1
Bernoulli equation
Daniel Bernoulli (1710-1782)
2
Bernoulli: preamble
Want to discuss the properties of a moving fluid.
Will do this initially under the simplest possible
conditions, leading to Bernoulli’s equation. The
following restrictions apply.
• Flow is inviscid, there are no viscous drag forces
• Heat conduction is not possible for an inviscid
flow
• The fluid is incompressible .
• The flow is steady (velocity pattern constant).
• The paths traveled by small sections of the fluid
are well defined.
• Will be implicitly using the Euler equations of
motion (discussed later)
3
Coordinates and streamlines
• Each piece of fluid has velocity v .
• Steady flow, nothing changes with time at given
location. All particles passing through (1) end
up at (2) with velocity v
• The trajectories followed by the particles are
called streamlines.
• Describe motion is terms of distance traveled
along streamline, s .
• Velocity given by |v| =
is n̂ .
ds
dt
. Normal to velocity
• Stream-line can bend, R is radius of curvature.
4
Coordinates and streamlines
Using body fixed coordinates. If the particles change
speed along stream-line, or if stream-line bends, then
accelerations must be present.
The tangential acceleration
dv
∂v ds
∂v
as =
=
=v
dt
∂s dt
∂s
The normal acceleration
v2
an =
R
The radius of curvature R changes along the
streamline.
5
Streamline coordinates
It is convenient to use a coordinate system defined in
terms of the flow streamlines. The coordinate along
the streamline is s and the coordinate normal to the
streamline is n . The unit vectors for the streamline
coordinates are ŝ and n̂ .
The direction of ŝ will be chosen to be in the same
direction as the velocity. So v = vŝ .
y
s = s2
n = n2
n = n1
s = s1
s=0
n=0
Streamlines
^
n
^
s
V
s
x
The flow plane is covered with an orthogonal curved
net of coordinate lines and v = v(s, n)ŝ and
ŝ = ŝ(s, n) for steady flow.
6
Forces on streamlines
Any particle travelling along the streamline will be
subjected to a number of forces.
The relevant Forces for Bernoulli’s equation are
gravity and pressure.
7
Streamlines F = ma
Will resolve forces in directions parallel ŝ and
perpendicular n̂ to particles motions. y is out of
page, z is down, x is horizontal.
X
∂v
δFs = δm as = δm v
∂s
∂v
= ρδV v
∂s
8
Streamlines F = ma
Resolve forces
δWs = δW sin θ = γδV sin θ
δWs would be zero for horizontal motion.
The pressure changes with height. Let p be pressure
in middle of fluid slab. Let p + δps be pressure in
front of slab and p − δps be pressure behind slab.
From Taylors series
∂p δs
δps =
∂s 2
The net pressure force
δFps
= (p − δps )δnδy − (p + δps )δnδy
∂p
∂p
= −2δps δnδy = − δsδnδy = − δV
∂s
∂s
Net force
δFs = δWs + δFps
∂p
=
−γ sin θ −
δV
∂s
9
Bernoulli equation
Equate two expressions for δFs
∂p
∂v
δFs = −γ sin θ −
δV = ρδV v
∂s
∂s
∂p
∂v
⇒ −γ sin θ −
= ρv
∂s
∂s
The change in fluid particle speed along a streamline
is accomplished by a combination of pressure and
gravity forces.
• Now use sin θ =
• And
v dv
ds
=
• And dp =
dz
ds
1 dv 2
2 ds
∂p
∂s ds
+
∂p
∂n dn
• Along streamline dn = 0
dz dp
−γ
−
ds ds
=
1 dv 2
ρ
2 ds
10
Bernoulli, compact expression
Now making the assumption that density is constant,
Bernoulli’s equation is obtained
1 dv 2
ρ
= 0
2 ds
1 2
ρv
= 0
2
1
γz + p + ρv 2 = Constant
2
dz dp
γ
+
+
ds
ds
d
γz + p +
ds
The constant density assumption (incompressible
flow) is good for liquids (sometimes gases at low
speed). Bernoulli’s equation presented in 1738
monograph Hydrodynamics by Daniel Bernoulli.
