2.7
Bernoulli’s equation for steady flows
The Euler equation for steady motion is
ρ(u · ∇)u = −∇p − ρ∇Φ,
assuming that the body forces possess a potential Φ. We can transform the term on the
right using the vector identity (see Appendix A)
u · ∇u = ∇( 21 u · u) − u × ω
where ω = ∇ × u is the vorticity. Thus we obtain
−u × ω = −∇(p/ρ + Φ + 21 u2 ).
(19)
To make the left hand side disappear we take the dot-product with u on both sides to
find
u · (u × ω) = 0 = u · ∇(p/ρ + Φ + 12 u2 ).
The definition of a streamline is
dx
= u, so
dt
d
dxi ∂
(p/ρ + Φ + 12 u2 ) =
(p/ρ + Φ + 21 u2 ) = u · ∇(p/ρ + Φ + 21 u2 ) = 0.
dt
dt ∂xi
Consequently,
p/ρ + Φ + 12 u2 = const,
along any streamline in the flow.
(20)
This equation can be seen as a statement of energy conservation, where u2 /2 is the kinetic
energy, and Φ the potential energy. Note: A very nice discussion of the Bernoulli effect
and of pressure forces is presented in the film: “Pressure fields and acceleration”, found
on Youtube.
Example 1: A three-dimensional source,
u=
m
r̂.
4πr 2
Evidently, stream lines are rays pointing radially outward from the center, and
1 2
m2
u =
.
2
32π 2r 4
According to (20),
m2 ρ
+ const,
32π 2 r 4
where the constant could in principle be different for each streamline. However, the
pressure must go to the same constant far from the source, so we reach the same conclusion
as before.
However, Bernoulli’s theorem is particularly powerful for more complicated flows, for
which simple solutions are not available. We give a few examples.
p=−
2.7.1
Example: Flow out of a tank
A tank of uniform cross section A0 has a small hole, with area Ae , and at a height h above
the base. The height of the fluid above the hole is H. Then, what is the flow speed out
area
Ao
patm
H
area
Ae
Uo
Ue
h
z
streamlines
of the drain ?
First, conservation of mass gives
ρA0 U0 = ρAe Ue ,
(21)
since the mass fluxes across two boundaries equal. Since the top fluid level decreases very
slowly, the flow is approximately steady. Second, we want to apply Bernoulli’s theorem
to a streamline that exits the hole. It is important to notice that any such streamline
must originate from the surface of the container (dashed lines). The potential in (20) is
Φ = gz. The pressure at the top surface and just outside the hole is p = patm .
Then
• At surface: patm + ρg(h + H) + 21 ρU02 = C
• At exit: patm + ρgh + 12 ρUe2 = C,
and along the same streamline the constant C is the same. Eliminating C and patm and
using (21) we find
Ue2 [1 − (Ae /A0 )2 ] = 2gH.
It is reasonable to assume that Ae ≪ A0 , so
p
Ue ≈ 2gH.
With this main result in hand, we can address two practical questions:
1) What is the draining time td for an initial filling height H0 to reach 0 ?
The equation of motion for the surface is
Ae p
dH
= −U0 ≈ −
2gH.
dt
A0
and so
td =
Z
0
H0
dt
A0
dH = −
dH
Ae
Z
0
H0
A0 p
dH
√
=
2H0 /g.
Ae
2gH
2) What is the horizontal distance x travelled by the jet?
The fluid particles making up the jet are subject to gravitational acceleration
√ only, and
so obey the same equations as a projectile firedp
horizontally with speed Ue = 2gH. The
time taken by a particle to hit the floor is t = 2h/g, and so
p
p
√
x = Ue t = 2gH 2h/g = 2 Hh.
Finally we can fix a total filling height (measured from the floor) to be Hf , so that
Hf = H + h. Where do we have to place h to maximise x? To answer this question, we
maximise
q
x = 2 (Hf − h)h
with respect to h. This gives h = 21 Hf , and so
xmax = Hf .
