If the velocity is approximated by its average value everywhere at the exit of the annulus, we get
ρ
v
2
π
D
0
2
4
D
1
2
D
0
2
−
1
⎛
⎝
⎜
⎞
⎠
⎟
=
−
F
x
=
F
.
Substituting the known average velocity obtained from the macroscopic balance, we have
F
=
ρ
u
0
2
D
1
2
D
0
2
−
1
⎛
⎝
⎜
⎞
⎠
⎟
π
D
0
2
4
,
where F as defined as the negative of F
X
if x points horizontally to the right.
Examination of the quasistatic assumption (not required)
To examine the assumption of quasistatic conditions used for the momentum balance, consider Eq.
11.2-17, the transport theorem or Leibnitz rule, with the function f replaced by
ρ
v
.
Then
d
dt
ρ
v
V(t)
∫
dV
=
ρ
∂
v
∂
t
V(t)
∫
dV
+
ρ
v
w
⋅
n
S(t)
∫
dS
.
The surface integral is non-zero only at the right side of the piston, where

v
=
w
=
−
u
0
e
x
(right edge of piston).
That term is therefore of magnitude
ρ
v
w
⋅
n
S(t)
∫
dS ~
ρ
u
0
2
π
D
0
2
4
.
If we compare this to the final result for F, we see that it is negligible if the piston and cylinder
diameters are close in size, or if
D
1
2
D
0
2
−
1
⎛
⎝
⎜
⎞
⎠
⎟
<<
1
.
The time derivative term is more difficult to estimate, but suppose we take L to be the length of piston
outside they cylinder.
Then the time scale for the process is of order L/u
0
.
Over most of the volume of
the control volume, the fluid velocity is zero.
However, it is reasonable to expect that in a region with
characteristic dimension L outside the cylinder, near the annulus and moving piston surface, the velocity
is not zero and changing.
Then we might estimate
ρ
∂
v
∂
t
V(t)
∫
dV ~
ρ
u
0
L / u
0
(
)
π
D
0
2
4
L
⎛
⎝
⎜
⎞
⎠
⎟
=
ρ
u
0
2
π
D
0
2
4
.
Again, this term is negligible compared to our result for F, if the cylinder and piston have nearly the
same diameter.
2.
Water is pumped into a cylindrical tank at a rate of 0.25m
3
/s as shown below.
The diameter of the
inlet pipe is 20cm, and the diameter of the tank is 1m.
Approximately what force is required to hold the
tank in position when h(t)=70cm, excluding the weight of the tank wall?

Solution
Consider the control volume shown below, with all surfaces fixed.
Because the top surface, at the air/water interface, is fixed, water flow through it, and the control
volume equals the volume water in the tank (excluding the pipe) only instantaneously.
From the
macroscopic momentum balance, i.e. Eq (11.3-3) in Deen, we find that integration of the total stress
over the tank surface yields
F
=
n
t
⋅
T
dA
S
T
∫
=
mg
+
n
I
⋅
T
dA
S
I
∫
−
ρ
v
v
⋅
n
dA
S
∫
−
d
dt
ρ
v
dV
V
∫
.
Here the normal vector to the tank surface n
T
is the negative of the normal vector n
to the control
volume surface at positions along the tank surface, and m is the mass of the contents of the control
volume, which is the liquid in the tank excluding the inlet pipe.
The contribution from the stresses
along vertical surface of the inlet pipe is neglected.
Only viscous stresses along vertical surfaces can

contribute to the z-component of force along a vertical surface, and viscous stresses are assumed to be
negligible.
We can account for the horizontal surface at the bottom of the inlet pipe in an approximate way by