If one has compressible fluid
Z
dp 1 2
+ v + gz = Constant
ρ
2
and knowledge of how ρ varies with p .
11
Forces normal to streamline
The acceleration normal to the streamline is an =
where R is the local radius of curvature of the
streamline.
X
(δm)v 2
ρ δV v 2
δFn =
=
R
R
v2
R
12
Forces normal to streamline
A change in stream direction occurs from pressure
and/or and gravity forces. Resolve forces
δWn = δW cos θ = γδV cos θ
δWn would be zero for vertical motion.
The pressure changes with height. Let p be pressure
in middle of fluid slab, p + δpn is pressure at top of
slab and p − δpn be pressure at bottom of slab.
From Taylors series
δpn =
∂p δn
∂n 2
The net pressure force, δFpn
δFpn
= (p − δpn )δs δy − (p + δpn )δs δy
∂p
∂p
= −2δpn δn δy = − δn δs δy = − δV
∂n
∂n
Need to combine pressure and weight forces to get
net Force
13
Forces normal to streamline
Combine weight and pressure forces
δFn = δWn + δFpn
∂p
ρδV v 2
=
−γ cos θ −
δV =
∂n
R
Pressure and weight forces imbalance produces the
curvature. For gas flows it is common to use
∂p
ρv 2
=−
∂n
R
The pressures increases
with distance away from
the center of curvature
∂p
is negative since
( ∂n
ρv 2 /R is positive).
For straight parallel streamlines (in gases),
No pressure change across streamlines
∂p
∂n
=0.
14
Forces normal to streamline
Will consider fluid parameters normal to stream line
ρv 2
∂p
+
=0
γ cos θ +
∂n
R
•
∂p
∂n
=
dp
dn
• cos θ =
since s is constant.
dz
dn
and so for incompressible flows
dz
ρv 2
dp
+γ
+
=0
dn
dn
R
dz
ρv 2
dp
+γ
+
=0
dn
dn
R
d
ρv 2
(p + γz) +
=0
dn
R
Z 2
v
dn = Constant
p + γz + ρ
R
For a compressible substance, the best reduction is
Z
Z 2
dp
v
+
+ gz = Constant
ρ
R
15
Interpretation for incompressible flows
Along the streamline
1 2
γz + p + ρv = Constant
2
Across the streamline
Z 2
v
p + γz + ρ
dn = Constant
R
The units of Bernoulli’s equations are J m−3 . This
is not surprising since both equations arose from an
integration of the equation of motion for the force
along the s and n directions.
The Bernoulli equation along the stream-line is a
statement of the work energy theorem. As the
particle moves, the pressure and gravitational forces
can do work, resulting in a change in the kinetic
energy.
16
Dynamic and static pressures
1 2
p + ρv + ρgz = constant
2
Static pressure is the pressure as measured moving
with the fluid. (e.g. static with fluid). This is the p
term in Bernoulli’s equation. Imagine moving along
the fluid with a pressure gauge.
Some times the ρgz term in Bernoulli’s equation is
called the hydrostatic pressure. (e.g. it is the change
in pressure due to change in elevation.)
Dynamic pressure is a pressure that occurs when
kinetic energy of the flowing fluid is converted into
pressure rise. This is the pressure associated with
the 12 ρv 2 term in Bernoulli’s equation.
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Dynamic and static pressures
The static pressure at 1 can be estimated by the
height of the column.
Open
p1 = γh3−1 + p3
= γh3−1 + γh4−3
H
(4)
h
= γh
h4-3
V
h3-1
(3)
ρ
(1)
(2)
V1 = V
V2 = 0
The dynamic pressure at 2 is estimated by
1 2
1 2
p2 + ρv2 + ρgz2 = p1 + ρv1 + ρgz1
2
2
1 2
p2 + 0 + ρgz1 = p1 + ρv1 + ρgz1 (v2 = 0)
2
1 2
p2 = p1 + ρv1
2
The additional pressure due to the dynamic pressure
will cause the fluid to rise a height of H > h .
The point (2) is called a stagnation point.
