3Dynamics
IB Fluid Dynamics
3.5 Momentum equation for inviscid (ν
= 0
) incompress-
ible fluid
We are now going to derive the Navier-Stokes equation in the special case where
viscosity is assumed to vanish. We will derive this by considering the change in
momentum.
Consider an arbitrary volume
D
with boundary
∂D
and outward pointing
normal n.
D
∂D
n
The total momentum of the fluid in D can change owing to four things:
(i) Momentum flux across the boundary ∂D;
(ii) Surface pressure forces;
(iii) Body forces;
(iv) Viscous surface forces.
We will ignore the last one. We can then write the rate of change of the total
momentum as
d
dt
Z
D
ρu dV = −
Z
∂D
ρu(u · n) dS −
Z
∂D
pn dS +
Z
D
f dV.
It is helpful to write this in suffix notation. In this case, the equation becomes
d
dt
Z
D
ρu
i
dV = −
Z
∂D
ρu
i
u
j
n
j
dS −
Z
∂D
−pn
i
dS +
Z
D
f
i
dV.
Just as in the case of mass conservation, we can use the divergence theorem to
write
Z
D
ρ
∂u
i
∂t
+ ρ
∂
∂x
j
(u
i
u
j
)
dV =
Z
D
−
∂p
∂x
i
+ f
i
dV.
Since D is arbitrary, we must have
ρ
∂u
i
∂t
+ ρ
∂u
i
∂x
j
+ ρu
i
∂u
j
∂x
j
= −
∂p
∂x
i
+ f
i
.
The last term on the left is the divergence of
u
, which vanishes by incompress-
ibility, and the remaining terms is just the material derivative of
u
. So we
get
Proposition (Euler momentum equation).
ρ
Du
Dt
= −∇p + f.
This is just the equation we get from the Navier-Stokes equation by ignoring
the viscous terms. However, we were able to derive this directly from momentum
conservation.
We can further derive some equations from this.
For conservative forces, we can write
f
=
−∇χ
, where
χ
is a scalar potential.
For example, gravity can be given by
f
=
ρg
=
∇
(
ρg · x
) (for
ρ
constant). So
χ = −ρg · x = gρz if g = (0, 0, −g).
In the case of a steady flow,
∂u
∂t
vanishes. Then the momentum equation
becomes
0 = −
Z
∂D
ρu(u · n) dS −
Z
∂D
pn dS −
Z
D
∇χ dV.
We can then convert this to
Proposition (Momentum integral for steady flow).
Z
∂D
(ρu(u · n) + pn + χn) dS = 0.
Alternatively, we notice the vector identity
u × (∇ × u) = ∇
1
2
|u|
2
− u · ∇u.
We use this to rewrite the Euler momentum equation as
ρ
∂u
∂t
+ ρ∇
1
2
|u|
2
− ρu × (∇ × u) = −∇p − ∇χ.
Dotting with u, the last term on the left vanishes, and we get
Proposition (Bernoulli’s equation).
1
2
ρ
∂|u|
2
∂t
= −u · ∇
1
2
ρ|u|
2
+ p + χ
.
Note also that this tells us that high velocity goes with low pressure; low
pressure goes with high velocity.
In the case where we have a steady flow, we know
H =
1
2
ρ|u|
2
+ p + χ
is constant along streamlines.
Even if the flow is not steady, we can still define the value
H
, and then we
can integrate Bernoulli’s equation over a volume D to obtain
d
dt
Z
D
1
2
ρ|u|
2
dV = −
Z
∂D
Hu · n dS.
So H is the transportable energy of the flow.
Example.
Consider a pipe with a constriction. Ignore the vertical pipes for the
moment — they are there just so that we can measure the pressure in the fluid,
as we will later see.
h
U
P
A
u, p, a
Suppose at the left, we have a uniform speed
U
, area
A
and pressure
P
. In the
middle constricted area, we have speed u, area a and pressure p.
By the conservation of mass, we have
q = UA = ua.
We apply Bernoulli along the central streamline, using the fact that
H
is constant
along streamlines. We can omit the body force
χ
=
ρgy
, since this is the same
at both locations. Then we get
1
2
ρU
2
+ P =
1
2
ρu
2
+ p.
Replacing our U with q/A and u with q/a, we obtain
1
2
ρ
q
A
2
+ P =
1
2
ρ
q
a
2
+ p.
Rearranging gives
1
A
2
−
1
a
2
q
2
=
2(p − P )
ρ
.
We see there is a difference in pressure due to the difference in area. This is
balanced by the difference in heights
h
. Using the
y
-momentum equation, we get
1
A
2
−
1
a
2
q
2
=
2(p − P )
ρ
= −2gh.
Then we obtain
q =
p
2gh
Aa
√
A
2
− a
2
.
Therefore we can measure
h
in order to find out the flow rate. This allows us to
measure fluid velocity just by creating a constriction and then putting in some
pipes.
Example
(Force on a fire hose nozzle)
.
Suppose we have a fire hose nozzle like
this:
(2)
(3)
(4)
(5)(1)
U
A
P
u, a, p = 0
We consider the steady-flow equation and integrate along the surface indicated
above. We integrate each section separately. The end (1) contributes
ρU(−U)A − P A.
On (2), everything vanishes. On (3), the first term vanishes since the velocity is
parallel to the surface. Then we get a contribution of
0 +
Z
nozzle
pn ·
ˆ
x dS. (3)
Similarly, everything in (4) vanishes. Finally, on (5), noting that p = 0, we get
ρu
2
a.
By the steady flow equation, we know these all sum to zero. Hence, the force on
the nozzle is just
F =
Z
nozzle
pn ·
ˆ
x dS = ρAU
2
− ρau
2
+ P A.
We can again apply Bernoulli along a streamline in the middle, which says
1
2
ρU
2
+ P =
1
2
ρu
2
.
So we can get
F = ρAU
2
− ρau
2
+
1
2
ρA(u
2
− U
2
) =
1
2
ρ
A
a
2
q
2
1 −
a
A
2
.
Let’s now put some numbers in. Suppose
A
= (0
.
1)
2
πm
2
and
a
= (0
.
05)
2
πm
2
.
So we get
A
a
= 4.
A typical fire hose has a flow rate of
q = 0.01 m
3
s
−1
.
So we get
F =
1
2
· 1000 ·
4
π/40
· 10
−4
·
3
4
2
≈ 14 N.