W-16
Chapter 6 Incompressible Inviscid Flow
*6.6 Unsteady Bernoulli Equation: Integration
of Euler's Equation Along a Streamline
It is not necessary to restrict the development of the Bernoulli equation to steady flow.
The purpose of this section is to develop the corresponding equation for unsteady flow
along a streamline and to illustrate its use.
The momentum equation for frictionless flow (Eq. 6.1) can be written (with ~
g in the
negative z direction) as
~
DV
1
5 2 rp 2 gk^
Dt
ρ
ð6:17Þ
Equation 6.17 is a vector equation. It can be converted to a scalar equation by taking
the dot product with d~
s , where d~
s is an element of distance along a streamline. Thus
~
DV
DV
@V
@V
1
d~
s5
ds 5 V
ds 1
ds 5 2 rp d~
s 2 gk^ d~
s
Dt
Dt
@s
@t
ρ
ð6:18Þ
Examining the terms in Eq. 6.18, we note that
@V
ds 5 dV
@s
rp d~
s 5 dp
j^ d~
s 5 dz
ðthe change in V along sÞ
ðthe change in pressure along sÞ
ðthe change in z along sÞ
Substituting into Eq. 6.18, we obtain
V dV 1
@V
dp
ds 5 2
2 g dz
@t
ρ
ð6:19Þ
Integrating along a streamline from point 1 to point 2 yields
Z
1
2
dp V22 2 V12
1
1 gðz2 2 z1 Þ 1
ρ
2
Z
2
1
@V
ds 5 0
@t
ð6:20Þ
For incompressible flow, the density is constant. For this special case, Eq. 6.20
becomes
V2
p1
p2
V2
1 1 1 gz1 5
1 2 1 gz2 1
ρ
2
ρ
2
Restrictions:
Z
1
2
@V
ds
@t
ð6:21Þ
(1) Incompressible flow.
(2) Frictionless flow.
(3) Flow along a streamline.
This is a form of the Bernoulli equation
R 2 for unsteady flows. It differs from the
Bernoulli equation (Eq. 6.8) by the factor 1 @V/@t ds. How can we explain this extra
term? The derivation above integrated the momentum equation along a streamline
from point 1 to point 2 so that we ended up with quantities with units of energy per
unit mass (force or momentum 3 distance 5 work or energy); hence, we can interpret
the factor as the work involved in the rate of increase of momentum of the fluid on the
streamline over time (as opposed to the change in momentum over distance, represented by the change in velocity from V1 to V2). Example 6.9 will demonstrate this
idea. Equation 6.21 may be applied to any flow in which the restrictions are compatible with the physical situation.
*This section may be omitted without loss of continuity in the text material.
6.6
E
xample
6.9
W-17
Unsteady Bernoulli Equation: Integration of Euler’s Equation Along a Streamline
UNSTEADY BERNOULLI EQUATION
A long pipe is connected to a large reservoir that initially
is filled with water to a depth of 3 m. The pipe is 150 mm
in diameter and 6 m long. Determine the flow velocity
leaving the pipe as a function of time after a cap is
removed from its free end.
1
h=3m
2
z
D = 150 mm
V2
Flow
Given:
Find:
L=6m
Pipe and large reservoir as shown.
V2(t).
Solution:
Apply the Bernoulli equation to the unsteady flow along a streamline from point 1 to point 2 .
Governing equation:
Assumptions:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
⯝ 0(5)
0(6)
2
2
2
p1
V
p
V
V
1 gz1 2 2 gz2 ds
2
2
t
1
冕
Incompressible flow.
Frictionless flow.
Flow along a streamline from 1 to 2 .
p1 5 p2 5 patm.
V12 C0:
z2 5 0.
z1 5 h 5 constant.
Neglect velocity in reservoir, except for small region near the inlet to the tube.
Then
V2
gz1 5 gh 5 2 1
2
In view of assumption (8), the integral becomes
Z 2
1
In the tube, V 5 V2 everywhere, so that
Z
0
L
@V
ds @t
@V
ds 5
@t
Z
L
0
Z
Z
L
0
2
1
@V
ds
@t
@V
ds
@t
dV2
dV2
ds 5 L
dt
dt
This is the rate of change over time of the momentum (per unit mass) within the pipe; in the long term it will
approach zero. Substituting gives
gh 5
V22
dV2
1L
2
dt
Separating variables, we obtain
dV2
dt
5
2L
2gh 2 V22
Integrating between limits V 5 0 at t 5 0 and V 5 V2 at t 5 t,
"
!#V2
Z V2
dV
1
V
t
21
pffiffiffiffiffiffiffiffi
5 pffiffiffiffiffiffiffiffi tanh
5
2
2gh
2
V
2L
2gh
2gh 0
0
W-18
Chapter 6 Incompressible Inviscid Flow
Since tanh21(0) 5 0, we obtain
!
1
V
t
2
pffiffiffiffiffiffiffiffi tanh 2 1 pffiffiffiffiffiffiffiffi 5
2L
2gh
2gh
For the given conditions,
pffiffiffiffiffiffiffiffi
2gh 5
t pffiffiffiffiffiffiffiffi
V2
pffiffiffiffiffiffiffiffi
2gh
5 tanh
2L
2gh
or
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m
2 3 9:81 2 3 3 m 5 7:67 m=s
s
and
t pffiffiffiffiffiffiffiffi
t
1
m
2gh 5 3
3 7:67
5 0:639t
2L
2 6m
s
The result is then V2 5 7.67 tanh (0.639t) m/s, as shown:
0
V2 (m/s)
8
6
4
V2 = 7.67 tanh (0.639 t)
2
0
1
2
3
t (s)
4
5
V2 ðtÞ
ß
Notes:
ü This p
roblem il
lustrates
unsteady
use of th
Bernoull
e
ü Initiall
y the hea i equation.
d
a
v
a
ilable at
1 is used
state
to accele
the pipe;
rate the
fluid in
eventuall
y
state 2
equals th the head at
ü This p
e
head at s
roblem is
somewha tate 1 .
tic excep
t un
t for the
initial ins realisthe asym
tants—
ptotic flo
w condit
ally corre
io
na
sponds to
a turbule ctunt flow!
The Exce
l workbo
ok fo
Example
a
llows exp r this
the effec
loratio
to
for this p f varying the para n of
roblem.
meters