18
The stagnation point
Stagnation point
Stagnation streamline
Stagnation point
(a)
(b)
When fluid flows around any stationary body, some
of the streamlines pass over and some pass under the
object. But there is always a stagnation point where
the stagnation streamline terminates. The
stagnation pressure is
pstagnation
1 2
= p + ρv
2
v is velocity at some point on stream-line away from
obstruction.
The total pressure, pT
pT
1 2
= p + ρv + γz
2
is sum of static, dynamic and hydrostatic pressures.
19
The pitot tube
Knowledge of static and stagnation pressures makes
it possible to determine the fluid velocity. Geometry
arranged so that elevation differences have little
impact. The free stream pressure is p .
(3)
Pressure measured
points 3 and 4
Stagnation pressure
at
(4)
1 2
p2 = p3 = p + ρv
2
(1)
V
p
(2)
Static pressure is just p1 = p ≈ p4 . Combining
equations
1 2
p = p4 = p3 − ρv
2
Rearranging leads to
s
v=
2(p3 − p4 )
ρ
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The pitot tube: complications
The main question is design of pitot tubes is where
to place the orifice to measure the static pressure.
American Blower company
V
The static pressure does
vary along the length of
the tube. More complicated analysis than
Bernoulli required here.
Also make mouth of tube smooth.
National Physical laboratory (England)
American Society of Heating & Ventilating Engineers
21
The free jet
The free jet result was first obtained in 1643 by
Evangelista Torricelli.
1 2
1 2
p1 + ρv1 + γz1 = p2 + ρv2 + γz2
2
2
1 2
ρv2
γz1 =
2
• p1 = patm = 0 ; gauge pressure
• v1 ≈ 0 ; large surface, so v1 << v2
• p2 ≈ p4 = patm = 0 ; streamlines parallel
between (2) and (4) . Radius of curvature is
infinite.
• z1 = h . (z2 = 0 )
22
The free jet
γz1
v2
1 2
= ρv2
2
s
2γh p
= 2gh
=
ρ
Outside nozzle, stream continues to fall and at (5)
p
2gh
v2 =
p
v5 =
2g(h + H)
√
Result v = 2gh is speed of freely falling body
starting from rest.
For the fluid, all the potential energy is converted to
kinetic energy when jet leaves tank. (assume no
viscous forces).
23
Free jet, fine details
Horizontal nozzle, velocity at center line v2 is
slightly smaller than v3
and slightly larger than
v1 .
For d ≪ h , OK to use
v2 as average velocity.
Streamlines cannot follow sharp corner exactly.
Would take an infinite pressure gradient to achieve
zero radius of curvature
(i.e. R = 0 ). Uniform velocity only occurs at a-a line.
Vena Contracta effect.
Jet diameter, dj
is
slightly smaller than
hole diameter dh .
24
Free jet, Vena contracta effect
The contraction coefficient, Cd = Aj /Ah is the ratio
of the jet area Aj , and hole area Ah .
25
Flow rates
How much water flows down a channel or through a
pipe?
The volume flow rate,
Q1 is defined as the volume of fluid that flows
past an imaginary (or
real) interface.
• Volume of fluid leaving δV = v1 δtA1
δV
= v1 A1
δt
• The volume flow rate Q = v1 A1
• Rate of volume change
• Mass of fluid leaving δm = ρv1 δtA1
dm
• Rate fluid leaving ṁ =
= ρv1 A1
dt
26
Equation of continuity
For a steady state situation, the mass of fluid going
into the tank must be the same as the mass of fluid
leaving the tank.
Mass of water in = Mass of water out
ρ1 A1 v1 = ρ2 A2 v2
This is the continuity equation and for
incompressible flow
A1 v1 = A2 v2
or Q1 = Q2
The equation of continuity and the Bernoulli’s
equation are used into conjunction to analyze many
flow situations.
27
Flow rate: Example 1
Given the water velocity
at (2) is 8.0 m/s and
the pipe diameter is 0.10
m , what are the volume
and mass flow rates?
Q = vA = vπ(d/2)2
= 8.0 × π0.0502 = 0.06283 m3 /s
The mass flow is just Q × ρ so
dm
dt
= 1000 × 0.06283 = 62.83 kg/s
28
Flow rate: Example 2
A stream of water d =
0.10 m
flows steadily
from a tank of diameter
D = 1.0 m
as shown
in the figure. What flowrate is needed from the
inlet to maintain a constant water volume in the
header tank depth? The
depth of water at the outlet is 2.0 m .