Experiment: The most reliable data was taken from the bottom hole. I re-measured
its height from the bottom, and found it to be h = 2.4cm. The length of the trajectory x
was measured as the filling height was decreasing:
Hf [cm]
10
9
8
7
p
2 (Hf − h)h [cm]
8.54
7.96
7.33
6.65
xexp [cm]
8
7.5
6.9
6.3
Great agreement!
2.7.2
Example: Flow through a slowly diverging channel
area
A2
area
A1
U1
streamline
U2
From mass conservation: A1 U1 = A2 U2 , and from Bernoulli’s equation (ignoring gravity):
1 2
1
ρU1 + p1 = const = ρU22 + p2 .
2
2
Hence, the pressure drop is
1
1
∆p = p2 − p1 = ρ(U12 − U22 ) = ρU12 (1 − A21 /A22 ) > 0
2
2
for a widening channel. In other words, fast flow implies low pressure and vice versa.
2.7.3
Example: The hydraulic jump
When a fast jet of water meets a plate, a thin jet of water spreads radially from the
point of impact. However, as seen in the Figure, at some finite radius R, the thickness of
the fluid layer increases suddenly: this is known as the hydraulic jump. Let us begin by
investigating the conditions which govern the neighborhood of the jump, using mass and
momentum conservation.
Q, U z
S
n
xx
xx
xx
xx
xx
xx
xx
xx
xx
xx
xx
xx
u1
xx
xx
xx
xx
xx
xx
xx
u2
xx xx
xx
g
h2 x
h1
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h(r)
H
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R
If u1 , h1 is the velocity of the flow and the height of the layer to the left, and u2 , h2 are
the corresponding expressions to the right of the jump, we can apply global mass balance
to the surface S shown as the dashed line:
Z
u · ndS = 0.
S
The argument u · n is zero everywhere except at the left face of S, where u · n = −u1 ,
and at the right, where u · n = u2 . Thus
u1 h1 = u2 h2 ,
after integrating over the faces.
The momentum theorem, on the other hand, reads
Z
ρu(u · n) + (p + ρgz)ndS = 0.
S
Now on the free surface, u · n = 0, and p = patm . To compute the integral over the left
and right hand faces of S, we need the pressure distribution. But since the flow is uniform
away from the jump, the pressure distribution is that of a staic fluid:
p1 = patm + ρg(h1 − z)
at the left face of S, and
p2 = patm + ρg(h2 − z)
to the right. Thus from the momentum balance in the horizontal (x̂)-direction, we get:
−
Z
h1
0
(ρu21 + patm + ρgh1 )dz = −h1 (ρu21 + patm + ρgh1 )
from the left face, and
h2 (ρu22 + patm + ρgh2 )
from the right face. It remains to calculate
Z
(patm + ρgh)nx dS
Sf
over the free surface Sf .
If the (unknown) profile is h(x), this integral is
Z x2
√
(patm + ρgh)nx 1 + h′2 dx,
x1
√
and nx = −h′ / 1 + h′2 . Thus the free surface contribution becomes
−
Z
x2
x1
′
(patm + ρgh)h dx = −
Z
h2
h1
(patm + ρgh)dh = (h1 − h2 )patm +
The sum of all three contributions must add up to zero, and thus
ρg 2
h1 − h22
2
h1 u21 + gh21 /2 = h2 u22 + gh22 /2
at the jump. Eliminating u2 using mass conservation, this turns into the jump condition
h1
gh22
h21
2
h1 u1 1 −
=
1− 2 .
(22)
h2
2
h2
Now we turn to radial flow for r < R, before the jump occurs. Let Q be the volume
flux of the jet impinging on the plate, and thus
Q = πrj2 U,
where U is the speed of the jet close to the point of impact, and rj its radius.
To find the flow speed in the layer, we employ an important argument, based on
the steady Bernoulli equation. Namely, the free surface is a streamline of the flow. We
also know that the pressure p = patm is constant along the free surface. Along such a
streamline, and neglecting small gravitational effects, we have
patm /ρ + u2 /2 = const,
which implies that the flow speed u along the surface must itself be a constant. But this
means that the flow speed along the surface of the layer is U. Since the flow in the layer
is uniform, the flow velocity in the layer is
u = Ur̂,
pointing radially outward.
