ECH 141
Problem Set #4
1. Euler’s First Law (or the Linear Momentum Principle, or Conservation of Linear Momentum) is
sometimes written in words as
The rate of change of
momentum of a fluid particle
⎧
⎨
⎩
⎫
⎬
⎭
=
The sum of the forces
acting on the particle
⎧
⎨
⎩
⎫
⎬
⎭
.
In a static fluid, the rate of change of momentum is zero, and the forces acting on a fluid particle are
generally the gravitational force and pressure, as described in lecture.
We showed therefore that, in a
static fluid,
∇
p
− ρ
g
=
0
is the equation that determines the pressure distribution.
a)
Consider a case where the static fluid is in a vehicle that undergoing a constant acceleration a
.
The
rate of change of momentum of a fluid particle with mass m is then ma
, or for an infinitesimally small
particle with volume dV, it would be
ρ
a
dV.
Derive the equation for the pressure distribution in a fluid
undergoing constant acceleration a
.
b)
Now consider a cylindrical glass of water with radius R on a horizontal turntable, rotating about its
axis of symmetry (the z-axis, where radial position r=0).
In the absence of rotation, the height of water
in the glass is H.
As you learned in physics, each position in an object undergoing pure rotation is
accelerating inward, with centripetal acceleration a
a
=
−ω
2
re
r
,
where r is the distance from the axis of rotation in a cylindrical coordinate system,
ω
is the rotational
velocity, and e
r
is a unit vector pointing away from the z-axis.
Use this expression for centripetal
acceleration with your result from part “a” to derive an expression for the height h(r) of the air-water
interface in a rotating, cylindrical glass of water.
Solution

ρ
a
V
m
∫
dV
=
ρ
g
V
m
∫
dV
−
∇
p
V
m
∫
dV
.
Since the material control volume is arbitrary, this integral balance can only be true if at every point the
equation
∇
p
=
−ρ
a
+
ρ
g
is satisfied.
b)
The vector equation from part (a) can be separated into components.
For the cylindrical glass of
water, it makes sense to use a cylindrical coordinate system.
Then writing out the equation yields
e
r
∂
∂
r
+
e
z
∂
∂
z
(
)
p
=
ρω
2
re
r
− ρ
ge
z
.
Dotting both sides of the equation with e
Z
gives
∂
p
∂
z
=
−ρ
g
and doing the same thing in the radial direction gives
∂
p
∂
r
=
ρω
2
r
.
The difference between this case and one where the fluid is not accelerating is that the top surface is not
flat, but rather has some height h(r).
At this height, the pressure equals atmospheric pressure.
Integrating the two equations, and noting that the derivatives are partial derivatives so that constants of
integration can be functions of the variable not being integrated, we find
p r,z
(
)
=