Can regard outlet as a free jet (note water level at
(1) is not going down).
p
√
2gh = 2 × 9.8 × 2.0 = 6.26 m/s
v2 =
⇒ Q2 = A2 v2 = π(0.050)2 6.26 = 0.0492 m3 /s
= Q1
29
Flow rate measurement
One way to measure flow-rate is to place a
constriction in a pipe. The resulting change in
velocity (continuity equation), leads to a pressure
difference. The absolute fluid velocity can be
determined from pressure difference between (1) and
(2) .
The Orifice, Nozzle and Venturi meters analysis here
ignores viscous, compressibility and other real-world
effects.
30
Flow rate measurement: 2
Want to determine flow rate, need v2
v1 A1 = v2 A2
A2
⇒ v1 =
v2
A1
1 2
1 2
p2 + ρv2 = p1 + ρv1
2
2
1
1
⇒ ρv22 − ρv12 = p1 − p2
2
2
1 2 1 A22 2
⇒ ρv2 − ρ 2 v2 = p1 − p2
2
2 A1
2(p1 − p2 )
2
⇒ v2 =
A22
ρ 1 − A2
1
So the flow rate is
v
u 2(p − p )
u
1
2
Q = A2 t A22
ρ 1 − A2
1
The pressure differences give the flow rate. Real
world flows are 1% to 40% smaller.
31
Bernoulli and Cavitation
The temperature at which water boils depends on
pressure.
T
10
(o C)
Q
pvap (kPa)
1.23
(1)
(2)
(3)
p
20
2.34
(Absolute
pressure)
Small Q
30
4.24
40
7.34
Moderate Q
pv
Large Q
0
Incipient cavitation
x
The process of cavitation involves
• Fluid velocity increases
• Pressure reduction
• If p < pvap , water boils
• Bubbles collapse when reach high pressure part
of fluid
32
Bernoulli and Cavitation
Pressure transients exceeding 100 MPa can be
produced. These transients can produce structural
damage to surfaces.
33
Sluice gate flow rate
The height of water in the channel can be used to
determine the flow rate of water out the reservoir.
v
u 2g(z − z )
p
u
1
2
≈ z2 b 2gz1
Q = z2 b t
z22
1 − z2
1
(b) is the width of the reservoir.
34
Venturi meter problem
Determine
flow
rate as a function
of the diameter of
the tube.
0.2 m
Q
D
0.1 m
Use Venturi meter equation
v
u 2(p − p )
u
1
2
Q = A2 t 2
A
ρ 1 − A22
1
• A2 = πD2 /4 m2
• p1 − p2 = γ × 0.20 = 9800 × 0.20 = 1960 Pa
• ρ = 1000 kgm−3
• 1 − A22 /A21 = 1 − D2 /0.102 = 1 − 100D2 m2
D2
Q = π
4
s
3920
2
=
πD
1000(1 − 100D2 )
s
0.245
(1 − 100D2
35
Sharp crested weir
• Between (1) and (2) pressure and gravitational
forces cause fluid to accelerate from v1 → v2 .
• p1 = γh and p2 ≈ 0 forces cause fluid to
accelerate from v1 → v2 .
• Assume flow is like free jet. Average velocity
√
across weir is C1 2gH , C1 = constant.
• Flow rate is
Q = (Hb) × C
p
p
2gH = C1 b 2gH 3
The parameter C1 is determined empirically.
36
The energy line and hydraulic grade line
Consider Bernoulli equation divided by γ = ρg
p 1 v2
+
+ z = H = Constant on streamline
γ 2 g
The dimensions of the equation are in length. There
is the pressure head, velocity head, and elevation
head. The sum, H is called the total head.
The energy line gives the total head available to a
fluid. It can be measured by measuring the
stagnation pressure with a pitot tube.
The Hydraulic grade line is the line produced from
the pressure and elevation heads. It is measured with
a static pitot tube.
37
The energy line and hydraulic grade line
• The energy line will be horizontal along the
stream line as long as Bernoulli assumptions are
valid.
