ProblemSet6_soln_18.pdf - ECH 141 Problem Set #6 1. A solid...

ECH 141
Problem Set #6
1. A solid piston with radius D
0
is pushed by a horizontal force F into a cylindrical cavity with diameter
D
1
as shown below. Use the macroscopic momentum balance to derive an expression for the relation
between the force F and the piston velocity u
0
if u
0
is constant. Except for the space occupied by the
piston, the entire cylindrical cavity as well as the region outside the cavity contains an incompressible
liquid.

Start with a mass balance (see also Deen Example 11.2-1). The macroscopic mass balance is
d
dt
ρ dV
V
a
(t)
∫
= − ρ n
A
a
(t)
∫
⋅ v − w
( )
dA
.
Use the control volume as shown above by the dashed line, with all boundaries fixed except for the part
along the right side of the piston. The outer edges, on the top, bottom and far right, can be arbitrarily far
from the piston. The density is constant, and the rate of change of the volume equals the rate at which
the piston enters the cylinder, i.e.
d
dt
ρ dV
V
a
(t)
∫
=
π
4
ρ D
0
2
u
0
.
The only place where the right side of the mass balance is non-zero is at the thin annulus where the fluid
exits the cylinder. The velocity of the piston is known, so substituting gives (remember the normal
vector points out of the control volume)
π
4
D
0
2
u
0
= v
π
4
D
0
2
⎛
⎝
⎜
⎞
⎠
⎟
D
1
2
D
0
2
− 1
⎛
⎝
⎜
⎞
⎠
⎟
.
We see that the average fluid velocity exiting the annulus is
v =
u
0
D
1
2
D
0
2
− 1
⎛
⎝
⎜
⎞
⎠
⎟
.
In the text, the same result is derived using a different control volume.
The macroscopic momentum balance is given by
d
dt
ρ v dV
V
a
(t)
∫
+ ρ v v − w
( )
⋅ n dA
A
a
(t)
∫
= ρ g dV
V
a
(t)
∫
+ n ⋅ T dA
A
a
(t)
∫ .
Note that, if the diameter of the cylinder is similar in magnitude to the diameter of the piston, then the
control volume used in Fig. 11.2 of Deen makes it difficult to extract the applied force, because the
stresses on the cylinder wall are not likely to be negligible relative to the stresses on the piston. The
control volume shown here, which is outside the cylinder, appears to be a better choice for the force
calculation on the piston.
We assume quasi-static conditions, so the time derivative term is negligible. Since we are interested in
the x-component of the force, dot with e
x
. Note also that v =w everywhere except at the opening of the
annulus, then

ρ v
x
v ⋅ n dA
A
annulus
∫
= e
x
⋅ n ⋅ T dA
A
a
(t)
∫
The surface integral on the right side is over the entire surface of the control volume. However, on the
surfaces other than the hydraulic ram, and the fluid exiting the annulus, it is reasonable to assume that
there is little flow and hence negligible stress. In the fluid exiting the annulus, we furthermore neglect
viscous stresses and assume that the exit pressure is the ambient pressure outside the cylinder. Then the
balance becomes
ρ v
x
v ⋅ n dA
A
annulus
∫
= e
x
⋅ n ⋅ T dA
A
ram
∫
Since the normal vector here points out of the control volume, at the ram surface it points into the ram.
The force being exerted on the ram is therefore, in terms of the normal n to the control volume,
F
x
= − e
x
⋅ n ⋅ T dA
A
ram
∫