Thus the total flux through the layer is Q = 2πrh(r)U, where h(r) is the thickness,
and the profile for r < R becomes
rj2
h= .
2r
In other words, just before the position r = R of the jump, the layer thickness is h1 =
rj2 /(2R), and u1 = U.
We conduct the experiment such that the jet falls into a container of depth H, which
guarantees that the thickness h2 = H of the outer layer can be controlled, see Figure. Now
we can use the jump condition (22). Defining the relative layer thickness as δ = h1 /H,
and the Froude number as
s
U2
,
Fr =
gH
we find
2 δ − δ2 F r2 = 1 − δ2.
Evidently, δ = 1 is a possible solution, corresponding to no jump occuring at all. The
other possible solution, corresponding to a jump, is
δ = −1 −
1
2F r 2
=
1 − 2F r 2
2F r 2 − 1
according to Vieta’s rule. Inserting the expression for h1 and solving for the position of
the jump, we finally obtain
rj2
2F r 2 − 1 .
(23)
R=
2H
What a beautifully simple formula!
For example, consider water
being poured at a rate Q from a height hf . Then the
p
speed upon impact is U = 2ghf according to Bernoulli, and the Froude number is
F r2 =
2hf
≫ 1.
H
By mass conservation, rj2 = Q/(πU), so we obtain
s
2hf
Q
R≈
.
2
πH
g
In our classroom experiment, the
pfilling height Hf of the bottle was about 10cm, which
produces an exit velocity Ue = 2gHf = 140 cm/s. With a cross sectional area of the
opening being A ≈ 0.28 cm2 , this produces a flow rate Q = 40 cm3 /s. With a total height
of fall hf = 21 cm, H = 1cm, and g = 981cm/s2 , this gives a radius of R ≈ 2.7cm. This
is realistic, but far off from the observed value of R ≈ 1cm. Note that the radius depends
strongly on the layer thickness H, which is easily confirmed experimentally; however, it
also makes the experiment somewhat difficult to control.
2.8
The Coanda effect
n
.
s
In deriving Bernoulli’s equation, we have integrated along flow lines. What about the
behavior of the pressure normal to the flow lines? To investigate this question let us
consider the two-dimensional curvilinear coordinate system spanned by the flow lines on
one hand, and the lines normal to them, see Figure. We can always choose the plane
such that locally the streamlines curve within the plane. Let s be the arclength along a
streamline, and n is chose such that it measures the arclength normal to it. Then the
gradient operator is
∂
∂
∇ = ŝ + n̂ ,
∂s
∂n
and
u = us ŝ,
since by definition the velocity points in the direction of the streamline.
The curvature κ of the streamline is defined by
∂ŝ
= κn̂.
∂s
so that the steady Euler equation becomes
ρus
∂p
∂p
∂us
ŝ + ρκu2s n̂ = − ŝ −
n̂.
∂s
∂s
∂n
The ŝ-component of this equation yields
ρ ∂u2s
∂p
=− ,
2 ∂s
∂s
which, upon integration, yields the steady Bernoulli equation.
The n̂-component, on the other hand, gives new information:
ρκu2s = −
∂p
.
∂n
(24)
In other words, the pressure decreases in the direction in which streamlines are curved.
This is the result of centrifugal forces experienced by fluid particles. Thus once one draws
the streamlines of a steady flow, using (20) and (24) one can deduce important qualitative
conclusions about the pressure distribution.
This so-called Coanda effect can be seen at work in many problems. For example,
place a finger at the side of a jet from a water faucet. As a result, the fluid has to go
around the finger, and the jet is deflected toward the finger, in the direction in which
streamlines are bent. Similarly, objects can be suspended in the air by a stream of air
directed at them obliquely.
y
n̂
x
Example:
Consider a vortex flow, whose streamlines are circles. Thus the flow field is of the
form u = f (r)θ̂, and the streamline curvature is κ = 1/r. The normal vector is pointing
radially inward. Then
∂p
f 2 (r)
∂p
=
=ρ
,
−
∂n
∂r
r
and the pressure will increase with increasing distance from the center. We will see below
that typically f (r) = A/r, which yields a pressure distribution
p(r) = −
A2
+ patm .
2r 2