• The hydraulic grade line will not be horizontal if
the fluid velocity changes along the stream line.
• If forces are present (this does occur in pipe
flows), then there will be a loss in energy and
the energy line will not be constant.
38
Example: EL and HGL
A scale drawing can be used to depict the pressure in
the tank/pipe system.
• The energy line is horizontal
• The elevation head at (2) is converted into
increased pressure head p2 /γ and velocity head
v22 /(2g) . The HGL decreases.
• At (3) , pressure is atmospheric. So the HGL to
the level of the pipe and the elevation head has
been converted entirely into a velocity head
v32 /(2g) .
39
EL and HGL
The EL and HGL can depict whether there is
positive pressure p > patm or negative pressure
p < patm .
• The water velocity will be constant in curved
pipe (equation of continuity).
• The pressure head will increase or decrease as
the elevation head changes. Useful to know for
leaking pipes.
40
Limitations on Bernoulli equation
A number of problems can invalidate the use of the
Bernoulli equation, these are compressibility effects,
rotational effects, unsteady effects.
Compressibility effects
When can compressibility effects impact on gas
flows? Consider stagnation point
• Stagnation pressure is greater than static
pressure by 12 ρv 2 (dynamic pressure), provided
ρ constant.
• ρ will not changes too much as long as dynamic
pressure is not too large when compared to
static pressure.
• So flows at low v will be incompressible
• But dynamic pressure increases as v 2 , so
compressibility effects most likely at high speed.
41
Compressibility effects
The isothermal model for an ideal gas, p = ρR∗ T
Z
dp 1 2
C =
+ v1 + gz1
ρ
2
Z
dp 1 2
C = RT
+ v1 + gz1
p
2
1 2
C = RT ln p1 + v1 + gz1
2
This can be used to get
RT
v22
v12
+ z1 +
ln(p1 /p2 ) =
+ z2
2g
g
2g
Now write as p1 /p2 = 1 + (p1 − p2 )/p2 = 1 + ∆p/p2
and use ln(1 + x) = x for x ≪ 1 .
v12
RT
+ z1 +
ln(1 + ∆p/p2 ) =
2g
g
v12
RT
+ z1 +
(∆p/p2 ) ≈
2g
g
v22
+ z2
2g
v22
+ z2
2g
This can be reduced to the standard Bernoulli
equation. Bernoulli recovered as long as pressure
differences are not large.
42
Compressibility effects: Isentropic flow
This is the situation that applies when there is no
heat transfer or friction during the flow (reasonable
for many gases). This gas law is p = ρk D where k
depends on specific heat capacities. Introduce the
Mach number, Ma = v/c (ratio of the flow speed to
the speed of sound). Consideration of the pressure
ratio between free stream and stagnation points
leads to
p2 − p1
p1
p2 − p1
p1
kMa21
=
incompressible
2
"
#
k
k−1
k−1
=
1+
− 1 compressible
2 +
Ma1
0.3
0.2
p2 – p1
______
Incompressible
(Eq. 3.26)
p1
The compressible and
incompressible expressions agree to 2% for
Ma < 0.3 .
Compressible
(Eq. 3.25)
0.1
k = 1.4
0
0
0.2
0.4
Ma1
0.6
0.8
43
Unsteady effects
Implicit in the discussion was an assumption that
the fluid flows along steady state streamlines, so
v = v(s) is a function of position along the stream
and does not contain any explicit time dependence.
If v = v(s, t) then then it would be necessary to
include this when integrating along the streamline.
Z t2
1 2
1 2
∂p
p1 + ρv1 + γz1 = p2 + ρv2 + γz2 + ρ
ds
2
2
t1 ∂s
The additional term does complicate matters and
can only be easily handled under restricted
circumstances. There are quasi-steady flows where
some time dependence exists, but Bernoulli’s
equations could be applied as if the flow were steady
(e.g. the draining of a tank).
44
Rotational effects
Model of wake behind insect
Y. D. Afanasyev, Memorial University of
Newfoundland
Bernoulli equation describes motion of fluid particles
along streamline. If particles spin about the
streamline then Bernoulli is no longer valid.
Need to characterize irrotational and rotational flows.