 + c
=0
∂t  ρ 0 ⋅ c 
∂s  ρ 0 ⋅ c 
∂s
∂U
∂U
1 ∂p
+U
+
=0
∂t
∂s ρ 0 ∂s
( 2.32 )
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
9
U n steady B ern ou lli E q u ati on
let us calculate the sum of the equations (2.32)



∂ p
∂ p
∂ p

+ U  + U 
+ U  + c 
+ U  = 0 ( 2.33 )
∂t  ρ 0 ⋅ c
∂s  ρ 0 ⋅ c
∂s  ρ 0 ⋅ c



receiving


∂ p
∂ p

+ U  + ( U + c ) 
+ U  = 0
∂t  ρ 0 ⋅ c
∂s  ρ 0 ⋅ c


( 2.34 )
and their difference


∂ p
∂ p

− U  + ( U − c ) 
− U  = 0
∂t  ρ 0 ⋅ c
∂s  ρ 0 ⋅ c


( 2.35 )
the equations (2.34) and (2.35) may be expressed in the common
form
DΨ
∂Ψ
∂Ψ
=0 ⇒
+a
= 0 where ψ = ψ ( s ,t )
Dt
∂t
∂s
( 2.36 )
as the so-called wave equation
Solution
it may be easily shown that any function ψ of the form
ψ ( s ,t ) = f ( s − a ⋅ t ) = f ( z ) where z = s − a ⋅ t
satisfies the equation (2.36)
( 2.37 )
Selected Problems of Real
F lu i ds
Advanced Fluid
M ech anics
10
U n steady B ern ou lli E q u ati on
proof
let us calculate the derivatives
∂ψ ∂ψ ∂z
∂ψ
=
⋅ =−a
∂t
∂z ∂t
∂z
∂ψ ∂ψ ∂z ∂ψ
=
⋅ =
∂s
∂z ∂s ∂z
and put them into (2.36)
∂Ψ
∂Ψ
∂ψ
∂ψ
+a
=−a
+a
=0
∂t
∂s
∂z
∂z
interpretation
The quantity Ψ is propagated in the flow with velocity a. In
other words Ψ is kept constant along the line meeting the
relation
ds/dt = a
(2.38)
Ψ ( s0 ,t0 ) = Ψ ( s1 ,t1 )
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
11
U n steady B ern ou lli E q u ati on
example:
Let us consider the flow through a pipe which is connected to a
large reservoir. Pressure over a free surface as well as this one
at the outlet from the pipe are both equal to p0. Suddenly flow
is stopped by the valve. What will happen with the pressures pus
and pds propagated up- and downstream the valve ?
data:
=
δ =
E =
k =
ρ0 =
H =
d
40 mm
5 mm
21 MN/cm2
210 kN/cm2
1000 kg/ m3
0.5 m
velocity inside the pipe before shutdown
U bs = 2 gH
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
12
U n steady B ern ou lli E q u ati on
At time t0 (just after the valve was shut down) in the
neighbourhood of the valve velocity of liquid elements fell
down to zero (U = U0 = 0), but the pressure was still p0. In the
same time liquid elements located further from the valve were
not yet retarded so their velocities became unchanged (U = Ubs)
p
± 2 gH = const
ρ0 ⋅ c
upstream the valve
pus
p
− 2 gH = 0
ρ0 ⋅ c
ρ0 ⋅ c
pus = p0 + ρ 0 ⋅ c 2 gH
pressure pus propagated upstream is called pressure wave or
water hammer due to its high values – it may cause damage of
the pipe (the problem of firemen)
downstream the valve
pds
p
+ 2 gH = 0
ρ0 ⋅ c
ρ0 ⋅ c
pds = p0 − ρ 0 ⋅ c 2 gH
due to great absolute values of the term ρ 0 ⋅ c 2 gH in
comparison to atmospheric pressure one may expect unrealistic
(negative) values of pds
pressure pds approaching vacuum may lead to undesired
phenomenon - cavitation
Advanced Fluid
M ech anics
Selected Problems of Real
F lu i ds
13
U n steady B ern ou lli E q u ati on
velocity of small disturbances
c=
1
ρ 0  +
1
k
1 d
⋅ 
E δ
= 1395 m / s
pressure change (maximal value)
∆p = pus − p0 = p0 − pds = ρ 0 ⋅ c 2 gH = 4.37 MPa
note:
The solution is independent of the lengths L1 and L2.
The approach presented here was derived for inviscid liquids.
Viscosity affects the phenomenon – pressure variations in
practical cases are smaller (water hammer becomes weaker)
due to dissipation processes