ECH 141
Problem Set #6
1.
Text 7-3
(Use only the macroscopic momentum balance, not the mechanical energy balance.
Your result will not
be the same as that obtained from the energy balance, because different assumptions are involved.
Assume quasi-steady conditions, neglect viscous stresses, and let the velocity at the exit to the annulus
between the ram and cylinder equal the average velocity, which must be found from a macroscopic
mass balance.)
Solution
Start with a mass balance.
The macroscopic mass balance is
d
dt
ρ
dV
V
a
(t)
∫
=
−
ρ
n
A
a
(t)
∫
⋅
v
−
w
(
)
dA
.
F
o
, u
0
D
1
D
0
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d
dt
ρ
dV
V
a
(t)
∫
=
π
4
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 ram 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 velocity exiting the annulus is
v
=
u
0
D
1
2
D
0
2
−
1
⎛
⎝
⎜
⎞
⎠
⎟
.
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)
∫
.
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
∫